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GATE ELECTRONICS & COMMUNICATION Topicwise Solved Paper Year 2013- 1996 By RK Kanodia & Ashish Murolia For more GATE Resources, Mock Test and Study material

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 1

UNIT 1

(C) x

(D) 1

2012

ENGINEERING MATHEMATICS 1.8

TWO MARKS

d 2 y (t) dy (t) Consider the differential equation +2 + y (t) = d (t) 2 dt dt dy with y (t) t = 0 =- 2 and =0 dt t = 0 dy The numerical value of is dt t = 0 (A) - 2 (B) - 1 (C) 0 (D) 1 -

2013 1.1

1.2

1.3

ONE MARK

1.9

The minimum eigen value of the following matrix is R3 5 2V S W S5 12 7W SS2 7 5WW T X (A) 0 (B) 1 (D) 3 (C) 2

1.10

2013 1.4

Let A be an m # n matrix and B an n # m matrix. It is given that determinant ^Im + AB h = determinant ^In + BAh, where Ik is the k # k identity matrix. Using the above property, the determinant of the matrix given below is V R S2 1 1 1W S1 2 1 1W S1 1 2 1W W S S1 1 1 2W X T (A) 2 (B) 5 (C) 8 (D) 16

2

t (D) x = 2

(C) x = t 2 1.6

1 - 2 . z+1 z+3 If C is a counter clockwise path in the z -plane such that z + 1 = 1, the value of 1 f (z) dz is 2pj C (A) - 2 (B) - 1 (C) 1 (D) 2

Click to Buy www.nodia.co.in probability that the number of required tosses is odd, is (A) 1/3 (B) 1/2 (C) 2/3 (D) 3/4 1.11

1.12

(D) 46

Given that

-5 -3 1 0 , the value of A3 is A=> and I = > H 2 0 0 1H (A) 15A + 12I (B) 19A + 30I (C) 17A + 15I (D) 17A + 21I 2011 1.13

ONE MARK

Consider a closed surface S surrounding volume V . If rv is the position vector of a point inside S , with nt the unit normal on S , the value of the integral ## 5rv $ nt dS is S

(A) 3V (C) 10V 1.14

Given f (z) =

If x = - 1, then the value of xx is (A) e- p/2 (B) e p/2

The maximum value of f (x) = x3 - 9x2 + 24x + 5 in the interval [1, 6] is (A) 21 (B) 25 (C) 41

(B) 5V (D) 15V dy = ky, y (0) = c is dx (B) x = kecy (D) y = ce-kx

The solution of the differential equation (A) x = ce-ky (C) y = cekx

#

1.7

A fair coin is tossed till a head appears for the first time. The

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ONE MARK

With initial condition x (1) = 0.5 , the solution of the differential equation t dx + x = t , is dt (A) x = t - 1 (B) x = t 2 - 1 2 2

The direction of vector A is radially outward from the origin, with A = krn . where r2 = x2 + y2 + z2 and k is a constant. The value of n for which d:A = 0 is (A) - 2 (B) 2 (C) 1 (D) 0

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TWO MARKS

2012 1.5

+

The maximum value of q until which the approximation sin q . q holds to within 10% error is (A) 10c (B) 18c (C) 50c (D) 90c

A polynomial f (x) = a 4 x 4 + a 3 x3 + a2 x2 + a1 x - a 0 with all coefficients positive has (A) no real roots (B) no negative real root (C) odd number of real roots (D) at least one positive and one negative real root

-

1.15

The value of the integral z = 1 is given by (A) 0 (C) 4/5

# c

- 3z + 4 dz where c is the circle (z 2 + 4z + 5) (B) 1/10 (D) 1

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

2011 1.16

1.17

Page 2

TWO MARKS

A numerical solution of the equation f (x) + x - 3 = 0 can be obtained using Newton- Raphson method. If the starting value is x = 2 for the iteration, the value of x that is to be used in the next step is (A) 0.306 (B) 0.739 (C) 1.694 (D) 2.306 The system of equations x+y+z = 6 x + 4y + 6y = 20 x + 4y + l z = m

2010 1.22

1.23

1.24

has NO solution for values of l and μ given by (A) l = 6, m = 20 (B) l = 6, m = Y 20 (C) l = Y 6, m = 20 1.18

2010

A fair coin is tossed independently four times. The probability of the event “the number of time heads shown up is more than the number of times tail shown up” (A) 1/16 (B) 1/3 (C) 1/4 (D) 5/16 v = xyatx + x 2 aty , then If A is

(A) 0 (C) 1 1.25

(D) 1/2 ONE MARKS

The eigen values of a skew-symmetric matrix are (A) always zero (B) always pure imaginary (C) either zero or pure imaginary (D) always real

1.26

The trigonometric Fourier series for the waveform f (t) shown below contains

1.27

(A) only (B) only (C) only (D) only 1.21

(B) minimum at x = e (D) minimum at x = e-1

# Av $ dlv over the path shown in the figure C

A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is (A) 2/36 (B) 2/6

(C) 5/12

1.20

If ey = x1/x , then y has a (A) maximum at x = e (C) maximum at x = e-1

(D) l = Y 6, m = 20

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1.19

TWO MARKS

cosine and zero values for the dc components cosine and a positive value for the dc components cosine and a negative value for the dc components sine and a negative value for the dc components

A function n (x) satisfied the differential equation d 2 n (x) n (x) - 2 =0 dx 2 L where L is a constant. The boundary conditions are : n (0) = K and n (3) = 0 . The solution to this equation is (B) n (x) = K exp (- x/ L ) (A) n (x) = K exp (x/L) 2 (D) n (x) = K exp (- x/L) (C) n (x) = K exp (- x/L)

(B) 2 3 (D) 2 3

The residues of a complex function 1 - 2z x (z) = z (z - 1) (z - 2) at its poles are (A) 1 , - 1 and 1 (B) 1 , - 1 and - 1 2 2 2 2 (D) 1 , - 1 and 3 (C) 1 , 1 and - 3 2 2 2 2 dy (x) Consider differential equation - y (x) = x , with the initial dx condition y (0) = 0 . Using Euler’s first order method with a step size of 0.1, the value of y (0.3) is (A) 0.01 (B) 0.031 (C) 0.0631 (D) 0.1 3s + 1 Given f (t) = L-1 ; 3 . If lim f (t) = 1, then the value t"3 s + 4s2 + (k - 3) s E of k is (A) 1 (B) 2 (C) 3 (D) 4

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1.29

ONE MARK

The order of the differential equation d2y dy 3 4 -t 2 + c dt m + y = e dt is (A) 1 (B) 2 (C) 3 (D) 4 A fair coin is tossed 10 times. What is the probability that only the first two tosses will yield heads? 2 2 (B) 10C2 b 1 l (A) c 1 m 2 2 10 10 (C) c 1 m (D) 10C 2 b 1 l 2 2

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

1.30

If f (z) = c 0 + c1 z-1 , then (A) 2pc1 (C) 2pjc1

unit circle

1.33

Match each differential equation curves from Group II Group I dy y A. = dx x dy y B. =dx x dy x C. = dx y dy D. =- x dx y (A) A - 2, B - 3, C - 3, D - 1 (B) A - 1, B - 3, C - 2, D - 1 (C) A - 2, B - 1, C - 3, D - 3 (D) A - 3, B - 2, C - 1, D - 2

1.34

1.35

1.38

TWO MARKS

Group II

1.37

Which of the following functions would have only odd powers of x in its Taylor series expansion about the point x = 0 ? (A) sin (x3) (B) sin (x2) (C) cos (x3) (D) cos (x2) Which of the following is a solution to the differential equation dx (t) + 3x (t) = 0 ? dt (A) x (t) = 3e - t (B) x (t) = 2e - 3t (D) x (t) = 3t2

(C) x (t) =- 23 t2

in Group I to its family of solution

2008 1.40

1. Circles

TWO MARKS

The recursion relation to solve x = e - x using Newton - Raphson method is

SPECIAL EDITION ( STUDY MATERIAL FORM ) At market Book is available in 3 volume i.e. in 3 book binding form. But at NODIA Online Store book is available in 10 book binding form. Each unit of Book is in separate binding.

2. Straight lines 3. Hyperbolas

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Click to Buy www.nodia.co.in (A) xn + 1 = e-x

(B) xn + 1 = xn - e-x x 2 - e-x (1 - xn) - 1 (D) xn + 1 = n xn - e-x

n

n

-x (C) xn + 1 = (1 + xn) e -x 1+e

n

n

The Eigen values of following matrix are V R S- 1 3 5 W S- 3 - 1 6 W SS 0 0 3 WW X T (A) 3, 3 + 5j, 6 - j (B) - 6 + 5j, 3 + j, 3 - j (C) 3 + j, 3 - j, 5 + j (D) 3, - 1 + 3j, - 1 - 3j ONE MARKS

p11 p12 are nonzero, p21 p22 G and one of its eigenvalue is zero. Which of the following statements is true? (B) p11 p22 - p12 p21 =- 1 (A) p11 p12 - p12 p21 = 1 (C) p11 p22 - p12 p21 = 0 (D) p11 p22 + p12 p21 = 0

1.41

1.42

For real values of x , the minimum value of the function

n

The residue of the function f (z) =

1.43

1.44

1.45

1 at z = 2 is (z + 2) 2 (z - 2) 2

(B) - 1 16 (D) 1 32

0 1 Consider the matrix P = = . The value of e p is - 2 - 3G 2e-2 - 3e-1 e-1 - e-2 e-1 + e-1 2e-2 - e-1 (A) > -2 (B) H >2e-1 - 4e2 3e-1 + 2e-2H 2e - 2e-1 5e-2 - e-1 5e-2 - e-1 3e-1 - e-2 (C) > -2 H 2e - 6e-1 4e-2 + 6-1

The system of linear equations

The equation sin (z) = 10 has (A) no real or complex solution (B) exactly two distinct complex solutions (C) a unique solution (D) an infinite number of complex solutions

n

(A) - 1 32 (C) 1 16

All the four entries of the 2 # 2 matrix P = =

4x + 2y = 7 2x + y = 6 has (A) a unique solution (B) no solution (C) an infinite number of solutions (D) exactly two distinct solutions 1.36

1.39

(x - p) 2 (D) - 1 + + ... 3!

2008

(B) 1 (D) 0

(C) 0.5

(B) 2p (1 + c0) (D) 2p (1 + c0)

The Taylor series expansion of sin x at x = p is given by x-p 2 (x - p ) (x - p) 2 (A) 1 + (B) - 1 + ... + ... 3! 3! (x - p) 2 (C) 1 + ... 3!

1.32

f (x) = exp (x) + exp (- x) is (A) 2

# 1 +zf (z) dz is given by

2009 1.31

Page 3

2e-1 - e-2 e-1 - e-2 (D) > H -1 -2 - 2e + 2e - e-1 + 2e-2

In the Taylor series expansion of exp (x) + sin (x) about the point x = p , the coefficient of (x - p) 2 is (A) exp (p) (B) 0.5 exp (p) (C) exp (p) + 1 (D) exp (p) - 1 The value of the integral of the function g (x, y) = 4x3 + 10y 4 along the straight line segment from the point (0, 0) to the point (1, 2) in the x - y plane is (A) 33 (B) 35 (C) 40 (D) 56 Consider points P and Q in the x - y plane, with P = (1, 0) and

#

Q

Q = (0, 1). The line integral 2 (xdx + ydy) along the semicircle P with the line segment PQ as its diameter (A) is - 1 (B) is 0

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(C) is 1 (D) depends on the direction (clockwise or anit-clockwise) of the semicircle 2007 1.46

2

#1 ydx

is

(B) 2.5 (D) 5.0

For x << 1, coth (x) can be approximated as (A) x (B) x2

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in (C) 1 x

(D) e - x

(A) jp (C) - p 1.55

2

For the function e - x , the linear approximation around x = 2 is (A) (3 - x) e - 2 (B) 1 - x 2 ) x @e - 2

2007

1.52

If the semi-circular control D of radius 2 is as shown in the figure, 1 then the value of the integral ds is 2 ( s 1 ) D

#

(B) 1 (D) not defined

Which one of following functions is strictly bounded? (B) ex (A) 1/x2

(C) 63 + 3 2 - (1 -

1.51

1.54

sin b q l 2 is lim q"0 q

(C) x2 1.50

(A) f1 (t) and f2 (t) are orthogonal (B) f1 (t) and f3 (t) are orthogonal (C) f2 (t) and f3 (t) are orthogonal D) f1 (t) and f2 (t) are orthonormal

(D) 12 x

(A) 0.5 (C) 2 1.49

Three functions f1 (t), f2 (t) and f3 (t) which are zero outside the interval [0, T] are shown in the figure. Which of the following statements is correct?

The following plot shows a function which varies linearly with x .

(A) 1.0 (C) 4.0

1.48

1.53

ONE MARK

The value of the integral I =

1.47

Page 4

(D) e - 2

d2 y The solution of the differential equation k2 2 = y - y2 under the dx boundary conditions (i) y = y1 at x = 0 and (ii) y = y2 at x = 3 , where k, y1 and y2 are constants, is

1.56

(A) y = (y1 - y2) exp a- x2 k + y2 k (C) y = ^y1 - y2h sinh a x k + y1 k

1.57

The equation x3 - x2 + 4x - 4 = 0 is to be solved using the Newton - Raphson method. If x = 2 is taken as the initial approximation of the solution, then next approximation using this method will be (A) 2/3 (B) 4/3 (C) 1 (D) 3/2

It is given that X1, X2 ...XM at M non-zero, orthogonal vectors. The dimension of the vector space spanned by the 2M vectors X1, X2,... XM , - X1, - X2,... - XM is (A) 2M (B) M + 1 (C) M (D) dependent on the choice of X1, X2,... XM

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TWO MARKS

(B) y = (y2 - y1) exp a- x k + y1 k (D) y = ^y1 - y2h exp a- x k + y2 k

(B) - jp (D) p

Consider the function f (x) = x2 - x - 2 . The maximum value of f (x) in the closed interval [- 4, 4] is (A) 18 (B) 10 (C) - 225 (D) indeterminate An examination consists of two papers, Paper 1 and Paper 2. The probability of failing in Paper 1 is 0.3 and that in Paper 2 is 0.2. Given that a student has failed in Paper 2, the probability of failing in Paper 1 is 0.6. The probability of a student failing in both the papers is (A) 0.5 (B) 0.18 (C) 0.12 (D) 0.06

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

2006

1.58

1.59

Page 5

ONE MARK

V R S1 1 1 W The rank of the matrix S1 - 1 0 W is SS1 1 1 WW X (B) 1 T (A) 0 (C) 2 (D) 3

1.67

1.62

## (4 # P) $ ds , where P is a vector, is equal to (A) # P $ dl (B) # 4#4# P $ dl (C) # 4# P $ dl (D) ### 4$ Pdv

60%

0.01

Y

30%

0.02

Z

10%

0.03

TWO MARKS

The eigenvalue and the corresponding eigenvector of 2 # 2 matrix are given by Eigenvalue Eigenvector 1 l1 = 8 v1 = = G 1 1 l2 = 4 v2 = = G -1 The matrix is 6 2 4 6 (A) = (B) = G 2 6 6 4G 2 4 4 8 (C) = (D) = G 4 2 8 4G

(A) 2 (C) 6 1.69

jp 2 jp (C) 2 The integral (A) 1 2 (C) 4 3

#0

p

1 dz is positive sense is z2 + 4 z-j = 2 (B) - p 2 (D) p 2

#

sin3 qdq is given by (B) 2 3 (D) 8 3

(B) 4 (D) 8

d2 y For the differential equation 2 + k2 y = 0 the boundary conditions dx are (i) y = 0 for x = 0 and (ii) y = 0 for x = a The form of non-zero solutions of y (where m varies over all integers) are (B) y = Am cos mpx (A) y = Am sin mpx a a m m

/ mp (C) y = / Am x a m

1.70

/ mpx (D) y = / Am e - a m

As x increased from - 3 to 3, the function f (x) =

ex 1 + ex

(A) monotonically increases (B) monotonically decreases (C) increases to a maximum value and then decreases (D) decreases to a minimum value and then increases

For the function of a complex variable W = ln Z (where, W = u + jv and Z = x + jy , the u = constant lines get mapped in Z -plane as (A) set of radial straight lines (B) set of concentric circles (C) set of confocal hyperbolas (D) set of confocal ellipses The value of the constant integral

4 2 For the matrix = the eigenvalue corresponding to the eigenvector 2 4G 101 =101G is

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(D) e u (t)

(A)

1.66

X

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t

2006

1.65

Probability of being supplied defective

SPECIAL EDITION ( STUDY MATERIAL FORM ) At market Book is available in 3 volume i.e. in 3 book binding form. But at NODIA Online Store book is available in 10 book binding form. Each unit of Book is in separate binding.

A solution for the differential equation xo (t) + 2x (t) = d (t) with initial condition x (0-) = 0 is (A) e - 2t u (t) (B) e2t u (t) (C) e u (t)

1.64

1.68

A probability density function is of the form p (x) = Ke- a x , x ! (- 3, 3) The value of K is (A) 0.5 (B) 1 (C) 0.5a (D) a

-t

1.63

% of Computer Supplied

Given that a computer is defective, the probability that was supplied by Y is (A) 0.1 (B) 0.2 (C) 0.3 (D) 0.4

(D) 4 (4$ P) - 4 2 P

1.61

Company

4#4# P , where P is a vector, is equal to (A) P # 4# P - 4 2 P (B) 4 2 P + 4 (4 # P) (C) 4 2 P + 4# P

1.60

Three companies X, Y and Z supply computers to a university. The percentage of computers supplied by them and the probability of those being defective are tabulated below

2005 1.71

The following differential equation has d2 y dy 3 3 c 2 m + 4 c m + y2 + 2 = x dt dt (A) degree = 2 , order = 1 (C) degree = 4 , order = 3

1.72

ONE MARK

(B) degree = 1, order = 2 (D) degree = 2 , order = 3

A fair dice is rolled twice. The probability that an odd number will follow an even number is (A) 1/2 (B) 1/6 (C) 1/3 (D) 1/4

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 1.73

A solution of the following differential equation is given by d2 y dy -5 + 6y = 0 2 dx dx (A) y = e2x + e-3x (B) y = e2x + e3x (C) y = e-2x + 33x (D) y = e-2x + e-3x 2005

1.74

1.75

1.77

In what range should Re (s) remain so that the Laplace transform of the function e(a + 2) t + 5 exits. (A) Re (s) > a + 2 (B) Re (s) > a + 7 (C) Re (s) < 2 (D) Re (s) > a + 5 The derivative of the symmetric function drawn in given figure will look like

Match the following and choose the correct combination: Group I Group 2 E. Newton-Raphson method 1. Solving nonlinear equations F. Runge-kutta method 2. Solving linear simultaneous equations G. Simpson’s Rule 3. Solving ordinary differential equations H. Gauss elimination 4. Numerical integration 5. Interpolation 6. Calculation of Eigenvalues (B) E - 1, F - 6, G - 4, H - 3 (A) E - 6, F - 1, G - 5, H - 3 (C) E - 1, F - 3, G - 4, H - 2 (D) E - 5, F - 3, G - 4, H - 1 -4 2 Given the matrix = , the eigenvector is 4 3G 3 4 (A) = G (B) = G 2 3 2 -1 (C) = G (D) = G -1 2

1.78

1.79

1 2 - 0.1 a and A - 1 = = 2 G. Then (a + b) = G 0 3 0 b (A) 7/20 (B) 3/20 (C) 19/60 (D) 11/20

Let, A = =

The value of the integral I = (A) 1 (C) 2

TWO MARKS

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1.76

Page 6

1.80

1 2p

Given an orthogonal matrix R1 1 1 1 V W S S1 1 - 1 - 1 W A =S 1 - 1 0 0W W S S0 0 1 1 W X T T 6AA @- 1 is V R1 S 4 0 0 0W S 0 14 0 0 W W (A) S 1 S0 0 2 0W S 0 0 0 12 W X T R1 0 0 0 V W S S0 1 0 0 W (C) S 0 0 1 0W W S S0 0 0 1 W X T

2

exp c- x m dx is 8 (B) p (D) 2p

#0

3

R1 S2 S0 (B) S S0 S0 TR 1 S4 S0 (D) S S0 S0 T

V 0 0 0W 1 W 2 0 0 W 1 0 2 0W 0 0 12 W XV 0 0 0W 1 W 4 0 0 W 0 14 0 W 0 0 14 W X

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 7

SOLUTIONS 1.1

Consider the given matrix be

R S2 S1 Im + AB = S S1 S1 T where m = 4 so, we obtain R S2 S1 AB = S S1 S1 T R S1 S1 =S S1 S1 T Hence, we get

Option (B) is correct. Here, as we know Lim sin q . 0 q"0

but for 10% error, we can check option (B) first, q = 18c = 18c # p = 0.314 180c sin q = sin 18c = 0.309 % error = 0.314 - 0.309 # 100% = 0.49% 0.309 Now, we check it for q = 50c

1 2 1 1

1 1 2 1

1 2 1 1

1 1 2 1

1 1 1 1

1 1 1 1

V 1W 1W 1WW 2W X V V R 1W S1 0 0 0W 1W S0 1 0 0W 1WW SS0 0 1 0WW 2W S0 0 0 1W X X T V R V 1W S1W 61 1 1 1@ 1W S1W = 1WW SS1WW 1W S1W X T X

q = 50c = 50c # p = 0.873 180c sin q = sin 50c = 0.77 % error = 0.77 - 0.873 =- 12.25% 0.873 so, the error is more than 10% . Hence, for error less than 10%, q = 18c can have the approximation

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sin q . q 1.2

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Option (A) is correct. For, a given matrix 6A@ the eigen value is calculated as

R V S1W S1W A = S W, B = 81 1 1 1B S1W S1W T X R V Therefore, BA = 81 1 1 1B S1W = 4 S1W S1W S W S1W From the given property T X Det ^Im + AB h = Det ^Im + BAh V V R _ ZR S2 1 1 1W b ]S1 0 0 0W b ]S0 1 0 0W S1 2 1 1W & Det S = Det [S + 4` W W 0 0 1 0W S1 1 2 1W b ]S b ] S0 0 0 1W S1 1 1 2W a \T X X T

A - lI = 0 where l gives the eigen values of matrix. Here, the minimum eigen value among the given options is l =0 We check the characteristic equation of matrix for this eigen value A - lI = A

(for l = 0 )

3 5 2 = 5 12 7 2 7 5 = 3 ^60 - 49h - 5 ^25 - 14h + 2 ^35 - 24h = 33 - 55 + 22 =0 Hence, it satisfied the characteristic equation and so, the minimum eigen value is

= 1+4 =5 Note : Determinant of identity matrix is always 1.

l =0 1.3

Option (D) is correct. Given, the polynomial

f ^x h = a 4 x 4 + a 3 x3 + a2 x2 + a1 x - a 0 Since, all the coefficients are positive so, the roots of equation is given by

f ^x h = 0 It will have at least one pole in right hand plane as there will be least one sign change from ^a1h to ^a 0h in the Routh matrix 1 st column. Also, there will be a corresponding pole in left hand plane i.e.; at least one positive root (in R.H.P) and at least one negative root (in L.H.P) Rest of the roots will be either on imaginary axis or in L.H.P 1.4

Option (B) is correct.

1.5

Option (D) is correct. t dx + x = t dt dx + x = 1 t dt dx + Px = Q (General form) dt Integrating factor, Solution has the form,

IF = e # = e = e lnt = t Pdt

1 # dt t

# ^Q # IF hdt + C x # t = # (1) (t) dt + C

x # IF =

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 8

dy (t) = e-t u (t) + te-t u (t) dt dy = e0 + 0 = 1 dt t = 0

2

xt = t + C 2 Taking the initial condition,

At t = 0+ ,

x (1) = 0.5 0.5 = 1 + C & C = 0 2

1.9

2

xt = t & x = t 2 2

So, 1.6

+

Option (C) is correct. f (z) = 1 2 pj

# f (z) dz C

1 - 2 z+1 z+3

= sum of the residues of the poles which lie inside the given closed region.

C & z+1 = 1 Only pole z =- 1 inside the circle, so residue at z =- 1 is. -z + 1 f (z) = (z + 1) (z + 3) (z + 1) (- z + 1) 2 = =1 = lim 2 z "- 1 (z + 1) (z + 3)

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in 1 2 pj

So 1.7

# f (z) dz C

n+2 = 0 1.10

p

xx = ^ei 2 h & ^ei 2 h = e- 2 1.8

p i

1.12

dy (t) dt

then,

sY (s) - y (0) =

So, 2

L

Y (s)

L

sY (s) - y (0)

max

= (2) 3 - 9 (2) 2 + 24 (2) + 5 = 8 - 36 + 48 + 5 = 25

Option (B) is correct. Characteristic equation.

5l + l2 + 6 = 0

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 l2 + 5l + 6 = 0 Since characteristic equation satisfies its own matrix, so

(- 2s - 3) s +2 (s2 + 2s + 1)

A2 + 5A + 6 = 0 & A2 =- 5A - 6I Multiplying with A A3 + 5A2 + 6A = 0 A3 + 5 (- 5A - 6I) + 6A = 0 A3 = 19A + 30I

2

= - 2s - 32 s + 2s + 4s + 2 (s + 2s + 1) 1 sY (s) - y (0) = s + 2 2 = s + 1 2 + (s + 1) (s + 1) (s + 1) 2 1 = 1 + s + 1 (s + 1) 2 Taking inverse Laplace transform

x = 4, x = 2

A - lI = 0 -5 - l -3 =0 2 -l

Y (s) [s2 + 2s + 1] = 1 - 2s - 4 Y (s) = 2- 2s - 3 s + 2s + 1 y (t)

f (x) = x3 - 9x2 + 24x + 5 df (x) = 3x2 - 18x + 24 = 0 dx df (x) = x2 - 6x + 8 = 0 dx

f (x)

2 6s Y (s) + 2s - 0@ + 2 6sY (s) + 2@ + Y (s) = 1

We know that, If,

=2 3

p

Option (D) is correct. d 2 y (t) 2dy (t) + + y (t) = d (t) dt dt 2 By taking Laplace transform with initial conditions dy 2 ;s Y (s) - sy (0) - dt E + 2 [sy (s) - y (0)] + Y (s) = 1 t=0 &

1 - 14

d 2 f (x) = 6x - 18 dx 2 d 2 f (x) For x = 2, = 12 - 18 =- 6 < 0 dx2 So at x = 2, f (x) will be maximum

- 1 = i = cos p + i sin p 2 2 p x

1 2

Option (B) is correct.

&

=1

x = ei 2

So,

H, TTH, TTTTH, ........... 3 5 P = 1 + b 1 l + b 1 l + ..... 2 2 2 P =

1.11

n =- 2

&

Option (C) is correct. Probability of appearing a head is 1/2. If the number of required tosses is odd, we have following sequence of events. Probability

Option (A) is correct. x=

Option (A) is correct. Divergence of A in spherical coordinates is given as d:A = 12 2 (r 2 Ar ) = 12 2 (krn + 2) r 2r r 2r = k2 (n + 2) rn + 1 r = k (n + 2) rn - 1 = 0 (given)

1.13

Option (D) is correct. From Divergence theorem, we have v $ nt ds v = #A v Adv ### 4$ s

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 9

The position vector

Thus

rv = ^utx x + uty y + utz z h

v = 5rv, thus Here, A

1.19

v 4$ A = cutx 2 + uty 2 + utz 2 m : ^utx x + uty y + utz z h 2x 2y 2z dy dz 5 = 3 # 5 = 15 = c dx + + dx dy dz m So, ## 5rv $ nt ds = ### 15 dv = 15V

1.20

P (E) = No. of favourable outcomes = 15 = 5 36 12 No. of total outcomes

Option (C) is correct. Eigen value of a Skew-symmetric matrix are either zero or pure imaginary in conjugate pairs. Option (C) is correct. For a function x (t) trigonometric fourier series is x (t) = Ao +

s

1.14

We have

#

Integrating or Since y (0) = c thus So, we get, or or 1.15

dy = ky dx dy = # k dx + A y

0

An = 2 # x (t) cos nwt dt T0 T 0

Bn = 2 # x (t) sin nwt dt T0 T For an even function x (t), Bn = 0 Since given function is even function so coefficient Bn = 0 , only cosine

ln y = kx + A ln c = A ln y = kx + ln c ln y = ln ekx + ln c y = cekx

0

Option (A) is correct. 3z + 4 dz where C is circle z = 1 C R Integrals is # 2 z 4z + 5 + C # f (z) dz = 0 if poles are outside C.

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C

2

z + 4z + 5 = 0 (z + 2) 2 + 1 = 0 Thus z1, 2 =- 2 ! j & z1, 2 > 1 So poles are outside the unit circle. Now

1.16

and constant are present in its fourier series representation. Constant term : 3T/4 T/4 3T/4 A0 = 1 # - 2AdtD x (t) dt = 1 : # Adt + # T -T/4 T -T/4 T/4 = 1 :TA - 2AT D =- A 2 2 T 2 Constant term is negative.

Option (C) is correct. f (x) = x + x - 3 = 0 f l (x) = 1 + 1 2 x Substituting x 0 = 2 we get f l (x 0) = 1.35355 and f (x 0) = 2 + We have

1.21

2 - 3 = 0.414

Newton Raphson Method x1 = x 0 -

f (x 0) f l (x 0)

Substituting all values we have x 1 = 2 - 0.414 = 1.694 1.3535 1.17

1.18

n=1

Ao = 1 # x (t) dt T0 " fundamental period T0 T

Where,

Option (C) is correct.

3

/ [An cos nwt + Bn sin nwt]

Option (B) is correct. Writing A: B we have R V S1 1 1 : 6 W S1 4 6 : 20W S W S1 4 l : m W T X Apply R 3 " R 3 - R2 R V 6 W S1 1 1 : S1 4 6 : 20 W S W S0 0 l - 6 : m - 20W T X For equation to have solution, rank of A and A: B must be same. Thus for no solution; l = 6, m ! 20 Option (C) is correct. Total outcome are 36 out of which favorable outcomes are : (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6); (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) which are 15.

Option (D) is correct. Given differential equation d 2 n (x) n (x) - 2 =0 dx 2 L lx Let n (x) = Ae lx So, Al2 elx - Ae2 = 0 L l2 - 12 = 0 & l = ! 1 L L Boundary condition, n (3) = 0 so take l =- 1 L x

So, 1.22

n (x) = Ae- L n (0) = Ae0 = K & A = K n (x) = Ke- (x/L)

Option (A) is correct. Given that

1

ey = x x

1

ln ey = ln x x or y = 1 ln x x 1 dy Now = 1 1 + ln x ^- x- x h = 12 - ln2 xx dx x x For maxima and minima : dy = 12 (1 - ln x) = 0 dx x ln x = 1 " x = e 1 or

2

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 10

From table, at x = 0.3, y (x = 0.3) = 0.031

d 2y =- 23 - ln x b- 23 l - 12 b 1 l dx 2 x x x x =- 22 + 2 ln3 x - 13 x x x 2 d x = -22 + 23 - 13 < 0 dy 2 at x = e e e e 1 So, y has a maximum at x = e Now

1.27

Option (D) is correct. Given that 3s + 1 f (t) = L - 1 ; 3 s + 4s 2 + (K - 3) s E lim f (t) = 1

1

1.23

t"3

By final value theorem

Option (D) is correct. According to given condition head should comes 3 times or 4 times 4 3 P (Heads comes 3 times or 4 times) = 4C 4 b 1 l + 4C 3 b 1 l b 1 l 2 2 2

lim f (t) = lim sF (s) = 1

t"3

1.24

Option (C) is correct. v = xyatx + x 2 aty A v = dxatx + dyaty dl v = # (xyatx + x 2 aty) : (dxatx + dyaty) # Av : dl C

1.28

C

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1.29

1.30

2/ 3

#1/

3

xdx +

1/ 3

#2/

3

3xdx +

#1

3

4 dy + 3

#3

1

1 dy 3

= 1 : 4 - 1 D + 3 :1 - 4 D + 4 [3 - 1] + 1 [1 - 3] 2 3 3 2 3 3 3 3

1.31

1 - 2z z (z - 1) (z - 2) Poles are located at z = 0, z = 1, and z = 2 At Z = 0 residues is R 0 = z : X (z) Z = 0 = 1 - 2 # 0 = 1 2 (0 - 1) (0 - 2)

= 1 - 2 # 2 =- 3 2 2 (2 - 1)

y2 y 4 - + ... 3! 5! Substituting x - p = y we get (x - p) 2 (x - p) 4 + ... f (x) =- 1 + 3! 5! or

h = 0.1, y (0) = 0

y

dy = x+y dx

0

0

0

f (x) = sin x x-p

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R2 = (z - 2) : X (z) z = 2

x

Option (D) is correct.

Substituting x - p = y ,we get sin (y + p) sin y == - 1 (sin y) f (y + p) = y y y 3 5 y y = - 1 cy - + - ...m y 3! 5!

R1 = (Z - 1) : X (Z ) Z = 1

Option (B) is correct. Taking step size

#

unit circle

We have

= 1-2#1 = 1 1 (1 - 2)

1.26

Option (C) is correct. Number of elements in sample space is 210 . Only one element 1 "H, H, T, T, T, T, T, T, T, T , is event. Thus probability is 10 2 Option (C) is correct. We have

= 2pjc1

X (z ) =

At z = 2 ,

Option (B) is correct. The highest derivative present in DE is of 2nd order.

unit circle

Option (C) is correct. Given function

at z = 1,

K =4

s"0

#

=1 1.25

or

lim

f (z) = c0 + c1 z - 1 z (1 + c0) + c1 1 + f (z) 1 + c0 + c1 z - 1 = = f1 (z) = z z z2 Since f1 (z) has double pole at z = 0 , the residue at z = 0 is z (1 + c0) + c1 Res f1 (z) z = 0 = lim z2 .f1 (z) = lim z2 . c m = c1 z"0 z"0 z2 Hence [1 + f (z)] dz = 2pj [Residue at z = 0 ] f1 (z) dz = z

# (xydx + x 2 dy) C

or

s : (3s + 1) =1 s3 + 4s2 + (K - 3) s s (3s + 1) lim 2 =1 s " 0 s [s + 4s + (K - 3)] 1 =1 K-3

or

= 1: 1 +4:1 :1 = 5 16 8 2 16

s"0

yi + 1 = yi + h

dy dx

y1 = 0 + 0.1 (0) = 0

0.1

0

0.1

y2 = 0 + 0.1 (0.1) = 0.01

0.2

0.01

0.21

y 3 = 0.01 + 0.21 # 0.1 = 0.031

0.3

0.031

1.32

f (y + p) =- 1 +

Option (A) is correct. (A)

dy y = dx x dy = dx y x

or

#

#

or

log y = log x + log c

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

or y = cx Thus option (A) and (C) may be correct. dy y (B) =dx x dy or =- dx y x

#

or

1.34

f'( xn) = 1 + e - x

#

1.41

1.37

1.38

Option (B) is correct. The given system is 4 2 x 7 =2 1G=y G = = 6 G 4 2 We have A == 2 1G 4 2 and =0 A = 2 1 4 2 7 Now C == G 2 1 6

1.40

=- 2 =- 1 32 64

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Option (D) is correct.

eP = L- 1 6(sI - A) - 1@ s 0 0 1 -1 -1 = L e= 0 s G =- 2 - 3Go s - 1 -1 -1 o = L e= 2 s + 3G

Rank of matrix r (A) < 2 Rank of matrix r (C) = 2

= L f> -1

1.43

2 4 cos x = 1 + x + x + ... 2! 4!

Thus

f'( x) = 1 + e

-x

2

= ex + sin x = ex + cos x = ex - sin x = e p - sin p = e p f"( p) Thus the coefficient of (x - p) 2 is 2!

Thus only (B) may be 1.44

f (x) = x - e - x

(x - a) 2 f"( a) + ... 2!

(x - p) f"( x)... f (x) = f (p) + x - p f'( p) + 1! 2!

Now

(D + 3) x (t) = 0

x = e-x

Hp

f (x) = f (a) + x - a f'( a) + 1! For x = p we have

Option (C) is correct.

1 (s + 1)( s + 2) s (s + 1)( s + 2)

Option (B) is correct. Taylor series is given as

3 5 sin x = x + x + x + ... 3! 5!

x (t) = Ce - 3t

s+3 (s + 1)( s + 2) -2 (s + 1)( s + 2)

2e - 1 - e - 2 e-1 - e-2 == G - 2e - 1 + 2e - 2 - e - 1 + 2e - 2

Option (B) is correct. dx (t) We have + 3x (t) = 0 dt

or

1 dn - 1 6(z - a) n f (z)@ z=a (n - 1)! dzn - 1

1 d (z - 2) 2 1 2 (2 - 1)! dz ; (z - 2) (z + 2) 2 Ez = a

Option (A) is correct.

We have

n

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Option (A) is correct. We have f (x) = ex + e-x For x > 0 , ex > 1 and 0 < e-x < 1 For x < 0 , 0 < ex < 1 and e-x > 1 Thus f (x) have minimum values at x = 0 and that is e0 + e-0 = 2 .

Since m =- 3 , solution.

n

-2 1 = d ; = dz (z + 2) 2 Ez = a ; (z + 2) 3 Ez = a

Option (A) is correct. sin z can have value between - 1 to + 1. Thus no solution.

or

(1 + xn) e - x 1 + e-x

Option (A) is correct.

Thus Res f (z) z = 2 =

Thus only sin (x3) will have odd power of x . 1.39

n

Here we have n = 2 and a = 2

p11 p22 - p12 p21 = 0

n

- xn

Option (D) is correct. Sum of the principal diagonal element of matrix is equal to the sum of Eigen values. Sum of the diagonal element is - 1 - 1 + 3 = 1.In only option (D), the sum of Eigen values is 1. Option (C) is correct. The product of Eigen value is equal to the determinant of the matrix. Since one of the Eigen value is zero, the product of Eigen value is zero, thus determinant of the matrix is zero.

n

xn + 1 = xn - xn - e- x = 1+e

Thus

Since r (A) ! r (C) there is no solution. 1.36

f (xn) = xn - e - x

Now

Hyperbola

Thus 1.35

The Newton-Raphson iterative formula is f (xn) xn + 1 = xn f'( xn)

Res f (z) z = a =

or 1.33

Straight Line

log y =- log x + log c log y = log 1 + log c x y =c x

or

Page 11

f (x) f'( x) f"( x) f"( p)

Option (A) is correct. The equation of straight line from (0, 0) to (1, 2) is y = 2x . Now g (x, y) = 4x3 + 10y 4 or, g (x, 2x) = 4x3 + 160x 4 Now

1

1

#0 g (x, 2x) = #0 (4x3 + 160x4) dx

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 12

= [x 4 + 32x5] 10 = 33 1.45

1.46

Option (B) is correct. Q

I =2

#P

=2

#1

0

(xdx + ydy) = 2 xdx + 2

Q

#P

xdx + 2

or

y = x+1

Now

I =

2

(x + 2

x k

y = (y1 - y2) e - + y2

2

#1 (x + 1) dx

=

9 4 E = - = 2.5 2 2

Option (C) is correct.

Thus

1.53

+3

1.54

#

1.55

x"3

lim e - x = 3

x"3

2

1.56

x"3

lim e - x = 1

Thus e - x is strictly bounded.

2

2

x"0

1.50

= (3 - x) e 1.51

Neglecting higher powers

-2

f'( x) = 2x - 1 = 0 " x = 1 2

Option (D) is correct. We have or A.E. or

Option (A) is correct. We have

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f (x) = e - x = e - (x - 2) - 2 = e - (x - 2) e - 2 (x - 2) 2 = ;1 - (x - 2) + ...E e - 2 2! = 61 - (x - 2)@ e - 2

Option (C) is correct. For two orthogonal vectors, we require two dimensions to define them and similarly for three orthogonal vector we require three dimensions to define them. 2M vectors are basically M orthogonal vector and we require M dimensions to define them.

f (x) = x2 - x + 2

Option (A) is correct. We have

Option (A) is correct. We know that 1 ds = 2pj [sum of residues] 2 s 1 D Singular points are at s = ! 1 but only s =+ 1 lies inside the given contour, Thus Residue at s =+ 1 is lim (s - 1) f (s) = lim (s - 1) 2 1 = 1 s"1 s"1 s -1 2 1 ds = 2pj 1 = pj `2j 2 s -1 D

#

Option (D) is correct.

lim e - x = 0

f (x) g (x) dx = 0

i.e. common area between f (x) and g (x) is zero.

coth x . 1 x

lim 12 = 3 x"0 x lim x2 = 3

Option (C) is correct. For two orthogonal signal f (x) and g (x)

#- 3

Option (A) is correct. q sin ^ q2 h sin ^ q2 h 1 lim sin ^ 2 h = 1 = 0.5 = lim = lim q"0 q " 0 2^ q h 2 q " 0 ^ q2 h q 2 2 We have,

Option (B) is correct. We have f (x) = x3 - x2 + 4x - 4 f'( x) = 3x2 - 2x + 4 Taking x0 = 2 in Newton-Raphosn method 23 - 22 + 4 (2) - 4 f (x0) x1 = x0 = 2=4 f'( x0) 3 3 (2) 2 - 2 (2) + 4

1) 2 2

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x k

From y (0) = y1 we get C1 + C2 = y1 - y2 From y (3) = y2 we get that C1 must be zero. Thus C2 = y1 - y2

1

as x << 1, cosh x . 1 and sinh x . x

1.49

x k

1.52

#1 ydx

- y22 1 c 2 m = y2 D2 - 12 k k

y = C1 e - + C2 e + y2

#0 ydy = 0

coth x = cosh x sinh x

1.48

Thus solution is

Q

#P ydy

Option (B) is correct. The given plot is straight line whose equation is x +y =1 -1 1

=; 1.47

P.I. =

f"( x) = 2

d2 y = y - y2 dx2 y d2 y y - 2 =- 22 2 k dx k k2

Since f"( x) = 2 > 0 , thus x = 1 is minimum point. The maximum 2 value in closed interval 6- 4, 4@ will be at x =- 4 or x = 4 Now maximum value = max [f (- 4), f (4)] = max (18, 10) = 18

D2 - 12 = 0 k D =! 1 k

1.57 x k

C.F. = C1 e - + C2 e

x k

Option (C) is correct. Probability of failing in paper 1 is Possibility of failing in Paper 2 is

P (A) = 0.3 P (B) = 0.2

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Now x = eu cos v and y = eu sin v Thus x 2 + y 2 = e 2u

Probability of failing in paper 1, when student has failed in paper 2 is P ^ BA h = 0.6 We know that (P + B) Pb A l = B P (B) or P (A + B) = P (B) P b A l = 0.6 # 0.2 = 0.12 B 1.58

1.59

Option (C) is correct. We have V V R R S1 1 1 W S1 1 1 W A = S1 - 1 0 W + S1 - 1 0 W SS1 1 1 WW SS0 0 0 WW X X T T Since one full row is zero, r (A) < 3 1 1 Now =- 2 ! 0 , thus r (A) = 2 1 -1

1.65

Option (D) is correct. We have 1 dz = 2 + z 4 z-j = 2

#

C

1.66

Option (C) is correct. I =

# p (x) dx 3

We know

-3

#

3

-3

-3

Keax dx +

1.67

=1

Ke- a x dx = 1

# Ke 3

- ax

0

dx = 1

1.68

Since x (0 -) = 0

1 s+2

Now taking inverse Laplace transform we have - 2t

x (t) = e u (t)

1.64

Option (A) is correct. Sum of the Eigen values must be equal to the sum of element of principal diagonal of matrix. 6 2 Only matrix = satisfy this condition. 2 6G Option (B) is correct. We have or or

W u + jv eu + jv eu e jv eu (cos v + j sin v)

= ln z = ln (x + jy) = x + jy = x + jy = x + jy

or or 1.69

4 2 2 4G

6A - lI @ [X] = 0

4 - l 2 101 0 = 2 4 - l G=101G = = 0 G

or

sX (s) + 2X (s) = 1

A ==

Now

sX (s) - x (0) + 2X (s) = 1 X (s) =

Option (C) is correct. We have

Taking Laplace transform both sides

1.63

`

P (y + d) = 0.3 # 0.02 = 0.006 P (d) = 0.6 # 0.1 + 0.3 # 0.02 + 0.1 # 0.03 = 0.015 y P a k = 0.006 = 0.4 d 0.015

K eax 0 + k e- ax 3 = 1 @0 a 6 @- 3 (- a) 6 K +K =1 or a a or K =a 2 Option (A) is correct. We have xo (t) + 2x (t) = s (t)

or

p

Option (D) is correct. Let d " defective and y " supply by Y P (y + d) y pa k = d P (d)

or

1.62

#0

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Option (C) is correct.

0

sin3 qdq

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## (4 # F) $ ds = # A $ dl

#

p

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Option (A) is correct. The Stokes theorem is

or

#0

3 sin q - sin 3q dq sin 3q = 3 sin q - 4 sin3 q j 4 p p = :- 3 cos qD = : ws3q D = 8 3 + 3 B - 8 1 + 1 B = 4 12 0 4 4 4 12 12 3 0

=

A # (B # C) = B (A $ C) - C (A $ B) 4#4# P = 4 (4$ P) - P (4$4)

Thus

z-j = 2

1 dz (z + 2i) (z - 2i)

P (0, 2) lies inside the circle z - j = 2 and P (0, - 2) does not lie. Thus By cauchy’s integral formula 2pi = p 1 = I = 2pi lim (z - 2i) z " 2i 2i + 2i 2 (z + 2i)( z - 2i)

R3 - R1

= 4 (4$ P) - 4 2 P

1.61

Equation of circle

#

#

Option (D) is correct. The vector Triple Product is Thus

1.60

Page 13

(101)( 4 - l) + 2 (101) = 0 l =6

Option (A) is correct. d2 y + k2 y = 0 2 dx or D2 y + k2 y = 0 The AE is m2 + k2 = 0 The solution of AE is m = ! ik Thus y = A sin kx + B cos kx From x = 0 , y = 0 we get B = 0 and x = a, y = 0 we get A sin ka = 0 or sin ka = 0 k = mpx a Thus y = Am sin ` mpx j a m We have

/

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 1.70

Page 14

(A - lI) Xi = 0 1 - (- 5) 2 x1 0 == G G G = = 4 8 - 4 x2 0

Option (A) is correct. x

e 1 + ex For x " 3, the value of f (x) monotonically increases. We have

1.71

1.72

1.73

f (x) =

1 2 x1 0 =0 0G=x G = = 0 G 2

Option (B) is correct. Order is the highest derivative term present in the equation and degree is the power of highest derivative term. Order = 2 , degree = 1 Option (D) is correct. Probability of coming odd number is 12 and the probability of coming even number is 12 . Both the events are independent to each other, thus probability of coming odd number after an even number is 12 # 12 = 14 .

x1 + 2x2 = 0 Let - x1 = 2 & x2 =- 1,

The A.E. is

1.78

1.79

1.77

Method-Solving nonlinear eq. Solving ordinary differential eq. Numerical Integration Solving linear simultaneous eq. -4 2 A == 4 3G

A - lI = 0 4-l 2 =0 4 3-l

or

(- 4 - l)(3 - l) - 8 = 0

or - 12 + l + l2 - 8 = 0 or l2 + l - 20 = 0 or l =- 5, 4 Eigen vector for l =- 5

1 2a - 0.1b 1 0 == G =0 3b 0 1G

Option (A) is correct. Gaussian PDF is f (x) = 1 2p s

#

3 - (x - m)2 2s2

-3

e

# f (x) dx 3

-3

Eigen values

dx

for - 3 # x # 3

=1

#

x2 8

or

1 2 2p 2

#

3 - x2 8

e

dx = 1

or

1 2p

#

3 - x2 8

dx = 1

0

0

e

Option (C) is correct. From orthogonal matrix [AAT ] = I

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 Since the inverse of I is I , thus [AAT ] -1 = I-1 = I

Characteristic equation is

or

or

Substituting m = 0 and s = 2 in above we get 3 1 e dx = 1 2p 2 - 3

1.80

Option (C) is correct. We have

1 0 2 - 0.1 1 a =0 3 G= 2 G = =0 1G 0 b

and

Option (C) is correct. For x > 0 the slope of given curve is negative. Only (C) satisfy this condition. " " " "

or

2a - 0.1 = 0 and 3b = 1 Thus solving above we have b = 1 and a = 1 3 60 Therefore a+b = 1 + 1 = 7 3 60 20

Option (A) is correct.

Option (C) is correct. Newton - Raphson Runge - kutta Method Simpson’s Rule Gauss elimination

AA - 1 = I

or

We have f (t) = e(a + 2) t + 5 = e5 .e(a + 2) t Taking Laplace transform we get 1 Thus Re (s) > (a + 2) F (s) = e5 ; s - (a + 2) E

1.76

Eigen vector

Option (A) is correct. We have

Now

The CF is yc = C1 e3x + C2 e2x Since Q = 0 , thus y = C1 e3x + C2 e2x Thus only (B) may be correct.

1.75

2 -1G

1 2 - 0.1 a -1 and A = = 2 G A == 0 3 G 0 b

d2 y dy -5 + 6y = 0 2 dx dx m2 - 5m + 6 = 0 m = 3, 2

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1.74

X ==

Thus

Option (B) is correct. We have

R2 - 4R1

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 15

UNIT 2 NETWORKS

(A) 125/100 and 80/100 (C) 100/100 and 100/100 2013 2.1

2.2

ONE MARK

2.6

Consider a delta connection of resistors and its equivalent star connection as shown below. If all elements of the delta connection are scaled by a factor k , k > 0 , the elements of the corresponding star equivalent will be scaled by a factor of

(A) k2

(B) k

(C) 1/k

(D)

The transfer function

k

V2 ^s h of the circuit shown below is V1 ^s h

(B) 100/100 and 80/100 (D) 80/100 and 80/100

Three capacitors C1 , C2 and C 3 whose values are 10 mF , 5 mF , and 2 mF respectively, have breakdown voltages of 10 V, 5 V and 2 V respectively. For the interconnection shown below, the maximum safe voltage in Volts that can be applied across the combination, and the corresponding total charge in mC stored in the effective capacitance across the terminals are respectively,

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2.3

(B) 3s + 6 s+2 (D) s + 1 s+2

A source vs ^ t h = V cos 100pt has an internal impedance of ^4 + j3h W . If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in W should be (A) 3 (B) 4 (C) 5 (D) 7 2013

2.4

(B) 7 and 119 (D) 7 and 80

TWO MARKS

Common Data For Q. 8 and 9:

In the circuit shown below, if the source voltage VS = 100+53.13c V then the Thevenin’s equivalent voltage in Volts as seen by the load resistance RL is

(A) 100+90c (C) 800+90c 2.5

(A) 2.8 and 36 (C) 2.8 and 32

(B) 800+0c (D) 100+60c

The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An ac voltage VWX1 = 100 V is applied across WX to get an open circuit voltage VYZ1 across YZ. Next, an ac voltage VYZ2 = 100 V is applied across YZ to get an open circuit voltage VWX2 across WX. Then, VYZ1 /VWX1 , VWX2 /VYZ2 are respectively,

Consider the following figure

2.7

2.8

2.9

The current IS in Amps in the voltage source, and voltage VS in Volts across the current source respectively, are (B) 8, - 10 (A) 13, - 20 (C) - 8, 20 (D) - 13, 20 The current in the 1W resistor in Amps is (A) 2 (B) 3.33 (C) 10 (D) 12 Two magnetically uncoupled inductive coils have Q factors q1 and q2

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 16

at the chosen operating frequency. Their respective resistances are R1 and R2 . When connected in series, their effective Q factor at the same operating frequency is (B) ^1/q1h + ^1/q2h (A) q1 + q2 (C) ^q1 R1 + q2 R2h / ^R1 + R2h (D) ^q1 R2 + q2 R1h / ^R1 + R2h

2012 2.10

In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0. The current i (t) for all t is

(A) zero

(A) 0.8 W (C) 2 W

ONE MARK 2.14

If VA - VB = 6 V then VC - VD is

(A) - 5 V (C) 3 V

(B) a step function

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(B) 1.4 W (D) 2.8 W

(B) 2 V (D) 6 V

Common Data For Q. 48 and 49 : With 10 V dc connected at port A in the linear nonreciprocal twoport network shown below, the following were observed : (i) 1 W connected at port B draws a current of 3 A (ii) 2.5 W connected at port B draws a current of 2 A

(C) an exponentially decaying function (D) an impulse function 2.11

2.12

The average power delivered to an impedance (4 - j3) W by a current 5 cos (100pt + 100) A is (A) 44.2 W (B) 50 W (C) 62.5 W (D) 125 W

2.15

In the circuit shown below, the current through the inductor is 2.16

With 10 V dc connected at port A, the current drawn by 7 W connected at port B is (A) 3/7 A (B) 5/7 A (C) 1 A (D) 9/7 A For the same network, with 6 V dc connected at port A, 1 W connected at port B draws 7/3 A. If 8 V dc is connected to port A , the open circuit voltage at port B is (A) 6 V (B) 7 V (C) 8 V (D) 9 V 2011

ONE MARK

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 2 A 1+j (C) 1 A 1+j (A)

2012 2.13

(B) - 1 A 1+j

2.17

In the circuit shown below, the Norton equivalent current in amperes with respect to the terminals P and Q is

(D) 0 A TWO MARKS

Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is (A) 6.4 - j 4.8 (C) 10 + j 0 2.18

(B) 6.56 - j 7.87 (D) 16 + j 0

In the circuit shown below, the value of RL such that the power

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 17

(A) i (t) = 15 exp (- 2 # 103 t) A (B) i (t) = 5 exp (- 2 # 103 t) A (C) i (t) = 10 exp (- 2 # 103 t) A (D) i (t) =- 5 exp (- 2 # 103 t) A

transferred to RL is maximum is

2010

(A) 5 W (C) 15 W 2.19

2.23

(B) 10 W (D) 20 W

For the two-port network shown below, the short-circuit ittance parameter matrix is

The circuit shown below is driven by a sinusoidal input vi = Vp cos (t/RC ). The steady state output vo is

4 (A) > -2 (A) (Vp /3) cos (t/RC ) (C) (Vp /2) cos (t/RC )

(B) (Vp /3) sin (t/RC ) (D) (Vp /2) sin (t/RC )

2011 2.20

TWO MARKS

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2.24

(B) 2.0+0c A (D) 3.2+0c A

In the circuit shown below, the network N is described by the following Y matrix: 0.1 S - 0.01 S . the voltage gain V2 is Y => 0.01 S 0.1 SH V1

2.22

0.5 S 1H

4 (D) > 2

2 S 4H

For parallel RLC circuit, which one of the following statements is NOT correct ? (A) The bandwidth of the circuit decreases if R is increased (B) The bandwidth of the circuit remains same if L is increased (C) At resonance, input impedance is a real quantity (D) At resonance, the magnitude of input impedance attains its minimum value. 2010

2.25

(A) 1/90 (C) –1/99

- 0.5 S 1H

1 (B) > - 0.5

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In the circuit shown below, the current I is equal to

(A) 1.4+0c A (C) 2.8+0c A

-2 S 4H

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1 (C) > 0.5

2.21

ONE MARK

TWO MARKS

In the circuit shown, the switch S is open for a long time and is closed at t = 0 . The current i (t) for t $ 0+ is

(B) –1/90 (D) –1/11

In the circuit shown below, the initial charge on the capacitor is 2.5 mC, with the voltage polarity as indicated. The switch is closed at time t = 0 . The current i (t) at a time t after the switch is closed is

(A) i (t) = 0.5 - 0.125e-1000t A (C) i (t) = 0.5 - 0.5e-1000t A 2.26

(B) i (t) = 1.5 - 0.125e-1000t A (D) i (t) = 0.375e-1000t A

The current I in the circuit shown is

(A) - j1 A

(B) j1 A

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(C) 0 A 2.27

Page 18

(D) 20 A

battery deliver during this talk-time?

In the circuit shown, the power supplied by the voltage source is

(A) 220 J (C) 13.2 kJ

(B) 12 kJ (D) 14.4 J

GATE 2009

(A) 0 W

(B) 5 W

(C) 10 W

(D) 100 W

GATE 2009 2.28

2.31

An AC source of RMS voltage 20 V with internal impedance Zs = (1 + 2j) W feeds a load of impedance ZL = (7 + 4j) W in the figure below. The reactive power consumed by the load is

ONE MARK

In the interconnection of ideal sources shown in the figure, it is known that the 60 V source is absorbing power.

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(A) 8 VAR (C) 28 VAR 2.32

2.33

Which of the following can be the value of the current source I ? (A) 10 A (B) 13 A (C) 15 A (D) 18 A If the transfer function of the following network is Vo (s) 1 = Vi (s) 2 + sCR

The value of the load resistance RL is (B) R (A) R 2 4 (C) R 2.30

(D) 2R

A fully charged mobile phone with a 12 V battery is good for a 10 minute talk-time. Assume that, during the talk-time the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V as shown in the figure. How much energy does the

(B) 16 VAR (D) 32 VAR

The switch in the circuit shown was on position a for a long time, and is move to position b at time t = 0 . The current i (t) for t > 0 is given by

(A) 0.2e-125t u (t) mA (C) 0.2e-1250t u (t) mA

2.29

TWO MARK

(B) 20e-1250t u (t) mA (D) 20e-1000t u (t) mA

In the circuit shown, what value of RL maximizes the power delivered to RL ?

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

2.34

(A) 2.4 W

(B) 8 W 3

(C) 4 W

(D) 6 W

The time domain behavior of an RL circuit is represented by L di + Ri = V0 (1 + Be-Rt/L sin t) u (t). dt For an initial current of i (0) = V0 , the steady state value of the R current is given by (A) i (t) " V0 (B) i (t) " 2V0 R R (C) i (t) " V0 (1 + B) (D) i (t) " 2V0 (1 + B) R R GATE 2008

2.35

Page 19

The component values are (A) L = 5 H, R = 0.5 W, C = 0.1 F (B) L = 0.1 H, R = 0.5 W, C = 5 F (C) L = 5 H, R = 2 W, C = 0.1 F (D) L = 0.1 H, R = 2 W, C = 5 F 2.39

The circuit shown in the figure is used to charge the capacitor C alternately from two current sources as indicated. The switches S1 and S2 are mechanically coupled and connected as follows: For 2nT # t # (2n + 1) T , (n = 0, 1, 2,..) S1 to P1 and S2 to P2 For (2n + 1) T # t # (2n + 2) T, (n = 0, 1, 2,...) S1 to Q1 and S2 to Q2

ONE MARK

In the following graph, the number of trees (P) and the number of cut-set (Q) are

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In the following circuit, the switch S is closed at t = 0 . The rate of change of current di (0+) is given by dt

(A) 0 (R + Rs) Is (C) L

(B) Rs Is L (D) 3

GATE 2008 2.37

Available Only at NODIA Online Store

(B) P = 2, Q = 6 (D) P = 4, Q = 10

TWO MARKS

The Thevenin equivalent impedance Zth between the nodes P and Q in the following circuit is

Assume that the capacitor has zero initial charge. Given that u (t) is a unit step function , the voltage vc (t) across the capacitor is given by (A)

3

/ (- 1) n tu (t - nT)

n=1

3

(B) u (t) + 2 / (- 1) n u (t - nT) n=1 3

(C) tu (t) + 2 / (- 1) n u (t - nT) (t - nT) (A) 1

2.38

(B) 1 + s + 1 s

2 (C) 2 + s + 1 (D) s2 + s + 1 s s + 2s + 1 The driving point impedance of the following network is given by Z (s) = 2 0.2s s + 0.1s + 2

n=1

(D) / 60.5 - e- (t - 2nT) + 0.5e- (t - 2nT) - T @ 3

n=1

Common Data For Q. 2.23 & 2.24 : The following series RLC circuit with zero conditions is excited by a unit impulse functions d (t).

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 2.40

For t > 0 , the output voltage vC ^ t h is

-1 (A) 2 ^e t - e t h (B) 2 te 2 t 3 3 1 -1 (C) 2 e 2 t cos c 3 t m (D) 2 e 2 t sin c 3 t m 2 2 3 3 For t > 0 , the voltage across the resistor is -1 2

2.41

Page 20

(A) 1 _e 3 (B) e (C)

3t 2

3 2

-e

-1t 2

(A) a low- filter (C) a band- filter

i

3 1 sin 3 t c 2 mG =cos c 2 t m 3 2 e -21 t sin 3 t c 2 m 3

-1 t 2

(B) a high- filter (D) a band-reject filter

GATE 2007 2.46

-1 (D) 2 e 2 t cos c 3 t m 2 3

TWO MARKS

Two series resonant filters are as shown in the figure. Let the 3-dB bandwidth of Filter 1 be B1 and that of Filter 2 be B2 . the value B1 is B2

Statement for linked Answers Questions 2.25 & 2.26: A two-port network shown below is excited by external DC source. The voltage and the current are measured with voltmeters V1, V2

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(A) 4 (C) 1/2 2.47

(B) 1 (D) 1/4

For the circuit shown in the figure, the Thevenin voltage and resistance looking into X - Y are

and ammeters. A1, A2 (all assumed to be ideal), as indicated

(A) (C) Under following conditions, the readings obtained are: (1) S1 -open, S2 - closed A1 = 0,V1 = 4.5 V,V2 = 1.5 V, A2 = 1 A (2) S1 -open, S2 - closed A1 = 4 A,V1 = 6 V,V2 = 6 V, A2 = 0 2.42

1.5 1.5G 4.5 1.5 G

1.5 (B) = 1.5 4.5 (D) = 1.5

(B) 4 V, 23 W (D) 4 V, 2 W

In the circuit shown, vC is 0 volts at t = 0 sec. For t > 0 , the capacitor current iC (t), where t is in seconds is given by

(A) 0.50 exp (- 25t) mA

(B) 0.25 exp (- 25t) mA

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-3 -1 (B) = 3 0.67 G 3 1 (D) = - 3 - 0.67 G

GATE 2007

2.45

4.5 4.5G 1.5 4.5G

The h -parameter matrix for this network is -3 3 (A) = - 1 0.67 G 3 3 (C) = 1 0.67 G

2.44

V, 2 W V, 23 W

The z -parameter matrix for this network is 1.5 (A) = 4.5 1.5 (C) = 1.5

2.43

2.48

4 3 4 3

(C) 0.50 exp (- 12.5t) mA ONE MARK

An independent voltage source in series with an impedance Zs = Rs + jXs delivers a maximum average power to a load impedance ZL when (B) ZL = Rs (A) ZL = Rs + jXs (C) ZL = jXs (D) ZL = Rs - jXs

2.49

(D) 0.25 exp (- 6.25t) mA

In the ac network shown in the figure, the phasor voltage VAB (in Volts) is

The RC circuit shown in the figure is (A) 0 (C) 12.5+30c

(B) 5+30c (D) 17+30c

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

GATE 2006 2.50

Page 21

TWO MARKS

A two-port network is represented by ABCD parameters given by V1 A B V2 = I G = =C D G=- I G 1 2 If port-2 is terminated by RL , the input impedance seen at port-1 is given by (B) ARL + C (A) A + BRL BRL + D C + DRL (C) DRL + A (D) B + ARL BRL + C D + CRL

2.51

In the two port network shown in the figure below, Z12 and Z21 and respectively

(A) re and br0 (C) 0 and bro 2.52

2.53

(B) 0 and - br0 (D) re and - br0

The first and the last critical frequencies (singularities) of a driving point impedance function of a ive network having two kinds of elements, are a pole and a zero respectively. The above property will be satisfied by (A) RL network only (B) RC network only (C) LC network only (D) RC as well as RL networks

(A) Rneg # Re Z1 (jw), 6w (C) Rneg # Im Z1 (jw), 6w

GATE 2005 2.56

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A 2 mH inductor with some initial current can be represented as shown below, where s is the Laplace Transform variable. The value of initial current is

(A) 0.5 A (C) 1.0 A

(B) 2.0 A (D) 0.0 A

2.57

In the figure shown below, assume that all the capacitors are initially uncharged. If vi (t) = 10u (t) Volts, vo (t) is given by

-t/0.004

(A) 8e Volts (C) 8u (t) Volts 2.55

ONE MARK

The condition on R, L and C such that the step response y (t) in the figure has no oscillations, is

L C (C) R $ 2 (D) R = 1 LC The ABCD parameters of an ideal n: 1 transformer shown in the figure are n 0 >0 x H (A) R $ 1 2

2.54

(B) Rneg # Z1 (jw) , 6w (D) Rneg # +Z1 (jw), 6w

(B) 8 (1 - e (D) 8 Volts

The value of x will be (A) n

-t/0.004

) Volts

A negative resistance Rneg is connected to a ive network N having driving point impedance as shown below. For Z2 (s) to be positive real,

L C L C

(C) n2 2.58

(B) 1 n (D) 12 n

In a series RLC circuit, R = 2 kW , L = 1 H, and C = 1 mF The 400 resonant frequency is (B) 1 # 10 4 Hz (A) 2 # 10 4 Hz p (C) 10 4 Hz

2.59

(B) R $

(D) 2p # 10 4 Hz

The maximum power that can be transferred to the load resistor RL from the voltage source in the figure is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 22

the figure, then the reading in the ideal voltmeter connected between a and b is

(A) 1 W (C) 0.25 W 2.60

(B) 10 W (D) 0.5 W

The first and the last critical frequency of an RC -driving point impedance function must respectively be (A) a zero and a pole (B) a zero and a zero (C) a pole and a pole (D) a pole and a zero

GATE 2005 2.61

(A) 10 3 90c A 2

(B) 10 3 - 90c A 2

(C) 5 60c A

(D) 5 - 60c A

Impedance Z as shown in the given figure is

(B) j9 W (D) j39 W

For the circuit shown in the figure, Thevenin’s voltage and Thevenin’s equivalent resistance at terminals a - b is

(A) 5 V and 2 W (C) 4 V and 2 W 2.64

The h parameters of the circuit shown in the figure are

For the circuit shown in the figure, the instantaneous current i1 (t) is

(A) j29 W (C) j19 W 2.63

2.65

(B) 0.138 V (D) 1 V

TWO MARKS

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2.62

(A) 0.238 V (C) - 0.238 V

(B) 7.5 V and 2.5 W (D) 3 V and 2.5 W

If R1 = R2 = R4 = R and R3 = 1.1R in the bridge circuit shown in

0.1 0.1 (A) = - 0.1 0.3G 30 20 (C) = 20 20G 2.66

A square pulse of 3 volts amplitude is applied to C - R circuit shown in the figure. The capacitor is initially uncharged. The output voltage V2 at time t = 2 sec is

(A) 3 V (C) 4 V GATE 2004 2.67

10 - 1 (B) = 1 0.05G 10 1 (D) = - 1 0.05G

(B) - 3 V (D) - 4 V ONE MARK

Consider the network graph shown in the figure. Which one of the following is NOT a ‘tree’ of this graph ?

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) a (C) c 2.68

Page 23

(B) b (D) d

The equivalent inductance measured between the terminals 1 and 2 for the circuit shown in the figure is

GATE 2004 2.72

(A) L1 + L2 + M (C) L1 + L2 + 2M 2.69

(B) L1 + L2 - M (D)L1 + L2 - 2M

The circuit shown in the figure, with R = 1 W, L = 1 H and C = 3 F 3 4 has input voltage v (t) = sin 2t . The resulting current i (t) is

TWO MARKS

For the lattice shown in the figure, Za = j2 W and Zb = 2 W . The z11 z12 values of the open circuit impedance parameters 6 z @ = = are z21 z22 G

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(A) (B) (C) (D) 2.70

5 sin (2t + 53.1c) 5 sin (2t - 53.1c) 25 sin (2t + 53.1c) 25 sin (2t - 53.1c)

For the circuit shown in the figure, the time constant RC = 1 ms. The input voltage is vi (t) = 2 sin 103 t . The output voltage vo (t) is equal to

1-j (A) = 1+j 1+j (C) = 1-j 2.73

(A) sin (103 t - 45c) (C) sin (103 t - 53c) 2.71

(B) sin (103 t + 45c) (D) sin (103 t + 53c)

1+j 1 + jG 1+j 1 - jG

1-j 1+j (B) = -1 + j 1 - j G 1 + j -1 + j (D) = -1 + j 1 + j G

The circuit shown in the figure has initial current iL (0-) = 1 A through the inductor and an initial voltage vC (0-) =- 1 V across the capacitor. For input v (t) = u (t), the Laplace transform of the current i (t) for t $ 0 is

For the R - L circuit shown in the figure, the input voltage vi (t) = u (t). The current i (t) is

s+2 s2 + s + 1 (D) 2 1 s +s+1 V (s) The transfer function H (s) = o of an RLC circuit is given by Vi (s) s s2 + s + 1 (C) 2 s - 2 s +s+1 (A)

2.74

(B)

106 s + 20s + 106 The Quality factor (Q-factor) of this circuit is (A) 25 (B) 50 H (s) =

2

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(C) 100 2.75

2.76

(D) 5000

(C) 100

For the circuit shown in the figure, the initial conditions are zero. Its V (s) is transfer function H (s) = c Vi (s)

1 (A) 2 s + 106 s + 106 103 (C) 2 s + 103 s + 106

Page 24

2.80

Consider the following statements S1 and S2 S1 : At the resonant frequency the impedance of a series RLC circuit is zero. S2 : In a parallel GLC circuit, increasing the conductance G results in increase in its Q factor.

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2.77

2.81

2.82

2.79

TWO MARKS

Twelve 1 W resistance are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is (A) 5 W (B) 1 W 6 (D) 3 W (C) 6 W 2 5 The current flowing through the resistance R in the circuit in the figure has the form P cos 4t where P is

(A) (0.18 + j0.72) (C) - (0.18 + j1.90)

(B) (0.46 + j1.90) (D) - (0.192 + j0.144)

The circuit for Q. 2.66 & 2.67 is given below. Assume that the switch S is in position 1 for a long time and thrown to position 2 at t = 0 .

ONE MARK

The minimum number of equations required to analyze the circuit shown in the figure is

(A) 3 (C) 6 2.78

2 (B) d 2i + 2 di + 2i (t) = cos t dt dt 2 (D) d 2i + 2 di + 2i (t) = sin t dt dt

GATE 2003

Which one of the following is correct? (A) S1 is FALSE and S2 is TRUE (B) Both S1 and S2 are TRUE (C) S1 is TRUE and S2 is FALSE (D) Both S1 and S2 are FALSE GATE 2003

The differential equation for the current i (t) in the circuit of the figure is

2 (A) 2 d 2i + 2 di + i (t) = sin t dt dt 2 (C) 2 d 2i + 2 di + i (t) = cos t dt dt

106 (B) 2 s + 103 s + 106 106 (D) 2 s + 106 s + 106

(D) 200

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(B) 4 (D) 7

A source of angular frequency 1 rad/sec has a source impedance consisting of 1 W resistance in series with 1 H inductance. The load that will obtain the maximum power transfer is (A) 1 W resistance (B) 1 W resistance in parallel with 1 H inductance (C) 1 W resistance in series with 1 F capacitor (D) 1 W resistance in parallel with 1 F capacitor A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100 . If each of R, L and C is doubled from its original value, the new Q of the circuit is (A) 25 (B) 50

2.83

2.84

At t = 0+ , the current i1 is (B) - V (A) - V R 2R (C) - V (D) zero 4R I1 (s) and I2 (s) are the Laplace transforms of i1 (t) and i2 (t) respectively. The equations for the loop currents I1 (s) and I2 (s) for the circuit shown in the figure, after the switch is brought from position 1 to position 2 at t = 0 , are V R + Ls + Cs1 - Ls I1 (s) s (A) > == G G 1 H= - Ls R + Cs I2 (s) 0 R + Ls + Cs1 - Ls I1 (s) - Vs (B) > = - Ls R + Cs1 H=I2 (s)G = 0 G

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 25

(A) 25 V (C) - 50 V

R + Ls + Cs1 - Ls I1 (s) - Vs (C) > = G G = = H - Ls R + Ls + Cs1 I2 (s) 0 V R + Ls + Cs1 - Cs I1 (s) s (D) > == G G 1 H= - Ls R + Ls + Cs I2 (s) 0

2.85

2.86

GATE 2002

The driving point impedance Z (s) of a network has the pole-zero locations as shown in the figure. If Z (0) = 3 , then Z (s) is

3 (s + 3) 2 (s + 3) (B) 2 s + 2s + 3 s + 2s + 2 3 (s + 3) 2 (s - 3) (C) 2 (D) 2 s + 2s + 2 s - 2s - 3 An input voltage v (t) = 10 2 cos (t + 10c) + 10 5 cos (2t + 10c) V is applied to a series combination of resistance R = 1 W and an inductance L = 1 H. The resulting steady-state current i (t) in ampere is (A) 10 cos (t + 55c) + 10 cos (2t + 10c + tan-1 2) (A)

(B) 10 cos (t + 55c) + 10 23 cos (2t + 55c) (C) 10 cos (t - 35c) + 10 cos (2t + 10c - tan-1 2)

2.87

2.90

2

(D) 10 cos (t - 35c) +

3 2

2.91

TWO MARKS

In the network of the fig, the maximum power is delivered to RL if its value is

(A) 16 W

(B) 40 W 3

(C) 60 W

(D) 20 W

If the 3-phase balanced source in the figure delivers 1500 W at

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cos (2t - 35c)

The impedance parameters z11 and z12 of the two-port network in the figure are

(A) (B) (C) (D)

(B) 50 V (D) 0 V

a leading power factor 0.844 then the value of ZL (in ohm) is approximately

z11 = 2.75 W and z12 = 0.25 W z11 = 3 W and z12 = 0.5 W z11 = 3 W and z12 = 0.25 W z11 = 2.25 W and z12 = 0.5 W

(A) 90+32.44c (C) 80+ - 32.44c

(B) 80+32.44c (D) 90+ - 32.44c

GATE 2001 GATE 2002 2.88

2.92

The Voltage e0 in the figure is

The dependent current source shown in the figure

(A) delivers 80 W (C) delivers 40 W 2.89

ONE MARK

ONE MARK

(A) 2 V (C) 4 V

(B) absorbs 80 W (D) absorbs 40 W

In the figure, the switch was closed for a long time before opening at t = 0 . The voltage vx at t = 0+ is

2.93

2.94

(B) 4/3 V (D) 8 V

If each branch of Delta circuit has impedance 3 Z , then each branch of the equivalent Wye circuit has impedance (B) 3Z (A) Z 3 (D) Z (C) 3 3 Z 3 The ittance parameter Y12 in the 2-port network in Figure is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 26 2.98

(A) - 0.02 mho (C) - 0.05 mho

(B) 0.1 mho (D) 0.05 mho

GATE 2001 2.95

The z parameters z11 and z21 for the 2-port network in the figure are

(A) z11 = 6 W; z21 = 16 W 11 11 (C) z11 = 6 W; z21 =- 16 W 11 11

TWO MARKS

The voltage e0 in the figure is

(B) z11 = 6 W; z21 = 4 W 11 11 (D) z11 = 4 W; z21 = 4 W 11 11

GATE 2000 2.99

(A) 48 V (C) 36 V

The circuit of the figure represents a

(B) 24 V (D) 28 V

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ONE MARK

(A) Low filter (C) band filter 2.100

(B) High filter (D) band reject filter

In the circuit of the figure, the voltage v (t) is

When the angular frequency w in the figure is varied 0 to 3, the locus of the current phasor I2 is given by (A) eat - ebt (C) aeat - bebt 2.101

(B) eat + ebt (D) aeat + bebt

In the circuit of the figure, the value of the voltage source E is

(A) - 16 V

(B) 4 V

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2.97

In the figure, the value of the load resistor RL which maximizes the power delivered to it is

(A) 14.14 W (C) 200 W

(B) 10 W (D) 28.28 W

GATE 2000 2.102

(D) 16 V TWO MARKS

Use the data of the figure (a). The current i in the circuit of the figure (b)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 27 2.107

(A) - 2 A (C) - 4 A

A Delta-connected network with its Wye-equivalent is shown in the given figure. The resistance R1, R2 and R3 (in ohms) are respectively

(A) 1.5, 3 and 9 (C) 9, 3 and 1.5

(B) 2 A (D) 4 A

(B) 3, 9 and 1.5 (D) 3, 1.5 and 9

GATE 1998 GATE 1999 2.103

ONE MARK

2.108

Identify which of the following is NOT a tree of the graph shown in the given figure is

2.109

(A) begh (C) abfg 2.104

ONE MARK

A network has 7 nodes and 5 independent loops. The number of branches in the network is (A) 13 (B) 12 (C) 11 (D) 10 The nodal method of circuit analysis is based on

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(B) defg (D) aegh

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A 2-port network is shown in the given figure. The parameter h21 for this network can be given by

(A) KVL and Ohm’s law (C) KCL and KVL 2.110

(A) - 1/2 (C) - 3/2

(B) + 1/2 (D) + 3/2

GATE 1999 2.105

2.111

The parallel RLC circuit shown in the figure is in resonance. In this circuit

(A) IR < 1 mA (C) IR + IC < 1 mA 2.112

2.106

Superposition theorem is NOT applicable to networks containing (A) nonlinear elements (B) dependent voltage sources (C) dependent current sources (D) transformers

TWO MARK

The Thevenin equivalent voltage VTH appearing between the terminals A and B of the network shown in the given figure is given by

(A) j16 (3 - j4) (C) 16 (3 + j4)

(B) KCL and Ohm’s law (D) KCL, KVL and Ohm’s law

(B) j16 (3 + j4) (D) 16 (3 - j4)

2.113

(B) IR + IL > 1 mA (D) IR + IC > 1 mA

0 - 1/2 The short-circuit ittance matrix a two-port network is > 1/2 0 H The two-port network is (A) non-reciprocal and ive (B) non-reciprocal and active (C) reciprocal and ive (D) reciprocal and active The voltage across the terminals a and b in the figure is

The value of R (in ohms) required for maximum power transfer in the network shown in the given figure is

(A) 2 (C) 8

(B) 4 (D) 16

(A) 0.5 V (C) 3.5 V 2.114

(B) 3.0 V (D) 4.0 V

A high-Q quartz crystal exhibits series resonance at the frequency

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

ws and parallel resonance at the frequency wp . Then (A) ws is very close to, but less than wp (B) ws << wp (C) ws is very close to, but greater than wp (D) ws >> wp GATE 1997 2.115

Page 28

(A) 10 V (C) 5 V 2.119

In the circuit of the figure is the energy absorbed by the 4 W resistor in the time interval (0, 3) is

ONE MARK

The current i4 in the circuit of the figure is equal to

(A) 36 Joules (C) 256 Joules 2.120

(A) 12 A (C) 4 A 2.116

(B) 15 V (D) None of the above

(B) 16 Joules (D) None of the above

In the circuit of the figure the equivalent impedance seen across terminals a, b, is

(B) - 12 A (D) None or these

The voltage V in the figure equal to

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(A) b 16 l W 3 (C) b 8 + 12j l W 3

(B) b 8 l W 3 (D) None of the above

GATE 1996 2.121

(A) 3 V (C) 5 V 2.117

ONE MARK

In the given figure, A1, A2 and A3 are ideal ammeters. If A2 and A3 read 3 A and 4 A respectively, then A1 should read

(B) - 3 V (D) None of these (A) 1 A

The voltage V in the figure is always equal to

(B) 5 A

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 (C) 7 A (A) 9 V (C) 1 V 2.118

(B) 5 V (D) None of the above

2.122

The voltage V in the figure is

The number of independent loops for a network with n nodes and b branches is (A) n - 1 (B) b - n (C) b - n + 1 (D) independent of the number of nodes GATE 1996

2.123

(D) None of these

TWO MARKS

The voltages VC1, VC2, and VC3 across the capacitors in the circuit in

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 29

the given figure, under steady state, are respectively.

(A) 80 V, 32 V, 48 V (C) 20 V, 8 V, 12 V

(B) 80 V, 48 V, 32 V (D) 20 V, 12 V, 8 V

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 30

SOLUTIONS 2.1

Option (B) is correct. In the equivalent star connection, the resistance can be given as Rb Ra RC = Ra + Rb + Rc Ra Rc RB = Ra + Rb + Rc Rb Rc RA = Ra + Rb + Rc

As the circuit open across RL so I2 = 0 or, j40I2 = 0 i.e., the dependent source in loop 1 is short circuited. Therefore, ^ j4h Vs VL1 = j4 + 3 j40 100 53.13c j4 + 3 40 90c 100 53.13c = 5 53.13c = 800 90c

VTh = 10 VL1 =

2.5

So, if the delta connection components Ra , Rb and Rc are scaled by a factor k then ^k Rb h^k Rc h RAl = kRa + kRb + kRc

Option (C) is correct. For the given transformer, we have V = 1.25 1 VWX

2 Rb Rc =k k Ra + Rb + Rc

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Since, So, or, at

= k RA hence, it is also scaled by a factor k 2.2

2.3

2.4

Option (D) is correct. For the given capacitance, C = 100mF in the circuit, we have the reactance. 1 10 4 = XC = 1 = sc s s # 100 # 10-6 So, 10 4 + 10 4 V2 ^s h = 4 s V1 ^s h 10 + 10 4 + 10 4 s s = s+1 s+2 Option (C) is correct. For the purely resistive load, maximum average power is transferred when 2 2 + XTh RL = RTh where RTh + jXTh is the equivalent thevinin (input) impedance of the circuit. Hence, we obtain RL = 42 + 32 5W

VYZ = 0.8 (attenuation factor) V VYZ = 0.8 1.25 = 1 ^ h^ h VWX VYZ = VWX V VWX = 100 V ; YZ = 100 100 VWX 1

1

1

at

VWZ = 100 V ; 2

VWX = 100 100 VYZ 2

2

2.6

Option (C) is correct. The quality factor of the inductances are given by q1 = wL1 R1 and q2 = wL2 R2 So, in series circuit, the effective quality factor is given by XLeq = wL 1 + wL 2 Q = Req R1 + R 2 q1 q wL 1 + wL 2 + 2 q R + q2 R2 R R R R R R 1 2 2 = 1 2 = 1 1 = 2 1 + 1 1 + 1 R1 + R 2 R 2 R1 R 2 R1

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Option (C) is correct.

Option (C) is correct. For evaluating the equivalent thevenin voltage seen by the load RL , we open the circuit across it (also if it consist dependent source). The equivalent circuit is shown below Consider that the voltage across the three capacitors C1 , C2 and C 3 are V1 , V2 and V3 respectively. So, we can write V2 = C 3 V3 C 2 ....(1)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

2.8

Since, Voltage is inversely proportional to capacitance Now, given that C1 = 10 mF ; ^V1hmax = 10V C2 = 5 mF ; ^V2hmax = 5 V C 3 = 2 mF ; ^V3hmax = 2V So, from Eq (1) we have V2 = 2 5 V3 for ^V3hmax = 2 We obtain, V2 = 2 # 2 = 0.8 volt < 5 5 i.e., V2 < ^V2hmax Hence, this is the voltage at C2 . Therefore, V3 = 2 volt V2 = 0.8 volt and V1 = V2 + V3 = 2.8 volt Now, equivalent capacitance across the terminal is Ceq = C 2 C 3 + C1 = 5 # 2 + 10 = 80 mF 5+2 7 C2 + C3 Equivalent voltage is (max. value) Vmax = V1 = 2.8 So, charge stored in the effective capacitance is Q = Ceq Vmax = b 80 l # ^2.8h = 32 mC 7 Option (D) is correct.

Page 31

I (s) =

vc (0) /s v (0) = c 1 + 1 1 + 1 C1 C 2 C1 s C 2 s

vC (0) = 12 V I (s) = b C1 C2 l (12 V) = 12Ceq C1 + C 2 Taking inverse Laplace transform for the current in time domain, i (t) = 12Ceq d (t) 2.11

Option (B) is correct. In phasor form,

Z = 4 - j 3 = 5 - 36.86cW I = 5 100c A

Average power delivered. Pavg. = 1 I 2 Z cos q = 1 # 25 # 5 cos 36.86c = 50 W 2 2

Alternate method: Z = (4 - j3) W ,

I = 5 cos (100pt + 100) A

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Click to Buy www.nodia.co.in 2 Pavg = 1 Re $ I Z . = 1 # Re "(5) 2 # (4 - j3), 2 2 = 1 # 100 = 50 W 2 2.12

At the node 1, voltage is given as V1 = 10 volt Applying KCL at node 1 IS + V1 + V1 - 2 = 0 2 1 10 10 IS + + - 2 = 0 2 1 IS =- 13 A Also, from the circuit, VS - 5 # 2 = V1 VS = 10 + V1 = 20 volt 2.9

2.10

(Impulse)

Option (C) is correct

Applying nodal analysis at top node. V1 + 1 0c V1 + 1 0c + = 1 0c 1 j1

Option (C) is correct. Again from the shown circuit, the current in 1 W resistor is I = V1 = 10 = 10 A 1 1 Option (D) is correct. The s -domain equivalent circuit is shown as below.

2.13

V1 (j 1 + 1) + j 1 + 1 0c = j1 V1 = - 1 1 + j1 1 V1 + 1 0c - 1 + j + 1 Current = I1 = j1 j1 j = = 1 A (1 + j) j 1 + j Option (A) is correct. We obtain Thevenin equivalent of circuit B .

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 32

So current in the branch will be IAB = 6 = 3 A 2 We can see, that the circuit is a one port circuit looking from terminal BD as shown below Thevenin Impedance :

ZTh = R Thevenin Voltage :

For a one port network current entering one terminal, equals the current leaving the second terminal. Thus the outgoing current from A to B will be equal to the incoming current from D to C as shown i.e. IDC = IAB = 3 A

VTh = 3 0c V Now, circuit becomes as

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The total current in the resistor 1 W will be (By writing KCL at node D ) I1 = 2 + IDC = 2+3 = 5A So, VCD = 1 # (- I1) =- 5 V 2.15

I1 = 10 - 3 2+R Power transfer from circuit A to B P = (I 12) 2 R + 3I1 2 P = :10 - 3D R + 3 :10 - 3D 2+R 2+R P = 49R 2 + 21 (2 + R) (2 + R) 49R + 21 (2 + R) P = (2 + R) 2 P = 42 + 70R2 (2 + R) 2 dP = (2 + R) 70 - (42 + 70R) 2 (2 + R) = 0 dR (2 + R) 4 (2 + R) [(2 + R) 70 - (42 + 70R) 2] = 0 140 + 70R - 84 - 140R = 0 56 = 70R R = 0.8 W

Option (C) is correct. When 10 V is connected at port A the network is

Current in the circuit,

2.14

Option (A) is correct. In the given circuit VA - VB = 6 V

Now, we obtain Thevenin equivalent for the circuit seen at load terminal, let Thevenin voltage is VTh, 10 V with 10 V applied at port A and Thevenin resistance is RTh .

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IL =

VTh,10 V RTh + RL

For RL = 1 W , IL = 3 A V 3 = Th,10 V RTh + 1 For RL = 2.5 W , IL = 2 A V = Th,10 V RTh + 2.5

...(i) ...(ii)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Dividing above two 3 = RTh + 2.5 2 RTh + 1

Page 33

ISC = 2.18

3RTh + 3 = 2RTh + 5 RTh = 2 W Substituting RTh into equation (i)

Option (C) is correct. Power transferred to RL will be maximum when RL is equal to the Thevenin resistance. We determine Thevenin resistance by killing all source as follows :

VTh,10 V = 3 (2 + 1) = 9 V Note that it is a non reciprocal two port network. Thevenin voltage seen at port B depends on the voltage connected at port A. Therefore we took subscript VTh,10 V . This is Thevenin voltage only when 10 V source is connected at input port A. If the voltage connected to port A is different, then Thevenin voltage will be different. However, Thevenin’s resistance remains same. Now, the circuit is as shown below :

RTH = 10 # 10 + 10 = 15 W 10 + 10 2.19

For RL = 7 W , 2.16

IL =

VTh,10 V = 9 = 1A 2 + RL 2 + 7

Option (A) is correct. The given circuit is shown below

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Option (B) is correct. Now, when 6 V connected at port A let Thevenin voltage seen at port B is VTh,6 V . Here RL = 1 W and IL = 7 A 3

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VTh, 6 V = RTh # 7 + 1 # 7 = 2 # 7 + 7 = 7 V 3 3 3 3 This is a linear network, so VTh at port B can be written as VTh = V1 a + b where V1 is the input applied at port A. We have V1 = 10 V , VTh,10 V = 9 V ...(i) ` 9 = 10a + b When V1 = 6 V , VTh, 6 V = 9 V ...(ii) ` 7 = 6a + b Solving (i) and (ii) a = 0.5 , b = 4 Thus, with any voltage V1 applied at port A, Thevenin voltage or open circuit voltage at port B will be So, VTh, V = 0.5V1 + 4 For V1 = 8 V VTh,8 V = 0.5 # 8 + 4 = 8 = Voc (open circuit voltage)

For parallel combination of R and C equivalent impedance is R$ 1 jwC R = Zp = 1 j 1 + wRC R+ jwC Transfer function can be written as R 1 + jwRC Vout = Z p = Vin Zs + Zp R R+ 1 + jwC 1 + jwRC

Option (A) is correct. Replacing P - Q by short circuit as shown below we have

Using current divider rule the current Isc is

jwRC jwRC + (1 + jwRC) 2 j Here w = 1 = 2 RC j + (1 + j) j = =1 3 (1 + j) 2 + j V = b p l cos (t/RC) 3 =

Vout Vin

1

2.17

25 (16 0 ) = (6.4 - j4.8) A 25 + 15 + j30

Thus 2.20

v out

Option (B) is correct. From star delta conversion we have

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Thus

R1 =

Page 34

Ra Rb 6.6 = = 2W Ra + Rb + Rc 6 + 6 + 6

Given circuit is as shown below

Here R1 = R 2 = R 3 = 2 W Replacing in circuit we have the circuit shown below :

By writing node equation at input port I1 = V1 + V1 - V2 = 4V1 - 2V2 0.5 0.5 By writing node equation at output port I2 = V2 + V2 - V1 =- 2V1 + 4V2 0.5 0.5 From (1) and (2), we have ittance matrix 4 -2 Y => - 2 4H

Now the total impedance of circuit is (2 + j4) (2 - j4) +2 = 7W Z = (2 + j4) (2 - j4) Current I = 14+0c = 2+0c 7 2.21

2.24

...(2)

Option (D) is correct. A parallel RLC circuit is shown below :

Option (D) is correct.

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in From given ittance matrix we get I1 = 0.1V1 - 0.01V2 and

...(1) ...(2)

I2 = 0.01V1 + 0.1V2 Now, applying KVL in outer loop; V2 =- 100I2 or I2 =- 0.01V2 From eq (2) and eq (3) we have

Now,

Z in =

At resonance

1 = wC wL Z in = 1 = R 1/R

(maximum at resonance)

Thus (D) is not true. Furthermore bandwidth is wB i.e wB \ 1 and is independent of L R , Hence statements A, B, C, are true. 2.25

Option (A) is correct. Here we take the current flow direction as positive. At t = 0- voltage across capacitor is -3 Q VC (0-) =- =- 2.5 # 10-6 =- 50 V C 50 # 10 + Thus VC (0 ) =- 50 V In steady state capacitor behave as open circuit thus

1 1 + 1 + jw C R jwL

Input impedance

So,

...(3)

- 0.01V2 = 0.01V1 + 0.1V2 - 0.11V2 = 0.01V1 V2 = - 1 11 V1 2.22

...(1)

Option (A) is correct. Let the current i (t) = A + Be-t/t t " Time constant When the switch S is open for a long time before t < 0 , the circuit is

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V (3) = 100 V VC (t) = VC (3) + (VC (0+) - VC (3)) e-t/RC = 100 + (- 50 - 100) e

-t 10 # 50 # 10-6

3

Now

= 100 - 150e- (2 # 10 t) ic (t) = C dV dt 3

= 50 # 10-6 # 150 # 2 # 103 e-2 # 10 t A 3

= 15e-2 # 10 t ic (t) = 15 exp (- 2 # 103 t) A 2.23

Option (A) is correct.

At t = 0 , inductor current does not change simultaneously, So the circuit is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 35

Applying nodal analysis VA - 10 + 1 + VA - 0 = 0 2 2 Current, Current is resistor (AB) i (0) = 0.75 = 0.375 A 2

2VA - 10 + 2 = 0 = V4 = 4 V I1 = 10 - 4 = 3 A 2

Current from voltage source is I 2 = I1 - 3 = 0 Since current through voltage source is zero, therefore power delivered is zero.

Similarly for steady state the circuit is as shown below 2.28

Option (A) is correct. Circuit is as shown below

i (3) = 15 = 0.5 A 3 -3 = 10-3 sec t = L = 15 # 10 Req 10 + (10 || 10)

Now and So, Hence 2.26

i (t) i (0) i (3) B i (t)

t

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= A + Be- 1 # 10 = A + Be-100t = A + B = 0.375 = A = 0.5 = 0.375 - 0.5 =- 0.125 = 0.5 - 0.125e-1000 t A -3

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Option (A) is correct. Circuit is redrawn as shown below

Where,

Z1 = jwL = j # 103 # 20 # 10-3 = 20j Z2 = R || XC 1 =- 20j XC = 1 = jwC j # 103 # 50 # 10-6 1 (- 20j) Z2 = R = 1W 1 - 20j

Since 60 V source is absorbing power. So, in 60 V source current flows from + to - ve direction So, I + I1 = 12

Voltage across Z2 VZ = 2

Z2 : 20 0 = Z1 + Z 2

=c

- 20j c 1 - 20j m 20j c 20j - 1 - 20j m

(- 20j) : 20 =- j 20j + 400 - 20j m

I = 12 - I1 I is always less then 12 A So, only option (A) satisfies this conditions.

: 20

2.29

Current in resistor R is j V I = Z =- =- j A 1 R 2

2.27

Option (A) is correct. The circuit can be redrawn as

Option (C) is correct. For given network we have (RL XC ) Vi V0 = R + (RL XC ) RL V0 (s) RL = 1 + sRL C = Vi (s) R RR R + L L sC + RL R+ 1 + sRL C =

RL 1 = R + RRL sC + RL R 1+ + RsC RL

But we have been given V (s) 1 = T . F. = 0 2 + sCR Vi (s) Comparing, we get 1 + R = 2 & RL = R RL 2.30

Option (C) is correct.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 36

The energy delivered in 10 minutes is E =

t

t

#0 VIdt = I #0Vdt

= I # Area

= 2 # 1 (10 + 12) # 600 = 13.2 kJ 2 2.31

Option (B) is correct. From given circuit the load current is 20+0c = = 20+0c IL = V Zs + ZL (1 + 2j) + (7 + 4j) 8 + 6j = 1 (8 - 6j) = 20+0c = 2+ - f where f = tan - 1 3 10+f 5 4

The open circuit voltage is From fig

The voltage across load is VL = IL ZL The reactive power consumed by load is

Vx =- 4 # 12.5 =- 50 V I2 = 100 + Vx = 100 - 50 = 12.5 A 4 4

Pr = VL IL* = IL ZL # IL* = ZL IL 2 2 = (7 # 4j) 20+0c = (7 + 4j) = 28 + 16j 8 + 6j

Isc = I1 + I2 = 25 A Rth = Voc = 100 = 4 W Isc 25

Thus average power is 28 and reactive power is 16. 2.32

Thus for maximum power transfer RL = Req = 4 W

Option (B) is correct.

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Voc = 100 V I1 = 100 = 12.5 A 8

2.34

2.35

-

At t = 0 , the circuit is as shown in fig below :

Option (A) is correct. Steady state all transient effect die out and inductor act as short circuits and forced response acts only. It doesn’t depend on initial current state. From the given time domain behavior we get that circuit has only R and L in series with V0 . Thus at steady state i (t) " i (3) = V0 R Option (C) is correct. The given graph is

There can be four possible tree of this graph which are as follows: V (0-) = 100 V Thus V (0+) = 100 V At t = 0+ , the circuit is as shown below There can be 6 different possible cut-set.

I (0+) = 100 = 20 mA 5k At steady state i.e. at t = 3 is I (3)= 0 Now

i (t) = I (0+) e-

t RCeq

u (t)

(0.5m + 0.3m) 0.2m = 0.16 m F 0.5m + 0.3m + 0.2m 1 1 = = 1250 RCeq 5 # 103 # 0.16 # 10-6 i (t) = 20e-1250t u (t) mA Ceq =

2.33

Option (C) is correct. For Pmax the load resistance RL must be equal to thevenin resistance Req i.e. RL = Req . The open circuit and short circuit is as shown below

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Option (B) is correct. Initially i (0-) = 0 therefore due to inductor i (0+) = 0 . Thus all current Is will flow in resistor R and voltage across resistor will be Is Rs . The voltage across inductor will be equal to voltage across Rs as no current flow through R.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Thus but Thus 2.37

Page 37

vL (0+) = Is Rs di (0+) vL (0+) = L dt

For 4T < t < 5T , capacitor will be charged from 0 V

di (0+) vL (0+) Is Rs = = L L dt

At t = 5T, Vc = T Volts Thus the output waveform is

Vc =

t

#4Tdt = t - 4T

Option (A) is correct. Killing all current source and voltage sources we have,

Only option C satisfy this waveform. 2.40

Zth = =

(1 + s) ( s1 + 1) (1 + s)( s1 + 1) (1 + s) + ( s1 + 1)

=

[ s1 + 1 + 1 + s] s + s1 + 1 + 1

or Zth = 1 Alternative : Here at DC source capacitor act as open circuit and inductor act as short circuit. Thus we can directly calculate thevenin Impedance as 1 W 2.38

Option (D) is correct.

Since Vs (t) = d (t) " Vs (s) = 1 and

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Z (s) = R 1 sL = 2 sC s + We have been given Z (s) = 2 0.2s s + 0.1s + 2 Comparing with given we get 1 = 0.2 or C = 5 F C 1 = 0.1 or R = 2 W RC 1 = 2 or L = 0.1 H LC 2.39

s C s RC

+

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1 LC

Option (C) is correct. Voltage across capacitor is t Vc = 1 idt C 0

Vc (s) =

Vt = 2 e- sin c 3 t m 2 3 t 2

2.41

t

#0 dt

Option (B) is correct. Let voltage across resistor be vR VR (s) s = 1 1 = VS (s) ( s + s + 1) (s2 + s + 1) Since vs = d (t) " Vs (s) = 1 we get s = VR (s) = 2 s 1 2 (s + s + 1) (s + 2 ) + 43

For 0 < t < T , capacitor will be charged from 0 V

#0

=

t

dt = t or

At t = T, Vc = T Volts For T < t < 2T , capacitor will be discharged from T volts as Vc = T -

Vc =

t

At t = 3T, Vc = T Volts For 3T < t < 4T , capacitor will be discharged from T Volts Vc = T At t = 4T, Vc = 0 Volts

t

#3Tdt = 4T - t

vR (t) = e- cos 1 2

t

#T dt = 2T - t

#2Tdt = t - 2T

(s + 12 ) (s + 12 ) 2 +

= e- 2 =cos

t

At t = 2T, Vc = 0 volts For 2T < t < 3T , capacitor will be charged from 0 V

3

2 Vc (s) = 2 = G 3 (s + 12 ) 2 + 43 Taking inverse Laplace transform we have

Here C = 1 F and i = 1 A. Therefore

Vc =

1 (s + s + 1) 2

or

#

Vc =

Option (D) is correct. Writing in transform domain we have 1 Vc (s) = 1 s = 2 1 Vs (s) ^ s + s + 1h (s + s + 1)

2.42

3 4

-

(s +

1 2 3 1 2 2) + 4

3 t-1 2 e- sin 3 t 2 2# 3 2 1 2

3 t - 1 sin 3 t 2 2 G 3

Option (C) is correct. From the problem statement we have = 6 = 1.5W z11 = v1 i1 i = 0 4 v = 4.5 = 4.5W z12 = 1 i2 i = 0 1 = 6 = 1.5W z21 = v2 i1 i = 0 4 = 1.5 = 1.5W z22 = v2 i2 i = 0 1 2

1

2

2

Thus z -parameter matrix is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

z11 z12 1.5 4.5 =z z G = =1.5 1.5 G 21 22 2.43

Option (A) is correct. From the problem statement we = h12 = v1 v2 i = 0 = h22 = i2 v2 i = 0 1

1

Page 38 2.46

have 4.5 = 3 1.5 1 = 0.67 1.5

From z matrix, we have v1 = z11 i1 + z12 i2 v2 = z21 i1 + z22 i2 If v2 = 0 i2 = - z21 = - 1.5 =- 1 = h Then 21 i1 z22 1.5

We know that bandwidth of series RLC circuit is R . Therefore L Bandwidth of filter 1 is B1 = R L1 Bandwidth of filter 2 is B2 = R = R = 4R L2 L1 L1 /4 B 1 Dividing above equation 1 = B2 4 2.47

or i2 =- i1 Putting in equation for v1, we get v1 i1

Option (D) is correct. Here Vth is voltage across node also. Applying nodal analysis we get

Vth + Vth + Vth - 2i = 2 2 1 1 But from circuit i = Vth = Vth 1

v1 = (z11 - z12) i1 = h11 = z11 - z12 = 1.5 - 4.5 =- 3

v2 = 0

Therefore Vth + Vth + Vth - 2Vth = 2 2 1 1

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or Vth = 4 volt From the figure shown below it may be easily seen that the short circuit current at terminal XY is isc = 2 A because i = 0 due to short circuit of 1 W resistor and all current will through short circuit.

Hence h -parameter will be h11 h12 -3 3 =h h G = =- 1 0.67 G 21 22 2.44

Option (D) is correct.

Option (D) is correct. According to maximum Power Transform Theorem ZL = Zs* = (Rs - jXs)

2.45

Option (C) is correct. At w " 3 , capacitor acts as short circuited and circuit acts as shown in fig below

Here we get V0 = 0 Vi At w " 0 , capacitor acts as open circuited and circuit look like as shown in fig below

Here we get also V0 = 0 Vi So frequency response of the circuit is as shown in fig and circuit is a Band filter.

Therefore 2.48

Rth = Vth = 4 = 2 W isc 2

Option (A) is correct. The voltage across capacitor is At t = 0+ , Vc (0+) = 0 At t = 3 , VC (3) = 5 V The equivalent resistance seen by capacitor as shown in fig is Req = 20 20 = 10kW

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Time constant of the circuit is t = Req C = 10k # 4m = 0.04 s Using direct formula

or

Vc (t) = VC (3) - [Vc (3) - Vc (0)] e-t/t = VC (3) (1 - e-t/t) + VC (0) e-t/t = 5 (1 - e-t/0.04) Vc (t) = 5 (1 - e-25t)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Now

IC (t) = C

Page 39

dVC (t) dt

Taking Laplace transform we get V (s) = sLI (s) - Li (0+) As per given in question - Li (0+) =- 1 mV Thus i (0+) = 1 mV = 0.5 A 2 mH

= 4 # 10-6 # (- 5 # 25e-25t) = 0.5e-25t mA 2.49

Option (D) is correct. Impedance

= (5 - 3j) (5 + 3j) = =

(5 - 3j) # (5 + 3j) 5 - 3j + 5 + 3 j

(5) 2 - (3j) 2 = 25 + 9 = 3.4 10 10

2.54

VAB = Current # Impedance = 5+30c # 34 = 17+30c 2.50

Option (D) is correct. The network is shown in figure below.

Now and also From (1) and (2) Thus

V1 = AV2 - BI2 I1 = CV2 - DI2 V2 =- I2 RL we get V1 = AV2 - BI2 CV2 - DI2 I1

...(1) ...(2) ...(3)

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Substituting value of V2 from (3) we get Input Impedance Zin = - A # I2 RL - BI2 - C # I2 RL - DI2 or Zin = ARL + B CRL + D 2.51

Option (B) is correct. At initial all voltage are zero. So output is also zero. Thus v0 (0+) = 0 At steady state capacitor act as open circuit.

Thus,

The equivalent resistance and capacitance can be calculate after killing all source

Option (B) is correct. The circuit is as shown below.

At input port V1 = re I1 At output port V2 = r0 (I2 - bI1) =- r0 bI1 + r0 I2 Comparing standard equation

= 1 4 = 0.8 kW = 4 1 = 5 mF = Req Ceq = 0.8kW # 5mF = 4 ms = v 0 (3) - [v 0 (3) - v 0 (0+)] e-t/t = 8 - (8 - 0) e-t/0.004 v0 (t) = 8 (1 - e-t/0.004) Volts

Req Ceq t v0 (t)

V1 = z11 I1 + z12 I2 V2 = z21 I1 + z22 I2 z12 = 0 and z21 =- r0 b 2.52

Option (B) is correct. For series RC network input impedance is Zins = 1 + R = 1 + sRC sC sC Thus pole is at origin and zero is at - 1 RC

2.55

Thus pole is at - 1 and zero is at infinity. RC Option (A) is correct. We know

v = Ldi dt

Option (A) is correct. Here Z2 (s) = Rneg + Z1 (s) or Z2 (s) = Rneg + Re Z1 (s) + j Im Z1 (s) For Z2 (s) to be positive real, Re Z2 (s) $ 0 Thus Rneg + Re Z1 (s) $ 0 or Re Z1 (s) $- Rneg But Rneg is negative quantity and - Rneg is positive quantity. Therefore

For parallel RC network input impedance is 1 R sC sC = Zin = 1 +R 1 + sRC sC

2.53

v0 (3) = 4 # vi = 4 # 10 = 8 5 5

or 2.56

Re Z1 (s) $ Rneg Rneg # Re Z1 (jw)

Option (C) is correct. Transfer function is

For all w.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 40

1 Y (s) 1 sC = = 2 1 U (s) s LC + scR + 1 R + sL + sC

For RC parallel network the driving point impedance is R 1 R Cs = Zinp = 1 1 + sRC R+ Cs

1 LC = s2 + R s + 1 L LC

2

Comparing with s + 2xwn s + = 0 we have Here 2xwn = R , L and wn = 1 LC Thus x = R LC = R C 2L 2 L For no oscillations, x $ 1 R C $1 Thus 2 L or 2.57

Here pole is s =- 1/RC and zero is at 3, therefore first critical frequency is a pole and last critical frequency is a zero.

wn2

R $2

2.61

i1 (t) + 5+0c = 10+60c or or 2.62

L C

Option (B) is correct. For given transformer

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2.58

2.63

V1 = AV2 + BI2 I1 = CV2 + DI2 n 0 A B =C D G = = 0 1 G n x = 1 n

Option (B) is correct. Open circuit at terminal ab is shown below

or Vab = 7.5 = Vth Short circuit at terminal ab is shown below

Option (B) is correct. Short circuit current from terminal ab is Isc = 1 + 10 = 3 A 5

Resonant frequency f0 =

1 == 2p LC 2p 3

4

= 10 # 20 = 10 Hz p 2p

1 1 # 1 # 10 - 6 400

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Option (C) is correct. Maximum power will be transferred when RL = Rs = 100W In this case voltage across RL is 5 V, therefore

Thus

Option (C) is correct. For stability poles and zero interlace on real axis. In RC series network the driving point impedance is Zins = R + 1 = 1 + sRC sC Cs Here pole is at origin and zero is at s =- 1/RC , therefore first critical frequency is a pole and last critical frequency is a zero.

Rth = Vth = 7.5 = 2.5 W Isc 3

Here current source being in series with dependent voltage source make it ineffective.

2 Pmax = V = 5 # 5 = 0.25 W R 100 2.60

Option (B) is correct. If L1 = j5W and L3 = j2W the mutual induction is subtractive because current enters from dotted terminal of j2W coil and exit from dotted terminal of j5W. If L2 = j2W and L3 = j2W the mutual induction is additive because current enters from dotted terminal of both coil. Thus Z = L1 - M13 + L2 + M23 + L3 - M31 + M32

Applying KCL at node we get Vab + Vab - 10 = 1 5 5

We have L = 1H and C = 1 # 10-6 400

2.59

i1 (t) = 10+60c - 5+0c = 5 + 5 3j - 5 i1 (t) = 5 3 +90c = 10 3 +90c 2

= j5 + j10 + j2 + j10 + j2 - j10 + j10 = j9

I2 = V1 = n V2 1 I1 or I1 = I2 and V1 = nV2 n Comparing with standard equation

Thus

Option (A) is correct. Applying KCL we get

2.64

Option (C) is correct. Here Va = 5 V because R1 = R2 and total voltage drop is 10 V. Now Vb = R3 # 10 = 1.1 # 10 = 5.238 V R3 + R4 2.1 V = Va - Vb = 5 - 5.238 =- 0.238 V

2.65

Option (D) is correct. For h parameters we have to write V1 and I2 in of I1 and V2 .

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 41

V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2 Applying KVL at input port V1 = 10I1 + V2 Applying KCL at output port V2 = I + I 1 2 20 or I2 =- I1 + V2 20

or

Taking inverse Laplace transform i (t) = 1 (1 - e-2t) u (t) 2

At t = 0 , i (t) = 0 At t = 12 , i (t) = 0.31 At t = 3 , i (t) = 0.5 Graph (C) satisfies all these conditions.

Thus from above equation we get h11 h12 10 1 =h h G = =- 1 0.05G 12 22 2.66

Option (B) is correct. RC = 0.1 # 10 - 6 # 103 = 10 - 4 sec Since time constant RC is very small, so steady state will be reached in 2 sec. At t = 2 sec the circuit is as shown in fig. Time constant

Z (s) = s + 2 V (s) 1 = I (s) = i s + 2 s (s + 2) I (s) = 1 ; 1 - 1 E 2 s s+2

Impedance

2.72

Option (D) is correct. We know that

where

V1 = z11 I1 + z12 I2 V2 = z11 I1 + z22 I2 z11 = V1 I1 I = 0

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Vc = 3 V V2 =- Vc =- 3 V 2.67

2.68

2.69

Option (B) is correct. For a tree there must not be any loop. So a, c, and d don’t have any loop. Only b has loop. Option (D) is correct. The sign of M is as per sign of L If current enters or exit the dotted terminals of both coil. The sign of M is opposite of L If current enters in dotted terminal of a coil and exit from the dotted terminal of other coil. Thus Leq = L1 + L2 - 2M Option (A) is correct. Here w = 2 and V = 1+0c Y = 1 + jwC + 1 R jwL = 3 + j2 # 3 + 1 1 = 3 + j4 j2 # 4 = 5+ tan - 1 4 = 5+53.11c 3 I = V * Y = (1+0c)( 5+53.1c) = 5+53.1c i (t) = 5 sin (2t + 53.1c)

Thus 2.70

Option (A) is correct. vi (t) =

Here w = 10 rad and Vi = Now

3

2 sin 10 t

3

2 +0c

1 jwC 1 V V0 = .Vt = + wCR i 1 1 j R+ jwC 1 1 + j # 103 # 10 - 3 = 1 - 45c v0 (t) = sin (103 t - 45c) =

2.71

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Option (C) is correct. Input voltage Taking Laplace transform

2 + 0c

vi (t) = u (t) Vi (s) = 1 s

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I1 = 0

Consider the given lattice network, when I2 = 0 . There is two similar path in the circuit for the current I1. So I = 1 I1 2

For z11 applying KVL at input port we get V1 = I (Za + Zb) Thus V1 = 1 I1 (Za + Zb) 2 z11 = 1 (Za + Zb) 2 For Z21 applying KVL at output port we get V2 = Za I1 - Zb I1 2 2 Thus V2 = 1 I1 (Za - Zb) 2 z21 = 1 (Za - Zb) 2 For this circuit z11 = z22 and z12 = z21. Thus V R S Za + Zb Za - Zb W z11 z12 2 2 =z z G = SS Za - Zb Za + Zb WW 21 22 S 2 2 W X T Here Za = 2j and Zb = 2W 1+j j-1 z11 z12 Thus =z z G = = j - 1 1 + j G 21 22

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 2.73

Page 42

Option (B) is correct. Applying KVL,

= minimum number of equation

Ldi (t) 1 + v (t) = Ri (t) + dt C

#0

3

Number of branches = b = 8 Number of nodes = n = 5 Minimum number of equation

i (t) dt

Taking L.T. on both sides, V (s) v (t) Hence

1 s 2 +1 s

or 2.74

I (s)

I (s) vc (0+) = RI (s) + LsI (s) - Li (0 ) + + sC sC = u (t) thus V (s) = 1 s I (s) 1 = I (s) + sI (s) - 1 + s s I (s) 2 = 6s + s + 1@ s = 2s+2 s +s+1 +

= 8-5+1 = 4 2.78

2.79

Thus Now 2.75

2.80

Option (C) is correct. Applying KVL we get, di (t) 1 + i (t) dt dt C di (t) + i (t) dt sin t = 2i (t) + 2 dt

#

#

or

Differentiating with respect to t , we get 2di (t) 2d2 i (t) cos t = + i (t) + dt dt2 2.81

Option (A) is correct. For current i there is 3 similar path. So current will be divide in three path

Option (D) is correct. V0 (s) Vi (s)

1 1 sC = = 2 1 s LC + sCR + 1 R + sL + sC 1 = 2 -2 -4 s (10 # 10 ) + s (10-4 # 10 4) + 1 106 = -6 2 1 = 2 10 s + s + 1 s + 106 s + 106 Option (D) is correct. Impedance of series RLC circuit at resonant frequency is minimum, not zero. Actually imaginary part is zero. Z = R + j ` wL - 1 j wC At resonance wL - 1 = 0 and Z = R that is purely resistive. wC Thus S1 is false Now quality factor Q =R C L Since G = 1 , Q = 1 C G L R If G - then Q . provided C and L are constant. Thus S2 is also false. 2.77

L =Q C 2

sin t = Ri (t) + L

2xw = 20 2x = 203 = 0.02 10 Q = 1 = 1 = 50 2x 0.02 H (s) =

2.76

L C

When R, L and C are doubled, 2L = 1 Q' = 1 2R 2C 2R Thus Q' = 100 = 50 2

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in 106 = 103

Option (B) is correct. Q = 1 R

s2 + 20s + 106 = 0

wn =

ZL = ZS* = Rs - jXs ZL = 1 - 1j

Thus

Option (B) is correct. Characteristics equation is Comparing with s2 + 2xwn s + wn2 = 0 we have

Option (C) is correct. For maximum power transfer

Option (B) is correct. Number of loops = b - n + 1

so, we get Vab - b i # 1l - b i # 1l - b 1 # 1l = 0 3 6 3

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 Vab = R = 1 + 1 + 1 = 5 W eq 6 i 3 6 3 2.82

2.83

Option ( ) is correct. Data are missing in question as L1 &L2 are not given. Option (A) is correct. At t = 0 - circuit is in steady state. So inductor act as short circuit and capacitor act as open circuit.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 43

Now

At t = 0 - ,

i1 (0 -) = i2 (0 -) = 0

vc (0 -) = V At t = 0+ the circuit is as shown in fig. The voltage across capacitor and current in inductor can’t be changed instantaneously. Thus

i1 = i2 =- V 2R

2.84

Option (C) is correct. When switch is in position 2, as shown in fig in question, applying KVL in loop (1), RI1 (s) + V + 1 I1 (s) + sL [I1 (s) - I2 (s)] = 0 s sC or I1 (s) 8R + 1 + sL B - I2 (s) sL = - V s sc

=

10 2 cos (t + 10c) 10 5 cos (2t + 10c) + 1+j 1 + j2

=

10 2 cos (t + 10c) 10 5 cos (2t + 10c) + 12 + 22 + tan-1 1 12 + 22 tan-1 2

10 2 cos (t + 10c) 10 5 cos (2t + 10c) + 2 + tan-1 45c 5 tan-1 2 i (t) = 10 cos (t - 35c) + 10 cos (2t + 10c - tan-1 2) =

2.87

At t = 0+ ,

Z1 = R + jw1 L = 1 + j1 Z2 = R + jw2 L = 1 + j2 v (t) v (t) i (t) = 1 + 2 Z1 Z2

Option (A) is correct. Using 3- Y conversion

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z11 I1 + z12 I2 = V1

Applying KVL in loop 2, sL [I2 (s) - I1 (s)] + RI2 (s) + 1 I2 (s) = 0 sC

2.85

2.86

Z12 I1 + Z22 I2 = V2 or - sLI1 (s) + 8R + sL + 1 BI2 (s) = 0 sc Now comparing with Z11 Z12 I1 V1 =Z Z G=I G = =V G 21 22 2 2 we get R V - sL SR + sL + 1 W I1 (s) -V sC S W sH = G = > SS - sL R + sL + 1 WW I2 (s) 0 sC T X Option (B) is correct. Zeros =- 3 Pole1 =- 1 + j Pole 2 =- 1 - j K (s + 3) Z (s) = (s + 1 + j)( s + 1 - j) K (s + 3) K (s + 3) = = 2 2 (s + 1) - j (s + 1) 2 + 1 From problem statement Z (0) w = 0 = 3 Thus 3K = 3 and we get K = 2 2 2 (s + 3) Z (s) = 2 s + 2s + 2

2 # 1 = 2 = 0.5 2+1+1 4 R2 = 1 # 1 = 1 = 0.25 2+1+1 4 R3 = 2 # 1 = 0.5 2+1+1 R1 =

Now the circuit is as shown in figure below.

Now

z11 = V1 I1

2.88

Option (A) is correct. Applying KCL at for node 2,

v (t) = 10 2 cos (t + 10c) + 10 5 cos (2t + 10c) 1 4444 2 4444 3 1 4444 4 2 4444 43 v1

= 2 + 0.5 + 0.25 = 2.75

z12 = R3 = 0.25

Option (C) is correct.

Thus we get w1 = 1 and w2 = 2

I2 = 0

v2

V2 + V2 - V1 = V1 5 5 5

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

or V2 = V1 = 20 V Voltage across dependent current source is 20 thus power delivered by it is PV2 # V1 = 20 # 20 = 80 W 5 5

Page 44

2.91

It deliver power because current flows from its +ive terminals. 2.89

Option (C) is correct. When switch was closed, in steady state, iL (0 -) = 2.5 A

Option (D) is correct. IP , VP " Phase current and Phase voltage IL, VL " Line current and line voltage Now VP = c VL m and IP = IL 3 So, Power = 3VP IL cos q 1500 = 3 c VL m (IL) cos q 3 also IL = c VL m 3 ZL 1500 = 3 c VL mc VL m cos q 3 3 ZL ZL =

As power factor is leading So, cos q = 0.844 " q = 32.44 As phase current leads phase voltage

At t = 0+ , iL (0+) = iL (0 -) = 2.5 A and all this current of will through 2 W resistor. Thus

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(400) 2 (.844) = 90 W 1500

ZL = 90+ - q = 90+ - 32.44c 2.92

Option (C) is correct. Applying KCL, we get e0 - 12 + e0 + e0 = 0 4 4 2+2 or

2.93

e0 = 4 V

Option (A) is correct. The star delta circuit is shown as below

Vx =- 2.5 # 20 =- 50 V 2.90

Option (A) is correct. For maximum power delivered, RL must be equal to Rth across same terminal.

Here Applying KCL at Node, we get 0.5I1 = Vth + I1 20 or but Thus

and

Vth + 10I1 = 0 I1 = Vth - 50 40 Vth + Vth - 50 = 0 4

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or Vth = 10 V For Isc the circuit is shown in figure below.

Now

2.94

but

ZA = ZB = ZC =

3Z 3Z = Z 3Z+ 3Z+ 3Z 3

Option (C) is correct. y11 y12 y1 + y3 - y3 =y y G = = - y y + y G 21 22 3 2 3 y12 =- y3 y12 =- 1 =- 0.05 mho 20

Isc = 0.5I1 - I1 =- 0.5I1 I1 =- 50 =- 1.25 A 40 Isc =- 0.5 # - 12.5 = 0.625 A Rth = Vth = 10 = 16 W Isc 0.625

ZAB = ZBC = ZCA = 3 Z ZAB ZCA ZA = ZAB + ZBC + ZCA ZAB ZBC ZB = ZAB + ZBC + ZCA ZBC ZCA ZC = ZAB + ZBC + ZCA

2.95

Option (D) is correct. We apply source conversion the circuit as shown in fig below.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 45

Now applying nodal analysis we have e0 - 80 + e0 + e0 - 16 = 0 6 10 + 2 12 4e0 = 112 e0 = 112 = 28 V 4

or

2.96

V0 = RL Vs RL + Rs At resonant frequency w = V0 = 0 .

(finite value)

1 circuit acts as shown in fig and LC

Option (A) is correct. jw C I2 = Em +01c = Em +0c 1 + jwCR2 R2 + jwC +90c + tan-1 wCR2 E m wC + (90c - tan-1 wCR2) I2 = 2 2 2 1 + w C R2 At w = 0 I2 = 0 and at w = 3, I2 = Em R2 Only fig. given in option (A) satisfies both conditions. +I2 =

2.97

Thus it is a band reject filter.

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Option (A) is correct.

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Xs = wL = 10 W For maximum power transfer RL = 2.98

Rs2 + Xs2 =

102 + 102 = 14.14 W

2.100

Option (C) is correct. Applying KVL in LHS loop or

E1 = 2I1 + 4 (I1 + I2) - 10E1 E1 = 6I1 + 4I2 11 11

Option (D) is correct. Applying KCL we get Now

2.101

Thus z11 = 6 11

iL = eat + ebt V (t) = vL = L diL = L d [eat + ebt] = aeat + bebt dt dt

Option (A) is correct. Going from 10 V to 0 V

Applying KVL in RHS loop E2 = 4 (I1 + I2) - 10E1 = 4 (I1 + I2) - 10 c 6I1 + 4I2 m 11 11 =- 16I1 + 4I2 11 11 Thus z21 =- 16 11 2.99

Option (D) is correct. At w = 0 , circuit act as shown in figure below.

or 2.102

V0 = RL Vs RL + Rs

(finite value)

At w = 3 , circuit act as shown in figure below:

2.103

10 + 5 + E + 1 = 0 E =- 16 V

Option (C) is correct. This is a reciprocal and linear network. So we can apply reciprocity theorem which states “Two loops A & B of a network N and if an ideal voltage source E in loop A produces a current I in loop B , then interchanging positions an identical source in loop B produces the same current in loop A. Since network is linear, principle of homogeneity may be applied and when volt source is doubled, current also doubles. Now applying reciprocity theorem i = 2 A for 10 V V = 10 V, i = 2 A V =- 20 V, i =- 4 A Option (C) is correct. Tree is the set of those branch which does not make any loop and

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 46

Req = 5W 20W + 4W

connects all the nodes. abfg is not a tree because it contains a loop l node (4) is not connected

Req = 5.20 + 4 = 4 + 4 = 8 W 5 + 20 2.107

2.104

Option (A) is correct. For a 2-port network the parameter h21 is defined as h21 = I2 I1 V = 0 (short circuit) 2

2.108

2.109

Applying node equation at node a we get

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Now and Thus

Va - V1 + Va - 0 + Va - 0 = 0 R R R & Va = V1 3Va = V1 3 V1 - V1 V 3 = 2V1 1 - Va = I1 = R R 3R 0 - V1 - V 0 V 3 = a 1 = I2 = R R 3R - V1 /3R - 1 I2 = h21 = = 2 I1 V = 0 2V1 /3R

2.110

2.111 2.112

2.113

Option (D) is correct. Delta to star conversion Rab Rac = 5 # 30 = 150 = 3 W R1 = 50 Rab + Rac + Rbc 5 + 30 + 15 Rab Rbc = 5 # 15 = 1.5 W R2 = Rab + Rac + Rbc 5 + 30 + 15 Rac Rbc = 15 # 30 = 9 W R3 = Rab + Rac + Rbc 5 + 30 + 15 Option (C) is correct. No. of branches = n + l - 1 = 7 + 5 - 1 = 11 Option (B) is correct. In nodal method we sum up all the currents coming & going at the node So it is based on KCL. Furthermore we use ohms law to determine current in individual branch. Thus it is also based on ohms law. Option (A) is correct. Superposition theorem is applicable to only linear circuits. Option (B) is correct. Option (B) is correct. For reciprocal network y12 = y21 but here y12 =- 12 ! y21 = 12 . Thus circuit is non reciprocal. Furthermore only reciprocal circuit are ive circuit. Option (C) is correct. Taking b as reference node and applying KCL at a we get Vab - 1 + Vab = 3 2 2 or or

2.114 2.115

Vab - 1 + Vab = 6 Vab = 6 + 1 = 3.5 V 2

Option (A) is correct. Option (B) is correct. The given figure is shown below.

2

2.105

Option (A) is correct. Applying node equation at node A Vth - 100 (1 + j0) Vth - 0 + =0 3 4j or or By simplifying

4jVth - 4j100 + 3Vth = 0 Vth (3 + 4j) = 4j100 4j100 Vth = 3 + 4j Vth =

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4j100 3 - 4j 3 + 4j # 3 - 4j

Vth = 16j (3 - j4) 2.106

Option (C) is correct. For maximum power transfer RL should be equal to RTh at same terminal. so, equivalent Resistor of the circuit is

Applying KCL at node a we have I = i 0 + i1 = 7 + 5 = 12 A

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

In the steady state condition all capacitors behaves as open circuit & Inductors behaves as short circuits as shown below :

Applying KCL at node f so 2.116

Page 47

I =- i 4 i 4 =- 12 amp

Option (A) is correct.

Thus voltage across capacitor C1 is VC = 100 # 40 = 80 V 10 + 40 1

so 2.117

2.118

2.119

Now the circuit faced by capacitor C2 and C 3 can be drawn as below :

V = 3 - 0 = 3 volt

Option (D) is correct. Can not determined V without knowing the elements in box. Option (A) is correct. The voltage V is the voltage across voltage source and that is 10 V. Option (B) is correct. Voltage across capacitor -t

VC (t) = VC (3) + (VC (0) - VC (3)) e RC Here VC (3) = 10 V and (VC (0) = 6 V. Thus -t

Now

= 10 - 10 + 4e Energy absorbed by resistor

-t RC

-t

= 4e

#0 3V RR(t) = #0 3 164e 4

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-t RC

-t

2

E 2.120

-t

VC (t) = 10 + (6 - 10) e RC = 10 - 4e RC = 10 - 4e 8 VR (t) = 10 - VC (t)

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=

-t

#0 3 4e 4

= 16 J

Option (B) is correct. It is a balanced whetstone bridge R1 R 3 b R2 = R 4 l so equivalent circuit is

Voltage across capacitor C2 and C 3 are VC = 80 C 3 = 80 # 3 = 48 volt 5 C2 + C3 VC = 80 C2 = 80 # 2 = 32 volt 5 C2 + C3 2

3

Zeq = (4W 8W) = 4 # 8 = 8 3 4+8 2.121

Option (B) is correct. Current in A2 , Inductor current can be Current in A 3 , Total current

I2 = 3 amp defined as I2 =- 3j I3 = 4 I1 = I 2 + I 3 I1 = 4 - 3j I =

2.122

2.123

(4) 2 + (3) 2 = 5 amp

Option (C) is correct. For a tree we have (n - 1) branches. Links are the branches which from a loop, when connect two nodes of tree. so if total no. of branches = b No. of links = b - (n - 1) = b - n + 1 Total no. of links in equal to total no. of independent loops. Option (B) is correct.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

UNIT 3

Page 47 3.6

In the CMOS circuit shown, electron and hole mobilities are equal, and M1 and M2 are equally sized. The device M1 is in the linear region if

ELECTRONICS DEVICES

2013 3.1

3.2

3.3

ONE MARK

In a forward biased pn junction diode, the sequence of events that best describes the mechanism of current flow is (A) injection, and subsequent diffusion and recombination of minority carriers (B) injection, and subsequent drift and generation of minority carriers (C) extraction, and subsequent diffusion and generation of minority carriers (D) extraction, and subsequent drift and recombination of minority carriers In IC technology, dry oxidation (using dry oxygen) as compared to wet oxidation (using steam or water vapor) produces (A) superior quality oxide with a higher growth rate (B) inferior quality oxide with a higher growth rate (C) inferior quality oxide with a lower growth rate (D) superior quality oxide with a lower growth rate

In the three dimensional view of a silicon n -channel MOS transistor shown below, d = 20 nm . The transistor is of width 1 mm . The depletion width formed at every p-n junction is 10 nm. The rela-

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Click to Buy www.nodia.co.in tive permittivity of Si and SiO 2 , respectively, are 11.7 and 3.9, and e0 = 8.9 # 10-12 F/m .

TWO MARKS

The small-signal resistance (i.e., dVB /dID ) in kW offered by the n-channel MOSFET M shown in the figure below, at a bias point of VB = 2 V is (device data for M: device transconductance parameter kN = mn C 0' x ^W/L h = 40 mA/V2 , threshold voltage VTN = 1 V , and neglect body effect and channel length modulation effects)

3.7

3.8

(A) 12.5 (C) 50 2012 3.5

(B) 1.875 V < Vin < 3.125 V (D) 0 < Vin < 5 V

Common Data For Q. 2 and 3 :

In a MOSFET operating in the saturation region, the channel length modulation effect causes (A) an increase in the gate-source capacitance (B) a decrease in the transconductance (C) a decrease in the unity-gain cutoff frequency (D) a decrease in the output resistance 2013

3.4

(A) Vin < 1.875 V (C) Vin > 3.125 V

(B) 25 (D) 100

The source-body junction capacitance is approximately (A) 2 fF (B) 7 fF (C) 2 pF (D) 7 pF 2011

TWO MARKS

The source of a silicon (ni = 1010 per cm3) n -channel MOS transistor has an area of 1 sq mm and a depth of 1 mm . If the dopant density in the source is 1019 /cm3 , the number of holes in the source region with the above volume is approximately (A) 107 (B) 100 (C) 10 (D) 0

The gate source overlap capacitance is approximately (A) 0.7 fF (B) 0.7 pF (C) 0.35 fF (D) 0.24 pF

3.9

3.10

ONE MARK

Drift current in the semiconductors depends upon (A) only the electric field (B) only the carrier concentration gradient (C) both the electric field and the carrier concentration (D) both the electric field and the carrier concentration gradient A Zener diode, when used in voltage stabilization circuits, is biased

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

in (A) reverse bias region below the breakdown voltage (B) reverse breakdown region (C) forward bias region (D) forward bias constant current mode 3.11

(C) epitaxial oxidation

3.16

3.17

TWO MARKS

Common Data For Q. 3.12 & 3.13 : The channel resistance of an N-channel JFET shown in the fig-

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In a uniformly doped BJT, assume that NE , NB and NC are the emitter, base and collector doping in atoms/cm3 , respectively. If the emitter injection efficiency of the BJT is close unity, which one of the following condition is TRUE (B) NE >> NB and NB > NC (A) NE = NB = NC (C) NE = NB and NB < NC (D) NE < NB < NC Compared to a p-n junction with NA = ND = 1014 /cm3 , which one of the following statements is TRUE for a p-n junction with NA = ND = 1020 /cm3 ? (A) Reverse breakdown voltage is lower and depletion capacitance is lower (B) Reverse breakdown voltage is higher and depletion capacitance is lower (C) Reverse breakdown voltage is lower and depletion capacitance is higher (D) Reverse breakdown voltage is higher and depletion capacitance is higher

The silicon sample with unit cross-sectional area shown below is in thermal equilibrium. The following information is given: T = 300 K electronic charge = 1.6 # 10-19 C , thermal voltage = 26 mV and electron mobility = 1350 cm2 / V-s

3.18

3.19

3.13

The channel resistance when VGS =- 3 V is (A) 360 W (B) 917 W (C) 1000 W (D) 3000 W

The magnitude of the electric field at x = 0.5 mm is (A) 1 kV/cm (B) 5 kV/cm (C) 10 kV/cm (D) 26 kV/cm The magnitude of the electron of the electron drift current density at x = 0.5 mm is (B) 1.08 # 10 4 A/m2 (A) 2.16 # 10 4 A/cm2

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The channel resistance when VGS = 0 V is (A) 480 W (B) 600 W (C) 750 W (D) 1000 W

(C) 4.32 # 103 A/cm2 2009

2010 3.14

3.15

TWO MARKS

Statements for Linked Answer Question : 3.10 & 3.11 :

ure below is 600 W when the full channel thickness (tch ) of 10 μm is available for conduction. The built-in voltage of the gate P+ N junction (Vbi ) is - 1 V . When the gate to source voltage (VGS ) is 0 V, the channel is depleted by 1 μm on each side due to the built in voltage and hence the thickness available for conduction is only 8 μm

3.12

(D) ion implantation

2010

A silicon PN junction is forward biased with a constant current at room temperature. When the temperature is increased by 10ºC, the forward bias voltage across the PN junction (A) increases by 60 mV (B) decreases by 60 mV (C) increases by 25 mV (D) decreases by 25 mV 2011

Page 48

ONE MARK

At room temperature, a possible value for the mobility of electrons in the inversion layer of a silicon n -channel MOSFET is (A) 450 cm2 / V-s (B) 1350 cm2 / V-s (C) 1800 cm2 / V-s (D) 3600 cm2 / V-s Thin gate oxide in a CMOS process in preferably grown using (A) wet oxidation (B) dry oxidation

3.20

3.21

(D) 6.48 # 102 A/cm2 ONE MARK

In an n-type silicon crystal at room temperature, which of the following can have a concentration of 4 # 1019 cm - 3 ? (A) Silicon atoms (B) Holes (C) Dopant atoms (D) Valence electrons The ratio of the mobility to the diffusion coefficient in a semiconductor has the units (A) V - 1 (B) cm.V1

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(C) V.cm - 1

(D) V.s

2009 3.22

Page 49 3.28

TWO MARKS

Consider the following two statements about the internal conditions in a n -channel MOSFET operating in the active region. S1 : The inversion charge decreases from source to drain S2 : The channel potential increases from source to drain. Which of the following is correct? (A) Only S2 is true (B) Both S1 and S2 are false (C) Both S1 and S2 are true, but S2 is not a reason for S1 (D) Both S1 and S2 are true, and S2 is a reason for S1

The drain current of MOSFET in saturation is given by ID = K (VGS - VT ) 2 where K is a constant. The magnitude of the transconductance gm is (A)

K (VGS - VT ) 2 VDS

(B) 2K (VGS - VT )

(C)

Id VGS - VDS

(D)

K (VGS - VT ) 2 VGS

2008 3.29

TWO MARKS

The measured trans conductance gm of an NMOS transistor operating in the linear region is plotted against the gate voltage VG at a constant drain voltage VD . Which of the following figures represents the expected dependence of gm on VG ?

Common Data For Q. 3.13 and 3.14 Consider a silicon p - n junction at room temperature having the following parameters: Doping on the n -side = 1 # 1017 cm - 3 Depletion width on the n -side = 0.1mm Depletion width on the p -side = 1.0mm Intrinsic carrier concentration = 1.4 # 1010 cm - 3 Thermal voltage = 26 mV Permittivity of free space = 8.85 # 10 Dielectric constant of silicon = 12 3.23

3.24

- 14

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F.cm - 1

The built-in potential of the junction (A) is 0.70 V (B) is 0.76 V (C) is 0.82 V (D) Cannot be estimated from the data given The peak electric field in the device is (A) 0.15 MV . cm - 1, directed from p -region to n -region (B) 0.15 MV . cm - 1, directed from n -region to p -region (C) 1.80 MV . cm - 1, directed from p-region to n -region (D) 1.80 MV . cm - 1, directed from n -region to p -region 2008

3.25

3.26

3.27

ONE MARK

Which of the following is NOT associated with a p - n junction ? (A) Junction Capacitance (B) Charge Storage Capacitance (C) Depletion Capacitance (D) Channel Length Modulations Which of the following is true? (A) A silicon wafer heavily doped with boron is a p+ substrate (B) A silicon wafer lightly doped with boron is a p+ substrate (C) A silicon wafer heavily doped with arsenic is a p+ substrate (D) A silicon wafer lightly doped with arsenic is a p+ substrate A silicon wafer has 100 nm of oxide on it and is furnace at a temperature above 1000c C for further oxidation in dry oxygen. The oxidation rate (A) is independent of current oxide thickness and temperature (B) is independent of current oxide thickness but depends on temperature (C) slows down as the oxide grows (D) is zero as the existing oxide prevents further oxidation

3.30

3.31

Silicon is doped with boron to a concentration of 4 # 1017 atoms cm3 . Assume the intrinsic carrier concentration of silicon to be 1.5 # 1010 / cm 3 and the value of kT/q to be 25 mV at 300 K. Compared to undopped silicon, the fermi level of doped silicon (A) goes down by 0.31 eV (B) goes up by 0.13 eV (C) goes down by 0.427 eV (D) goes up by 0.427 eV The cross section of a JFET is shown in the following figure. Let Vc be - 2 V and let VP be the initial pinch -off voltage. If the width W is doubled (with other geometrical parameters and doping levels remaining the same), then the ratio between the mutual trans conductances of the initial and the modified JFET is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 50

(C) P - 2, Q - 2, R - 1, S- -2 (D) P - 2, Q - 1, R - 2, S - 2 3.36

(A) 4 (C) e 3.32

1 - 2/Vp o 1 - 1/2Vp

1 - 2/Vp (B) 1 e 2 1 - 1/2Vp o 1 - (2 - Vp ) (D) 1 - [1 (2 Vp )]

Consider the following assertions. S1 : For Zener effect to occur, a very abrupt junction is required. S2 : For quantum tunneling to occur, a very narrow energy barrier is required. Which of the following is correct ? (A) Only S2 is true (B) S1 and S2 are both true but S2 is not a reason for S1

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(A) P - 3, Q - 1, R - 4, S - 2 (B) P - 1, Q - 4, R - 3, S - 2 (C) P - 3, Q - 4, R - 1, S - 2 (D) P - 3, Q - 2, R - 1, S - 4 3.37

3.38

3.33

3.34

2007 3.35

TWO MARKS

Group I lists four types of p - n junction diodes. Match each device in Group I with one of the option in Group II to indicate the bias condition of the device in its normal mode of operation. Group - I Group-II (P) Zener Diode (1) Forward bias (Q) Solar cell (2) Reverse bias (R) LASER diode (S) Avalanche Photodiode (A) P - 1, Q - 2, R - 1, S - 2 (B) P - 2, Q - 1, R - 1, S - 2

The DC current gain (b) of a BJT is 50. Assuming that the emitter injection efficiency is 0.995, the base transport factor is (A) 0.980 (B) 0.985 (C) 0.990 (D) 0.995

The figure shows the high-frequency capacitance - voltage characteristics of Metal/Sio 2 /silicon (MOS) capacitor having an area of 1 # 10 - 4 cm 2 . Assume that the permittivities (e0 er ) of silicon and Sio2 are 1 # 10 - 12 F/cm and 3.5 # 10 - 13 F/cm respectively.

ONE MARK

The electron and hole concentrations in an intrinsic semiconductor are ni per cm3 at 300 K. Now, if acceptor impurities are introduced with a concentration of NA per cm3 (where NA >> ni , the electron concentration per cm3 at 300 K will be) (A) ni (B) ni + NA 2 (D) ni (C) NA - ni NA + In a p n junction diode under reverse biased the magnitude of electric field is maximum at (A) the edge of the depletion region on the p-side (B) the edge of the depletion region on the n -side (C) the p+ n junction (D) the centre of the depletion region on the n -side

A p+ n junction has a built-in potential of 0.8 V. The depletion layer width a reverse bias of 1.2 V is 2 mm. For a reverse bias of 7.2 V, the depletion layer width will be (A) 4 mm (B) 4.9 mm (C) 8 mm (D) 12 mm

Common Data For Q. 2.29, 2.30 and 2.31 :

(C) S1 and S2 and are both true but S2 is not a reason for S1 (D) Both S1 and S2 are false 2007

Group I lists four different semiconductor devices. match each device in Group I with its charactecteristic property in Group II Group-I Group-II (P) BJT (1) Population iniversion (Q) MOS capacitor (2) Pinch-off voltage (R) LASER diode (3) Early effect (S) JFET (4) Flat-band voltage

3.39

The gate oxide thickness in the MOS capacitor is (A) 50 nm (B) 143 nm (C) 350 nm (D) 1 mm

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 3.40

3.41

The maximum depletion layer width in silicon is (A) 0.143 mm (B) 0.857 mm (C) 1 mm (D) 1.143 mm Consider the following statements about the C - V characteristics plot : S1 : The MOS capacitor has as n -type substrate S2 : If positive charges are introduced in the oxide, the C - V polt will shift to the left. Then which of the following is true? (A) Both S1 and S2 are true (B) S1 is true and S2 is false

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 51

(C) S1 is false and S2 is true (D) Both S1 and S2 are false

(B) E - 3, F - 4, G - 1, H - 3 (C) E - 2, F - 4, G - 1, H - 2 (D) E - 1, F - 3, G - 2, H - 4

2006 3.42

3.43

3.44

3.45

ONE MARK

The values of voltage (VD) across a tunnel-diode corresponding to peak and valley currents are Vp, VD respectively. The range of tunneldiode voltage for VD which the slope of its I - VD characteristics is negative would be (A) VD < 0 (B) 0 # VD < Vp (C) Vp # VD < Vv (D) VD $ Vv The concentration of minority carriers in an extrinsic semiconductor under equilibrium is (A) Directly proportional to doping concentration (B) Inversely proportional to the doping concentration (C) Directly proportional to the intrinsic concentration (D) Inversely proportional to the intrinsic concentration Under low level injection assumption, the injected minority carrier current for an extrinsic semiconductor is essentially the (A) Diffusion current (B) Drift current (C) Recombination current (D) Induced current The phenomenon known as “Early Effect” in a bipolar transistor refers to a reduction of the effective base-width caused by (A) Electron - hole recombination at the base (B) The reverse biasing of the base - collector junction (C) The forward biasing of emitter-base junction (D) The early removal of stored base charge during saturation-tocut off switching 2006

3.46

3.48

(B) vR =+ 5 (D) - 5 # vR < 0

The majority carriers in an n-type semiconductor have an average drift velocity v in a direction perpendicular to a uniform magnetic field B . The electric field E induced due to Hall effect acts in the direction (A) v # B (B) B # v (C) along v (D) opposite to v Find the correct match between Group 1 and Group 2 Group 1 Group 2 E - Varactor diode 1. Voltage reference F - PIN diode 2. High frequency switch G - Zener diode 3. Tuned circuits H - Schottky diode 4. Current controlled attenuator (A) E - 4, F - 2, G - 1, H - 3

A heavily doped n - type semiconductor has the following data: Hole-electron ratio : 0.4 Doping concentration : 4.2 # 108 atoms/m3 Intrinsic concentration : 1.5 # 10 4 atoms/m 3 The ratio of conductance of the n -type semiconductor to that of the intrinsic semiconductor of same material and ate same temperature is given by (A) 0.00005 (B) 2000 (C) 10000 (D) 20000 2005

3.50

ONE MARK

The bandgap of Silicon at room temperature is (A) 1.3 eV (B) 0.7 eV

SPECIAL EDITION ( STUDY MATERIAL FORM ) At market Book is available in 3 volume i.e. in 3 book binding form. But at NODIA Online Store book is available in 10 book binding form. Each unit of Book is in separate binding. Available Only at NODIA Online Store

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TWO MARKS

In the circuit shown below, the switch was connected to position 1 at t < 0 and at t = 0 , it is changed to position 2. Assume that the diode has zero voltage drop and a storage time ts . For 0 < t # ts, vR is given by (all in Volts)

(A) vR =- 5 (C) 0 # vR < 5 3.47

3.49

3.52

A Silicon PN junction at a temperature of 20c C has a reverse saturation current of 10 pico - Ameres (pA). The reserve saturation current at 40cC for the same bias is approximately (A) 30 pA (B) 40 pA (C) 50 pA (D) 60 pA The primary reason for the widespread use of Silicon in semiconductor device technology is (A) abundance of Silicon on the surface of the Earth. (B) larger bandgap of Silicon in comparison to Germanium. (C) favorable properties of Silicon - dioxide (SiO2) (D) lower melting point 2005

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(D) 1.4 eV

TWO MARKS

A Silicon sample A is doped with 1018 atoms/cm 3 of boron. Another sample B of identical dimension is doped with 1018 atoms/cm 3 phosphorus. The ratio of electron to hole mobility is 3. The ratio of conductivity of the sample A to B is (A) 3 (B) 1 3 (C) 2 (D) 3 3 2 A Silicon PN junction diode under reverse bias has depletion region of width 10 mm. The relative permittivity of Silicon, er = 11.7 and the permittivity of free space e0 = 8.85 # 10 - 12 F/m. The depletion capacitance of the diode per square meter is (A) 100 mF (B) 10 mF (C) 1 mF (D) 20 mF A MOS capacitor made using p type substrate is in the accumulation mode. The dominant charge in the channel is due to the presence of

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) holes (C) positively charged icons 3.56

Page 52

(B) electrons (D) negatively charged ions

For an n -channel MOSFET and its transfer curve shown in the figure, the threshold voltage is

(C) a CMOS inverter (D) a BJT inverter 3.61

(A) 1 V and the device is in active region (B) - 1 V and the device is in saturation region (C) 1 V and the device is in saturation region (D) - 1 V and the device is an active region 2004

(A) 56 mA (C) 60 mA 2004 ONE MARK

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in 3.57

3.58

The impurity commonly used for realizing the base region of a silicon n - p - n transistor is (A) Gallium (B) Indium (C) Boron (D) Phosphorus

3.62

3.63

3.64

If for a silicon npn transistor, the base-to-emitter voltage (VBE ) is 0.7 V and the collector-to-base voltage (VCB) is 0.2 V, then the transistor is operating in the (A) normal active mode (B) saturation mode (C) inverse active mode (D) cutoff mode 3.65

3.59

3.60

Assuming VCEsat = 0.2 V and b = 50 , the minimum base current (IB) required to drive the transistor in the figure to saturation is

Consider the following statements S1 and S2. S1 : The b of a bipolar transistor reduces if the base width is increased. S2 : The b of a bipolar transistor increases if the dopoing concentration in the base is increased. Which remarks of the following is correct ? (A) S1 is FALSE and S2 is TRUE (B) Both S1 and S2 are TRUE (C) Both S1 and S2 are FALSE (D) S1 is TRUE and S2 is FALSE

TWO MARKS

In an abrupt p - n junction, the doping concentrations on the p side and n -side are NA = 9 # 1016 /cm 3 respectively. The p - n junction is reverse biased and the total depletion width is 3 mm. The depletion width on the p -side is (A) 2.7 mm (B) 0.3 mm (C) 2.25 mm (D) 0.75 mm The resistivity of a uniformly doped n -type silicon sample is 0.5W mc. If the electron mobility (mn) is 1250 cm 2 /V-sec and the charge of an electron is 1.6 # 10 - 19 Coulomb, the donor impurity concentration (ND) in the sample is (A) 2 # 1016 /cm 3 (B) 1 # 1016 /cm 3 (C) 2.5 # 1015 /cm 3 (D) 5 # 1015 /cm 3 Consider an abrupt p - n junction. Let Vbi be the built-in potential of this junction and VR be the applied reverse bias. If the junction capacitance (Cj ) is 1 pF for Vbi + VR = 1 V, then for Vbi + VR = 4 V, Cj will be (A) 4 pF (B) 2 pF (C) 0.25 pF (D) 0.5 pF Consider the following statements Sq and S2. S1 : The threshold voltage (VT ) of MOS capacitor decreases with increase in gate oxide thickness. S2 : The threshold voltage (VT ) of a MOS capacitor decreases with increase in substrate doping concentration. Which Marks of the following is correct ?

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 (A) S1 is FALSE and S2 is TRUE (B) Both S1 and S2 are TRUE (C) Both S1 and S2 are FALSE (D) S1 is TRUE and S2 is FALSE

Given figure is the voltage transfer characteristic of

3.66

(A) an NOMS inverter with enhancement mode transistor as load (B) an NMOS inverter with depletion mode transistor as load

(B) 140 mA (D) 3 mA

3.67

The drain of an n-channel MOSFET is shorted to the gate so that VGS = VDS . The threshold voltage (VT ) of the MOSFET is 1 V. If the drain current (ID) is 1 mA for VGS = 2 V, then for VGS = 3 V, ID is (A) 2 mA (B) 3 mA (C) 9 mA (D) 4 mA The longest wavelength that can be absorbed by silicon, which has

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 53

the bandgap of 1.12 eV, is 1.1 mm. If the longest wavelength that can be absorbed by another material is 0.87 mm, then bandgap of this material is (A) 1.416 A/cm 2 (B) 0.886 eV (C) 0.854 eV 3.68

(D) 0.706 eV

The neutral base width of a bipolar transistor, biased in the active region, is 0.5 mm. The maximum electron concentration and the diffusion constant in the base are 1014 / cm 3 and Dn = 25 cm 2 / sec respectively. Assuming negligible recombination in the base, the collector current density is (the electron charge is 1.6 # 10 - 19 Coulomb) (A) 800 A/cm 2 (B) 8 A/cm 2 (C) 200 A/cm 2 (D) 2 A/cm 2 2003

3.69

3.70

3.71

n -type silicon is obtained by doping silicon with (A) Germanium (B) Aluminium (C) Boron (D) Phosphorus

3.73

3.75

3.76

A particular green LED emits light of wavelength 5490 Ac. The

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The intrinsic carrier concentration of silicon sample at 300 K is 1.5 # 1016 /m 3 . If after doping, the number of majority carriers is 5 # 1020 /m 3 , the minority carrier density is (A) 4.50 # 1011/m 3 (B) 3.333 # 10 4 /m 3

energy bandgap of the semiconductor material used there is (Plank’s constant = 6.626 # 10 - 34 J - s ) (A) 2.26 eV (B) 1.98 eV (C) 1.17 eV (D) 0.74 eV

-5

(D) 3.00 # 10 /m 3

Choose proper substitutes for X and Y to make the following statement correct Tunnel diode and Avalanche photo diode are operated in X bias ad Y bias respectively (A) X: reverse, Y: reverse (B) X: reverse, Y: forward (C) X: forward, Y: reverse (D) X: forward, Y: forward For an n - channel enhancement type MOSFET, if the source is connected at a higher potential than that of the bulk (i.e. VSB > 0 ), the threshold voltage VT of the MOSFET will (A) remain unchanged (B) decrease (C) change polarity (D) increase

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TWO MARKS

An n -type silicon bar 0.1 cm long and 100 mm2 i cross-sectional area has a majority carrier concentration of 5 # 1020 /m 2 and the carrier mobility is 0.13 m2 /V-s at 300 K. If the charge of an electron is 1.5 # 10 - 19 coulomb, then the resistance of the bar is (A) 106 Ohm (B) 10 4 Ohm (C) 10 - 1 Ohm

At 300 K, for a diode current of 1 mA, a certain germanium diode requires a forward bias of 0.1435 V, whereas a certain silicon diode requires a forward bias of 0.718 V. Under the conditions state above, the closest approximation of the ratio of reverse saturation current in germanium diode to that in silicon diode is (A) 1 (B) 5 (D) 8 # 103 (C) 4 # 103

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2003 3.74

3.77

3.78

The Bandgap of silicon at 300 K is (A) 1.36 eV (B) 1.10 eV (C) 0.80 eV (D) 0.67 eV

(C) 5.00 # 10 /m 3

Group 2 1. Heavy doping 2. Coherent radiation 3. Spontaneous emission 4. Current gain

(A) P - 1, Q - 2, R - 4, S - 3 (B) P - 2, Q - 3, R - 1, S - 4 (C) P - 3 Q - 4, R - 1, S - 2 (D) P - 2, Q - 1, R - 4, S - 3

ONE MARK

20

3.72

Group 1 P. LED Q. Avalanche photo diode R. Tunnel diode S. LASER

3.81

(D) 10 - 4 Ohm

The electron concentration in a sample of uniformly doped n -type silicon at 300 K varies linearly from 1017 /cm 3 at x = 0 to 6 # 1016 / cm 3 at x = 2mm . Assume a situation that electrons are supplied to keep this concentration gradient constant with time. If electronic charge is 1.6 # 10 - 19 coulomb and the diffusion constant Dn = 35 cm 2 /s, the current density in the silicon, if no electric field is present, is (A) zero (B) -112 A/cm 2 (C) +1120 A/cm 2 (D) -1120 A/cm 2 Match items in Group 1 with items in Group 2, most suitably.

When the gate-to-source voltage (VGs) of a MOSFET with threshold voltage of 400 mV, working in saturation is 900 mV, the drain current is observed to be 1 mA. Neglecting the channel width modulation effect and assuming that the MOSFET is operating at saturation, the drain current for an applied VGS of 1400 mV is (A) 0.5 mA (B) 2.0 mA (C) 3.5 mA (D) 4.0 mA If P is ivation, Q is n -well implant, R is metallization and S is source/drain diffusion, then the order in which they are carried out in a standard n -well CMOS fabrication process, is (B) Q - S - R - P (A) P - Q - R - S (C) R - P - S - Q (D) S - R - Q - P The action of JFET in its equivalent circuit can best be represented as a (A) Current controlled current source (B) Current controlled voltage source (C) Voltage controlled voltage source (D) Voltage controlled current source 2002

3.82

ONE MARK

In the figure, silicon diode is carrying a constant current of 1 mA. When the temperature of the diode is 20cC, VD is found to be 700 mV. If the temperature rises to 40cC, VD becomes approximately equal to

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 54

1998 3.89

(A) 740 mV (C) 680 mV 3.83

(B) 660 mV (D) 700 mV 3.90

If the transistor in the figure is in saturation, then

The electron and hole concentrations in a intrinsic semiconductor are ni and pi respectively. When doped with a p-type material, these change to n and p, respectively, Then (B) n + ni = p + pi (A) n + p = ni + pi (C) npi = ni p (D) np = ni pi The fT of a BJT is related to its gm, C p and C m as follows + Cm gm gm (C) fT = + Cm

(A) fT =

3.91

(A) (B) (C) (D)

IC IC IC IC

is always equal to bdc IB is always equal to - bde IB is greater than or equal to bdc IB is less than or equal to bdc IB

3.92

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in 2001 3.84

3.85

MOSFET can be used as a (A) current controlled capacitor (C) current controlled inductor

3.93

ONE MARK

(B) voltage controlled capacitor (D) voltage controlled inductor

3.94

The effective channel length of MOSFET in saturation decreases with increase in (A) gate voltage (B) drain voltage (C) source voltage (D) body voltage 1999

3.86

The early effect in a bipolar junction transistor is caused by (A) fast turn-on (B) fast turn-off (C) large collector-base reverse bias (D) large emitter-base forward bias 1999

3.87

3.88

ONE MARK

An npn transistor (with C = 0.3 pF ) has a unity-gain cutoff frequency fT of 400 MHz at a dc bias current Ic = 1 mA . The value of its Cm (in pF) is approximately (VT = 26 mV) (A) 15 (B) 30 (C) 50 (D) 96

A long specimen of p-type semiconductor material (A) is positively charged (B) is electrically neutral (C) has an electric field directed along its length (D) acts as a dipole Two identical FETs, each characterized by the parameters gm and rd are connected in parallel. The composite FET is then characterized by the parameters g g (A) m and 2rd (B) m and rd 2 2 2 (C) 2gm and rd (D) 2gm and 2rd 2 q The units of are kT (A) V (B) V-1 (C) J (D) J/K ONE MARK

For a MOS capacitor fabricated on a p-type semiconductor, strong inversion occurs when (A) surface potential is equal to Fermi potential (B) surface potential is zero (C) surface potential is negative and equal to Fermi potential in magnitude (D) surface potential is positive and equal to twice the Fermi po-

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014

TWO MARKS

An n -channel JEFT has IDSS = 2 mA and Vp =- 4 V . Its transconductance gm (in milliohm) for an applied gate-to-source voltage VGS of - 2 V is (A) 0.25 (B) 0.5 (C) 0.75 (D) 1.0

2p (C p + C m) gm gm (D) fT = 2p (C p + C m) (B) fT =

The static characteristic of an adequately forward biased p-n junction is a straight line, if the plot is of (A) log I vs log V (B) log I vs V (C) I vs log V (D) I vs V

1997 3.95

ONE MARK

tential 3.96

The intrinsic carrier density at 300 K is 1.5 # 1010 /cm3 , in silicon. For n -type silicon doped to 2.25 # 1015 atoms/cm3 , the equilibrium electron and hole densities are (A) n = 1.5 # 1015 /cm3, p = 1.5 # 1010 /cm3 (B) n = 1.5 # 1010 /cm3, p = 2.25 # 1015 /cm3 (C) n = 2.25 # 1015 /cm3, p = 1.0 # 1015 /cm3 (D) n = 1.5 # 1010 /cm3, p = 1.5 # 1010 /cm3

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

1996 3.97

3.98

3.99

3.100

3.102

3.103

ONE MARK

The p-type substrate in a conventional pn -junction isolated integrated circuit should be connected to (A) nowhere, i.e. left floating (B) a DC ground potential (C) the most positive potential available in the circuit (D) the most negative potential available in the circuit If a transistor is operating with both of its junctions forward biased, but with the collector base forward bias greater than the emitter base forward bias, then it is operating in the (A) forward active mode (B) reverse saturation mode (C) reverse active mode (D) forward saturation mode The common-emitter short-circuit current gain b of a transistor (A) is a monotonically increasing function of the collector current IC (B) is a monotonically decreasing function of IC (C) increase with IC , for low IC , reaches a maximum and then decreases with further increase in IC (D) is not a function of IC A n -channel silicon (Eg = 1.1 eV) MOSFET was fabricated using n +poly-silicon gate and the threshold voltage was found to be 1 V. Now, if the gate is changed to p+ poly-silicon, other things remaining the same, the new threshold voltage should be (A) - 0.1 V (B) 0 V (C) 1.0 V (D) 2.1 V 1996

3.101

Page 55

TWO MARKS

In a bipolar transistor at room temperature, if the emitter current is doubled the voltage across its base-emitter junction (A) doubles (B) halves (C) increases by about 20 mV (D) decreases by about 20 mV An npn transistor has a beta cut-off frequency fb of 1 MHz and common emitter short circuit low-frequency current gain bo of 200 it unity gain frequency fT and the alpha cut-off frequency fa respectively are (B) 200 MHz, 199 MHz (A) 200 MHz, 201 MHz (C) 199 MHz, 200 MHz (D) 201 MHz, 200 MHz A silicon n MOSFET has a threshold voltage of 1 V and oxide thickness of Ao. [er (SiO 2) = 3.9, e0 = 8.854 # 10-14 F/cm, q = 1.6 # 10-19 C] The region under the gate is ion implanted for threshold voltage tailoring. The dose and type of the implant (assumed to be a sheet charge at the interface) required to shift the threshold voltage to - 1 V are (B) 1.08 # 1012 /cm2 , n-type (A) 1.08 # 1012 /cm2 , p-type (C) 5.4 # 1011 /cm2 , p-type (D) 5.4 # 1011 /cm2 , n-type

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 56

SOLUTIONS

= 10-8 cm2 # 10-4 cm = 10-12 cm3 So total no. of holes is, p = p 0 # V = 10 # 10-12 = 10-11 Which is approximately equal to zero.

3.1

3.2

3.3

Option (A) is correct. The potential barrier of the pn junction is lowered when a forward bias voltage is applied, allowing electrons and holes to flow across the space charge region (Injection) when holes flow from the p region across the space charge region into the n region, they become excess minority carrier holes and are subject to diffuse, drift and recombination processes.

Option (D) is correct. In a MOSFET operating in the saturation region, the channel length modulation effect causes a decrease in output resistance.

Option (A) is correct. Given, VB = 2V VTN = 1V So, we have Drain voltage

2 or, ID = kN ^VB - 1h Differentiating both side with respect to ID 1 = kN 2 ^VB - 1hdVB dID

Since, Hence, we obtain

VBQ = 2 volt (at D.C. Voltage) dVB = 1 dID 2kN ^VB - 1h 1 = 2 # 40 # 10-6 # ^2 - 1h = 12.5 # 103 W = 12.5 kW

Option (D) is correct. For the semiconductor,

Volume of given device, = 1 mm 2 # 1 mm

Since all the parameters of PMOS and NMOS are equal. So, mn = mp COX bW l = COX bW l = COX bW l L M2 L L M1 Given that M1 is in linear region. So, we assume that M2 is either in cutoff or saturation. Case 1 : M2 is in cut off So, I 2 = I1 = 0 Where I1 is drain current in M1 and I2 is drain current in M2 . m C 2 Since, I1 = p OX bW l82VSD ^VSG - VTp h - V SD B 2 L m C 2 0 = p OX bW l [2VSD ^VSG - VTp h - V SD & ] 2 L Solving it we get, 2 ^VSG - VTp h = VSD 2 ^5 - Vin - 1h = 5 - VD Vin = VD + 3 2

&

VD = 2 volt VG = 2 volt VS = 0 (Ground) Therefore, VGS = 2 > VTN and VDS = 2 > VGS - VTN So, the MOSFET is in the saturation region. Therefore, drain current is ID = kN ^VGS - VTN h2

3.5

Option (A) is correct. Given the circuit as below :

Option (D) is correct. In IC technology, dry oxidation as compared to wet oxidation produces superior quality oxide with a lower growth rate

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in 3.4

3.6

n 0 p 0 = n i2 2 20 p 0 = n i = 1019 = 10 per cm3 n 0 10 V = Area # depth

& For So, So for the NMOS

I1 = 0 , VD = 5 V Vin = 5 + 3 = 4 V 2

VGS = Vin - 0 = 4 - 0 = 4 V and VGS > VTn So it can’t be in cutoff region. Case 2 : M2 must be in saturation region. So, I1 = I 2

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 mp COX W mn COX W 2 2 2 (VSG - VTp) VSD - V SD @ = 2 L (VGS - VTn) 2 L6 2 2 (VSG - VTp) VSD - V SD = (VGS - VTn) 2 & 2 (5 - Vin - 1) (5 - VD) - (5 - VD) 2 = (Vin - 0 - 1) 2 & 2 (4 - Vin) (5 - VD) - (5 - VD) 2 = (Vin - 1) 2 Substituting VD = VDS = VGS - VTn and for N -MOS & VD = Vin - 1 & 2 (4 - Vin) (6 - Vin) - (6 - Vin) 2 = (Vin - 1) 2 & 48 - 36 - 8Vin =- 2Vin + 1 & 6Vin = 11 & Vin = 11 = 1.833 V 6

&

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 57

So for M2 to be in saturation Vin < 1.833 V or Vin < 1.875 V 3.7

Option (B) is correct. Gate source overlap capacitance. Co = dWeox e0 (medium Sio 2 ) tox = 0.69 # 10-15 F

3.8

rl = 3.13

-9 -6 -12 = 20 # 10 # 1 # 10 #-93.9 # 8.9 # 10 1 # 10

3.9

3.10

Option (C) is correct. At VGS = 0 V , Thus

3.14

Option (B) is correct. Source body junction capacitance. Cs = Aer e0 d A = (0.2 mm + 0.2 mm + 0.2 mm) # 1 mm + 2 (0.2 mm # 0.2 mm) = 0.68 mm2 d = 10 nm (depletion width of all junction) -12 8.9 # 10-12 Cs = 0.68 # 10 # 11.7 -# 9 10 # 10 = 7 # 10-15 F

3.15

3.16

Option (B) is correct. Dry oxidation is used to achieve high quality oxide growth. Option (B) is correct. Emitter injection efficiency is given as 1 g = 1 + NB NE SPECIAL EDITION ( STUDY MATERIAL FORM )

At market Book is available in 3 volume i.e. in 3 book binding form. But at NODIA Online Store book is available in 10 book binding form. Each unit of Book is in separate binding.

Option (C) is correct. Drift current Id = qnmn E It depends upon Electric field E and carrier concentration n

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Option (B) is correct. Zener diode operates in reverse breakdown region.

3.17

3.12

Option (D) is correct. For every 1c C increase in temperature, forward bias voltage across diode decreases by 2.5 mV. Thus for 10c C increase, there us 25 mV decreases. Option (B) is correct. Full channel resistance is r L r # = 600 W W#a If VGS is applied, Channel resistance is r L where b = a c1 rl = # W#b Pinch off voltage, qN Vp = D a2 2e If depletion on each side is d = 1 μm at VGS = 0 . qN Vj = D d2 2e qND qN or 1 = D (1 # 10-6) 2 & = 1012 2e 2e Now from equation (2), we have 12

...(1)

g = 1, NE >> NB

VGS Vp m

Option (C) is correct. Reverse bias breakdown or Zener effect occurs in highly doped PN junction through tunneling mechanism. In a highly doped PN junction, the conduction and valence bands on opposite sides of the junction are sufficiently close during reverse bias that electron may tunnel directly from the valence band on the p-side into the conduction band on n -side. Breakdown voltage VB \ 1 NA ND So, breakdown voltage decreases as concentration increases Depletion capacitance 1/2 ees NA ND C =' 1 2 (Vbi + VR) (NA + ND) Thus C \ NA ND Depletion capacitance increases as concentration increases

3.18

...(2)

Option (C) is correct. Sample is in thermal equilibrium so, electric field 1 = 10 kV/cm 1 mm Option (A) is correct. Electron drift current density E =

3.19

Jd = ND mn eE = 1016 # 1350 # 1.6 # 10-19 # 10 # 1013 = 2.16 # 10 4 A/cm2 3.20

-6 2

Vp = 10 # (5 # 10 )

or Vp =- 25 V At VGS =- 3 V ; - 3 mm = 3.26 mm b = 5 b1 - 25 l

since 2b = 8 mm b = 4 mm rL a = 600 5 = 750 W rl = #4 Wa # b

Option (A) is correct. At room temperature mobility of electrons for Si sample is given mn = 1350 cm2 /Vs . For an n -channel MOSFET to create an inversion layer of electrons, a large positive gate voltage is to be applied. Therefore, induced electric field increases and mobility decreases. So, Mobility mn < 1350 cm2 /Vs for n -channel MOSFET

To achieve

3.11

rL rL a = 600 5 = # 3.26 = 917 W W # b Wa # b

3.21

Option (C) is correct. Only dopant atoms can have concentration of 4 # 1019 cm - 3 in n type silicon at room temperature. Option (A) is correct. 2 Unit of mobility mn is = cm V. sec

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 58

2 Unit of diffusion current Dn is = cm sec 2 2 m Thus unit of n is = cm / cm = 1 = V-1 V $ sec sec V Dn 3.22

3.23

Option (D) is correct. Both S1 and S2 are true and S2 is a reason for S1.

17 E2 - E1 = 25 # 10-3 e ln 4 # 10 10 = 0.427 eV 1.5 # 10 Hence fermi level goes down by 0.427 eV as silicon is doped with boron. 3.31

VP = VP1 VP1 = W12 = W2 Now VP2 W22 (2W) 2 or 4VP1 = VP2 Initial transconductance gm = Kn ;1 - Vbi - VGS E Vp Let

NA WP = ND WN 17 -6 NA = ND WN = 1 # 10 # 0.1-6# 10 = 1 # 1016 WP 1 # 10

The built-in potential is D Vbi = VT 1n c NA N n i2 m 17 16 # 10 = 0.760 = 26 # 10-3 ln e 1 # 10 # 1 10 o (1.4 # 10 ) 2 3.24

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3.25

3.26

3.27

3.28

3.29

3.30

For first condition gm1 0 - (- 2) = Kn =1 = Kn ;1 VP1 G

Option (B) is correct. The peak electric field in device is directed from p to n and is

MV/cm

Option (D) is correct. Channel length modulation is not associated with a p - n junction. It is being associated with MOSFET in which effective channel length decreases, producing the phenomenon called channel length modulation.

gm2 = Kn =1 -

Hence 3.32 3.33

NA = 4 # 1017 ni = 1.5 # 1010

Option (A) is correct. Option (D) is correct. As per mass action law

Thus or 3.34

3.35

p = NA nNA = ni2 2 n = ni NA

Option (C) is correct. The electric field has the maximum value at the junction of p+ n . Option (B) is correct. Zener diode and Avalanche diode works in the reverse bias and laser diode works in forward bias. In solar cell diode works in forward bias but photo current is in reverse direction. Thus Zener diode : Reverse Bias

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Option (B) is correct. gm = 2ID = 2 K (VGS - VT ) 2 = 2K (VGS - VT ) 2VGS 2VGS

E2 - E1 = kT ln NA ni

2 4VP1 E

np = ni2 If acceptor impurities are introduces

Option (D) is correct. Oxidation rate is zero because the existing oxide prevent the further oxidation.

Option (C) is correct.

0 - (- 2) = K2 ;1 VP2 G

1 - 2/VP1 gm1 =f p gm2 1 - 1/ (2VP1) VP = VP1

Dividing

Option (A) is correct. Trivalent impurities are used for making p - type semiconductors. So, Silicon wafer heavily doped with boron is a p+ substrate.

Solar Cell : Forward Bias Laser Diode : Forward Bias Avalanche Photo diode : Reverse Bias 3.36

Which is straight line.

2 VP1 E

For second condition

from p to n E =- eND xn es from n to p = eND xn es -19 17 -5 # 1 # 10 = 0.15 = 1.6 # 10 # 1 #-10 14 8.85 # 10 # 12

Option (C) is correct. As VD = constant Thus gm \ (VGS - VT )

2 VP = eW ND es

Pinch off voltage

Option (B) is correct. We know that or

Option (C) is correct.

Option (C) is correct. In BJT as the B-C reverse bias voltage increases, the B-C space charge region width increases which xB (i.e. neutral base width) > A change in neutral base width will change the collector current. A reduction in base width will causes the gradient in minority carrier concentration to increase, which in turn causes an increased in the diffusion current. This effect si known as base modulation as early effect.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

In JFET the gate to source voltage that must be applied to achieve pinch off voltage is described as pinch off voltage and is also called as turn voltage or threshold voltage. In LASER population inversion occurs on the condition when concentration of electrons in one energy state is greater than that in lower energy state, i.e. a non equilibrium condition. In MOS capacitor, flat band voltage is the gate voltage that must be applied to create flat ban condition in which there is no space charge region in semiconductor under oxide. Therefore BJT : Early effect MOS capacitor : Flat-band voltage LASER diode : Population inversion JFET : Pinch-off voltage 3.37

Page 59

Now = 0.857 mm 3.41

3.42

C2 = 7 pF 6 - 12 -4 D2 = e0 er2 A = 1 # 710 #- 1210 = 6 # 10 - 4 cm C2 7 6 # 10

Option (C) is correct. Depletion region will not be formed if the MOS capacitor has n type substrate but from C-V characteristics, C reduces if V is increased. Thus depletion region must be formed. Hence S1 is false If positive charges is introduced in the oxide layer, then to equalize the effect the applied voltage V must be reduced. Thus the C - V plot moves to the left. Hence S2 is true. Option (C) is correct. For the case of negative slope it is the negative resistance region

Option (A) is correct. W = K V + VR Now 2m = K 0.8 + 1.2 From above two equation we get 0.8 + 7.2 = 0.8 + 1.2 W2 = 4 m m

W = 2m or 3.38

8 =2 2

Option (B) is correct. a=

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b = 50 = 50 b + 1 50 + 1 51

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Current Gain = Base Transport Factor # Emitter injection Efficiency or 3.39

3.40

a = b1 # b2 50 b1 = a = = 0.985 51 # 0.995 b2

Option (A) is correct. At low voltage when there is no depletion region and capacitance is decide by SiO2 thickness only, C = e0 er1 A D -13 10-4 = 50 nm or D = e0 er1 A = 3.5 # 10 -# 12 C 7 # 10

3.43

Option (B) is correct. The construction of given capacitor is shown in fig below

Option (A) is correct. For n -type p is minority carrier concentration np = ni2 np = Constant p \ 1 n

Since ni is constant

Thus p is inversely proportional to n . 3.44

3.45

When applied voltage is 0 volts, there will be no depletion region and we get C1 = 7 pF When applied voltage is V , a depletion region will be formed as shown in fig an total capacitance is 1 pF. Thus CT = 1 pF or CT = C1 C2 = 1 pF C1 + C2 1 = 1 + 1 or C1 C2 CT Substituting values of CT and C1 we get

3.46

Option (A) is correct. Diffusion current, since the drift current is negligible for minority carrier. Option (B) is correct. In BJT as the B-C reverse bias voltage increases, the B-C space charge region width increases which xB (i.e. neutral base width) > A change in neutral base width will change the collector current. A reduction in base width will causes the gradient in minority carrier concentration to increases, which in turn causes an increases in the diffusion current. This effect si known as base modulation as early effect. Option (A) is correct. For t < 0 diode forward biased and VR = 5 . At t = 0 diode abruptly changes to reverse biased and current across resistor must be 0. But in storage time 0 < t < ts diode retain its resistance of forward

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 60

sp = pqmp sp m = p =1 3 sn mn

biased. Thus for 0 < t < ts it will be ON and VR =- 5 V 3.47

Option (B) is correct. According to Hall effect the direction of electric field is same as that of direction of force exerted. E =- v # B E = B#v

or 3.48

3.54

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3.55

3.56

Option (D) is correct. mP We have = 0.4 mn

3.57

3.58

3.59

mp mn h

Option (C) is correct. For silicon at 0 K,

3.60 3.61

At T = 300 K, Eg300 = 1.21 - 3.6 # 10 - 4 # 300 = 1.1 eV This is standard value, that must be ed. Option (B) is correct. The reverse saturation current doubles for every 10cC rise in temperature as follows : (T - T1)/10

I0 (T) = I 01 # 2 Thus at 40c C, I0 = 40 pA

sn = nqmn

b =

a 1-a

Option (C) is correct. Option (A) is correct. Applying KVL we get

or Now 3.62

IC = VCC - VCE = 3 - 0.2 = 2.8 mA RC 1k IB = IC = 2.8m = 56 mA 50 b

Option (B) is correct. We know that Wp NA = Wn ND

Option (A) is correct. Silicon is abundant on the surface of earth in the from of SiO2 . Option (B) is correct.

Option (D) is correct.

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EgT = Eg0 - 3.6 # 10 T

3.53

Option (A) is correct. Here emitter base junction is forward biased and base collector junction is reversed biased. Thus transistor is operating in normal active region.

VCC - IC RC - VCE = 0

-4

3.52

Option (C) is correct. Trivalent impurities are used for making p type semiconductor. Boron is trivalent.

Thus

Eg0 = 1.21 eV At any temperature

3.51

Option (C) is correct. From the graph it can be easily seen that Vth = 1 V Now VGS = 3 - 1 = 2 V

a -" b a ." b . If the base width increases, recombination of carrier in base region increases and a decreases & hence b decreases. If doping in base region increases, recombination of carrier in base increases and a decreases thereby decreasing b . Thus S1 is true and S2 is false.

Conductance of n type semiconductor sn = nqmn Conductance of intrinsic semiconductor

3.50

Option (B) is correct. In accumulation mode for NMOS having p -substrate, when positive voltage is applied at the gate, this will induce negative charge near p - type surface beneath the gate. When VGS is made sufficiently large, an inversion of electrons is formed and this in effect forms and n - channel.

We have

si = ni q (mn + mp) nmn sn = n Ratio is = n ( ) n 1 s m + m i i n p i^ + 8 4 . 4 2 10 # = = 2 # 10 1.5 # 10 4 (1 + 0.4)

C = e0 er A d C = e0 er = 8.85 # 10-12 # 11.7 = 10.35 m F d A 10 # 10-6

and VDS = 5 - 1 = 4 V Since VDS > VGS $ VDS > VGS - Vth Thus MOSFET is in saturation region.

Schottky diode : High frequency switch 3.49

Option (B) is correct.

or

Option (B) is correct. The varacter diode is used in tuned circuit as it can provide frequently stability. PIN diode is used as a current controlled attenuator. Zener diode is used in regulated voltage supply or fixed voltage reference. Schottkey diode has metal-semiconductor function so it has fast switching action so it is used as high frequency switch Varactor diode : Tuned circuits PIN Diode : Current controlled attenuator Zener diode : Voltage reference

(n = p)

3 m # 1016 = 0.3 m m Wp = Wn # ND = NA 9 # 1016 Option (B) is correct. Conductivity s = nqun

or 3.63

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

r = 1 = 1 nqmn s n = 1 qrmn 16 = 10 /cm 3

or resistivity Thus

At T = 300 K, Eg300 = 1.21 - 3.6 # 10 - 4 # 300 = 1.1 eV This is standard value, that must be ed.

1 1.6 # 10 # 0.5 # 1250 For n type semiconductor n = ND =

3.64

- 19

Page 61

3.71

np = ni2 2 16 .5 # 1016 = 4.5 # 1011 p = ni = 1.5 # 10 # 120 n 5 # 10

Option (D) is correct. We know that eeS NA ND Cj = ; 2 (Vbi + VR)( NA + ND) E 1 Cj \ (Vbi + VR) C j2 (Vbi + VR) 1 1 =1 = = C j1 (Vbi + VR) 2 4 2 C Cj2 = j1 = 1 = 0.5 pF 2 2 1 2

Thus Now or 3.65

3.66

Option (C) is correct. Increase in gate oxide thickness makes difficult to induce charges in channel. Thus VT increases if we increases gate oxide thickness. Hence S1 is false. Increase in substrate doping concentration require more gate voltage because initially induce charges will get combine in substrate. Thus VT increases if we increase substrate doping concentration. Hence S2 is false.

3.72

Option (D) is correct.

3.74

Option (A) is correct.

2 ID2 = (3 - 1) = 4 ID1 (2 - 1) 2 ID2 = 4IDI = 4 mA

or

Eg \ 1 l Eg2 = l1 = 1.1 Eg1 0.87 l2 Eg2 = 1.1 # 1.12 = 1.416 eV 0.87

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3.70

From above relation we have R = = 106 W 3.75

EgT = Eg0 - 3.6 # 10 - 4 T

dn = 6 # 1016 - 1017 =- 2 # 1020 dx 2 # 10 - 4 - 0 Jn = nqme E + Dn q dn dx

Since no electric field is present, E = 0 and we get So, Jn = qDn dn dx = 1.6 # 10 - 19 # 35 # (- 2 # 1020) =- 1120 A/cm 2 3.76

3.77

Option (C) is correct. LED works on the principal of spontaneous emission. In the avalanche photo diode due to the avalanche effect there is large current gain. Tunnel diode has very large doping. LASER diode are used for coherent radiation. Option (C) is correct.

We know that I = Io `e h V - 1j where h = 1 for germanium and h = 2 silicon. As per question VD1

si

T

Io `e e - 1j = Io `e hV - 1j VDsi

VDGe

hVT

n

Option (D) is correct. Pentavalent make n -type semiconductor and phosphorous is pentavalent. Option (C) is correct. For silicon at 0 K Eg0 = 1.21 eV At any temperature

1 0.1 # 10 - 2 = 20 nqmn A 5 # 10 # 1.6 # 10 - 19 # 0.13 # 100 # 10 - 12

Option (D) is correct.

Now

Option (B) is correct. Concentration gradient dn = 1014 = 2 # 1018 -4 dx 0.5 # 10 q = 1.6 # 10 - 19 C Dn = 25 1014 dn = dx 0.5 # 10 - 4 JC = qDn dn = 1.6 # 10 - 19 # 25 # 2 # 1018 = 8 A/cm 2 dx

rl , r = 1 and a = nqun A s

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Option (A) is correct.

Thus

3.69

R =

We that

Option (D) is correct. We know that

or

3.68

Option (C) is correct. Tunnel diode shows the negative characteristics in forward bias. It is used in forward bias. Avalanche photo diode is used in reverse bias.

3.73

ID = K (VGS - VT ) 2 (VGS2 - VT ) 2 I DS Thus = IDI (VGS1 - VT ) 2 Substituting the values we have

3.67

Option (A) is correct. By Mass action law

T

Ge

VDsi

or

si

T

DGe

si

3.78

0.718

Io # 10 - 1 = 4 # 103 = eVhV - 1 = e 2 #026.1435 Io e 26 # 10 - 1 e hV - 1 T

-3

-3

Option (A) is correct.

In eV

-34 8 Eg = hc = 6.626 # 10 # -310# 10 = 3.62 J l 54900 # 10 -19 E (J) Eg (eV) = g = 3.62 # 10-19 = 2.26 eV e 1.6 # 10

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 62

Alternatively

At a given value of vBE , increasing the reverse-bias voltage on the collector-base junction and thus increases the width of the depletion region of this junction. This in turn results in a decrease in the effective base width W . Since IS is inversely proportional to W , IS increases and that iC increases proportionally. This is early effect.

1.24 Eg = 1.24 eV = = 2.26 eV l (mm) 5490 # 10-4 mm 3.79

Option (D) is correct. We know that ID = K (VGS - VT ) 2 2 ID2 = (VGS2 - VT ) Thus ID1 (VGS1 - VT ) 2 Substituting the values we have

or 3.80

2 ID2 = (1.4 - 0.4) = 4 ID1 (0.9 - 0.4) 2 ID2 = 4IDI = 4 mA

3.88

Option (D) is correct. For a JFET in active region we have 2 IDS = IDSS c1 - VGS m VP

Option (B) is correct. At constant current the rate of change of voltage with respect to temperature is dV =- 2.5 mV per degree centigrade dT Here Thus Therefore,

3.83

3.84

3 T = T2 - T1 = 40 - 20 = 20cC 3 VD =- 2.5 # 20 = 50 mV VD = 700 - 50 = 650 mV

or or

3.85

3.89

3.86

Option (C) is correct.

C m 15.3 # 10-12 - 0.3 # 10-12 = 15 # 10-12 15 pF

Option (D) is correct. For any semiconductor (Intrinsic or extrinsic) the product n p remains constant at a given temperature so here np = ni pi

3.90

Option (D) is correct. fT =

3.91

gm 2p (C p + C m)

Option (B) is correct. For a Forward Bias p-n junction, current equation I = I 0 (eV/kT - 1) or or

I + 1 = eV/kT I0 kT log b I + 1l = V I0

So if we plot log I vs V we get a straight line. 3.92

3.93 3.94

Option (B) is correct. A specimen of p - type or n - type is always electrical neutral. Option (C) is correct. Option (B) is correct. The unit of q is e and unit of kT is eV. Thus unit of e/kT is

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Option (B) is correct. The metal area of the gate in conjunction with the insulating dielectric oxide layer and semiconductor channel, form a parallel plate capacitor. It is voltage controlled capacitor because in active region the current voltage relationship is given by Option (D) is correct. In MOSFET the body (substrate) is connected to power supply in such a way to maintain the body (substrate) to channel junction in cutoff condition. The resulting reverse bias voltage between source and body will have an effect on device function. The reverse bias will widen the depletion region resulting the reduction in channel length.

gm = IC = 1 26 VT gm fT = 2p (C p + C m) 1/26 400 = 2p (0.3 # 10-12 + C m) 1 = 15.3 # 10-12 (0.3 # 10-12 + C m) = 2p # 26 # 400

or

Option (D) is correct. Condition for saturation is IC < bIB

IDS = K (VGS - VT ) 2

Option (A) is correct.

Now

From above equation it is clear that the action of a JFET is voltage controlled current source. 3.82

Option (B) is correct. For an n -channel JEFT trans-conductance is -3 (- 2) gm = - 2IDSS b1 - VGS l = - 2 # 2 # 10 =1 -4 VP VP (- 4)G = 10-3 # 1 = 0.5 mho 2

We have

Option (B) is correct. In n -well CMOS fabrication following are the steps : (A) n - well implant (B) Source drain diffusion (C) Metalization (D) ivation

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3.87

e/eV = V-1 . 3.95 3.96

Option (D) is correct. Option (C) is correct. We have

ni = 1.5 # 1010 /cm3 Nd = 2.25 # 1015 atoms/cm3 For n type doping we have electron concentration n - Nd = 2.25 # 1015 atom/cm3 For a given temperature np = n i2 Hole concentration

2 (1.5 # 1010) 2 p = ni = = 1.0 # 105 /cm3 n 2.25 # 1015

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

3.97

3.98

Page 63

Option (D) is correct. In p n -junction isolated circuit we should have high impedance, so that p n junction should be kept in reverse bias. (So connect p to negative potential in the circuit) Option (B) is correct.

If both junction are forward biased and collector base junction is more forward biased then IC will be flowing out wards (opposite direction to normal mode) the collector and it will be in reverse saturation mode. 3.99

Option (C) is correct. For normal active mode we have b = IC IB For small values of IC , if we increases IC , b also increases until we reach (IC ) saturation. Further increases in IC (since transistor is in saturation mode know) will increases IB and b decreases.

3.100

Option (C) is correct. For a n -channel mosfet thresholds voltage is given by VTN = VGS - VDS (sat) for p-channel [p+ polysilicon used in gate]

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VTP = VSD (sat) - VGS so VTP =- VDS (sat) + VGS so threshold voltage will be same. 3.101

Option (C) is correct. Emitter current is given by IE = I 0 (eV /kT - 1) IE = I 0 eV /kT eV VBE = kT ln b IE l I0 (VBE ) 1 = kT ln b IE 1 l I0 (VBE ) 2 = kT ln b IE 2 l I0 (VBE ) 2 - (VBE ) 1 = kT ;ln b IE 2 lE = kT ln b 2IE 1 l IE 1 IE 1 BE

or or Now

or

BE

BE

/kT

>> 1

Now if emitter current is double i.e. IE 2 = 2IE1 (VBE ) 2 = (VBE ) 1 + (25 # 0.60) m volt = (VBE ) 1 + 15 m volt Thus if emitter current is doubled the base emitter junction voltage is increased by 15 mV. 3.102

Option (A) is correct. Unity gain frequency is given by fT = fB # b = 106 # 200 = 200 MHz a-cutoff frequency is given by f fb fa = b = = fb (b + 1) 1-a b 1b+1 = 106 # (200 + 1) = 201 MHz

3.103

Option (A) is correct.

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 64

UNIT 4 ANALOG CIRCUITS

2013 4.1

ONE MARK

In the circuit shown below what is the output voltage ^Vouth if a silicon transistor Q and an ideal op-amp are used?

(A) 125 and 125 (C) 250 and 125 4.4

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(B) - 0.7 V (D) + 15 V

In a voltage-voltage as shown below, which one of the following statements is TRUE if the gain k is increased?

(A) The input es (B) The input increases (C) The input decreases (D) The input es

impedance increases and output impedance decreas-

impedance decreases and output impedance also

4.3

In the circuit shown below, the knee current of the ideal Zener dioide is 10 mA. To maintain 5 V across RL , the minimum value of RL in W and the minimum power rating of the Zener diode in mW , respectively, are

In the circuit shown below the op-amps are ideal. Then, Vout in Volts is

(A) 4 (C) 8

impedance decreases and output impedance increas-

TWO MARKS

(B) 32 (D) 200

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impedance increases and output impedance also

4.6

2013

The ac schematic of an NMOS common-source state is shown in the figure below, where part of the biasing circuits has been omitted for simplicity. For the n -channel MOSFET M, the transconductance gm = 1 mA/V , and body effect and channel length modulation effect are to be neglected. The lower cutoff frequency in HZ of the circuit is approximately at

(A) 8 (C) 50 4.5

(B) 125 and 250 (D) 250 and 250

(B) 6 (D) 10

In the circuit shown below, Q1 has negligible collector-to-emitter saturation voltage and the diode drops negligible voltage across it under forward bias. If Vcc is + 5 V , X and Y are digital signals with 0 V as logic 0 and Vcc as logic 1, then the Boolean expression for Z is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 65

The current in the circuit is (A) 10 mA (C) 6.67 mA (A) XY (C) XY 4.7

(B) XY (D) XY

4.11

The diodes and capacitors in the circuit shown are ideal. The voltage v (t) across the diode D1 is

A voltage 1000 sin wt Volts is applied across YZ . Assuming ideal diodes, the voltage measured across WX in Volts, is

(A) cos (wt) - 1

(A) sin wt (C) ^sin wt - sin wt h /2 4.8

(B) _sin wt + sin wt i /2 (D) 0 for all t

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In the circuit shown below, the silicon npn transistor Q has a very high value of b . The required value of R2 in kW to produce IC = 1 mA is

(A) 20 (C) 40

(C) 1 - cos (wt)

(B) 100 W (D) 10.1 kW

2012

ONE MARK

The current ib through the base of a silicon npn transistor is 1 + 0.1 cos (10000pt) mA At 300 K, the rp in the small signal model of the transistor is

(D) 1 - sin (wt)

The impedance looking into nodes 1 and 2 in the given circuit is

(A) 50 W (C) 5 kW

(B) 30 (D) 50

2012

(B) sin (wt)

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4.12

4.9

(B) 9.3 mA (D) 6.2 mA

4.13

The circuit shown is a

1 rad/s (R1 + R2) C (B) high filter with f3dB = 1 rad/s R1 C (C) low filter with f3dB = 1 rad/s R1 C 1 (D) high filter with f3dB = rad/s (R1 + R2) C (A) low filter with f3dB =

(A) 250 W (C) 25 W 4.10

(B) 27.5 W (D) 22.5 W

The i -v characteristics of the diode in the circuit given below are v - 0.7 A, v $ 0.7 V i = * 500 0A v < 0. 7 V

4.14

The voltage gain Av of the circuit shown below is

TWO MARKS

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 66

and the threshold voltage VT = 1 V . The voltage Vx at the source of the upper transistor is

(A) Av . 200 (C) Av . 20

(B) Av . 100 (D) Av . 10

(A) 1 V (C) 3 V

2011 4.15

ONE MARK

4.18

In the circuit shown below, capacitors C1 and C2 are very large and are shorts at the input frequency. vi is a small signal input. The gain magnitude vo at 10 M rad/s is vi

(B) 2 V (D) 3.67 V

For the BJT, in the circuit shown below, Q1 b = 3, VBEon = 0.7 V, VCEsat = 0.7 V . The switch is initially closed. At time t = 0 , the switch is opened. The time t at which Q1 leaves the active region is

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(B) 25 ms (D) 100 ms

For a BJT, the common base current gain a = 0.98 and the collector base junction reverse bias saturation current ICO = 0.6 mA . This BJT is connected in the common emitter mode and operated in the active region with a base drive current IB = 20 mA . The collector current IC for this mode of operation is (A) 0.98 mA (B) 0.99 mA (C) 1.0 mA (D) 1.01 mA

Statement for Linked Answer Questions: 4.6 & 4.7 (A) maximum (C) unity 4.16

The circuit below implements a filter between the input current ii and the output voltage vo . Assume that the op-amp is ideal. The filter implemented is a

(A) low filter (C) band stop filter 2011 4.17

(B) minimum (D) zero

(B) band filter (D) high filter TWO MARKS

In the circuit shown below, for the MOS transistors, mn Cox = 100 mA/V 2

In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage Vt = kT/q = 25 mV . The small signal input vi = Vp cos ^wt h where Vp = 100 mV.

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 4.20

4.21

The bias current IDC through the diodes is (A) 1 mA (B) 1.28 mA (C) 1.5 mA (D) 2 mA

2010

The ac output voltage vac is (A) 0.25 cos ^wt h mV (C) 2 cos (wt) mV

Consider the common emitter amplifier shown below with the following circuit parameters: b = 100, gm = 0.3861 A/V, r0 = 259 W, RS = 1 kW, RB = 93 kW, RC = 250 kW, RL = 1 kW, C1 = 3 and C2 = 4.7 mF

TWO MARKS

Common Data For Q. 4.11 & 4.12 :

(B) 1 cos (wt) mV (D) 22 cos (wt) mV

2010 4.22

Page 67

ONE MARK

The amplifier circuit shown below uses a silicon transistor. The capacitors CC and CE can be assumed to be short at signal frequency and effect of output resistance r0 can be ignored. If CE is disconnected from the circuit, which one of the following statements is true

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(A) The input resistance Ri increases and magnitude of voltage gainAV decreases (B) The input resistance Ri decreases and magnitude of voltage gain AV increases (C) Both input resistance Ri and magnitude of voltage gain AV decreases (D) Both input resistance Ri and the magnitude of voltage gain AV increases 4.23

In the silicon BJT circuit shown below, assume that the emitter area of transistor Q1 is half that of transistor Q2

The value of current Io is approximately (A) 0.5 mA (B) 2 mA (C) 9.3 mA (D) 15 mA 4.24

Assuming the OP-AMP to be ideal, the voltage gain of the amplifier shown below is

(A) - R2 R1 R || R 3 (C) - 2 R1

(B) - R 3 R1 (D) -b R2 + R 3 l R1

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4.26

4.27

The resistance seen by the source vS is (A) 258 W (B) 1258 W (C) 93 kW (D) 3 The lower cut-off frequency due to C2 is (A) 33.9 Hz (B) 27.1 Hz (C) 13.6 Hz (D) 16.9 Hz The transfer characteristic for the precision rectifier circuit shown below is (assume ideal OP-AMP and practical diodes)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

2009 4.28

4.30

In the circuit below, the diode is ideal. The voltage V is given by

(A) min (Vi, 1) (C) min (- Vi, 1) 4.29

TWO MARKS

Page 68

(B) max (Vi, 1) (D) max (- Vi, 1)

4.31

In the following a stable multivibrator circuit, which properties of v0 (t) depend on R2 ?

4.32

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For small increase in VG beyond 1V, which of the following gives the correct description of the region of operation of each MOSFET (A) Both the MOSFETs are in saturation region (B) Both the MOSFETs are in triode region (C) n-MOSFETs is in triode and p -MOSFET is in saturation region (D) n- MOSFET is in saturation and p -MOSFET is in triode region Estimate the output voltage V0 for VG = 1.5 V. [Hints : Use the appropriate current-voltage equation for each MOSFET, based on the answer to Q.4.16] (B) 4 + 1 (A) 4 - 1 2 2 (C) 4 - 3 (D) 4 + 3 2 2 In the circuit shown below, the op-amp is ideal, the transistor has VBE = 0.6 V and b = 150 . Decide whether the in the circuit is positive or negative and determine the voltage V at the output of the op-amp.

(A) Positive , V = 10 V (B) Positive , V = 0 V (C) Negative , V = 5 V (D) Negative , V = 2 V 4.33

(A) Only the frequency (B) Only the amplitude (C) Both the amplitude and the frequency (D) Neither the amplitude nor the frequency

Statement for Linked Answer Question 4.16 and 4.17 Consider for CMOS circuit shown, where the gate voltage v0 of the n-MOSFET is increased from zero, while the gate voltage of the p -MOSFET is kept constant at 3 V. Assume, that, for both transistors, the magnitude of the threshold voltage is 1 V and the product of the trans-conductance parameter is 1mA. V - 2

A small signal source Vi (t) = A cos 20t + B sin 106 t is applied to a transistor amplifier as shown below. The transistor has b = 150 and hie = 3W . Which expression best approximate V0 (t)

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 69

(A) 0 V (C) 0.7 V 4.37

(B) 0.1 V (D) 1.1 V

The OPAMP circuit shown above represents a

(A) V0 (t) =- 1500 (A cos 20t + B sin 106 t) (B) V0 (t) = - 1500( A cos 20t + B sin 106 t) (C) V0 (t) =- 1500B sin 106 t (D) V0 (t) =- 150B sin 106 t 2008 4.34

ONE MARK

In the following limiter circuit, an input voltage Vi = 10 sin 100pt is applied. Assume that the diode drop is 0.7 V when it is forward biased. When it is forward biased. The zener breakdown voltage is 6.8 V The maximum and minimum values of the output voltage respectively are

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Click to Buy www.nodia.co.in (A) high filter (C) band filter (A) 6.1 V, - 0.7 V (C) 7.5 V, - 0.7 V 2008 4.35

(B) 0.7 V, - 7.5 V (D) 7.5 V, - 7.5 V

4.38

TWO MARSK

Two identical NMOS transistors M1 and M2 are connected as shown below. Vbias is chosen so that both transistors are in saturation. The equivalent gm of the pair is defied to be 2Iout at constant Vout 2Vi The equivalent gm of the pair is

For the circuit shown in the following figure, transistor M1 and M2 are identical NMOS transistors. Assume the M2 is in saturation and the output is unloaded.

4.39

4.36

(B) low filter (D) band reject filter

The current Ix is related to Ibias as (B) Ix = Ibias (A) Ix = Ibias + Is (D) Ix = Ibias - Is (C) Ix = Ibias - cVDD - Vout m RE Consider the following circuit using an ideal OPAMP. The I-V V characteristic of the diode is described by the relation I = I 0 _eV - 1i where VT = 25 mV, I0 = 1m A and V is the voltage across the diode (taken as positive for forward bias). For an input voltage Vi =- 1 V , the output voltage V0 is t

(A) the sum of individual gm ' s of the transistors (B) the product of individual gm ’s of the transistors (C) nearly equal to the gm of M1 g (D) nearly equal to m of M2 g0 Consider the Schmidt trigger circuit shown below A triangular wave which goes from -12 to 12 V is applied to the inverting input of OPMAP. Assume that the output of the OPAMP swings from +15 V to -15 V. The voltage at the non-inverting input switches between

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) - 12V to +12 V (C) -5 V to +5 V

Page 70

(B) -7.5 V to 7.5 V (D) 0 V and 5 V

Statement for Linked Answer Question 3.26 and 3.27: In the following transistor circuit, VBE = 0.7 V, r3 = 25 mV/IE , and b and all the capacitances are very large (A) -2 V (C) -0.5 V 4.45

4.40

The value of DC current IE is (A) 1 mA (C) 5 mA

(B) 2 mA (D) 10 mA

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in 4.41

4.43

In the Op-Amp circuit shown, assume that the diode current follows the equation I = Is exp (V/VT ). For Vi = 2V, V0 = V01, and for Vi = 4V, V0 = V02 . The relationship between V01 and V02 is

(A) V02 = 2 Vo1 (C) Vo2 = Vo1 1n2

ONE MARK

The correct full wave rectifier circuit is

4.47

(B) Vo2 = e2 Vo1 (D) Vo1 - Vo2 = VT 1n2

In the CMOS inverter circuit shown, if the trans conductance parameters of the NMOS and PMOS transistors are W kn = kp = mn Cox Wn = mCox p = 40mA/V2 Ln Lp and their threshold voltages ae VTHn = VTHp = 1 V the current I is

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In a transconductance amplifier, it is desirable to have (A) a large input resistance and a large output resistance (B) a large input resistance and a small output resistance (C) a small input resistance and a large output resistance (D) a small input resistance and a small output resistance 2007

4.44

(B) saturation (D) reverse active

The mid-band voltage gain of the amplifier is approximately (A) -180 (B) -120 (C) -90 (D) -60 2007

4.42

For the BJT circuit shown, assume that the b of the transistor is very large and VBE = 0.7 V. The mode of operation of the BJT is

(A) cut-off (C) normal active 4.46

(B) -1 V (D) 0.5 V

TWO MARKS

For the Op-Amp circuit shown in the figure, V0 is

(A) 0 A (C) 45 mA 4.48

(B) 25 mA (D) 90 mA

For the Zener diode shown in the figure, the Zener voltage at knee is 7 V, the knee current is negligible and the Zener dynamic resistance

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 71

is 10 W. If the input voltage (Vi) range is from 10 to 16 V, the output voltage (V0) ranges from

(A) 0 Volt (C) 9.45 Volts 4.54

(A) 7.00 to 7.29 V (C) 7.14 to 7.43 V

(B) 6.3 Volt (D) 10 Volts

For the circuit shown below, assume that the zener diode is ideal with a breakdown voltage of 6 volts. The waveform observed across R is

(B) 7.14 to 7.29 V (D) 7.29 to 7.43 V

Statement for Linked Answer Questions 4.35 & 4.36: Consider the Op-Amp circuit shown in the figure.

4.49

4.50

The transfer function V0 (s)/ Vi (s) is (A) 1 - sRC (B) 1 + sRC 1 + sRC 1 - sRC 1 1 (C) (D) 1 - sRC 1 + sRC

4.52

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ONE MARK

The input impedance (Zi) and the output impedance (Z0) of an ideal trans-conductance (voltage controlled current source) amplifier are (B) Zi = 0, Z0 = 3 (A) Zi = 0, Z0 = 0 (C) Zi = 3, Z0 = 0 (D) Zi = 3, Z0 = 3 An n-channel depletion MOSFET has following two points on its ID - VGs curve: (i) VGS = 0 at ID = 12 mA and (ii) VGS =- 6 Volts at ID = 0 mA Which of the following Q point will given the highest trans conductance gain for small signals? (B) VGS =- 3 Volts (A) VGS =- 6 Volts (C) VGS = 0 Volts (D) VGS = 3 Volts 2006

4.53

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If Vi = V1 sin (wt) and V0 = V2 sin (wt + f), then the minimum and maximum values of f (in radians) are respectively (B) 0 and p (A) - p and p 2 2 2 (C) - p and 0 (D) - p and 0 2 2006

4.51

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TWO MARKS

For the circuit shown in the following figure, the capacitor C is initially uncharged. At t = 0 the switch S is closed. The Vc across the capacitor at t = 1 millisecond is In the figure shown above, the OP-AMP is supplied with ! 15V .

Common Data For Q. 4.41, 4.42 and 4.43 : In the transistor amplifier circuit shown in the figure below, the transistor has the following parameters: bDC = 60 , VBE = 0.7V, hie " 3 The capacitance CC can be assumed to be infinite. In the figure above, the ground has been shown by the symbol 4

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

4.55

4.56

4.57

Under the DC conditions, the collector-or-emitter voltage drop is (A) 4.8 Volts (B) 5.3 Volts (C) 6.0 Volts (D) 6.6 Volts

Page 72

4.61

If bDC is increased by 10%, the collector-to-emitter voltage drop (A) increases by less than or equal to 10% (B) decreases by less than or equal to 10% (C) increase by more than 10% (D) decreases by more than 10% The small-signal gain of the amplifier vc is vs

4.62

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in (A) -10 (C) 5.3

(B) 10 kW

(C) 40 kW

(D) infinite

The effect of current shunt in an amplifier is to (A) increase the input resistance and decrease the output resistance (B) increases both input and output resistance (C) decrease both input and output resistance (D) decrease the input resistance and increase the output resistance The cascade amplifier is a multistage configuration of (A) CC - CB (B) CE - CB (C) CB - CC (D) CE - CC 2005

4.63

(B) -5.3 (D) 10 4.64

Common Data For Q. 4.44 & 4.45:

(A) 30 kW 4

A regulated power supply, shown in figure below, has an unregulated input (UR) of 15 Volts and generates a regulated output Vout . Use the component values shown in the figure.

In an ideal differential amplifier shown in the figure, a large value of (RE ). (A) increase both the differential and common - mode gains. (B) increases the common mode gain only. (C) decreases the differential mode gain only. (D) decreases the common mode gain only. For an npn transistor connected as shown in figure VBE = 0.7 volts. Given that reverse saturation current of the junction at room temperature 300 K is 10 - 13 A, the emitter current is

(A) 30 mA (C) 49 mA 4.65

4.58

4.59

The power dissipation across the transistor Q1 shown in the figure is (A) 4.8 Watts (B) 5.0 Watts (C) 5.4 Watts (D) 6.0 Watts

(B) 39 mA (D) 20 mA

The voltage e0 is indicated in the figure has been measured by an

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 ideal voltmeter. Which of the following can be calculated ?

If the unregulated voltage increases by 20%, the power dissipation across the transistor Q1 (A) increases by 20% (B) increases by 50% (C) remains unchanged (D) decreases by 20% 2005

4.60

TWO MARKS

ONE MARK

The input resistance Ri of the amplifier shown in the figure is (A) Bias current of the inverting input only (B) Bias current of the inverting and non-inverting inputs only

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 73

(C) Input offset current only (D) Both the bias currents and the input offset current 4.66

The Op-amp circuit shown in the figure is filter. The type of filter and its cut. Off frequency are respectively

4.70

(A) high , 1000 rad/sec. (C) high , 1000 rad/sec 4.67

(B) Low , 1000 rad/sec (D) low , 10000 rad/sec

The circuit using a BJT with b = 50 and VBE = 0.7V is shown in the figure. The base current IB and collector voltage by VC and respectively

4.71

Zi and Z0 of the circuit are respectively (B) 2 MW and 20 kW (A) 2 MW and 2 kW 11 (C) infinity and 2 MW (D) infinity and 20 kW 11 ID and VDS under DC conditions are respectively (A) 5.625 mA and 8.75 V (B) 1.875 mA and 5.00 V (C) 4.500 mA and 11.00 V (D) 6.250 mA and 7.50 V

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(A) 43 mA and 11.4 Volts (C) 45 mA and 11 Volts 4.68

The Zener diode in the regulator circuit shown in the figure has a Zener voltage of 5.8 volts and a zener knee current of 0.5 mA. The maximum load current drawn from this current ensuring proper functioning over the input voltage range between 20 and 30 volts, is

(A) 23.7 mA (C) 13.7 mA 4.69

(B) 40 mA and 16 Volts (D) 50 mA and 10 Volts

(B) 14.2 mA (D) 24.2 mA

Both transistors T1 and T2 show in the figure, have a b = 100 , threshold voltage of 1 Volts. The device parameters K1 and K2 of T1 and T2 are, respectively, 36 mA/V2 and 9 mA/V 2 . The output voltage Vo i s

(A) 1 V (C) 3 V

(B) 2 V (D) 4 V

Common Data For Q. 4.58, 4.59 and 4.60 : Given, rd = 20kW , IDSS = 10 mA, Vp =- 8 V

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4.73

Transconductance in milli-Siemens (mS) and voltage gain of the amplifier are respectively (A) 1.875 mS and 3.41 (B) 1.875 ms and -3.41 (C) 3.3 mS and -6 (D) 3.3 mS and 6 Given the ideal operational amplifier circuit shown in the figure indicate the correct transfer characteristics assuming ideal diodes with zero cut-in voltage.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

2004 4.74

4.75

4.76

Page 74

ONE MARK

An ideal op-amp is an ideal (A) voltage controlled current source (B) voltage controlled voltage source (C) current controlled current source (D) current controlled voltage source Voltage series (also called series-shunt ) results in (A) increase in both input and output impedances (B) decrease in both input and output impedances (C) increase in input impedance and decrease in output impedance (D) decrease in input impedance and increase in output impedance The circuit in the figure is a

(A) 1 mF 2p 1 mF (C) 2p 6 4.79

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(A) low- filter (C) band- filter 2004 4.77

4.78

(B) high- filter (D) band-reject filter

4.81

TWO MARKS

A bipolar transistor is operating in the active region with a collector current of 1 mA. Assuming that the b of the transistor is 100 and the thermal voltage (VT ) is 25 mV, the transconductance (gm) and the input resistance (rp) of the transistor in the common emitter configuration, are (A) gm = 25 mA/V and rp = 15.625 kW (B) gm = 40 mA/V and rp = 4.0 kW (C) gm = 25 mA/V and rp = 2.5 k W (D) gm = 40 mA/V and rp = 2.5 kW The value of C required for sinusoidal oscillations of frequency 1 kHz in the circuit of the figure is

(D) 2p 6 mF

In the op-amp circuit given in the figure, the load current iL is

(A) - Vs R2 (C) - Vs RL 4.80

(B) 2p mF

(B) Vs R2 (D) Vs R1

In the voltage regulator shown in the figure, the load current can vary from 100 mA to 500 mA. Assuming that the Zener diode is ideal (i.e., the Zener knee current is negligibly small and Zener resistance is zero in the breakdown region), the value of R is

(A) 7 W (B) 70 W (D) 14 W (C) 70 W 3 In a full-wave rectifier using two ideal diodes, Vdc and Vm are the dc and peak values of the voltage respectively across a resistive load. If PIV is the peak inverse voltage of the diode, then the appropriate

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4.82

relationships for this rectifier are (A) Vdc = Vm , PIV = 2Vm (B) Idc = 2 Vm , PIV = 2Vm p p (C) Vdc = 2 Vm , PIV = Vm (D) Vdc Vm , PIV = Vm p p Assume that the b of transistor is extremely large and VBE = 0.7V, IC and VCE in the circuit shown in the figure

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 75

resistance of the current-shunt negative amplifier using the above amplifier with a factor of 0.2, is (B) 1 kW (A) 1 kW 5 11 (C) 5 kW 4.89

(A) IC = 1 mA, VCE = 4.7 V (C) IC = 1 mA, VCE = 2.5 V

(B) IC = 0.5 mA, VCE = 3.75 V (D) IC = 0.5 mA, VCE = 3.9 V

2003 4.83

4.84

In the amplifier circuit shown in the figure, the values of R1 and R2 are such that the transistor is operating at VCE = 3 V and IC = 1.5 mA when its b is 150. For a transistor with b of 200, the operating point (VCE , IC ) is

ONE MARK

Choose the correct match for input resistance of various amplifier configurations shown below : Configuration Input resistance CB : Common Base LO : Low CC : Common Collector MO : Moderate CE : Common Emitter HI : High (A) CB - LO, CC - MO, CE - HI (B) CB - LO, CC - HI, CE - MO (C) CB - MO, CC - HI, CE - LO (D) CB - HI, CC - LO, CE - MO

(A) (2 V, 2 mA) (C) (4 V, 2 mA)

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The circuit shown in the figure is best described as a

(A) bridge rectifier (C) frequency discriminator

The oscillator circuit shown in the figure has an ideal inverting amplifier. Its frequency of oscillation (in Hz) is

(B) ring modulator (D) voltage double

If the input to the ideal comparators shown in the figure is a sinusoidal signal of 8 V (peak to peak) without any DC component, then the output of the comparators has a duty cycle of 1 (2p 6 RC) 1 (C) ( 6 RC) (A) (A) 1/2 (C) 1/6

4.86

4.87

(B) 1/3 (D) 1/2

4.91

If the differential voltage gain and the common mode voltage gain of a differential amplifier are 48 dB and 2 dB respectively, then common mode rejection ratio is (A) 23 dB (B) 25 dB (C) 46 dB (D) 50 dB Generally, the gain of a transistor amplifier falls at high frequencies due to the (A) internal capacitances of the device (B) coupling capacitor at the input (C) skin effect (D) coupling capacitor at the output 2003

TWO MARKS

An amplifier without has a voltage gain of 50, input resistance of 1 k W and output resistance of 2.5 kW. The input

(B)

1 (2pRC)

(D)

6 (2pRC)

The output voltage of the regulated power supply shown in the figure is

(A) 3 V (C) 9 V 4.92

4.88

(B) (3 V, 2 mA) (D) (4 V, 1 mA)

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4.90

4.85

(D) 11 kW

(B) 6 V (D) 12 V

If the op-amp in the figure is ideal, the output voltage Vout will be equal to

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) 1 V (C) 14 V 4.93

4.94

Page 76

(B) 6 V (D) 17 V

Three identical amplifiers with each one having a voltage gain of 50, input resistance of 1 kW and output resistance of 250 W are cascaded. The opened circuit voltages gain of the combined amplifier is (A) 49 dB (B) 51 dB (C) 98 dB (D) 102 dB An ideal sawtooth voltages waveform of frequency of 500 Hz and amplitude 3 V is generated by charging a capacitor of 2 mF in every cycle. The charging requires (A) Constant voltage source of 3 V for 1 ms

2002 4.98

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in (B) Constant voltage source of 3 V for 2 ms (C) Constant voltage source of 1 mA for 1 ms (D) Constant voltage source of 3 mA for 2 ms 2002 4.95

4.96

4.97

In a negative amplifier using voltage-series (i.e. voltagesampling, series mixing) . (A) Ri decreases and R0 decreases (B) Ri decreases and R0 increases (C) Ri increases and R0 decreases (D) Ri increases and R0 increases (Ri and R0 denote the input and output resistance respectively)

The circuit in the figure employs positive and is intended to generate sinusoidal oscillation. If at a frequency V (f) 1 = +0c, then to sustain oscillation at this frequency f0, B (f) = 3 f V0 (f) 6

(A) R2 = 5R1 (C) R2 = R1 6

ONE MARK 4.99

4.100

A 741-type opamp has a gain-bandwidth product of 1 MHz. A noninverting amplifier suing this opamp and having a voltage gain of 20 dB will exhibit a -3 dB bandwidth of (A) 50 kHz (B) 100 kHz (D) 1000 kHz (C) 1000 kHz 7.07 17 Three identical RC-coupled transistor amplifiers are cascaded. If each of the amplifiers has a frequency response as shown in the figure, the overall frequency response is as given in

TWO MARKS

(B) R2 = 6R1 (D) R2 = R1 5

An amplifier using an opamp with a slew-rate SR = 1 V/m sec has a gain of 40 dB. If this amplifier has to faithfully amplify sinusoidal signals from dc to 20 kHz without introducing any slew-rate induced distortion, then the input signal level must not exceed. (A) 795 mV (B) 395 mV (C) 79.5 mV (D) 39.5 mV A zener diode regulator in the figure is to be designed to meet the specifications: IL = 10 mA V0 = 10 V and Vin varies from 30 V to 50 V. The zener diode has Vz = 10 V and Izk (knee current) =1 mA. For

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 satisfactory operation

(A) R # 1800W (C) 3700W # R # 4000W 4.101

(B) 2000W # R # 2200W (D) R $ 4000W

The voltage gain Av = v0 of the JFET amplifier shown in the figure vt is IDSS = 10 mA Vp =- 5 V(Assume C1, C2 and Cs to be very large

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 77

(D) Pz = 115 mW, PT = 11.9 W 4.107

(A) +16 (C) +8

(B) -16 (D) -6

4

2001 4.102

The current gain of a BJT is (A) gm r0 (C) gm rp

4.103

4.104

4.106

(A) Hartely oscillator with foscillation = 79.6 MHz (B) Colpitts oscillator with foscillation = 50.3 MHz (C) Hartley oscillator with foscillation = 159.2 MHz (D) Colpitts oscillator with foscillation = 159.3 MHz

ONE MARK

gm r g (D) m rp

(B)

4.108

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Consider the following two statements : Statement 1 : A stable multi vibrator can be used for generating square wave. Statement 2: Bistable multi vibrator can be used for storing binary information. (A) Only statement 1 is correct (B) Only statement 2 is correct (C) Both the statements 1 and 2 are correct (D) Both the statements 1 and 2 are incorrect

of 100.

TWO MARKS -14

-13

An npn BJT has gm = 38 mA/V, C m = 10 F, C p = 4 # 10 and DC current gain b0 = 90 . For this transistor fT and fb are (A) fT = 1.64 # 108 Hz and fb = 1.47 # 1010 Hz (B) fT = 1.47 # 1010 Hz and fb = 1.64 # 108 Hz (C) fT = 1.33 # 1012 Hz and fb = 1.47 # 1010 Hz (D) fT = 1.47 # 1010 Hz and fb = 1.33 # 1012 Hz

The inverting OP-AMP shown in the figure has an open-loop gain

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Thee ideal OP-AMP has the following characteristics. (B) Ri = 0, A = 3, R0 = 0 (A) Ri = 3, A = 3, R0 = 0 (C) Ri = 3, A = 3, R0 = 3 (D) Ri = 0, A = 3, R0 = 3

2001 4.105

The oscillator circuit shown in the figure is

The closed-loop gain V0 is Vs (A) - 8 (C) - 10

F,

4.109

The transistor shunt regulator shown in the figure has a regulated output voltage of 10 V, when the input varies from 20 V to 30 V. The relevant parameters for the zener diode and the transistor are : Vz = 9.5 , VBE = 0.3 V, b = 99 , Neglect the current through RB . Then the maximum power dissipated in the zener diode (Pz ) and the transistor (PT ) are

In the figure assume the OP-AMPs to be ideal. The output v0 of the circuit is

(A) 10 cos (100t) (C) 10 - 4 2000 4.110

(A) Pz = 75 mW, PT = 7.9 W (B) Pz = 85 mW, PT = 8.9 W (C) Pz = 95 mW, PT = 9.9 W

(B) - 9 (D) - 11

t

#0 cos (100t) dt

(B) 10

t

#0 cos (100t) dt

(D) 10 - 4 d cos (100t) dt ONE MARK

In the differential amplifier of the figure, if the source resistance of the current source IEE is infinite, then the common-mode gain is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) zero (C) indeterminate 4.111

4.116

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4.112

4.113

4.114

(A) precision integrator (B) Hartely oscillator (C) Butterworth high filter (D) Wien-bridge oscillator

(B) infinite (D) Vin1 + Vin2 2VT

In the circuit of the figure, V0 is

(A) -1 V (C) +1 V

Page 78

(A) square wave (C) parabolic wave 4.117

(B) 2 V (D) +15 V

Introducing a resistor in the emitter of a common amplifier stabilizes the dc operating point against variations in (A) only the temperature (B) only the b of the transistor (C) both temperature and b (D) none of the above The current gain of a bipolar transistor drops at high frequencies because of (A) transistor capacitances (B) high current effects in the base (C) parasitic inductive elements (D) the Early effect

Assume that the op-amp of the figure is ideal. If vi is a triangular wave, then v0 will be

The most commonly used amplifier is sample and hold circuits is (A) a unity gain inverting amplifier (B) a unity gain non-inverting amplifier (C) an inverting amplifier with a gain of 10 (D) an inverting amplifier with a gain of 100 2000

4.118

(B) triangular wave (D) sine wave

TWO MARKS

In the circuit of figure, assume that the transistor is in the active region. It has a large b and its base-emitter voltage is 0.7 V. The value of Ic is

If the op-amp in the figure, is ideal, then v0 is

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 (A) zero (C) - (V1 + V2) sin wt 4.115

(B) (V1 - V2) sin wt (D) (V1 + V2) sin wt

The configuration of the figure is a

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 79 4.125

A dc power supply has a no-load voltage of 30 V, and a full-load voltage of 25 V at a full-load current of 1 A. Its output resistance and load regulation, respectively, are (A) 5 W and 20% (B) 25 W and 20% (C) 5 W and 16.7% (D) 25 W and 16.7% 1998

4.126

(A) Indeterminate since Rc is not given (B) 1 mA (C) 5 mA (D) 10 mA 4.119

ONE MARK

The circuit of the figure is an example of of the following type

If the op-amp in the figure has an input offset voltage of 5 mV and an open-loop voltage gain of 10000, then v0 will be (A) current series (C) voltage series (A) 0 V (C) + 15 V or -15 V

4.120

4.121

4.122

4.124

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ONE MARK

The first dominant pole encountered in the frequency response of a compensated op-amp is approximately at (A) 5 Hz (B) 10 kHz (C) 1 MHz (D) 100 MHz Negative in an amplifier (A) reduces gain (B) increases frequency and phase distortions (C) reduces bandwidth (D) increases noise

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4.128

In the cascade amplifier shown in the given figure, if the commonemitter stage (Q1) has a transconductance gm1 , and the common base stage (Q2) has a transconductance gm2 , then the overall transconductance g (= i 0 /vi) of the cascade amplifier is

(A) gm1 g (C) m1 2 4.123

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(B) 5 mV (D) +50 V or -50 V

1999

(B) current shunt (D) voltage shunt

(B) gm2 g (D) m2 2

4.129

In a differential amplifier, CMRR can be improved by using an increased (A) emitter resistance (B) collector resistance (C) power supply voltages (D) source resistance From a measurement of the rise time of the output pulse of an amplifier whose is a small amplitude square wave, one can estimate the following parameter of the amplifier (A) gain-bandwidth product (B) slow rate (C) upper 3–dB frequency (D) lower 3–dB frequency The emitter coupled pair of BJT’s given a linear transfer relation between the differential output voltage and the differential output voltage and the differential input voltage Vid is less a times the thermal voltage, where a is (A) 4 (B) 3 (C) 2 (D) 1

Crossover distortion behavior is characteristic of (A) Class A output stage (B) Class B output stage (C) Class AB output stage (D) Common-base output stage

In a shunt-shunt negative amplifier, as compared to the basic amplifier (A) both, input and output impedances,decrease (B) input impedance decreases but output impedance increases (C) input impedance increase but output (D) both input and output impedances increases.

1999

1998

4.130

TWO MARK

An amplifier has an open-loop gain of 100, an input impedance of 1 kW,and an output impedance of 100 W. A network with a factor of 0.99 is connected to the amplifier in a voltage series mode. The new input and output impedances, respectively, are (A) 10 W and 1W (B) 10 W and 10 kW (C) 100 kW and 1 W (D) 100 kW and 1 kW

4.131

TWO MARKS

A multistage amplifier has a low- response with three real poles at s =- w1 - w2 and w3 . The approximate overall bandwidth B of the amplifier will be given by (B) 1 = 1 + 1 + 1 (A) B = w1 + w2 + w3 w1 w2 w3 B (C) B = (w1 + w2 + w3) 1/3

(D) B =

w12 + w22 + w23

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 4.132

4.133

4.134

One input terminal of high gain ground and a sinusoidal voltage output of comparator will be (A) a sinusoid (C) a half rectified sinusoid

Page 80

comparator circuit is connected to is applied to the other input. The (B) a full rectified sinusoid (D) a square wave

In a series regulated power supply circuit, the voltage gain Av of the ‘’ transistor satisfies the condition (B) 1 << Av < 3 (A) Av " 3 (C) Av . 1 (D) Av << 1 For full wave rectification, a four diode bridge rectifier is claimed to have the following advantages over a two diode circuit : (A) less expensive transformer, (B) smaller size transformer, and (C) suitability for higher voltage application. Of these, (A) only (1) and (2) are true (B) only (1) and (3) are true

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(A) decrease the voltage gain and decrease the input impedance (B) increase the voltage gain and decrease the input impedance (C) decrease the voltage gain and increase the input impedance (D) increase the voltage gain and increase the input impedance 4.138

4.139

(C) only (2) and (3) are true (D) (1), (2) as well as (3) are true 4.135

4.136

4.137

In a common emitter BJT amplifier, the maximum usable supply voltage is limited by (A) Avalanche breakdown of Base-Emitter junction (B) Collector-Base breakdown voltage with emitter open (BVCBO) (C) Collector-Emitter breakdown voltage with base open (BVCBO) (D) Zener breakdown voltage of the Emitter-Base junction 1997

In the MOSFET amplifier of the figure is the signal output V1 and V2 obey the relationship

(A) V1 = V2 2

(B) V1 =-V2 2

(C) V1 = 2V2

(D) V1 =- 2V2

4.140

4.141

ONE MARK

In the BJT amplifier shown in the figure is the transistor is based in the forward active region. Putting a capacitor across RE will

TWO MARKS

In the circuit of in the figure is the current iD through the ideal diode (zero cut in voltage and forward resistance) equals

(A) 0 A (C) 1 A

For small signal ac operation, a practical forward biased diode can be modelled as (A) a resistance and a capacitance in series (B) an ideal diode and resistance in parallel (C) a resistance and an ideal diode in series (D) a resistance 1997

A cascade amplifier stags is equivalent to (A) a common emitter stage followed by a common base stage (B) a common base stage followed by an emitter follower (C) an emitter follower stage followed by a common base stage (D) a common base stage followed by a common emitter stage

(B) 4 A (D) None of the above

The output voltage V0 of the circuit shown in the figure is

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) - 4 V (C) 5 V 4.142

A half wave rectifier uses a diode voltage is Vm sin wt and the load is given by (A) Vm 2 RL 2 V m (C) p

(B) 6 V (D) - 5.5 V with a forward resistance Rf . The resistance is RL . The DC current Vm p (R f + RL) (D) Vm RL (B)

1996 4.143

ONE MARK

4.147

(B) 0.5 gm1 (D) 0.5 gm2

Value of R in the oscillator circuit shown in the given figure, so chosen that it just oscillates at an angular frequency of w. The value of w and the required value of R will respectively be

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(B) 0.4 A (D) 0.4 mamp p

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The circuit shown in the figure is that of

(A) a non-inverting amplifier (C) an oscillator

(B) an inverting amplifier (D) a Schmitt trigger

1996 4.145

(A) gm1 (C) gm2

In the circuit of the given figure, assume that the diodes are ideal and the meter is an average indicating ammeter. The ammeter will read

(A) 0.4 2 A (C) 0.8 A p 4.144

Page 81

TWO MARKS

In the circuit shown in the given figure N is a finite gain amplifier with a gain of k , a very large input impedance, and a very low output impedance. The input impedance of the amplifier with the impedance Z connected as shown will be (A) 105 rad/ sec, 2 # 10 4 W (C) 2 # 10 4 rad/ sec, 105 W 4.148

(A) Z b1 - 1 l k Z (C) (k - 1) 4.146

(B) Z (1 - k) (D) Z (1 - k)

(B) 2 # 10 4 rad/ sec, 2 # 10 4 W (D) 105 rad/ sec, 105 W

A zener diode in the circuit shown in the figure is has a knee current of 5 mA, and a maximum allowed power dissipation of 300 mW . What are the minimum and maximum load currents that can be drawn safely from the circuit, keeping the output voltage V0 constant at 6 V?

A Darlington stage is shown in the figure. If the transconductance of c Q1 is gm1 and Q2 is gm2 , then the overall transconductance gmc ;T i cc E vbe is given by

(A) 0 mA, 180 mA (C) 10 mA, 55 mA

(B) 5 mA, 110 mA (D) 60 mA, 180 mA

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 82

***********

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 83

SOLUTIONS 4.1

(1) Since, voltage across zener diode is 5 V so, current through 100 W resistor is obtained as Is = 10 - 5 = 0.05 A 100 Therefore, the load current is given by IL = 5 RL Since, for proper operation, we must have IZ $ Iknes So, from Eq. (1), we write 0.05 A - 5 $ 10 mA RL 50 mA - 5 $ 10 mA RL 40 mA $ 5 RL -3 40 # 10 $ 5 RL

Option (B) is correct. For the given ideal op-amp, negative terminal will be also ground (at zero voltage) and so, the collector terminal of the BJT will be at zero voltage. i.e., VC = 0 volt The current in 1 kW resistor is given by I = 5 - 0 = 5 mA 1 kW This current will flow completely through the BJT since, no current will flow into the ideal op-amp ( I/P resistance of ideal opamp is infinity). So, for BJT we have VC = 0 VB = 0 IC = 5 mA i.e.,the base collector junction is reverse biased (zero voltage) therefore, the collector current (IC ) can have a value only if baseemitter is forward biased. Hence, VBE = 0.7 volts & VB - VE = 0.7 & 0 - Vout = 0.7 or, Vout =- 0.7 volt

4.2

4.3

Option (A) is correct. The i/p voltage of the system is given as Vin = V1 + Vf = V1 + k Vout = V1 + k A 0 V1 ^Vout = A 0 V1h = V1 ^1 + k A 0h Therefore, if k is increased then input voltage is also increased so, the input impedance increases. Now, we have Vout = A 0 V1 Vin = A0 ^1 + k A 0h A 0 Vin = ^1 + k A 0h Since, Vin is independent of k when seen from output mode, the output voltage decreases with increase in k that leads to the decrease of output impedance. Thus, input impedance increases and output impedance decreases.

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Click to Buy www.nodia.co.in RL 1 -3 # 5 40 # 10 5 # RL 40 # 10-3 or, 125 W # RL Therefore, minimum value of RL = 125 W Now, we know that power rating of Zener diode is given by PR = VZ IZ^maxh IZ^maxh is maximum current through zener diode in reverse bias. Maximum currrent through zener diode flows when load current is zero. i.e., IZ^maxh = Is = 10 - 5 = 0.05 100 Therefore, PR = 5 # 0.05 W = 250 mW 4.4

Option (A) is correct. For the given circuit, we obtain the small signal model as shown in figure below :

Option (B) is correct.

We obtain the node voltage at V1 as V1 + V1 + gm Vi = 0 RD R + 1 L sC & From the circuit, we have or,

Is = IZ + I L IZ = Is - I L

V1 =

- gm Vi 1 + 1 RD R + 1 L sC

Therefore, the output voltage V0 is obtained as

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

V0 = V1 RL RL + 1 sC =

RL RL + 1 sC

4.6

- gm Vi J K 1 1 K RD + K RL + 1 sC L

so, the transfer function is V0 = - RD RL sCgm Vi 1 + sC ^RD + RL h 1 Then, we have the pole at w = C ^RD + RL h It gives the lower cutoff frequency of transfer function. 1 i.e., w0 = C ^RD + RL h 1 or, f0 = 2pC ^RD + RL h 1 = 2p # 10-6 # 20 # 103 = 7.97 . 8 Hz 4.5

Page 84

N O O O P

Logic 0 means voltage is v = 0 volt and logic 1 means voltage is 5 volt For x = 0 , y = 0 , Transistor is at cut off mode and diode is forward biased. Since, there is no drop across forward biased diode. So, Z =Y=0 For x = 0 , y = 1, Again Transistor is in cutoff mode, and diode is forward biased. with no current flowing through resistor. So, Z =Y=1 For x = 1, y = 0 , Transistor is in saturation mode and so, z directly connected to ground irrespective of any value of Y . i.e., Z = 0 (ground) Similarly for X = Y = 1 Z = 0 (ground) Hence, from the obtained truth table, we get Z =XY

Option (C) is correct.

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Option (B) is correct. For the given circuit, we can make the truth table as below X Y Z 0 0 0 0 1 1 1 0 0 1 1 0

4.7

Option (D) is correct. Given, the input voltage VYZ = 100 sin wt

For + ve half cycle VYZ > 0 i.e., VY is a higher voltage than VZ So, the diode will be in cutoff region. Therefore, there will no voltage difference between X and W node. i.e., VWX = 0 Now, for - ve half cycle all the four diodes will active and so, X and W terminal is short circuited For the given ideal op-Amps we can assume V 2- = V 2+ = V2 (ideal) V 1+ = V 1- = V1 (ideal) So, by voltage division V1 = Vout # 1 2 Vout = 2V1 and, as the I/P current in Op-amp is always zero therefore, there will be no voltage drop across 1 KW in II op-amp i.e., V2 = 1 V Therefore, V1 - V2 = V2 - ^- 2h 1 1 & V1 - 1 = 1 + 2 or, V1 = 4 Hence, Vout = 2V1 = 8 volt

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 i.e., Hence, 4.8

VWX = 0 VWX = 0 for all t

Option (C) is correct. The equivalent circuit can be shown as

VTh = VCC

R2 R1 + R 2

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 85

3R2 R1 + R 2 and RTh = R2 R1 R 2 + R1 Since, IC = bIB has b . 3 (very high) so, IB is negative in comparison to IC . Therefore, we can write the base voltage VB = VTh So, VTh - 0.7 - IC RE = 0 =

or, or, or, or, Hence, 4.9

4.10

3R2 - 0.7 - 10-3 500 = 0 ^ h^ h R1 + R 2 3R2 = 0.7 + 0.5 60 kW + R2

The peak rectifier adds + 1 V to peak voltage, so overall peak voltage lowers down by - 1 volt. So, vo = cos wt - 1 4.12

3R2 = ^60 kWh^1.2h + 1.2R2

Option (A) is correct. We put a test source between terminal 1, 2 to obtain equivalent impedance

1.8R2 = ^60 kWh # ^1.2h R2 = 60 # 1.2 = 40 kW 1.8

Option (C) is correct. Given ib = 1 + 0.1 cos (1000pt) mA So, IB = DC component of ib = 1 mA In small signal model of the transistor bVT VT " Thermal voltage rp = IC IC = I = VT = VT B I b IC /b B = VT IB So, rp = 25 mV = 25 W VT = 25 mV, IB = 1 mA 1 mA Option (D) is correct. Let v > 0.7 V and diode is forward biased. By applying Kirchoff’s voltage law

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10 - i # 1k - v = 0 10 - :v - 0.7 D (1000) - v = 0 500

So, 4.11

10 - (v - 0.7) # 2 - v = 0 10 - 3v + 1.4 = 0 v = 11.4 = 3.8 V > 0.7 3 i = v - 0.7 = 3.8 - 0.7 = 6.2 mA 500 500

ZTh = Vtest Itest Applying KCL at top right node Vtest + Vtest - 99I = I test b 9 k + 1k 100 Vtest + Vtest - 99I = I test b 10 k 100 ...(i)

(Assumption is true)

Option (A) is correct. The circuit composed of a clamper and a peak rectifier as shown.

But

Clamper clamps the voltage to zero voltage, as shown 4.13

Ib =- Vtest =-Vtest 9k + 1k 10k

Substituting Ib into equation (i), we have Vtest + Vtest + 99Vtest = I test 10 k 100 10 k 100Vtest + Vtest = I test 10 # 103 100 2Vtest = I test 100 ZTh = Vtest = 50 W Itest Option (B) is correct. First we obtain the transfer function.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 86

IC - IE = 13.7 - VC = (b + 1) IB 12k 13.7 - VC = 100I ...(ii) B 12 # 103 Solving equation (i) and (ii), IB = 0.01 mA Small Signal Analysis : Transforming given input voltage source into equivalent current source.

0 - Vi (jw) 0 - Vo (jw) + =0 1 +R R2 1 jwC Vo (jw) - Vi (jw) = 1 +R R2 1 j wC Vi (jw) R2 R1 - j 1 wC 1 " 3, so V = 0 o wC

Vo (jw) =At w " 0 (Low frequencies),

This is a shunt-shunt amplifier. Given parameters, rp = VT = 25 mV = 2.5 kW IB 0.01 mA b 100 = 0.04 s gm = = rp 2.5 # 1000 Writing KCL at output node v0 + g v + v0 - vp = 0 m p RC RF v 0 : 1 + 1 D + v p :gm - 1 D = 0 RC RF RF Substituting RC = 12 kW, RF = 100 kW, gm = 0.04 s v 0 (9.33 # 10-5) + v p (0.04) = 0 v 0 =- 428.72Vp ...(i) Writing KCL at input node vi = v p + v p + v p - vo Rs Rs rp RF vi = v 1 + 1 + 1 - v 0 p: Rs Rs rp RF D RF vi = v (5.1 10-4) - v 0 # p Rs RF Substituting Vp from equation (i)

At w " 3 (higher frequencies) 1 " 0, so V (jw) =- R2 V (jw) o R1 i wC

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in The filter es high frequencies so it is a high filter. H (jw) = Vo = - R2 Vi R1 - j 1 wC R R 2 2 = H (3) = R1 R1 At 3 dB frequency, gain will be

6H (3)@

2 times of maximum gain

H ^ jw0h = 1 H (3) 2 R2 R2 1 = b l 2 R1 R 12 + 21 2 w0 C

So,

2R 12 = R 12 +

vi = - 5.1 # 10-4 v - v 0 0 428.72 Rs RF

1 w C2 2 0

vi =- 1.16 # 10-6 v 0 - 1 # 10-5 v 0 10 # 103

R = 21 2 w C w0 = 1 R1 C 2 1

4.14

Option (D) is correct. DC Analysis :

Rs = 10 kW

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 (source resistance) vi =- 1.116 # 10-5 10 # 103 1 - 8.96 Av = v 0 = 3 vi 10 # 10 # 1.116 # 10-5 4.15

Using KVL in input loop, VC - 100IB - 0.7 = 0 VC = 100IB + 0.7

...(i)

Option (A) is correct. For the parallel RLC circuit resonance frequency is, 1 wr = 1 = = 10 M rad/s -6 LC 10 # 10 # 1 # 10-9 Thus given frequency is resonance frequency and parallel RLC circuit has maximum impedance at resonance frequency

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 87

Gain of the amplifier is gm # (ZC RL) where ZC is impedance of parallel RLC circuit. At w = wr , ZC = R = 2 kW = ZC max . Hence at this frequency (wr ), gain is Gain w = w = gm (ZC RL) = gm (2k 2k) = gm # 103 which is maximum. Therefore gain is maximum at wr = 10 M rad/ sec . r

4.16

In active region, for common emitter amplifier, Substituting ICO 4.19

Option (D) is correct. The given circuit is shown below :

...(1) IC = bIB + (1 + b) ICO = 0.6 mA and IB = 20 mA in above eq we have, IC = 1.01 mA

Option (C) is correct. In active region VBEon = 0.7 V Emitter voltage VE = VB - VBEon =- 5.7 V V - (- 10) - 5.7 - (- 10) Emitter Current = = 1 mA IE = E 4.3k 4.3k Now IC . IE = 1 mA Applying KCL at collector i1 = 0.5 mA i1 = C dVC dt VC = 1 # i1 dt = i1 t C C

Since or From diagram we can write Ii = Vo + Vo R1 sL1

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Transfer function

or

H (s) = Vo = sR1 L1 I1 R1 + sL1 jwR1 L1 H (jw) = R 1 + j wL 1

At w = 0 At w = 3 4.17

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H (jw) = 0 H (jw) = R1 = constant .

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Option (C) is correct. Given circuit is shown below.

with time, the capacitor charges and voltage across collector changes from 0 towards negative. When saturation starts, VCE = 0.7 & VC =+ 5 V (across capacitor) Thus from (1) we get, + 5 = 0.5 mA T 5 mA

For transistor M2 , VGS = VG - VS = Vx - 0 = Vx VDS = VD - VS = Vx - 0 = Vx Since VGS - VT = Vx - 1 < VDS , thus M2 is in saturation. By assuming M1 to be in saturation we have IDS (M ) = IDS (M ) mn C 0x m C (4) (5 - Vx - 1) 2 = n 0x 1 (Vx - 1) 2 2 2 1

or Taking positive root,

At Vx = 3 V for M1,VGS is true and Vx = 3 V . 4.18

...(1)

Option (D) is correct. We have Now

2

or 4.20

4 (4 - Vx ) 2 = (Vx - 1) 2 2 (4 - Vx ) = ! (Vx - 1) 4.21

8 - 2Vx = Vx - 1 Vx = 3 V = 5 - 3 = 2 V < VDS . Thus our assumption

a = 0.98 b = a = 4.9 1-a

-6 T = 5 # 5 # 10 = 50 m sec -3 0.5 # 10

Option (A) is correct. The current flows in the circuit if all the diodes are forward biased. In forward biased there will be 0.7 V drop across each diode. 12.7 - 4 (0.7) Thus = 1 mA IDC = 9900 Option (B) is correct. The forward resistance of each diode is r = VT = 25 mV = 25 W IC 1 mA 4 (r) Thus Vac = Vi # e 4 (r) + 9900 o = 100 mV cos (wt) 0.01 = 1 cos (wt) mV

4.22

Option (A) is correct.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

The equivalent circuit of given amplifier circuit (when CE is connected, RE is short-circuited)

Page 88 4.24

Option (A) is correct. The circuit is as shown below :

Input impedance Ri = RB || r p Voltage gain AV = gm RC Now, if CE is disconnected, resistance RE appears in the circuit So, or 4.25

Input impedance

Input resistance seen by source vs R in = vs = Rs + Rs || rs is = (1000 W) + (93 kW || 259 W) = 1258 W 4.26

Input impedance increases

4.23

Option (B) is correct. By small signal equivalent circuit analysis

R in = RB || [rp + (b + 1)] RE

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in AV =

Voltage gain

0 - Vi + 0 - Vo = 0 R1 R2 Vo =- R2 R1 Vi

gm RC 1 + gm R E

fo =

Voltage gain decreases.

2

1

1 = 271 Hz 2 # 3.14 # 1250 # 4.7 # 10-6 Lower cut-off frequency f fL . o = 271 = 27.1 Hz 10 10 =

2

The circuit is as shown below : 4.27

VB =- 10 - (- 0.7) =- 9.3 V Collector current I1 =

0 - (- 9.3) = 1 mA (9.3 kW)

b 1 = 700 (high), So IC . IE Applying KCL at base we have 1 - IE = IB + IB 1 - (b 1 + 1) IB = IB + IB 1

1

2

1

2

1 = (700 + 1 + 1)

IB + IB 2

1

2

2

IB . 2 702 2

I 0 = IC = b 2 : IB = 715 # 2 . 2 mA 702 2

2

1 2p (RC + RL) C2

fo

Option (B) is correct. Since, emitter area of transistor Q1 is half of transistor Q2 , so current IE = 1 IE and IB = 1 IB 2 2 1

Option (B) is correct. Cut-off frequency due to C2

Option (B) is correct. The circuit is as shown below

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 89

I = 20 - 0 + Vi - 0 = 5 + Vi 4R R R If I > 0, diode D2 conducts So, for 5 + VI > 0 & VI > - 5, D2 conducts 2 Equivalent circuit is shown below

IC = 10 - 5 = 1 mA 5k

Thus

Current

IE = IC VE = IE RE = 1m # 1.4k = 1.4V = 0.6 + 1.4 = 2V Thus the is negative and output voltage is V = 2V . 4.33

Option (D) is correct. The output voltage is V0 = Ar Vi .-

hfe RC Vi hie

Here RC = 3 W and hie = 3 kW V0 . - 150 # 3k Vi 3k

Thus Output is Vo = 0 . If I < 0 , diode D2 will be off 5 + VI < 0 & V < - 5, D is off I 2 R The circuit is shown below

.- 150 (A cos 20t + B sin 106 t) Since coupling capacitor is large so low frequency signal will be filtered out, and best approximation is V0 .- 150B sin 106 t 4.34

0 - Vi + 0 - 20 + 0 - Vo = 0 R 4R R

4.28

4.29

4.30

4.31 4.32

or

Vo =- Vi - 5

At Vi =- 5 V, At Vi =- 10 V,

Vo = 0 Vo = 5 V

Option (A) is correct. Let diode be OFF. In this case 1 A current will flow in resistor and voltage across resistor will be V = 1.V Diode is off, it must be in reverse biased, therefore Vi - 1 > 0 " Vi > 1 Thus for Vi > 1 diode is off and V = 1V Option (B) and (C) doesn’t satisfy this condition. Let Vi < 1. In this case diode will be on and voltage across diode will be zero and V = Vi Thus V = min (Vi, 1)

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Click to Buy www.nodia.co.in For the positive half of Vi , the diode D1 is forward bias, D2 is reverse bias and the zener diode is in breakdown state because Vi > 6.8 . Thus output voltage is V0 = 0.7 + 6.8 = 7.5 V For the negative half of Vi, D2 is forward bias thus Then V0 =- 0.7 V 4.35

Option (C) is correct.

Option (B) is correct. By Current mirror,

^Lh Ix = W 2 Ibias ^ L h1 Since MOSFETs are identical, W W Thus b L l =b L l 2 2 W

Option (A) is correct. The R2 decide only the frequency. Option (D) is correct. For small increase in VG beyond 1 V the n - channel MOSFET goes into saturation as VGS "+ ive and p - MOSFET is always in active region or triode region.

Option (C) is correct.

Hence 4.36

Ix = Ibias

Option (B) is correct. The circuit is using ideal OPAMP. The non inverting terminal of OPAMP is at ground, thus inverting terminal is also at virtual ground.

Option (D) is correct. The circuit is shown in fig below

Thus current will flow from -ive terminal (0 Volt) to -1 Volt source. Thus the current I is 0 - (- 1) I = = 1 100k 100k The voltage at non inverting terminal is 5 V because OP AMP is ideal and inverting terminal is at 5 V.

The current through diode is I = I 0 _eV - 1i V

t

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 90

Now VT = 25 mV and I0 = 1 mA Thus or Now V 4.37

V I = 10-6 8e 25 # 10 - 1B = 1 5 10 V = 0.06 V V0 = I # 4k + V = 1 # 4k + 0.06 = 0.1 100k -3

The Thevenin resistance and voltage are VTH = 10 # 9 = 3 V 10 + 20 and total RTH = 10k # 20k = 6.67 kW 10k + 20k

Option (B) is correct. The circuit is using ideal OPAMP. The non inverting terminal of OPAMP is at ground, thus inverting terminal is also at virtual ground.

Since b is very large, therefore IB is small and can be ignored Thus IE = VTH - VBE = 3 - 0.7 = 1 mA RE 2.3k 4.41

Option (D) is correct. The small signal model is shown in fig below

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gm =

Vo =- gm Vp # (3k 3k ) =- 1 Vin (1.5k) 25

Thus we can write vi = R1 + sL or

-v

R2 sR2 C2 + 1

v0 =R2 vi (R1 + sL)( sR2 C2 + 1)

or 4.42

and from this equation it may be easily seen that this is the standard form of T.F. of low filter K H (s) = (R1 + sL)( sR2 C2 + 1) and form this equation it may be easily seen that this is the standard form of T.F. of low filter H (s) = 2 K as + bs + b 4.38

4.39

4.43

Option ( ) is correct. The current in both transistor are equal. Thus gm is decide by M1. Hence (C) is correct option. Option (C) is correct. Let the voltage at non inverting terminal be V1, then after applying KCL at non inverting terminal side we have 15 - V1 + V0 - V1 = V1 - (- 15) 10 10 10 or V1 = V0 3

IC = 1m = 1 A/V VT 25m 25

Vp = Vin

=- 60Vin Am = Vo =- 60 Vin

Option (C) is correct. The circuit shown in (C) is correct full wave rectifier circuit.

Option (A) is correct. In the transconductance amplifier it is desirable to have large input resistance and large output resistance.

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Option (C) is correct. We redraw the circuit as shown in fig.

If V0 swings from -15 to +15 V then V1 swings between -5 V to +5 V. 4.40

IC . IE

Option (A) is correct. For the given DC values the Thevenin equivalent circuit is as follows Applying voltage division rule v+ = 0.5 V

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

v+ = vv- = 0.5 V i = 1 - 0.5 = 0.5 mA 1k i = 0.5 - v0 = 0.5 mA 2k

We know that Thus Now and

v0 = 0.5 - 1 =- 0.5 V

or 4.45

Page 91

Option (B) is correct. If we assume b very large, then IB = 0 and IE = IC ; VBE = 0.7 V. We assume that BJT is in active, so applying KVL in Base-emitter loop IE = 2 - VBE = 2 - 0.7 = 1.3 mA RE 1k

The range of current through 200 kW is 3 = 15 mA to 9 = 45 mA 200k 200k The range of variation in output voltage 15m # RZ = 0.15 V to 45m # RZ = 0.45 Thus the range of output voltage is 7.15 Volt to 7.45 Volt 4.49

V+ =

4.46

Option (D) is correct. Here the inverting terminal is at virtual ground and the current in resistor and diode current is equal i.e.

or

Click to Buy www.nodia.co.in V0 = 1 - sRC Vi 1 + sRC 4.50

Option (C) is correct. V0 = H (s) = 1 - sRC Vi 1 + sRC 1 - jwRC H (jw) = 1 + jwRC

For the first condition VD = 0 - Vo1 = VT 1n 4 Is R

+H (jw) = f =- tan - 1 wRC - tan - 1 wRC

Subtracting above equation Vo1 - Vo2 = VT 1n 4 - VT 1n 2 Is R Is R Vo1 - Vo2 = VT 1n 4 = VT 1n2 2

Option (D) is correct. We have Vthp = Vthp = 1 V WP W and = N = 40mA/V2 LP LN From figure it may be easily seen that Vas for each NMOS and PMOS is 2.5 V mA Thus ID = K (Vas - VT ) 2 = 40 2 (2.5 - 1) 2 = 90 m A V

4.48

1 V 1 + sCR i

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VD = 0 - Vo1 = VT 1n 2 Is R

4.47

1 V 1 + sCR i

T

For the first condition

or

Vi =

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Vi = I eV /V s R VD = VT 1n Vi Is R D

R + sC1

Applying voltage division rule (V + Vi) V+ = R1 (V0 + Vi) = o R1 + R1 2 (Vo + Vi) 1 or V = 1 + sCR i 2 Vo =- 1 + 2 or Vi 1 + sRC

IR = ID or

1 sC

V- = V+ =

Now

Since b is very large, we have IE = IC , thus IC = 1.3 mA Now applying KVL in collector-emitter loop 10 - 10IC - VCE - IC = 0 or VCE =- 4.3 V Now VBC = VBE - VCE = 0.7 - (- 4.3) = 5 V Since VBC > 0.7 V, thus transistor in saturation.

Option (A) is correct. The voltage at non-inverting terminal is

Minimum value, Maximum value, 4.51

4.52 4.53

fmin fmax

=- 2 tan - 2 wRC = - p (at w " 3) = 0( at w = 0)

Option (D) is correct. In the transconductance amplifier it is desirable to have large input impedance and large output impedance. Option (C) is correct. Option (D) is correct. The voltage at inverting terminal is V- = V+ = 10 V Here note that current through the capacitor is constant and that is I = V- = 10 = 10 mA 1k 1k

Option (C) is correct. We have VZ = 7 volt, VK = 0, RZ = 10W Circuit can be modeled as shown in fig below

Thus the voltage across capacitor at t = 1 msec is 1m 1m VC = 1 Idt = 1 10mdt C 0 1m 0 Im 4 = 10 dt = 10 V

#

#

#0

4.54

Since Vi is lies between 10 to 16 V, the range of voltage across 200 kW V200 = Vi - VZ = 3 to 9 volt

Option (A) is correct. In forward bias Zener diode works as normal diode. Thus for negative cycle of input Zener diode is forward biased and it conducts giving VR = Vin .

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

For positive cycle of input Zener diode is reversed biased when 0 < Vin < 6 , Diode is OFF and VR = 0 when Vin > 6 Diode conducts and voltage across diode is 6 V. Thus voltage across is resistor is VR = Vin - 6 Only option (B) satisfy this condition. 4.55

Page 92

Now VCE = 15 - 9 = 6 V The power dissipated in transistor is P = VCE IC = 6 # 0.9 = 5.4 W 4.59

Option (C) is correct. The circuit under DC condition is shown in fig below

Option (B) is correct. If the unregulated voltage increase by 20%, them the unregulated voltage is 18 V, but the VZ = Vin = 6 remain same and hence Vout and IC remain same. There will be change in VCE Thus, VCE - 18 - 9 = 9 V IC = 0.9 A Power dissipation P = VCE IC = 9 # 0.9 = 8.1 W Thus % increase in power is 8.1 - 5.4 # 100 = 50% 5.4

4.60

Applying KVL we have VCC - RC (IC + IB) - VCE = 0

...(1)

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4.61

Option (B) is correct. Since the inverting terminal is at virtual ground, the current flowing through the voltage source is Is = Vs 10k Vs = 10 kW = R or in Is Option (D) is correct. The effect of current shunt in an amplifier is to decrease the input resistance and increase the output resistance as : Rif = Ri 1 + Ab Rof = R0 (1 + Ab) Ri " Input resistance without Rif " Input resistance with .

where and VCC - RB IB - VBE = 0 Substituting IC = bIB in (1) we have VCC - RC (bIB + IB) - VCE = 0 Solving (2) and (3) we get VCE = VCC - VCC - VBE RB 1+ RC (1 + b) Now substituting values we get 12 - 0.7 = 5.95 V VCE = 12 53 1+ 1 + (1 + 60) 4.56

...(2) ...(3)

4.62

...(4)

4.63

ACM =- RC 2RE

b' = 110 # 60 = 66 100

And differential mode gain

Substituting b' = 66 with other values in (iv) in previous solutions 12 - 0.7 = 5.29 V VCE = 12 53 1+ 1 + (1 + 66) Thus change is

4.58

Option (D) is correct. Common mode gain

Option (B) is correct. We have

4.57

Option (B) is correct. The CE configuration has high voltage gain as well as high current gain. It performs basic function of amplifications. The CB configuration has lowest Ri and highest Ro . It is used as last step to match a very low impedance source and to drain a high impedance load Thus cascade amplifier is a multistage configuration of CE-CB

=- gm RC

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= 5.29 - 59.5 # 100 =- 4.3% 5.95

Thus only common mode gain depends on RE and for large value of RE it decreases.

Option (A) is correct. Option (C) is correct. The Zener diode is in breakdown region, thus V+ = VZ = 6 V = Vin R We know that Vo = Vin c1 + f m R1 or Vout = Vo = 6`1 + 12k j = 9 V 24k

The current in 12 kW branch is negligible as comparison to 10 W. Thus Current IC . IE . = Vout = 9 = 0.9 A RL 10

4.64

Option (C) is correct.

IE = Is `e nV - 1j VBE

T

= 10 4.65

- 13

0.7

c e1 # 26 # 10 - 1m = 49 mA -3

Option (C) is correct. The circuit is as shown below

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 93

IL = 24.2 - 0.5 = 23.7 mA

or 4.69 4.70

Option (D) is correct. Option (B) is correct. The small signal model is as shown below

Writing equation for I- have e 0 - V- = I 1M or e0 = I- (1M) + VWriting equation for I+ we have 0 - V+ = I+ 1M

...(1)

or V+ = - I+ (1M) Since for ideal OPAMP V+ = V- , from (1) and (2) we have

...(2)

From the figure we have Zin = 2 MW

e0 = I- (1M) - I + (1M) = (I- - I+) (1M) = IOS (1M) Thus if e0 has been measured, we can calculate input offset current IOS only. 4.66

4.67

Option (C) is correct. At low frequency capacitor is open circuit and voltage acr s noninverting terminal is zero. At high frequency capacitor act as short circuit and all input voltage appear at non-inverting terminal. Thus, this is high circuit. The frequency is given by 1 = 1000 w = 1 = 3 RC 1 # 10 # 1 # 10 - 6 rad/sec

Z0 = rd RD = 20k 2k = 20 kW 11

and 4.71

Option (A) is correct. The circuit in DC condition is shown below

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Option (B) is correct. The circuit under DC condition is shown in fig below

Since the FET has high input resistance, gate current can be neglect and we get VGS =- 2 V Since VP < VGS < 0 , FET is operating in active region 2 (- 2) 2 Now ID = IDSS c1 - VGS m = 10 c1 (- 8) m VP = 5.625 mA Now

Applying KVL we have VCC - RB IB - VBE - RE IE = 0 or VCC - RB IB - VBE - RE (b + 1) IB = 0 Since IE = IB + bIB or IB = VCC - VBE RB + (b + 1) RE 20 - 0.7 = = 40m A 430k + (50 + 1)1 k Now 4.68

= 8.75 V 4.72

Option (B) is correct. The transconductance is gm = or, The gain is

IC = bIB = 50 # 40m = 2 mA VC = VCC - RC IC = 20 - 2m # 2k = 16 V

Option (A) is correct. The maximum load current will be at maximum input voltage i.e. Vmax = 30 V i.e. Vmax - VZ = I + I L Z 1k 30 - 5.8 = I = 0.5 m or L 1k

VDS = VDD - ID RD = 20 - 5.625 m # 2 k

So, 4.73

VP

2 ID IDSS

= 2 5.625mA # 10mA = 1.875 mS 8 A =- gm (rd RD) = 1.875ms # 20 K =- 3.41 11

Option (B) is correct. Only one diode will be in ON conditions When lower diode is in ON condition, then Vu = 2k Vsat = 2 10 = 8 V 2.5k 2.5 when upper diode is in ON condition

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 94

Vu = 2k Vsat = 2 (- 10) =- 5 V 2.5k 4 4.74

4.75

or 2V+ - Vo + IL R2 = 0 Since V- = V+ , from (1) and (2) we have

Option (B) is correct. An ideal OPAMP is an ideal voltage controlled voltage source. Option (C) is correct. In voltage series feed back amplifier, input impedance increases by factor (1 + Ab) and output impedance decreases by the factor (1 + Ab).

Vs + IL R2 = 0

4.77

IL =- Vs R2

or 4.80

Rif = Ri (1 + Ab) Ro Rof = (1 + Ab) 4.76

...(2)

Option (D) is correct. If IZ is negligible the load current is 12 - Vz = I L R as per given condition 100 mA # 12 - VZ # 500 mA R At IL = 100 mA 12 - 5 = 100 mA R

Option (A) is correct. This is a Low filter, because V0 = 0 At w = 3 Vin V0 = 1 and at w = 0 Vin

VZ = 5 V

R = 70W

or

At IL = 500 mA 12 - 5 = 500 mA R

VZ = 5 V

or R = 14 W Thus taking minimum we get

Option (D) is correct. When IC >> ICO

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R = 14 W 4.81 4.82

Option (B) is correct. Option (C) is correct. The Thevenin equivalent is shown below

IC = 1mA = 0.04 = 40 mA/V VT 25mV b rp = = 100 - 3 = 2.5 kW gm 40 # 10

gm =

4.78

4.79

Option (A) is correct. The given circuit is wein bridge oscillator. The frequency of oscillation is 2pf = 1 RC 1 or = 1 m C = 1 = 3 3 2pRf 2p 2p # 10 # 10

VT =

R1 V = 1 #5 = 1 V R1 + R2 C 4+1

Since b is large is large, IC . IE , IB . 0 and IE = VT - VBE = 1 - 0.7 = 3 mA RE 300 VCE = 5 - 2.2kIC - 300IE = 5 - 2.2k # 1m - 300 # 1m = 2.5 V

Now

Option (A) is correct. The circuit is as shown below 4.83

Option (B) is correct. For the different combinations the table is as follows

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CE

CC

CB

We know that for ideal OPAMP

Ai

High

High

Unity

V- = V+ Applying KCL at inverting terminal V- - Vs + V- - V0 = 0 R1 R1

Av

High

Unity

High

Ri

Medium

High

Low

Ro

Medium

Low

High

or 2V- - Vo = Vs Applying KCL at non-inverting terminal V+ V - Vo + IL + + =0 R2 R2

...(1)

4.84

4.85

Option (D) is correct. This circuit having two diode and capacitor pair in parallel, works as voltage doubler. Option (B) is correct.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 95

If the input is sinusoidal signal of 8 V (peak to peak) then

or

Vi = 4 sin wt The output of comparator will be high when input is higher than Vref = 2 V and will be low when input is lower than Vref = 2 V. Thus the waveform for input is shown below

In second case IB2 Thus 4.90

4.91

From fig, first crossover is at wt1 and second crossover is at wt2 where 4 sin wt1 = 2V wt1 = sin - 1 1 = p 2 6 wt2 = p - p = 5p 6 6 5p p -6 Duty Cycle = 6 =1 2p 3

Thus

4.86

Thus the output of comparators has a duty cycle of 1 . 3 Option (C) is correct. CMMR = Ad Ac

R2 = 2kW IB1 = IC1 = 1.5m = 0.01 mA 150 b1 will we equal to IB1 as there is no in R1. IC2 = b2 IB2 = 200 # 0.01 = 2 mA VCE2 = VCC - IC2 R2 = 6 - 2m # 2 kW = 2 V

Option (A) is correct. The given circuit is a R - C phase shift oscillator and frequency of its oscillation is 1 f = 2p 6 RC Option (C) is correct. If we see th figure we find that the voltage at non-inverting terminal is 3 V by the zener diode and voltage at inverting terminal will be 3 V. Thus Vo can be get by applying voltage division rule, i.e. 20 V = 3 20 + 40 o V0 = 9 V

or

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Option (B) is correct. The circuit is as shown below

20 log CMMR = 20 log Ad - 20 log Ac = 48 - 2 = 46 dB Where Ad "Differential Voltage Gain and AC " Common Mode Voltage Gain

or

4.87

Option (B) is correct. The gain of amplifier is Ai =

8 (3) = 8 kW 1+8 3 V+ = V- = 8 V 3

- gm gb + jwC

V+ =

Thus the gain of a transistor amplifier falls at high frequencies due to the internal capacitance that are diffusion capacitance and transition capacitance. 4.88

4.89

Now applying KCL at inverting terminal we get V- - 2 + V- - Vo = 0 1 5

Option (A) is correct. We have Ri = 1kW, b = 0.2, A = 50 Ri Thus, Rif = = 1 kW (1 + Ab) 11 Option (A) is correct. The DC equivalent circuit is shown as below. This is fixed bias circuit operating in active region.

4.93

Option (C) is correct. The equivalent circuit of 3 cascade stage is as shown in fig.

or

1k 50V1 = 40V1 1k + 0.25k 1k V3 = 50V2 = 40V2 1k + 0.25k V2 =

In first case VCC - IC1 R2 - VCE1 = 0 6 - 1.5mR2 - 3 = 0

Vo = 6V- - 10 = 6 # 8 - 10 = 6 V 3

or

Similarly

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

or or or 4.94

Page 96

V3 = 40 # 40V1 Vo = 50V3 = 50 # 40 # 40V1 AV = Vo = 50 # 40 # 40 = 8000 V1

Thus from above equation for sustained oscillation 6 = 1 + R2 R1

20 log AV = 20 log 8000 = 98 dB

Option (D) is correct. If a constant current is made to flow in a capacitor, the output voltage is integration of input current and that is sawtooth waveform as below : t VC = 1 idt C 0 The time period of wave form is T = 1 = 1 = 2 m sec f 500

4.99

20 # 10 1 idt 6 2 # 10 0 or i (2 # 10 - 3 - 0) = 6 # 10 - 6 or i = 3 mA Thus the charging require 3 mA current source for 2 msec.

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Slew Rate or

4.100

-6

Option (A) is correct. The circuit is shown as below

Option (C) is correct. In voltage-amplifier or voltage-series amplifier, the Ri increase and Ro decrease because I For satisfactory operations Vin - V0 R When Vin = 30 V, 30 - 10 R 20 or R

Option (B) is correct. Let x be the gain and it is 20 db, therefore

6

Hz

1 2n

-1

1 23

or Thus R # 1818W

$ 11 mA

R # 3636W

Option (D) is correct. We have

1

2 2 - 1 = 0.5 kHz

Option (A) is correct. As per Barkhousen criterion for sustained oscillations Ab $ 1 and phase shift must be or 2pn .

$ (10 + 1) mA

40 $ 11 # 10 - 3 R

4.101

fHn = fH

[IZ + IL = I]

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-1

The higher cutoff frequency is

> IZ + IL

50 - 10 $ (10 + 1) mA R

Option (A) is correct. In multistage amplifier bandwidth decrease and overall gain increase. From bandwidth point of view only options (A) may be correct because lower cutoff frequency must be increases and higher must be decreases. From following calculation we have We have fL = 20 Hz and fH = 1 kHz For n stage amplifier the lower cutoff frequency is fL 20 f = = = 39.2 . 40 Ln

= IZ + IL

R # 1818 W

or when Vin = 50 V

6

BW = 10 = 10 = 105 Hz = 100 kHz Gain 10

4.98

dVO = AV Vm w = AV Vm 2pf c dt m max Vm = SR AV V2pf

1 10 # 100 # 2p # 20 # 103 or VM = 79.5 mV =

20 log x = 20 or x = 10 Since Gain band width product is 106 Hz, thus So, bandwidth is

4.97

VO = VV Vi = Vm sin wt dVO = A V w cos wt V m dt

Now

Rif = Ri (1 + Ab) Ro Rof = (1 + Ab) 4.96

Option (C) is correct. Let the gain of OPAMP be AV then we have 20 log AV = 40 dB or AV = 100 Let input be Vi = Vm sin wt then we have

-3

#

3=

R2 = 5R1

or

#

Thus

VO (f) = 1 + R2 Vf (f) R1 V (f) b (f) = 1 +0 = f 6 VO (f) A=

Now from circuit

Now and Thus

IDSS VG VS VGS

= 10 mA and VP =- 5 V =0 = ID RS = 1 # 2.5W = 2.5 V = VG - VS = 0 - 2.5 =- 2.5 V

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

gm = 2IDSS 81 - ` - 2.5 jB = 2 mS VP -5 AV = V0 =- gm RD Vi

Now

9 = 1 # 10 = 50.3 MHz 2p 10

Option (C) is correct. The current gain of a BJT is

4.108

hfe = gm rp 4.103

4.105

4.106

Option (D) is correct. The circuit is as shown below

Option (A) is correct. The ideal op-amp has following characteristic :

and 4.104

1 2p LCeq Ceq = C1 C2 = 2 # 2 = 1 pF C1 + C2 4 1 f = 2p 10 # 10 - 6 # 10 - 12 f =

=- 2ms # 3k =- 6

So, 4.102

Page 97

Ri " 3 R0 " 0 A"3

Option (C) is correct. Both statements are correct because (1) A stable multivibrator can be used for generating square wave, because of its characteristic (2) Bi-stable multivibrator can store binary information, and this multivibrator also give help in all digital kind of storing. Option (B) is correct. If fT is the frequency at which the short circuit common emitter gain attains unity magnitude then gm 38 # 10 - 3 = fT = 2p (Cm + ) 2p # (10 - 14 + 4 # 10 - 13) or = 1.47 # 1010 Hz If fB is bandwidth then we have 10 f fB = T = 1.47 # 10 = 1.64 # 108 Hz 90 b

Let V- be the voltage of inverting terminal, since non inverting terminal a at ground, the output voltage is

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Click to Buy www.nodia.co.in Vo = AOL VNow applying KCL at inverting terminal we have V- - Vs + V- - V0 = 0 R1 R2

Option (C) is correct. If we neglect current through RB then it can be open circuit as shown in fig.

4.109

Maximum power will dissipate in Zener diode when current through it is maximum and it will occur at Vin = 30 V I = Vin - Vo = 30 - 10 = 1 A 20 20

or

I IC + IZ = bIB + IZ = bIZ + IZ = (b + 1) IZ IZ = I = 1 = 0.01 A 99 + 1 b+1

...(1) ...(2)

From (1) and (2) we have VO = A = - R2 CL Vs R - R2 + R1 ROL Substituting the values we have - 10k =- 1000 . - 11 ACL = 10 89 k 1 k + 1k 100k Option (A) is correct. The first OPAMP stage is the differentiator and second OPAMP stage is integrator. Thus if input is cosine term, output will be also cosine term. Only option (A) is cosine term. Other are sine term. However we can calculate as follows. The circuit is shown in fig

Since IC = bIB since IB = IZ

Power dissipated in zener diode is PZ = VZ IZ = 9.5 # 0.01 = 95 mW IC = bIZ = 99 # 0.1 = 0.99 A VCE = Vo = 10 V Power dissipated in transistor is PT = VC IC = 10 # 0.99 = 9.9 W 4.107

Option (B) is correct. From the it may be easily seen that the tank circuit is having 2-capacitors and one-inductor, so it is colpits oscillator and frequency is

Applying KCL at inverting terminal of first OP AMP we have V1 = - wjL = - 100 # 10 # 10 - 3 = - 1 R 10 10 VS - jVS or V1 = = j cos 100t 10 Applying KCL at inverting terminal of second OP AMP we have VO = - 1/jwC 100 V1 1 == j10 j100 # 10 # 10 - 6 # 100

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

or 4.110

Page 98

V0 = j10V2 = j10 (- j cos 100t) V0 = 10 cos 100t

Option (A) is correct. Common mode gain is AC = aRC REE Since source resistance of the current source is infinite REE = 3 , common mode gain AC = 0

4.111

4.112

Option (D) is correct. In positive feed back it is working as OP-AMP in saturation region, and the input applied voltage is +ve. So, V0 =+ Vsat = 15 V Option (C) is correct. With the addition of RE the DC abis currents and voltages remain closer to the point where they were set by the circuit when the outside condition such as temperature and transistor parameter b change.

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R1 V = 5 # 15 = 5 V R1 + R2 C 10 + 5 Since b is large is large, IC . IE , IB . 0 and IE = VT - VBE RE 4.3 = 5 - 0.7 = = 10 mA 0.430KW 0.430kW VT =

4.119

4.120 4.121

4.122

Option (A) is correct. At high frequency

and

Option (A) is correct. Option (A) is correct. Negative in amplifier reduces the gain of the system. Option (A) is correct. By drawing small signal equivalent circuit

gm + jw (C) 1 Ai \ Capacitance 1 Ai a frequency Ai =-

or,

Option (C) is correct. The output voltage will be input offset voltage multiplied by open by open loop gain. Thus So V0 = 5mV # 10, 000 = 50 V But V0 = ! 15 V in saturation condition So, it can never be exceeds ! 15 V So, V0 = ! Vset = ! 15V

' gbc

Thus due to the transistor capacitance current gain of a bipolar transistor drops. 4.114

4.115

4.116

4.117

4.118

by applying KCL at E2

Option (C) is correct. As OP-AMP is ideal, the inverting terminal at virtual ground due to ground at non-inverting terminal. Applying KCL at inverting terminal sC (v1 sin wt - 0) + sC (V2 sin wt - 0) + sC (Vo - 0) = 0 or Vo =- (V1 + V2) sin wt

gm1 Vp 1

Vp = gm2 Vp rp 2

2

2

at C2 from eq (1) and (2)

i 0 =- gm2 Vp

2

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Option (D) is correct. There is R - C , series connection in parallel with parallel R - C combination. So, it is a wein bridge oscillator because two resistors R1 and R2 is also in parallel with them.

gm1 Vp +

Option (A) is correct. The given circuit is a differentiator, so the output of triangular wave will be square wave.

1

i 0 =- i 0 gm2 rp 2

gm1 Vp =- i 0 :1 + 1 D gm2 rp 1

2

Option (B) is correct. In sampling and hold circuit the unity gain non-inverting amplifier is used.

gm2 rp = b >> 1 so gm1 Vp =- i 0 i 0 =- g m1 Vp i0 = g a Vp = Vi m1 Vi Option (B) is correct. Crossover behavior is characteristic of calss B output stage. Here 2 2

1

Option (D) is correct. The Thevenin equivalent is shown below

1

1

4.123

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

transistor are operated one for amplifying +ve going portion and other for -ve going portion. 4.124

Option (C) is correct. In Voltage series mode input impedance is given by R in = Ri (1 + bv Av) where bv = factor , Av = openloop gain and Ri = Input impedance So, R in = 1 # 103 (1 + 0.99 # 100) = 100 kW Similarly output impedance is given by R0 ROUT = R 0 = output impedance (1 + bv Av) 100 Thus = 1W ROUT = (1 + 0.99 # 100)

4.125

4.127

4.128

4.133

4.134

4.135

Option (D) is correct. This is a voltage shunt as the samples a portion of output voltage and convert it to current (shunt). Option (A) is correct. In a differential amplifier CMRR is given by (1 + b) IQ R 0 CMRR = 1 ;1 + E 2 VT b So where R 0 is the emitter resistance. So CMRR can be improved by increasing emitter resistance.

Option (C) is correct. In series voltage regulator the transistor is in common collector configuration having voltage gain close to unity. Option (D) is correct. In bridge rectifier we do not need central tap transformer, so its less expensive and smaller in size and its PIV (Peak inverse voltage) is also greater than the two diode circuit, so it is also suitable for higher voltage application. Option (C) is correct. In the circuit we have V2 = IS # RD 2 and

Option (B) is correct. Regulation = Vno - load - Vfuel - load Vfull - load = 30 - 25 # 100 = 20% 25 Output resistance = 25 = 25 W 1

4.126

Page 99

V1 = IS # RD V2 = 1 2 V1 V1 = 2V2

4.136

Option (C) is correct.

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Option (C) is correct. The equivalent circuit of given amplifier circuit (when CE is connected, RE is short-circuited)

Option (C) is correct. We know that rise time (tr ) is tr = 0.35 fH where fH is upper 3 dB frequency. Thus we can obtain upper 3 dB frequency it rise time is known.

4.129

4.130

Input impedance Ri = RB || r p Voltage gain AV = gm RC Now, if CE is disconnected, resistance RE appears in the circuit

Option (D) is correct. In a BJT differential amplifier for a linear response Vid < VT . Option (D) is correct. In a shunt negative amplifier. Input impedance Ri R in = (1 + bA) where

Ri = input impedance of basic amplifier b = factor A = open loop gain

Input impedance R in = RB || [rp + (b + 1)] RE Input impedance increases gm RC Voltage gain Voltage gain decreases. AV = 1 + gm R E

So, R in < Ri Similarly ROUT =

R0 (1 + bA)

ROUT < R 0 Thus input & output impedances decreases. 4.131 4.132

Option (A) is correct. Option (D) is correct. Comparator will give an output either equal to + Vsupply or - Vsupply . So output is a square wave.

4.138

4.139

Option (A) is correct. In common emitter stage input impedance is high, so in cascaded amplifier common emitter stage is followed by common base stage. Option (C) is correct. We know that collect-emitter break down voltage is less than compare to collector base breakdown voltage. BVCEO < BVCBO both avalanche and zener break down. Voltage are higher than

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 100

2Va - 4 + Va - V0 = 0 V0 = 3Va - 4 Va - V0 + Va - 0 = 0 100 10

BVCEO .So BVCEO limits the power supply. 4.140

Option (C) is correct.

If we assume consider the diode in reverse bias then Vn should be greater than VP .

So

VP < Vn by calculating VP = 10 # 4 = 5 Volt 4+4

V0 =- 5.5 Volts 4.142

Vn = 2 # 1 = 2 Volt here VP > Vn (so diode cannot be in reverse bias mode).

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apply node equation at node a Va - 10 + Va + Va = 2 1 4 4

so current

4.141

6Va - 10 = 8 Va = 3 Volt Ib = 0 - 3 + 10 - 3 4 4 6 10 = 1 amp Ib = 4

Option (D) is correct. By applying node equation at terminal (2) and (3) of OP -amp

Va - V0 + 10Va = 0 11Va = V0 Va = V0 11 V0 = 3V0 - 4 11 8V0 =- 4 11

Option (B) is correct. Circuit with diode forward resistance looks

So the DC current will IDC = 4.143

Vm p (R f + RL)

Option (D) is correct. For the positive half cycle of input diode D1 will conduct & D2 will be off. In negative half cycle of input D1 will be off & D2 conduct so output voltage wave from across resistor (10 kW) is –

Ammeter will read rms value of current so I rms = Vm (half wave rectifier) pR 4 = = 0.4 mA p (10 kW) p 4.144

4.145

Option (D) is correct. In given circuit positive is applied in the op-amp., so it works as a Schmitt trigger. Option (D) is correct.

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 Gain with out factor is given by V0 = kVi after connecting impedance Z

Va - Q Va - V0 =0 + 5 10

given input impedance is very large, so after connecting Z we have Ii = Vi - V0 V0 = kVi Z

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Ii = Vi - kVi Z Zin = Vi = Z Ii (1 - k)

input impedance 4.146 4.147

Page 101

Option (A) is correct. Option (A) is correct. For the circuit, In balanced condition It will oscillated at a frequency w 1 = 1 = 105 rad/ sec -3 -6 LC 10 # 10 # .01 # 10 In this condition R1 = R 3 R2 R4 5 =R 100 1 =

R = 20 kW = 2 # 10 4 W 4.148

Option (C) is correct. V0 kept constant at so current in 50 W resistor

V0 = 6 volt I = 9-6 50 W

I = 60 m amp Maximum allowed power dissipation in zener PZ = 300 mW Maximum current allowed in zener PZ = VZ (IZ ) max = 300 # 10-3 & = 6 (IZ ) max = 300 # 10-3 & = (IZ ) max = 50 m amp Given knee current or minimum current in zener In given circuit

(IZ ) min = 5 m amp I = IZ + I L I L = I - IZ (IL) min = I - (IZ ) max = (60 - 50) m amp = 10 m amp (IL) max = I - (IZ ) min = (60 - 5) = 55 m amp

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

UNIT 5

Page 101

2012 5.4

ONE MARK

Consider the given circuit

DIGITAL CIRCUITS

2013 5.1

5.2

A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles (A) and AND gate (B) an OR gate (C) an XOR gate (D) a NAND gate For 8085 microprocessor, the following program is executed. MVI A, 05H; MVI B, 05H; PTR: ADD B; DCR B; JNZ PTR; ADI 03H; HLT; At the end of program, accumulator contains (A) 17H (B) 20H (C) 23H (D) 05H 2013

5.3

ONE MARK

TWO MARKS

In this circuit, the race around (A) does not occur (B) occur when CLK = 0 (C) occur when CLK = 1 and A = B = 1 (D) occur when CLK = 1 and A = B = 0 5.5

The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input A is greater than the 2-bit input B . The number of combinations for which the output is logic 1, is

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There are four chips each of 1024 bytes connected to a 16 bit address bus as shown in the figure below, RAMs 1, 2, 3 and 4 respectively are mappped to addresses

In the circuit shown

(A) Y = A B + C (C) Y = (A + B ) C 5.7

5.8

(B) Y = (A + B) C (D) Y = AB + C

/

In the sum of products function f (X, Y, Z) = (2, 3, 4, 5), the prime implicants are (A) XY, XY (B) XY, X Y Z , XY Z (C) XY Z , XYZ, XY (D) XY Z , XYZ, XY Z , XY Z 2012

(A) 0C00H-0FFFH, 1C00H-1FFFH, 2C00H-2FFFH, 3C00H3FFFH (B) 1800H-1FFFH, 2800H-2FFFH, 3800H-3FFFH, 4800H-4FFFH (C) 0500H-08FFH, 1500H-18FFH, 3500H-38FFH, 5500H-58FFH (D) 0800H-0BFFH, 1800H-1BFFH, 2800H-2BFFH, 3800H-3BFFH

(B) 6 (D) 10

TWO MARKS

In the CMOS circuit shown, electron and hole mobilities are equal, and M1 and M2 are equally sized. The device M1 is in the linear region if

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) Vin < 1.875 V (C) Vin > 3.125 V 5.9

(A) changed from “0” to “1” (C) changed in either direction

(B) 1.875 V < Vin < 3.125 V (D) 0 < Vin < 5 V

The state transition diagram for the logic circuit shown is

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2011 5.10

Page 102

5.12

The logic function implemented by the circuit below is (ground implies a logic “0”)

(A) F = AND ^P, Q h (C) F = XNOR ^P, Q h 2011 5.13

(B) changed from “1” to “0” (D) not changed

(B) F = OR ^P, Q h (D) F = XOR ^P, Q h TWO MARKS

The output of a 3-stage Johnson (twisted ring) counter is fed to a digital-to analog (D/A) converter as shown in the figure below. Assume all states of the counter to be unset initially. The waveform which represents the D/A converter output Vo is

ONE MARK

The output Y in the circuit below is always ‘1’ when

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(A) two or more of the inputs P, Q, R are ‘0’ (B) two or more of the inputs P, Q, R are ‘1’ (C) any odd number of the inputs P, Q, R is ‘0’ (D) any odd number of the inputs P, Q, R is ‘1’ 5.11

When the output Y in the circuit below is “1”, it implies that data has

5.14

Two D flip-flops are connected as a synchronous counter that goes through the following QB QA sequence 00 " 11 " 01 " 10 " 00 " .... The connections to the inputs DA and DB are (A) DA = QB, DB = QA

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

5.15

(B) DA = Q A, DB = Q B (C) DA = (QA Q B + Q A QB), DB = QA (D) DA = (QA QB + Q A Q B), DB = Q B

(A) A = 1, B = 1, C = 0 (C) A = 0, B = 1, C = 0

An 8085 assembly language program is given below. Assume that the carry flag is initially unset. The content of the accumulator after the execution of the program is

2010

(A) 8CH (C) 23H

5.19

(B) 64H (D) 15H

2010 5.16

Page 103

(B) A = 1, B = 0, C = 0 (D) A = 0, B = 0, C = 1 TWO MARKS

Assuming that the flip-flop are in reset condition initially, the count sequence observed at QA , in the circuit shown is

(A) 0010111... (C) 0101111...

(B) 0001011... (D) 0110100....

ONE MARK

Match the logic gates in Column A with their equivalents in Column B

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(A) P-2, Q-4, R-1, S-3 (C) P-2, Q-4, R-3, S-1 5.17

The Boolean function realized by the logic circuit shown is

(B) P-4, Q-2, R-1, S-3 (D) P-4, Q-2, R-3, S-1 (A) F = Sm (0, 1, 3, 5, 9, 10, 14) (C) F = Sm (1, 2, 4, 5, 11, 14, 15)

In the circuit shown, the device connected Y5 can have address in the range 5.21

For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is

(A) 00H (C) 67H (A) 2000 - 20FF (C) 2E00 - 2EFF

(B) 2D00 - 2DFF (D) FD00 - FDFF

2009 5.22

5.18

For the output F to be 1 in the logic circuit shown, the input combination should be

(B) F = Sm (2, 3, 5, 7, 8, 12, 13) (D) F = Sm (2, 3, 5, 7, 8, 9, 12)

(B) 45H (D) E7H ONE MARK

The full form of the abbreviations TTL and CMOS in reference to logic families are (A) Triple Transistor Logic and Chip Metal Oxide Semiconductor (B) Tristate Transistor Logic and Chip Metal Oxide Semiconductor (C) Transistor Transistor Logic and Complementary Metal Oxide Semiconductor (D) Tristate Transistor Logic and Complementary Metal Oxide

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 104

Silicon 5.23

In a microprocessor, the service routine for a certain interrupt starts from a fixed location of memory which cannot be externally set, but the interrupt can be delayed or rejected Such an interrupt is (A) non-maskable and non-vectored (B) maskable and non-vectored (C) non-maskable and vectored (D) maskable and vectored

(1) push buttons pressed/not pressed in equivalent to logic 1/0 respectively. (2) a segment glowing/not glowing in the display is equivalent to logic 1/0 respectively. 5.27

5.28

2009 5.24

5.25

TWO MARKS

If X = 1 in logic equation 6X + Z {Y + (Z + XY )}@ {X + X (X + Y)} = 1 , then (A) Y = Z (B) Y = Z (C) Z = 1 (D) Z = 0 What are the minimum number of 2- to -1 multiplexers required to generate a 2- input AND gate and a 2- input Ex-OR gate (A) 1 and 2 (B) 1 and 3

5.29

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in (C) 1 and 1 5.26

If segments a to g are considered as functions of P1 and P2 , then which of the following is correct (A) g = P 1 + P2, d = c + e (B) g = P1 + P2, d = c + e (C) g = P1 + P2, e = b + c (D) g = P1 + P2, e = b + c What are the minimum numbers of NOT gates and 2 - input OR gates required to design the logic of the driver for this 7 - Segment display (A) 3 NOT and 4 OR (B) 2 NOT and 4 OR (C) 1 NOT and 3 OR (D) 2 NOT and 3 OR Refer to the NAND and NOR latches shown in the figure. The inputs (P1, P2) for both latches are first made (0, 1) and then, after a few seconds, made (1, 1). The corresponding stable outputs (Q1, Q2) are

(A) NAND: first (0, 1) then (0, 1) NOR: first (1, 0) then (0, 0) (B) NAND : first (1, 0) then (1, 0) NOR : first (1, 0) then (1, 0) (C) NAND : first (1, 0) then (1, 0) NOR : first (1, 0) then (0, 0) (D) NAND : first (1, 0) then (1, 1) NOR : first (0, 1) then (0, 1)

(D) 2 and 2

What are the counting states (Q1, Q2) for the counter shown in the figure below

2008 5.30

(A) 11, 10, 00, 11, 10,... (C) 00, 11, 01, 10, 00...

TWO MARKS

The logic function implemented by the following circuit at the terminal OUT is

(B) 01, 10, 11, 00, 01... (D) 01, 10, 00, 01, 10...

Statement for Linked Answer Question 5.18 & 5.19 : Two products are sold from a vending machine, which has two push buttons P1 and P2 . When a buttons is pressed, the price of the corresponding product is displayed in a 7 - segment display. If no buttons are pressed, '0' is displayed signifying ‘Rs 0’. If only P1 is pressed, ‘2’ is displayed, signifying ‘Rs. 2’ If only P2 is pressed ‘5’ is displayed, signifying ‘Rs. 5’ If both P1 and P2 are pressed, 'E' is displayed, signifying ‘Error’ The names of the segments in the 7 - segment display, and the glow of the display for ‘0’, ‘2’, ‘5’ and ‘E’ are shown below.

Consider

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 (A) P NOR Q (C) P OR Q 5.31

5.32

(B) P NAND Q (D) P AND Q

The two numbers represented in signed 2’s complement form are P + 11101101 and Q = 11100110 . If Q is subtracted from P , the value obtained in signed 2’s complement is (A) 1000001111 (B) 00000111 (C) 11111001 (D) 111111001 Which of the following Boolean Expressions correctly represents the

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 105 3

relation between P, Q, R and M1

per the relation VDAC = 2n - 1 bn Volts, where b3 (MSB), b1, b2 and 0 The counter starts from the b0 (LSB) are the countern =outputs. clear state.

/

(A) M1 = (P OR Q) XOR R (B) M1 = (P AND Q) X OR R (C) M1 = (P NOR Q) X OR R (D) M1 = (P XOR Q) XOR R 5.33

For the circuit shown in the figure, D has a transition from 0 to 1 after CLK changes from 1 to 0. Assume gate delays to be negligible Which of the following statements is true

5.35

The stable reading of the LED displays is

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(A) Q goes to 1 at the CLK transition and stays at 1 (B) Q goes to 0 at the CLK transition and stays 0 (C) Q goes to 1 at the CLK tradition and goes to 0 when D goes to 1 (D) Q goes to 0 at the CLK transition and goes to 1 when D goes to 1 5.34

For each of the positive edge-triggered J - K flip flop used in the following figure, the propagation delay is 3 t .

Click to Buy www.nodia.co.in (A) 06 (C) 12 5.36

5.37

(B) 07 (D) 13

The magnitude of the error between VDAC and Vin at steady state in volts is (A) 0.2 (B) 0.3 (C) 0.5 (D) 1.0 For the circuit shown in the following, I0 - I3 are inputs to the 4:1 multiplexers, R(MSB) and S are control bits. The output Z can be represented by

Which of the following wave forms correctly represents the output at Q1 ?

(A) PQ + PQS + QRS (B) PQ + PQR + PQS (C) PQR + PQR + PARS + QRS (D) PQR + PQRS + PQRS + QRS 5.38

Statement For Linked Answer Question 5.26 & 5.27 : In the following circuit, the comparators output is logic “1” if V1 > V2 and is logic "0" otherwise. The D/A conversion is done as

An 8085 executes the following instructions 2710 LXI H, 30A0 H 2713 DAD H 2414 PCHL All address and constants are in Hex. Let PC be the contents of the program counter and HL be the contents of the HL pair just after executing PCHL. Which of the following statements is correct ? PC = 30A0H PC = 2715H (B) (A) HL = 2715H HL = 30A0H

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

PC = 6140H (C) HL = 6140H

Page 106

(C) Q1: normal active; Q2: cut-off; Q3: cut-off; Q4: saturation (D) Q1: saturation; Q2: saturation; Q3: saturation; Q4: normal active

PC = 6140H (D) HL = 2715H 5.44

2007 5.39

5.40

X = 01110 and Y = 11001 are two 5-bit binary numbers represented in two’s complement format. The sum of X and Y represented in two’s complement format using 6 bits is (A) 100111 (B) 0010000 (C) 000111 (D) 101001 The Boolean function Y = AB + CD is to be realized using only 2 input NAND gates. The minimum number of gates required is (A) 2 (B) 3 (C) 4 (D) 5 2007

5.41

ONE MARK

(A) (B) (C) (D)

TWO MARKS

The Boolean expression Y = ABC D + ABCD + ABC D + ABC D can be minimized to

5.45

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in (A) Y = ABC D + ABC + AC D (B) Y = ABC D + BCD + ABC D (C) Y = ABCD + BC D + ABC D (D) Y = ABCD + BC D + ABC D 5.42

5.43

P = 1, Q = 0; P = 1, Q = 0; P = 1, Q = 0; P = 1, Q = 0;

P = 1, Q = 0; P = 1, Q = 0 or P = 0, Q = 1 P = 0, Q = 1; or P = 0, Q = 1; P = 0, Q = 1 P = 1, Q = 1; P = 1, Q = 0 or P = 0, Q = 1 P = 1, Q = 1; P = 1, Q = 1

An 8255 chip is interfaced to an 8085 microprocessor system as an I/O mapped I/O as show in the figure. The address lines A0 and A1 of the 8085 are used by the 8255 chip to decode internally its thee ports and the Control . The address lines A3 to A7 as well as the IO/M signal are used for address decoding. The range of addresses for which the 8255 chip would get selected is

(A) F8H - FBH (C) F8H - FFH

(B) F8GH - FCH (D) F0H - F7H

Statement for Linked Answer Question 5.37 and 5.38 :

In the following circuit, X is given by

(A) (B) (C) (D)

The following binary values were applied to the X and Y inputs of NAND latch shown in the figure in the sequence indicated below : X = 0,Y = 1; X = 0, Y = 0; X = 1; Y = 1 The corresponding stable P, Q output will be.

In the Digital-to-Analog converter circuit shown in the figure below, VR = 10V and R = 10kW

X = ABC + ABC + ABC + ABC X = ABC + ABC + ABC + ABC X = AB + BC + AC X = AB + BC + AC

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The circuit diagram of a standard TTL NOT gate is shown in the figure. Vi = 25 V, the modes of operation of the transistors will be 5.46

5.47

The current is (A) 31.25mA (C) 125mA

(B) 62.5mA (D) 250mA

The voltage V0 is (A) - 0.781 V (C) - 3.125 V

(B) - 1.562 V (D) - 6.250 V

Statement for Linked Answer Questions 5.39 & 5.40 : (A) Q1: revere active; Q2: normal active; Q3: saturation; Q4: cut-off (B) Q1: revere active; Q2: saturation; Q3: saturation; Q4: cut-off

An 8085 assembly language program is given below. Line 1: MVI A, B5H

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

2: 3: 4: 5: 6: 7: 8: 5.48

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MVI B, OEH XRI 69H ADD B ANI 9BH I 9FH STA 3010H HLT

- out shift s loaded with the data shown are used to feed the data to a full adder. Initially, all the flip - flops are in clear state. After applying two clock pulse, the output of the full-adder should be

The contents of the accumulator just execution of the ADD instruction in line 4 will be (A) C3H (B) EAH (C) DCH (D) 69H After execution of line 7 of the program, the status of the CY and Z flags will be (B) CY = 0, Z = 1 (A) CY = 0, Z = 0 (C) CY = 1, Z = 0 (D) CY = 1, Z = 1

(A) S = 0, C0 = 0 (C) S = 1, C0 = 0 5.54

For the circuit shown, the counter state (Q1 Q0) follows the sequence

(B) S = 0, C0 = 1 (D) S = 1, C0 = 1

A new Binary Coded Pentary (B) number system is proposed in which every digit of a base-5 number is represented by its corresponding 3-bit binary code. For example, the base-5 number

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(A) 00, 01, 10, 11, 00 (C) 00, 01, 11, 00, 01 2006 5.51

24 will be represented by its B code 010100. In this numbering system, the B code 10001001101 corresponds of the following number is base-5 system (A) 423 (B) 1324 (C) 2201 (D) 4231

ONE MARK

The number of product in the minimized sum-of-product expression obtained through the following K - map is (where, "d" denotes don’t care states)

(A) 2 (C) 4

5.55

TWO MARKS

In the figure shown above, the ground has been shown by the symbol 4

An I/O peripheral device shown in Fig. (b) below is to be interfaced to an 8085 microprocessor. To select the I/O device in the I/O address range D4 H - D7 H, its chip-select (CS ) should be connected to the output of the decoder shown in as below :

5.56

(A) output 7 (C) output 2 5.53

A 4 - bit D/A converter is connected to a free - running 3 - big UP counter, as shown in the following figure. Which of the following waveforms will be observed at V0 ?

(B) 3 (D) 5

2006 5.52

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(B) 00, 01, 10, 00, 01 (D) 00, 10, 11, 00, 10

(B) output 5 (D) output 0

For the circuit shown in figures below, two 4 - bit parallel - in serial

Following is the segment of a 8085 assembly language program LXI SP, EFFF H CALL 3000 H : : : 3000 H LXI H, 3CF4

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

PUSH PSW SPHL POP PSW RET On completion of RET execution, the contents of SP is (A) 3CF0 H (B) 3CF8 H (C) EFFD H (D) EFFF H 5.57

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2005 5.61

TWO MARKS

The transistors used in a portion of the TTL gate show in the figure have b = 100 . The base emitter voltage of is 0.7 V for a transistor in active region and 0.75 V for a transistor in saturation. If the sink current I = 1 A and the output is at logic 0, then the current IR will be equal to

Two D - flip - flops, as shown below, are to be connected as a synchronous counter that goes through the following sequence 00 " 01 " 11 " 10 " 00 " ... The inputs D0 and D1 respectively should be connected as,

(A) Q 1 and Q0 (C) Q1 Q0 and Q 1 Q0

(A) 0.65 mA (C) 0.75 mA

(B) Q 0 and Q1 (D) Q 1 Q 0 and Q1 Q0

5.62

(B) 0.70 mA (D) 1.00 mA

The Boolean expression for the truth table shown is

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The point P in the following figure is stuck at 1. The output f will be

(A) B (A + C)( A + C ) (C) B (A + C )( A + C) 5.63

(A) ABC (C) ABC

(B) A (D) A

2005 5.59

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5.64

The present output Qn of an edge triggered JK flip-flop is logic 0. If J = 1, then Qn + 1 (A) Cannot be determined (B) Will be logic 0 (C) will be logic 1 (D) will rave around The given figure shows a ripple counter using positive edge triggered flip-flops. If the present state of the counter is Q2 Q1 Q0 = 001 then is next state Q2 Q1 Q will be

ONE MARK

Decimal 43 in Hexadecimal and BCD number system is respectively (A) B2, 0100 011 (B) 2B, 0100 0011 (C) 2B, 0011 0100 (D) B2, 0100 0100

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The Boolean function f implemented in the figure using two input multiplexes is

(A) ABC + ABC (C) ABC + ABC

(B) B (A + C )( A + C) (D) B (A + C)( A + C )

(A) 010 (C) 100

(B) ABC + ABC (D) ABC + ABC 5.65

(B) 111 (D) 101

What memory address range is NOT represents by chip # 1 and chip # 2 in the figure A0 to A15 in this figure are the address lines and CS means chip select.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

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(C) 5 5.71

5.72

(A) 0100 - 02FF (C) F900 - FAFF

(B) 1500 - 16FF (D) F800 - F9FF

Consider an 8085 microprocessor system.

5.67

The following program starts at location 0100H. LXI SP, OOFF LXI H, 0701 MVI A, 20H SUB M The content of accumulator when the program counter reaches 0109 H is (A) 20 H (B) 02 H (C) 00 H (D) FF H If in addition following code exists from 019H onwards, ORI 40 H ADD M What will be the result in the accumulator after the last instruction is executed ? (A) 40 H (B) 20 H (C) 60 H (D) 42 H 2004

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Choose the correct one from among the alternatives A, B, C, D after matching an item from Group 1 most appropriate item in Group 2. Group 1 Group 2 P. Shift 1. Frequency division Q. Counter 2. Addressing in memory chips R. Decoder 3. Serial to parallel data conversion (B) P - 3, Q - 1, R - 2 (A) P - 3, Q - 2, R - 1 (C) P - 2, Q - 1, R - 3 (D) P - 1, Q - 2, R - 2 The figure the internal schematic of a TTL AND-OR-OR-Invert (AOI) gate. For the inputs shown in the figure, the output Y is

(A) 0 (C) AB

Statement For Linked Answer Questions 5.57 & 5.58 : 5.66

(D) 7

(B) 1 (D) AB

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Click to Buy www.nodia.co.in 2004 5.73

5.74

TWO MARKS

11001, 1001, 111001 correspond to the 2’s complement representation of which one of the following sets of number (A) 25,9, and 57 respectively (B) -6, -6, and -6 respectively (C) -7, -7 and -7 respectively (D) -25, -9 and -57 respectively In the modulo-6 ripple counter shown in figure, the output of the 2- input gate is used to clear the J-K flip-flop The 2-input gate is

ONE MARK

A master - slave flip flop has the characteristic that (A) change in the output immediately reflected in the output (B) change in the output occurs when the state of the master is affected (C) change in the output occurs when the state of the slave is affected (D) both the master and the slave states are affected at the same time The range of signed decimal numbers that can be represented by 6-bits 1’s complement number is (A) -31 to +31 (B) -63 to +63 (C) -64 to +63 (D) -32 to +31 A digital system is required to amplify a binary-encoded audio signal. The should be able to control the gain of the amplifier from minimum to a maximum in 100 increments. The minimum number of bits required to encode, in straight binary, is (A) 8 (B) 6

(A) a NAND gate (C) an OR gate 5.75

5.76

5.77

(B) a NOR gate (D) a AND gare

The minimum number of 2- to -1 multiplexers required to realize a 4- to -1 multiplexers is (A) 1 (B) 2 (C) 3 (D) 4 The Boolean expression AC + BC is equivalent to (A) AC + BC + AC (B) BC + AC + BC + ACB (C) AC + BC + BC + ABC (D) ABC + ABC + ABC + ABC A Boolean function f of two variables x and y is defined as follows : f (0, 0) = f (0, 1) = f (1, 1) = 1; f (1, 0) = 0 Assuming complements of x and y are not available, a minimum cost solution for realizing f using only 2-input NOR gates and 2-

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 110

input OR gates (each having unit cost) would have a total cost of (A) 1 unit (B) 4 unit (C) 3 unit (D) 2 unit 5.78

The 8255 Programmable Peripheral Interface is used as described below. (i) An A/D converter is interface to a microprocessor through an 8255. The conversion is initiated by a signal from the 8255 on Port C. A signal on Port C causes data to be stobed into Port A. (ii) Two computers exchange data using a pair of 8255s. Port A works as a bidirectional data port ed by appropriate handshaking signals. The appropriate modes of operation of the 8255 for (i) and (ii) would be (A) Mode 0 for (i) and Mode 1 for (ii) (B) Mode 1 for (i) and Mode 2 for (ii) (C) Mode for (i) and Mode 0 for (ii) (D) Mode 2 for (i) and Mode 1 for (ii)

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in 5.79

5.80

5.81

The number of memory cycles required to execute the following 8085 instructions (i) LDA 3000 H (ii) LXI D, FOF1H would be (A) 2 for (i) and 2 for (ii) (B) 4 for (i) and 3 for (ii) (C) 3 for (i) and 3 for (ii) (D) 3 for (i) and 4 for (ii) Consider the sequence of 8085 instructions given below LXI H, 9258 MOV A, M CMA MOV M, A Which one of the following is performed by this sequence ? (A) Contents of location 9258 are moved to the accumulator (B) Contents of location 9258 are compared with the contents of the accumulator (C) Contents of location 8529 are complemented and stored in location 8529 (D) Contents of location 5892 are complemented and stored in location 5892 It is desired to multiply the numbers 0AH by 0BH and store the result in the accumulator. The numbers are available in s B and C respectively. A part of the 8085 program for this purpose is given below : MVI A, 00H LOOP --------------HLT END The sequence of instructions to complete the program would be

(A) JNX LOOP, ADD B, DCR C (B) ADD B, JNZ LOOP, DCR C (C) DCR C, JNZ LOOP, ADD B (D) ADD B, DCR C, JNZ LOOP 2003 5.82

5.83

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5.85

5.86

The number of distinct Boolean expressions of 4 variables is (A) 16 (B) 256 (C) 1023 (D) 65536 The minimum number of comparators required to build an 8-bits flash ADC is (A) 8 (B) 63 (C) 255 (D) 256 The output of the 74 series of GATE of TTL gates is taken from a BJT in (A) totem pole and common collector configuration (B) either totem pole or open collector configuration (C) common base configuration (D) common collector configuration Without any additional circuitry, an 8:1 MUX can be used to obtain (A) some but not all Boolean functions of 3 variables (B) all functions of 3 variables but non of 4 variables (C) all functions of 3 variables and some but not all of 4 variables (D) all functions of 4 variables A 0 to 6 counter consists of 3 flip flops and a combination circuit of 2 input gate (s). The common circuit consists of (A) one AND gate (B) one OR gate (C) one AND gate and one OR gate (D) two AND gates 2003

5.87

ONE MARK

TWO MARKS

The circuit in the figure has 4 boxes each described by inputs P, Q, R and outputs Y, Z with Y = P 5 Q 5 R and Z = RQ + PR + QP The circuit acts as a

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(A) 4 bit adder giving P + Q (B) 4 bit subtractor giving P - Q

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

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(C) 4 bit subtractor giving Q-P (D) 4 bit adder giving P + Q + R 5.88

5.89

5.90

Propagation delay is minimum

If the function W, X, Y and Z are as follows W = R + PQ + RS X = PQRS + PQRS + PQRS Y = RS + PR + PQ + P .Q Z = R + S + PQ + P .Q .R + PQ .S Then, (A) W = Z, X = Z (B) W = Z, X = Y (C) W = Y (D) W = Y = Z

The correct column is (A) P (C) R 5.92

ECL

TTL

TTL

(B) Q (D) S

The circuit shown in figure converts

A 4 bit ripple counter and a bit synchronous counter are made using flip flops having a propagation delay of 10 ns each. If the worst case delay in the ripple counter and the synchronous counter be R and S respectively, then (A) R = 10 ns, S = 40 ns (B) R = 40 ns, S = 10 ns (C) R = 10 ns S = 30 ns (D) R = 30 ns, S = 10 ns In the circuit shown in the figure, A is parallel-in, parallel-out 4 bit , which loads at the rising edge of the clock C . The input lines are connected to a 4 bit bus, W . Its output acts at input to a 16 # 4 ROM whose output is floating when the input to a partial table of the contents of the ROM is as follows Data

0011

1111

0100

1010

1011

1000

0010

1000

Address

0

2

4

6

8

10

11

14

The clock to the is shown, and the data on the W bus at time t1 is 0110. The data on the bus at time t2 is

(A) BCD to binary code (C) Excess -3 to gray code

(A) 1111 (C) 1000

(B) Binary to excess - 3 code (D) Gray to Binary code

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Click to Buy www.nodia.co.in 5.93

5.94

5.91

CMOS

In an 8085 microprocessor, the instruction CMP B has been executed while the content of the accumulator is less than that of B . As a result (A) Carry flag will be set but Zero flag will be reset (B) Carry flag will be rest but Zero flag will be set (C) Both Carry flag and Zero flag will be rest (D) Both Carry flag and Zero flag will be set The circuit shown in the figure is a 4 bit DAC

The input bits 0 and 1 are represented by 0 and 5 V respectively. The OP AMP is ideal, but all the resistance and the 5 v inputs have a tolerance of ! 10%. The specification (rounded to nearest multiple of 5%) for the tolerance of the DAC is (A) ! 35% (B) ! 20% (C) ! 10% (D) ! 5%

(B) 1011 (D) 0010

The DTL, TTL, ECL and CMOS famil GATE of digital ICs are compared in the following 4 columns (P)

(Q)

(R)

(S)

Fanout is minimum

DTL

DTL

TTL

CMOS

Power consumption is minimum

TTL

CMOS

ECL

DTL

2002 5.95

5.96

ONE MARK

4 - bit 2’s complement representation of a decimal number is 1000. The number is (A) +8 (B) 0 (C) -7 (D) -8 If the input to the digital circuit (in the figure) consisting of a

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 112

cascade of 20 XOR - gates is X , then the output Y is equal to

(A) 0 (C) X 5.97

The number of comparators required in a 3-bit comparators type ADC (A) 2 (B) 3 (C) 7 (D) 8 2002

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(B) 1 (D) X

TWO MARKS

(A) gray code numbers (C) excess - 3 code numbers 5.101

The circuit in the figure has two CMOS NOR gates. This circuit functions as a:

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(B) 2 4 2 1 BCD numbers (D) none of the above

Consider the following assembly language program MVI B, 87H MOV A, B START : JMP NEXT MVI B, 00H XRA B OUT PORT1 HLT NEXT : XRA B JP START OUT PORT2 HTL The execution of above program in an 8085 microprocessor will result in (A) an output of 87H at PORT1 (B) an output of 87H at PORT2 (C) infinite looping of the program execution with accumulator data remaining at 00H (D) infinite looping of the program execution with accumulator data alternating between 00H and 87H 2001

(A) flip-flop (C) Monostable multivibrator 5.99

(B) Schmitt trigger (D) astable multivibrator

The gates G1 and G2 in the figure have propagation delays of 10 ns and 20 ns respectively. If the input V1, makes an output change from logic 0 to 1 at time t = t0 , then the output waveform V0 is

5.102

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ONE MARKS

The 2’s complement representation of -17 is (A) 101110 (B) 101111 (C) 111110 (D) 110001 For the ring oscillator shown in the figure, the propagation delay of each inverter is 100 pico sec. What is the fundamental frequency of the oscillator output

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5.100

If the input X3, X2, X1, X0 to the ROM in the figure are 8 4 2 1 BCD numbers, then the outputs Y3, Y2, Y1, Y0 are (A) 10 MHz (C) 1 GHz 5.104

(B) 100 MHz (D) 2 GHz

Ab 8085 microprocessor based system uses a 4K # 8 bit RAM whose starting address is AA00H. The address of the last byte in this RAM is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) OFFFH (C) B9FFH 2001 5.105

Page 113

(B) 1000H (D) BA00H TWO MARKS

In the TTL circuit in the figure, S2 and S0 are select lines and X7 and X0 are input lines. S0 and X0 are LSBs. The output Y is

(A) 5 V; 3 V; 7 V (C) 5 V; 5 V; 5 V

(B) 4 V; 3 V; 4 V (D) 4 V; 4 V; 4 V

2000 5.109

(A) indeterminate (C) A 5 B 5.106

(B) A 5 B (D) C (A 5 B ) + C (A 5 B)

5.110

In the figure, the LED

ONE MARKS

An 8 bit successive approximation analog to digital communication has full scale reading of 2.55 V and its conversion time for an anaput of 1 V is 20 ms. The conversion time for a 2 V input will be (A) 10 ms (B) 20 ms (C) 40 ms (D) 50 ms The number of comparator in a 4-bit flash ADC is (A) 4 (B) 5

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Click to Buy www.nodia.co.in (A) emits light when both S1 and S2 are closed (B) emits light when both S1 and S2 are open (C) emits light when only of S1 and S2 is closed (D) does not emit light, irrespective of the switch positions. 5.107

(C) 15 5.111

For the logic circuit shown in the figure, the required input condition (A, B, C) to make the output (X) = 1 is

The digital block in the figure is realized using two positive edge triggered D-flip-flop. Assume that for t < t0, Q1 = Q2 = 0 . The circuit in the digital block is given by (A) 1,0,1 (C) 1,1,1 5.112

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5.114

In the DRAM cell in the figure, the Vt of the NMOSFET is 1 V. For the following three combinations of WL and BL voltages.

5.115

(B) 0,0,1 (D) 0,1,1

The number of hardware interrupts (which require an external signal to interrupt) present in an 8085 microprocessor are (A) 1 (B) 4 (C) 5 (D) 13 In the microprocessor, the RST6 instruction transfer the program execution to the following location : (A)30 H (B) 24 H (C) 48 H (D) 60 H 2000

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(D) 16

TWO MARKS

The contents of (B) and accumulator (A) of 8085 microprocessor are 49J are 3AH respectively. The contents of A and status of carry (CY) and sign (S) after execution SUB B instructions are (A) A = F1, CY = 1, S = 1 (B) A = 0F, CY = 1, S = 1 (C) A = F0, CY = 0, S = 0 (D) A = 1F, CY = 1, S = 1 For the logic circuit shown in the figure, the simplified Boolean expression for the output Y is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 114

(A) y = AB (C) y = A + B 5.120

(A) A + B + C (C) B 5.116

(B) A (D) C

For the 4 bit DAC shown in the figure, the output voltage V0 is

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(A) 10 V (C) 4 V

(B) 5 V (D) 8 V

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in 5.117

A sequential circuit using D flip-flop and logic gates is shown in the figure, where X and Y are the inputs and Z is the inputs. The circuit is

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(A) S - R (B) S - R (C) J - K (D) J - K 5.118

A Darlington emitter follower circuit is sometimes used in the output stage of a TTL gate in order to (A) increase its IOL (B) reduce its IOH (C) increase its speed of operation (D) reduce power dissipation Commercially available ECL gears use two ground lines and one negative supply in order to (A) reduce power dissipation (B) increase fan-out (C) reduce loading effect (D) eliminate the effect of power line glitches or the biasing circuit The resolution of a 4-bit counting ADC is 0.5 volts. For an anaput of 6.6 volts, the digital output of the ADC will be (A) 1011 (B) 1101 (C) 1100 (D) 1110 1999

5.123

Flip-Flop with inputs X = R and Y = S Flip-Flop with inputs X = S and Y = R Flip-Flop with inputs X = J and Y = K Flip-Flop with input X = K and Y = J

The minimized form of the logical expression (ABC + ABC + ABC + ABC ) is (A) AC + BC + AB (B) AC + BC + AB (C) AC + BC + AB (D) AC + BC + AB For a binary half-subtractor having two inputs A and B, the correct set of logical expressions for the outputs D (= A minus B) and X (= borrow) are (A) D = AB + AB, X = AB (B) D = AB + AB + AB , X = AB (C) D = AB + AB , X = AB (D) D = AB + AB , X = AB The ripple counter shown in the given figure is works as a

(A) mod-3 up counter (C) mod-3 down counter 5.126

(B) 1.0 kHz (D) 0.77 kHz

1999 5.119

The logical expression y = A + AB is equivalent to

ONE MARK

TWO MARKS

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In the figure, the J and K inputs of all the four Flip-Flips are made high. The frequency of the signal at output Y is

(A) 0.833 kHz (C) 0.91 kHz

(B) y = AB (D) y = A + B

(B) mod-5 up counter (D) mod-5 down counter

If CS = A15 A14 A13 is used as the chip select logic of a 4 K RAM in an 8085 system, then its memory range will be (A) 3000 H - 3 FFF H (B) 7000 H - 7 FFF H (C) 5000 H - 5 FFF H and 6000 H - 6 FFF H (D) 6000 H - 6 FFF H and 7000 H - 7 FFF H

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

1998 5.127

ONE MARK

Page 115 5.135

The minimum number of 2-input NAND gates required to implement of Boolean function Z = ABC , assuming that A, B and C are available, is (A) two (B) three (C) five (D) six 5.136

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5.129

The noise margin of a TTL gate is about (A) 0.2 V (B) 0.4 V (C) 0.6 V (D) 0.8 V In the figure is A = 1 and B = 1, the input B is now replaced by a sequence 101010....., the output x and y will be

(A) (B) (C) (D) 5.130

5.131

5.133

5.134

An instruction used to set the carry Flag in a computer can be classified as (A) data transfer (B) arithmetic (C) logical (D) program control The figure is shows a mod-K counter, here K is equal to

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fixed at 0 and 1, respectively x = 1010.....while y = 0101...... x = 1010.....and y = 1010...... fixed at 1 and 0, respectively

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An equivalent 2’s complement representation of the 2’s complement number 1101 is (A) 110100 (B) 01101 (C) 110111 (D) 111101 The threshold voltage for each transistor in the figure is 2 V. For this circuit to work as an inverter, Vi must take the values

(A) - 5 V and 0 V (C) - 0 V and 3 V 5.132

5.137

For the identity AB + AC + BC = AB + AC , the dual form is (A) (A + B) (A + C) (B + C) = (A + B) (A + C) (B) (A + B ) (A + C ) (B + C ) = (A + B ) (A + C ) (C) (A + B) (A + C) (B + C) = (A + B ) (A + C ) (D) AB + AC + BC = AB + AC

(A) 1 (C) 3 5.138

(B) - 5 V and 5 V (D) 3 V and 5 V

An I/O processor control the flow of information between (A) cache memory and I/O devices (B) main memory and I/O devices (C) two I/O devices (D) cache and main memories

The current I through resistance r in the circuit shown in the figure is

(A) - V 12R (C) V 6R 5.139

(B) 2 (D) 4

(B) V 12R (D) V 3T

The K -map for a Boolean function is shown in the figure is the number of essential prime implicates for this function is

Two 2’s complement number having sign bits x and y are added and the sign bit of the result is z . Then, the occurrence of overflow is indicated by the Boolean function (A) xyz (B) x y z (C) x yz + xyz (D) xy + yz + zx The advantage of using a dual slope ADC in a digital voltmeter is that (A) its conversion time is small (B) its accuracy is high (C) it gives output in BCD format (D) it does not require a

(A) 4 (C) 6

(B) 5 (D) 8

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

1997 5.140

5.141

5.142

Page 116

ONE MARK

Each cell of a static Random Access Memory contains (A) 6 MOS transistors (B) 4 MOS transistors and 2 capacitors (C) 2 MOS transistors and 4 capacitors (D) 1 MOS transistors and 1 capacitors

(C) 30 H to 33 H 5.147

A 2 bit binary multiplier can be implemented using (A) 2 inputs ANSs only (B) 2 input XORs and 4 input AND gates only (C) Two 2 inputs NORs and one XNO gate (D) XOR gates and shift s In standard TTL, the ‘totem pole’ stage refers to (A) the multi-emitter input stage (B) the phase splitter (C) the output buffer (D) open collector output stage

5.144

The inverter 74 ALSO4 has the following specifications IOH max =- 0.4 A, IOL max = 8 mA, IIH max = 20 mA, IIL max =- 0.1 mA The fan out based on the above will be (A) 10 (B) 20 (C) 60 (D) 100

5.146

ADDRESS (HEX)

INSTRUCTION

6010

LXI H, 8 A 79 H

6013

MOV A, L

6015

ADDH

6016

DAA

6017

MOV H, A

6018

PCHL

5.148

A signed integer has been stored in a byte using the 2’s complement format. We wish to store the same integer in a 16 bit word. We should (A) copy the original byte to the less significant byte of the word and fill the more significant with zeros (B) copy the original byte to the more significant byte of the word and fill the less significant byte with zeros (C) copy the original byte to the less significant byte of the word and make each fit of the more significant byte equal to the most significant bit of the original byte (D) copy the original byte to the less significant byte as well as the more significant byte of the word 1997

5.149

TWO MARKS

For the NMOS logic gate shown in the figure is the logic function implemented is

The output of the logic gate in the figure is

(A) 0 (C) A 5.145

The following instructions have been executed by an 8085 mP

From which address will the next instruction be fetched ? (A) 6019 (B) 6379 (C) 6979 (D) None of the above

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(D) 70 H to 73 H

(B) 1 (D) F

In an 8085 mP system, the RST instruction will cause an interrupt (A) only if an interrupt service routine is not being executed (B) only if a bit in the interrupt mask is made 0 (C) only if interrupts have been enabled by an EI instruction (D) None of the above

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The decoding circuit shown in the figure is has been used to generate the active low chip select signal for a microprocessor peripheral. (The address lines are designated as AO to A7 for I/O address)

(A) ABCDE (C) A : (B + C) + D : E 5.150

The peripheral will correspond to I/O address in the range (A) 60 H to 63 H (B) A4 to A 7H

(B) (AB + C ) : (D + E ) (D) (A + B ) : C + D : E

In a J–K flip-flop we have J = Q and K = 1. Assuming the flip flop was initially cleared and then clocked for 6 pulses, the sequence at the Q output will be

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) 010000 (C) 010010 5.151

5.152

5.154

5.155

5.156

5.158

5.159

5.160

(D) 13

The following sequence of instructions are executed by an 8085 microprocessor: 1000 LXI SP, 27 FF 1003 CALL 1006 1006 POP H The contents of the stack pointer (SP) and the HL, pair on completion of execution of these instruction are (A) SP = 27 FF, HL = 1003 (B) SP = 27 FD, HL = 1003 (C) SP = 27 FF, HL = 1006 (D) SP = 27 FD, HL = 1006

The boolean function A + BC is a reduced form of (A) AB + BC (B) (A + B) : (A + C) (C) AB + ABC (D) (A + C) : B ONE MARK

Schottky clamping is resorted in TTl gates (A) to reduce propagation delay (B) to increase noise margins (C) to increase packing density (D) to increase fan-out A pulse train can be delayed by a finite number of clock periods using (A) a serial-in serial-out shift (B) a serial-in parallel-out shift (C) a parallel-in serial-out shift (D) a parallel-in parallel-out shift A 12-bit ADC is operating with a 1 m sec clock period and the total conversion time is seen to be 14 m sec . The ADC must be of the (A) flash type (B) counting type (C) intergrating type (D) successive approximation type The total number of memory accesses involved (inclusive of the opcode fetch) when an 8085 processor executes the instruction LDA 2003 is (A) 1 (B) 2 (C) 3 (D) 4 1996

5.157

(C) 8

The gate delay of an NMOS inverter is dominated by charge time rather than discharge time because (A) the driver transistor has larger threshold voltage than the load transistor (B) the driver transistor has larger leakage currents compared to the load transistor (C) the load transistor has a smaller W/L ratio compared to the driver transistor (D) none of the above

1996 5.153

(B) 011001 (D) 010101

Page 117

TWO MARKS

A dynamic RAM cell which hold 5 V has to be refreshed every 20 m sec, so that the stored voltage does not fall by more than 0.5 V . If the cell has a constant discharge current of 1 pA, the storage capacitance of the cell is (A) 4 # 10-6 F (B) 4 # 10-9 F (C) 4 # 10-12 F (D) 4 # 10-15 F A 10-bit ADC with a full scale output voltage of 10.24 V is designed to have a ! LSB/2 accuracy. If the ADC is calibrated at 25c C and the operating temperature ranges from 0c C to 25c C , then the maximum net temperature coefficient of the ADC should not exceed (A) ! 200 mV/cC (B) ! 400 mV/cC (C) ! 600 mV/cC (D) ! 800 mV/cC A memory system of size 26 K bytes is required to be designed using memory chips which have 12 address lines and 4 data lines each. The number of such chips required to design the memory system is (A) 2 (B) 4

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 118

SOLUTIONS 5.1

For S1 S 0 = 0 0 We have A13 = A12 = 0 and for I/p = 1we obtain A10 = 1 or A10 = 0 A11 = 1 A14 = 1 or A14 = 0 A15 = 1 or A15 = 0 Since, A 0 - A 9 can have any value 0 or 1 Therefore, we have the address range as

Option (C) is correct. Let A denotes the position of switch at ground floor and B denotes the position of switch at upper floor. The switch can be either in up position or down position. Following are the truth table given for different combinations of A and B

A15 A14 A13 A12 A11 A10 A 9

A6

A5

A4

A3

A2

A1

A0

B

Y(Bulb)

From

0

0

0

0

1

0

0

0

0

0

0

0

0

0

0

0

up(1)

up(1)

OFF(0)

to

0

0

0

0

1

0

1

1

1

1

1

1

1

1

1

1

Down(0)

Down(0)

OFF(0)

up(1)

Down(0)

ON(1)

In Hexadecimal & 0800 H to 0BFFH Similarly, for chip 2, we obtain the range as follows E = 1 for S1 S 0 = 0 1 so, A13 = 0 and A12 = 1 and also the I/P = 1 for A10 = 0 , A11 = 1, A14 = 0 , A15 = 0 so, the fixed I/ps are

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Output of ADD B (Stored value at A)

05

05 + 05

04

05 + 05 + 04

03

05 + 05 + 04 + 03

02

05 + 05 + 04 + 03 + 02

01

05 + 05 + 04 + 03 + 02 + 01

00

System is out of loop

i.e., A = 05 + 05 + 04 + 03 + 02 + 01 = 144 At this stage, the 8085 microprocessor exits from the loop and reads the next instruction. i.e., the accumulator is being added to 03 H. Hence, we obtain A = A + 03 H = 14 + 03 = 17 H Option (D) is correct. For chip-1, we have the following conclusions: it is enable when (i) S1 S 0 = 0 0 and (ii) Input = 1

A13

A12

A11

A10

0

0

0

1

1

0

A8

A7

A6

A5

A4

A3

A2

A1

A0

From

0

0

0

1

1

0

0

0

0

0

0

0

0

0

0

0

to

0

0

0

1

1

0

1

1

1

1

1

1

1

1

1

1

In hexadecimal it is from 1800 H to 1BFFH . There is no need to obtain rest of address ranged as only (D) is matching to two results.

i.e., the XOR gate

Content in B

A14

A15 A14 A13 A12 A11 A10 A 9

Y = A5B Option (A) is correct. The program is being executed as follows MVI A, 0.5H; A = 05H MVI B, 0.5H; B = 05H At the next instruction, a loop is being introduced in which for the instruction “DCR B” if the result is zero then it exits from loop so, the loop is executed five times as follows :

A15

Therefore, the address range is

position leads to the ON state of bulb. Hence, from the truth table, we get

5.3

A7

A

Down(0) up(1) ON(1) When the switches A and B are both up or both down, output will be zero (i.e. Bulb will be OFF). Any of the switch changes its

5.2

A8

5.4

Option (A) is correct. The given circuit is

Condition for the race-around It occurs when the output of the circuit (Y1, Y2) oscillates between

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 ‘0’ and ‘1’ checking it from the options. 1. Option (A): When CLK = 0 Output of the NAND gate will be A1 = B1 = 0 = 1. Due to these input to the next NAND gate, Y2 = Y1 : 1 = Y1 and Y1 = Y2 : 1 = Y2 . If Y1 = 0 , Y2 = Y1 = 1 and it will remain the same and doesn’t oscillate. If Y2 = 0 , Y1 = Y2 = 1 and it will also remain the same for the clock period. So, it won’t oscillate for CLK = 0 . So, here race around doesn’t occur for the condition CLK = 0 . 2. Option (C): When CLK = 1, A = B = 1 A1 = B1 = 0 and so Y1 = Y2 = 1

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 119

And it will remain same for the clock period. So race around doesn’t occur for the condition. 3. Option (D): When CLK = 1, A = B = 0 So, A1 = B1 = 1 And again as described for Option (B) race around doesn’t occur for the condition. 5.5

mp COX W 2 2V V - VTp h - V SD B 2 b L l8 SD ^ SG m C 2 0 = p OX bW l [2VSD ^VSG - VTp h - V SD ] 2 L

I1 =

Since, & Solving it we get,

2 ^VSG - VTp h = VSD 2 ^5 - Vin - 1h = 5 - VD Vin = VD + 3 2

&

Option ( ) is correct. &

I1 = 0 , VD = 5 V Vin = 5 + 3 = 4 V 2

For So, So for the NMOS

Y = 1, when A > B

VGS = Vin - 0 = 4 - 0 = 4 V and VGS > VTn So it can’t be in cutoff region. Case 2 : M2 must be in saturation region. So, I1 = I 2

A = a1 a 0, B = b1 b 0 a1

a0

b1

b0

Y

0

1

0

0

1

1

0

0

0

1

1

0

0

1

1

1

1

0

0

1

1

1

0

1

1

1

1

1

0

1

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Total combination = 6 5.6

5.7

Option (A) is correct. Parallel connection of MOS & OR operation Series connection of MOS & AND operation The pull-up network acts as an inverter. From pull down network we write Y = (A + B) C = (A + B) + C = A B + C

mp COX W mn COX W 2 2 2 (VSG - VTp) VSD - V SD @ = 2 L (VGS - VTn) 2 L6 &

Option (A) is correct. Prime implicants are the that we get by solving K-map

2 (5 - Vin - 1) (5 - VD) - (5 - VD) 2 = (Vin - 0 - 1) 2 & 2 (4 - Vin) (5 - VD) - (5 - VD) 2 = (Vin - 1) 2 Substituting VD = VDS = VGS - VTn and for N -MOS & VD = Vin - 1 &

& & &

Option (A) is correct. Given the circuit as below :

2 (4 - Vin) (6 - Vin) - (6 - Vin) 2 = (Vin - 1) 2 48 - 36 - 8Vin =- 2Vin + 1

6Vin = 11 & Vin = 11 = 1.833 V 6 So for M2 to be in saturation Vin < 1.833 V or Vin < 1.875 V

F = XY + XY 1prime 44 2 44 3 implicants 5.8

2 2 (VSG - VTp) VSD - V SD = (VGS - VTn) 2

5.9

Option (D) is correct. Let Qn + 1 is next state and Qn is the present state. From the given below figure.

If A = 0, If A = 1, So state diagram is Since all the parameters of PMOS and NMOS are equal. So, mn = mp COX bW l = COX bW l = COX bW l L M2 L L M1 Given that M1 is in linear region. So, we assume that M2 is either in cutoff or saturation. Case 1 : M2 is in cut off So, I 2 = I1 = 0 Where I1 is drain current in M1 and I2 is drain current in M2 .

5.10

D = Y = AX 0 + AX1 Qn + 1 = D = AX 0 + AX1 Qn + 1 = A Qn + AQn X 0 = Q , X1 = Q Qn + 1 = Qn (toggle of previous state) Qn + 1 = Qn

Option (B) is correct. The given circuit is shown below:

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 120

0

1

1

0

1 QB ^t + 1h

0

0

0

(PQ QR ) PR = (PQ + QR PR ) = PQ + QR + PR = PQ + QR + PR If any two or more inputs are ‘1’ then output y will be 1. 5.11

QB ^t + 1h = Q A

Option (A) is correct. For the output to be high, both inputs to AND gate should be high. The D-Flip Flop output is the same, after a delay. Let initial input be 0; (Consider Option A) st then Q = 1 (For 1 D-Flip Flop). This is given as input to 2nd FF. Let the second input be 1. Now, considering after 1 time interval; The output of 1st Flip Flop is 1 and 2nd FF is also 1. Thus Output

DA = Q A Q B + QA QB 5.15

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RRC Thus A = 23 H

= 1. 5.12

5.16

Option (D) is correct.

Option (C) is correct. Initially Carry Flag, C = 0 MVI A, 07 H ; A = 0000 0111 RLC ; Rotate left without carry. A = 0000 1110 MVO B, A ; B = A = 0000 1110 RLC ; A = 0001 1100 RLC ; A = 0011 1000 ADD B ; A = 0011 1000 + 0000 1110 ; 0100 0110 ; ; Rotate Right with out carry, A = 0010 0011

Option ( ) is correct.

F = S 1 S 0 I 0 + S 1 S 0 I 1 + S 1 S 0 I 2 + S1 S 0 I 3 I0 = I3 = 0 ( S1 = P, S 0 = Q ) F = PQ + PQ = XOR (P, Q) 5.13

Option (A) is correct. All the states of the counter are initially unset.

5.17

State Initially are shown below in table :

5.14

Q2

Q1

Q0

0

0

0

0

1

0

0

4

1

1

0

6

1

1

1

7

0

1

1

3

0

0

1

1

0

0

0

0

Option (B) is correct. Since G2 is active low input, output of NAND gate must be 0

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 G2 = A15 : A14 A13 A12 A11 = 0 So, A15 A14 A13 A12 A11 = 00101 To select Y5 Decoder input ABC = A 8 A9 A10 = 101 Address range A15 A14 A13 A12 A11 A10 A 9 A 8 ...............A 0 0011101........A 0 S S 2 D ^2D00 - 2DFF h

Option (D) is correct. The sequence is QB QA 00 " 11 " 01 " 10 " 00 " ... QB

QA

QB (t + 1)

QA (t + 1)

0

0

1

1

1

1

0

1

5.18

Option (A) (B) (C) are correct. In the circuit F = (A 5 B) 9 (A 9 B) 9 C

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

For two variables A5B = A9B So, (A 5 B) 9 (A 9 B) = 0 (always) F = 09C = 0$C+1$C = C So, F = 1 when C = 1 or C = 0 5.19

Option (D) is correct. Let QA (n), QB (n), QC (n) are present states and QA (n + 1), QB (n + 1), QC (n + 1) are next states of flop-flops. In the circuit QA (n + 1) = QB (n) 9 QC (n) QB (n + 1) QA (n) QC (n + 1) QB (n) Initially all flip-flops are reset 1st clock pulse

Page 121

A = A 5 B = 00100010 5 01000101 = 01100111 = 674 5.22

5.23

5.24

QA = 0 9 0 = 1 QB = 0 QC = 0

5.25

2 nd clock pulse

Option (C) is correct. TTL " Transistor - Transistor logic CMOS " Complementary Metal Oxide Semi-conductor Option (D) is correct. Vectored interrupts : Vectored interrupts are those interrupts in which program control transferred to a fixed memory location. Maskable interrupts : Maskable interrupts are those interrupts which can be rejected or delayed by microprocessor if it is performing some critical task. Option (D) is correct. We have 6X + Z {Y + (Z + XY )}@[X + Z (X + Y)] = 1 Substituting X = 1 and X = 0 we get [1 + Z {Y + (Z + 1Y )}][ 0 + Z (1 + Y)] = 1 or [1][ Z (1)] = 1 1 + A = 1 and 0 + A = A or Z =1)Z=0 Option (A) is correct.

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QA = 0 9 0 = 1 QB = 1 QC = 0 3 rd clock pulse

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QA = 1 9 0 = 0 QB = 1 QC = 1

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4 th clock pulse

The AND gate implementation by 2:1 mux is as follows QA = 1 9 1 = 1 QB = 0 QC = 1

So, sequence 5.20

QA = 01101.......

Y = AI 0 + AI1 = AB

Option (D) is correct. Output of the MUX can be written as

The EX - OR gate implementation by 2:1 mux is as follows

F = I 0 S 0 S1 + I1 S 0 S1 + I 2 S 0 S 1 + I 3 S 0 S 1 Here, I 0 = C, I1 = D, I2 = C , I 3 = CD and S 0 = A, S1 = B So, F = C A B + D A B + C A B + C DA B Writing all SOP F = A B C D + A B C D + A BCD + A B C D 1 44 2 44 3 1 44 2 44 3 S 1 44 2 4 43 m m m m 3

7

2

5

+A B C D + A B C D + ABC D 1 44 2 4 4 3 1 44 2 44 3 S m m m 9

F = / m (2, 3, 5, 7, 8, 9, 12) 5.21

8

12

Y = BI0 + BI1 = AB + BA 5.26

Option (A) is correct. The given circuit is as follows.

Option (C) is correct. By executing instruction one by one MVI A, 45 H & MOV 45 H into accumulator, A = 45 H STC & Set carry, C = 1 CMC & Complement carry flag, C = 0 RAR & Rotate accumulator right through carry The truth table is as shown below. Sequence is 00, 11, 10, 00 ...

A = 00100010 XRA B & XOR A and B

CLK

J1

K1

Q1

J2

K2

Q2

1

1

1

0

1

1

0

2

1

1

1

1

1

1

3

0

0

1

0

1

0

4

1

1

0

1

1

0

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 5.27

Page 122

From the figure shown below it may be easily seen upper MOSFET are shorted and connected to Vdd thus OUT is 1 only when the node S is 0,

Option (B) is correct. The given situation is as follows

The truth table is as shown below P1

P2

a

b

c

d

e

f

g

0

0

1

1

1

1

1

1

0

0

1

1

0

1

1

0

1

1

1

0

1

1

0

1

1

0

1

1

1

1

0

0

1

1

1

1

Since the lower MOSFETs are shorted to ground, node S is 0 only when input P and Q are 1. This is the function of AND gate. 5.31

From truth table we can write a =1 b = P 1 P 2 + P1 P 2 = P 2

and

P - Q = (- 19) - (- 26) = 7 Thus 7 signed two’s complements form is

1 NOT Gate

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1 NOT Gate

c = P1 P2 = P1 + P2

1 OR GATE

1 OR GATE f = P1 P2 = P1 + P2 1 OR GATE g = P1 P2 = P1 + P2 Thus we have g = P1 + P2 and d = 1 = c + e . It may be observed easily from figure that Led g does not glow only when both P1 and P2 are 0. Thus

Option (B) is correct. MSB of both number are 1, thus both are negative number. Now we get 11101101 = (- 19) 10 and 11100110 = (- 26) 10

(7) 10 = 00000111 5.32

Option (D) is correct. The circuit is as shown below

So and 5.33

X = PQ Y = (P + Q) Z = PQ (P + Q) = (P + Q )( P + Q) = PQ + PQ = P 5 Q M1 = Z 5 R = (P 5 Q) 5 R

Option (A) is correct. The circuit is as shown below

g = P1 + P2 LED d is 1 all condition and also it depends on d = c+e 5.28

5.29

Option (D) is correct. As shown in previous solution 2 NOT gates and 3-OR gates are required.

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Option (C) is correct. For the NAND latche the stable states are as follows

The truth table is shown below. When CLK make transition Q goes to 1 and when D goes to 1, Q goes to 0 5.34

For the NOR latche the stable states are as follows

5.30

Option (D) is correct.

Option (B) is correct. Since the input to both JK flip-flop is 11, the output will change every time with clock pulse. The input to clock is

The output Q0 of first FF occurs after time 3 T and it is as shown below

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 123

Thus after execution above instruction contests of PC and HL are same and that is 6140H 5.39

The output Q1 of second FF occurs after time 3 T when it gets input (i.e. after 3 T from t1) and it is as shown below

5.40 5.35

Option (D) is correct. VDAC =

We have

/ 2n - 1bn = 2- 1b0 + 20 b1 + 21b2 + 22 b3

or VDAC = 0.5b0 + b1 + 2b2 + 4b3 The counter outputs will increase by 1 from 0000 till Vth > VDAC .

Option (B) is correct. Y = AB + CD = AB .CD This is SOP form and we require only 3 NAND gate

3

n=0

Option (C) is correct. MSB of Y is 1, thus it is negative number and X is positive number Now we have X = 01110 = (14) 10 and Y = 11001 = (- 7) 10 X + Y = (14) + (- 7) = 7 In signed two’s complements from 7 is (7) 10 = 000111

5.41

Option (A) is correct. The circuit is as shown below

The output of counter and VDAC is as shown below Clock

b3 b3 b2 b0

VDAC

1

0001

0

2

0010

0.5

3

0011

1

4

0100

1.5

5

0101

2

6

0110

2.5

7

0111

3

8

1000

3.5

9

1001

4

10

1010

4.5

11

1011

5

12

1100

5.5

13

1101

6

14

1110

6.5

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Y = AB + AB

and when VADC = 6.5 V (at 1101), the output of AND is zero and the counter stops. The stable output of LED display is 13. 5.36

5.37

Option (B) is correct. The VADC - Vin at steady state is = 6.5 - 6.2 = 0.3V

and

5.42

5.43

5.44

Z = PQ + PQS + QRS 5.38

Option (C) is correct. 2710H LXI H, 30A0H 2713H DAD H 2714H PCHL

; Load 16 bit data 30A0 in HL pair ; 6140H " HL ; Copy the contents 6140H of HL in PC

Option (D) is correct. Y = ABCD + ABCD + ABC D + ABC D = ABCD + ABC D + ABC D + ABC D = ABCD + ABC D + BC D (A + A) = ABCD + ABC D + BC D

Option (A) is correct. Z = I0 RS + I1 RS + I2 RS + I3 RS = (P + Q ) RS + PRS + PQRS + PRS = PRS + QRS + PRS + PQRS + PRS The k - Map is as shown below

X = YC + YC = (AB + AB ) C + (AB + AB ) C = (AB + AB) C + (AB + AB ) C = ABC + ABC + ABC + ABC

5.45

A+A = 1

Option (B) is correct. In given TTL NOT gate when Vi = 2.5 (HIGH), then Q1 " Reverse active Q2 " Saturation Q3 " Saturation Q4 " cut - off region Option (C) is correct. For X = 0, Y = 1 For X = 0, Y = 0 For X = 1, Y = 1

P = 1, Q = 0 P = 1, Q = 1 P = 1, Q = 0 or P = 0, Q = 1

Option (C) is correct. Chip 8255 will be selected if bits A3 to A7 are 1. Bit A0 to A2 can

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

be 0 or. 1. Thus address range is 11111000 11111111 5.46

Page 124

; Since 8 AH < 9BH, CY = 1 ; Store the contents of A to location 3010

7 : STA 3010 H H

F8H FFH

Option (B) is correct. Since the inverting terminal is at virtual ground the resistor network can be reduced as follows

8 : HLT ; Stop Thus the contents of accumulator after execution of ADD instruction is EAH. 5.49

5.50

Option (C) is correct. The CY = 1 and Z = 0 Option (A) is correct. For this circuit the counter state (Q1, Q0) follows the sequence 00, 01, 10, 00 ... as shown below Clock

D1 D0

Q1 Q0

Q1 NOR Q0

00

1

1st

01

10

0

2nd

10

01

0

3rd

00

00

0

The current from voltage source is

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5.51

I = VR = 10 = 1 mA R 10k This current will be divide as shown below

5.52

5.53

-3 i = I = 1 # 10 = 62.5 m A 16 16

Now 5.47

5.48

Option (C) is correct. The net current in inverting terminal of OP - amp is I- = 1 + 1 = 5I 4 16 16 So that V0 =- R # 5I =- 3.125 16 Option (B) is correct. Line 1 : MVI A, B5H 2 : MVI B, 0EH 3 : XRI 69H 4 : ADDB 5 : ANI 9BH 6 : I 9FH

; Move B5H to A ; Move 0EH to B ; [A] XOR 69H and store in A ; Contents of A is CDH ; Add the contents of A to contents of B and ; store in A, contents of A is EAH ; [a] AND 9BH, and store in A, ; Contents of A is 8 AH ; Compare 9FH with the contents of A

5.54

Option (A) is correct. As shown below there are 2 in the minimized sum of product expression. 1

0

0

1

0

d

0

0

0

0

d

1

1

0

0

1

Option (B) is correct. The output is taken from the 5th line. Option (D) is correct. After applying two clock poles, the outputs of the full adder is S = 1 , C0 = 1 A B Ci S Co 1st 1 0 0 0 1 2nd 1 1 1 1 1 Option (D) is correct.

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 100010011001 S S S S 4 2 3 1 5.55

Option (B) is correct. In this the diode D2 is connected to the ground. The following table shows the state of counter and D/A converter Q2 Q1 Q0

D3 = Q2

D2 = 0

D1 = Q1

D0 = Q0

Vo

000

0

0

0

0

0

001

0

0

0

1

1

010

0

0

1

0

2

011

0

0

1

1

3

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

100

1

0

0

0

8

101

1

0

0

1

9

110

1

0

1

0

10

111

1

0

1

1

11

000

0

0

0

0

0

001

0

0

0

1

1

Page 125

The circuit is as shown below

Thus option (B) is correct 5.56

5.57

5.58

Option (B) is correct. LXI, EFFF H ; Load SP with data EFFH CALL 3000 H ; Jump to location 3000 H : : : 3000H LXI H, 3CF4 ; Load HL with data 3CF4H PUSH PSW ; Store contnets of PSW to Stack POP PSW ; Restore contents of PSW from stack PRE ; stop Before instruction SPHL the contents of SP is 3CF4H. After execution of POP PSW, SP + 2 " SP After execution of RET, SP + 2 " SP Thus the contents of SP will be 3CF4H + 4 = 3CF8H

If output is at logic 0, the we have V0 = 0 which signifies BJT Q3 is in saturation and applying KVL we have or or 5.62

Option (A) is correct. We have f = ABC + ABC = B (AC + AC ) = B (A + C)( A + C )

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Option (A) is correct. The inputs D0 and D1 respectively should be connected as Q1 and Q0 where Q0 " D1 and Q1 " D0 Option (D) is correct. If the point P is stuck at 1, then output f is equal to A

VBE3 = IR # 1k 0.75 = IR # 1k IR = 0.75 mA

Click to Buy www.nodia.co.in 5.63

Option (C) is correct. Characteristic equation for a jk flip-flop is written as Where So,

5.59

Option (B) is correct. Dividing 43 by 16 we get 2 16 43 32

g

11 11 in decimal is equivalent is B in hexamal. Thus 4310 * 2B16 Now 410 * 01002 310 * 00112 Thus 4310 * 01000011BCD 5.60

5.64

5.65 5.66

5.67

5.61

Option (C) is correct.

= JQ n + K Qn is the present output is next output = 10 + K : 0

Qn = 0

=1

Option (C) is correct. Since T2 T1 T0 is at 111, at every clock Q2 Q1 Q0 will be changes. Ir present state is 011, the next state will be 100. Option (D) is correct. Option (C) is correct. 0100H LXI SP, 00FF 0103H LXI H, 0701 0106H MVI A, 20H 0108 H SUB M

; Load SP with 00FFG ; Load HL with 0107H ; Move A with 20 H ; Subtract the contents of memory ; location whose address is stored in HL ; from the A and store in A 0109H ORI 40H ; 40H OR [A] and store in A 010BH ADD M ; Add the contents of memeory location ; whose address is stored in HL to A ; and store in A HL contains 0107H and contents of 0107H is 20H Thus after execution of SUB the data of A is 20H - 20H = 00

Option (A) is correct. The diagram is as shown in fig

f' = BC + BC f = f'A + f'0 = f'A = ABC + ABC

Qn + 1 Qn Qn + 1 Qn + 1 Qn + 1

Option (C) is correct. Before ORI instruction the contents of A is 00H. On execution the ORI 40H the contents of A will be 40H 00H = 00000000 40H = 01000000 ORI 01000000

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

After ADD instruction the contents of memory location whose address is stored in HL will be added to and will be stored in A 40H + 20 H = 60 H 5.68

Page 126

multiplexers is required. 5.76

AC + BC = AC1 + BC 1 = AC (B + B ) + BC (A + A) = ACB + ACB + BC A + BC A

Option (C) is correct. A master slave D-flip flop is shown in the figure. 5.77

In the circuit we can see that output of flip-flop call be triggered only by transition of clock from 1 to 0 or when state of slave latch is affected. 5.69

5.78

Option (A) is correct. The range of signed decimal numbers that can be represented by n - bits 1’s complement number is - (2n - 1 - 1) to + (2n - 1 - 1). Thus for n = 6 we have

Range =- (26 - 1 - 1) to + (26 - 1 - 1) =- 31 to + 31

5.71

5.72

5.73

5.74

5.75

5.79

2n $ 100 n $7

Option (B) is correct. Shift " Serial to parallel data conversion Counter " Frequency division Decoder " Addressing in memory chips. Option (A) is correct. For the TTL family if terminal is floating, then it is at logic 1. Thus Y = (AB + 1) = AB .0 = 0

5.80

Option (C) is correct. 11001 1001 111001 00110 0110 000110 +1 +1 +1 00111 0111 000111 7 7 7 Thus 2’s complement of 11001, 1001 and 111001 is 7. So the number given in the question are 2’s complement correspond to -7. Option (C) is correct. In the modulo - 6 ripple counter at the end of sixth pulse (i.e. after 101 or at 110) all states must be cleared. Thus when CB is 11 the all states must be cleared. The input to 2-input gate is C and B and the desired output should be low since the CLEAR is active low Thus when C and B are 0, 0, then output must be 0. In all other case the output must be 1. OR gate can implement this functions. Option (C) is correct. Number of MUX is 4 = 2 and 2 = 1. Thus the total number 3 2 3

Option (D) is correct. For 8255, various modes are described as following. Mode 1 : Input or output with hand shake In this mode following actions are executed 1. Two port (A & B) function as 8 - bit input output ports. Each port uses three lines from C as a hand shake signal Input & output data are latched.

Form (ii) the mode is 1. Mode 2 : Bi-directional data transfer This mode is used to transfer data between two computer. In this mode port A can be configured as bidirectional port. Port A uses five signal from port C as hand shake signal. For (1), mode is 2

Option (D) is correct. The minimum number of bit require to encode 100 increment is or

Option (D) is correct. We have f (x, y) = xy + xy + xy = x (y + y) + xy = x + xy or f (x, y) = x + y Here compliments are not available, so to get x we use NOR gate. Thus desired circuit require 1 unit OR and 1 unit NOR gate giving total cost 2 unit.

2. 3.

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5.70

Option (D) is correct.

Option (B) is correct. LDA 16 bit & Load accumulator directly this instruction copies data byte from memory location (specified within the instruction) the accumulator. It takes 4 memory cycle-as following. 1. in instruction fetch 2. in reading 16 bit address 1. in copying data from memory to accumulator LXI D, (F0F1) 4 & It copies 16 bit data into pair D and E. It takes 3 memory cycles. Option (A) is correct. LXI H, 9258H MOV A, M CMa

; 9258H " HL ; (9258H) " A ; A"A

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 MOV M, A ; A"M This program complement the data of memory location 9258H. 5.81

Option (D) is correct. MVI A, 00H ; Clear accumulator LOOP ADD B ; Add the contents of B to A DCR C ; Decrement C JNZ LOOP ; If C is not zero jump to loop HLT END This instruction set add the contents of B to accumulator to contents of C times.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 5.82

16

2 = 2 = 65536

5.85

5.86

5.87

Thus W = Z and X = Z 5.89

Option (C) is correct. In the flash analog to digital converter, the no. of comparators is equal to 2n - 1, where n is no. of bit.s So,

5.84

= R + S + PQ + PR + PQS + QRS

Option (D) is correct. The number of distinct boolean expression of n variable is 22n . Thus 24

5.83

Page 127

2n - 1 = 28 - 1 = 255

Option (B) is correct. When output of the 74 series gate of TTL gates is taken from BJT then the configuration is either totem pole or open collector configuration . Option (D) is correct. A 2n: 1 MUX can implement all logic functions of (n + 1) variable without andy additional circuitry. Here n = 3 . Thus a 8 : 1 MUX can implement all logic functions of 4 variable. Option (D) is correct. Counter must be reset when it count 111. This can be implemented by following circuitry

5.90

Option (C) is correct. After t = t1, at first rising edge of clock, the output of shift is 0110, which in input to address line of ROM. At 0110 is applied to . So at this time data stroed in ROM at 1010 (10), 1000 will be on bus. When W has the data 0110 and it is 6 in decimal, and it’s data value at that add is 1010

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Option (B) is correct. We have Y = P5Q5R

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Z = RQ + PR + QP Here every block is a full subtractor giving P - Q - R where R is borrow. Thus circuit acts as a 4 bit subtractor giving P - Q . 5.88

Option (B) is correct. Propagation delay of flip flop is tpd = 10 nsec Propagation delay of 4 bit ripple counter R = 4tpd = 40 ns and in synchronous counter all flip-flop are given clock simultaneously, so S = tpd = 10 ns

then 1010 i.e. 10 is acting as odd, at time t2 and data at that movement is 1000.

Option (A) is correct. 5.91

W = R + PQ + RS X = PQRS + PQRS + PQRS Y = RS + PR + PQ + PQ = RS + PR $ PQ $ PQ = RS + (P + R )( P + Q)( P + Q) = RS + (P + PQ + PR + QR )( P + Q) = RS + PQ + QR (P + P ) + QR = RS + PQ + QR

5.92

5.93

Z = R + S + PQ + PQR + PQS = R + S + PQ $ PQR $ PQS 5.94

Option (B) is correct. The DTL has minimum fan out and CMOS has minimum power consumption. Propagation delay is minimum in ECL. Option (D) is correct. Let input be 1010; output will be 1101 Let input be 0110; output will be 0100 Thus it convert gray to Binary code. Option (A) is correct. CMP B & Compare the accumulator content with context of B If A < R CY is set and zero flag will be reset. Option (A) is correct. Vo =- V1 :R bo + R b1 + R b2 + R b 3D 4R R 2R 4R

= R + S + (P + Q )( P + Q + R)( P + Q + S) = R + S + PQ + PQ + PQS + PR + PQR + PRS + PQ + PQS + PQR + QRS

Exact value when V1 = 5 , for maximum output VoExact =- 5 :1 + 1 + 1 + 1 D =- 9.375 2 4 8

= R + S + PQ + PQS + PR + PQR + PRS

Maximum Vout due to tolerance Vo max =- 5.5 :110 + 110 + 110 + 110 D 90 2 # 90 4 # 90 8 # 90

+ PQS + PQR + QRS

Tolerance

= R + S + PQ (1 + S) + PR (1 + P ) + PRS + PQS + PQR + QRS = R + S + PQ + PR + PRS + PQS + PQR + QRS

5.95

5.96

= R + S + PQ + PR (1 + Q ) + PQS + QRS

=- 12.604 = 34.44% = 35%

Option (D) is correct. If the 4- bit 2’s complement representation of a decimal number is 1000, then the number is -8 Option (B) is correct. Output of 1 st XOR = = X $ 1 + X $ 1 = X Output of 2 nd XOR = X X + XX = 1

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 128

Its 1’s complement is 101110 So 2’s compliment is

So after 4,6,8,...20 XOR output will be 1. 5.97

5.98

Option (C) is correct. In the comparator type ADC, the no. of comparators is equal to 2n - 1 , where n is no. of bit.s So, 23 - 1 = 7 Option (C) is correct. The circuit is as shown below

+

101111 5.103

5.104

The circuit shown is monostable multivibrator as it requires an external triggering and it has one stable and one quasistable state. 5.99

Option (B) is correct. They have prorogation delay as respectively, G1 " 10 nsec G2 " 20 nsec

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in For abrupt change in Vi from 0 to 1 at time t = t0 we have to assume the output of NOR then we can say that option (B) is correct waveform.

5.100

5.101

Option (B) is correct. Let X3 X2 X1 X0 be 1001 then Y3 Y2 Y1 Y0 will be 1111. Let X3 X2 X1 X0 be 1000 then Y3 Y2 Y1 Y0 will be 1110 Let X3 X2 X1 X0 be 0110 then Y3 Y2 Y1 Y0 will be 1100 So this converts 2-4-2-1 BCD numbers. Option (B) is correct. MVI B, 87H MOV A, B START : JMP NEXT XRA B JP START

NEXT :

JMP NEXT XRA

JP START OUT PORT2 5.102

; B = 87 ; A = B = 87 ; Jump to next ; A 5 B " A, ; A = 00, B = 87 ; Since A = 00 is positive ; so jump to START ;Jump to NEXT ; unconditionally ; B ; A 5 B " A, A = 87 , ; B = 87 H ; will not jump as D7 , of A is 1 ; A = 87 " PORT2

Option (B) is correct. The two’s compliment representation of 17 is 17 = 010001

101110 1

Option (C) is correct. The propagation delay of each inverter is tpd then The fundamental frequency of oscillator output is 1 = 1 GHz f = 1 = 2ntpd 2 # 5 # 100 # 10 - 12 Option (C) is correct. 4K # 8 bit means 102410 location of byte are present Now 102410 * 1000H It starting address is AA00H then address of last byte is AA00H + 1000H - 0001H = B9FFH

5.105

Option (D) is correct.

or 5.106

5.107

Y = I0 + I3 + I5 + I6 = C BA + C AB + CBA + CBA Y = C (A 5 B ) + C (A 5 B)

Option (D) is correct. For the LED to glow it must be forward biased. Thus output of NAND must be LOW for LED to emit light. So both input to NAND must be HIGH. If any one or both switch are closed, output of AND will be LOW. If both switch are open, output of XOR will be LOW. So there can’t be both input HIGH to NAND. So LED doesn’t emit light. Option (C) is correct. The output of options (C) satisfy the given conditions

5.108

Option (B) is correct.

5.109

Option (B) is correct.

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 Conversion time of successive approximate analog to digital converters is independent of input voltage. It depends upon the number of bits only. Thus it remains unchanged. 5.110

5.111

Option (C) is correct. In the flash analog to digital converter, the no. of comparators is equal to 2n - 1, where n is no. of bits. So, 2 4 - 1 = 15 Option (D) is correct. As the output of AND is X = 1, the all input of this AND must be 1. Thus ...(1) AB + AB = 1

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 129

...(2) BC + BC = 1 ...(3) C =1 From (2) and (3), if C = 1, then B = 1 If B = 1, then from (1) A = 0 . Thus A = 0, B = 1 and C = 1 5.112

5.113

5.114

Option (C) is correct. Interrupt is a process of data transfer by which an external device can inform the processor that it is ready for communication. 8085 microprocessor have five interrupts namely TRAP, INTR, RST 7.5, RST 6.5 and RST 5.5

Option (A) is correct. Accumulator contains A = 49 H B = 3 AH SUB B = A minus B

Carry so here outputA Carry CY Sign flag S

Vo = 8 # 5 = 5V 8 5.117

Option (D) is correct. The truth table is shown below Z = XQ + YQ Comparing from the truth table of J - K FF Y = J, X =K X

Y

Z

0

0

Q

0

1

0

1

0

1

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Click to Buy www.nodia.co.in 1

=1 =0F =1 =1

5.118

Option (C) is correct. The circuit is as shown below :

5.119

5.120

Y = B + (B + C ) = B (B + C ) = B Option (B) is correct. The circuit is as shown below

5.121 5.122

1

Q1

Option (B) is correct. In the figure the given counter is mod-10 counter, so frequency of output is 10k = 1k 10 Option (D) is correct. We have y = A + AB we know from Distributive property Thus

5.116

...(2)

From (1) and (2) we have

Option (A) is correct. For any RST instruction, location of program transfer is obtained in following way. RST x & (x ) 8) 10 " convert in hexadecimal So for RST 6 & (6 ) 8) 10 = (48) 10 = (30) H

A = 49 H = 01001001 B = 3 AH = 00111010 2’s complement of (- B) = 11000110 A - B = A + (- B) 010 010 01 & +1 1 0 0 0 1 1 0 0 0 0 0 1111

5.115

Now applying voltage divider rule V- = 1k V% = 1 Vo 1k + 7k 8

x + yz = (x + y) (x + z) y = (A + A) (A + B) = A + B

Option (C) is correct. Darligton emitter follower provides a low output impedance in both logical state (1 or 0). Due to this low output impedance, any stray capacitance is rapidly charged and discharged, so the output state changes quickly. It improves speed of operation. Option (D) is correct. Option (B) is correct. For ADC we can write Anaput = (decimal eq of digital output) # resol 6.6 = (decimal eq. of digital output) # 0.5 6.6 = decimal eq of digital. output 0.5 13.2 = decimal equivalent of digital output so output of ADC is = 1101.

The voltage at non-inverting terminal is V+ = 1 + 1 = 5 8 2 8 V- = V+ = 5 8

5.123

...(1)

Option (A) is correct. We use the K -map as below.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 130

address range A15 A14 A13 A12 initial 1 1 1 address final 1 1 1 address so address range is (7 0

So given expression equal to = AC + BC + AB 5.124

Option (C) is correct. For a binary half-subtractor truth table si given below.

5.127

(111) A11 A10 A 9 A 8 A7 A6 A5 A 4 A 3 A2 A1 A 0

0 0 0 0 0 0 0 0 0 0 0 0 0 &7000H 1 1 1 1 1 1 1 1 1 1 1 11 &7FFFH 0 0 H – 7 F F F H)

Option (C) is correct. Given boolean function is Z = ABC Z = ABC = ACB = AC + B

Now

Thus Z = AC + B we have Z = X + Y (1 NOR gate) where X = AC (1 NAND gate) To implement a NOR gate we required 4 NAND gates as shown below in figure.

from truth table we can find expressions of D & X D = A 5 B = AB + AB X = AB

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here total no. of NAND gates required = 4+1 = 5 5.128

Option (D) is correct. From the given figure we can write the output

Option (B) is correct. For TTL worst cases low voltages are VOL (max) = 0.4 V VIL (max) = 0.8 V Worst case high voltages are VOH (min) = 2.4 V VIH (min) = 2 V The difference between maximum input low voltage and maximum output low voltage is called noise margin. It is 0.4 V in case of TTL.

5.129

For the state 010 all preset = 1 and output QA QB QC = 111 so here total no. of states = 5 (down counter) 5.126

Option (B) is correct. We have 4 K RAM (12 address lines)

Option (D) is correct. From the figure we can see If A =1 then y =1 If A =1 then also y =1

B=0 x=0 B=1 x=0

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 so for sequence B = 101010....output x and y will be fixed at 0 and 1 respectively. 5.130

5.131 5.132

so here chip select logic CS = A15 A14 A13

5.133

Option (D) is correct. Given 2’s complement no. 1101; the no. is 0011 for 6 digit output we can write the no. is – 000011 2’s complement representation of above no. is 111101 Option (A) is correct. Option (B) is correct. An I/O Microprocessor controls data flow between main memory and the I/O device which wants to communicate. Option (D) is correct.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 5.134

5.135

Option (B) is correct. Dual slope ADC is more accurate.

The given gate is ex-OR so output

Option (A) is correct. Dual form of any identity can be find by replacing all AND function to OR and vice-versa. so here dual form will be

Here input

(A + B) (A + C) (B + C) = (A + B) (A + C) 5.136

5.137

Option (B) is correct. Carry flag will be affected by arithmetic instructions only. Option (C) is correct. This is a synchronous counter. we can find output as QA QB 0 0 1 0 0 1 0 0 h So It counts only three states. It is a mod-3 counter. K =3

5.138 5.139

Page 131

Option (B) is correct. Option (A) is correct. Essential prime implicates for a function is no. of that we get by solving K -map. Here we get 4 when solve the K -map.

F = AB + AB B = 0 so, F = A1 + A0 = A

5.145

Option (C) is correct. EI = Enabled Interput flag ,RST will cause an Interrupt only it we enable EI .

5.146

5.147

Option (A) is correct. Here only for the range 60 to 63 H chipselect will be 0, so peripheral will correspond in this range only chipselect = 1 for rest of the given address ranges. Option (B) is correct. By executing instructions one by one LXI H, 8A79 H (Load HL pair by value 8A79) H = 8AH L = 79 H MOV A, L (copy contain of L to accumulator)

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Click to Buy www.nodia.co.in A = 79 H ADDH (add contain of H to accumulator) A = 79 H = 0 1111 0 0 1 H = 8AH = add 1 0 0 0 1 0 1 0 =A= 0 0 0 0 0 0 11 Carry = 1 DAA (Carry Flag is set, so DAA adds 6 to high order four bits) y = B D + A C D + C AB + CA B so no of prime implicates is 4 5.140 5.141

Option (A) is correct. Option (B) is correct. For a 2 bit multiplier B1 # A1 A 0 B1

B0 A0 A0 B0

# A1 B1 A1 B 0 C3 C2 C1 C0 This multiplication is identical to AND operation and then addition. 5.142

5.143

H = 63 H PCHL (Load program counter by HL pair) PC = 6379 H 5.148 5.149

Option (B) is correct. Consider high output state fan out = IOH max = 400 mA = 20 IIH max 20 mA Consider low output state fan out = IOL max = 8 mA = 80 IIL max 0.1 mA Option (A) is correct.

Option (C) is correct. Option (C) is correct. NMOS In parallel makes OR Gate & in series makes AND so here we can have F = A (B + C) + DE we took complement because there is another NMOS given above (works as an inverter)

Option (C) is correct. In totem pole stage output resistance will be small so it acts like a output buffer.

Thus fan out is 20 5.144

0 1111 0 0 1 DAA add 1 0 0 0 1 0 1 0 A = 0 0 0 0 0 0 1 1 = 63 H MOV H, A (copy contain of A to H)

5.150

Option (D) is correct. For a J -K flip flop we have characteristic equation as Q (t + 1) = JQ (t) + KQ (t) Q (t) & Q (t + 1) are present & next states. In given figure J = Q (t), K = 1 so Q (t + 1) = Q (t) Q (t) + 0Q (t) Q (t + 1) = Q (t)[complement of previous state] we have initial input Q (t) = 0

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 132

so for 6 clock pulses sequence at output Q will be 010101 5.151 5.152

Option (B) is correct. By distributive property in boolean algebra we have (A + BC) = (A + B) (A + C) (A + B) (A + C) = AA + AC + AB + BC = A (1 + C) + AB + BC = A + AB + BC = A (1 + B) + BC = A + BC

5.153

5.154

Option (A) is correct. The current in a p n junction diode is controlled by diffusion of majority carriers while current in schottky diode dominated by the flow of majority carrier over the potential barrier at metallurgical junction. So there is no minority carrier storage in schottky diode, so switching time from forward bias to reverse bias is very short compared to p n junction diode. Hence the propagation delay will reduces.

= 200 mV/cC 5.159

10.24 2 # 1024 # (50 - 25) cC

Option (D) is correct. 210 # 8 = 13 No. of chips = 26 # 12 2 #4

5.160

Option (C) is correct. Given instruction set 1000 LXI SP 27FF 1003 CALL 1006 1006 POP H First Instruction will initialize the SP by a value 27FF SP ! 27FF CALL 1006 will “Push PC” and Load PC by value 1006 PUSH PC will store value of PC in stack PC = 1006

Option (B) is correct.

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5.155

Tcoff =

or

Option (C) is correct.

Option (D) is correct. The total conversion time for different type of ADC are given as– t is clock period For flash type & 1t Counter type & (2n - t) = 4095 m sec

now POP H will be executed which load HL pair by stack values HL = 1006 and SP = SPl + 2 SP = SPl + 2 = SP - 2 + 2 = SP SP = 27FF

n = no.of bits Integrating type conver time > 4095 m sec successive approximation type nt = 12 m sec here n = 12 so nt = 12 12t = 12 so this is succ. app. type ADC. 5.156

5.157

Option (D) is correct. LDA 2003 (Load accumulator by a value 2003 H) so here total no. of memory access will be 4. 1 = Fetching instruction 2 = Read the value from memory 1 = write value to accumulator Option (D) is correct. Storage capacitance -12 C = i = 1 # 10 5 - 0.5 dv b dt l b 20 10-3 l # = 1 # 10

5.158

-12

# 20 # 10 4.5

-3

= 4.4 # 10-15 F

Option (A) is correct.

or

Accuracy ! 1 LSB = Tcoff # DT 2 1 10.24 = T coff # DT 2 # 210

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

UNIT 6

Page 133

2013 6.8

SIGNALS & SYSTEMS 6.9

2013 6.1

6.2

ONE MARK

Two systems with impulse responses h1 ^ t h and h2 ^ t h are connected in cascade. Then the overall impulse response of the cascaded system is given by (A) product of h1 ^ t h and h2 ^ t h (B) sum of h1 ^ t h and h2 ^ t h (C) convolution of h1 ^ t h and h2 ^ t h (D) subtraction of h2 ^ t h from h1 ^ t h The impulse response of a system is h ^ t h = tu ^ t h. For an input u ^t - 1h, the output is 2 t ^t - 1h (A) t u ^ t h (B) u ^t - 1h 2 2

^t - 1h2 (C) u ^t - 1h 2 6.3

6.4

6.5

6.6

6.7

2 (D) t - 1 u ^t - 1h 2

6.10

The impulse response of a continuous time system is given by h ^ t h = d ^t - 1h + d ^t - 3h. The value of the step response at t = 2 is (A) 0 (B) 1 (C) 2 (D) 3 A system described by the differential equation 2 dy dy + 5 + 6y ^ t h = x ^ t h. Let x ^ t h be a rectangular pulse given by dt dt2 1 0 <2 x^t h = * 0 otherwise dy Assuming that y ^0 h = 0 and = 0 at t = 0 , the Laplace transdt form of y ^ t h is 1 - e-2s e-2s (B) (A) s ^s + 2h^s + 3h s ^s + 2h^s + 3h -2s e 1 - e-2s (C) (D) ^s + 2h^s + 3h ^s + 2h^s + 3h A system described by a linear, constant coefficient, ordinary, first

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For a periodic signal v ^ t h = 30 sin 100t + 10 cos 300t + 6 sin ^500t + p/4h, the fundamental frequency in rad/s (A) 100 (B) 300 (C) 500 (D) 1500

order differential equation has an exact solution given by y ^ t h for t > 0 , when the forcing function is x ^ t h and the initial condition is y ^0 h. If one wishes to modify the system so that the solution becomes - 2y ^ t h for t > 0 , we need to (A) change the initial condition to - y ^0 h and the forcing function to 2x ^ t h (B) change the initial condition to 2y ^0 h and the forcing function to - x ^ t h (C) change the initial condition to j 2 y ^0 h and the forcing function to j 2 x ^ t h (D) change the initial condition to - 2y ^0 h and the forcing function to - 2x ^ t h

A band-limited signal with a maximum frequency of 5 kHz is to be sampled. According to the sampling theorem, the sampling frequency which is not valid is (A) 5 kHz (B) 12 kHz (C) 15 kHz (D) 20 kHz Which one of the following statements is NOT TRUE for a continuous time causal and stable LTI system? (A) All the poles of the system must lie on the left side of the jw axis (B) Zeros of the system can lie anywhere in the s-plane (C) All the poles must lie within s = 1 (D) All the roots of the characteristic equation must be located on the left side of the jw axis.

TWO MARKS

Assuming zero initial condition, the response y ^ t h of the system given below to a unit step input u ^ t h is

The DFT of a vector 8a b c dB is the vector 8a b g dB . Consider the product V R Sa b c d W Sd a b c W 8p q r sB = 8a b c dBSc d a b W W S Sb c d aW X T The DFT of the vector 8p q r sB is a scaled version of (B) 9 a b (A) 9a2 b2 g2 d2C g dC (C) 8a + b b + d d + g g + aB (D) 8a b g dB

(A) u ^ t h 2 (C) t u ^ t h 2

2012

(B) tu ^ t h (D) e-t u ^ t h 6.12

Let g ^ t h = e- pt , and h ^ t h is a filter matched to g ^ t h. If g ^ t h is applied as input to h ^ t h, then the Fourier transform of the output is (A) e- pf (B) e- pf /2 2

2

(C) e- p f

6.11

2

(D) e-2pf

2

ONE MARK

The unilateral Laplace transform of f (t) is 2 1 . The unilateral s +s+1 Laplace transform of tf (t) is (A) - 2 s (B) - 2 2s + 1 2 (s + s + 1) 2 (s + s + 1) (C) 2 s (D) 2 2s + 1 2 2 (s + s + 1) (s + s + 1)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 6.13

If x [n] = (1/3) n - (1/2) n u [n], then the region of convergence (ROC) of its z -transform in the z -plane will be (A) 1 < z < 3 (B) 1 < z < 1 3 3 2 (C) 1 < z < 3 (D) 1 < z 2 3 2012

6.14

6.19

6.20

TWO MARKS

The input x (t) and output y (t) of a system are related as y (t) =

If the unit step response of a network is (1 - e- at), then its unit impulse response is (A) ae- at (B) a-1 e- at (C) (1 - a-1)e- at (D) (1 - a) e- at

t

-3

The Fourier transform of a signal h (t) is H (jw) = (2 cos w) (sin 2w) /w . The value of h (0) is (A) 1/4 (B) 1/2 (C) 1 (D) 2

2011 6.21

6.22

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in 6.16

A system is defined by its impulse response h (n) = 2n u (n - 2). The system is (A) stable and causal (B) causal but not stable (C) stable but not causal (D) unstable and non-causal

# x (t) cos (3t) dt . The system is

(A) time-invariant and stable (B) stable and not time-invariant (C) time-invariant and not stable (D) not time-invariant and not stable 6.15

Page 134

Let y [n] denote the convolution of h [n] and g [n], where h [n] = (1/2) n u [n] and g [n] is a causal sequence. If y [0] = 1 and y [1] = 1/2, then g [1] equals (A) 0 (B) 1/2 (C) 1

Two systems H1 (Z ) and H2 (Z ) are connected in cascade as shown below. The overall output y (n) is the same as the input x (n) with a one unit delay. The transfer function of the second system H2 (Z ) is

1 - 0.6z-1 z-1 (1 - 0.4z-1) z-1 (1 - 0.4z-1) (C) (1 - 0.6z-1)

(D) 3/2

2011

An input x (t) = exp (- 2t) u (t) + d (t - 6) is applied to an LTI system with impulse response h (t) = u (t) . The output is (A) [1 - exp (- 2t)] u (t) + u (t + 6) (B) [1 - exp (- 2t)] u (t) + u (t - 6) (C) 0.5 [1 - exp (- 2t)] u (t) + u (t + 6) (D) 0.5 [1 - exp (- 2t)] u (t) + u (t - 6)

(A)

6.23

ONE MARK

TWO MARKS

(B)

z-1 (1 - 0.6z-1) (1 - 0.4z-1)

(D)

1 - 0.4 z-1 z-1 (1 - 0.6z-1)

The first six points of the 8-point DFT of a real valued sequence are 5, 1 - j 3, 0, 3 - j 4, 0 and 3 + j 4 . The last two points of the DFT are respectively (A) 0, 1 - j 3 (B) 0, 1 + j 3 (C) 1 + j3, 5 (D) 1 - j 3, 5

2

6.17

The differential equation 100 ddty - 20 dy dt + y = x (t) describes a system with an input x (t) and an output y (t). The system, which is initially relaxed, is excited by a unit step input. The output y ^ t h can be represented by the waveform 2

2010 6.24

ONE MARK

The trigonometric Fourier series for the waveform f (t) shown below contains

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6.18

The trigonometric Fourier series of an even function does not have the (A) dc term (B) cosine (C) sine (D) odd harmonic

(A) only cosine and zero values for the dc components

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 135

(B) only cosine and a positive value for the dc components (C) only cosine and a negative value for the dc components (D) only sine and a negative value for the dc components 6.25

6.26

6.27

Consider the z -transform x (z) = 5z2 + 4z-1 + 3; 0 < z < 3. The inverse z - transform x [n] is (A) 5d [n + 2] + 3d [n] + 4d [n - 1] (B) 5d [n - 2] + 3d [n] + 4d [n + 1] (C) 5u [n + 2] + 3u [n] + 4u [n - 1] (D) 5u [n - 2] + 3u [n] + 4u [n + 1] Two discrete time system with impulse response h1 [n] = d [n - 1] and h2 [n] = d [n - 2] are connected in cascade. The overall impulse response of the cascaded system is (B) d [n - 4] (A) d [n - 1] + d [n - 2] (C) d [n - 3] (D) d [n - 1] d [n - 2] For a N -point FET algorithm N = 2m which one of the following statements is TRUE ? (A) It is not possible to construct a signal flow graph with both input and output in normal order (B) The number of butterflies in the m th stage in N/m (C) In-place computation requires storage of only 2N data (D) Computation of a butterfly requires only one complex multiplication. 2010

6.28

6.29

6.30

6.32

6.33

3s + 1 Given f (t) = L ; 3 . If lim f (t) = 1, then the value t"3 s + 4s2 + (k - 3) s E of k is (A) 1 (B) 2 (C) 3 (D) 4

The transfer function of a discrete time LTI system is given by 2 - 3 z-1 4 H (z) = 3 -1 1 - z + 1 z-2 8 4 Consider the following statements: S1: The system is stable and causal for ROC: z > 1/2 S2: The system is stable but not causal for ROC: z < 1/4 S3: The system is neither stable nor causal for ROC: 1/4 < z < 1/2 Which one of the following statements is valid ? (A) Both S1 and S2 are true (B) Both S2 and S3 are true (C) Both S1 and S3 are true (D) S1, S2 and S3 are all true

The Fourier series of a real periodic function has only (P) cosine if it is even (Q) sine if it is even (R) cosine if it is odd (S) sine if it is odd

The ROC of z -transform of the discrete time sequence n n x (n) = b 1 l u (n) - b 1 l u (- n - 1) is 3 2 (B) z > 1 (A) 1 < z < 1 2 3 2

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-1

A continuous time LTI system is described by d 2 y (t) dy (t) dx (t) + 4x (t) +4 + 3y (t) = 2 2 dt dt dt Assuming zero initial conditions, the response y (t) of the above system for the input x (t) = e-2t u (t) is given by (B) (e-t - e-3t) u (t) (A) (et - e3t) u (t) (C) (e-t + e-3t) u (t) (D) (et + e3t) u (t)

A function is given by f (t) = sin2 t + cos 2t . Which of the following is true ? (A) f has frequency components at 0 and 1 Hz 2p (B) f has frequency components at 0 and 1 Hz p (C) f has frequency components at 1 and 1 Hz p 2p (D) f has frequency components at 0.1 and 1 Hz p 2p

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TWO MARKS

2009 6.31

Which of the above statements are correct ? (A) P and S (B) P and R (C) Q and S (D) Q and R

(D) 2 < z < 3

2009 6.34

TWO MARKS

Given that F (s) is the one-side Laplace transform of f (t), the Laplace transform of

t

#0 f (t) dt is

(A) sF (s) - f (0) (C) 6.35

6.36

#0

s

F (t) dt

(B) 1 F (s) s (D) 1 [F (s) - f (0)] s

A system with transfer function H (z) has impulse response h (.) defined as h (2) = 1, h (3) =- 1 and h (k) = 0 otherwise. Consider the following statements. S1 : H (z) is a low- filter. S2 : H (z) is an FIR filter. Which of the following is correct? (A) Only S2 is true (B) Both S1 and S2 are false (C) Both S1 and S2 are true, and S2 is a reason for S1 (D) Both S1 and S2 are true, but S2 is not a reason for S1 Consider a system whose input x and output y are related by the equation y (t) =

# x (t - t) g (2t) dt where h (t) is shown in the graph. 3

-3

ONE MARK

Which of the following four properties are possessed by the system ? BIBO : Bounded input gives a bounded output.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Causal : The system is causal, LP : The system is low . LTI : The system is linear and time-invariant. (A) Causal, LP (B) BIBO, LTI (C) BIBO, Causal, LTI (D) LP, LTI 6.37

6.38

comes zero are (A) p, 2p (C) 0, p 6.43

The 4-point Discrete Fourier Transform (DFT) of a discrete time sequence {1,0,2,3} is (A) [0, - 2 + 2j , 2, - 2 - 2j ] (B) [2, 2 + 2j , 6, 2 - 2j ] (C) [6, 1 - 3j , 2, 1 + 3j ] (D) [6, - 1 + 3j , 0, - 1 - 3j ] An LTI system having transfer function s +s 2+s 1+ 1 and input x (t) = sin (t + 1) is in steady state. The output is sampled at a rate ws rad/s to obtain the final output {x (k)}. Which of the following is true ? (A) y (.) is zero for all sampling frequencies ws (B) y (.) is nonzero for all sampling frequencies ws (C) y (.) is nonzero for ws > 2 , but zero for ws < 2 (D) y (.) is zero for ws > 2 , but nonzero for w2 < 2 2

2008

6.40

6.42

6.44

6.45

ONE MARK

The input and output of a continuous time system are respectively denoted by x (t) and y (t). Which of the following descriptions corresponds to a causal system ? (B) y (t) = (t - 4) x (t + 1) (A) y (t) = x (t - 2) + x (t + 4) (C) y (t) = (t + 4) x (t - 1) (D) y (t) = (t + 5) x (t + 5)

A discrete time linear shift - invariant system has an impulse response h [n] with h [0] = 1, h [1] =- 1, h [2] = 2, and zero otherwise The system is given an input sequence x [n] with x [0] = x [2] = 1, and zero otherwise. The number of nonzero samples in the output sequence y [n], and the value of y [2] are respectively (A) 5, 2 (B) 6, 2 (C) 6, 1 (D) 5, 3

The signal x (t) is described by 1 for - 1 # t # + 1 x (t) = ) 0 otherwise Two of the angular frequencies at which its Fourier transform be-

{x (n)} is a real - valued periodic sequence with a period N . x (n) and X (k) form N-point Discrete Fourier Transform (DFT) pairs. N-1 The DFT Y (k) of the sequence y (n) = 1 / x (r) x (n + r) is N r=0 N-1 (B) 1 / X (r) X (k + r) (A) X (k) 2 N r=0 (D) 0

Statement for Linked Answer Question 6.31 and 6.32: In the following network, the switch is closed at t = 0- and the sampling starts from t = 0 . The sampling frequency is 10 Hz.

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TWO MARKS

A linear, time - invariant, causal continuous time system has a rational transfer function with simple poles at s =- 2 and s =- 4 and one simple zero at s =- 1. A unit step u (t) is applied at the input of the system. At steady state, the output has constant value of 1. The impulse response of this system is (A) [exp (- 2t) + exp (- 4t)] u (t) (B) [- 4 exp (- 2t) - 12 exp (- 4t) - exp (- t)] u (t) (C) [- 4 exp (- 2t) + 12 exp (- 4t)] u (t) (D) [- 0.5 exp (- 2t) + 1.5 exp (- 4t)] u (t)

Let x (t) be the input and y (t) be the output of a continuous time system. Match the system properties P1, P2 and P3 with system relations R1, R2, R3, R4 Properties Relations P1 : Linear but NOT time - invariant R1 : y (t) = t2 x (t) P2 : Time - invariant but NOT linear R2 : y (t) = t x (t) P3 : Linear and time - invariant R3 : y (t) = x (t) R4 : y (t) = x (t - 5) (A) (P1, R1), (P2, R3), (P3, R4) (B) (P1, R2), (P2, R3), (P3, R4) (C) (P1, R3), (P2, R1), (P3, R2) (D) (P1, R1), (P2, R2), (P3, R3)

N-1 (C) 1 / X (r) X (k + r) N r=0

The impulse response h (t) of a linear time invariant continuous time system is described by h (t) = exp (at) u (t) + exp (bt) u (- t) where u (- t) denotes the unit step function, and a and b are real constants. This system is stable if (A) a is positive and b is positive (B) a is negative and b is negative (C) a is negative and b is negative (D) a is negative and b is positive 2008

6.41

(B) 0.5p, 1.5p (D) 2p, 2.5p

2

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in 6.39

Page 136

6.46

6.47

The samples x (n), n = (0, 1, 2, ...) are given by (A) 5 (1 - e-0.05n) (B) 5e-0.05n (C) 5 (1 - e-5n) (D) 5e-5n The expression and the region of convergence of the z -transform of the sampled signal are 5z , z < e-0.05 (A) 5z 5 , z < e-5 (B) z e-0.05 z-e 5z , z > e-0.05 (C) (D) 5z -5 , z > e-5 -0.05 z-e z-e

Statement for Linked Answer Question 6.33 & 6.34: The impulse response h (t) of linear time - invariant continuous

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 137

time system is given by h (t) = exp (- 2t) u (t), where u (t) denotes the unit step function. 6.48

6.49

The frequency response H (w) of this system in of angular frequency w, is given by H (w) 1 (B) sin w (A) w 1 + j2w jw 1 (C) (D) 2 + jw 2 + jw

#

#

6.58

The output of this system, to the sinusoidal input x (t) = 2 cos 2t for all time t , is (A) 0 (B) 2-0.25 cos (2t - 0.125p) (C) 2-0.5 cos (2t - 0.125p) (D) 2-0.5 cos (2t - 0.25p) 6.59

2007 6.50

ONE MARK

1 If the Laplace transform of a signal Y (s) = , then its final s (s - 1) value is (A) - 1 (B) 0 (C) 1 (D) Unbounded 2007

6.51

TWO MARKS

The 3-dB bandwidth of the low- signal e-t u (t), where u (t) is the unit step function, is given by (B) 1 (A) 1 Hz 2 - 1 Hz 2p 2p (C) 3

6.52

(A) 5 (C) 16p 6.53

6.54

6.55

p

-p

6.56

6.57

Click to Buy www.nodia.co.in 1 sin t - p ` 4j 2 (C) 1 e-t sin t 2 (A)

(B) 10p (D) 5 + j10p

(B) non-causal system (D) low- system

The frequency response of a linear, time-invariant system is given by H (f) = 1 + j510pf . The step response of the system is t (A) 5 (1 - e-5t) u (t) (B) 5 61 - e- 5@ u (t) t (C) 1 (1 - e-5t) u (t) (D) 1 ^1 - e- 5 h u (t) 2 5

6.60

6.61

6.62

ONE MARK

Let x (t) * X (jw) be Fourier Transform pair. The Fourier Transform of the signal x (5t - 3) in of X (jw) is given as j3w j3w jw jw (A) 1 e- 5 X b l (B) 1 e 5 X b l 5 5 5 5 jw jw (C) 1 e-j3w X b l (D) 1 e j3w X b l 5 5 5 5 The Dirac delta function d (t) is defined as 1 t=0 (A) d (t) = ) 0 otherwise

(B)

1 sin t + p ` 4j 2

(D) sin t - cos t

2006 -1

2006

<2 3

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X (e jw) dw is

The z -transform X (z) of a sequence x [n] is given by X [z] = 1 -0.25z . It is given that the region of convergence of X (z) includes the unit circle. The value of x [0] is (A) - 0.5 (B) 0 (C) 0.25 (D) 05 A Hilbert transformer is a (A) non-linear system (C) time-varying system

<3

In the system shown below, x (t) = (sin t) u (t) In steady-state, the response y (t) will be

(D) 1 Hz

#

1 < z < 2 then the 3 3

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A 5-point sequence x [n] is given as x [- 3] = 1, x [- 2] = 1, x [- 1] = 0, x [0] = 5 and x [1] = 1. Let X (eiw) denoted the discrete-time Fourier transform of x [n]. The value of

3 t=0 (B) d (t) = ) 0 otherwise 3 1 t=0 (C) d (t) = ) and d (t) dt = 1 -3 0 otherwise 3 3 t=0 (D) d (t) = ) and d (t) dt = 1 -3 0 otherwise If the region of convergence of x1 [n] + x2 [n] is region of convergence of x1 [n] - x2 [n] includes (A) 1 < z < 3 (B) 2 < z 3 3 (C) 3 < z < 3 (D) 1 < z 2 3

6.63

TWO MARKS

Consider the function f (t) having Laplace transform F (s) = 2 w0 2 Re [s] > 0 s + w0 The final value of f (t) would be (A) 0 (B) 1 (C) - 1 # f (3) # 1 (D) 3 A system with input x [n] and output y [n] is given as y [n] = (sin 56 pn) x [n] . The system is (A) linear, stable and invertible (B) non-linear, stable and non-invertible (C) linear, stable and non-invertible (D) linear, unstable and invertible The unit step response of a system starting from rest is given by c (t) = 1 - e-2t for t $ 0 . The transfer function of the system is 1 (B) 2 (A) 2+s 1 + 2s (C) 1 (D) 2s 2+s 1 + 2s The unit impulse response of a system is f (t) = e-t, t $ 0 . For this system the steady-state value of the output for unit step input is equal to (A) - 1 (B) 0 (C) 1 (D) 3

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

2005 6.64

6.65

Page 138

ONE MARK

Choose the function f (t); - 3 < t < 3 for which a Fourier series cannot be defined. (A) 3 sin (25t) (B) 4 cos (20t + 3) + 2 sin (710t) (C) exp (- t ) sin (25t) (D) 1

(B) 41 (C) 42 (D) 82 2005 6.70

The function x (t) is shown in the figure. Even and odd parts of a unit step function u (t) are respectively,

TWO MARKS

The output y (t) of a linear time invariant system is related to its input x (t) by the following equations y (t)= 0.5x (t - td + T) + x (t - td ) + 0.5x (t - td + T) The filter transfer function H (w) of such a system is given by (A) (1 + cos wT) e-jwt d

(B) (1 + 0.5 cos wT) e-jwt (C) (1 - cos wT) e-jwt (D) (1 - 0.5 cos wT) e-jwt

d

d

d

(A) 1 , 1 x (t) 2 2 (C) 1 , - 1 x (t) 2 2

(B) - 1 , 1 x (t) 2 2 (D) - 1 , - 1 x (t) 2 2

6.71

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6.66

6.67

The region of convergence of z - transform of the sequence 5 n 6 n b 6 l u (n) - b 5 l u (- n - 1) must be (B) z > 5 (A) z < 5 6 6 (C) 5 < z < 6 (D) 6 < z < 3 6 5 5

Match the following and choose the correct combination. Group 1 E. Continuous and aperiodic signal F. Continuous and periodic signal G. Discrete and aperiodic signal H. Discrete and periodic signal Group 2 1. Fourier representation is continuous and aperiodic 2. Fourier representation is discrete and aperiodic 3. Fourier representation is continuous and periodic 4. Fourier representation is discrete and periodic (A) (B) (C) (D)

6.72

Which of the following can be impulse response of a causal system ?

E - 3, E - 1, E - 1, E - 2,

F - 2, F - 3, F - 2, F - 1,

G - 4, G - 2, G - 3, G - 4,

H-1 H-4 H-4 H-3

A signal x (n) = sin (w0 n + f) is the input to a linear time- invariant system having a frequency response H (e jw). If the output of the system Ax (n - n0) then the most general form of +H (e jw) will be (A) - n0 w0 + b for any arbitrary real (B) - n0 w0 + 2pk for any arbitrary integer k (C) n0 w0 + 2pk for any arbitrary integer k (D) - n0 w0 f

Statement of linked answer question 6.59 and 6.60 : A sequence x (n) has non-zero values as shown in the figure.

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6.68

Let x (n) = ( 12 ) n u (n), y (n) = x2 (n) and Y (e jw) be the Fourier transform of y (n) then Y (e j0) (A) 1 4 (B) 2 (C) 4 (D) 4 3

6.69

The power in the signal s (t) = 8 cos (20p - p2 ) + 4 sin (15pt) is (A) 40

x ( n2 - 1),

6.73

The sequence y (n) = * 0,

For n even will be For n odd

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 139

The system is stable only if (A) a = 2 , b < 2 (C) a < 2 , any value of b 6.81

(B) a > 2, b > 2 (D) b < 2 , any value of a

The impulse response h [n] of a linear time invariant system is given as -2 2

n = 1, - 1

h [ n] = * 4 2 n = 2, - 2 0 otherwise If the input to the above system is the sequence e jpn/4 , then the output is (A) 4 2 e jpn/4 (B) 4 2 e-jpn/4 (C) 4e jpn/4 (D) - 4e jpn/4 6.82

Let x (t) and y (t) with Fourier transforms F (f) and Y (f) respectively be related as shown in Fig. Then Y (f) is

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6.74

6.75

The Fourier transform of y (2n) will be (A) e-j2w [cos 4w + 2 cos 2w + 2] (B) cos 2w + 2 cos w + 2 -jw (C) e [cos 2w + 2 cos w + 2] (D) e-j2w [cos 2w + 2 cos + 2] For a signal x (t) the Fourier transform is X (f). Then the inverse Fourier transform of X (3f + 2) is given by j4pt (A) 1 x` t j e j3pt (B) 1 x` t j e - 3 2 2 3 3 (C) 3x (3t) e-j4pt

(D) x (3t + 2)

2004 6.76

6.77

6.78

ONE MARK

The impulse response h [n] of a linear time-invariant system is given by h [n] = u [n + 3] + u [n - 2) - 2n [n - 7] where u [n] is the unit step sequence. The above system is (A) stable but not causal (B) stable and causal (C) causal but unstable (D) unstable and not causal

The Fourier transform of a conjugate symmetric function is always (A) imaginary (B) conjugate anti-symmetric (C) real (D) conjugate symmetric

6.80

6.83

A causal LTI system is described by the difference equation 2y [n] = ay [n - 2] - 2x [n] + bx [n - 1]

(B) - 1 X (f/2) e j2pf 2

(C) - X (f/2) e j2pf

(D) - X (f/2) e-j2pf ONE MARK

The Laplace transform of i (t) is given by 2 I (s) = s (1 + s) At t " 3, The value of i (t) tends to (A) 0 (B) 1 (C) 2 (D) 3

6.84

The Fourier series expansion of a real periodic signal with fundamental frequency f0 is given by gp (t) = cn e j2pf t . It is given n =- 3 that c3 = 3 + j5 . Then c-3 is (A) 5 + j3 (B) - 3 - j5 (C) - 5 + j3 (D) 3 - j5

/

TWO MARKS

Consider the sequence x [n] = [- 4 - j51 + j25]. The conjugate antisymmetric part of the sequence is (B) [- j2.5, 1, j2.5] (A) [- 4 - j2.5, j2, 4 - j2.5] (C) [- j2.5, j2, 0] (D) [- 4, 1, 4]

(A) - 1 X (f/2) e-jpf 2

2003

The z -transform of a system is H (z) = z -z0.2 . If the ROC is z < 0.2 , then the impulse response of the system is (A) (0.2) n u [n] (B) (0.2) n u [- n - 1] (C) - (0.2) n u [n] (D) - (0.2) n u [- n - 1]

2004 6.79

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6.85

6.86

0

Let x (t) be the input to a linear, time-invariant system. The required output is 4p (t - 2). The transfer function of the system should be (A) 4e j4pf (B) 2e-j8pf (C) 4e-j4pf (D) 2e j8pf A sequence x (n) with the z -transform X (z) = z 4 + z2 - 2z + 2 - 3z-4 is applied as an input to a linear, time-invariant system with the impulse response h (n) = 2d (n - 3) where

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

1, n = 0 d (n) = ) 0, otherwise The output at n = 4 is (A) - 6 (B) zero (C) 2 (D) - 4 2003 6.87

Page 140

(A) e f u (f) (C) e f u (- f) 6.93

TWO MARKS

Let P be linearity, Q be time-invariance, R be causality and S be stability. A discrete time system has the input-output relationship, x (n) n$1 y (n) = *0, n=0 x (n + 1) n # - 1 where x (n) is the input and y (n) is the output. The above system has the properties (A) P, S but not Q, R (B) P, Q, S but not R (C) P, Q, R, S

6.94

6.95

(D) Q, R, S but not P 6.96

The system under consideration is an RC low- filter (RC-LPF) with R = 1 kW and C = 1.0 mF. Let H (f) denote the frequency response of the RC-LPF. Let f1 be H (f1) the highest frequency such that 0 # f # f1 $ 0.95 . Then f1 H (0) (in Hz) is (A) 324.8 (B) 163.9 (C) 52.2 (D) 104.4

6.97

6.98

Let tg (f) be the group delay function of the given RC-LPF and f2 = 100 Hz. Then tg (f2) in ms, is (A) 0.717 (B) 7.17 (C) 71.7 (D) 4.505 2002

6.90

6.91

6.92

TWO MARKS

The Laplace transform of continuous - time signal x (t) is X (s) = . If the Fourier transform of this signal exists, the x (t) is (A) e2t u (t) - 2e-t u (t) (B) - e2t u (- t) + 2e-t u (t) (C) - e2t u (- t) - 2e-t u (t) (D) e2t u (- t) - 2e-t u (t)

5-s s2 - s - 2

If the impulse response of discrete - time system is h [n] =- 5n u [- n - 1], then the system function H (z) is equal to (A) - z and the system is stable z-5 (B) z and the system is stable z-5 (C) - z and the system is unstable z-5 (D) z and the system is unstable z-5 2001

Common Data For Q. 6.73 & 6.74 :

6.89

A linear phase channel with phase delay Tp and group delay Tg must have (A) Tp = Tg = constant (B) Tp \ f and Tg \ f (C) Tp = constant and Tg \ f ( f denote frequency) (D) Tp \ f and Tp = constant 2002

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6.88

(B) e-f u (f) (D) e-f u (- f)

ONE MARK

The transfer function of a system is given by H (s) = 2 1 . The s (s - 2) impulse response of the system is (B) (t * e2t) u (t) (A) (t2 * e-2t) u (t) (C) (te-2 t) u (t) (D) (te-2t) u (t) The region of convergence of the z - transform of a unit step function is (A) z > 1 (B) z < 1 (C) (Real part of z ) > 0 (D) (Real part of z ) < 0 Let d (t) denote the delta function. The value of the integral 3 d (t) cos b 3t l dt is 2 -3

#

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ONE MARK

Convolution of x (t + 5) with impulse function d (t - 7) is equal to (A) x (t - 12) (B) x (t + 12) (C) x (t - 2) (D) x (t + 2)

(A) 1 (C) 0

Which of the following cannot be the Fourier series expansion of a periodic signal? (A) x (t) = 2 cos t + 3 cos 3t (B) x (t) = 2 cos pt + 7 cos t (C) x (t) = cos t + 0.5 (D) x (t) = 2 cos 1.5pt + sin 3.5pt

6.99

1 The Fourier transform F {e-1 u (t)} is equal to . Therefore, 1 + j2pf 1 is F' 1 + j2pt 1

6.100

(B) - 1 (D) p2

If a signal f (t) has energy E , the energy of the signal f (2t) is equal to (A) 1 (B) E/2 (C) 2E (D) 4E 2001

TWO MARKS

The impulse response functions of four linear systems S1, S2, S3, S4 are given respectively by

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 141

h1 (t) = 1, h2 (t) = u (t), u (t) and h3 (t) = t+1 h 4 (t) = e-3t u (t) where u (t) is the unit step function. Which of these systems is time invariant, causal, and stable? (A) S1 (B) S2 (C) S3 (D) S4 2000 6.101

ONE MARK

L [f (t)] = s2+ 2 , s +1 t h (t) = f (t) g (t - t) dt . 0 L [h (t)] is

Given

that

L [g (t)] =

#

2 (A) s + 1 s+3

(C) 6.102

(B)

s2 + 1 + s+2 (s + 3)( s + 2) s2 + 1

s2 + 1 (s + 3) (s + 2)

and

1 s+3

6.107

(D) None of the above 2

The Fourier Transform of the signal x (t) = e-3t is of the following form, where A and B are constants : (A) Ae-B f (B) Ae-Bf (C) A + B f 2 (D) Ae-Bf 2

6.103

6.104

2000 6.105

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A system with an input x (t) and output y (t) is described by the relations : y (t) = tx (t). This system is (A) linear and time - invariant (B) linear and time varying (C) non - linear and time - invariant (D) non - linear and time - varying A linear time invariant system has an impulse response e2t, t > 0 . If the initial conditions are zero and the input is e3t , the output for t > 0 is (A) e3t - e2t (B) e5t (C) e3t + e2t (D) None of these TWO MARKS

Click to Buy www.nodia.co.in df (w) f (w0) , dw w = w w0 df (w) (C) wo , d (w) w = w f (wo)

(A) -

6.106

and and and and

n-2 n-3 n-2 n-2

(D) wo f (wo),

o

d2 f (w0) dw2

#

wo

-3

6.108

6.109

ONE MARK

The z -transform F (z) of the function f (nT) = anT is (A) z T (B) z T z-a z+a z z (C) (D) z - a-T z + a-T If [f (t)] = F (s), then [f (t - T)] is equal to (A) esT F (s) (B) e-sT F (s) F (s) F (s) (C) (D) sT 1-e 1 - e-sT A signal x (t) has a Fourier transform X (w). If x (t) is a real and odd function of t , then X (w) is (A) a real and even function of w (B) a imaginary and odd function of w (C) an imaginary and even function of w (D) a real and odd function of w 1999

6.111

TWO MARKS

The Fourier series representation of an impulse train denoted by s (t) =

3

/ d (t - nT0) is given by

n =- 3

Let u (t) be the step function. Which of the waveforms in the figure corresponds to the convolution of u (t) - u (t - 1) with u (t) - u (t - 2) ?

3 j2pnt (A) 1 / exp T0 n =- 3 T0 3 jpnt (C) 1 / exp T0 n =- 3 T0 6.112

w = wo

f (l)

1999

6.110

n-3 n-2 n-1 n-4

(B) f (wo), -

0

One period (0, T) each of two periodic waveforms W1 and W2 are shown in the figure. The magnitudes of the nth Fourier series coefficients of W1 and W2 , for n $ 1, n odd, are respectively proportional to

(A) (B) (C) (D)

A system has a phase response given by f (w), where w is the angular frequency. The phase delay and group delay at w = w0 are respectively given by

3 jpnt (B) 1 / exp T0 n =- 3 T0 3 j2pnt (D) 1 / exp T0 n =- 3 T0

The z -transform of a signal is given by

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

C (z) =

1997

1z-1 (1 - z-4) 4 (1 - z-1) 2

Its final value is (A) 1/4 (C) 1.0

6.120

(B) zero (D) infinity

1998 6.113

6.114

6.115

Page 142

ONE MARK

The function f (t) has the Fourier Transform g (w). The Fourier Transform ff (t) g (t) e = (A) 1 f (w) 2p

ONE MARK

If F (s) = 2 w 2 , then the value of Limf (t) t"3 s +w (A) cannot be determined (B) is zero (C) is unity (D) is infinite The trigonometric Fourier series of a even time function can have only (A) cosine (B) sine (C) cosine and sine (D) d.c and cosine

(C) 2pf (- w) 6.121

A periodic signal x (t) of period T0 is given by

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6.123

-3

(B) 1 f (- w) 2p (D) None of the above

The Laplace Transform of eat cos (at) is equal to (s - a) (s + a) (A) (B) 2 2 (s - a) + a (s - a) 2 + a2 1 (C) (D) None of the above (s - a) 2 1996

6.122

3

# g (t) e-jwt dt o is

ONE MARK

The trigonometric Fourier series of an even function of time does not have the (A) dc term (B) cosine (C) sine (D) odd harmonic The Fourier transform of a real valued time signal has (A) odd symmetry (B) even symmetry (C) conjugate symmetry (D) no symmetry

1, t < T1 x (t) = * 0, T1 < t < T0 2 The dc component of x (t) is (A) T1 T0 (C) 2T1 T0 6.116

6.117

The unit impulse response of a linear time invariant system is the unit step function u (t). For t > 0 , the response of the system to an excitation e-at u (t), a > 0 will be (A) ae-at (B) (1/a) (1 - e-at) (C) a (1 - e-at) (D) 1 - e-at The z-transform of the time function (A) z - 1 z

k=0

(B)

z z-1

(z - 1) 2 (D) z

A 22 + A 32 + ..... A 22 + A 32 + ..... (D) c m A1 A 12 + A 22 + A 32 + .... The Fourier transform of a function x (t) is X (f). The Fourier dX (t) transform of will be df dX (f) (B) j2pfX (f) (A) df X (f) (D) (C) jfX (f) jf (C)

6.119

3

/ d (n - k) is

z (z - 1) 2 A distorted sinusoid has the amplitudes A1, A2, A 3, .... of the fundamental, second harmonic, third harmonic,..... respectively. The total harmonic distortion is 2 2 (B) A 2 + A 3 + ..... (A) A2 + A 3 + .... A1 A1 (C)

6.118

(B) T1 2T0 (D) T0 T1

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 143

SOLUTIONS 6.1

Given, the maximum frequency of the band-limited signal fm = 5 kHz According to the Nyquist sampling theorem, the sampling frequency must be greater than the Nyquist frequency which is given as fN = 2fm = 2 # 5 = 10 kHz So, the sampling frequency fs must satisfy

Option (C) is correct. If the two systems with impulse response h1 ^ t h and h2 ^ t h are connected in cascaded configuration as shown in figure, then the overall response of the system is the convolution of the individual impulse responses.

fs $ fN fs $ 10 kHz only the option (A) doesn’t satisfy the condition therefore, 5 kHz is not a valid sampling frequency. 6.5

6.2

Option (C) is correct. Given, the input

x ^ t h = u ^t - 1h It’s Laplace transform is -s X ^s h = e s

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The impulse response of system is given

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h^t h = t u^t h Its Laplace transform is H ^s h = 12 s Hence, the overall response at the output is Y ^s h = X ^s h H ^s h -s =e3 s its inverse Laplace transform is ^t - 1h2 y^t h = u ^t - 1h 2 6.3

system stability and causality both. But, Option (C) is not true for the stable system as, S = 1 have one pole in right hand plane also. 6.6

Option (A) is correct. Given, the signal

v ^ t h = 30 sin 100t + 10 cos 300t + 6 sin ^500t + p4 h

So we have w1 = 100 rad/s w2 = 300 rad/s w3 = 500 rad/s Therefore, the respective time periods are T1 = 2p = 2p sec w1 100 T2 = 2p = 2p sec w2 300 T3 = 2p sec 500 So, the fundamental time period of the signal is LCM ^2p, 2p, 2ph L.C.M. ^T1, T2 T3h = HCF ^100, 300, 500h or, T0 = 2p 100 Hence, the fundamental frequency in rad/sec is w0 = 2p = 100 rad/s 10 6.4

Option (A) is correct.

Option (C) is correct. For a system to be casual, the R.O.C of system transfer function H ^s h which is rational should be in the right half plane and to the right of the right most pole. For the stability of LTI system. All poles of the system should lie in the left half of S -plane and no repeated pole should be on imaginary axis. Hence, options (A), (B), (D) satisfies an LTI

Option (B) is correct. The Laplace transform of unit step fun n is U ^s h = 1 s So, the O/P of the system is given as Y ^s h = b 1 lb 1 l s s = 12 s For zero initial condition, we check dy ^ t h u^t h = dt & U ^s h = SY ^s h - y ^0 h & U ^s h = s c 12 m - y ^0 h s 1 or, U ^s h = s ^y ^0 h = 0h Hence, the O/P is correct which is Y ^s h = 12 s its inverse Laplace transform is given by y ^ t h = tu ^ t h

6.7

No Option is correct. The matched filter is characterized by a frequency response that is given as H ^ f h = G * ^ f h exp ^- j2pfT h f where g^t h G^f h Now, consider a filter matched to a known signal g ^ t h. The fourier transform of the resulting matched filter output g 0 ^ t h will be G0 ^ f h = H^ f hG^ f h

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 144

ordinary, first order differential equation with forcing function x ^ t h is y ^ t h so, we can define a function relating x ^ t h and y ^ t h as below dy P + Qy + K = x ^ t h dt

= G * ^ f h G ^ f h exp ^- j2pfT h = G ^ f h 2 exp ^- j2pfT h

T is duration of g ^ t h Assume exp ^- j2pfT h = 1 So, G0 ^ f h = G_ f i 2 Since, the given Gaussian function is g ^ t h = e- pt Fourier transform of this signal will be f g ^ t h = e- pt e- pf = G ^ f h Therefore, output of the matched filter is 2 G 0 ^ f h = e- pf

where P , Q , K are constant. Taking the Laplace transform both the sides, we get ....(1) P sY ^s h - Py ^0 h + Q Y ^s h = X ^s h Now, the solutions becomes y1 ^ t h =- 2y ^ t h or, Y1 ^s h =- 2Y ^s h So, Eq. (1) changes to

2

2

2

2

6.8

Option (B) is correct. Given, the impulse response of continuous time system h ^ t h = d ^t - 1h + d ^t - 3h From the convolution property, we know x ^ t h * d ^t - t 0h = x ^t - t 0h So, for the input x ^ t h = u ^ t h (Unit step fun n ) The output of the system is obtained as

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in y^t h = u^t h * h^t h = u ^ t h * 6d ^t - 1h + d ^t - 3h@ = u ^t - 1h + u ^t - 3h at t = 2

6.9

y ^2 h = u ^2 - 1h + u ^2 - 3h =1

Option (B) is correct. Given, the differential equation d2y dy 2 + 5 dt + 6y ^ t h = x ^ t h dt Taking its Laplace transform with zero initial conditions, we have s2 Y ^s h + 5sY ^s h + 6Y ^s h = X ^s h

....(1) Now, the input signal is

1 0 <2 0 otherwise i.e., x ^ t h = u ^ t h - u ^t - 2h Taking its Laplace transform, we obtain -2s X ^s h = 1 - e s s -2s = 1-e s Substituting it in equation (1), we get 2s X ^s h Y ^s h = 2 = 21 - e s + 5s + 6 s ^s + 5s + 6h x^t h = *

= 6.10

1 - e-2s s ^s + 2h^s + 3h

Option (D) is correct. The solution of a system described by a linear, constant coefficient,

P sY1 ^s h - P y1 ^0 h + Q Y1 ^s h = X1 ^s h or, ....(2) - 2PSY ^s h - P y1 ^0 h - 2QY1 ^s h = X1 ^s h Comparing Eq. (1) and (2), we conclude that X1 ^s h =- 2X ^s h y1 ^0 h =- 2y ^0 h Which makes the two equations to be same. Hence, we require to change the initial condition to - 2y ^0 h and the forcing equation to - 2x ^ t h 6.11

Option (A) is correct. Given, the DFT of vector 8a b c dB as D.F.T. %8a b c dB/ = 8a b g dB Also, we have V R Sa b c d W Sd a b c W ...(1) 8p q r sB = 8a b c dBSc d a b W W S Sb c d aW X T For matrix circular convolution, we know Rh h h VRx V S 0 2 1WS 0W x 6n@ * h 6n@ = Sh1 h 0 h2WSx1W SSh h h WWSSx WW 2 1 0 1 T XT X where "x 0, x1, x2, are three point signals for x 6n@ and similarly for h 6n@, h 0 , h1 and h2 are three point signals. Comparing this transformation to Eq(1), we get VT R Sa d c W Sb a d W 6p q r s@ = Sc b aW 8a b c dB W S Sd c b W X T = 6a b c d @T * 6a b c d @T

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Now, we know that So,

R V R V Sa W Sa W Sb W Sb W =S W * S W Sc W Sc W Sd W Sd W T X T X x1 6n@ * x2 6n@ = X1DFT 6k @ X2, DFT 6k @ R V Sa W Sb W Sc W S W Sd W T X

*

R V R V Sa W SaW Sb W SbW Sc W = Sg W S W S W Sd W Sd W T X T X

*

R V SaW SbW Sg W S W Sd W T X

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 145

= 9a2 b 2 g2 d 2C 6.12

Option (D) is correct. Using s -domain differentiation property of Laplace transform. If

f (t)

F (s) dF (s) L tf (t) ds 2s + 1 = L [tf (t)] = - d ; 2 1 ds s + s + 1E (s2 + s + 1) 2

So, 6.13

L

h (0) = h1 (0) + h2 (0) = 1 + 1 = 1 2 2

Option (C) is correct. n n x [ n] = b 1 l - b 1 l u [ n ] 3 2 n -n n x [n] = b 1 l u [n] + b 1 l u [- n - 1] - b 1 l u (n) 3 3 2 Taking z -transform X 6z @ 3

3

So, inverse Fourier transform of H (jw) h (t) = h1 (t) + h2 (t)

6.16

y [n] = h [n] * g [n] = For causal sequence,

3

1 n -n 1 -n -n 1 n -n = b 3 l z u [n ] + b 3 l z u [- n - 1] b 2 l z u [ n] n =- 3 n =- 3 n =- 3

/

/

/

-1

3

=

3

/

3

/

/ b 31z l + / b 13 z l n

n=0

m

-

m=1

/ b 21z l

n

Taking m =- n

n=0

1 44 2 44 3 II

14 42 4 43 I

3

14 42 4 43 III

1 < 1 or z > 1 3 3z Series II converges if 1 z < 1 or z < 3 3 Series III converges if 1 < 1 or z > 1 2z 2 Region of convergence of X (z) will be intersection of above three So, ROC : 1 < z < 3 2 Option (D) is correct. y (t) =

Click to Buy www.nodia.co.in For n = 0 ,

For n = 1,

# x (t) cos (3t) dt -3

# d (t) cos (3t) dt -3

y (t, t 0) = Delayed output,

From equation (i),

# d (t - t ) cos (3t) dt 0

-3

y (t - t 0) = u (t - t 0) y (t, t 0) ! y (t - t 0)

= u (t) cos (3t 0)

#

As t " 3, y (t) " 3 (unbounded) 6.15

So, 6.17

System is not time invariant.

Stability : Consider a bounded input x (t) = cos 3t t t 1 - cos 6t = 1 y (t) = cos2 3t = 2 2 -3 -3

#

y [1] y [1] 1 2

= u (t) cos (0) = u (t)

For a delayed input (t - t 0) output is t

t

-3

-3

y [0] = h [0] g [0] + h [1] g [- 1] + ........... y [0] = h [0] g [0] g [- 1] = g [- 2] = ....0 ...(i) y [0] = h [0] g [0] = h [1] g [1] + h [1] g [0] + h [1] g [- 1] + .... = h [1] g [1] + h [1] g [0] 1 = 1 g [1] + 1 g [0] h [1] = b 1 l = 1 2 2 2 2 1 = g [1] + g [0] g [1] = 1 - g [0] y [0] 1 = =1 g [0] = h [0] 1 g [1] = 1 - 1 = 0

Option (A) is correct. d2 y dy We have 100 2 - 20 + y = x (t) dt dt Applying Laplace transform we get 100s2 Y (s) - 20sY (s) + Y (s) = X (s) Y (s) 1 or = H (s) = X (s) 100s2 - 20s + 1 1/100 A = 2 = s - (1/5) s + 1/100 s2 + 2xwn s + w2 Here wn = 1/10 and 2xwn =- 1/5 giving x =- 1 Roots are s = 1/10, 1/10 which lie on Right side of s plane thus unstable.

# 1dt - 12 # cos 6t dt t

/ h [n] g [n - k]

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Time Invariance : Let, x (t) = d (t) y (t) =

3

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t

t

/ h [n] g [n - k] k =- 3

y [n] = h [n] g [n] + h [n] g [n - 1] + h [n] g [n - 2] + .....

Series I converges if

6.14

y [n] =

3

k=0

3

1 n -n 1 -n -n 1 n -n = b3l z + b3l z b2l z n=0 n =- 3 n=0

/

Option (A) is correct. Convolution sum is defined as

System is not stable.

Option (C) is correct. (2 cos w) (sin 2w) H (jw) = = sin 3w + sin w w w w We know that inverse Fourier transform of sin c function is a rectangular function. 6.18

6.19

Option (C) is correct. For an even function Fourier series contains dc term and cosine term (even and odd harmonics). Option (B) is correct. Function h (n) = an u (n) stable if a < 1 and Unstable if a H 1 We We have h (n) = 2n u (n - 2);

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 146

Here a = 2 therefore h (n) is unstable and since h (n) = 0 for n < 0 Therefore h (n) will be causal. So h (n) is causal and not stable. 6.20

Constant term is negative. 6.25

Option (A) is correct.

6.21

6.26

Option (D) is correct. We have x (t) = exp (- 2t) m (t) + s (t - 6) and h (t) = u (t) Taking Laplace Transform we get X (s) = b 1 + e-6s l and H (s) = 1 s+2 s Now

or Thus

Y (s) = H (s) X (s) -6s 1 = 1 : 1 + e-6sD = +e s s+2 s s (s + 2)

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or Now

6.27

6.28

6.23

6.24

x (t) = Ao + Where,

3

/ [An cos nwt + Bn sin nwt]

n=1

Ao 1 # x (t) dt T0 T

T0 "fundamental period

h [n] = d [n - 3]

Option (D) is correct. For an N-point FET algorithm butterfly operates on one pair of samples and involves two complex addition and one complex multiplication. Option (D) is correct. 3s + 1 f (t) = L - 1 ; 3 s + 4s 2 + (k - 3) s E lim f (t) = 1

t"3

By final value theorem lim f (t) = lim sF (s) = 1

t"3

-1

Option (C) is correct. For a function x (t) trigonometric fourier series is

X (z) = 5z + 4z + 3 x [n] = 5d [n + 2] + 4d [n - 1] + 3d [n]

=z

and

Y (z) = H (z) = z-1 X (z)

Option (B) is correct. For 8 point DFT, x* [1] = x [7]; x* [2] = x [6]; x* [3] = x [5] and it is conjugate symmetric about x [4], x [6] = 0 ; x [7] = 1 + j3

ad [n ! a]

-1

-3

We have

y (n) = x (n - 1) Y (z) = z-1 X (z)

H1 (z) H2 (z) = z 1 - 0.4z-1 -1 c 1 - 0.6z-1 m H2 (z) = z z-1 (1 - 0.6z-1) H2 (z) = (1 - 0.4z-1)

-2

= z :z or,

6.29

s"0

or

s. (3s + 1) =1 s + 4s2 + (k - 3) s s (3s + 1) lim 2 =1 s " 0 s [s + 4s + (k - 3)] 1 =1 k-3

or

k =4

Option (B) is correct. or

2

H (z ) = H1 (z ) : H2 (z )

or 6.22

Inverse Z - transform

Option (C) is correct. We have h1 [n] = d [n - 1] or H1 [Z ] = Z - 1 and h 2 [n] = d [n - 2] or H2 (Z ) = Z - 2 Response of cascaded system -1

-6s 1 +e Y (s) = 1 2s 2 (s + 2) s

y (t) = 0.5 [1 - exp (- 2t)] u (t) + u (t - 6)

aZ ! a

We know that Given that Inverse z-transform

Impulse response = d (step response) dt = d (1 - e- at) dt = 0 + ae- at = ae- at

Option (A) is correct.

lim s"0

3

Option (B) is correct. System is described as d 2 y (t) dt (t) dx (t) + 4x (t) +4 + 3y (t) = 2 2 dt dt dt Taking Laplace transform on both side of given equation s 2 Y (s) + 4sY (s) + 3Y (s) = 2sX (s) + 4X (s) (s 2 + 4s + 3) Y (s) = 2 (s + 2) X (s) s Transfer function of the system

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0

An = 2 # x (t) cos nwt dt T0 T Bn = 2 # x (t) sin nwt dt T0 T For an even function x (t), Bn = 0 Since given function is even function so coefficient Bn = 0 , only cosine and constant are present in its fourier series representation 3T/4 Constant term A0 = 1 # x (t) dt T -T/4 T/4 3T/4 = 1 : # Adt + # - 2AdtD T -T/4 T/4 = 1 :TA - 2AT D =- A 2 2 T 2 and

H (s) =

0

0

Input or,

2 (s + 2) Y (s) 2 (s + 2) = = (s + 3) (s + 1) X (s) s 2 + 4s + 3

x (t) = e-2t u (t) X (s) = 1 (s + 2)

Y (s) = H (s) : X (s) 2 (s + 2) Y (s) = : 1 (s + 3) (s + 1) (s + 2) By Partial fraction Y (s) = 1 - 1 s+1 s+3 Taking inverse Laplace transform Output

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 147

y (t) = (e-t - e-3t) u (t) 6.30

Option (C) is correct. We have 2 - 34 z - 1 1 - 34 z - 1 + 18 z - 2 By partial fraction H (z ) can be written as 1 1 H (z ) = -1 1 -1 + ^1 - 2 z h ^1 - 14 z h For ROC : z > 1/2 H (z) =

It has the finite magnitude values. So it is a finite impulse response filter. Thus S2 is true but it is not a low filter. So S1 is false. 6.36

n n h [n] = b 1 l u [n] + b 1 l u [n], n > 0 2 4

1 = an u [n], z > a 1 - z -1 Thus system is causal. Since ROC of H (z ) includes unit circle, so it is stable also. Hence S1 is True For ROC : z < 1 4 n n h [n] =-b 1 l u [- n - 1] + b 1 l u (n), z > 1 , z < 1 2 2 4 4 System is not causal. ROC of H (z ) does not include unity circle, so it is not stable and S 3 is True 6.31

6.32

Option (A) is correct. The Fourier series of a real periodic function has only cosine if it is even and sine if it is odd.

6.37

Option (B) is correct. Here h (t) ! 0 for t < 0 . Thus system is non causal. Again any bounded input x (t) gives bounded output y (t). Thus it is BIBO stable. Here we can conclude that option (B) is correct. Option (D) is correct. We have x [n] = {1, 0, 2, 3) and N = 4 X [k ] =

Click to Buy www.nodia.co.in For N = 4 ,

n =- 1

n

n=0 n=3

Now

n =- 3 n =- 1

1 z-1 < 1 " 3 1 z-1 > 1 " 2

First term gives Second term gives

#

Here function is defined for 0 < t < t , Thus t F (s) L f (t) s 0

#

6.35

Option (A) is correct. We have h (2) = 1, h (3) =- 1 otherwise h (k) = 0 . The diagram of response is as follows :

3

/ x [ n] 3

/ x [n] e

-jpn/2

n=0

= x [0] + x [1] e-jp/2 + x [2] e-jp + x [3] e-jp3/2 = 1 + 0 - 2 + j3 =- 1 + j3 X [2] =

3

/ x [n] e

-jpn

n=0

= x [0] + x [1] e-jp + x [2] e-j2p + x [3] e-jp3 = 1+0+2-3 = 0

1< z 3 1> z 2

#

X [0] =

x [1] =

n

-

k = 0, 1,... 3

= 1+0+2+3 = 6

1 n -n b2l z

Option (B) is correct. By property of unilateral Laplace transform t F (s) 1 0 L f (t) dt f (t) dt + s s -3 -3

-j2pnk/4

n=0

X [3] =

3

/ x [n] e

-j3pn/2

n=0

Thus its ROC is the common ROC of both . that is 1< z <1 3 2 6.34

3

/ x [n] e

= x [0] + x [1] + x [2] + x [3]

= / b 1 z-1 l - / b 1 z-1 l 3 2 n=0 n =- 3 n

X [k ] =

n=0

Taking z transform we have

/ b 13 l z-n - /

k = 0, 1...N - 1

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The function has a DC term and a cosine function. The frequency of cosine is w = 2 = 2pf " f = 1 Hz p The given function has frequency component at 0 and 1 Hz. p Option (A) is correct. n n x [n] = b 1 l u (n) - b 1 l u (- n - 1) 3 2 n=3

-j2pnk/N

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Option (B) is correct. Given function is

X (z) =

/ x [n] e

n=0

f (t) = sin2 t + cos 2t = 1 - cos 2t + cos 2t = 1 + 1 cos 2t 2 2 2

6.33

N-1

= x [0] + x [1] e-j3p/2 + x [2] e-j3p + x [3] e-j9p/2 = 1 + 0 - 2 - j3 =- 1 - j3 Thus [6, - 1 + j3, 0, - 1 - j3] 6.38 6.39

Option (A) is correct. Option (C) is correct. The output of causal system depends only on present and past states only. In option (A) y (0) depends on x (- 2) and x (4). In option (B) y (0) depends on x (1). In option (C) y (0) depends on x (- 1). In option (D) y (0) depends on x (5). Thus only in option (C) the value of y (t) at t = 0 depends on x (- 1) past value. In all other option present value depends on future value.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 6.40

6.41

Page 148

The y (t) = t x (t) is not linear, thus option (B) is wrong and (a) is correct. We can see that R1: y (t) = t2 x (t) Linear and time variant. R2: y (t) = t x (t) Non linear and time variant. R3: y (t) = x (t) Non linear and time invariant R4: y (t) = x (t - 5) Linear and time invariant

Option (D) is correct. We have h (t) = eat u (t) + e bt u (- t) This system is stable only when bounded input has bounded output For stability at < 0 for t > 0 that implies a < 0 and bt > 0 for t > 0 that implies b > 0 . Thus, a is negative and b is positive. Option (C) is correct. K (s + 1) , and R (s) = 1 G (s) = (s + 2)( s + 4) s K (s + 1) C (s) = G (s) R (s) = s (s + 2)( s + 4) K = K + - 3K 8s 4 (s + 2) 8 (s + 4) c (t) = K :1 + 1 e-2t - 3 e-4tD u (t) 8 4 8

Thus

6.45

Given :

Hence 6.46

1 x (t) = ) 0 Fourier transform is

#- 33e-jwt x (t) dt

=

for - 1 # t # + 1 otherwise

x (n) = 5e 6.47

6.48

Option (C) is correct.

= 6.49

#- 33 h (t) e-jwt dt #0 3e-2t e-jwt dt = #0 3e-(2 + jw)t dt

=

1 (2 + jw)

Option (D) is correct. H (jw) =

1 (2 + jw)

The phase response at w = 2 rad/sec is +H (jw) =- tan-1 w =- tan-1 2 =- p =- 0.25p 2 2 4

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 Magnitude response at w = 2 rad/sec is 1 = 1 H (jw) = 22 + w2 2 2 Input is x (t) = 2 cos (2t) Output is = 1 # 2 cos (2t - 0.25p) 2 2 = 1 cos [2t - 0.25p] 2

3

/ x (k) h (n - k)

3

/ x (k) h (2 - k) k =- 3

Option (B) is correct. Mode function are not linear. Thus y (t) = x (t) is not linear but this functions is time invariant. Option (A) and (B) may be correct.

= 5e-0.05n For t > 0

Option (C) is correct. Since x (n) = 5e-0.05n u (n) is a causal signal Its z transform is 1 5z = X (z) = 5 : 1 - e-0.05 z-1 D z - e-0.05 Its ROC is e-0.05 z-1 > 1 " z > e-0.05

H (jw) =

= x (0) h (2 - 0) + x (1) h (2 - 1) + x (2) h (2 - 2) = h (2) + 0 + h (0) = 1 + 2 = 3 There are 5 non zero sample in output sequence and the value of y [2] is 3.

-n 2 # 10

h (t) = e-2t u (t)

k =- 3

6.44

I = I (0+) e-t/RC I (0+) = V = 5 = 25mA R 200k

t 2

Option (D) is correct. Given h (n) = [1, - 1, 2] x (n) = [1, 0, 1] y (n) = x (n)* h (n) The length of y [n] is = L1 + L2 - 1 = 3 + 3 - 1 = 5

y (2) =

X (k) 2

= VR # R = 5e- V Here the voltages across the resistor is input to sampler at frequency of 10 Hz. Thus

1

y (n) = x (n) * h (n) =

DFT

t 2

This is zero at w = p and w = 2p 6.43

y (n) = rxx (n)

I = 25e- m A

#-1 e-jwt 1dt

= 1 [e-jwt]-11 - jw = 1 (e-jw - e jw) = 1 (- 2j sin w) - jw - jw = 2 sin w w

r=0

RC = 200k # 10m = 2 sec

Option (A) is correct. We have

N-1

/x (r) x (n + r)

Option (B) is correct. Current through resistor (i.e. capacitor) is Here,

h (t) = L-1 G (s) = (- 4e-2t + 12e-4t) u (t) 6.42

y (n) = 1 N

It is Auto correlation.

At steady-state , c (3) = 1 K = 1 or K = 8 Thus 8 8 (s + 1) Then, = 12 - 4 G (s) = (s + 4) (s + 2) (s + 2)( s + 4)

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Option (A) is correct.

6.50

Option (D) is correct. Y (s) =

1 s (s - 1) Final value theorem is applicable only when all poles of system lies in left half of S -plane. Here s = 1 is right s -plane pole. Thus it is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 149

a t = 0 . The area of dirac delta function is unity.

unbounded. 6.51

Option (A) is correct.

6.58

-t

x (t) = e u (t) Taking Fourier transform X (jw) = 1 1 + jw X (jw) = 1 2 1+w Magnitude at 3dB frequency is

1 2

1 = 1 2 1 + w2 w = 1 rad f = 1 Hz 2p

Thus or or 6.52

6.59

Option (B) is correct. For discrete time Fourier transform (DTFT) when N " 3 p x [ n] = 1 X (e jw) e jwn dw 2p - p

6.60

Option (D) is correct. The ROC of addition or subtraction of two functions x1 (n) and x2 (n) is R1 + R2 . We have been given ROC of addition of two function and has been asked ROC of subtraction of two function. It will be same. Option (A) is correct. As we have x (t) = sin t , Now H (s) = 1 s+1 or H (jw) = 1 = 1 jw + 1 j+1 or H (jw) = 1 + - 45c 2 Thus y (t) = 1 sin (t - p4 ) 2 Option (C) is correct. F (s) =

#

Putting n = 0 we get = 1 2p

#

p

-p

x [0] = 1 2p X (e jw) dw

or 6.53

#

p

-p

#

p

-p

X (e jw) e jw0 dw

Available Only at NODIA Online Store jw

X (e ) dw = 2px [0] = 2p # 5 = 10p

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Option (B) is correct. 0.5 1 - 2z-1 Since ROC includes unit circle, it is left handed system x (n) =- (0.5) (2) -n u (- n - 1) x (0) = 0 If we apply initial value theorem x (0) = lim X (z) = lim 0.5 -1 = 0.5 z"3 z " 31 - 2z That is wrong because here initial value theorem is not applicable because signal x (n) is defined for n < 0 .

6.55

L-1 F (s) = sin wo t f (t) = sin wo t Thus the final value is - 1 # f (3) # 1 6.61

Let Now 6.62

Option (A) is correct. A Hilbert transformer is a non-linear system.

H (s)

6.57

Step response

Y (s)

or

Y (s)

or

y (t)

x (n) = d (n) y (n) = sin 0 = 0 (bounded)

BIBO stable

Option (B) is correct. c (t) = 1 - e-2t Taking Laplace transform C (s) 2 = C (s) = #s = 2 s (s + 2) s+2 U (s)

Option (B) is correct. 5 1 + j10pf 5 = 5 = = 1 1 + 5s 5^s + 15 h s + =1 a 1 s ^s + 5 h =1 11 =5- 51 s ^s + 5 h s s+ 5 -t/5 = 5 (1 - e ) u (t)

Option (C) is correct. y (n) = b sin 5 pn l x (n) 6

H (f) =

6.56

w0 s + w2 2

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X (z) =

6.54

thus w = 1

6.63

1 s+1 L x (t) = u (t) X (s) = 1 s Y (s) = H (s) X (s) = 1 # 1 = 1 - 1 s+1 s s s+1 h (t) = e-t

1 5

Option (A) is correct. F x (t) X (jw) Using scaling we have F 1 X jw x (5t) 5 c 5 m Using shifting property we get F 1 X jw e- j35w x ;5 bt - 3 lE 5 5 b5l Option (D) is correct. Dirac delta function d (t) is defined at t = 0 and it has infinite value

Option (C) is correct. L

H (s) =

y (t) = u (t) - e-t In steady state i.e. t " 3, y (3) = 1 6.64

6.65

Option (C) is correct. Fourier series is defined for periodic function and constant. 3 sin (25t) is a periodic function. 4 cos (20t + 3) + 2 sin (710t) is sum of two periodic function and also a periodic function. e- t sin (25t) is not a periodic function, so FS can’t be defined for it. 1 is constant Option (A) is correct.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

g (t) + g (- t) 2 g (t) - g (- t) odd{g (t)} = 2

6.70

Ev{g (t)} =

Here Thus

Page 150

g (t) = u (t) u (t) + u (- t) = ue (t) = 2 u (t) - u (- t) = uo (t) = 2

Option (A) is correct. y (t) = 0.5x (t - td + T) + x (t - td ) + 0.5x (t - td - T) Taking Fourier transform we have Y (w)

1 2 x (t) 2

= 0.5e

-jw (- td + T)

or

X (w) + e-jwt X (w) + 0.5e-jw (- t - T) X (w) Y (w) = e-jwt [0.5e jwT + 1 + 0.5e-jwT ] X (w) d

d

d

= e-jwt [0.5 (e jwT + e-jwT ) + 1] d

6.66

Option (C) is correct. Here x1 (n) = ` 5 jn u (n) 6 1 X1 (z) = 1 - ^ 65 z-1h

= e-jwt [cos wT + 1] d

6.71

x2 (n) =-` 6 jn u (- n - 1) 5 1 X1 (z) = 1 1 - ^ 65 z-1h

ROC : R2 " z < 6 5 Thus ROC of x1 (n) + x2 (n) is R1 + R2 which is 5 < z < 6 6 5

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in 6.67

6.68

6.72

Option (C) is correct. For continuous and aperiodic signal Fourier representation is continuous and aperiodic. For continuous and periodic signal Fourier representation is discrete and aperiodic. For discrete and aperiodic signal Fourier representation is continuous and periodic. For discrete and periodic signal Fourier representation is discrete and periodic. Option (B) is correct.

Y (e jw) = Ae-jw n X (e jw) Y (e jw) or = Ae-jw n H (e jw) = X (e jw) Thus +H (e jw) =- wo no For LTI discrete time system phase and frequency of H (e jw) are periodic with period 2p. So in general form o

n x (n) = b 1 l u (n) 2 2n

Y (e j0) =

n

n=3

/ ` 14 j

n=0

or

Y (e j0) =

1 1-

1 4

=

n=3

...(1)

/ b 14 l e-jwn n

n=0

n =- 3

or

6.73

2 n n y (n) = ;b 1 l E u (n) = b 1 l u (n) 2 4

/ y (n) e-jwn

1 3 4 = 1 +b1l +b1l+b1l +b1l 4 4 4 4

=4 3

n=2 n = 3, n=4 n = 5,

Option (A) is correct.

s (t) = 8 cos ` p - 20pt j + 4 sin 15pt 2

= 8 sin 20pt + 4 sin 15pt Here A1 = 8 and A2 = 4 . Thus power is 2 2 2 2 P = A1 + A2 = 8 + 4 = 40 2 2 2 2

Option (A) is correct. From x (n) = [ 12 , 1, 2, 1, 1, 12 ] y (n) = x ^ n2 - 1h, n even = 0 , for n odd n =- 2 , y (- 2) = x ( -22 - 1) = x (- 2) = 12 n =- 1, y (- 1) = 0 n = 0, y (0) = x ( 20 - 1) = x (- 1) = 1 n = 1, y (1) = 0

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Alternative : Taking z transform of (1) we get 1 Y (z) = 1 - 14 z-1 Substituting z = e jw we have 1 Y (e jw) = 1 - 14 e-jw Y (e j0) = 1 1 = 4 3 1- 4 6.69

o

q (w) =- no wo + 2pk

y (n) = x2 (n) = b 1 l u2 (n) 2

n=3

o

o

Option (D) is correct.

Y (e jw) =

d

y (n) = Ax (n - no) Taking Fourier transform

Option (D) is correct. For causal system h (t) = 0 for t # 0 . Only (D) satisfy this condition.

or

Y (w) = e-jwt (cos wT + 1) X (w)

H (w) =

or

ROC : R1 " z > 5 6

6.74

y (2) y (3) y (4) y (5) y (6)

= x ( 22 - 1) = x (0) = 2 =0 = x ( 24 - 1) = x (1) = 1 =0 = x ( 26 - 1) = x (2) = 12

n=6 Hence y (n) = 1 d (n + 2) + d (n) + 2d (n - 2) + d (n - 4) + 1 d (n - 6) 2 2 Option (C) is correct. Here y (n) is scaled and shifted version of x (n) and again y (2n) is scaled version of y (n) giving

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 151

z (n) = y (2n) = x (n - 1) = 1 d (n + 1) + d (n) + 2d (n - 1) + d (n - 2) + 1 d (n - 3) 2 2 Taking Fourier transform. Z (e jw) = 1 e jw + 1 + 2e-jw + e-2jw + 1 e-3jw 2 2 = e-jw b 1 e2jw + e jw + 2 + e-jw + 1 e-2jw l 2 2 =e

6.80

Option (B) is correct. F

x (at)

F

xb 1 f l 3

F

X (f)

Using scaling we have

Thus

1 X f a ca m 3X (3f)

1e 3

Thus

1 e-jp 3 6.76

x (t) = X (f + f0) F X (3f + 2) xb 1 t l 3 F t xb 1 t l 3X (3 (f + 23 )) 3 F t xb 1 t l X [3 (f + 23 )] 3

-j 4 pt 3 2 3

4 3

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Option (A) is correct. 3

/

A system is stable if

a <2

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-j2pf0 t

e-j2p

-

Option (C) is correct. We have 2y (n) = ay (n - 2) - 2x (n) + bx (n - 1) Taking z transform we get

or

Using shifting property we get e

4 - j 25 ]

2j

2Y (z) = aY (z) z-2 - 2X (z) + bX (z) z-1 Y (z) bz-1 - 2 or ...(i) =c m X (z) 2 - az-2 z ( b - z) or H (z) = 22 a (z - 2 ) It has poles at ! a/2 and zero at 0 and b/2 . For a stable system poles must lie inside the unit circle of z plane. Thus a <1 2

e2jw + e-2jw + e jw + 2 + e-jw b l 2

x (t)

x (n) - x* (- n) 2

= [- 4 - j 25 ,

Z (e jw) = e-jw [cos 2w + 2 cos w + 2]

or 6.75

-jw

xcas (n) =

h (n) < 3. The plot of given h (n) is

n =- 3

But zero can lie anywhere in plane. Thus, b can be of any value. 6.81

Option (D) is correct. x (n) = e jpn/4 h (n)

We have and

= 4 2 d (n + 2) - 2 2 d (n + 1) - 2 2 d (n - 1) + 4 2 d (n - 2) y (n) = x (n)* h (n)

Now 3

/

Thus

6

/

h (n) =

= h (n)

z H (z) = z - 0.2

z < 0.2

6.79

p 4

1 1 - az-1 h [n] =- (0.2) n u [- n - 1]

Option (A) is correct. We have x (n) = [- 4 - j5, 1 + 2j,

4]

1 - 2j, - 4 + j5] -

+ ej

p 4

(n + 1)

- 2 2 ej

@ - 2 2 6e j

(n - 2)

p 4

p 4

(n - 1)

(n + 1)

+ 4 2 ej

+ ej

p 4

p 4

(n - 2)

@

(n - 1)

p 2

p 2

p 2

p 4

p 4

r 4

6.82

Option (B) is correct. From given graph the relation in x (t) and y (t) is y (t) =- x [2 (t + 1)] x (t)

F

x (at)

F

x (2t)

F

X (f)

Using scaling we have

x *( n) = [- 4 + j5, 1 - 2j, 4] x *( - n) = [4,

(n + 2)

p 4

p 4

-

- 2 2 ej

= 4 2 e j n 6e j + e-j @ - 2 2 e j n 6e j + e-j @ = 4 2 e j n [0] - 2 2 e j n [2 cos p4 ] or y (n) =- 4e j n

z

Option (C) is correct. The Fourier transform of a conjugate symmetrical function is always real.

(n + 2)

p 4

p 4

- an u [- n - 1] * Thus

y (n) = x (n + 2) h (- 2) + x (n + 1) h (- 1) + x (n - 1) h (1) + x (n - 2) h (2)

= 4 2 6e j

Option (D) is correct. We know that

6.78

= 4 2 ej

/ x (n - k) h (k) k =- 2

or

= 1+1+1+1+2+2+2+2+2 = 15 < 3 Hence system is stable but h (n) ! 0 for n < 0 . Thus it is not causal. 6.77

/

2

x (n - k) h (k) =

k =- 3

n =- 3

n =- 3

3

Thus

Using shifting property we get

1 X f a ca m 1X f 2 c2m

p 4

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 152

x (t - t0) = e-j2pft X (f) 0

x [2 (t + 1)]

F

- x [2 (t + 1)]

F

Thus

6.83

or or or

Option (C) is correct. From the Final value theorem we have lim i (t) = lim sI (s) = lim s

t"3 6.84

Now

j2pf f f e-j2pf (- 1) 1 X b l = e X b l 2 2 2 2 j2p f f -e Xc m 2 2

s"0

s"0

2 = lim 2 =2 s (1 + s) s " 0 (1 + s)

or

Option (D) is correct. Here C3 = 3 + j5 For real periodic signal

or or

Ck*

Thus 6.85

C-k = C-3 = Ck = 3 - j5

or Thus

Option (C) is correct. 6.89

y (t) = 4x (t - 2) Taking Fourier transform we get Y (e j2pf ) = 4e-j2pf2 X (e j2pf )

Thus 6.86

Y (e j2pf ) = 4e-4jpf X (e j2pf ) H (e j2pf ) = 4e-4jpf

Option (B) is correct. We have h (n) = 3d (n - 3) or Taking z transform H (z) = 2z-3 4 2 -4 X (z) = z + z - 2z + 2 - 3z Now Y (z) = H (z) X (z) = 2z-3 (z 4 + z2 - 2z + 2 - 3z-4) = 2 (z + z-1 - 2z-2 + 2z-3 - 3z-7) Taking inverse z transform we have

6.88

Option (C) is correct. The frequency response of RC-LPF is 1 H (f) = 1 + j2pfRC

1 1 + jwRC

q (w) =- tan-1 wRC dq (w) RC tg == dw 1 + w2 R2 C2 10-3 = = 0.717 ms 2 1 + 4p # 10 4 # 10-6 6.90

6.91

Option (C) is correct. If x (t)* h (t) = g (t) Then x (t - t1)* h (t - t2) = y (t - t1 - t2) Thus x (t + 5)* d (t - 7) = x (t + 5 - 7) = x (t - 2) Option (B) is correct. In option (B) the given function is not periodic and does not satisfy Dirichlet condition. So it cant be expansion in Fourier series. x (t) = 2 cos pt + 7 cos t T1 = 2p = 2 w T2 = 2p = 2p 1 T1 = 1 = irrational T2 p

6.92

Option (C) is correct. From the duality property of fourier transform we have If Then

x (t)

FT

X (f)

X (t)

FT

x (- f)

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Option (A) is correct. System is non causal because output depends on future value For n # 1 y (- 1) = x (- 1 + 1) = x (0) Time varying y (n - n0) = x (n - n0 + 1) Depends on Future y (n) = x (n + 1) i.e. None causal y (1) = x (2) For bounded input, system has bounded output. So it is stable. y (n) = x (n) for n $ 1 = 0 for n = 0 = x (x + 1) for n #- 1 So system is linear.

f1 max

f1 # 52.2 Hz = 52.2 Hz

Option (A) is correct.

Time Shifting property

y (n) = 2[ d (n + 1) + d (n - 1) - 2d (n - 2) + 2d (n - 3) - 3d (n - 7)] At n = 4 , y (4) = 0 6.87

1 + 4p2 f12 R2 C2 # 1.108 4p2 f12 R2 C2 # 0.108 2pf1 RC # 0.329 f1 # 0.329 2pRC f1 # 0.329 2pRC f1 # 0.329 2p1k # 1m

H (w) =

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H (0) = 1 H (f1) 1 = $ 0.95 H (0) 1 + 4p2 f12 R2 C2

Therefore if Then 6.93

e-t u (t) 1 1 + j2pt

FT

1 1 + j2pf

FT

e f u (- f)

Option (A) is correct.

and Thus

q (w) =- wt0 - q (w) = t0 tp = w dq (w) = t0 tg =dw tp = tg = t0 = constant

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 6.94

Option (*) is correct. 5-s = -2 + 1 X (s) = 2 5 - s = s+1 s-2 ( s 1 )( s 2 ) + s -s-2 Here three ROC may be possible. Re (s) < - 1 Re (s) rel="nofollow"> 2 - 1 < Re (s) < 2 Since its Fourier transform exits, only - 1 < Re (s) < 2 include imaginary axis. so this ROC is possible. For this ROC the inverse Laplace transform is

Page 153

and

Option (B) is correct. For left sided sequence we have z

Thus

- 5n u (- n - 1)

z

or

- 5n u (- n - 1)

z

1 1 - az-1 1 1 - 5z-1 z z-5

6.100

where z < 5

Since ROC is z < 5 and it include unit circle, system is stable. Alternative : h (n) =- 5n u (- n - 1) 3

/ h (n) z

H (z) =

-n

-1

/-5 z

=

n =- 3

n -n

-1

/ (5z

=-

n =- 3

-3

/ (5z

-1 -m

)

= 1-

n =- 1

3

/ (5

-1

6.96

)

5-1 z < 1 or z < 5

Thus 6.102

6.103

x1 (t)* x2 (t)

3

/ x (n) .z

-n

=

3

/ 1.z

-n

n=0

n =- 3

=

3

/ (z

-1 n

)

n=0

n=0

Option (A) is correct.

6.104

# d (t) = 1 3

6.99

# d (t) x (t) = # x (0) d (t) dt 3

-3

-3

Thus system is linear

Option (A) is correct. We have

h (t) = e2t

LS

and

x (t) = e3t

LS

-3

Let x (t) = cos ( 23 t), then x (0) = 1 3

Option (B) is correct. Let x (t) = ax1 (t) + bx2 (t) ay1 (t) = atx1 (t) by2 (t) = btx2 (t) Adding above both equation we have

y (t - t0) = (t - t0) x (t - t0) which is not equal. Thus system is time varying.

and this is possible when z-1 < 1. Thus ROC is z-1 < 1 or z >1 We know that d (t) x (t) = x (0) d (t) and

p2 f2 a

FT

yd (d) = tx (t - t0) If output is delayed then we have

3

/ (z-1) n < 3

Now

Option (B) is correct. Since normalized Gaussion function have Gaussion FT p eThus e-at a

ay1 (t) + by2 (t) = atx1 (t) + btx2 (t) = t [ax1 (t) + bx2 (t)] = tx (t) or ay1 (t) + by2 (t) = y (t) If input is delayed then we have

H (z) is convergent if

6.98

f (t)* g (t) = h (t) H (s) = F (s) # G (s) s2 + 1 = 1 H (s) = s2+ 2 # + + ( s 2 )( s 3 ) s+3 s +1

2

Option (A) is correct. We have h (n) = u (n) H (z) =

3

L

L 1 # 1 (t * e2t) u (t) 2 s 2 s Here we have used property that convolution in time domain is multiplication in s - domain

6.97

[f (p)] 2 dp = E 2 -3

#

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m=0

LT

-3

dp 1 = 2 2

[f (p)] 2

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-1 n

Option (B) is correct. 1 = 12 # 1 2 s-2 s s (s - 2)

X1 (s) X2 (s)

3

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z) -m

1 , 1 - 5-1 z = 1- 5 = z z-5 5-z

#

Option (B) is correct.

n =- 3

= 1-

[f (2t)] 2 dt

Option (B) is correct. Since h1 (t) ! 0 for t < 0 , thus h1 (t) is not causal h2 (t) = u (t) which is always time invariant, causal and stable. u (t) is time variant. h3 (t) = 1+t h 4 (t) = e-3t u (t) is time variant.

Let n =- m, then H (z) =-

3

[f (t)] 2 dt

h (t) = f (t)* g (t) We know that convolution in time domain is multiplication in s domain.

where z < a

where z < 5

-3

-3

E1 =

6.101

- an u (- n - 1)

3

Substituting 2t = p we get

x (t) = [- 2e-t u (t) - 2e2t u (- t)] 6.95

# E1 = # E =

=

# d (t) dt = 1 3

Now output is

-3

Option (B) is correct. Let E be the energy of f (t) and E1 be the energy of f (2t), then

Thus

1 s-2 X (s) = 1 s-3

H (s) =

Y (s) = H (s) X (s) = 1 # 1 = 1 - 1 s-2 s-3 s-3 s-2 y (t) = e3t - e2t

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 6.105

z-1 (1 - z-4) (z - 1) z"1 4 (1 - z-1) 2 z-1 z-4 (z 4 - 1) (z - 1) = lim z"1 4z-2 (z - 1) 2 -3 (z - 1) (z + 1) (z 2 + 1) (z - 1) = lim z z"1 4 (z - 1) 2 -3 = lim z (z + 1) (z2 + 1) = 1 z"1 4

Option (C) is correct. We know that for a square wave the Fourier series coefficient sin nw t Cnsq = At nw t2 T 2 1 Cnsq \ n 0

Thus

= lim ...(i)

0

If we integrate square wave, triangular wave will be obtained, Hence Cntri \ 12 n 6.106

Page 154

6.113

Option (B) is correct. u (t) - u (t - 1) = f (t)

L

u (t) - u (t - 2) = g (t)

L

L

f (t)* g (t)

w s2 + w2 lim f (t) final value theorem states that: t"3 lim f (t) = lim sF (s) t"3

t"3

6.114

So

x (t) = A 0 +

3

/ An cos nwt

n=1

6.115

Option (C) is correct. Given periodic signal 1, t < T1 x (t) = * 0, T1 < t < T0 2 The figure is as shown below.

Option (A) is correct. Option (A) is correct. We have f (nT) = anT Taking z -transform we get 3

/ anT z-n

3

/ (aT ) n z-n

=

n =- 3

= z T z-a

=

3

n=0

n =- 3

T n

/ b az

l

For x (t) fourier series expression can be written as x (t) = A 0 +

Option (B) is correct. If L [f (t)] = F (s) Applying time shifting property we can write

6.110

Option (A) is correct.

6.111

Option (A) is correct.

n=1

where dc term T /2 A 0 = 1 # x (t) dt = 1 # x (t) dt T0 T T0 -T /2 T T = 1 : # x (t) dt + # x (t) dt + T0 -T /2 -T = 1 60 + 2T1 + 0@ T0 A 0 = 2T1 T0 0

0

Option (C) is correct. Given z transform

0

1

0

C (z) =

z-1 (1 - z-4) 4 (1 - z-1) 2

Applying final value theorem lim f (n) = lim (z - 1) f (z)

n"3

z"1

lim (z - 1) F (z) = lim (z - 1) z"1

3

/ [An cos nwt + Bn sin nwt]

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L [f (t - T)] = e-sT F (s)

6.112

n=1

Series will contain only DC & cosine .

L

F (z) =

6.109

3

/ [An cos nwt + Bn sin nwt]

For even function x (t), Bn = 0

f (t)* g (t) = t - (t - 2) u (t - 2) - (t - 1) u (t - 1) + (t - 3) u (t - 3) The graph of option (B) satisfy this equation.

6.108

Option (D) is correct. Trigonometric Fourier series of a function x (t) is expressed as : x (t) = A 0 +

-2s -s -3s = 12 - e 2 - e 2 + e 2 s s s s Taking inverse Laplace transform we have

6.107

s"0

It must be noted that final value theorem can be applied only if poles lies in –ve half of s -plane. Here poles are on imaginary axis (s1, s2 = ! jw) so can not apply final value theorem. so lim f (t) cannot be determined.

F (s) G (s)

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F (s) =

We have

F (s) = 1 [1 - e-s] s G (s) = 1 [1 - e-2s] s

= 12 [1 - e-s] [1 - e-2s] s = 12 [1 - e-2s - e-s + e-3s] s

or

Option (A) is correct.

z"1

-1

6.116

-4

z (1 - z ) 4 (1 - z-1) 2

1

1

#T

T0 /2

x (t) dtD

1

Option (B) is correct. The unit impulse response of a LTI system is u (t) Let h (t) = u (t)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Taking LT we have H (s) = 1 s If the system excited with an input x (t) = e-at u (t), a > 0 , the response Y (s) = X (s) H (s) 1 (s + a) Y (s) = 1 1 = 1 :1 - 1 D a s s+a (s + a) s

Page 155 6.123

Option (C) is correct. The conjugation property allows us to show if x (t) is real, then X (jw) has conjugate symmetry, that is [ x (t) real] X (- jw) = X)(jw) Proof :

X (s) = L [x (t)] =

so

replace w by - w then

Taking inverse Laplace, the response will be y (t) = 1 61 - e-at@ a 6.117

x [ n] = X (z) =

X)(jw) = = if x (t) real x)(t) = x (t)

/ d (n - k) / x [n] z-n

=

k=0

then 3

3

/ ; / d (n - k) z-nE

Option (B) is correct.

6.119

Option (B) is correct. F

x (t) X (f) by differentiation property; dx (t) F; = jwX (w) dt E dx (t) or F; = j2pfX (f) dt E 6.120

F

so

2pf (- w) 3

# g (t) e-jwt dt = 2pf (- w)

F [g (t)] =

-3 6.121

Option (B) is correct. Given function Now

x (t) = eat cos (at) L s cos (at) s2 + a2 x (t)

L

X (s)

s0 t

e x (t)

L

eat cos (at)

L

X (s - s 0) (s - a) (s - a) 2 + a2

If then so 6.122

3

# x)(t) e jwt dt -3

3

# x (t) e jwt dt = X (- jw) -3

shifting in s-domain

Option (C) is correct. For a function x (t), trigonometric fourier series is : x (t) = A 0 +

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We have f (t) g (w) by duality property of fourier transform we can write F

-3

=

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Option (C) is correct.

g (t)

X)(jw) =

)

# x (t) e-jwt dtG

n =- 3 k = 0

Since d (n - k) defined only for n = k so 3 1 = z X (z) = / z-k = (z - 1) (1 - 1/z) k=0 6.118

3

# x (t) e jwt dt -3 3

3

k=0 3

-3

X (- jw) =

Option (B) is correct. We have

3

# x (t) e-jwt dt

X (jw) =

3

/ [An cos nwt + Bn sin nwt]

n=1

A 0 = 1 # x (t) dt T0 =Fundamental period T0 T An = 2 # x (t) cos nwtdt T0 T Bn = 2 # x (t) sin nwtdt T0 T For an even function x (t), coefficient Bn = 0 for an odd function x (t), A0 = 0 where

0

0

0

An = 0 so if x (t) is even function its fourier series will not contain sine .

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 156

UNIT 7

approximate value of Ka that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is

CONTROL SYSTEMS

2013 7.1

(A) 1 (C) 10

ONE MARK

The Bode plot of a transfer function G ^s h is shown in the figure below.

7.3

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s+1 5s2 + 6s + 2 (C) 2 s + 1 s + 4s + 2

The state diagram of a system is shown below. A system is o = AX + Bu ; described by the state-variable equations X y = CX + Du

7.2

and - 8 dB at 1 rad/s and is negative for all w. Then G ^s h is (B) 392.8 s (D) 322 s TWO MARKS

The open-loop transfer function of a dc motor is given as w ^s h = 10 . When connected in as shown below, the Va ^s h 1 + 10s

s+1 s 2 + 6s + 2 (D) 2 1 5s + 6s + 2 (B)

Statement for Linked Answer Questions 4 and 5:

7.5

2013

The signal flow graph for a system is given below. The transfer Y ^s h function for this system is U ^s h

(A)

7.4

The gain _20 log G ^s h i is 32 dB 10 rad/s respectively. The phase (A) 39.8 s (C) 32 s

(B) 5 (D) 100

The state-variable equations of the system shown in the figure above are o = >- 1 0 H X + >- 1H u o = >- 1 0 H X + >- 1H u X X (A) (B) 1 -1 1 -1 -1 1 y = 61 - 1@X + u y = 6- 1 - 1@X + u o = >- 1 0 H X + >- 1H u o = >- 1 - 1H X + >- 1H u X X (C) (D) -1 -1 1 0 -1 1 y = 6- 1 - 1@X - u y = 61 - 1@ X - u The state transition matrix eAt of the system shown in the figure

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 above is e-t 0 (A) > -t -tH te e

e-t 0 (B) > H -t - te e-t

e-t 0 (C) > -t -tH e e

e-t - te-t (D) > H 0 e-t

2012 7.6

ONE MARK

(s2 + 9) (s + 2) (s + 1) (s + 3) (s + 4) is excited by sin (wt). The steady-state output of the system is zero at A system with transfer function G (s) =

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) w = 1 rad/s (C) w = 3 rad/s

(B) w = 2 rad/s (D) w = 4 rad/s

2012 7.7

Page 157

TWO MARKS

The system shown below oscillates at 2 rad/s when

2011 7.11

(A) K = 2 and a = 0.75 (C) K = 4 and a = 0.5 7.8

(B) K = 3 and a = 0.75 (D) K = 2 and a = 0.5

The state variable description of an LTI system is given by Jxo1N J 0 a1 0NJx1N J0N K O K OK O K O Kxo2O = K 0 0 a2OKx2O + K0O u Kxo O Ka 0 0OKx 3O K 1O 3 3 L P L PL P L P Jx1N K O y = _1 0 0iKx2O Kx 3O L P where y is the output and u is the input. The system is controllable for (B) a1 = 0, a2 ! 0, a 3 ! 0 (A) a1 ! 0, a2 = 0, a 3 ! 0 (C) a1 = 0, a 3 ! 0, a 3 = 0 (D) a1 ! 0, a2 ! 0, a 3 = 0 2011

7.9

The block diagram of a system with one input u and two outputs y1 and y2 is given below.

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ONE MARK

The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by

s (s + 1) (s + 2) (s + 3) (s + 1) (B) G ^s h H ^s h = k s (s + 2) (s + 3) 2 1 (C) G ^s h H ^s h = k s (s - 1) (s + 2) (s + 3) (s + 1) (D) G ^s h H ^s h = k s (s + 2) (s + 3) (A) G ^s h H ^s h = k

7.10

TWO MARKS

For the transfer function G (jw) = 5 + jw , the corresponding Nyquist plot for positive frequency has the form

A state space model of the above system in of the state vector x and the output vector y = [y1 y2]T is (A) xo = [2] x + [1] u ; y = [1 2] x 1 (B) xo = [- 2] x + [1] u; y = > H x 2 -2 0 1 (C) xo = > x + > H u ; y = 81 2B x H 0 -2 1 2 0 1 1 (D) xo = > H x + > H u ; y = > H x 0 2 1 2

Common Data For Q. 7.4 & 7.5 100 . s (s + 10) 2 The plant is placed in a unity negative configuration as shown in the figure below. The input-output transfer function of a plant H (s) =

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

7.12

7.13

Page 158

The gain margin of the system under closed loop unity negative is (A) 0 dB (B) 20 dB (C) 26 dB (D) 46 dB

(A) 10s + 1 0.1s + 1 (C) 100s 10s + 1

The signal flow graph that DOES NOT model the plant transfer function H (s) is

2010 7.17

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7.14

1 s+1 (C) 2 (D) 2 s+1 s+3 Y (s) A system with transfer function has an output = s X (s) s + p p y (t) = cos a2t - k 3 for the input signal x (t) = p cos a2t - p k. Then, the system param2 eter p is (A) 3 (B) 2/ 3 (A) 0

7.15

(C) 1 7.16

7.19

The state variable representation of the system can be 0 -1 1 1 1 0 o= > xo = > x +> Hu H x x u + H > H (A) (B) 2 -1 0 -1 0 2 yo = 80 0.5B x yo = [0 0.5] x 1 1 0 xo = > x +> Hu H (C) (D) 2 -1 0 yo = 80.5 0.5B x The transfer function of the system is (A) s2+ 1 (B) s +1 (C) 2 s + 1 (D) s +s+1

-1 xo = > -1 yo = 80.5

1 0 x +> Hu H 0 2 0.5B x

s-1 s2+1 s-1 s2+s+1

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(B)

(D)

A unity negative closed loop system has a plant with the transfer function G (s) = 2 1 and a controller Gc (s) in the s + 2s + 2 feed forward path. For a unit set input, the transfer function of the controller that gives minimum steady state error is (A) Gc (s) = s + 1 (B) Gc (s) = s + 2 s+2 s+1 (s + 1) (s + 4) (C) Gc (s) = (D) Gc (s) = 1 + 2 + 3s s (s + 2) (s + 3)

The signal flow graph of a system is shown below:

ONE MARK

The transfer function Y (s) /R (s) of the system shown is

TWO MARKS

Common Data For Q. 7.10 & 7.11 :

7.18

2010

(B) 100s + 1 0.1s + 1 (D) 0.1s + 1 10s + 1

2009 7.20

ONE MARK

The magnitude plot of a rational transfer function G (s) with real coefficients is shown below. Which of the following compensators has such a magnitude plot ?

3 /2

For the asymptotic Bode magnitude plot shown below, the system transfer function can be

(A) Lead compensator

(B) Lag compensator

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(C) PID compensator 7.21

Page 159

(D) Lead-lag compensator

(C) 6 dB and 90c (D) 3 dB and 90c

Consider the system dx = Ax + Bu with A = =1 0G and B = = p G 0 1 q dt where p and q are arbitrary real numbers. Which of the following statements about the controllability of the system is true ? (A) The system is completely state controllable for any nonzero values of p and q (B) Only p = 0 and q = 0 result in controllability (C) The system is uncontrollable for all values of p and q (D) We cannot conclude about controllability from the given data 2009

7.22

2008 7.26

ONE MARKS

Step responses of a set of three second-order underdamped systems all have the same percentage overshoot. Which of the following diagrams represents the poles of the three systems ?

TWO MARKS

The configuration and the pole-zero locations of 2 G (s) = s2 - 2s + 2 s + 2s + 2 are shown below. The root locus for negative values of k , i.e. for - 3 < k < 0 , has breakaway/break-in points and angle of departure at pole P (with respect to the positive real axis) equal to

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7.23

(A) ! 2 and 0c

(B) ! 2 and 45c

(C) ! 3 and 0c

(D) ! 3 and 45c

The unit step response of an under-damped second order system has steady state value of -2. Which one of the following transfer functions has theses properties ? (B) 2 - 3.82 (A) 2 - 2.24 s + 1.91s + 1.91 s + 2.59s + 1.12 - 382 (C) 2 - 2.24 (D) 2 s - 2.59s + 1.12 s - 1.91s + 1.91

Common Data For Q. 7.16 and 7.17 : The Nyquist plot of a stable transfer function G (s) is shown in the figure are interested in the stability of the closed loop system in the configuration shown.

7.27

7.24

7.25

The pole-zero given below correspond to a

Which of the following statements is true ? (A) G (s) is an all- filter (B) G (s) has a zero in the right-half plane (C) G (s) is the impedance of a ive network (D) G (s) is marginally stable The gain and phase margins of G (s) for closed loop stability are (A) 6 dB and 180c (B) 3 dB and 180c

(A) Law filter (C) Band filter

(B) High filter (D) Notch filter

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

2008 7.28

Page 160

TWO MARKS

Group I lists a set of four transfer functions. Group II gives a list of possible step response y (t). Match the step responses with the corresponding transfer functions.

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(A) P - 3, Q - 1, R - 4, S - 2 (C) P - 2, Q - 1, R - 4, S - 2 7.29

(B) P - 3, Q - 2, R - 4, S - 1 (D) P - 3, Q - 4, R - 1, S - 2

A signal flow graph of a system is given below

The set of equalities that corresponds to this signal flow graph is Jx1N R b - g 0 VJx1N R0 0 V W u1 WK O S K O S (A) d K x2O = S g a 0 WK x2O+ S0 1 We o u2 dt K O S x3 S- a b 0 WWK x3O SS1 0 WW L P L P Jx1N RT0 a g XVJx1N TR1 0 XV W u1 W S K O K O S (B) d K x2O = S0 - a - g WK x2O+ S0 1 We o u2 dt K O S x3 S0 b - b WWK x3O SS0 0 WW L P TR L P Jx1N - a b 0 VXJx1N RT1 0 VX W u1 WK O S K O S (C) d K x2O = S- b - g 0 WK x2O+ S0 1 We o u2 dt K O S x3 S a g 0 WWK x3O SS0 0 WW L P TR L P Jx1N - a 0 b XVJx1N TR1 0XV W u1 WK O S K O S (D) d K x2O = S g 0 a WK x2O+ S0 1 We o u2 dt K O S x3 S- b 0 - a WWK x3O SS0 0 WW L P T X XL P T

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A certain system has transfer function G (s) = 2 s + 8 s + as - 4 where a is a parameter. Consider the standard negative unity configuration as shown below

Which of the following statements is true? (A) The closed loop systems is never stable for any value of a (B) For some positive value of a, the closed loop system is stable,

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 161

but not for all positive values. (C) For all positive values of a, the closed loop system is stable. (D) The closed loop system stable for all values of a, both positive and negative. 7.31

7.32

7.33

The number of open right half plane of 10 is G (s) = 5 4 3 s + 2s + 3s + 6s2 + 5s + 3 (A) 0 (B) 1 (C) 2 (D) 3

7.37

The magnitude of frequency responses of an underdamped second order system is 5 at 0 rad/sec and peaks to 10 at 5 2 rad/sec. 3 The transfer function of the system is 500 (B) 2 375 (A) 2 s + 5s + 75 s + 10s + 100 720 (C) 2 (D) 2 1125 s + 12s + 144 s + 25s + 225 Group I gives two possible choices for the impedance Z in the diagram. The circuit elements in Z satisfy the conditions R2 C2 > R1 C1. The transfer functions V0 represents a kind of controller. Vi

5 (s + 5)( s2 + s + 1) The second-order approximation of T (s) using dominant pole concept is 5 1 (B) (A) (s + 5)( s + 1) (s + 5)( s + 1) (C) 2 5 (D) 2 1 s +s+1 s +s+1 The open-loop transfer function of a plant is given as G (s) = s 1- 1 . If the plant is operated in a unity configuration, then the lead compensator that an stabilize this control system is 10 (s + 4) 10 (s - 1) (B) (A) s+2 s+2 10 (s + 2) 2 (s + 2) (C) (D) s + 10 s + 10 T (s) =

7.38

2

A unity control system has an open-loop transfer function K G (s) = 2 s (s + 7s + 12) The gain K for which s = 1 + j1 will lie on the root locus of this

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Match the impedances in Group I with the type of controllers in Group II

system is (A) 4 (C) 6.5 7.39

(A) Q - 1, R - 2 (C) Q - 2, R - 3 2007 7.34

7.35

The asymptotic Bode plot of a transfer function is as shown in the figure. The transfer function G (s) corresponding to this Bode plot is

(B) Q - 1, R - 3 (D) Q - 3, R - 2 ONE MARK

If the closed-loop transfer function of a control system is given as s-5 , then It is T (s) (s + 2)( s + 3) (A) an unstable system (B) an uncontrollable system (C) a minimum phase system (D) a non-minimum phase system 2007

1 1 (B) (s + 1)( s + 20) s (s + 1)( s + 20) 100 100 (C) (D) s (s + 1)( s + 20) s (s + 1)( 1 + 0.05s) The state space representation of a separately excited DC servo motor dynamics is given as (A)

7.40

dw dt dio dt

> H = =- 1 - 10G=ia G + =10Gu

TWO MARKS

A control system with PD controller is shown in the figure. If the velocity error constant KV = 1000 and the damping ratio z = 0.5 , then the value of KP and KD are

7.36

The transfer function of a plant is

(B) KP = 100, KD = 0.9 (D) KP = 10, KD = 0.9

-1

1 w

0

where w is the speed of the motor, ia is the armature current and w (s) of the motor u is the armature voltage. The transfer function U (s) is 10 s2 + 11s + 11 (C) 2 10s + 10 s + 11s + 11 (A)

(A) KP = 100, KD = 0.09 (C) KP = 10, KD = 0.09

(B) 5.5 (D) 10

1 s2 + 11s + 11 (D) 2 1 s + s + 11 (B)

Statement for linked Answer Question 8.33 & 8.34 :

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Consider a linear system whose state space representation is 1 x (t) = Ax (t). If the initial state vector of the system is x (0) = = G, 2 e-2x then the system response is x (t) = > H. If the itial state vector - 2e-2t 1 of the system changes to x (0) = = G, then the system response -2 e-t becomes x (t) = > -tH -e

Page 162 7.47

7.48 7.41

7.42

The eigenvalue and eigenvector pairs (li vi) for the system are 1 1 1 1 (A) e- 1 = Go and e- 2 = Go (B) e- 1, = Go and e2, = Go -1 -2 -1 -2 1 1 1 1 (C) e- 1, = Go and e- 2, = Go (D) e- 2 = Go and e1, = Go -1 -2 -1 -2 The system matrix A is 0 1 (A) = - 1 1G 2 1 (C) = - 1 - 1G

Statement for Linked Answer Questions 7.41 & 7.42 :

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a2 + 4b , a2 + 4b

The Nyquist plot of G (jw) H (jw)for a closed loop control system, es through (- 1, j0) point in the GH plane. The gain margin of the system in dB is equal to (A) infinite (B) greater than zero (C) less than zero

7.46

(D)

(D) zero

The positive values of K and a so that the system shown in the figures below oscillates at a frequency of 2 rad/sec respectively are

(A) 1, 0.75 (C) 1, 1

(B) 2, 0.75 (D) 2, 2

With the value of a set for a phase - margin of p , the value of unit 4 - impulse response of the open - loop system at t = 1 second is equal to (A) 3.40 (B) 2.40 (C) 1.84 (D) 1.74 ONE MARK

A linear system is equivalently represented by two sets of state equations : Xo = AX + BU and Wo = CW + DU The eigenvalues of the representations are also computed as [l] and [m]. Which one of the following statements is true ? (A) [l] = [m] and X = W (B) [l] = [m] and X ! W (C) [l] ! [m] and X = W (D) [l] = [m] and X ! W

TWO MARKS

a2 - 4b , a2 - 4b

Consider a unity - gain control system whose open - loop 1 transfer function is : G (s) = as + s2 The value of a so that the system has a phase - margin equal to p 4 is approximately equal to (A) 2.40 (B) 1.40 (C) 0.84 (D) 0.74

2005 7.51

Consider two transfer functions G1 (s) = 2 1 and s + as + b s . G2 (s) = 2 s + as + b The 3-dB bandwidths of their frequency responses are, respectively (A) a2 - 4b , a2 + 4b (B) a2 + 4b , a2 - 4b (C)

7.45

7.50

The open-loop function of a unity-gain control system is given by K G (s) = (s + 1)( s + 2) The gain margin of the system in dB is given by (A) 0 (B) 1 (C) 20 (D) 3

7.44

7.49

ONE MARK

2006

A linear system is described by the following state equation 0 1 Xo (t) = AX (t) + BU (t), A = = - 1 0G The state transition matrix of the system is cos t sin t - cos t sin t (A) = (B) = G - sin t cos t - sin t - cos t G - cos t - sin t cos t - sin t (C) = (D) = G - sin t cos t cos t sin t G

1 1 (B) = - 1 - 2G 0 1 (D) = - 2 - 3G

2006

The transfer function of a phase lead compensator is given by Gc (s) = 1 + 3Ts where T > 0 The maximum phase shift provide by 1 + Ts such a compensator is (B) p (A) p 3 2 (C) p (D) p 4 6

7.52

Which one of the following polar diagrams corresponds to a lag

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 7.53

Despite the presence of negative , control systems still have problems of instability because the (A) Components used have non- linearities (B) Dynamic equations of the subsystem are not known exactly. (C) Mathematical analysis involves approximations. (D) System has large negative phase angle at high frequencies. 2005

7.54

Page 163

TWO MARKS

The polar diagram of a conditionally stable system for open loop gain K = 1 is shown in the figure. The open loop transfer function of the system is known to be stable. The closed loop system is stable for

Statement for Linked Answer Question 40 and 41 :

7.59

The open loop transfer function of a unity system is given by -2s G (s) = 3e s (s + 2) The gain and phase crossover frequencies in rad/sec are, respectively

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7.55

7.56

7.57

7.58

(A) K < 5 and 1 < K < 1 (B) K < 1 and 1 < K < 5 2 8 8 2 (C) K < 1 and 5 < K (D) K > 1 and 5 > K 8 8 In the derivation of expression for peak percent overshoot - px Mp = exp e o # 100% 1 - x2 Which one of the following conditions is NOT required ? (A) System is linear and time invariant (B) The system transfer function has a pair of complex conjugate poles and no zeroes. (C) There is no transportation delay in the system. (D) The system has zero initial conditions. A ramp input applied to an unity system results in 5% steady state error. The type number and zero frequency gain of the system are respectively (A) 1 and 20 (B) 0 and 20 (D) 1 and 1 (C) 0 and 1 20 20 A double integrator plant G (s) = K/s2, H (s) = 1 is to be compensated to achieve the damping ratio z = 0.5 and an undamped natural frequency, wn = 5 rad/sec which one of the following compensator Ge (s) will be suitable ? (A) s + 3 (B) s + 99 s + 99 s+3 (C) s - 6 (D) s - 6 s + 8.33 s K (1 - s) An unity system is given as G (s) = . s (s + 3) Indicate the correct root locus diagram.

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Click to Buy www.nodia.co.in (A) 0.632 and 1.26 (C) 0.485 and 0.632 7.60

(B) 0.632 and 0.485 (D) 1.26 and 0.632

Based on the above results, the gain and phase margins of the system will be (B) 7.09 dB and 87.5c (A) -7.09 dB and 87.5c (C) 7.09 dB and - 87.5c (D) - 7.09 and - 87.5c 2004

7.61

7.62

ONE MARK

The gain margin for the system with open-loop transfer function 2 (1 + s) , is G (s) H (s) = s2 (A) 3 (B) 0 (C) 1 (D) - 3 K Given G (s) H (s) = .The point of intersection of the s (s + 1)( s + 3) asymptotes of the root loci with the real axis is (A) - 4 (B) 1.33 (C) - 1.33 (D) 4 2004

7.63

TWO MARKS

Consider the Bode magnitude plot shown in the fig. The transfer function H (s) is

(A)

(s + 10) (s + 1)( s + 100)

(B)

10 (s + 1) (s + 10)( s + 100)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

7.64

7.65

7.66

102 (s + 1) 103 (s + 100) (C) (D) (s + 10)( s + 100) (s + 1)( s + 10) A causal system having the transfer function H (s) = 1/ (s + 2) is excited with 10u (t). The time at which the output reaches 99% of its steady state value is (A) 2.7 sec (B) 2.5 sec (C) 2.3 sec (D) 2.1 sec A system has poles at 0.1 Hz, 1 Hz and 80 Hz; zeros at 5 Hz, 100 Hz and 200 Hz. The approximate phase of the system response at 20 Hz is (A) - 90c (B) 0c (C) 90c (D) - 180c

(D) controllable and observable 7.71

(A)

1 - (be + cf + dg) abcd

abcd 1 - (be + cf + dg) + bedg -2 2 If A = = , then sin At is 1 - 3G (C)

7.69

7.70

1 0 , the state transition matrix eAt is given by 0 1G

0 e-t (A) > -t H e 0

et 0 (B) = G 0 et

e-t 0 (C) > H 0 e-t

0 et (D) = t G e 0

2003 7.72

ONE MARK

Fig. shows the Nyquist plot of the open-loop transfer function G (s) H (s) of a system. If G (s) H (s) has one right-hand pole, the closed-loop system is

bedg 1 - (be + cf + dg) 1 - (be + cf + dg) + bedg (D) abcd

(A) always stable (B) unstable with one closed-loop right hand pole (C) unstable with two closed-loop right hand poles (D) unstable with three closed-loop right hand poles 7.73

A PD controller is used to compensate a system. Compared to the uncompensated system, the compensated system has (A) a higher type number (B) reduced damping (C) higher noise amplification (D) larger transient overshoot

(B)

2003 7.74

sin (- 4t) + 2 sin (- t) - 2 sin (- 4t) + 2 sin (- t) (A) 1 = G 3 - sin (- 4t) + sin (- t) 2 sin (- 4t) + sin (- t) sin (- 2t) sin (2t) (B) = sin (t) sin (- 3t)G sin (4t) + 2 sin (t) 2 sin (- 4t) - 2 sin (- t) (C) 1 = 2 sin (4t) + sin (t) G 3 sin (- 4t) + sin (t) cos (- t) + 2 cos (t) 2 cos (- 4t) + 2 cos (- t) (D) 1 = G 3 - cos (- 4t) + cos (- t) - 2 cos (- 4t) + cos (t) 7.68

Given A = =

Consider the signal flow graph shown in Fig. The gain x5 is x1

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7.67

Page 164

The open-loop transfer function of a unity system is K G (s) = 2 s (s + s + 2)( s + 3) The range of K for which the system is stable is (B) 13 > K > 0 (A) 21 > K > 0 4 (C) 21 < K < 3 (D) - 6 < K < 3 4 For the polynomial P (s) = s2 + s 4 + 2s3 + 2s2 + 3s + 15 the number of roots which lie in the right half of the s -plane is (A) 4 (B) 2 (C) 3 (D) 1 The state variable equations of a system are : xo1 =- 3x1 - x2 = u, xo2 = 2x1 and y = x1 + u . The system is (A) controllable but not observable (B) observable but not controllable (C) neither controllable nor observable

The signal flow graph of a system is shown in Fig. below. The transfer function C (s)/ R (s) of the system is

(A)

6 s2 + 29s + 6

(B)

s (s + 2) s + 29s + 6

(D)

6s s2 + 29s + 6

s (s + 27) s + 29s + 6 K The root locus of system G (s) H (s) = has the breaks (s + 2)( s + 3) (C)

7.75

TWO MARKS

2

2

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(B) (- 2.548, 0) (D) (- 0.784, 0)

The approximate Bode magnitude plot of a minimum phase system is shown in Fig. below. The transfer function of the system is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 165 7.82

7.83

(s + 0.1) 3 (s + 0.1) 3 7 (B) 10 (s + 10)( s + 100) (s + 10) 2 (s + 100) 2 (s + 0.1) (s + 0.1) 3 (C) (D) (s + 10) 2 (s + 100) (s + 10)( s + 100) 2 A second-order system has the transfer function C (s) = 2 4 R (s) s + 4s + 4 With r (t) as the unit-step function, the response c (t) of the system is represented by (A) 108

7.77

The phase margin of a system with the open - loop transfer function (1 - s) G (s) H (s) = (1 + s)( 2 + s) (A) 0c (B) 63.4c (C) 90c (D) 3 The transfer function Y (s)/ U (s) of system described by the state equation xo (t) =- 2x (t) + 2u (t) and y (t) = 0.5x (t) is 1 (A) 0.5 (B) (s - 2) (s - 2) 1 (C) 0.5 (D) (s + 2) (s + 2) 2002

7.84

7.85

TWO MARKS

The system shown in the figure remains stable when (A) k < - 1 (B) - 1 < k < 3 (C) 1 < k < 3 (D) k > 3 The transfer function of a system is G (s) =

100 . For a (s + 1)( s + 100)

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7.78

7.79

The gain margin and the phase margin of system with 8 are G (s) H (s) = (s + 100) 3 (A) dB, 0c (B) 3, 3 (C) 3, 0c (D) 88.5 dB, 3

2002

7.81

7.86

(C) unstable

ONE MARK

Consider a system with transfer function G (s) = 2s + 6 . Its ks + s + 6 damping ratio will be 0.5 when the value of k is (B) 3 (A) 2 6 (D) 6 (C) 1 6 Which of the following points is NOT on the root locus of a system k with the open-loop transfer function G (s) H (s) = s (s + 1)( s + 3) (A) s =- j 3 (B) s =- 1.5 (C) s =- 3 (D) s =- 3

The characteristic polynomial of a system is The system is (A) stable

7.87

et (B) = G t t (D) = t G te

(B) 4 sec (D) 0.01 sec q (s) = 2s5 + s 4 + 4s3 + 2s2 + 2s + 1

The zero-input response of a system given by the state-space equation x1 (0) 1 1 0 x1 xo1 =xo G = =1 1G=x G and =x (0)G = = 0 G is 2 2 2 tet (A) = G t et (C) = t G te

7.80

(A)100 sec (C) 1 sec

The system with the open loop transfer 1 has a gain margin of G (s) H (s) = 2 s ( s ) + s + 1 (A) - 6 db (B) 0 db (C) 35 db (D) 6 db 2001

7.88

7.89

(B) marginally stable (D) oscillatory function

ONE MARK

The Nyquist plot for the open-loop transfer function G (s) of a unity negative system is shown in the figure, if G (s) has no pole in the right-half of s -plane, the number of roots of the system characteristic equation in the right-half of s -plane is (A) 0 (B) 1 (C) 2 (D) 3 The equivalent of the block diagram in the figure is given is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

7.90

7.91

Page 166

If the characteristic equation of a closed - loop system is s2 + 2s + 2 = 0 , then the system is (A) overdamped (B) critically damped (C) underdamped (D) undamped The root-locus diagram for a closed-loop system is shown in the figure. The system is overdamped.

7.93

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Z3 (s) - Z3 (s) , Z1 (s) + Z3 (s) + Z4 (s) Z1 (s) + Z3 (s) - Z3 (s) - Z3 (s) (B) , Z2 (s) - Z3 (s) + Z4 (s) Z1 (s) + Z3 (s) Z3 (s) Z3 (s) (C) , Z2 (s) + Z3 (s) + Z4 (s) Z1 (s) + Z3 (s) - Z3 (s) Z3 (s) (D) , Z2 (s) - Z3 (s) + Z4 (s) Z1 (s) + Z3 (s) The open-loop DC gain of a unity negative system with closed-loop transfer function 2 s + 4 is s + 7s + 13 (B) 4 (A) 4 9 13 (A)

(C) 4 7.94

The control system in the figure is stable

(A) for all K $ 0 (C) only if 0 # K < 1 2000 7.95

2001 7.92

ONE MARK

TWO MARKS

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(B) only if 1 < k < 5 (D) if 0 # k < 1 or k > 5 TWO MARK

(B) only if K $ 0 (D) only if 0 # K # 1

An amplifier with resistive negative has tow left half plane poles in its open-loop transfer function. The amplifier (A) will always be unstable at high frequency (B) will be stable for all frequency (C) may be unstable, depending on the factor (D) will oscillate at low frequency. 2000

(A) only if 0 # k # 1 (C) only if k > 5

(D) 13

7.96

An electrical system and its signal-flow graph representations are shown the figure (A) and (B) respectively. The values of G2 and H , respectively are

1 A system described by the transfer function H (s) = 3 2 + a + ks + 3 s s is stable. The constraints on a and k are. (A) a > 0, ak < 3 (B) a > 0, ak > 3 (C) a < 0, ak > 3 (D) a > 0, ak < 3 1999

7.97

ONE MARK

For a second order system with the closed-loop transfer function T (s) = 2 9 s + 4s + 9 the settling time for 2-percent band, in seconds, is (A) 1.5 (B) 2.0

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(C) 3.0 7.98

(D) 4.0

the eigen values of the closed-loop system will be (A) 0, - 1, - 2 (B) 0, - 1, - 3 (C) - 1, - 1, - 2 (D) 0, - 1, - 1

The gain margin (in dB) of a system a having the loop transfer function G (s) H (s) = (A) 0 (C) 6

7.99

Page 167

2 is s (s + 1) (B) 3 (D) 3

1998

The system modeled described by the state equations is 0 1 0 X => x + > Hu H 2 -3 1

7.105

Y = 81 1B x

7.100

7.101

7.102

7.103

7.104

7.106

The phase margin (in degrees) of a system having the loop transfer function G (s) H (s) = 2 3 is s (s + 1) (A) 45c (B) - 30c (C) 60c (D) 30c 1999

The number of roots of s3 + 5s2 + 7s + 3 = 0 in the left half of the s -plane is (A) zero (B) one (C) two

(A) controllable and observable (B) controllable, but not observable (C) observable, but not controllable (D) neither controllable nor observable

TWO MARKS

An amplifier is assumed to have a single-pole high-frequency transfer function. The rise time of its output response to a step function input is 35 n sec . The upper 3 dB frequency (in MHz) for the amplifier to as sinusoidal input is approximately at (A) 4.55 (B) 10 (C) 20 (D) 28.6 If the closed - loop transfer function T (s) of a unity negative system is given by an - 1 s + an T (s) = n n-1 s + a1 s + .... + an - 1 s + an then the steady state error for a unit ramp input is (A) an (B) an an - 1 an - 2 (C) an - 2 (D) zero an - 2 Consider the points s1 =- 3 + j4 and s2 =- 3 - j2 in the s-plane. Then, for a system with the open-loop transfer function G (s) H (s) = K 4 (s + 1) (A) s1 is on the root locus, but not s2 (B) s2 is on the root locus, but not s1 (C) both s1 and s2 are on the root locus (D) neither s1 nor s2 is on the root locus For the system described by the state equation R 0 1 0V R0V S W S W xo = S 0 0 1W x + S0W u SS0.5 1 2WW SS1WW T X T X If the control signal u is given by u = [- 0.5 - 3 - 5] x + v , then

ONE MARK

(D) three

The transfer function of a tachometer is of the form (A) Ks (B) K s K (C) K (D) (s + 1) s (s + 1)

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7.108

7.109

7.110

7.111

Consider a unity control system with open-loop transfer K . function G (s) = s (s + 1) The steady state error of the system due to unit step input is (A) zero (B) K (C) 1/K (D) infinite The transfer function of a zero-order-hold system is (A) (1/s) (1 + e-sT ) (B) (1/s) (1 - e-sT ) (C) 1 - (1/s) e-sT (D) 1 + (1/s) e-sT In the Bode-plot of a unity control system, the value of phase of G (jw) at the gain cross over frequency is - 125c. The phase margin of the system is (A) - 125c (B) - 55c (C) 55c (D) 125c Consider a control system with loop transfer function K (1 + 0.5s) G (s) H (s) = s (1 + s) (1 + 2s) The type of the closed loop system is (A) zero (B) one (C) two (D) three The transfer function of a phase lead controller is 1 + 3Ts . The 1 + Ts

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

maximum value of phase provided by this controller is (A) 90c (B) 60c (C) 45c (D) 30c 7.112

7.113

Page 168

(C) 0 (D) None of the above ***********

The Nyquist plot of a phase transfer function g (jw) H (jw) of a system encloses the (–1, 0) point. The gain margin of the system is (A) less than zero (B) zero (C) greater than zero (D) infinity 2s2 + 6s + 5 (s + 1) 2 (s + 2) The characteristic equation of the system is (A) 2s2 + 6s + 5 = 0 (B) (s + 1) 2 (s + 2) = 0 (C) 2s2 + 6s + 5 + (s + 1) 2 (s + 2) = 0 The transfer function of a system is

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in (D) 2s2 + 6s + 5 - (s + 1) 2 (s + 2) = 0 7.114

In a synchro error detector, the output voltage is proportional to [w (t)] n, where w (t) is the rotor velocity and n equals (A) –2 (B) –1 (C) 1 (D) 2 1997

7.115

ONE MARK

In the signal flow graph of the figure is y/x equals

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(A) 3 (B) 5 2 (C) 2 (D) None of the above 7.116

A certain linear time invariant system has the state and the output equations given below 1 - 1 X1 0 Xo1 > o H = >0 1 H>X H + >1H u 2 X2 y = 81 1B: X1 D X2 If X1 (0) = 1, X2 (0) =- 1, u (0) = 0, then (A) 1 (B) –1

dy dt

is t=0

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 169

SOLUTIONS 7.1

Pk2 = ^1 h^s-1h^1 h^1 h = s-1 since, all the loops are touching to the paths Pk1 and Pk2 so, D k 1 = Dk 2 = 1 Now, we have D = 1 - (sum of individual loops) + (sum of product of nontouching loops) Here, the loops are

Option (B) is correct. From the given plot, we obtain the slope as 20 log G2 - 20 log G1 Slope = log w2 - log w1

L1 = ^- 4h^1 h =- 4 L2 = ^- 4h^s-1h = 4s-1 L 3 = ^- 2h^s-1h^s-1h =- 2s-2 L 4 = ^- 2h^s-1h^1 h =- 2s-1 As all the loop L1, L2, L 3 and L 4 are touching to each other so,

From the figure

and

20 log G2 20 log G1 w1 w2

=- 8 dB = 32 dB = 1 rad/s = 10 rad/s

D = 1 - ^L1 + L2 + L 3 + L 4h = 1 - ^- 4 - 4s-1 - 2s-2 - 2s-1h = 5 + 6s1 + 2s2

So, the slope is

Slope = - 8 - 32 log 10 - log 1 =- 40 dB/decade Therefore, the transfer function can be given as G ^s h = k2 S at w = 1 G ^ jwh = k 2 = k w In decibel, 20 log G ^ jwh = 20 log k = 32 32

7.2

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or, k = 10 = 39.8 Hence, the Transfer function is G ^s h = k2 = 392.8 s s Option (C) is correct. Given, open loop transfer function G ^s h = 10Ka = Ka 1 1 + 10s s + 10 By taking inverse Laplace transform, we have g ^ t h = e- t 20

From Mason’s gain formulae Y ^s h = SPk Dk D U ^s h s-2 + s-1 = 5 + 6s-1 + 2s-2 = 2s+1 5s + 6s + 2

1 10

Comparing with standard form of transfer function, Ae-t/t , we get the open loop time constant, tol = 10 Now, we obtain the closed loop transfer function for the given system as G ^s h 10Ka H ^s h = = 1 + G ^s h 1 + 10s + 10Ka Ka = s + ^Ka + 101 h By taking inverse Laplace transform, we get h ^ t h = ka .e-^k + ht So, the time constant of closed loop system is obtained as tcl = 1 1 ka + 10 or, tcl = 1 ka (approximately) a

1 10

Now, given that ka reduces open loop time constant by a factor of 100. i.e., tcl = tol 100 1 or, = 10 100 ka Hence, ka = 10 7.3

Option (A) is correct. For the given SFG, we have two forward paths Pk1 = ^1 h^s-1h^s-1h^1 h = s-2

7.4

Option (A) is correct. For the shown state diagram we can denote the states x1 , x2 as below

So, from the state diagram, we obtain xo1 =- x1 - u xo2 =- x2 + ^1 h^- 1h^1 h^- 1h u + ^- 1h^1 h^- 1h x1 xo2 =- x2 + x1 + u and

y = ^- 1h^1 h x2 + ^- 1h^1 h^- 1h x1 + ^1 h^- 1h^1 h^- 1h^1 h u = x1 - x 2 + u Hence, in matrix form we can write the state variable equations - 1 0 x1 -1 xo1 > o H = > 1 - 1H >x H + > 1 H u x2 2 x1 and y = 81 - 1B > H + u x2 which can be written in more general form as -1 0 -1 Xo = > X +> H 1 - 1H 1 y = 81 - 1B X + u

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 7.5

Page 170

Option (A) is correct. From the obtained state-variable equations We have -1 0 A => 1 - 1H

s = j k+2 jw = j k + 2

S+1 0 SI - A = > - 1 S + 1H

So,

1 >S + 1 0 H ^S + 1h2 1 S + 1 R 1 V S 0 W S+1 W =S 1 1 W S S^S + 1h2 S + 1W T X Hence, the state transition matrix is obtained as eAt = L-1 ^SI - Ah-1 V_ ZR 1 ]]S 0 Wbb S+1 W = L-1 [S 1 1 W` S ]S^S + 1h2 S + 1Wb \T Xa

^SI - Ah-1 =

and

7.8

e-1 0 = > -t -tH te e

Option (C) is correct.

(s2 + 9) (s + 2) (s + 1) (s + 3) (s + 4) (- w2 + 9) (jw + 2) = (jw + 1) (jw + 3) (jw + 4) The steady state output will be zero if G (s) =

G (jw) = 0 -w 2 + 9 = 0 7.7

&

w = 3 rad/s

7.9

7.10

Option (A) is correct. K (s + 1) [R (s) - Y (s)] s + as2 + 2s + 1 K (s + 1) K (s + 1) R (s) = 3 Y (s) ;1 + 3 E 2 2 1 s + as2 + 2s + 1 s + as + s + Y (s) [s3 + as2 + s (2 + k) + (1 + k)] = K (s + 1) R (s) Y (s) Transfer Function, H (s) = R (s) K (s + 1) = 3 s + as2 + s (2 + k) + (1 + k) Routh Table : Y (s) =

3

7.11

Option (D) is correct. General form of state equations are given as xo = Ax + Bu yo = Cx + Du For the given problem R 0 a 0V R0V 1 S W S W A = S 0 0 a2W, B = S0W SSa SS1WW 0 0WW 3 RT 0 a 0VXR0V R 0VT X 1 S WS W S W AB = S 0 0 a2WS0W = Sa2W SSa 0 0WWSS1WW SS 0WW 3 RT 0 0XT aX1 a2VWTRS0XVW RSa1 a2VW S 0 0WS0W = S 0W A2 B = Sa2 a 3 SS 0 a a 0WWSS1WW SS 0WW 3 1 T XT X T X For controllability it is necessary that following matrix has a tank of n = 3 . R0 0 a a V 1 2W S 0W U = 6B : AB : A2 B@ = S0 a2 SS1 0 0WW So, a2 ! 0 T X a 3 may be zero or not. a1 a 2 ! 0 & a1 ! 0 Option (B) is correct. For given plot root locus exists from - 3 to 3, So there must be odd number of poles and zeros. There is a double pole at s =- 3 Now poles = 0, - 2, - 3, - 3 zeros =- 1 k (s + 1) Thus transfer function G (s) H (s) = s (s + 2) (s + 3) 2 Option (A) is correct. We have G (jw) = 5 + jw Here s = 5 . Thus G (jw) is a straight line parallel to jw axis. Option (B) is correct.

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Now

For oscillation,

a (2 + K) - (1 + K) =0 a a = K+1 K+2

w = k + 2 = 2 (Oscillation frequency) k =2 a = 2 + 1 = 3 = 0.75 2+2 4

and

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7.6

A (s) = as2 + (k + 1) = 0 s2 =- k + 1 = - k + 1 (k + 2) =- (k + 2) a (k + 1)

Auxiliary equation

dy x = y1 and xo = 1 dx y1 x 1 y = > H = > H = > Hx y2 2x 2 y1 = 1 u s+2 y1 (s + 2) yo1 + 2y1 xo + 2x xo xo

=u =u =u =- 2x + u = [- 2] x + [1] u Drawing SFG as shown below

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 171

Y (s) = 1 R (s) s + 1 7.15

Y (s) = s X (s) s + p jw H (jw) = jw + p H (s) =

xo1 = [- 2] x1 + [1] u y1 = x1 ; y2 = 2x1

Thus

y1 1 y = > H = > H x1 y2 2

Amplitude Response H (jw) =

x1 = x

Here 7.12

Option (C) is correct.

Phase Response

100 s (s + 10) 2 100 Now G (jw) H (jw) = jw (jw + 10) 2 If wp is phase cross over frequency +G (jw) H (jw) = 180c G (s) H (s) =

We have

- 180c = 100 tan-1 0 - tan-1 3 - 2 tan-1 a

Thus or or or or Now

Option (B) is correct. Transfer function is given as

Input Output wp 10 k

- 180c =- 90 - 2 tan-1 (0.1wp) 45c = tan-1 (0.1wp) tan 45c 0.1wp = 1 wp = 10 rad/se 100 G (jw) H (jw) = w (w2 + 100)

7.14

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100 = 1 10 (100 + 100) 20

1 = p

or or Alternative :

Option (D) is correct. From option (D) TF = H (s) 100 100 = ! s (s + 10) 2 s (s2 + 100)

So,

Option (B) is correct. From the given block diagram

or 7.16

H (s) = Y (s) - E (s) $

1 s+1

E (s) = R (s) - H (s) = R (s) - Y (s) +

E (s) (s + 1)

1 = R (s) - Y (s) s + 1D sE (s) = R (s) - Y (s) (s + 1) E (s) Y (s) = s+1

E (s) :1 -

From (1) and (2) Transfer function

sY (s) = R (s) - Y (s) (s + 1) Y (s) = R (s)

y (t) = H (jw) x (t - qh) = cos a2t - p k 3 w H (jw) = p = w2+ p 2

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Gain Margin =- 20 log 10 G (jw) H (jw) =- 20 log 10 b 1 l 20 = 26 dB 7.13

qh (w) = 90c - tan-1 a w k p x (t) = p cos a2t - p k 2

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At w = wp G (jw) H (jw) =

w w +p2 2

...(1)

2 , (w = 2 rad/ sec) 4+p2 4p 2 = 4 + p 2 & 3p 2 = 4 p = 2/ 3 qh = 9- p - a- p kC = p 3 2 6 p = p - tan-1 w apk 6 2 tan-1 a w k = p - p = p p 2 6 3 w = tan p = 3 a3k p 2 = 3 , (w = 2 rad/ sec) p p = 2/ 3

Option (A) is correct. Initial slope is zero, so K = 1 At corner frequency w 1 = 0.5 rad/ sec , slope increases by + 20 dB/ decade, so there is a zero in the transfer function at w 1 At corner frequency w 2 = 10 rad/ sec , slope decreases by - 20 dB/ decade and becomes zero, so there is a pole in transfer function at w2 K a1 + s k w1 Transfer function H (s) = s a1 + w 2 k

...(2) 7.17

Option (D) is correct. Steady state error is given as eSS = lim s"0

1 a1 + s k (1 + 10s) 0.1 = = s (1 + 0.1s) + 1 a 0.1 k

sR (s) 1 + G (s) GC (s)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

R (s) = 1 s

Page 172

D = 1 - [L1 + L2] = 1 - :- 1 - 12 D = 1 + 1 + 12 s s s s

(unit step unit)

1 1 + G (s) GC (s) 1 = lim s"0 G (s) 1+ 2 C s + 2s + 2 eSS will be minimum if lim GC (s) is maximum s"0 In option (D) lim GC (s) = lim 1 + 2 + 3s = 3 s s"0 s"0 So, eSS = lim 1 = 0 (minimum) s"0 3 eSS = lim s"0

7.18

Option (D) is correct. Assign output of each integrator by a state variable

So,

7.20

7.21

Option (C) is correct. This compensator is roughly equivalent to combining lead and lad compensators in the same design and it is referred also as PID compensator. Option (C) is correct. Here

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xo1 =- x1 + x2 xo2 =- x1 + 2u y = 0.5x1 + 0.5x2 State variable representation -1 1 0 xo = > x + > Hu H -1 0 2 yo = [0.5 0.5] x 7.19

Option (C) is correct. By masson’s gain formula

Transfer function Y (s) = H (s) = U (s) Forward path given

D1 = 1, D2 = 2 Y (s) = P1 D 1 + P2 D 2 H (s) = D U (s) 1 :1+1:1 2 (1 + s) s =s = 2 1 1 (s + s + 1) 1+ + 2 s s

0 p and B = = G G 1 q 0 p p == G G G = 1 q q p q S = 8B AB B = = q pG

1 0 1 AB = = 0 A ==

S = pq - pq = 0 Since S is singular, system is completely uncontrollable for all values of p and q . 7.22

Option (B) is correct. The characteristic equation is 1 + G (s) H (s) = 0 K (s2 - 2s + 2) or 1+ =0 s 2 + 2s + 2 or s2 + 2s + 2 + K (s2 - 2s + 2) = 0 2 or K =- s2 + 2s + 2 s - 2s + 2 For break away & break in point differentiating above w.r.t. s we have 2 2 dK =- (s - 2s + 2)( 2s + 2) - (s + 2s + 2)( 2s - 2) = 0 ds (s2 - 2s + 2) 2

Thus

(s2 - 2s + 2)( 2s + 2) - (s2 + 2s + 2)( 2s - 2) = 0

s =! 2 Let qd be the angle of departure at pole P , then or

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/ PK DK D

P1 (abcdef ) = 2 # 1 # 1 # 0.5 = 12 s s s P2 (abcdef ) = 2 # 1 # 1 # 0.5 3 Loop gain L1 (cdc) =- 1 s L2 (bcdb) = 1 # 1 # - 1 = -21 s s s

- qd - qp1 + qz1 + qz2 = 180c - qd = 180c - (- qp1 + qz1 + q2)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 173

= 180c - (90c + 180 - 45c) =- 45c 7.23

1 + G (s) H (s) = 0 1+ 2 s+8 =0 s + as - 4 or s 2 + as - 4 + s + 8 = 0 or s2 + (a + 1) s + 4 = 0 This will be stable if (a + 1) > 0 " a > - 1. Thus system is stable for all positive value of a.

Option (B) is correct. For under-damped second order response kwn2 where x < 1 s2 + 2xwn s + wn2 Thus (A) or (B) may be correct For option (A) wn = 1.12 and 2xwn = 2.59 " x = 1.12 For option (B) wn = 1.91 and 2xwn = 1.51 " x = 0.69 T (s) =

7.24

7.25

7.26

7.27

7.31

Option (B) is correct. The plot has one encirclement of origin in clockwise direction. Thus G (s) has a zero is in RHP.

1 + G (s) = 0 or s5 + 2s 4 + 3s3 + 6s2 + 5s + 3 = 0 Substituting s = z1 we have

Option (C) is correct. The Nyzuist plot intersect the real axis ate - 0.5. Thus G. M. =- 20 log x =- 20 log 0.5 = 6.020 dB And its phase margin is 90c. Option (C) is correct. Transfer function for the given pole zero plot is: (s + Z1)( s + Z2) (s + P1)( s + P2) From the plot Re (P1 and P2 )>(Z1 and Z2 ) So, these are two lead compensator. Hence both high filters and the system is high filter.

Option (C) is correct. The characteristic equation is

3z5 + 5z 4 + 6z3 + 3z2 + 2z + 1 = 0 The routh table is shown below. As there are tow sign change in first column, there are two RHS poles.

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Option (C) is correct. Percent overshoot depends only on damping ratio, x . 2

Mp = e- xp 1 - x If Mp is same then x is also same and we get x = cos q Thus q = constant The option (C) only have same angle. 7.28

Option (D) is correct. P = 2 25 2xwn = 0, x = 0 " Undamped s + 25 2

6 Q= 2 s + 20s + 62 R= S= 7.29

2

2xwn = 20, x > 1 " Overdamped 2xwn = 12, x = 1 " Critically

72 s + 7s + 72

2xwn = 7, x < 1 " underdamped

2

Graph 4 7.32

6 s + 12s + 62 2

Graph 3

Graph 1

z5

3

6

2

z4

5

3

1

z3

21 5

7 5

z2

4 3

3

z1

- 74

z0

1

Option (C) is correct. For underdamped second order system the transfer function is Kwn2 s2 + 2xwn s + wn2 It peaks at resonant frequency. Therefore T (s) =

Graph 2

wr = wn 1 - 2x2

Resonant frequency

Option (C) is correct. We labeled the given SFG as below :

and peak at this frequency mr =

7.30

From this SFG we have xo1 =- gx1 + bx3 + m1 xo2 = gx1 + ax3 xo3 =- bx1 - ax3 + u2 R V R VR V R V Sx1 W S- g 0 b WSx1 W S0 1 W u1 Sx2 W = S g 0 a WSx2 W+ S0 0 We o Thus SSx WW SS- b 0 - a WWSSx WW SS1 0 WW u2 3 3 X T X T XT X T Option (C) is correct. The characteristic equation of closed lop transfer function is

5 2x 1 - x2

We have wr = 5 2 , and mr = 10 . Only options (A) satisfy these 3 values. wn = 10, x = 1 2 where wr = 10 1 - 2` 1 j = 5 2 4 and Hence satisfied mr = 1 5 1 = 10 22 1- 4 3 7.33

Option (B) is correct. The given circuit is a inverting amplifier and transfer function is Vo = - Z = - Z (sC1 R1 + 1) R R1 Vi sC R + 1 1

1

1

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(sC2 R2 + 1) sC2 (sC2 R2 + 1) (sC1 R1 + 1) PID Controller =# sC2 R1 R2 = (sC2 R2 + 1) (sC1 R1 + 1) R2 =# (sC2 R2 + 1) R1

Z =

For Q ,

Vo Vi For R,

Z Vo Vi

Page 174

7.38

1 + G (s) = 0 K or 1+ =0 2 s (s + 7s + 12) or s (s2 + 7s + 12) + K = 0 Point s =- 1 + j lie on root locus if it satisfy above equation i.e

Since R2 C2 > R1 C1, it is lag compensator. 7.34

7.35

Option (D) is correct. In a minimum phase system, all the poles as well as zeros are on the left half of the s -plane. In given system as there is right half zero (s = 5), the system is a non-minimum phase system.

(- 1 + j)[( - 1 + j) 2 + 7 (- 1 + j) + 12) + K] = 0 or K =+ 10 7.39

Option (B) is correct. We have Kv = lim sG (s) H (s) s"0

1000 = lim s

or

s"0

(Kp + KD s) 100 = Kp s (s + 100)

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Hence 7.40

or and

Kp = 100

or

2xwn = 10 + 100KD KD = 0.9

or or

Option (D) is correct. We have

or 7.41

2

In given transfer function denominator is (s + 5)[( s + 0.5) +

3 4]

. We can see easily that pole at s =- 0.5 ! j 23 is dominant then pole at s =- 5 . Thus we have approximated it. 7.37

Option (A) is correct. 1 = 1 2 ( s 1 )( s - 1) + s -1 The lead compensator C (s) should first stabilize the plant i.e. 1 term. From only options (A), C (s) can remove this remove (s - 1) term G (s) =

Thus

satisfies.

dw dt dia dt

> H = =- 1 - 10G=in G + =10Gu -1

1 w

0

dw =- w + i n dt dia =- w - 10i + 10u a dt

sw (s) =- w (s) = Ia (s) or (s + 1) w (s) = Ia (s) Taking Laplace transform (ii) we get

or s2 + (10 + 100KD) s + 10 4 = 0 Comparing with s2 + 2xwn + wn2 = 0 we get

5 T (s) = (s + 5)( s2 + s + 1) 5 = = 2 1 5`1 + s j (s2 + s + 1) s +s+1 5

100 s (s + 1)( 1 + .05s)

...(1) ...(2)

Taking Laplace transform (i) we get

Now characteristics equation is

7.36

G (s) =

Option (A) is correct. We have

(K + KD s) 100 = Kp 1000 = lims " 0 s p s (s + 100)

or

K =5

Thus

1 + G (s) H (s) = 0

or

Option (D) is correct. At every corner frequency there is change of -20 db/decade in slope which indicate pole at every corner frequency. Thus K G (s) = s (1 + s)`1 + s j 20 Bode plot is in (1 + sT) form = 60 dB = 1000 20 log K w w = 0. 1

Now characteristics equations is

1 + G (s) H (s) = 0 (100 + KD s) 100 1+ =0 s (s + 10)

Option (D) is correct. For ufb system the characteristics equation is

10 (s - 1) 1 G (s) C (s) = # (s + 1)( s - 1) (s + 2) 10 Only option (A) = (s + 1)( s + 2)

sIa (s) =- w (s) - 10Ia (s) + 10U (s) w (s) = (- 10 - s) Ia (s) + 10U (s) = (- 10 - s)( s + 1) w (s) + 10U (s) w (s) =- [s2 + 11s + 10] w (s) + 10U (s) (s2 + 11s + 11) w (s) = 10U (s) w (s) = 2 10 U (s) (s + 11s + 11)

...(3)

From (3)

Option (A) is correct.

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 We have Let

xo (t) = Ax (t) p q A == r sG

1 For initial state vector x (0) = = G the system response is -2 e-2t x (t) = > H - 2e-2t Thus

e-2t > d (- 2e-2t)H dt d dt

t=0

or

- 2e-2 (0)

p q

==

p q 1 r s G=- 2G

1

> 4e-2 (0) H = =r s G=- 2G

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

-2 p - 2q = 4 G = = r - 2s G p - 2q =- 2 and r - 2s = 4 1 For initial state vector x (0) = = G the system response is -1 e-t x (t) = > -tH -e

We get

Thus

e-t > d (- e-t)H dt d dt

t=0

==

Page 175

Given system is 2nd order and for 2nd order system G.M. is infinite. 7.44

...(i)

7.45

7.46

p q 1 - e- (0) > e- (0) H = =r s G=- 1G -1 p-q = 1G = = r - s G

The characteristic equation lI - A = 0 l -1 =0 2 l+3

Option (D) is correct. If the Nyquist polt of G (jw) H (jw) for a closed loop system through (- 1, j0) point, the gain margin is 1 and in dB GM =- 20 log 1 = 0 dB

p q 1 r s G=- 1G

We get p - q =- 1 and r - s = 1 Solving (1) and (2) set of equations we get p q 0 1 =r s G = =- 2 - 3G

Option (D) is correct.

Option (B) is correct. The characteristics equation is 1 + G (s) H (s) = 0 1+

K (s + 1) =0 s + as2 + 2s + 1 3

s3 + as2 + (2 + K) s + K + 1 = 0 The Routh Table is shown below. For system to be oscillatory stable

...(2)

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or l (l + 3) + 2 = 0 or l =- 1, - 2 Thus Eigen values are - 1 and - 2 Eigen vectors for l1 =- 1

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(l1 I - A) X1 = 0 or

l1 - 1 x11 = 2 l + 3G=x G = 0 1 21 - 1 - 1 x11 = 2 2 G=x G = 0 21

or

as2 + K + 1 = 0 At 2 rad/sec we have s = jw " s2 =- w2 =- 4 , Thus - 4a + K + 1 = 0 Solving (i) and (ii) we get K = 2 and a = 0.75 .

7.47

7.43

Option (D) is correct.

2+K

s2

a

1+K

s1

(1 + K) a - (1 + K) a

s0

1+K

Option (D) is correct. The transfer function of given compensator is Gc (s) = 1 + 3Ts 1 + Ts Comparing with Gc (s) = 1 + aTs we get a = 3 1 + Ts The maximum phase sift is

Option (D) is correct. As shown in previous solution the system matrix is 0 1 A == - 2 - 3G

or 7.48

...(2)

1

s3

or - x11 - x21 = 0 or x11 + x21 = 0 We have only one independent equation x11 =- x21 . Let x11 = K, then x21 =- K , the Eigen vector will be x12 K 1 =x G = =- 2K G = K =- 2G 22 7.42

...(1)

Then we have

or - x11 - x21 = 0 or x11 + x21 = 0 We have only one independent equation x11 =- x21 . Let x11 = K , then x21 =- K , the Eigen vector will be 1 x11 K =x G = =- K G = K =- 1G 21 Now Eigen vector for l2 =- 2 (l2 I - A) X2 = 0 l2 - 1 x12 or = 2 l + 3G=x G = 0 2 22 - 2 - 1 x11 or = 2 1 G=x G = 0 21

a (2 + K) - (K + 1) =0 a a = K+1 K+2

fmax = tan-1 a - 1 2 a = tan-1 3 - 1 = tan-1 1 2 3 3 fmax = p 6

Option (A) is correct.

T>0

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(sI - A) = = (sI - A)

-1

s 0 0 1 s -1 -= == G G 0 s -1 0 1 sG

s s -1 s +1 1 = 2 G = > -1 = s +1 1 s s +1 2

2

f (t) = eAt = L-1 [(sI - A)] -1 7.49

1 s2 + 1 s s2 + 1

T (s) = 1 + sT 1 + sbT

H

or

and At w = 0 , At w = 0 ,

cos t sin t == - sin t cos t G

p = p + tan-1 (w a) - p g 4 p = tan-1 (w a) g 4

At w = 3 ,

7.54 7.55

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+T (jw) = tan-1 (wT) - tan-1 (wbT) T (jw) = 1 +T (jw) =- tan-1 0 = 0 T (jw) = 1 b

At w = 3 , 7.53

b > 1; T > 0

1 + w2 T2 1 + w2 b2 T2

T (jw) =

Option (C) is correct. 1 G (s) = as + We have s2 +G (jw) = tan-1 (wa) - p Since PM is p i.e. 45c, thus 4 p = p + +G (jw ) w " Gain cross over Frequeng g 4 cy or

Page 176

+T (jw) = 0

Option (A) is correct. Despite the presence of negative , control systems still have problems of instability because components used have nonlinearity. There are always some variation as compared to ideal characteristics. Option (B) is correct. Option (C) is correct. The peak percent overshoot is determined for LTI second order closed loop system with zero initial condition. It’s transfer function is wn2 s2 + 2xwn s + wn2 Transfer function has a pair of complex conjugate poles and zeroes. T (s) =

7.56

Option (A) is correct. For ramp input we have R (s) = 12 s Now ess = lim sE (s) s"0

or awg = 1 At gain crossover frequency G (jwg) = 1 Thus or or 7.50

s"0

1 + a2 wg2 =1 wg2 1 + 1 = wg2 wg = (2)

or

C (s) = G (s) R (s) = G (s) = 0.84s2 + 1 = 12 + 0.84 s s s Taking inverse Laplace transform

7.52

c (t) = (t + 0.84) u (t) c (1 sec) = 1 + 0.84 = 1.84

Option (C) is correct. We have where l is set of Eigen values Xo = AX + BU and where m is set of Eigen values Wo = CW + DU If a liner system is equivalently represented by two sets of state equations, then for both sets, states will be same but their sets of Eigne values will not be same i.e. X = W but l ! m Option (D) is correct. The transfer function of a lag network is

Finite

kv is finite for type 1 system having ramp input.

1 4

G (s) = 0.84s2 + 1 s Due to ufb system H (s) = 1 and due to unit impulse response R (s) = 1, thus

7.51

But

(as awg = 1)

Option (C) is correct. For a = 0.84 we have

At t = 1,

R (s) 1 = lim 1 + G (s) s " 0 s + sG (s) ess = lim 1 = 5% = 1 s " 0 sG (s) 20 kv = 1 = lim sG (s) = 20 s"0 ess = lim s

7.57 7.58

Option (A) is correct. Option (C) is correct. Any point on real axis of s - is part of root locus if number of OL poles and zeros to right of that point is even. Thus (B) and (C) are possible option. The characteristics equation is 1 + G (s) H (s) = 0

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1+

K (1 - s) =0 s (s + 3) 2 K = s + 3s 1-s

For break away & break in point dK = (1 - s)( 2s + 3) + s2 + 3s = 0 ds or - s2 + 2s + 3 = 0 which gives s = 3 , - 1 Here - 1 must be the break away point and 3 must be the break in point.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 7.59

Page 177

Option (D) is correct. -2s G (s) = 3e s (s + 2) -2jw or G (jw) = 3e jw (jw + 2) 3 G (jw) = w w2 + 4 Let at frequency wg the gain is 1. Thus 3 =1 wg (wg2 + 4)

7.62

7.63

or or or Now

wg4 +

4wg2

-9 = 0 wg2 = 1.606 wg = 1.26 rad/sec +G (jw) =- 2w - p - tan-1 w 2 2

Let at frequency wf we have +GH =- 180c w - p =- 2wf - p - tan-1 f 2 2 w or 2wf + tan-1 f = p 2 2 w w 3 or 2wf + c f - 1 ` f j m = p 2 2 3 2 or

or 7.60

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Click to Buy www.nodia.co.in At w = 1 change in slope is +20 dB " 1 zero at w = 1 At w = 10 change in slope is - 20 dB " 1 poles at w = 10 At w = 100 change in slope is - 20 dB " 1 poles at w = 100 K (s + 1) Thus T (s) = s s + 1) ( 10 + 1)( 100 Now 20 log10 K =- 20 " K = 0.1 0.1 (s + 1) 100 (s + 1) Thus = T (s) = s s ( 10 + 1)( 100 + 1) (s + 10)( s + 100)

wf = 0.63 rad

1 2

7.64

The phase at gain cross over frequency is w +G (jwg) =- 2wg - p - tan-1 g 2 2 =- 2 # 1.26 - p - tan-1 1.26 2 2 or 7.61

Option (C) is correct. The given bode plot is shown below

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5wf wf3 =p 2 24 2 5wf .p 2 2

Option (D) is correct. The gain at phase crossover frequency wf is 3 3 = G (jwg) = wf (wf2 + 4) 0.63 (0.632 + 4) or G (jwg) = 2.27 G.M. =- 20 log G (jwg) - 20 log 2.26 =- 7.08 dB Since G.M. is negative system is unstable.

w2 = 3 G (jw) H (jw) = 2 1 + w2 Thus gain margin is = 1 = 0 and in dB this is - 3 . 3 Option (C) is correct. Centroid is the point where all asymptotes intersects. SReal of Open Loop Pole - SReal Part of Open Loop Pole s = SNo.of Open Loop Pole - SNo.of Open Loop zero = - 1 - 3 =- 1.33 3

C (s) = H (s) $ R (s) =

=- 4.65 rad or - 266.5c PM = 180c + +G (jwg) = 180c - 266.5c =- 86.5c

Option (D) is correct. The open loop transfer function is 2 (1 + s) G (s) H (s) = s2 Substituting s = jw we have 2 (1 + jw) ...(1) G (jw) H (jw) = - w2 +G (jw) H (jw) =- 180c + tan-1 w The frequency at which phase becomes - 180c, is called phase crossover frequency. Thus - 180 =- 180c + tan-1 wf or tan-1 wf = 0 or wf = 0 The gain at wf = 0 is

Option (C) is correct. We have r (t) = 10u (t) or R (s) = 10 s Now H (s) = 1 s+2

or

1 $ 10 10 s + 2 s s (s + 2)

C (s) = 5 - 5 s s+2

c (t) = 5 [1 - e-2t] The steady state value of c (t) is 5. It will reach 99% of steady state value reaches at t , where or or or 7.65

5 [1 - e-2t] = 0.99 # 5 1 - e-2t = 0.99 e-2t = 0.1 - 2t = ln 0.1 t = 2.3 sec

Option (A) is correct. Approximate (comparable to 90c) phase shift are Due to pole at 0.01 Hz " - 90c Due to pole at 80 Hz " - 90c Due to pole at 80 Hz " 0 Due to zero at 5 Hz " 90c

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 178

Due to zero at 100 Hz " 0 Due to zero at 200 Hz " 0 Thus approximate total - 90c phase shift is provided. 7.66

We have only one independent equation x12 = 2x22 Let x22 = K , then x12 = 2K . Thus Eigen vector will be x12 2K 2 =x G = = K G = K = 1 G 22

Option (C) is correct. Mason Gain Formula T (s) =

Spk 3 k 3

Digonalizing matrix

In given SFG there is only one forward path and 3 possible loop. p1 = abcd 31 = 1 3= 1 - (sum of indivudual loops) - (Sum of two non touching loops)

Now Diagonal matrix of sin At is D where sin (l1 t) 0 sin (- 4t) 0 D == == G 0 sin (l2 t) 0 sin (l2 t)G

L1 L2 = bedg

Now matrix

C (s) p1 3 1 = 1 - (be + cf + dg) + bedg R (s)

=

abcd 1 - (be + cf + dg) + bedg

- sin (- 4t) - 2 sin (- t) 2 sin (- 4t) - 2 sin (- t) =-` 1 j= sin (- 4t) + 2 sin (t) - 2 sin (- 4t) - sin (- t)G 3 - sin (- 4t) - 2 sin (- t) 2 sin (- 4t) - 2 sin (- t) =-` 1 j= sin (- 4t) - sin (- t) - 2 sin (- 4t) + 2 sin (- t)G 3 sin (- 4t) + 2 sin (- t) - 2 sin (- 4t) + 2 sin (- t) = ` 1 j= Gs 3 - sin (- 4t + sin (- t) 2 sin (- 4t) + sin (- t) 7.68

Option (A) is correct. We have

-2 2 A == 1 - 3G

l + 2 -2 =0 -1 l + 3

or (l + 2)( l + 3) - 2 = 0 or l2 + 5l + 4 = 0 Thus l1 =- 4 and l2 =- 1 Eigen values are - 4 and - 1. Eigen vectors for l1 =- 4

s4

Now Eigen vector for l2 =- 1 (l2 I - A) X2 = 0 l2 + 2 - 2 x12 or = - 1 l + 3G=x G = 0 2 22

1

5

K

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(l1 I - A) X1 = 0 l1 + 2 - 2 x11 = 1 l + 3G=x G = 0 1 21 - 2 - 2 x11 =- 1 - 1G=x G = 0 21

or - 2x11 - 2x21 = 0 or x11 + x21 = 0 We have only one independent equation x11 =- x21 . Let x21 = K , then x11 =- K , the Eigen vector will be -K -1 x11 =x G = = K G = K = 1 G 21

1 + G (s)

K =0 s (s2 + 2s + 2)( s + 3) s 4 + 4s3 + 5s2 + 6s + K = 0 The routh table is shown below. For system to be stable, (21 - 4K) 0 < K and 0 < 2/7 This gives 0 < K < 21 4 1+

[lI - A] = 0

or

Option (A) is correct. For ufb system the characteristic equation is 1 + G (s) = 0

Characteristic equation is

or

B = sin At = MDM-1 - 1 2 sin (- 4t) 0 1 -2 =-` 1 j= G G = = 1 1 0 sin (- t) - 1 - 1G 3

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in

7.67

-1 2 x11 x12 = M == x21 x22 G = 1 1G 1 -2 M-1 = ` - 1 j= G 3 -1 -1

Now

= 1 - (L1 + L2 + L3) + (L1 L3) Non touching loop are L1 and L3 where

Thus

1 - 2 x12 =- 1 2 G=x G = 0 22

or

7.69

s3

4

6

s2

7 2

K

s1

21 - 4K 7/2

0

s0

K

0

Option (B) is correct. We have P (s) = s5 + s 4 + 2s3 + 3s + 15 The routh table is shown below. If e " 0+ then 2e +e 12 is positive and -15e2-e +2412e - 144 is negative. Thus there are two sign change in first column. Hence system has 2 root 2

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 179

on RHS of plane.

7.70

s5

1

2

3

s4

1

2

15

s3

e

- 12

0

s2

2e + 12 e

15

0

s1

2

-15e - 24e - 144 2e + 12

s0

0

7.75

Option (D) is correct. 1 - 3 - 1 x1 x1 We have =x G = = 2 0 G=x G + = 0 Gu 2 2 1 x1 and Y = [1 0]= G + = G u 2 x2 -3 -1 1 Here , B = = G and C = [1 0] A == G 2 0 0 The controllability matrix is 1 -3 QC = [B AB ] = = 0 2G det QC ! 0 The observability matrix is

Thus controllable

Q0 = [CT AT CT ] 1 -3 == !0 0 - 1G det Q0 ! 0 7.71

where L1 and L3 are non-touching C (s) This R (s) p1 3 1 = 1 - (loop gain) + pair of non - touching loops ^ s +s27 h = = 1 - ^ -s3 - 24s - s2 h + -s2 . -s3 s (s + 27) = 2 s + 29s + 6

which gives

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Thus observable

The location of poles on s - plane is

1 0 (s - 1) s-1 1 = = 0 (s - 1)G > 0 (s - 1) 2

0 1 s-1

H

Option (A) is correct.

Since breakpoint must lie on root locus so s =- 0.748 is possible. 7.76

Option (D) is correct. Mason Gain Formula Spk 3 k 3

In given SFG there is only forward path and 3 possible loop. p1 = 1 31 = 1 + 3 + 24 = s + 27 s s s L1 = - 2 , L2 = - 24 and L3 = - 3 s s s

Option (A) is correct. The given bode plot is shown below

At w = 0.1 change in slope is + 60 dB " 3 zeroes at w = 0.1 At w = 10 change in slope is - 40 dB " 2 poles at w = 10 At w = 100 change in slope is - 20 dB " 1 poles at w = 100 K ( 0s.1 + 1) 3 Thus T (s) = s s + 1) ( 10 + 1) 2 ( 100 Now 20 log10 K = 20 or K = 10 10 ( 0s.1 + 1) 3 108 (s + 0.1) 3 Thus = T (s) = s s + 1) (s + 10) 2 (s + 100) ( 10 + 1) 2 ( 100

Option (C) is correct. PD Controller may accentuate noise at higher frequency. It does not effect the type of system and it increases the damping. It also reduce the maximum overshoot.

T (s) =

s = - 10 ! 100 - 72 =- 0.784, - 2.548 6

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Z = P-N N " Net encirclement of (- 1 + j0) by Nyquist plot, P " Number of open loop poles in right hand side of s - plane Z " Number of closed loop poles in right hand side of s - plane Here N = 1 and P = 1 Thus Z =0 Hence there are no roots on RH of s -plane and system is always stable.

7.74

K =- s (s2 + 5s2 + 6s) dK =- (3s2 + 10s + 6) = 0 ds

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et 0 eAt = L-1 [(sI - A)] -1 = = G 0 et

7.73

1 + G (s) H (s) = 0 K =0 1+ s (s + 2)( s + 3)

or

s 0 1 0 s-1 0 -= == (sI - A) = = G G 0 s 0 1 0 s - 1G

7.72

s + 27

Option (D) is correct. We have or

Option (B) is correct.

(sI - A) -1 =

^ s h 1 + 29s + s62

7.77

Option (B) is correct. The characteristics equation is s2 + 4s + 4 = 0 Comparing with s2 + 2xwn + wn2 = 0

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

we get Thus

7.78 7.79

2xwn = 4 and wn2 = 4 x =1 ts = 4 = 4 = 2 xwn 1#2

Critically damped

or or

Option (C) is correct. We have 1 xo1 =xo G = =1 2 1 A == 1 s (sI - A) = = 0

7.84

0 x1 x1 (0) 1 and = = 1G=x2 G x2 (0)G = 0 G 0 1G 0 1 0 s-1 0 -= G = = G s 1 1 - 1 s - 1G

7.85

1 (s - 1) 0 s-1 1 (sI - A) = > H = > +1 (s - 1) 2 + 1 (s - 1) (s - 1) t e 0 L-1 [(sI - A) -1] = eAt = = t t G te e et 0 1 et x (t) = eAt # [x (t0)] = = t t G= G = = t G te e 0 te

0 2

1 s-1

H

7.86

2

ks + s + 6 = 0 s2 + 1 s + 6 = 0 K K

Comparing with s2 + 2xwn s + wn2 = 0 we have we get 2xwn = 1 and wn2 = 6 K K or 2 # 0.5 # 6 Kw = 1 K 6 = 1 & K =1 or 6 K K2

7.82

7.83

or or Now

y (t) = 0.5x (t)

Option (B) is correct. Routh table is shown below. Here all element in 3rd row are zero, so system is marginal stable. s5

2

4

2

s4

1

2

1

s3

0

0

0

s1 s0 Given x = 0.5

Option (D) is correct. From the expression of OLTF it may be easily see that the maximum magnitude is 0.5 and does not become 1 at any frequency. Thus gain cross over frequency does not exist. When gain cross over frequency does not exist, the phase margin is infinite.

sX (s) =- 2X (s) + 2U (s) (s + 2) X (s) = 2U (s) 2U (s) X (s) = (s + 2)

Overdamped

s2

Option (B) is correct. Any point on real axis lies on the root locus if total number of poles and zeros to the right of that point is odd. Here s =- 1.5 does not lie on real axis because there are total two poles and zeros (0 and - 1) to the right of s =- 1.5 .

Option (D) is correct. We have xo (t) =- 2x (t) + 2u (t) Taking Laplace transform we get

Option (B) is correct. The characteristics equation is

For overdamped system settling time can be determined by the dominant pole of the closed loop system. In given system dominant pole consideration is at s =- 1. Thus 1 =1 and Ts = 4 = 4 sec T T

Option (C) is correct. The characteristics equation is or

Option (D) is correct. From Mason gain formula we can write transfer function as K Y (s) K s = = 3 R (s) 1 - ( s + -sK ) s - 3 (3 - K) For system to be stable (3 - K) < 0 i.e. K > 3

(s + 1)( s + 100) = 0 s2 + 101s + 100 = 0 Comparing with s2 + 2xwn + wn2 = 0 we get 2xwn = 101 and wn2 = 100 Thus x = 101 20

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7.81

Y (s) = 0.5X (s) 0.5 # 2U (s) Y (s) = s+2 Y (s) 1 = U (s) (s + 2)

Option (B) is correct.

-1

7.80

Page 180

...(i)

7.87

Option (B) is correct. The open loop transfer function is 1 G (s) H (s) = 2 s (s + s + 1) Substituting s = jw we have 1 G (jw) H (jw) = jw (- w2 + jw + 1)

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 +G (jw) H (jw) =- p - tan-1 w 2 2 (1 - w ) The frequency at which phase becomes - 180c, is called phase crossover frequency. wf Thus - 180 =- 90 - tan-1 1 - wf2 wf or - 90 =- tan-1 1 - wf2 or 1 - w2f = 0 wf = 1 rad/sec The gain margin at this frequency wf = 1 is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 181

GM =- 20 log10 G (jwf) H (jwf)

Comparing (1) and (6) we have Z3 (s) H = Z1 (s) + Z3 (s)

= 20 log10 (wf (1 - w2f) 2 + w2f =- 20 log 1 = 0 7.88

Option (A) is correct. Z = P-N N " Net encirclement of (- 1 + j0) by Nyquist plot, P " Number of open loop poles in right had side of s - plane Z " Number of closed loop poles in right hand side of s - plane Here N = 0 (1 encirclement in CW direction and other in CCW) and P = 0 Thus Z = 0 Hence there are no roots on RH of s - plane.

7.89

7.90

7.93

or or

Option (D) is correct. Take off point is moved after G2 as shown below

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Option (C) is correct. The characteristics equation is

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2 and x = 1 2 Since x < 1 thus system is under damped Option (D) is correct. If roots of characteristics equation lie on negative axis at different positions (i.e. unequal), then system response is over damped. From the root locus diagram we see that for 0 < K < 1, the roots are on imaginary axis and for 1 < K < 5 roots are on complex plain. For K > 5 roots are again on imaginary axis. Thus system is over damped for 0 # K < 1 and K > 5 .

= G1 Vi (s) + HI2 (s) = G2 I1 (s) = G3 I2 (s) KVL in given block diagram we have

Vi (s) = I1 (s) Z1 (s) + [I1 (s) - I2 (s)] Z3 (s) 0 = [I2 (s) - I1 (s)] Z3 (s) + I2 (s) Z2 (s) + I2 (s) Z4 (s) From (4) we have or Vi (s) = I1 (s)[ Z1 (s) + Z3 (S)] - I2 (s) Z3 (S) Z3 (s) 1 or + I2 I1 (s) = Vi Z1 (s) + Z3 (s) Z1 (s) + Z3 (s)

1 + G (s) H (s) = 0 K (s - 2) 1+ (s - 2) = 0 (s + 2) 2 or (s + 2) 2 + K (s - 2) 2 = 0 or (1 + K) s2 + 4 (1 - K) s + 4K + 4 = 0 Routh Table is shown below. For System to be stable 1 + k > 0 , and 4 + 4k > 0 and 4 - 4k > 0 . This gives - 1 < K < 1 As per question for 0 # K < 1

...(1) ...(2) ...(3) ...(4) ...(5) 7.95

...(6)

From (5) we have or

I1 (s) Z3 (S) = I2 (s)[ Z2 (s) + Z3 (s) + Z4 (s)] I1 (s) Z3 (s) Is (s) = Z3 (s) + Z2 (s) + Z4 (s)

Comparing (2) and (7) we have Z3 (s) G2 = Z3 (s) + Z2 (s) + Z4 (s)

7.96

...(7)

Option (C) is correct. From the Block diagram transfer function is G (s) T (s) = 1 + G (s) H (s) K (s - 2) Where G (s) = (s + 2) and H (s) = (s - 2) The Characteristic equation is

Option (C) is correct. From SFG we have I1 (s) I2 (s) V0 (s) Now applying

G (0) = 4 9

Thus

wn =

7.92

1 = s2 + 7s + 13 - 1 = s2 + 6s + 9 G (s) s+4 s+4 G (s) = 2 s + 4 s + 6s + 9

For DC gain s = 0 , thus

s2 + 2s + 2 = 0 Comparing with s2 + 2xwn + wn2 = 0 we get 2xwn = 2 and wn2 = 2

7.91

Option (B) is correct. For unity negative system the closed loop transfer function is G (s) s+4 , CLTF = = G (s) " OL Gain 1 + G (s) s2 + 7s + 13 2 1 + G (s) or = s + 7s + 13 G (s) s+4

s2

1+k

4 + 4k

s1

4 - 4k

0

s0

4 + 4k

Option (B) is correct. It is stable at all frequencies because for resistive network factor is always less than unity. Hence overall gain decreases. Option (B) is correct. The characteristics equation is s2 + as2 + ks + 3 = 0 The Routh Table is shown below For system to be stable a > 0 and aK - 3 > 0 a Thus a > 0 and aK > 3

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

s3

1

K

s2

a

3

s1

aK - 3 a

0

s0

3

Page 182

The observability matrix is 1 2 !0 Q0 = [CT AT CT ] = = 1 - 2G 7.100

Option (D) is correct. G (s) H (s) = 2 3 s (s + 1)

we have 7.97

Option (B) is correct. Closed loop transfer function is given as T (s) = 2 9 s + 4s + 9 by comparing with standard form we get natural freq.

G (jw) H (jw) at w = w = 1 g

x =

wg =

g

g

= 180 - 150 = 30c

7.101

Option (D) is correct. Given loop transfer function is 2 s (s + 1)

2 jw (jw + 1) Phase cross over frequency can be calculated as G (jw) H (jw) =

1 G (jw) H (jwp) o

A ==

Here input

0 0 1 , B = = G and C = [1 1] G 1 2 -3

The controllability matrix is QC = [B AB ] = = det QC ! 0

7.103

0 1 1 - 3G

G (s) =

R (s) = 12 (unit Ramp) s 1 so E (s) = lim 12 s " 0 s 1 + G (s) n a1 sn - 1 + .... + an - 2 s2 = lim 12 s + s"0 s sn + a1 sn - 1 + .... + an = an - 2 an

2 =0 wp w2p + 1 G.M. = 20 log 10 b 1 l = 3 0

Option (A) is correct. Here

Option (B) is correct.

7.104

Thus controllable

an - 1 s + an sn + a1 sn - 1 + ....an - 2 s2

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1 at w = wp G (jw) H (jw) G

G (jwp) H (jwp) =

7.99

G (s) H (s) =

an - 1 s + an sn + a1 sn - 1 + ....an - 2 s2 Steady state error is given by 1 E (s) = lim R (s) s"0 1 + G (s) H (s) for unity feed back H (s) = 1 Thus

Gain margin

so

Option (C) is correct. Closed-loop transfer function is given by an - 1 s + an T (s) = n s + a1 sn - 1 + ... + an - 1 s + an an - 1 s + an n n-1 2 = s + a1 s + ...an - 2 s an - 1 s + an 1+ n s + a1 sn - 1 + ...an - 2 s2

For unity feed back H (s) = 1

p

G.M. = 20 log 10 e

Option (B) is correct.

Thus

f (w) at w = w =- 180c f (w) =- 90c - tan-1 (w) - 90c - tan-1 (wp) =- 180c tan-1 (wp) = 90c wp = 3 20 log 10 =

3 f (w) at w = w =- 90 - tan-1 (wg)

=- 90 - tan-1 3 =- 90 - 60 =- 150 Phase margin = 180 + f (w) at w = w

7.102

G (s) H (s) =

w1, w2 = ! 3

which gives

4 =4 =2 ts = 4 = xwn 3 # 2/3 2

So here

2 3 =1 w w2 + 1 12 = w2 (w2 + 1) w4 + w2 - 12 = 0 (w2 + 4) (w2 - 3) = 0 w2 = 3 and w2 =- 4

or

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2 3 jw (jw + 1)

Gain cross over frequency

4 = 2/3 2#3 for second order system the setting time for 2-percent band is given by

7.98

G (jw) H (jw) =

or

wA2 = 9 wn = 3 2xwn = 4 damping factor

Thus observable

det Q0 ! 0

7.105

Option (A) is correct.

7.106

Option (A) is correct.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 183

By applying Routh’s criteria 3

Phase is f (w) = tan-1 (3Tw) - tan-1 (Tw) w f (w) = tan-1 ; 3Tw - T 1 + 3T 2 w2 E f (w) = tan-1 ; 2Tw2 2 E 1 + 3T w For maximum value of phase df (w) =0 dw

2

s + 5s + 7s + 3 = 0 s3

1

s2

5

s1

7#5-3 5

s0

3

7 3 =

32 5

0

There is no sign change in the first column. Thus there is no root lying in the left-half plane. 7.107

7.108

Option (A) is correct. Techometer acts like a differentiator so its transfer function is of the form ks . Option (A) is correct. Open loop transfer function is K G (s) = s (s + 1) Steady state error sR (s) E (s) = lim s " 0 1 + G (s) H (s) R (s) = input

Where

H (s) = 1 (unity )

R (s) = 1 s

7.109

1 = 3T 2 w2 Tw = 1 3 So maximum phase is or

s1 s (s + 1) s so =0 = lim 2 E (s) = lim s"0 s"0 s + s + K K 1+ s (s + 1) Option (B) is correct. Fig given below shows a unit impulse input given to a zero-order hold circuit which holds the input signal for a duration T & therefore, the output is a unit step function till duration T .

fmax = tan-1 ; 2Tw2 2 E at Tw = 1 1 + 3T w 3 R V 1 S 2 W -1 3 W = tan-1 1 = 30c = tan S ; 3E SS1 + 3 # 1 WW 3 T X

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7.114

Option (A) is correct. G (jw) H (jw) enclose the (- 1, 0) point so here G (jwp) H (jwp) > 1 wp = Phase cross over frequency 1 Gain Margin = 20 log 10 G (jwp) H (jwp) so gain margin will be less than zero. Option (B) is correct. The denominator of Transfer function is called the characteristic equation of the system. so here characteristic equation is (s + 1) 2 (s + 2) = 0

h (t) = u (t) - u (t - T) Taking Laplace transform we have H (s) = 1 - 1 e-sT = 1 61 - e-sT @ s s s 7.110

7.111

7.112

7.115

7.116

Option (C) is correct. Phase margin = 180c + qg where qg = value of phase at gain crossover frequency. Here qg =- 125c so P.M = 180c - 125c = 55c Option (B) is correct. Open loop transfer function is given by K (1 + 0.5s) G (s) H (s) = s (1 + s) (1 + 2s) Close looped system is of type 1. It must be noted that type of the system is defined as no. of poles lies at origin in OLTF. lying Option (D) is correct. Transfer function of the phase lead controller is 1 + (3Tw) j T.F = 1 + 3Ts = 1+s 1 + (Tw) j

Option (C) is correct. In synchro error detector, output voltage is proportional to [w (t)], where w (t) is the rotor velocity so here n = 1 Option (C) is correct. By masson’s gain formulae / Dk Pk y = x D Forward path gain

so gain 7.117

P1 = 5 # 2 # 1 = 10 D = 1 - (2 # - 2) = 1 + 4 = 5 D1 = 1 y = 10 # 1 = 2 5 x

Option (C) is correct. By given matrix equations we can have Xo1 = dx1 = x1 - x2 + 0 dt Xo2 = dx2 = 0 + x2 + m dt x1 y = [1 1] > H = x1 + x2 x2 dy = dx1 + dx2 dt dt dt

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 184

dy = x1 + m dt dy dt

= x1 (0) + m (0) t=0

= 1+0 = 0

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

UNIT 8

Page 184 8.6

COMMUNICATION SYSTEMS

The optimum threshold to achieve minimum bit error rate (BER) is (A) 1 (B) 4 2 5 (C) 1

(D) 3 2

2012 2013 8.1

ONE MARK

The bit rate of a digital communication system is R kbits/s . The modulation used is 32-QAM. The minimum bandwidth required for ISI free transmission is (A) R/10 Hz (B) R/10 kHz (C) R/5 Hz (D) R/5 kHz 2013

8.2

8.7

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8.8

P ^3V F 2U h is

8.3

8.4

(B) 1/2 (D) 5/9

Consider two identically distributed zero-mean random variables U and V . Let the cumulative distribution functions of U and 2V be F ^x h and G ^x h respectively. Then, for all values of x (A) F ^x h - G ^x h # 0 (B) F ^x h - G ^x h $ 0 (C) ^F (x) - G (x)h .x # 0 (D) ^F (x) - G (x)h .x $ 0

8.9

8.10

Let U and V be two independent and identically distributed random variables such that P ^U =+ 1h = P ^U =- 1h = 1 . The entropy 2 H ^U + V h in bits is (A) 3/4 (B) 1 (C) 3/2 (D) log 2 3

Bits 1 and 0 are transmitted with equal probability. At the receiver, the pdf of the respective received signals for both bits are as shown below.

If the detection threshold is 1, the BER will be (A) 1 (B) 1 2 4 (C) 1 (D) 1 8 16

(A) 6000/p, 0

(B) 6400/p, 0

(C) 6400/p, 20/ (p 2 )

(D) 6000/p, 20/ (p 2 )

In a baseband communications link, frequencies upto 3500 Hz are used for signaling. Using a raised cosine pulse with 75% excess bandwidth and for no inter-symbol interference, the maxi mum possible signaling rate in symbols per second is (A) 1750 (B) 2625 (C) 4000 (D) 5250 A source alphabet consists of N symbols with the probability of the first two symbols being the same. A source encoder increases the probability of the first symbol by a small amount e and decreases that of the second by e. After encoding, the entropy of the source (A) increases (B) remains the same (C) increases only if N = 2 (D) decreases Two independent random variables X and Y are uniformly distributed in the interval 6- 1, 1@. The probability that max 6X, Y @ is less than 1/2 is (A) 3/4 (B) 9/16 (C) 1/4 (D) 2/3 2012

TWO MARKS

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Common Data for Questions 5 and 6:

8.5

The power spectral density of a real process X (t) for positive frequencies is shown below. The values of E [X 2 (t)] and E [X (t)] , respectively, are

TWO MARKS

Let U and V be two independent zero mean Gaussain random variables of variances 1 and 1 respectively. The probability 9 4

(A) 4/9 (C) 2/3

ONE MARK

8.11

8.12

A BPSK scheme operating over an AWGN channel with noise power spectral density of N 0 /2, uses equiprobable signals s1 (t) = 2E sin (wc t) and s2 (t) =- 2E sin (wc t) over the symbol T T interval (0, T). If the local oscillator in a coherent receiver is ahead in phase by 45c with respect to the received signal, the probability of error in the resulting system is (A) Q c 2E m (B) Q c E m N0 N0 E E (C) Q c (D) Q c 2N 0 m 4N 0 m A binary symmetric channel (BSC) has a transition probability of 1/8. If the binary symbol X is such that P (X = 0) = 9/10, then the

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 185

probability of error for an optimum receiver will be (B) 63/80 (A) 7/80 (C) 9/10 (D) 1/10 8.13

(D) P-2, Q-4, R-3, S-1 2011

The signal m (t) as shown is applied to both a phase modulator (with k p as the phase constant) and a frequency modulator (with k f as the frequency constant) having the same carrier frequency.

The ratio k p /k f (in rad/Hz) for the same maximum phase deviation is (A) 8p (B) 4p (C) 2p (D) p

Statement for Linked Answer Question 14 and 15 :

8.14

8.15

The transfer function of a compensator is given as Gc (s) = s + a s+b Gc (s) is a lead compensator if (A) a = 1, b = 2 (B) a = 3, b = 2 (D) a = 3, b = 1 (C) a =- 3, b =- 1

8.18

6 rad/s

X (t) is a stationary random process with auto-correlation function RX (t) = exp (- pt 2). This process is ed through the system shown below. The power spectral density of the output process Y (t) is

(A) (4p 2 f 2 + 1) exp (- pf 2) (C) (4p 2 f 2 + 1) exp (- pf ) 8.19

(B) (4p 2 f 2 - 1) exp (- pf 2) (D) (4p 2 f 2 - 1) exp (- pf )

A message signal m (t) = cos 2000pt + 4 cos 4000pt modulates the

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Click to Buy www.nodia.co.in carrier c (t) = cos 2pfc t where fc = 1 MHz to produce an AM signal. For demodulating the generated AM signal using an envelope detector, the time constant RC of the detector circuit should satisfy (A) 0.5 ms < RC < 1 ms (B) 1 μs << RC < 0.5 ms (C) RC << 1 μs (D) RC >> 0.5 ms

The phase of the above lead compensator is maximum at (A) 2 rad/s (B) 3 rad/s (C)

TWO MARKS

(D) 1/ 3 rad/s

Common Data For Q. 8.5 & 8.6 2011 8.16

8.17

ONE MARK

A four-phase and an eight-phase signal constellation are shown in the figure below.

An analog signal is band-limited to 4 kHz, sampled at the Nyquist rate and the samples are quantized into 4 levels. The quantized levels are assumed to be independent and equally probable. If we transmit two quantized samples per second, the information rate is (A) 1 bit/sec (B) 2 bits/sec (C) 3 bits/sec (D) 4 bits/sec The Column -1 lists the attributes and the Column -2 lists the modulation systems. Match the attribute to the modulation system that best meets it. Column -1

Column -2

P.

Power efficient transmission of 1. signals

Conventional AM

Q.

Most bandwidth efficient 2. transmission of voice signals

FM

R.

Simplest receiver structure

S.

Bandwidth efficient transmission 4. of signals with significant dc component

(A) P-4, Q-2, R-1, S-3 (B) P-2, Q-4, R-1, S-3 (C) P-3, Q-2, R-1, S-4

3.

8.20

8.21

VSB SSB-SC

For the constraint that the minimum distance between pairs of signal points be d for both constellations, the radii r 1 , and r 2 of the circles are (B) r 1 = 0.707d, r 2 = 1.932d (A) r 1 = 0.707d, r2 = 2.782d (C) r 1 = 0.707d, r 2 = 1.545d (D) r 1 = 0.707d, r 2 = 1.307d Assuming high SNR and that all signals are equally probable, the additional average transmitted signal energy required by the 8-PSK signal to achieve the same error probability as the 4-PSK signal is (A) 11.90 dB (B) 8.73 dB (C) 6.79 dB (D) 5.33 dB 2010

8.22

ONE MARK

Suppose that the modulating signal is m (t) = 2 cos (2pfm t) and the

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 186

carrier signal is xC (t) = AC cos (2pfC t), which one of the following is a conventional AM signal without over-modulation (A) x (t) = AC m (t) cos (2pfC t) (B) x (t) = AC [1 + m (t)] cos (2pfC t) (C) x (t) = AC cos (2pfC t) + AC m (t) cos (2pfC t) 4

gain and cut-off frequency 1 MHz. Let Yk represent the random variable y (tk ). Yk = Nk , if transmitted bit bk = 0 Yk = a + Nk if transmitted bit bk = 1 Where Nk represents the noise sample value. The noise sample has a probability density function, PNk (n) = 0.5ae- a n (This has mean zero and variance 2/a 2 ). Assume transmitted bits to be equiprobable and threshold z is set to a/2 = 10-6 V .

(D) x (t) = AC cos (2pfm t) cos (2pfC t) + AC sin (2pfm t) sin (2pfC t) 8.23

Consider an angle modulated signal x (t) = 6 cos [2p # 106 t + 2 sin (800pt)] + 4 cos (800pt) The average power of x (t) is (A) 10 W (B) 18 W (C) 20 W (D) 28 W

8.24

Consider the pulse shape s (t) as shown below. The impulse response h (t) of the filter matched to this pulse is 8.25

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8.26

8.27

8.28

The value of the parameter a (in V - 1 ) is (A) 1010 (B) 107 (C) 1.414 # 10-10 (D) 2 # 10-20 The probability of bit error is (A) 0.5 # e-3.5 (C) 0.5 # e-7

(B) 0.5 # e-5 (D) 0.5 # e-10

The Nyquist sampling rate for the signal sin (500pt) sin (700) pt is given by s (t) = # pt pt (A) 400 Hz (B) 600 Hz (C) 1200 Hz (D) 1400 Hz X (t) is a stationary process with the power spectral density Sx (f ) > 0 , for all f . The process is ed through a system shown below

Let Sy (f ) be the power spectral density of Y (t). Which one of the following statements is correct (A) Sy (f ) > 0 for all f (B) Sy (f ) = 0 for f > 1 kHz

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 (C) Sy (f ) = 0 for f = nf0, f0 = 2 kHz kHz, n any integer (D) Sy (f ) = 0 for f = (2n + 1) f0 = 1 kHz , n any integer 2009 2010

TWO MARKS

Statement for linked Answer Question : 8.10 & 8.11 : Consider a baseband binary PAM receiver shown below. The additive channel noise n (t) is with power spectral density Sn (f ) = N 0 /2 = 10-20 W/Hz . The low- filter is ideal with unity

8.29

ONE MARK

For a message siganl m (t) = cos (2pfm t) and carrier of frequency fc , which of the following represents a single side-band (SSB) signal ? (B) cos (2pfc t) (A) cos (2pfm t) cos (2pfc t) (C) cos [2p (fc + fm) t] (D) [1 + cos (2pfm t) cos (2pfc t)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

2009 8.30

TWO MARKS

Consider two independent random variables X and Y with identical distributions. The variables X and Y take values 0, 1 and 2 with probabilities 12 , 14 and 1 respectively. What is the conditional 4 probability P (X + Y = 2 X - Y = 0) ? (A) 0 (C) 1/6

8.31

8.33

2008 8.37

TWO MARKS

The probability density function (pdf) of random variable is as shown below

(B) 1/16 (D) 1 The corresponding commutative distribution function CDF has the form

A discrete random variable X takes values from 1 to 5 with probabilities as shown in the table. A student calculates the mean X as 3.5 and her teacher calculates the variance of X as 1.5. Which of the following statements is true ? k

1

2

3

4

5

P (X = k)

0.1

0.2

0.3

0.4

0.5

(A) Both the student and the teacher are right (B) Both the student and the teacher are wrong (C) The student is wrong but the teacher is right (D) The student is right but the teacher is wrong 8.32

Page 187

A message signal given by m (t) = ( 12 ) cos w1 t - ( 12 ) sin w2 t amplitude - modulated with a carrier of frequency wC to generator s (t)[ 1 + m (t)] cos wc t . What is the power efficiency achieved by this modulation scheme ? (B) 11.11% (A) 8.33% (C) 20% (D) 25%

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A communication channel with AWGN operating at a signal to noise ration SNR >> 1 and bandwidth B has capacity C1. If the SNR is doubled keeping constant, the resulting capacity C2 is given by (B) C2 . C1 + B (A) C2 . 2C1 (C) C2 . C1 + 2B (D) C2 . C1 + 0.3B

Common Data For Q. 8.19 & 8.20 : The amplitude of a random signal is uniformly distributed between -5 V and 5 V. 8.34

If the signal to quantization noise ratio required in uniformly quantizing the signal is 43.5 dB, the step of the quantization is approximately (A) 0.033 V (B) 0.05 V (C) 0.0667 V (D) 0.10 V 8.38

8.35

If the positive values of the signal are uniformly quantized with a step size of 0.05 V, and the negative values are uniformly quantized with a step size of 0.1 V, the resulting signal to quantization noise ration is approximately (A) 46 dB (B) 43.8 dB (C) 42 dB (D) 40 dB 2008

8.36

8.39

ONE MARK

Consider the amplitude modulated (AM) signal Ac cos wc t + 2 cos wm t cos wc t . For demodulating the signal using envelope detector, the minimum value of Ac should be (A) 2 (B) 1 (C) 0.5 (D) 0

8.40

A memory less source emits n symbols each with a probability p. The entropy of the source as a function of n (A) increases as log n (B) decreases as log ( n1 ) (C) increases as n (D) increases as n log n Noise with double-sided power spectral density on K over all frequencies is ed through a RC low filter with 3 dB cut-off frequency of fc . The noise power at the filter output is (A) K (B) Kfc (D) 3 (C) kpfc Consider a Binary Symmetric Channel (BSC) with probability of error being p. To transmit a bit, say 1, we transmit a sequence of three 1s. The receiver will interpret the received sequence to represent 1 if at least two bits are 1. The probability that the transmitted bit will be received in error is (B) p3 (A) p3 + 3p2 (1 - p) (D) p3 + p2 (1 - p) (C) (1 - p3)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 8.41

8.42

8.43

Page 188

(A) E [X2] - E2 [X] (C) E [X2]

Four messages band limited to W, W, 2W and 3W respectively are to be multiplexed using Time Division Multiplexing (TDM). The minimum bandwidth required for transmission of this TDM signal is (A) W (B) 3W (C) 6W (D) 7W Consider the frequency modulated signal 10 cos [2p # 105 t + 5 sin (2p # 1500t) + 7.5 sin (2p # 1000t)] with carrier frequency of 105 Hz. The modulation index is (A) 12.5 (B) 10 (C) 7.5 (D) 5 The signal cos wc t + 0.5 cos wm t sin wc t is (A) FM only (B) AM only (C) both AM and FM (D) neither AM nor FM

2007 8.50

8.51

8.52

Common Data For Q. 8.29, 8.30 and 8.31 : A speed signal, band limited to 4 kHz and peak voltage varying between +5 V and - 5 V, is sampled at the Nyquist rate. Each

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(B) E [X2] + E2 [X] (D) E2 [X]

A Hilbert transformer is a (A) non-linear system (C) time-varying system

(B) non-causal system (D) low- system

In delta modulation, the slope overload distortion can be reduced by (A) decreasing the step size (B) decreasing the granular noise (C) decreasing the sampling rate (D) increasing the step size The raised cosine pulse p (t) is used for zero ISI in digital communications. The expression for p (t) with unity roll-off factor is given by sin 4pWt p (t) = 4pWt (1 - 16W2 t2) The value of p (t) at t = 1 is 4W (A) - 0.5 (C) 0.5

8.53

TWO MARKS

(B) 0 (D) 3

In the following scheme, if the spectrum M (f) of m (t) is as shown, then the spectrum Y (f) of y (t) will be

sample is quantized and represented by 8 bits. 8.44

8.45

8.46

If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is (A) 64 kHz (B) 32 kHz (C) 8 kHz (D) 4 kHz Assuming the signal to be uniformly distributed between its peak to peak value, the signal to noise ratio at the quantizer output is (A) 16 dB (B) 32 dB (C) 48 dB (D) 4 kHz Assuming the signal to be uniformly distributed between its peak to peak value, the signal to noise ratio at the quantizer output is (A) 1024 (B) 512 (C) 256 (D) 64 2007

8.47

8.48

If S (f) is the power spectral density of a real, wide-sense stationary random process, then which of the following is ALWAYS true? (B) S (f) $ 0 (A) S (0) # S (f) (D)

3

#- 3 S (f) df = 0

If E denotes expectation, the variance of a random variable X is given by

During transmission over a certain binary communication channel,

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ONE MARK

If R (t) is the auto correlation function of a real, wide-sense stationary random process, then which of the following is NOT true (A) R (t) = R (- t) (B) R (t) # R (0) (C) R (t) =- R (- t) (D) The mean square value of the process is R (0)

(C) S (- f) =- S (f) 8.49

8.54

bit errors occur independently with probability p. The probability of AT MOST one bit in error in a block of n bits is given by (A) pn (B) 1 - pn (C) np (1 - p) n - 1 + (1 + p) n 8.55

(D) 1 - (1 - p) n

In a GSM system, 8 channels can co-exist in 200 kHz bandwidth using TDMA. A GSM based cellular operator is allocated 5 MHz bandwidth. Assuming a frequency reuse factor of 1 , i.e. a five-cell 5 repeat pattern, the maximum number of simultaneous channels that can exist in one cell is (A) 200 (B) 40 (C) 25 (D) 5

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

8.56

In a Direct Sequence CDMA system the chip rate is 1.2288 # 106 chips per second. If the processing gain is desired to be AT LEAST 100, the data rate (A) must be less than or equal to 12.288 # 103 bits per sec (B) must be greater than 12.288 # 103 bits per sec (C) must be exactly equal to 12.288 # 103 bits per sec (D) can take any value less than 122.88 # 103 bits per sec

Common Data For Q. 8.41 & 8.42 :

Page 189

2006 8.61

8.58

If these constellations are used for digital communications over an AWGN channel, then which of the following statements is true ? (A) Probability of symbol error for Constellation 1 is lower (B) Probability of symbol error for Constellation 1 is higher (C) Probability of symbol error is equal for both the constellations (D) The value of N0 will determine which of the constellations has a lower probability of symbol error

8.62

8.60

The values of a and b are (B) a = 1 and b = 3 (A) a = 1 and b = 1 12 5 40 6 (C) a = 1 and b = 1 (D) a = 1 and b = 1 4 16 3 24 Assuming that the reconstruction levels of the quantizer are the mid-points of the decision boundaries, the ratio of signal power to quantization noise power is (B) 64 (A) 152 3 9 (C) 76 (D) 28 3

A signal with bandwidth 500 Hz is first multiplied by a signal g (t) where 3

/(- 1)k d (t - 0.5 # 10- 4 k)

R =- 3

The resulting signal is then ed through an ideal low filter with bandwidth 1 kHz. The output of the low filter would be (A) d (t) (B) m (t) (C) 0 (D) m (t) d (t) The minimum sampling frequency (in samples/sec) required to

SPECIAL EDITION ( STUDY MATERIAL FORM ) At market Book is available in 3 volume i.e. in 3 book binding form. But at NODIA Online Store book is available in 10 book binding form. Each unit of Book is in separate binding. Available Only at NODIA Online Store

Click to Buy www.nodia.co.in reconstruct the following signal from its samples without distortion 3 2 x (t) = 5` sin 2p100t j + 7` sin 2p100t j would be pt pt 3 3 (B) 4 # 10 (A) 2 # 10 3 (C) 6 # 10 (D) 8 # 103 8.64

The minimum step-size required for a Delta-Modulator operating at 32k samples/sec to track the signal (here u (t) is the unit-step function) x (t) = 125[ u (t) - u (t - 1) + (250t)[ u (t - 1) - u (t - 2)] so that slope-overload is avoided, would be (A) 2 - 10 (B) 2 - 8 (C) 2 - 6

8.65

8.66

8.59

TWO MARKS

g (t) =

Statement for Linked Answer Question 8.44 & 8.45 : An input to a 6-level quantizer has the probability density function f (x) as shown in the figure. Decision boundaries of the quantizer are chosen so as to maximize the entropy of the quantizer output. It is given that 3 consecutive decision boundaries are’ - 1'.'0' and ' 1' .

(D) A (w) = C, f (w) = kw- 1

2006

8.63

The if ratio or the average energy of Constellation 1 to the average energy of Constellation 2 is (A) 4a2 (B) 4 (C) 2 (D) 8

A low- filter having a frequency response H (jw) = A (w) e jf (w) does not produce any phase distortions if (B) A (w) = Cw2, f (w) = kw (A) A (w) = Cw3, f (w) = kw3 (C) A (w) = Cw, f (w) = kw2

Two 4-array signal constellations are shown. It is given that f1 and f2 constitute an orthonormal basis for the two constellation. Assume that the four symbols in both the constellations are equiprobable. Let N0 denote the power spectral density of white 2 Gaussian noise.

8.57

ONE MARK

8.67

(D) 2 - 4

A zero-mean white Gaussian noise is es through an ideal low filter of bandwidth 10 kHz. The output is then uniformly sampled with sampling period ts = 0.03 msec. The samples so obtained would be (A) correlated (B) statistically independent (C) uncorrelated (D) orthogonal A source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit rate is (A) 6000 bits/sec (B) 4500 bits/sec (C) 3000 bits/sec (D) 1500 bits/sec The diagonal clipping in Amplitude Demodulation (using envelop detector) can be avoided it RC time-constant of the envelope detector satisfies the following condition, (here W is message bandwidth and w is carrier frequency both in rad/sec) (B) RC > 1 (A) RC < 1 W W

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

8.68

Page 190

(C) RC < 1 (D) RC > 1 w w A uniformly distributed random variable X with probability density function fx (x) = 1 pu (x + 5) - u (x - 5)] 10

signal and modulation index would be (A) t, 0.5 (B) t, 1.0 (C) t, 2.0 (D) t2, 0.5

where u (.) is the unit step function is ed through a transformation given in the figure below. The probability density function of the transformed random variable Y would be

The following two question refer to wide sense stationary stochastic process

Common Data For Q. 8.58 & 8.59 :

8.73

(A) fy (y) = 1 [u (y + 2.5) - u (y - 2.25)] 5 (B) fy (y) = 0.5d (y) + 0.5d (y - 1) (C) fy (y) = 0.25d (y + 2.5) + 0.25d (y - 2.5) + 5d (y) (D) fy (y) = 0.25d (y + 2.5) + 0.25d (y - 2.5) + 1 [u (y + 2.5) - u (y - 2.5)] 10

8.74

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in 8.69

Consider the following Amplitude Modulated (AM) signal, where fm < B XAM (t) = 10 (1 + 0.5 sin 2pfm t) cos 2pfc t 8.75

8.76

(A) 3

8.70

2 (C) 3 (D) 3 12 L A message signal with bandwidth 10 kHz is Lower-Side Band SSB modulated with carrier frequency fc1 = 106 Hz. The resulting signal is then ed through a Narrow-Band Frequency Modulator with carrier frequency fc2 = 109 Hz. The bandwidth of the output would be (B) 2 # 106 Hz (A) 4 # 10 4 Hz (D) 2 # 1010 Hz (C) 2 # 109 Hz

The parameters of the system obtained in previous Q would be (A) first order R-L low filter would have R = 4W L = 1H (B) first order R-C high filter would have R = 4W C = 0.25F (C) tuned L-C filter would have L = 4H C = 4F (D) series R-L-C low filter would have R = 1W , L = 4H , C = 4F

Common Data For Q. 8.60 & 8.61 :

In the following figure the minimum value of the constant "C" , which is to be added to y1 (t) such that y1 (t) and y2 (t) are different , is

(B) 3 2

It is desired to generate a stochastic process (as voltage process) with power spectral density S (w) = 16/ (16 + w2) by driving a Linear-Time-Invariant system by zero mean white noise (As voltage process) with power spectral density being constant equal to 1. The system which can perform the desired task could be (A) first order low R-L filter (B) first order high R-C filter (C) tuned L-C filter (D) series R-L-C filter

The average side-band power for the AM signal given above is (A) 25 (B) 12.5 (C) 6.25 (D) 3.125 The AM signal gets added to a noise with Power Spectral Density Sn (f) given in the figure below. The ratio of average sideband power to mean noise power would be :

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Common Data For Q. 8.56 & 8.57 : Let g (t) = p (t)*( pt), where * denotes convolution & p (t) = u (t) - u (t - 1) lim with u (t) being the unit step function z"3

8.71

8.72

The impulse response of filter matched s (t) = g (t) - d (1 - 2)* g (t) is given as : (A) s (1 - t) (B) - s (1 - t) (C) - s (t) (D) s (t)

to

the

signal

An Amplitude Modulated signal is given as xAM (t) = 100 [p (t) + 0.5g (t)] cos wc t in the interval 0 # t # 1. One set of possible values of modulating

(A)

25 8N0 B

(B)

25 4N0 B

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(C)

25 2N0 B

(D) 25 N0 B

2005 8.77

Page 191

the probability density function as shown in the figure. The mean square value of v is ONE MARK

Find the correct match between group 1 and group 2. Group 1 Group 2 W. Phase modulation P. {1 + km (t) A sin (wc t)} Q. km (t) A sin (wc t) X. Frequency modulation R. A sin {wc t + km (t)} Y. Amplitude modulation t S. A sin ; wc t + k m (t) dt E Z. DSB-SC modulation

(A) 4 (C) 8

#- 3

(A) (B) (C) (D) 8.78

P - Z, Q - Y, R - X, S - W P - W, Q - X, R - Y, S - Z P - X, Q - W, R - Z, S - Y P - Y, Q - Z, R - W, S - X

Which of the following analog minimum transmitted power and (A) VSB (C) SSB 2005

8.79

8.80

8.83

modulation scheme requires the minimum channel bandwidth ? (B) DSB-SC (D) AM TWO MARKS

A device with input X (t) and output y (t) is characterized by: Y (t) = x2 (t). An FM signal with frequency deviation of 90 kHz and modulating signal bandwidth of 5 kHz is applied to this device. The bandwidth of the output signal is (A) 370 kHz (B) 190 kHz (C) 380 kHz (D) 95 kHz

(B) 6 (D) 9

A carrier is phase modulated (PM) with frequency deviation of 10 kHz by a single tone frequency of 1 kHz. If the single tone frequency is increased to 2 kHz, assuming that phase deviation remains unchanged, the bandwidth of the PM signal is (A) 21 kHz (B) 22 kHz (C) 42 kHz (D) 44 kHz

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Click to Buy www.nodia.co.in Common Data For Q. 8.69 and 8.70 : Asymmetric three-level midtread quantizer is to be designed assuming equiprobable occurrence of all quantization levels.

A signal as shown in the figure is applied to a matched filter. Which of the following does represent the output of this matched filter ?

8.84

8.85

If the probability density function is divide into three regions as shown in the figure, the value of a in the figure is (B) 2 (A) 1 3 3 (C) 1 (D) 1 2 4 The quantization noise power for the quantization region between - a and + a in the figure is (B) 1 (A) 4 9 81 (C) 5 (D) 2 81 81 2004

8.86

8.81

8.82

Noise with uniform power spectral density of N0 W/Hz is ed though a filter H (w) = 2 exp (- jwtd ) followed by an ideal filter of bandwidth B Hz. The output noise power in Watts is (A) 2N0 B (B) 4N0 B (C) 8N0 B (D) 16N0 B An output of a communication channel is a random variable v with

In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor (B) 12 (A) 8 6 (C) 16

8.87

ONE MARK

(D) 8

An AM signal is detected using an envelop detector. The carrier frequency and modulating signal frequency are 1 MHz and 2 kHz respectively. An appropriate value for the time constant of the envelop detector is (A) 500m sec (B) 20m sec

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(C) 0.2m sec 8.88

8.89

8.90

Page 192

symbols are represented as shown in the figure. The source output is transmitted using two modulation schemes, namely Binary PSK (BPSK) and Quadrature PSK (QPSK). Let B1 and B2 be the bandwidth requirements of the above rectangular pulses is 10 kHz, B1 and B2 are

(D) 1m sec

An AM signal and a narrow-band FM signal with identical carriers, modulating signals and modulation indices of 0.1 are added together. The resultant signal can be closely approximated by (A) broadband FM (B) SSB with carrier (C) DSB-SC (D) SSB without carrier In the output of a DM speech encoder, the consecutive pulses are of opposite polarity during time interval t1 # t # t2 . This indicates that during this interval (A) the input to the modulator is essentially constant (B) the modulator is going through slope overload (C) the accumulator is in saturation (D) the speech signal is being sampled at the Nyquist rate

(A) B1 = 20 kHz, B2 = 20 kHz (C) B1 = 20 khz, B2 = 10 kHz 8.94

The distribution function Fx (x) of a random variable x is shown in the figure. The probability that X = 1 is

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2004 8.91

8.92

8.93

(B) 0.25 (D) 0.30 TWO MARKS

A 1 mW video signal having a bandwidth of 100 MHz is transmitted to a receiver through cable that has 40 dB loss. If the effective oneside noise spectral density at the receiver is 10 - 20 Watt/Hz, then the signal-to-noise ratio at the receiver is (A) 50 dB (B) 30 dB (C) 40 dB (D) 60 dB

(A) 1 sec - 1 (B) - 1 sec - 1 2 (C) - 1 sec - 1 (D) 1 sec - 1 2 A source produces binary data at the rate of 10 kbps. The binary

(B)

1 + sin (2p # 106 t)

5 - sin (2p - 106 t) 5 + cos (2p # 106 t) (D) 4 4 Two sinusoidal signals of same amplitude and frequencies 10 kHz and 10.1 kHz are added together. The combined signal is given to an ideal frequency detector. The output of the detector is (A) 0.1 kHz sinusoid (B) 20.1 kHz sinusoid (C) a linear function of time (D) a constant (C)

8.96

Consider the signal x (t) shown in Fig. Let h (t) denote the impulse response of the filter matched to x (t), with h (t) being non-zero only in the interval 0 to 4 sec. The slope of h (t) in the interval 3 < t < 4 sec is

A 100 MHz carrier of 1 V amplitude and a 1 MHz modulating signal of 1 V amplitude are fed to a balanced modulator. The ourput of the modulator is ed through an ideal high- filter with cutoff frequency of 100 MHz. The output of the filter is added with 100 MHz signal of 1 V amplitude and 90c phase shift as shown in the figure. The envelope of the resultant signal is

(A) constant

8.95

(A) zero (C) 0.55

(B) B1 = 10 kHz, B2 = 20 kHz (D) B1 = 10 kHz, B2 = 10 kHz

Consider a binary digital communication system with equally likely 0’s and 1’s. When binary 0 is transmitted the detector input can lie between the levels - 0.25 V and + 0.25 V with equl probability : when binary 1 is transmitted, the voltage at the detector can have any value between 0 and 1 V with equal probability. If the detector has a threshold of 0.2 V (i.e., if the received signal is greater than 0.2 V, the bit is taken as 1), the average bit error probability is (A) 0.15 (B) 0.2 (C) 0.05 (D) 0.5

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 8.97

8.98

A random variable X with uniform density in the interval 0 to 1 is quantized as follows : If 0 # X # 0.3 , xq = 0 If 0.3 < X # 1, xq = 0.7 where xq is the quantized value of X. The root-mean square value of the quantization noise is (A) 0.573 (B) 0.198 (D) 0.266 (C) 2.205 Choose the current one from among the alternative A, B, C, D after matching an item from Group 1 with the most appropriate item in Group 2. Group 1 Group 2

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

1. 2. 3. 4.

P. Slope overload Q. m-law R. Envelope detector S. Hilbert transform T. Hilbert transform U. Matched filter (A) 1 - T, 2 - P, 3 - U, 4 - S (B) 1 - S, 2 - U, 3 - P, 4 - T (C) 1 - S, 2 - P, 3 - U, 4 - Q (D) 1 - U, 2 - R, 3 - S, 4 - Q 8.99

8.100

Page 193

Common Data For Q. 8.90 & 8.91 :

FM DM PSK PCM

Three analog signals, having bandwidths 1200 Hz, 600 Hz and 600 Hz, are sampled at their respective Nyquist rates, encoded with 12 bit words, and time division multiplexed. The bit rate for the multiplexed. The bit rate for the multiplexed signal is (A) 115.2 kbps (B) 28.8 kbps (C) 57.6 kbps (D) 38.4 kbps Consider a system shown in the figure. Let X (f) and Y (f) and denote the Fourier transforms of x (t) and y (t) respectively. The ideal HPF has the cutoff frequency 10 kHz.

X (t) is a random process with a constant mean value of 2 and the auto correlation function Rxx (t) = 4 (e - 0.2 t + 1). 8.105

Let X be the Gaussian random variable obtained by sampling the process at t = ti and let 3 Q (a) = - 1 e dy a 2p The probability that 6x # 1@ is (B) Q (0.5) (A) 1 - Q (0.5) 1 (C) Q c (D) 1 - Q c 1 m m 2 2 2 2 Let Y and Z be the random variable obtained by sampling X (t) at t = 2 and t = 4 respectively. Let W = Y - Z . The variance of W is (A) 13.36 (B) 9.36 (C) 2.64 (D) 8.00

#

8.106

8.107

x2 2

A sinusoidal signal with peak-to-peak amplitude of 1.536 V is quantized into 128 levels using a mid-rise uniform quantizer. The

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The positive frequencies where Y (f) has spectral peaks are (A) 1 kHz and 24 kHz (B) 2 kHz and 244 kHz (C) 1 kHz and 14 kHz (D) 2 kHz and 14 kHz 2003 8.101

8.102

8.103

ONE MARK

(B) 48 # 10 - 6 V2

(B) 12 # 10 - 6 V2

(D) 3.072 V

Let x (t) = 2 cos (800p) + cos (1400pt). x (t) is sampled with the rectangular pulse train shown in the figure. The only spectral components (in kHz) present in the sampled signal in the frequency range 2.5 kHz to 3.5 kHz are

The input to a coherent detector is DSB-SC signal plus noise. The noise at the detector output is (A) the in-phase component (B) the quadrature - component (C) zero (D) the envelope The noise at the input to an ideal frequency detector is white. The detector is operating above threshold. The power spectral density of the noise at the output is (A) raised - cosine (B) flat (C) parabolic (D) Gaussian

(A) 2.7, 3.4 (C) 2.6, 2.7, 3.3, 3.4, 3.6 8.109

At a given probability of error, binary coherent FSK is inferior to binary coherent PSK by. (A) 6 dB (B) 3 dB (C) 2 dB (D) 0 dB 2003

8.104

8.108

quantization-noise power is (A) 0.768 V

(B) 3.3, 3.6 (D) 2.7, 3.3

A DSB-SC signal is to be generated with a carrier frequency fc = 1 MHz using a non-linear device with the input-output characteristic V0 = a0 vi + a1 vi3 where a0 and a1 are constants. The output of the non-linear device can be filtered by an appropriate band- filter. Let Vi = Aci cos (2pfi ct) + m (t) is the message signal. Then the value of fci (in MHz) is (A) 1.0 (B) 0.333 (B) 0.5 (D) 3.0

TWO MARKS

Let X and Y be two statistically independent random variables uniformly distributed in the ranges (- 1, 1) and (- 2, 1) respectively. Let Z = X + Y . Then the probability that (z #- 1) is (A) zero (B) 1 6 (C) 1 (D) 1 3 12

Common Data For Q. 8.95 & 8.96 : Let m (t) = cos [(4p # 103) t] be the message signal & c (t) = 5 cos [(2p # 106 t)] be the carrier. 8.110

c (t) and m (t) are used to generate an AM signal. The modulation index of the generated AM signal is 0.5. Then the quantity Total sideband power is Carrier power

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) 1 2 (C) 1 3 8.111

8.112

(B) 1 4 (D) 1 8

c (t) and m (t) are used to generated an FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM signal, then the coefficient of the term cos [2p (1008 # 103 t)] in the FM signal (in of the Bessel coefficients) is (B) 5 J8 (3) (A) 5J4 (3) 2 (C) 5 J8 (4) (D) 5J4 (6) 2 Choose the correct one from among the alternative A, B, C, D after matching an item in Group 1 with most appropriate item in Group 2. Group 1 Group 2 P. Ring modulator 1. Clock recovery Q. VCO 2. Demodulation of FM R. Foster-Seely discriminator 3. Frequency conversion

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(A) P - 1; Q - 3; R - 2; S - 4 (C) P - 6; Q - 1; R - 3; S - 2 8.113

8.114

8.115

8.116

4. Summing the two inputs 5. Generation of FM 6. Generation of DSB-Sc (B) P - 6; Q = 5; R - 2; S - 3 (D) P - 5; Q - 6; R - 1; S - 3

Page 194

(A) r = 0.5, S is required (C) r = 0.5, S is not required 8.117

The input to a linear delta modulator having a step-size 3= 0.628 is a sine wave with frequency fm and peak amplitude Em . If the sampling frequency fx = 40 kHz, the combination of the sine-wave frequency and the peak amplitude, where slope overload will take place is Em fm (A) 0.3 V 8 kHz (B) 1.5 V 4 kHz (C) 1.5 V 2 kHz (D) 3.0 V 1 kHz If S represents the carrier synchronization at the receiver and r represents the bandwidth efficiency, then the correct statement for the coherent binary PSK is

A signal is sampled at 8 kHz and is quantized using 8 - bit uniform quantizer. Assuming SNRq for a sinusoidal signal, the correct statement for PCM signal with a bit rate of R is (A) R = 32 kbps, SNRq = 25.8 dB (B) R = 64 kbps, SNRq = 49.8 dB (C) R = 64 kbps, SNRq = 55.8 dB (D) R = 32 kbps, SNRq = 49.8 dB 2002

8.118

8.119

ONE MARK

A 2 MHz sinusoidal carrier amplitude modulated by symmetrical square wave of period 100 m sec . Which of the following frequencies will NOT be present in the modulated signal ? (A) 990 kHz (B) 1010 kHz (C) 1020 kHz (D) 1030 kHz Consider a sample signal y (t) = 5 # 10 - 6 # (t)

+3

/ d (t - nTs)

n =- 3

where x (t) = 10 cos (8p # 103) t and Ts = 100m sec. When y (t) is ed through an ideal low filter with a cutoff frequency of 5 KHz, the output of the filter is (A) 5 # 10 - 6 cos (8p # 103) t (b) 5 # 10 - 5 cos (8p # 103) t (C) 5 # 10 - 1 cos (8p # 103) t 8.120

8.121

A superheterodyne receiver is to operate in the frequency range 550 kHz - 1650 kHz, with the intermediate frequency of 450 kHz. Let R = Cmax /Cmin denote the required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, then (B) R = 2.10, I - 1150 (A) R = 4.41, I = 1600 (C) R = 3.0, I = 600 (D) R = 9.0, I = 1150 If Eb , the energy per bit of a binary digital signal, is 10 - 5 wattsec and the one-sided power spectral density of the white noise, N0 = 10 - 6 W/Hz, then the output SNR of the matched filter is (A) 26 dB (B) 10 dB (C) 20 dB (D) 13 dB

(B) r = 1.0, S is required (D) r = 1.0, S is not required

(D) 10 cos (8p # 103) t

For a bit-rate of 8 Kbps, the best possible values of the transmitted frequencies in a coherent binary FSK system are (A) 16 kHz and 20 kHz (C) 20 kHz and 32 kHz (C) 20 kHz and 40 kHz (D) 32 kHz and 40 kHz The line-of-sight communication requires the transmit and receive antennas to face each other. If the transmit antenna is vertically polarized, for best reception the receiver antenna should be (A) horizontally polarized (B) vertically polarized (C) at 45c with respect to horizontal polarization (D) at 45c with respect to vertical polarization 2002

8.122

TWO MARKS

An angle-modulated signal is given by s (t) = cos 2p (2 # 106 t + 30 sin 150t + 40 cos 150t).

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 The maximum frequency and phase deviations of s (t) are (A) 10.5 kHz, 140p rad (B) 6 kHz, 80p rad (C) 10.5 kHz, 100p rad (D) 7.5 kHz, 100p rad 8.123

In the figure m (t) = 2 sin 2pt , s (t) = cos 200pt and n (t) = sin 199pt t t . The output y (t) will be

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

8.124

8.125

(A) sin 2pt (B) sin 2pt + sin pt cos 3pt t t t (C) sin 2pt + sin 0.5pt cos 1.5pt (D) sin 2pt + sin pt cos 0.75pt t t t t 3 A signal x (t) = 100 cos (24p # 10 ) t is ideally sampled with a sampling period of 50m sec ana then ed through an ideal low filter with cutoff frequency of 15 kHz. Which of the following frequencies is/are present at the filter output ? (A) 12 kHz only (B) 8 kHz only (C) 12 kHz and 9 kHz (D) 12 kHz and 8 kHz If the variance ax2 of d (n) = x (n) - x (n - 1) is one-tenth the variance ax2 of stationary zero-mean discrete-time signal x (n), then R (k) the normalized autocorrelation function xx 2 at k = 1 is (A) 0.95 (B) 0.90 ax (C) 0.10

8.127

of K0 m (t) is greater than 1. Which of the following could be the detector output ? (A) Ac m (t) (B) Ac2 [1 + Ka m (t)] 2 (C) [Ac (1 + Ka m (t)] (D) Ac [1 + Ka m (t)] 2 8.133

ONE MARK

A bandlimited signal is sampled at the Nyquist rate. The signal can be recovered by ing the samples through (A) an RC filter (B) an envelope detector (C) a PLL (D) an ideal low- filter with the appropriate bandwidth

The frequency range for satellite communication is (A) 1 KHz to 100 KHz (B) 100 KHz to 10 KHz (C) 10 MHz to 30 MHz (D) 1 GHz to 30 GHz 2000

8.134

(D) 0.05

2001 8.126

Page 195

8.135

TWO MARKS

In a digital communication system employing Frequency Shift Keying (FSK), the 0 and 1 bit are represented by sine waves of 10 KHz and 25 KHz respectively. These waveforms will be orthogonal for a bit interval of (A) 45m sec (B) 200m sec (C) 50m sec (D) 250m sec A message m (t) bandlimited to the frequency fm has a power of Pm

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The PDF of a Gaussian random variable X is given by (x - 4) px (x) = 1 e - 18 . The probability of the event {X = 4} is 3 2p 1 1 (B) (A) 2 3 2p (C) 0 (D) 1 4 2

2001 8.128

8.129

8.130

8.131

The Nyquist sampling interval, for the signal sin c (700t) + sin c (500t) is (A) 1 sec (B) p sec 350 350 (C) 1 sec (D) p sec 700 175 During transmission over a communication channel, bit errors occur independently with probability p. If a block of n bits is transmitted, the probability of at most one bit error is equal to (B) p + (n - 1)( 1 - p) (A) 1 - (1 - p) n n-1

n

(D) (1 - p) + np (1 - p)

n-1

8.136

8.137

8.138

In a FM system, a carrier of 100 MHz modulated by a sinusoidal signal of 5 KHz. The bandwidth by Carson’s approximation is 1 MHz. If y (t) = (modulated waveform) 3 , than by using Carson’s approximation, the bandwidth of y (t) around 300 MHz and the and the spacing of spectral components are, respectively. (A) 3 MHz, 5 KHz (B) 1 MHz, 15 KHz (C) 3 MHz, 15 KHz (D) 1 MHz, 5 KHz ONE MARK

The input to a channel is a band signal. It is obtained by linearly modulating a sinusoidal carrier with a single-tone signal. The output of the channel due to this input is given by y (t) = (1/100) cos (100t - 10-6) cos (106 t - 1.56) The group delay (tg) and the phase delay (t p) in seconds, of the channel are (A) tg = 10-6, t p = 1.56 (B) tg = 1.56, t p = 10-6 (C) tg = 108, t p = 1.56 # 10-6 (D) tg = 108, t p = 1.56

ONE MARK

The amplitude modulated waveform s (t) = Ac [1 + Ka m (t)] cos wc t is fed to an ideal envelope detector. The maximum magnitude

(B) Pm 4

2 2 (C) Pm sin q (D) Pm cos q 4 4 The Hilbert transform of cos w1 t + sin w2 t is (A) sin w1 t - cos w2 t (B) sin w1 t + cos w2 t (C) cos w1 t - sin w2 t (D) sin w1 t + sin w2 t

1999

The PSD and the power of a signal g (t) are, respectively, Sg (w) and Pg . The PSD and the power of the signal ag (t) are, respectively, (A) a2 Sg (w) and a2 Pg (B) a2 Sg (w) and aPg (C) aSg (w) and a2 Pg (D) aSg (w) and aPs 2000

8.132

(A) Pm cos q 2

TWO MARKS

A video transmission system transmits 625 picture frames per second. Each frame consists of a 400 # 400 pixel grid with 64 intensity levels per pixel. The data rate of the system is (A) 16 Mbps (B) 100 Mbps (C) 600 Mbps (D) 6.4 Gbps

(C) np (1 - p)

. The power of the output signal in the figure is

8.139

A modulated signal is given by s (t) = m1 (t) cos (2pfc t) + m2 (t) sin (2pfc t)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

where the baseband signal m1 (t) and m2 (t) have bandwidths of 10 kHz, and 15 kHz, respectively. The bandwidth of the modulated signal, in kHz, is (A) 10 (B) 15 (C) 25 (D) 30 8.140

A modulated signal is given by s (t) = e-at cos [(wc + Dw) t] u (t), where a wc and Dw are positive constants, and wc >> Dw . The complex envelope of s (t) is given by (A) exp (- at) exp [j (wc + Dw) t] u (t) (B) exp (- at) exp (jDwt) u (t) (C) exp (jDwt) u (t) (D) exp [jwc + Dw) t] 1999

8.141

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8.142

8.143

8.144

(D) 1000

8.146

8.148

8.149

The input to a matched filter is given by 6 -4 sec "10 0 sin (2p # 10 t) 0 < 1 < 10 s (t) = otherwise The peak amplitude of the filter output is (A) 10 volts (B) 5 volts (C) 10 millivolts (D) 5 millivolts

8.151

8.153

8.154

The image channel selectivity of superheterodyne receiver depends upon (A) IF amplifiers only (B) RF and IF amplifiers only (C) Preselector, RF and IF amplifiers (D) Preselector, and RF amplifiers only In a PCM system with uniform quantisation, increasing the number of bits from 8 to 9 will reduce the quantisation noise power by a factor of (A) 9 (B) 8 (C) 4 (D) 2 Flat top sampling of low signals (A) gives rise to aperture effect (B) implies oversampling (C) leads to aliasing (D) introduces delay distortion A DSB-SC signal is generated using the carrier cos (we t + q) and modulating signal x (t). The envelope of the DSB-SC signal is (A) x (t) (B) x (t) (C) only positive portion of x (t) (D) x (t) cos q Quadrature multiplexing is (A) the same as FDM (B) the same as TDM (C) a combination of FDM and TDM (D) quite different from FDM and TDM The Fourier transform of a voltage signal x (t) is X (f). The unit of X (f) is (A) volt (B) volt-sec (C) volt/sec (D) volt 2 Compression in PCM refers to relative compression of (A) higher signal amplitudes (B) lower signal amplitudes (C) lower signal frequencies (D) higher signal frequencies For a give data rate, the bandwidth B p of a BPSK signal and the bandwidth B 0 of the OOK signal are related as (A) B p = B 0 (B) B p = B 0 2 4 (C) B p = B 0

Four independent messages have bandwidths of 100 Hz, 200 Hz and 400 Hz , respectively. Each is sampled at the Nyquist rate, and the samples are time division multiplexed (TDM) and transmitted. The transmitted sample rate (in Hz) is (A) 1600 (B) 800 (C) 400 (D) 200

8.155

(D) B p = 2B 0

The spectral density of a real valued random process has

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 (A) an even symmetry (C) a conjugate symmetry

ONE MARK

The amplitude spectrum of a Gaussian pulse is (A) uniform (B) a sine function (C) Gaussian (D) an impulse function The ACF of a rectangular pulse of duration T is (A) a rectangular pulse of duration T (B) a rectangular pulse of duration 2T (C) a triangular pulse of duration T (D) a triangular pulse of duration 2T

8.150

8.152

The peak-to-peak input to an 8-bit PCM coder is 2 volts. The signal power-to-quantization noise power ratio (in dB) for an input of 0.5 cos (wm t) is (A) 47.8 (B) 49.8 (C) 95.6 (D) 99.6

1998 8.145

8.147

TWO MARKS

The Nyquist sampling frequency (in Hz) of a signal given by 6 # 10 4 sin c2 (400t) * 106 sin c3 (100t) is (A) 200 (B) 300

(C) 500

Page 196

8.156

The probability density function of the envelope of narrow band Gaussian noise is (A) Poisson (B) Gaussian (C) Rayleigh (D) Rician 1997

8.157

(B) an odd symmetry (D) no symmetry

ONE MARK

The line code that has zero dc component for pulse transmission of random binary data is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 197

(A) Non-return to zero (NRZ) (B) Return to zero (RZ) (C) Alternate Mark Inversion (AM) (D) None of the above 8.158

8.159

from (A) IF stages only (B) RF stages only (C) detector and RF stages only (D) detector RF and IF stages 1996

2

A probability density function is given by p (x) = Ke-x /2 - 3 < x < 3 . The value of K should be 2 (B) (A) 1 p 2p (C) 1 (D) 1 2 p p 2 A deterministic signal has the power spectrum given in the figure is, The minimum sampling rate needed to completely represent this signal is

8.164

The number of bits in a binary PCM system is increased from n to n + 1. As a result, the signal to quantization noise ratio will improve by a factor (A) n + 1 (B) 2(n + 1)/n n (C) 22 (n + 1)/n

8.165

8.166

TWO MARKS

(D) which is independent of n

The auto correlation function of an energy signal has (A) no symmetry (B) conjugate symmetry (C) odd symmetry (D) even symmetry An FM signal with a modulation index 9 is applied to a frequency tripler. The modulation index in the output signal will be

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(A) 1 kHz (C) 3 kHz 8.160

8.161

8.163

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A communication channel has first order low transfer function. The channel is used to transmit pulses at a symbol rate greater than the half-power frequency of the low function. Which of the network shown in the figure is can be used to equalise the received pulses?

The power spectral density of a deterministic signal is given by [sin (f) /f 2] where f is frequency. The auto correlation function of this signal in the time domain is (A) a rectangular pulse (B) a delta function (C) a sine pulse (D) a triangular pulse 1996

8.162

(B) 2 kHz (D) None of these

ONE MARK

A rectangular pulse of duration T is applied to a filter matched to this input. The out put of the filter is a (A) rectangular pulse of duration T (B) rectangular pulse of duration 2T (C) triangular pulse (D) sine function The image channel rejection in a superheterodyne receiver comes

(A) 0 (C) 9

(B) 3 (D) 27

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 198

SOLUTIONS 8.1

Option (B) is correct. In ideal Nyquist Channel, bandwidth required for ISI (Inter Symbol reference) free transmission is W = Rb 2 Here, the used modulation is 32 - QAM (Quantum Amplitude modulation i.e., q = 32 or 2v = 32 v = 5 bits So, the signaling rate (sampling rate) is Rb = R 5 (R " given bit rate)

for which we have X = 2U = 0 Y = 2V = 0 and X2 = 4U2 = 1 also, Y2 = 9V2 = 1 Therefore, X - Y is also a normal random variable with X-Y = 0 Hence, P ^X - Y $ 0h = P ^X - Y # 0h = 1 2 or, we can say P ^2U - 3V # 0h = 1 2 Thus, P ^3V $ 2U h = 1 2 8.3

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8.2

Hence, for ISI free transmission, minimum bandwidth is W = Rb = R kHz 2 10 Option (B) is correct. Given, random variables U and V with mean zero and variances 1 4 and 1 9 i.e., U =V=0 su2 = 1 4 1 2 and sv = 9 so, P ^U $ 0h = 1 2 1 and P ^V $ 0h = 2 The distribution is shown in the figure below

8.4

Option (C) is correct. The mean of random variables U and V are both zero i.e., U =V=0 Also, the random variables are identical i.e., fU ^u h = fV ^v h or, FU ^u h = FV ^v h i.e., their cdf are also same. So, FU ^u h = F2V ^2v h i.e., the cdf of random variable 2V will be also same but for any instant 2V $ U Therefore, G ^x h = F ^x h but, x G ^x h $ xF ^x h or, 6F ^x h - G ^x h@x # 0 Option (C) is correct.

P ^U =+ 1h = P ^U =- 1h = 1 2 where U is a random variable which is identical to V i.e., P ^V =+ 1h = P ^V =- 1h = 1 2 So, random variable U and V can have following values U =+ 1, - 1; V =+ 1, - 1 Therefore the random variable U + V can have the following values, U+V - 2 When U = V =- 1 = *0 When U = 1,V = 1 or u =- 1, v = 1 2 When U = V = 1 Hence, we obtain the probabilities for U + V as follows Given,

P ^U + V h

U+V

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 1 1=1 2#2 4 1 1 1 1 1 b2 # 2l+b2 # 2l = 2

-2 0 fu ^u h = fv ^v h =

1 e -u 2s 2p su2

2 u

1 e -v 2s 2p sv2 We can express the distribution in standard form by assuming X = u - 0 = u = 2U su Y2 v 0 and = v = 3V Y = sv Y3 2 v

1 1=1 2#2 4 Therefore, the entropy of the ^U + V h is obtained as 2

H ^U + V h =

/ P^U + V h log

1 ' P ^U + V h 1 = 1 log 2 4 + 1 log 2 2 + 1 log 2 4 2 4 4 2 1 2 = + + 4 2 4 2

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 199

= 1 ;2 b 1 # 1 # 103 # 6 l + 400E p 2 = 1 66000 + 400@ = 6400 p p E [X (t)] is the absolute value of mean of signal X (t) which is also equal to value of X (w) at (w = 0). From given PSD

=3 2 8.5

Option (D) is correct. For the shown received signal, we conclude that if 0 is the transmitted signal then the received signal will be also zero as the threshold is 1 and the pdf of bit 0 is not crossing 1. Again, we can observe that there is an error when bit 1 is received as it crosses the threshold. The probability of error is given by the area enclosed by the 1 bit pdf (shown by shaded region)

SX (w) w = 0 = 0 SX (w) = X (w) 2 = 0 X (w) 2w = 0 = 0 X (w) w = 0 = 0 8.8

P (error when bit 1 received) = 1 # 1 # 0.25 = 1 8 2 or P b received 1 l = 1 8 transmitted 0 Since, the 1 and 0 transmission is equiprobable: i.e., P ^ 0 h = P ^1 h = 1 2 Hence bit error rate (BER) is BER 0 received received 1 P 1 = Pb l P 0 + P b transmitted 0l ^ h transmitted 1 ^ h = 0+1 #1 8 2 1 = 16 8.6

8.7

Option (B) is correct. The optimum threshold is the threshold value for transmission as obtained at the intersection of two pdf. From the shown pdf. We obtain at the intersection (transmitted, received) = b 4 , 1 l 5 5 we can obtain the intersection by solving the two linear eqs pdf of received bit 0 x+y = 1 y = 0.5 x pdf of received bit 1 2 4 Hence for threshold = , we have 5 BER = P b received 1 l P ^0 h + P b received 0 l P ^1 h transmitted 0 transmitted 1 = b1 # 1 # 1l# 1 +b1 # 4 # 1l# 1 2 5 2 2 2 5 5 2 = 1 <(BER for threshold = 1) 20 Hence, optimum threshold is 4 5 Option (A) is correct. The mean square value of a stationary process equals the total area under the graph of power spectral density, that is E [X 2 (t)] = or, or,

#S 3

-3

E [X 2 (t)] = 1 2p

X

Option (C) is correct. For raised cosine spectrum transmission bandwidth is given as a " Roll of factor BT = W (1 + a) BT = Rb (1 + a) Rb " Maximum signaling rate 2

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Click to Buy www.nodia.co.in 3500 = Rb (1 + 0.75) 2 Rb = 3500 # 2 = 4000 1.75 8.9

Option (D) is correct. Entropy function of a discrete memory less system is given as H =

3

-3

X

k

k

where Pk is probability of symbol Sk . For first two symbols probability is same, so N-1

H = P1 log b 1 l + P2 log b 1 l + Pk log b 1 l P1 P2 Pk k=3

/

=-e P1 log P1 + P2 log P2 +

N-1

/ P log P o k

k

k=3

=-e 2P log P +

N-1

/ P log P o k

k

(P1 = P2 = P)

k=3

P1 = P + e, P2 = P - e

Now, So,

Hl =-=(P + e) log (P + e) + (P - e) log (P - e) +

N-1

/ P log P G k

k

k=3

By comparing, 8.10

Hl < H , Entropy of source decreases.

Option (B) is correct. Probability density function of uniformly distributed variables X and Y is shown as

(w) dw

3 E [X 2 (t)] = 2 # 1 SX (w) dw (Since the PSD is even) 2p 0 = 1 [area under the triangle + integration of delta function] p

/ P log b P1 l

k=0

(f ) df

#S

N-1

#

P &[max (x, y)] < 1 0 2

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Since X and Y are independent. P &[max (x, y)] < 1 0 2 PbX < 1 l 2 Similarly for Y : P bY < 1 l 2 So P &[max (x, y)] < 1 0 2

= P b X < 1 l P bY < 1 l 2 2 = shaded area = 3 4 =3 4 =3#3= 9 4 4 16

Alternate Method: From the given data since random variables X and Y lies in the interval [- 1, 1] as from the figure X , Y lies in the region of the square ABCD .

Page 200

where 0 # t # T , E is the transmitted energy per bit. General function of local oscillator 2 sin (w t), 0 # t < T f1 (t) = c T But here local oscillator is ahead with 45c. so, 2 sin (w t + 45c) f1 (t) = c T The coordinates of message points are s11 = =

T

# s (t) f (t) dt #

1

2E sin w t c T

T

0

=

2E T

=

2E T

= 1 T

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1

0

E

Similarly, Signal space diagram

2 sin (w t + 45c) dt c T

T

# sin (w t) sin (w t + 45c) dt c

0

2 T

#

0

T

#

0

T

c

1 [sin 45c + sin (2w t + 45c)] dt c 2

1 dt + 1 E Tsin (2w t + 45c) dt c T 0 2 1444444 42 4 44444 3 0 = E 2

#

s21 =-

E 2

Now here the two message points are s11 and s21 . The error at the receiver will be considered. When : (i) s11 is transmitted and s21 received (ii) s21 is transmitted and s11 received So, probability for the 1st case will be as : P b s21 received l = P (X < 0) (as shown in diagram) s11 transmitted = P _ E/2 + N < 0i = P _N < - E/2 i Taking the Gaussian distribution as shown below :

8.11

Probability for max 6X, Y @ < 1/2 : The points for max 6X, Y @ < 1/2 will be inside the region of square AEFG . So, P &max 6X, Y @ < 1 0 = Area of 4AEFG 2 Area of square ABCD 3 3 #2 2 = = 9 2#2 16 Option (B) is correct. In a coherent binary PSK system, the pair of signals s1 (t) and s2 (t) used to represent binary system 1 and 0 respectively. s1 (t) = 2E sin wc t T s2 (t) =- 2E sin wc t T

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 201

Note: For phase lead compensator zero is nearer to the origin as compared to pole, so option (C) can not be true.

Mean of the Gaussian distribution =

E/2 Variance = N 0 2 Putting it in the probability function : P bN < -

Taking,

E = 2l

#

0

=

#

0

1

`x + E/2 j

2N 0 /2

f = tan-1 a w k - tan-1 a w k a b 1/a 1/b df = =0 2 2 dw 1 +awk 1 +awk a b 1 + w2 = 1 + 1 w2 a ab2 b b a2 1 - 1 = w2 1 - 1 a b ab b a b l

dx

0

-3

x + E/2 =t N 0 /2 N 0 dt 2

dx = So, P _N < - E/2 i =

#

E/N 0

w = ab = 1 # 2 =

E N0 m

1 e- t2 dt Q c 2p 2

3

8.16

where Q is error function. Since symbols are equiprobable in the 2 nd case So, P b s11 received l = Q c E m N0 s21 transmitted So the average probability of error = 1 ;P b s21 received l + P b s11 received lE 2 s11 transmitted s21 transmitted = 1 =Q c 2 8.12 8.13

Option (A) is correct.

2

e-

2p N 0 2 2 1 e- `x + NE/2 j dx pN 0

-3

8.15

E +Q c N0 m

E =Q c N 0 mG

E N0 m

Option (D) is correct. Quantized 4 level require 2 bit representation i.e. for one sample 2 bit are required. Since 2 sample per second are transmitted we require 4 bit to be transmitted per second.

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Option ( ) is correct. Option (B) is correct. General equation of FM and PM waves are given by fFM (t) = Ac cos ;wc t + 2pk f

8.17

t

# m (t) dtE 0

fPM (t) = Ac cos [wc t + k p m (t)] For same maximum phase deviation. t

k p [m (t)] max = 2pk f ; m (t) dtE

# 0

max

k p # 2 = 2pk f [x (t)] max where,

x (t) =

2 rad/ sec

t

# m (t) dt

8.18

0

Option (B) is correct. In FM the amplitude is constant and power is efficient transmitted. No variation in power. There is most bandwidth efficient transmission in SSB- SC. because we transmit only one side band. Simple Diode in Non linear region ( Square law ) is used in conventional AM that is simplest receiver structure. In VSB dc. component exists. Option (A) is correct. We have Sx (f) = F {Rx (t)} = F {exp (- pt2)} 2

= e- pf The given circuit can be simplified as

So,

Power spectral density of output is

[x (t)] max = 4 k p # 2 = 2pk f # 4 kp = 4p kf

Sy (f) = G (f) 2 Sx (f) = j2pf - 1 2 e- pf

= ( (2pf) 2 + 1) 2 e- pf or

8.14

Option (A) is correct. jw + a GC (s) = s + a = s+b jw + b Phase lead angle f = tan-1 a w k - tan-1 a w k a b Jw - wN -1 K a b O = tan-1 w (b - a) f = tan c ab + w 2 m 2 O KK 1+ w O ab P L For phase-lead compensation f > 0 b-a > 0 b >a

2

8.19

Sy (f) = (4p2 f 2 + 1) e- pf

2

Option (B) is correct. Highest frequency component in m (t) is fm = 4000p/2p = 2000 Hz Carrier frequency fC = 1 MHz For Envelope detector condition 1/fC << RC << 1/fm 1 μs << RC << 0.5 ms

8.20

2

Option (D) is correct. Four phase signal constellation is shown below

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 202

Where m < 1, for no over modulation. In option (C) x (t) = AC :1 + 1 m (t)D cos (2pfC t) 4 Thus m = 1 < 1 and this is a conventional AM-signal without over4 modulation 8.23

2

2 1

d = r +r d2 = 2r 12

Now

2 1

Option (B) is correct. (6) 2 = 18 W P = 2

Power

r1 = d/ 2 = 0.707d

8.24

Option (C) is correct. Impulse response of the matched filter is given by h (t) = S (T - t)

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8.25

q = 2p = 2p = p 8 4 M Applying Cooine law we have

1, H (f ) = * 0,

d2 = r 22 + r 22 - 2r 22 cos p 4 = 2r 22 - 2r 22 1/ 2 = (2 d r2 = = 1.3065d 2- 2

or 8.21

Option (B) is correct. Let response of LPF filters

Noise variance (power) is given as 2 ) r 22

P = s2 =

#0

Option (D) is correct. Here Pe for 4 PSK and 8 PSK is same because Pe depends on d . Since Pe is same, d is same for 4 PSK and 8 PSK. or 8.26

#0

fo

2 H (f ) No df = 22 (given) a

2 # 10-20 df = 22 a -20 6 2 # 10 # 10 = 22 a 2 a = 1014 a = 107

1 # 106

Option (D) is correct. Probability of error is given by

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Additional Power SNR = (SNR) 2 - (SNR) 1 = 10 log b ES2 l - 10 log b ES1 l No No = 10 log b ES2 l ES1

Pe = 1 [P (0/1) + P (1/0)] 2 P (0/1) =

Additional SNR = 5.33 dB Option (C) is correct. Conventional AM signal is given by x (t) = AC [1 + mm (t)] cos (2pfC t)

a/2

#- 3 0.5e- a n - a dn = 0.5e-10

where a = 2 # 10-6 V and a = 107 V - 1 P (1/0) =

2 = 10 log a r2 k & 20 log a r2 k = 20 log 1.3065d r1 r1 0.707d

8.22

f < 1 MHz elsewhere

#a/32 0.5e- a n dn

= 0.5e-10

Pe = 0.5e-10 8.27

Option (C) is correct. S (t) = sin c (500t) sin c (700t) S (f ) is convolution of two signals whose spectrum covers f 1 = 250 Hz

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 203

and f 2 = 350 Hz . So convolution extends f = 25 + 350 = 600 Hz Nyquist sampling rate

X2 = Sxi2 pi (x) = 1 # 0.1 + 4 # 0.2 + 9 # 0.4 + 16 # 0.2 + 25 # 0.1 = 0.1 + 0.8 + 3.6 + 3.2 + 2.5 = 10.2

N = 2f = 2 # 600 = 1200 Hz 8.28

Option (D) is correct. For the given system, output is written as y (t) = d [x (t) + x (t - 0.5)] dt dx (t) dx (t - 0.5) + y (t) = dt dt Taking Laplace on both sides of above equation

sx2 = X2 - ^X h2 = 10.2 - (3) 2 = 1.2

Variance 8.32

Option (C) is correct. m (t) = 1 cos w1 t - 1 sin w2 t 2 2 Modulation index

Y (s) = sX (s) + se-0.5s X (s) Y (s) = s (1 + e-0.5s) H (s) = X (s) H (f ) = jf (1 + e-0.5 # 2pf ) = jf (1 + e- pf ) Power spectral density of output SY (f ) = H (f ) 2 SX (f ) = f 2 (1 + e- pf ) 2 SX (f ) For SY (f ) = 0 , 1 + e- pf = 0 f = (2n + 1) f0 or f0 = 1 KHz 8.29

Option (C) is correct. cos (2pfm t) cos (2pfc t) cos (2pfc t) cos [2p (fc + fm) t] [1 + cos (2pfm t) cos (2pfc t)]

8.30

$ $ $ $

DSB suppressed carrier Carrier Only USB Only USB with carrier

=

Click to Buy www.nodia.co.in Option (B) is correct. C1 = B log2 `1 + S j N . B log2 ` S j N

As S >> 1 N

If we double the S ratio then N

C2 . B log2 ` 2S j N

. B log2 2 + B log2 S . B + C1 N

8.34

P (A + B) P (B)

Event P (A + B) happen when X + Y = 2 and X - Y = 0 . It is only the case when X = 1 and Y = 1. Thus P (A + B) = 1 # 1 = 1 4 4 16 Now event P (B) happen when X - Y = 0 It occurs when X = Y , i.e. X = 0 and Y = 0 or X = 1 and Y = 1 or X = 2 and Y = 2 Thus P (B) = 1 # 1 + 1 # 1 + 1 # 1 = 6 2 2 4 4 4 4 16 1/16 P (A + B) Now = =1 6 P (B) 6/16 8.31

# 100% = 20%

We have

X+Y = 2 $ A X-Y = 0 $ B P (X + Y = 2 X - Y = 0) =

2

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p (X = 0) = p (Y = 0) = 1 2 p (X = 1) = p (Y = 1) = 1 4 p (X = 2) = p (Y = 2) = 1 4 Let and Now

1 2 2j 1 2 + 2j

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8.33

Option (C) is correct. We have

`

`

sAM (t) = [1 + m (t)] cos wc t m (t) max = Vc 2 2 m = `1j +`1j = 1 2 2 2 2 h = m # 100% m2 + 2

Option (B) is correct. The mean is X = Sxi pi (x) = 1 # 0.1 + 2 # 0.2 + 3 # 0.4 + 4 # 0.2 + 5 # 0.1 = 0.1 + 0.4 + 1.2 + 0.8 + 0.5 = 3.0

Option (C) is correct. We have SNR = 1.76 + 6n or 43.5 = 1.76 + 6n 6n = 43.5 + 1.76 6n = 41.74 $ n . 7 No. of quantization level is 27 = 128 Step size required is = VH - VL = 128

5 - (- 5) = 10 128 128

= .078125 . .0667 8.35

Option (B) is correct. For positive values step size s+ = 0.05 V For negative value step size s- = 0.1 V No. of quantization in + ive is = 5 = 5 = 100 s+ 0.05 Thus 2n + = 100 $ n+ = 7 No. of quantization in - ve

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 204

Q1 = 5 = 5 = 50 s0.1

= 1 [2W + 2W + 4W + 6W] = 7W 2

2n = 50 $ n - = 6 -

Thus

8.42

+ S ` N j+ = 1.76 + 6n = 1.76 + 42 = 43.76 dB S ` N j- = 1.76 + 6n = 1.76 + 36 = 37.76 dB

Best 8.36

Maximum frequency deviation is

S ` N j0 = 43.76 dB

3wmax = 2p (5 # 1500 + 7.5 # 1000) 3 fmax = 15000 3f Modulation index is = max = 15000 = 10 fm 1500

Option (A) is correct. We have xAM (t) = Ac cos wc + 2 cos wm t cos wc t = AC c1 + 2 cos wm t m cos wc t Ac

For demodulation by envelope demodulator modulation index must be less than or equal to 1. 2 #1 Thus Ac

8.43

Option (C) is correct.

8.44

Option (B) is correct. fm = 4 KHz fs = 2fm = 8 kHz Bit Rate Rb = nfs = 8 # 8 = 64 kbps The minimum transmission bandwidth is BW = Rb = 32 kHz 2

Ac $ 2 Hence minimum value of Ac = 2

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8.38

8.45

8.46

m

bits

8.47

p1 = p2 = ... = pn = 1 n

8.48

H =

n

/ n1 log n = log n

i=1 8.39

Option (C) is correct. PSD of noise is

8.49

N0 = K 2

...(1)

The 3-dB cut off frequency is fc = 1 2pRC

8.50

...(2)

= N0 = c N0 m 1 = Kpfc 4RC 2 2RC Option (D) is correct. At receiving end if we get two zero or three zero then its error. Let p be the probability of 1 bit error, the probability that transmitted bit error is = Three zero + two zero and single one = 3 C3 p3 + 3C2 p2 (1 - p) = p3 + p2 (1 - p) 8.41

Option (D) is correct. Bandwidth of TDM is = 1 (sum of Nyquist Rate) 2

Option (C) is correct. Autocorrelation is even function. Option (B) is correct. Power spectral density is non negative. Thus it is always zero or greater than zero. Option (A) is correct. The variance of a random variable x is given by E [X2] - E2 [X] Option (A) is correct.

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Output noise power is

8.40

Option (B) is correct. Noise \ 12 L To reduce quantization noise by factor 4, quantization level must be two times i.e. 2L . Now L = 2n = 28 = 256 Thus 2L = 512

i=1

Since

We have n = 8

As

Option (A) is correct. The entropy is

/ pi log2 p1i

Option (C) is correct. S0 c N m = 1.76 + 6n dB 0

= 1.76 + 6 # 8 = 49.76 dB

Option (A) is correct. CDF is the integration of PDF. Plot in option (A) is the integration of plot given in question.

H =

Option (B) is correct. We have qi = 2p105 t + 5 sin (2p1500t) + 7.5 sin (2p1000t) wi = dqi = 2p105 + 10p1500 cos (2p1500t) + 15p1000 cos (2p1000t) dt

A Hilbert transformer is a non-linear system. 8.51

8.52

Option (D) is correct. Slope overload distortion can be reduced by increasing the step size 3 $ slope of x (t) Ts Option (C) is correct. We have

p (t) =

sin (4pWt) 4pWt (1 - 16W2 t2)

at t = 1 it is 0 form. Thus applying L' Hospital rule 0 4W 4pW cos (4pWt) p( ) = 4pW [1 - 48W2 t2] 1 4W

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

= 8.53

Page 205

cos (4pWt) = cos p = 0.5 1-3 1 - 48W2 t2

Option (B) is correct. The block diagram is as shown below

6 Rb max = RC = 1.2288 # 10 = 12.288 # 103 bps Gmin 100 8.57

Option (B) is correct. Energy of constellation 1 is Eg1 2

2 a) 2 + ( 2 a) 2 + (- 2 2 a) 2 = 2a2 + 2a2 + 2a2 + 8a2 = 16a2 Energy of constellation 2 is = (0) + (-

2

2 a) + (-

Eg2 = a2 + a2 + a2 + a2 = 4a2 2 E Ratio = g1 = 16a2 = 4 Eg2 4a Here

M1 (f) = Mt (f)

8.58

Y1 (f) = M (f) c e

j2 p B

Y2 (f) = M1 (f) c e

j 2p B

Noise Power is same for both which is N0 . 2 Thus probability of error will be lower for the constellation 1 as it has higher signal energy.

+ e -j 2 p B m 2

- e -j2pB m 2

Y (f) = Y1 (f) + Y2 (f) All waveform is shown below

Option (A) is correct.

8.59

Option (A) is correct.

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Click to Buy www.nodia.co.in Area under the pdf curve must be unity Thus 2a + 4a + 4b = 1 ...(1) 2a + 8b = 1 For maximum entropy three region must by equivaprobable thus ...(2) 2a = 4b = 4b From (1) and (2) we get b = 1 and a = 1 12 6 8.60 8.61

8.54

Option (C) is correct. By Binomial distribution the probability of error is n

r

n-r

pe = Cr p (1 - p) Probability of at most one error = Probability of no error + Probability of one error = n C0 p0 (1 - p) n - 0 + n C1 p1 (1 - p) n - 1

Option (*) is correct. Option (B) is correct. A LPF will not produce phase distortion if phase varies linearly with frequency. i.e.

8.62

f (w) \ w f (w) = kw

Option (B) is correct. Let m (t) is a low signal, whose frequency spectra is shown below

= (1 - p) n + np (1 - p) n - 1 8.55

8.56

Option (B) is correct. Bandwidth allocated for 1 Channel = 5 M Hz Average bandwidth for 1 Channel 5 = 1 MHz 5 Total Number of Simultaneously Channel = 1M # 8 = 40 Channel 200k Option (A) is correct. Chip Rate RC = 1.2288 # 106 chips/sec Data Rate Rb = RC G Since the processing gain G must be at least 100, thus for Gmin we get

Fourier transform of g (t) 3

1 d (f - 20 # 103 k) 0.5 # 10-4 k =- 3 Spectrum of G (f ) is shown below G (t) =

/

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 206

= 0.25 log2 4 + 0.25 log2 4 + 0.5 log2 2 = 0.5 + 0.5 + 1 = 3 bits/symbol 2 2 Rb = 3000 symbol/sec = Rb H = 3 # 3000 = 4500 bits/sec 2

Average bit rate

Now when m (t) is sampled with above signal the spectrum of sampled signal will look like.

When sampled signal is ed through a LP filter of BW 1 kHz, only m (t) will remain. 8.63

Option (C) is correct.

8.67

We can say that RC depends on W , thus RC < 1 W 8.68 8.69

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8.70

Option (B) is correct. We have

8.71

x (t) = 125t [u (t) - u (t - 1)] + (250 - 125t) [u (t - 1) - u (t - 2)] The slope of expression x (t) is 125 and sampling frequency fs is 32 # 1000 samples/sec. Let 3 be the step size, then to avoid slope overload 3 $ slope x (t) Ts 3 fc $ slope x (t) 3# 32000 $ 125 3 $ 125 32000 Option (A) is correct. The sampling frequency is fs =

1 = 33 kHz 0.03m

Option (B) is correct. We have p1 = 0.25 , p2 = 0.25 and p3 = 0.5 H =

3

/ p1 log2 p11

Option (C) is correct. After the SSB modulation the frequency of signal will be fc - fm i.e. 1000 - 10 kHz . 1000 kHz The bandwidth of FM is

Option (A) is correct. We have p (t) = u (t) - u (t - 1) g (t) = p (t)* p (t) s (t) = g (t) - d (t - 2)* g (t) = g (t) - g (t - 2) All signal are shown in figure below :

The impulse response of matched filter is h (t) = s (T - t) = s (1 - t) Here T is the time where output SNR is maximum.

Since fs $ 2fm , the signal can be recovered and are correlated. 8.66

Option (B) is correct. When 3 /2 is added to y (t) then signal will move to next quantization level. Otherwise if they have step size less than 3 then they will be 2 on the same quantization level.

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3 = 2- 8 8.65

Option (B) is correct.

BW = 2 (b + 1) 3 f For NBFMb << 1, thus BWNBFM . 2 3 f = 2 (109 - 106) . 2 # 109

The highest frequency signal in x (t) is 1000 # 3 = 3 kHz if expression is expanded. Thus minimum frequency requirement is f = 2 # 3 # 103 = 6 # 103 Hz 8.64

Option (A) is correct. The diagonal clipping in AM using envelop detector can be avoided if 1 << RC < 1 wc W 1 $ Wm sin Wt But from RC 1 + m cos Wt

bits/symbol

i=1

= p1 log2 1 + p2 log2 1 + p3 log2 1 p1 p2 p3 = 0.25 log2 1 + 0.25 log2 1 + 0.5 log2 1 0.25 0.25 0.5

8.72

Option (A) is correct. We have xAM (t) = 10 [P (t) + 0.5g (t)] cos wc t where p (t) = u (t) - u (t - 1) and g (t) = r (t) - 2r (t - 1) + r (t - 2) For desired interval 0 # t # 1, p (t) = 1 and g (t) = t , Thus we have, xAM (t) = 100 (1 - 0.5t) cos wc t

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

8.73

Page 207

Hence modulation index is 0.5

The transfer function of matched filter is

Option (A) is correct. We know that SYY (w) = H (w) 2 .SXX (w) Now SYY (w) = 16 2 and SXX (w) = 1 white noise 16 + w

h (t) = x (t - t) = x (2 - t) The output of matched filter is the convolution of x (t) and h (t) as shown below

16 = H (w) 2 16 + w2 4 H (w) = 16 + w2 H (s) = 4 4+s

Thus or or

which is a first order low RL filter. 8.74

Option (A) is correct. R = 4 4+s R + sL

We have

8.81

Option (B) is correct. H (f) = 2e - jwt

We have

R L

= 4 4+s +s Comparing we get L = 1 H and R = 4W or

8.75

8.76

R L

Option (C) is correct. We have xAM (t) = 10 (1 + 0.5 sin 2pfm t) cos 2pfc t The modulation index is 0.5 Carrier power

Pc =

(10) 2 = 50 2

Side band power

Ps =

(10) 2 = 50 2

Side band power

2 (0.5) 2 (50) = 6.25 Ps = m Pc = 2 2

Option (B) is correct. Mean noise power = Area under the PSD curve = 4 ; 1 # B # No E = BNo 2 2

H (f) = 2 G0 (f) = H (f) 2 Gi (f)

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8.82

$ Amplitude modulation $ DSB-SC modulation $ Phase Modulation $ Frequency Modulation

Option (A) is correct. Let x (t) be the input signal where x (t) = cos (cos t + b1 cos wm t) cos (2wc t + 2b1 cos wm t) y (t) = x2 (t) = 1 + 2 2 3f Here = 90 = 18 b = 2b1 and b1 = fm 5 BW = 2 (b + 1) fm = 2 (2 # 18 + 1) # 5 = 370 kHz

8.80

Now mean square value is sv2 =

Option (C) is correct.

+3

#- 3

v2 p (v) dv

v2 ` v j dv 8 3 4 v = c 8 m dv = 8 0 =

#0

4

as p (v) = 1 v 8

#

8.83

Option (C) is correct. VSB $ fm + fc DSB - SC $ 2fm SSB $ fm AM $ 2fm Thus SSB has minimum bandwidth and it require minimum power.

8.79

Option (C) is correct. As the area under pdf curve must be unity 1 (4 # k) = 1 $ k = 1 2 2

Option (D) is correct. {1 + km (t)} A sin (wc t) dm (t) Asin (wc t) A sin {cos t + km (t)} A sin [wct + k] t- 3 m (t) dt

8.78

= 4No W/Hz = 4No # B

The noise power is

The ratio of average sideband power to mean noise power is Side Band Power = 6.25 = 25 Noise Power N0 B 4No B

8.77

d

Option (D) is correct. The phase deviation is b =

3f = 10 = 10 fm 1

If phase deviation remain same and modulating frequency is changed BW = 2 (b + 1) fm' = 2 (10 + 1) 2 = 44 kHz 8.84

8.85

Option (B) is correct. As the area under pdf curve must be unity and all three region are equivaprobable. Thus are under each region must be 13 . 2a # 1 = 1 $ a = 2 3 3 4 Option (A) is correct. Nq =

+a

#- a

x2 p (x) dx = 2

Substituting a = 2 we have 3 Nq = 4 81

#0

3 a 3 x $ 1 dx = 1 ; x E = a 2 3 0 6 4

a 2

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 8.86

Option (C) is correct. When word length is 6 2#6 12 S ` N jN = 6 = 2 = 2

8.90

Option (B) is correct. Carrier frequency Modulating frequency

or 8.91

Option (A) is correct. The SNR at transmitter is

10 - 3 = 109 10 - 20 # 100 # 106 In dB SNRtr = 10 log 109 = 90 dB Cable Loss = 40 db At receiver after cable loss we have SNRRc = 90 - 40 = 50 dB

fc = 1 # 106 Hz

For an envelope detector

8.92

2pfc > 1 > 2pfm Rc 1 < RC < 1 2pfc 2pfm

1 < RC < 1 2pfc 2pfm 1 1 < RC < 6 2p10 2 # 103 1.59 # 10 - 7 < RC < 7.96 # 10 - 5 so, 20 msec sec best lies in this interval.

Option (B) is correct. The impulse response of matched filter is h (t) = x (T - t) Since here T = 4 , thus

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h (t) = x (4 - t) The graph of h (t) is as shown below.

From graph it may be easily seen that slope between 3 < t < 4 is - 1. 8.93

Option (B) is correct. SAM (t) = Ac [1 + 0.1 cos wm t] cos wm t sNBFM (t) = Ac cos [wc t + 0.1 sin wm t] s (t) = SAM (t) + SNB fm (t)

For QPSK, Thus

+ Ac cos wc t cos (0.1 sin wm t) - Ac sin wc t. sin (0.1 sin wm t) As 0.1 sin wm t ,+ 0.1 to - 0.1 so, cos (0.1 sin wm t) . 1 As when q is small cos q . 1 and sin q , q, thus

- Ac 0.1 sin wm t sin wc t = 2Ac cos wc t + 0.1Ac cos (wc + wm) t 1 44 2 44 3 1 4444 4 2 4444 43 cosec USB

Thus it is SSB with carrier. Option (A) is correct. Consecutive pulses are of same polarity when modulator is in slope overload. Consecutive pulses are of opposite polarity when the input is constant.

M = 4 = 2n $ n = 2 B2 = 2Rb = 10 kHz 2

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sin (0.1 sin wm t) = 0.1 sincos wc t cos wm t + Ac cos wc t

Option (C) is correct. The required bandwidth of M array PSK is BW = 2Rb n where 2n = M and Rb is bit rate For BPSK, M = 2 = 2n $ n = 1 Thus B1 = 2Rb = 2 # 10 = 20 kHz 1

= Ac [1 + 0.1 cos wm t] cos wc t + Ac cos (wc t + 0.1 sin wm t) = Ac cos wc t + Ac 0.1 cos wm t cos wc t

8.89

P (X = 1) = P (X = 1+) - P (X = 1 -) = 0.55 - 0.25 = 0.30

SNRtr = Ptr NB

fm = 2 # 103 Hz

8.88

Option (D) is correct. F (x1 # X < x2) = p (X = x2) - P (X = x1)

When word length is 8 2#8 16 S ` N jN = 8 = 2 = 2 S 16 ^ N hN = 8 Now = 212 = 2 4 = 16 S ^ N hN = 6 2 Thus it improves by a factor of 16. 8.87

Page 208

Option (C) is correct. We have MHz

fc = 100 MHz = 100 # 106 and fm = 1

= 1 # 106 The output of balanced modulator is VBM (t) = [cos wc t][ cos wc t] = 1 [cos (wc + wm) t + cos (wc - wm) t] 2 If VBM (t) is ed through HPF of cut off frequency fH = 100 # 106 , then only (wc + wm) es and output of HPF is VHP (t) = 1 cos (wc + wm) t 2

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 209

V0 (t) = VHP (t) + sin (2p # 100 # 106) t

Now

Probability of error of 1 P (0 # X # 0.2) = 0.2 Probability of error of 0 :

= 1 cos [2p100 # 106 + 2p # 1 # 106 t] + sin (2p # 100 # 106) t 2 = 1 cos [2p108 + 2p106 t] + sin (2p108) t 2

P (0.2 # X # 0.25) = 0.05 # 2 = 0.1 P (0 # X # 0.2) + P (0.2 # X # 0.25) 2 = 0.2 + 0.1 = 0.15 0

Average error =

= 1 [cos (2p108 t) t cos (2p106 t)] - sin [2p108 t sin (2p106 t) + sin 2p108 t] 2 8.97

= 1 cos (2p106 t) cos 2p108 t + `1 - 1 sin 2p106 t j sin 2p108 t 2 2 This signal is in form

Option (B) is correct. The square mean value is

#- 3 (x - xq) 2 f (x) dx

=

#0 (x - xq) 2 f (x) dx

= A cos 2p108 t + B sin 2p108 t Now

GATE

ELECTRONICS & COMMUNICATION by RK Kanodia & Ashish Murolia in 3 Volumes

Fully Revised

=

=

=

8.95

1 cos2 (2p106 t) + 1 + 1 sin2 (2p106 t) - sin (2p106 t) 4 4 1 + 1 - sin (2p106 t) = 4 5 - sin (2p106 t) = 4

Option (A) is correct. The pdf of transmission of 0 and 1 will be as shown below :

3 0.3 3 2 1 = ; x E + ; x + 0.49x - 14 x E 3 0 3 2 0.3

s2 = 0.039 RMS =

8.98

8.99

s2 =

0.039 = 0.198

Option (C) is correct. FM DM PSK PCM

$ Capture effect $ Slope over load $ Matched filter $ m - law

Option (C) is correct. Since fs = 2fm , the signal frequency and sampling frequency are as follows fm1 = 1200 Hz $ 2400 samples per sec fm2 = 600 Hz $ 1200 samples per sec fm3 = 600 Hz $ 1200 samples per sec Thus by time division multiplexing total 4800 samples per second will be sent. Since each sample require 12 bit, total 4800 # 12 bits per second will be sent Thus bit rate Rb = 4800 # 12 = 57.6 kbps

3

8.96

0.1

#0.3 (x - 0.7) 2 f (x) dx

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or

Option (A) is correct. s (t) = A cos [2p10 # 10 t] + A cos [2p10.1 # 103 t] 1 Here T1 = = 100m sec 10 # 103 1 and = 99m sec T2 = 10.1 # 103 Period of added signal will be LCM [T1, T2] Thus T = LCM [100, 99] = 9900m sec Thus frequency f = 1 = 0.1 kHz 9900m

(x - 0) 2 f (x) dx +

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A2 + B2

2 6 6 2 1 1 ` 2 cos (2p10 t)j + `1 - 2 sin (2p10 t j

0. 3

1

SPECIAL EDITION ( STUDY MATERIAL FORM ) At market Book is available in 3 volume i.e. in 3 book binding form. But at NODIA Online Store book is available in 10 book binding form. Each unit of Book is in separate binding.

Every Unit Contain All Questions of Last 15 Year Papers Purchase from Nodia Online Store at Maximum Discount with free Shipping The envelope of this signal is =

#0

3

s2 =

8.100

Option (B) is correct. The input signal X (f) has the peak at 1 kHz and - 1 kHz. After balanced modulator the output will have peak at fc ! 1 kHz i.e. : 10 ! 1 $ 11 and 9 kHz 10 ! (- 1) $ 9 and 11 kHz 9 kHz will be filtered out by HPF of 10 kHz. Thus 11 kHz will remain. After ing through 13 kHz balanced modulator signal will have 13 ! 11 kHz signal i.e. 2 and 24 kHz. Thus peak of Y (f) are at 2 kHz and 24 kHz.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 8.101

8.102

8.103

Option (A) is correct. The input is a coherent detector is DSB - SC signal plus noise. The noise at the detector output is the in-phase component as the quadrature component nq (t) of the noise n (t) is completely rejected by the detector.

8.106

Ed 2h m

Since Pe of Binary FSK is 3 dB inferior to binary PSK

8.107

Option (D) is correct. The pdf of Z will be convolution of pdf of X and pdf of Y as shown below. Now

p [Z # z] = p [Z #- 2] =

W = Y-Z E [W2] = E [Y - Z] 2 = E [Y2] + E [Z2] - 2E [YZ] = sw2 E [X2 (t)] = Rx (10) = 4 [e - 0.2 0 + 1] = 4 [1 + 1] = 8 E [Y2] = E [X2 (2)] = 8 E [Z2] = E [X2 (4)] = 8 E [YZ] = RXX (2) = 4 [e-0.2 (4 - 2) + 1] = 6.68 E [W2] = sw2 = 8 + 8 - 2 # 6.68 = 2.64

Option (B) is correct. Pe = 1 erfc c 2

Option (C) is correct.

We have

Option (C) is correct. The noise at the input to an ideal frequency detector is white. The PSD of noise at the output is parabolic We have

8.104

Page 210

Option (C) is correct. Step size d =

2mp = 1.536 = 0.012 V L 128

2 (0.012) 2 Quantization Noise power = d = 12 12

z

#- 3 fZ (z) dz

= 12 # 10-6 V2

-2

#- 3fZ (z) dz

8.108

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Option (D) is correct. The frequency of pulse train is f 1- 3 = 1 k Hz 10 The Fourier Series coefficient of given pulse train is -T /2 Cn = 1 Ae-jnw t dt To -T /2

#

o

o

o

= 1 To

#

-To /6

-To /6

Ae-jhw t dt o

A [e-jw t] --TT //66 To (- jhwo) A = (e-jw t - e jhw T /6) (- j2pn) = A (e jhp/3 - e-jhp/3) j2pn or Cn = A sin ` np j pn 3 From Cn it may be easily seen that 1, 2, 4, 5, 7, harmonics are present and 0, 3, 6, 9,.. are absent. Thus p (t) has 1 kHz, 2 kHz, 4 kHz, 5 kHz, 7 kHz,... frequency component and 3 kHz, 6 kHz.. are absent. The signal x (t) has the frequency components 0.4 kHz and 0.7 kHz. The sampled signal of x (t) i.e. x (t)* p (t) will have 1 ! 0.4 and 1 ! 0.7 kHz 2 ! 0.4 and 2 ! 0.7 kHz 4 ! 0.4 and 4 ! 0.7 kHz =

= Area [z #- 2] = 1 # 1 #1 = 1 2 6 12

o

o o

o

o

o

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Thus in range of 2.5 kHz to 3.5 kHz the frequency present is 2 + 0.7 = 2.7 kHz 4 - 0.7 = 3.3 kHz

Option (D) is correct. We have

RXX (t) = 4 (e - 0.2 t + 1) RXX (0) = 4 (e - 0.2 0 + 1) = 8 = s2

or mean Now

s =2 2 m =0 P (x # 1) = Fx (1)

Given

X-m at x = 1 s m = 1 - Qc 1 - 0 m = 1 - Qc 1 m 2 2 2 2

= 1 - Qc

8.109

Option (C) is correct. vi = Ac1 cos (2pfc t) + m (t) v0 = ao vi + avi3 v0 = a0 [Ac' cos (2pfc' t) + m (t)] + a1 [Ac' cos (2pfc' t) + m (t)] 3 = a0 Ac' cos (2pfc' t) + a0 m (t) + a1 [(Ac' cos 2pfc' t) 3

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 211

then for coherent binary PSK r = 0.5 and s is required. + (Ac' cos (2pfc') t) 2 m (t)

+

3Ac' cos (2pfc' t) m2 (t)

3

+ m (t)]

= a0 Ac' cos (2pfc' t) + a0 m (t) + a1 (Ac' cos 2fc' t) 3

+ 3a1 Ac'2 ;

1 + cos (4pfc' t) E m (t) 2

= 3a1 Ac' cos (2pfc' t) m2 (t) + m3 (t) The term 3a1 Ac' ( cos 42pf t ) m (t) is a DSB-SC signal having carrier frequency 1. MHz. Thus 2fc' = 1 MHz or fc' = 0.5 MHz

8.117

= 1.76 + 6.02 # 8 = 49.8 dB 8.118

' c

8.110

Option (D) is correct. 2 PT = Pc c1 + a m 2

2 P (0.5) 2 Psb = Pc a = c 2 2 Psb = 1 Pc 8

or 8.111

Option (D) is correct. AM Band width = 2fm Peak frequency deviation = 3 (2fm) = 6fm 6f Modulation index b = m = 6 fm

The FM signal is represented in of Bessel function as xFM (t) = Ac

3

Option (B) is correct. Bit Rate = 8k # 8 = 64 kbps (SNR)q = 1.76 + 6.02n dB Option (C) is correct. The frequency of message signal is fc = 1000 kHz 1 The frequency of message signal is 1 = 10 kHz fm = 100 # 10 - 6 Here message signal is symmetrical square wave whose FS has only odd harmonics i.e. 10 kHz, 30 kHz 50 kHz. Modulated signal contain fc ! fm frequency component. Thus modulated signal has fc ! fm = (1000 ! 10) kH = 1010 kHz, 990 kHz fc ! 3fm = (1000 ! 10) kH = 1030 kHz, 970 kHz

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/Jn (b) cos (wc - nwn) t

n =- 3

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3

wc + nwm = 2p (1008 # 10 ) 2p10 + n4p # 103 = 2p (1008 # 103), n = 4 Thus coefficient = 5J4 (6) 6

8.112

8.113

Option (B) is correct. Ring modulation $ Generation of DSB - SC VCO $ Generation of FM Foster seely discriminator $ Demodulation of fm mixer $ frequency conversion

Thus, there is no 1020 kHz component in modulated signal. 8.119

We have

8.114

10 cos (8p # 103) t # 5 # 10 - 6 100 # 10 - 6 = 5 # 10 - 1 cos (8p # 103) t =

8.120

8.121

Option (D) is correct. Eb = 10 - 6 watt-sec No = 10 - 5 W/Hz o (SNR) matched filler = E = N

106 = .05 2 # 10 - 5 2 (SNR)dB = 10 log 10 (0.05) = 13 dB o

8.115

Option (B) is correct. 3 fs 2pfm This is satisfied with Em = 1.5 V and fm = 4 kHz For slopeoverload to take place Em $

8.116

Option (A) is correct. If s " carrier synchronization at receiver r " represents bandwidth efficiency

+3

/ d (t - nTs)

n =- 3

fmax = 1650 + 450 = 2100 kHz fmin = 550 + 450 = 1000 kHz 1 or f = 2p LC frequency is minimum, capacitance will be maximum

or

y (t) = 5 # 10 - 6 x (t)

x (t) = 10 cos (8p # 103) t Ts = 100m sec The cut off fc of LPF is 5 kHz We know that for the output of filter x (t) y (t) = Ts

Option (A) is correct.

f2 = (2.1) 2 R = Cmax = max 2 Cmin fmin R = 4.41 fi = fc + 2fIF = 700 + 2 (455) = 1600 kHz

Option (C) is correct.

8.122

Option (C) is correct. Transmitted frequencies in coherent BFSK should be integral of bit rate 8 kHz. Option (B) is correct. For best reception, if transmitting waves are vertically polarized, then receiver should also be vertically polarized i.e. transmitter and receiver must be in same polarization. Option (D) is correct. s (t) = cos 2p (2 # 106 t + 30 sin 150t + 40 cos 150t) = cos {4p106 t + 100p sin (150t + q)} Angle modulated signal is s (t) = A cos {wc t + b sin (wm t + q)} Comparing with angle modulated signal we get Phase deviations b = 100p Frequency deviations 3 f = bfm = 100p # 150 = 7.5 kHz 2p

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 212 8.126

8.123

Option (*) is correct. We have m (t) s (t) = y1 (t) 2 sin (2pt) cos (200pt) = t sin (202pt) - sin (198pt) = t

8.127

Option (A) is correct. For any PDF the probability at mean is 1 . Here given PDF is 2 Gaussian random variable and X = 4 is mean.

8.128

y1 (t) + n (t) = y2 (t) = sin 202pt - sin 198pt + sin 199pt t t y2 (t) s (t) = u (t)

8.129

=

Option (A) is correct. An ideal low - filter with appropriate bandwidth fm is used to recover the signal which is sampled at nyquist rate 2fm .

[sin 202pt - sin 198pt + sin 199pt] cos 200pt t

Option (C) is correct. We require 6 bit for 64 intensity levels because 64 = 26 Data Rate = Frames per second # pixels per frame # bits per pixel = 625 # 400 # 400 # 6 = 600 Mbps sec Option (C) is correct. We have sin c (700t) + sin c (500t) =

Here the maximum frequency component is 2pfm = 700p i.e. fm = 350 Hz Thus Nyquist rate fs = 2fm = 2 (350) = 700 Hz Thus sampling interval = 1 sec 700

= 1 [sin (402pt) + sin (2pt) - {sin (398pt) - sin (2pt)} 2 + sin (399pt) - sin (pt)]

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8.130

Option (D) is correct. Probability of error = p Probability of no error = q = (1 - p) Probability for at most one bit error = Probability of no bit error

After filtering sin (2pt) + sin (2pt) - sin (pt) y (t) = 2t sin (2pt) + 2 sin (0.5t) cos (1.5pt) = 2t sin sin . t 2 p 0 5pt cos 1.5pt = + t 2t 8.124

+ probability of 1 bit error = (1 - p) n + np (1 - p) n - 1 8.131

and power is Pg = 1 2p

3 fm = 24p10 = 12 kHz 2p Ts = 50m sec " fs = 1 = 1 # 106 = 20 50 Ts

d (n) = x (n) - x (n - 1) E [d (n)] 2 = E [x (n) - x (n - 1)] 2 or E [d (n)] 2 = E [x (n)] 2 + E [x (n - 1)] 2 - 2E [x (n) x (n - 1)] or sd2 = sx2 + sx2 - 2Rxx (1) 2 As we have been given sd2 = sx , therefore 10

or

G (w) Sg (w) = G (w) 2

Now PSD of ag (t) is

or

sx2 = s2 + s2 - 2R (1) x x xx 10 2Rxx (1) = 19 sx2 10 Rxx = 19 = 0.95 20 sx2

ag (t)

3

#- 3Sg (w) dw

FT

aG (w)

Sag (w) = a (G (w)) 2 = a2 G (w) 2 Sag (w) = a2 Sg (w)

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Option (A) is correct.

or

FT

g (t) then PSD of g (t) is

After sampling signal will have fs ! fm frequency component i.e. 32 and 12 kHz At filter output only 8 kHz will be present as cutoff frequency is 15 kHz. 8.125

Option (A) is correct. If

Option (B) is correct. The signal frequency is

kHz

sin (700pt) sin (500pt) + 700pt 500pt

Similarly 8.132

as k = 1 8.133

8.134

Pag = a2 Pg

Option (C) is correct. The envelope of the input signal is [1 + ka m (t)] that will be output of envelope detector. Option (D) is correct. Frequency Range for satellite communication is 1 GHz to 30 GHz, Option (B) is correct. Waveform will be orthogonal when each bit contains integer number of cycles of carrier. Bit rate Rb = HCF (f1, f2) = HCF (10k, 25k)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

= 5 kHz Tb = 1 = 1 = 0.2 msec = 200 msec Rb 5k

Thus bit interval is 8.135

Page 213 8.141

g (t) = 6 # 10 sin c (400t) ) 10 sin c3 (100t) Let g1 (t) = 6 # 10 4 sin c2 (400t) g2 (t) = (106) sin c3 (100t) We know that g1 (t) ) g2 (t) ? G1 (w) G2 (w) occupies minimum of Bandwidth of G1 (w) or G2 (w) Band width of G1 (w) = 2 # 400 = 800 rad/ sec or = 400 Hz Band width of G2 (w) = 3 # 100 = 300 rad/ sec or 150 Hz Sampling frequency = 2 # 150 = 300 Hz

Option (D) is correct.

4

Pm = m2 (t)

We have The input to LPF is

x (t) = m (t) cos wo t cos (wo t + q) m (t) [cos (2wo t + q) + cos q] 2 m (t) cos (2wo t + q) m (t) cos q = + 2 2 =

The output of filter will be y (t) =

m (t) cos q 2

Power of output signal is

8.142

2 Py = y2 (t) = 1 m2 (t) cos2 q = Pm cos q 4 4 8.136

Option (A) is correct. Hilbert transformer always adds - 90c to the positive frequency component and 90c to the negative frequency component. Hilbert Trans form

Thus 8.137

cos wt " sin wt sin wt " cos wt cos w1 t + sin w2 t " sin w1 t - cos w2 t

Option (A) is correct. We have

8.139

8.140

2

6

Option (B) is correct. For a sinusoidal input SNR (dB) is PCM is obtained by following formulae.

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x (t) = Ac cos {wc t + b sin wm t} y (t) = {x (t)} 3

= Ac2 cos (3wc t + 3b sin wm t) + 3 cos (wc t + b sin wm t) Thus the fundamental frequency doesn’t change but BW is three times. BW = 2 (3 f') = 2 (3 f # 3) = 3 MHz 8.138

Option (B) is correct. Given function

SNR (dB) = 1.8 + 6n n is no. of bits n =8 SNR (dB) = 1.8 + 6 # 8 = 49.8

Here So, 8.143

Option (D) is correct. We know that matched filter output is given by g 0 (t) = t = T0

Option (C) is correct. Option (C) is correct. This is Quadrature modulated signal. In QAM, two signals having bandwidth. B 1 & B 2 can be transmitted simultaneous over a bandwidth of (B 1 + B 2) Hz so B.W. = (15 + 10) = 25 kHz Option (B) is correct. A modulated signal can be expressed in of its in-phase and quadrature component as

6g 0 (t)@max = =

# g (l) g (T - t + l) dl at 3

-3

0

# g (l) g (l) dl = # 3

3

-3

-3

#

1 # 10-4

0

g 2 (t) dt

[10 sin (2p # 106) 2] dt

[g 0 (t)] max = 1 # 100 # 10-4 = 5 mV 2 8.144

Option (B) is correct. Sampling rate must be equal to twice of maximum frequency. f s = 2 # 400 = 800 Hz

8.145

S (t) = S1 (t) cos (2pfc t) - SQ (t) sin (2pfc t) Here S (t) -at at = [e sDwt cos wc t - e sin Dwt sin wc t] m (t)

Option (C) is correct. The amplitude spectrum of a gaussian pulse is also gaussian as shown in the fig. -y 2 fY (y) = 1 exp c 2 m 2p

= [e-at cos Dwt] cos 2pfc t - [e-at sin Dwt] sin 2pfc t = S1 (t) cos 2pfc t - SQ (t) sin 2pfc t Complex envelope of s (t) is S (t) = S1 (t) + jSQ (t) = e-at cos Dwt + je-at sin Dwt = e-at [cos Dwt + j sin Dwt] = exp (- at) exp (jDwt) m (t) 8.146

Option (C) is correct.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 214 2#8 (SNR) 1 = 22 # 9 = 22 = 1 4 (SNR) 2 2 so SNR will increased by a factor of 4

Let the rectangular pulse is given as

8.149

Auto correlation function is given by T/2 Rxx (t) = 1 x (t) x (t - t) dt T -T/2 When x (t) is shifted to right (t > 0), x (t - t) will be shown as dotted line.

8.150

#

8.151

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8.152

8.153

8.154

8.155

8.156 8.157

Rxx (t) = 1 T

#

T +t 2

A2 dt

T - +t 2

2 2 = A :T + T - tD = A :T - tD 2 T 2 T 2 (t) can be negative or positive, so generalizing above equations 2 Rxx (t) = A :T - t D T 2 Rxx (t) is a regular pulse of duration T .

8.147

8.148

Option (A) is correct. In flat top sampling an amplitude distortion is produced while reconstructing original signal x (t) from sampled signal s (t). High frequency of x (t) are mostly attenuated. This effect is known as aperture effect. Option (A) is correct. Carrier C (t) = cos (we t + q) Modulating signal = x (t) DSB - SC modulated signal = x (t) c (t) = x (t) cos (we t + q) envelope = x (t) Option (D) is correct. In Quadrature multiplexing two baseband signals can transmitted or modulated using I 4 phase & Quadrature carriers and its quite different form FDM & TDM. Option (A) is correct. Fourier transform perform a conversion from time domain to frequency domain for analysis purposes. Units remain same. Option (A) is correct. In PCM, SNR is depends an step size (i.e. signal amplitude) SNR can be improved by using smaller steps for smaller amplitude. This is obtained by compressing the signal. Option (C) is correct. Band width is same for BPSK and APSK(OOK) which is equal to twice of signal Bandwidth. Option (A) is correct. The spectral density of a real value random process symmetric about vertical axis so it has an even symmetry. Option (A) is correct. Option (C) is correct. It is one of the advantage of bipolar signalling (AMI) that its spectrum has a dc null for binary data transmission PSD of bipolar signalling is

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Option (B) is correct. Selectivity refers to select a desired frequency while rejecting all others. In super heterodyne receiver selective is obtained partially by RF amplifier and mainly by IF amplifier. Option (C) is correct. In PCM, SNR a 22n so if bit increased from 8 to 9

8.158

Option (A) is correct. Probability Density function (PDF) of a random variable x defined as

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Px (x) = K =

so here 8.159

8.160

8.161

1 e-x /2 2p 1 2p

Page 215

factor of 22 (n + 1)/n

2

8.165

Option (C) is correct. Here the highest frequency component in the spectrum is 1.5 kHz [at 2 kHz is not included in the spectrum] Minimum sampling freq. = 1.5 # 2 = 3 kHz

R (t) = put Let

Option (B) is correct. We need a high filter for receiving the pulses. Option (D) is correct. Power spectral density function of a signal g (t) is fourier transform of its auto correlation function

8.162

Option (C) is correct. For a signal g (t), its matched filter response given as h (t) = g (T - t) so here g (t) is a rectangular pulse of duration T .

3

# x (t) x (t + t) dt -3 3

R (- t) = # x (t) x (t - t) dt -3 t-t = a dt = da

R (- t) = change variable a " t

F

Sg (w) Rg (t) here Sg (w) = sin c2 (f) so Rg (t) is a triangular pulse. f [triang.] = sin c2 (f)

Option (D) is correct. The auto correlation of energy signal is an even function. auto correlation function is gives as

R (- t) =

3

# x (a + t) x (a) da -3

3

# x (t) x (t + t) dt = R (t) -3

R (- t) = R (t) even function 8.166

Option (D) is correct.

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output of matched filter y (t) = g (t) ) h (t)

if we shift g (- t) for convolution y (t) increases first linearly then decreases to zero.

8.163

8.164

Option (C) is correct. The difference between incoming signal frequency (fc) and its image frequency (fc) is 2I f (which is large enough). The RF filter may provide poor selectivity against adjacent channels separated by a small frequency differences but it can provide reasonable selectivity against a station separated by 2I f . So it provides adequate suppression of image channel. Option (C) is correct. In PCM SNR is given by SNR = 3 22n 2 if no. of bits is increased from n to (n + 1) SNR will increase by a

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 216

UNIT 9

p # 10 ^x + z h (C) 45c and E 0 ^atx - atz h e-j 3 2 V/m 2 4

ELECTROMAGNETICS

p # 10 z (D) 60c and E 0 ^atx - atz h e-j 3 V/m 2 The expression for Evr is p # 10 ^x - z h (A) 0.23 E 0 ^atx + atz h e-j 3 2 V/m 2 4

9.5

4

2013 9.1

v ^rvh. The closed loop line integral A v : dlv Consider a vector field A can be expressed as vh : dsv over the closed surface bounded by the loop (A) ^d # A vh dv over the closed volume bounded by the loop (B) ^d : A

p # 10 z (B) - E 0 ^atx + atz h e j 3 V/m 2

(C)

p # 10 ^x + z h 3 (D) E 0 ^atx + atz h e-j V/m 2

4

#

(D) 9.2

ONE MARK

## ### ### ^d : Avhdv over the open volume bounded by the loop ## ^d # Avh : dsv over the open surface bounded by the loop

p # 10 ^x - z h (C) 0.44 E 0 ^atx + atz h e-j 3 2 V/m 2 4

4

v = xatx + yaty + zatz is The divergence of the vector field A (A) 0 (B) 1/3

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2012 9.6

-

(D) 3

The return loss of a device is found to be 20 dB. The voltage standing wave ratio (VSWR) and magnitude of reflection coefficient are respectively (A) 1.22 and 0.1 (B) 0.81 and 0.1 (C) – 1.22 and 0.1 (D) 2.44 and 0.2 9.8

2013

TWO MARKS

Statement for Linked Answer Questions 52 and 53: A monochromatic plane wave of wavelength l = 600 mm is propagating in the direction as shown in the figure below. Evi , Evr and Evt denote incident, reflected, and transmitted electric field vectors associated with the wave.

A plane wave propagating in air with E = (8ax + 6ay + 5az ) e j (wt + 3x - 4y) V/m is incident on a perfectly conducting slab positioned at x # 0 . The E field of the reflected wave is (A) (- 8ax - 6ay - 5az ) e j (wt + 3x + 4y) V/m (B) (- 8ax + 6ay - 5az ) e j (wt + 3x + 4y) V/m (C) (- 8ax - 6ay - 5az ) e j (wt - 3x - 4y) V/m (D) (- 8ax + 6ay - 5az ) e j (wt - 3x - 4y) V/m

9.7 9.3

ONE MARK

The electric field of a uniform plane electromagnetic wave in free space, along the positive x direction is given by E = 10 (ay + jaz ) e-j 25x . The frequency and polarization of the wave, respectively, are (A) 1.2 GHz and left circular (B) 4 Hz and left circular (C) 1.2 GHz and right circular (D) 4 Hz and right circular A coaxial-cable with an inner diameter of 1 mm and outer diameter of 2.4 mm is filled with a dielectric of relative permittivity 10.89. -9 Given m0 = 4p # 10-7 H/m, e0 = 10 F/m , the characteristic 36p impedance of the cable is (A) 330 W (C) 143.3 W

9.9

(B) 100 W (D) 43.4 W

The radiation pattern of an antenna in spherical co-ordinates is given by F (q) = cos 4 q ; 0 # q # p/2 . The directivity of the antenna is

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 (A) 10 dB (C) 11.5 dB 2012 9.10 9.4

The angle of incidence qi and the expression for Evi are p # 10 ^x + 2h (A) 60c and E 0 ^atx - atz h e-j 3 2 V/m 2 4

p # 10 z (B) 45c and E 0 ^atx + atz h e-j 3 V/m 2 4

(B) 12.6 dB (D) 18 dB TWO MARKS

A transmission line with a characteristic impedance of 100 W is used to match a 50 W section to a 200 W section. If the matching is to be done both at 429 MHz and 1 GHz, the length of the transmission line can be approximately (A) 82.5 cm (b) 1.05 m (C) 1.58 cm (D) 1.75 m

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 9.11

Page 217

The magnetic field among the propagation direction inside a rectangular waveguide with the cross-section shown in the figure is

(A) 0.8 # 108 m/s (C) 1.6 # 108 m/s

Hz = 3 cos (2.094 # 102 x) cos (2.618 # 102 y) cos (6.283 # 1010 t - bz) 9.16

The phase velocity v p of the wave inside the waveguide satisfies (A) v p > c (B) v p = c (C) 0 < v p < c (D) v p = 0

Statement for Linked Answer Question 7 and 8 : An infinitely long uniform solid wire of radius a carries a uniform dc current of density J 9.12

9.13

The magnetic field at a distance r from the center of the wire is proportional to (A) r for r < a and 1/r 2 for r > a (B) 0 for r < a and 1/r for r > a (C) r for r < a and 1/r for r > a (D) 0 for r < a and 1/r 2 for r > a A hole of radius b (b < a) is now drilled along the length of the wire at a distance d from the center of the wire as shown below.

(B) 1.2 # 108 m/s (D) 3 # 108 m/s

The modes in a rectangular waveguide are denoted by TE mn where TM mn m and n are the eigen numbers along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is TRUE? (A) The TM 10 mode of the waveguide does not exist (B) The TE 10 mode of the waveguide does not exist (C) The TM 10 and the TE 10 modes both exist and have the same cut-off frequencies (D) The TM 10 and the TM 01 modes both exist and have the same cut-off frequencies 2011

9.17

TWO MARKS

A current sheet Jv = 10uty A/m lies on the dielectric interface x = 0 between two dielectric media with er 1 = 5, mr 1 = 1 in Region-1

SPECIAL EDITION ( STUDY MATERIAL FORM ) At market Book is available in 3 volume i.e. in 3 book binding form. But at NODIA Online Store book is available in 10 book binding form. Each unit of Book is in separate binding. Available Only at NODIA Online Store

Click to Buy www.nodia.co.in (x < 0) and er2 = 2, mr2 = 2 in Region-2 (x 2 0). If the magnetic field in Region-1 at x = 0- is Hv1 = 3utx + 30uty A/m the magnetic field in Region-2 at x = 0+ is

The magnetic field inside (A) uniform and depends (B) uniform and depends (C) uniform and depends (D) non uniform 2011 9.14

the hole is only on d only on b on both b and d

(A) Hv2 = 1.5utx + 30uty - 10utz A/m (B) Hv2 = 3utx + 30uty - 10utz A/m (C) Hv2 = 1.5utx + 40uty A/m (D) Hv2 = 3utx + 30uty + 10utz A/m ONE MARK

Consider the following statements regarding the complex Poynting vector Pv for the power radiated by a point source in an infinite homogeneous and lossless medium. Re(Pv ) denotes the real part of Pv, S denotes a spherical surface whose centre is at the point source, and nt denotes the unit surface normal on S . Which of the following statements is TRUE? (A) Re(Pv ) remains constant at any radial distance from the source (B) Re(Pv ) increases with increasing radial distance from the

9.18

9.19

source (C)

##s Re (Pv) : nt dS

remains constant at any radial distance from

##s Re (Pv) : nt dS

decreases with increasing radial distance from

the source (D)

the source 9.15

A transmission line of characteristic impedance 50 W is terminated by a 50 W load. When excited by a sinusoidal voltage source at 10 GHz, the phase difference between two points spaced 2 mm apart on the line is found to be p/4 radians. The phase velocity of the wave along the line is

A transmission line of characteristic impedance 50 W is terminated in a load impedance ZL . The VSWR of the line is measured as 5 and the first of the voltage maxima in the line is observed at a distance of l/4 from the load. The value of ZL is (A) 10 W (B) 250 W (C) (19.23 + j 46.15) W (D) (19.23 - j 46.15) W The electric and magnetic fields for a TEM wave of frequency 14 GHz in a homogeneous medium of relative permittivity er and relative permeability mr = 1 are given by Ev = E p e j (wt - 280py) utz V/m and Hv = 3e j (wt - 280py) utx A/m . Assuming the speed of light in free space to be 3 # 108 m/s , the intrinsic impedance of free space to be 120p , the relative permittivity er of the medium and the electric field amplitude E p are (B) er = 3, E p = 360p (A) er = 3, E p = 120p (C) er = 9, E p = 360p (D) er = 9, E p = 120p 2010

9.20

If the scattering matrix [S ] of a two port network is

ONE MARK

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 218

0.2+0c 0.9+90c , then the network is [S ] = > 0.9+90c 0.1+90cH (A) lossless and reciprocal (B) lossless but not reciprocal (C) not lossless but reciprocal (D) neither lossless nor reciprocal 9.21

9.22

A transmission line has a characteristic impedance of 50 W and a resistance of 0.1 W/m . If the line is distortion less, the attenuation constant(in Np/m) is (A) 500 (B) 5 (C) 0.014 (D) 0.002 The electric field component of a time harmonic plane EM wave traveling in a nonmagnetic lossless dielectric medium has an amplitude of 1 V/m. If the relative permittivity of the medium is 4, the magnitude of the time-average power density vector (in W/m2 ) is (A) 1 (B) 1 30p 60p (C) 1 (D) 1 120p 240p

(A) 1.00 (C) 2.50

(B) 1.64 (D) 3.00

2009 9.26

ONE MARK

Two infinitely long wires carrying current are as shown in the figure below. One wire is in the y - z plane and parallel to the y - axis. The other wire is in the x - y plane and parallel to the x - axis. Which components of the resulting magnetic field are non-zero at the origin ?

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in 2010 9.23

v = xyatx + x 2 aty , then If A is

TWO MARKS

(A) x, y, z components (C) y, z components

# Av $ dlv over the path shown in the figure

o

C

9.27

(A) 0 (C) 1 9.24

9.25

In the circuit shown, all the transmission line sections are lossless. The Voltage Standing Wave Ration(VSWR) on the 60 W line is

Which of the following statements is true regarding the fundamental mode of the metallic waveguides shown ?

(A) Only P has no cutoff-frequency (B) Only Q has no cutoff-frequency (C) Only R has no cutoff-frequency (D) All three have cutoff-frequencies

(B) 2 3 (D) 2 3

A plane wave having the electric field components Evi = 24 cos ^3 # 108 - by h atx V/m and traveling in free space is incident normally on a lossless medium with m = m0 and e = 9e0 which occupies the region y $ 0 . The reflected magnetic field component is given by (A) 1 cos (3 # 108 t + y) atx A/m 10p (B) 1 cos (3 # 108 t + y) atx A/m 20p (C) - 1 cos (3 # 108 t + y) atx A/m 20p (D) - 1 cos (3 # 108 t + y) atx A/m 10p

(B) x, y components (D) x, z components

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If a vector field V is related to another vector field A through V = 4# A , which of the following is true? (Note : C and SC refer to any closed contour and any surface whose boundary is C . ) (B) A $ dl = (A) V $ dl = V $ dS A $ dS (C)

9.29

TWO MARKS

#S #C #S #C #C #C #C D # V $ dl = #S #C D # A $ d S (D) #C D # V $ dl = #S #CV $ d S

A transmission line terminates in two branches, each of length l , 4 as shown. The branches are terminated by 50W loads. The lines are lossless and have the characteristic impedances shown. Determine the impedance Zi as seen by the source.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 219

(C) 35 dB

(D) 45 dB

2007 9.38

(A) 200W (C) 50W 9.30

A magnetic field in air is measured to be y xt m B = B0 c 2 x 2 yt - 2 x + y2 x +y What current distribution leads to this field ? [Hint : The algebra is trivial in cylindrical coordinates.] t t (A) J = B0 z c 2 1 2 m, r ! 0 (B) J =- B0 z c 2 2 2 m, r ! 0 m0 x + y m0 x + y t (C) J = 0, r ! 0 (D) J = B0 z c 2 1 2 m, r ! 0 m0 x + y 2008

9.31

9.32

(B) 100W (D) 25W

ONE MARK

For a Hertz dipole antenna, the half power beam width (HPBW) in the E -plane is (A) 360c (B) 180c (C) 90c (D) 45c

ONE MARK

A plane wave of wavelength l is traveling in a direction making an angle 30c with positive x - axis and 90c with positive y - axis. The E field of the plane wave can be represented as (E0 is constant) p p 3p 3p t 0 e j c wt - l x - l z m t 0 e jc wt - l x - l z m (A) E = yE (B) E = yE t 0 e jc wt + (C) E = yE

9.39

3 p x+ p z l l m

p

t 0 e jc wt - l x + (D) E = yE

If C is code curve enclosing a surface S , then magnetic field intensity H , the current density j and the electric flux density D are related by 2D (A) H $ ds = c j + 2t m $ d t S c

##

(B) (C)

##

#S H $ d l = ##S c j + 22Dt m $ dS

##S H $ dS = #C c j + 22Dt m $ d t

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For static electric and magnetic fields in an inhomogeneous sourcefree medium, which of the following represents the correct form of Maxwell’s equations ? (A) 4$ E = 0 , 4# B = 0 (B) 4$ E = 0 , 4$ B = 0 (C) 4# E = 0 , 4# B = 0 (D) 4# E = 0 , 4$ B = 0

#C H $ d l # = ##S c j + 22Dt m $ ds

(D)

c

2007 2008 9.33

9.34

9.35

9.36

9.37

TWO MARKS

9.40

A rectangular waveguide of internal dimensions (a = 4 cm and b = 3 cm) is to be operated in TE11 mode. The minimum operating frequency is (A) 6.25 GHz (B) 6.0 GHz (C) 5.0 GHz (D) 3.75 GHz One end of a loss-less transmission line having the characteristic impedance of 75W and length of 1 cm is short-circuited. At 3 GHz, the input impedance at the other end of transmission line is (A) 0 (B) Resistive (C) Capacitive (D) Inductive

At 20 GHz, the gain of a parabolic dish antenna of 1 meter and 70% efficiency is (A) 15 dB (B) 25 dB

TWO MARKS

The E field in a rectangular waveguide of inner dimension a # b is given by 2 wm E = 2 ` l j H0 sin ` 2px j sin (wt - bz) yt a h 2

Where H0 is a constant, and a and b are the dimensions along the x - axis and the y - axis respectively. The mode of propagation in the waveguide is (A) TE20 (B) TM11 (C) TM20 (D) TE10 9.41

A uniform plane wave in the free space is normally incident on an infinitely thick dielectric slab (dielectric constant e = 9 ). The magnitude of the reflection coefficient is (A) 0 (B) 0.3 (C) 0.5 (D) 0.8 In the design of a single mode step index optical fibre close to upper cut-off, the single-mode operation is not preserved if (A) radius as well as operating wavelength are halved (B) radius as well as operating wavelength are doubled (C) radius is halved and operating wavelength is doubled (D) radius is doubled and operating wavelength is halved

3pz l m

A load of 50 W is connected in shunt in a 2-wire transmission line of Z0 = 50W as shown in the figure. The 2-port scattering parameter matrix (s-matrix) of the shunt element is

(A) >

1 2 1 1 2 2

(C) >

2 3 1 2 3 3

- 12

- 13

9.42

H

0 1 (B) = 1 0G

H

(D) > -

1 4 3 4

- 43 1 4

H

The parallel branches of a 2-wirw transmission line re terminated in 100W and 200W resistors as shown in the figure. The characteristic impedance of the line is Z0 = 50W and each section has a length of l . The voltage reflection coefficient G at the input is 4

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 220

(C) 2 2006 9.47

(A) - j 7 5 (C) j 5 7 9.43

(B) - 5 7 (D) 5 7

The H field (in A/m) of a plane wave propagating in free space is given by H = xt 5 3 cos (wt - bz) + yt`wt - bz + p j . h0 2 The time average power flow density in Watts is h (A) 0 (B) 100 100 h0

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in (C) 9.44

9.45

9.48

50h20

(D) 50 h0

9.50

An air-filled rectangular waveguide has inner dimensions of 3 cm # 2 cm. The wave impedance of the TE20 mode of propagation in the waveguide at a frequency of 30 GHz is (free space impedance h0 = 377 W ) (A) 308 W (B) 355 W (C) 400 W (D) 461 W A l dipole is kept horizontally at a height of l0 above a perfectly 2 2 conducting infinite ground plane. The radiation pattern in the lane of the dipole (E plane) looks approximately as

9.51

ONE MARK

The electric field of an electromagnetic wave propagation in the positive direction is given by E = atx sin (wt - bz) + aty sin (wt - bz + p/2). The wave is (A) Linearly polarized in the z -direction (B) Elliptically polarized (C) Left-hand circularly polarized (D) Right-hand circularly polarized A transmission line is feeding 1 watt of power to a horn antenna having a gain of 10 dB. The antenna is matched to the transmission line. The total power radiated by the horn antenna into the free space is (A) 10 Watts (B) 1 Watts (C) 0.1 Watts (D) 0.01 Watt 2006

9.49

(D) 3

TWO MARKS

When a planes wave traveling in free-space is incident normally on a medium having the fraction of power transmitted into the medium is given by (B) 1 (A) 8 2 9 (C) 1 (D) 5 3 6 A medium of relative permittivity er2 = 2 forms an interface with free - space. A point source of electromagnetic energy is located in the medium at a depth of 1 meter from the interface. Due to the total internal reflection, the transmitted beam has a circular crosssection over the interface. The area of the beam cross-section at the interface is given by (A) 2p m 2 (B) p2 m 2 (C) p m 2 (D) p m 2 2 A rectangular wave guide having TE10 mode as dominant mode is having a cut off frequency 18 GHz for the mode TE30 . The inner broad - wall dimension of the rectangular wave guide is (B) 5 cm (A) 5 cm 3 (D) 10 cm (C) 5 cm 2

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 9.52

9.46

A medium is divide into regions I and II about x = 0 plane, as shown in the figure below.

A right circularly polarized (R) plane wave is incident at an angle 60c to the normal, on an air-dielectric interface. If the reflected wave is linearly polarized, the relative dielectric constant xr2 is.

(A)

2

(B)

3

An electromagnetic wave with electric field E1 = 4atx + 3aty + 5atz is incident normally on the interface from region I . The electric file E2 in region II at the interface is (B) 4atx + 0.75aty - 1.25atz (A) E2 = E1

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(C) 3atx + 3aty + 5atz 9.53

(D) - 3atx + 3aty + 5atz

2005

ONE MARK

The magnetic field intensity vector of a plane wave is given by H (x, y, z, t) = 10 sin (50000t + 0.004x + 30) aty

Refractive index of glass is 1.5. Find the wavelength of a beam of light with frequency of 1014 Hz in glass. Assume velocity of light is 3 # 108 m/s in vacuum (A) 3 mm (B) 3 mm (C) 2 mm (D) 1 mm 2005

9.56

9.57

9.58

Voltage standing wave pattern in a lossless transmission line with characteristic impedance 50 and a resistive load is shown in the figure.

(D) 20p2

where aty , denotes the unit vector in y direction. The wave is propagating with a phase velocity. (B) - 3 # 108 m/s (A) 5 # 10 4 m/s (C) - 1.25 # 107 m/s (D) 3 # 108 m/s 9.55

Statement of Linked Answer Questions 9.46 & 9.47 :

A mast antenna consisting of a 50 meter long vertical conductor operates over a perfectly conducting ground plane. It is base-fed at a frequency of 600 kHz. The radiation resistance of the antenna is Ohms is 2 2 (A) 2p (B) p 5 5 2 (C) 4p 5

9.54

Page 221

9.59

9.60

The value of the load resistance is (A) 50 W (B) 200 W (C) 12.5 W (D) 0 The reflection coefficient is given by (A) - 0.6 (B) - 1 (C) 0.6 (D) 0

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TWO MARKS

Which one of the following does represent the electric field lines for the mode in the cross-section of a hollow rectangular metallic waveguide ?

Characteristic impedance of a transmission line is 50 W. Input impedance of the open-circuited line when the transmission line a short circuited, then value of the input impedance will be. (A) 50 W (B) 100 + j150W (C) 7.69 + j11.54W (D) 7.69 - j11.54W

9.61

Many circles are drawn in a Smith Chart used for transmission line calculations. The circles shown in the figure represent

(A) Unit circles (B) Constant resistance circles (C) Constant reactance circles (D) Constant reflection coefficient circles. 2004 9.62

Two identical and parallel dipole antennas are kept apart by a distance of l in the H - plane. They are fed with equal currents 4 but the right most antenna has a phase shift of + 90c. The radiation pattern is given as.

9.63

The phase velocity of an electromagnetic wave propagating in a hollow metallic rectangular waveguide in the TE10 mode is (A) equal to its group velocity (B) less than the velocity of light in free space (C) equal to the velocity of light in free space (D) greater than the velocity of light in free space Consider a lossless antenna with a directive gain of + 6 dB. If 1 mW of power is fed to it the total power radiated by the antenna will be (A) 4 mW (B) 1 mW (C) 7 mW (D) 1/4 mW 2004

9.64

ONE MARK

TWO MARKS

A parallel plate air-filled capacitor has plate area of 10 - 4 m 2 and plate separation of 10 - 3 m. It is connect - ed to a 0.5 V, 3.6 GHz

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

source. The magnitude of the displacement current is ( e = F/m) (A) 10 mA (B) 100 mA (C) 10 A (D) 1.59 mA 9.65

1 36p

Page 222

10 - 9

the medium should be (A) 120p W (C) 600p W 9.70

Consider a 300 W, quarter - wave long (at 1 GHz) transmission line as shown in Fig. It is connected to a 10 V, 50 W source at one end and is left open circuited at the other end. The magnitude of the voltage at the open circuit end of the line is

(B) 60p W (D) 24p W

A lossless transmission line is terminated in a load which reflects a part of the incident power. The measured VSWR is 2. The percentage of the power that is reflected back is (A) 57.73 (B) 33.33 (C) 0.11 (D) 11.11 2003

9.71

(A) 10 V (C) 60 V 9.66

(B) 5 V (D) 60/7 V

9.72

In a microwave test bench, why is the microwave signal amplitude modulated at 1 kHz

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in

9.73

9.68

If E = (atx + jaty) e jkz - kwt and H = (k/wm) (aty + katx ) e jkz - jwt , the timeaveraged Poynting vector is (A) null vector (B) (k/wm) atz (C) (2k/wm) atz (D) (k/2wm) atz Consider an impedance Z = R + jX marked with point P in an impedance Smith chart as shown in Fig. The movement from point P along a constant resistance circle in the clockwise direction by an angle 45c is equivalent to

9.74

9.75

(B) Ampere/meter (D) Ampere-meter

The depth of penetration of electromagnetic wave in a medium having conductivity s at a frequency of 1 MHz is 25 cm. The depth of penetration at a frequency of 4 MHz will be (A) 6.25 dm (B) 12.50 cm (C) 50.00 cm (D) 100.00 cm 2003

(A) To increase the sensitivity of measurement (B) To transmit the signal to a far-off place (C) To study amplitude modulations (D) Because crystal detector fails at microwave frequencies 9.67

The unit of 4# H is (A) Ampere (C) Ampere/meter 2

ONE MARK

TWO MARKS

Medium 1 has the electrical permittivity e1 = 1.5e0 farad/m and occupies the region to the left of x = 0 plane. Medium 2 has the electrical permittivity e2 = 2.5e0 farad/m and occupies the region to the right of x = 0 plane. If E1 in medium 1 is E1 = (2ux - 3uy + 1uz ) volt/m, then E2 in medium 2 is (A) (2.0ux - 7.5uy + 2.5uz ) volt/m (B) (2.0ux - 2.0uy + 0.6uz ) volt/m (C) (2.0ux - 3.0uy + 1.0uz ) volt/m (D) (2.0ux - 2.0uy + 0.6uz ) volt/m If the electric field intensity is given by E = (xux + yuy + zuz ) volt/m, the potential difference between X (2, 0, 0) and Y (1, 2, 3) is (A) + 1 volt (B) - 1 volt (C) + 5 volt (D) + 6 volt A uniform plane wave traveling in air is incident on the plane boundary between air and another dielectric medium with er = 4 . The reflection coefficient for the normal incidence, is (A) zero (B) 0.5+180c (B) 0.333+0c (D) 0.333+180c

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(A) adding an inductance in series with Z (B) adding a capacitance in series with Z (C) adding an inductance in shunt across Z (D) adding a capacitance in shunt across Z 9.69

A plane electromagnetic wave propagating in free space is incident normally on a large slab of loss-less, non-magnetic, dielectric material with e > e0 . Maxima and minima are observed when the electric field is measured in front of the slab. The maximum electric field is found to be 5 times the minimum field. The intrinsic impedance of

9.77

If the electric field intensity associated with a uniform plane electromagnetic wave traveling in a perfect dielectric medium is given by E (z, t) = 10 cos (2p107 t - 0.1pz) V/m, then the velocity of the traveling wave is (B) 2.00 # 108 m/sec (A) 3.00 # 108 m/sec (C) 6.28 # 107 m/sec (D) 2.00 # 107 m/sec A short - circuited stub is shunt connected to a transmission line as shown in fig. If Z0 = 50 ohm, the ittance Y seen at the junction of the stub and the transmission line is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 223

(C) elliptically polarized 9.84

(D) unpolarized

Distilled water at 25c C is characterized by s = 1.7 # 10 - 4 mho/m and e = 78eo at a frequency of 3 GHz. Its loss tangent tan d is ( e = 10 36p F/m) -9

(A) 1.3 # 10-5 (C) 1.3 # 10-4 /78 9.85

(B) 1.3 # 10-3 (D) 1.3 # 10-5 /78e0

The electric field on the surface of a perfect conductor is 2 V/m. The conductor is immersed in water with e = 80eo . The surface charge density on the conductor is ( e = 10 36p F/m) 2 (A) 0 C/m (B) 2 C/m 2 -9

(A) (0.01 - j0.02) mho (C) (0.04 - j0.02) mho 9.78

9.79

(C) 1.8 # 10 - 11 C/m 2

(B) (0.02 - j0.01) mho (D) (0.02 + j0) mho

9.86

A rectangular metal wave guide filled with a dielectric material of relative permittivity er = 4 has the inside dimensions 3.0 cm # 1.2 cm. The cut-off frequency for the dominant mode is (A) 2.5 GHz (B) 5.0 GHz (C) 10.0 GHz (D) 12.5 GHz Two identical antennas are placed in the q = p/2 plane as shown in Fig. The elements have equal amplitude excitation with 180c polarity difference, operating at wavelength l. The correct value of the magnitude of the far-zone resultant electric field strength normalized with that of a single element, both computed for f = 0 , is

(D) 1.41 # 10 - 9 C/m 2

A person with receiver is 5 Km away from the transmitter. What is the distance that this person must move further to detect a 3-dB decrease in signal strength (A) 942 m (B) 2070 m (C) 4978 m (D) 5320 m

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A transmission line is distortonless if (B) RL = GC (A) RL = 1 GC (C) LG = RC

(A) 2 cos b 2ps l l (C) 2 cos a ps k l

(B) 2 sin b 2ps l l (D) 2 sin a ps k l

9.88

2002 9.80

9.81

9.82

The VSWR can have any value between (A) 0 and 1 (B) - 1 and + 1 (C) 0 and 3 (D) 1 and 3

9.89

In in impedance Smith movement along a constant resistance circle gives rise to (A) a decrease in the value of reactance (B) an increase in the value of reactance (C) no change in the reactance value (D) no change in the impedance The phase velocity for the TE10 waveguide is (c is the velocity of (A) less than c (C) greater than c 2002

9.83

ONE MARK

9.90

-mode in an air-filled rectangular plane waves in free space) (B) equal to c (D) none of these TWO MARKS jp/2

jwt - jkz

t )e A plane wave is characterized by E = (0.5xt + ye wave is (A) linearly polarized (B) circularly polarized

. This

(D) RG = LC

2 2 If a plane electromagnetic wave satisfies the equal d E2x = c2 d E2x , dZ dt the wave propagates in the (A) x - direction (B) z - direction (C) y - direction (D) xy plane at an angle of 45c between the x and z direction

The plane velocity of wave propagating in a hollow metal waveguide is (A) grater than the velocity of light in free space (B) less than the velocity of light in free space (C) equal to the velocity of light free space (D) equal to the velocity of light in free The dominant mode in a rectangular waveguide is TE10 , because this mode has (A) the highest cut-off wavelength (B) no cut-off (C) no magnetic field component (D) no attenuation 2001

9.91

ONE MARK

TWO MARKS

A material has conductivity of 10 - 2 mho/m and a relative permittivity of 4. The frequency at which the conduction current in the medium is equal to the displacement current is

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) 45 MHz (C) 450 MHz 9.92

9.93

9.94

(l1 /l2) is (A) er1 /er2

(B) 90 MHz (D) 900 MHz

A uniform plane electromagnetic wave incident on a plane surface of a dielectric material is reflected with a VSWR of 3. What is the percentage of incident power that is reflected ? (A) 10% (B) 25% (C) 50% (D) 75% A medium wave radio transmitter operating at a wavelength of 492 m has a tower antenna of height 124. What is the radiation resistance of the antenna? (A) 25 W (B) 36.5 W (C) 50 W (D) 73 W In uniform linear array, four isotropic radiating elements are spaced l apart. The progressive phase shift between required for forming 4 the main beam at 60c off the end - fire is : (A) - p (B) - p2 radians (C) - p4 radians (D) - p8 radians

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in 2000 9.95

9.96

9.97

A TEM wave is incident normally upon a perfect conductor. The E and H field at the boundary will be respectively, (A) minimum and minimum (B) maximum and maximum (C) minimum and maximum (D) maximum and minimum If the diameter of a l dipole antenna is increased from l to l 2 100 50 , then its

9.99

9.100

(D)

9.102

Two coaxial cable 1 and 2 are filled with different dielectric constants er1 and er2 respectively. The ratio of the wavelength in the cables

For an 8 feet (2.4m) parabolic dish antenna operating at 4 GHz, the minimum distance required for far field measurement is closest to (A) 7.5 cm (B) 15 cm (C) 15 m (D) 150 m ONE MARK

An electric field on a place is described by its potential V = 20 (r-1 + r-2) where r is the distance from the source. The field is due to (A) a monopole (B) a dipole (C) both a monopole and a dipole (D) a quadruple

9.103

9.104

9.105

Assuming perfect conductors of a transmission line, pure TEM propagation is NOT possible in (A) coaxial cable (B) air-filled cylindrical waveguide (C) parallel twin-wire line in air (D) semi-infinite parallel plate wave guide Indicate which one of the following will NOT exist in a rectangular resonant cavity. (A) TE110 (B) TE 011 (C) TM110 (D) TM111 Identify which one of the following will NOT satisfy the wave equation. (A) 50e j (wt - 3z) (B) sin [w (10z + 5t)] 2 (C) cos (y + 5t) (D) sin (x) cos (t) 1999

9.106

9.107

TWO MARKS

In a twin-wire transmission line in air, the adjacent voltage maxima are at 12.5 cm and 27.5 cm . The operating frequency is (A) 300 MHz (B) 1 GHz (C) 2 GHz (D) 6.28 GHz A transmitting antenna radiates 251 W isotropically. A receiving antenna, located 100 m away from the transmitting antenna, has an effective aperture of 500 cm2 . The total received by the antenna is

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 (A) 10 mW (C) 20 mW 9.108

2

A rectangular waveguide has dimensions 1 cm # 0.5 cm. Its cut-off frequency is (A) 5 GHz (B) 10 GHz (C) 15 GHz (D) 12 GHz

(B) er2 /er1 (D) er2 /er1

1999

TWO MARKS

A uniform plane wave in air impings at 45c angle on a lossless dielectric material with dielectric constant dr . The transmitted wave propagates is a 30c direction with respect to the normal. The value of dr is (B) 1.5 (A) 1.5 (C) 2

9.101

(B) bandwidth decrease (D) gain decreases

2000 9.98

(C) er1 /er2

ONE MARK

The magnitudes of the open-circuit and short-circuit input impedances of a transmission line are 100 W and 25 W respectively. The characteristic impedance of the line is, (A) 25 W (B) 50 W (C) 75 W (D) 100 W

(A) bandwidth increases (C) gain increases

Page 224

9.109

(B) 1 mW (D) 100 mW

In air, a lossless transmission line of length 50 cm with L = 10 mH/m , C = 40 pF/m is operated at 25 MHz . Its electrical path length is (A) 0.5 meters (B) l meters (C) p/2 radians (D) 180 deg rees A plane wave propagating through a medium [er = 8, vr = 2, and s = 0] t - (z/3) sin (108 t - bz) V/m . The has its electric field given by Ev = 0.5Xe wave impedance, in ohms is (A) 377 (B) 198.5+180c (D) 133.3 (C) 182.9+14c

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

1998 9.110

9.111

9.112

ONE MARK

Page 225 9.119

The intrinsic impedance of copper at high frequencies is (A) purely resistive (B) purely inductive (C) complex with a capacitive component (D) complex with an inductive component The Maxwell equation V # H = J + 2D is based on 2t (A) Ampere’s law (B) Gauss’ law (C) Faraday’s law (D) Coulomb’s law

9.120

All transmission line sections shown in the figure is have a characteristic impedance R 0 + j 0 . The input impedance Zin equals

9.121

An antenna in free space receives 2 mW of power when the incident electric field is 20 mV/m rms. The effective aperture of the antenna is (A) 0.005 m2 (B) 0.05 m2 (C) 1.885 m2 (D) 3.77 m2 The maximum usable frequency of an ionospheric layer at 60c incidence and with 8 MHz critical frequency is (A) 16 MHz (B) 16 MHz 3 (C) 8 MHz (D) 6.93 MHz A loop is rotating about they y -axis in a magnetic field Bv = B 0 cos (wt + f) avx T. The voltage in the loop is (A) zero

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(A) 2 R 0 3 (C) 3 R 0 2

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(B) R 0 (D) 2R 0

1998 9.113

9.114

9.115

9.116

9.117

9.118

(B) due to rotation only (C) due to transformer action only (D) due to both rotation and transformer action

TWO MARKS

The time averages Poynting vector, Ev = 24e j (wt + bz) avy V/m in free space is (A) - 2.4 avz (B) p (C) 4.8 avz (D) p

in W/m2 , for a wave with

9.122

2.4 av p z - 4.8 avz p

The wavelength of a wave with propagation constant (0.1p + j0.2p) m-1 is 2 m (A) (B) 10 m 0.05 (C) 20 m (D) 30 m

1997 9.123

The depth of penetration of wave in a lossy dielectric increases with increasing (A) conductivity (B) permeability (C) wavelength (D) permittivity The polarization of Ev = E 0 e j^wt + bz h ^avx + avy h is (A) linear (C) left hand circular

wave

with

electric

field

vector

9.124

(B) elliptical (D) right hand circular

The vector H in the far field of an antenna satisfies (A) d $ Hv = 0 and d # Hv = 0 (B) d $ Hv ! 0 and d # Hv ! 0 (C) d $ Hv = 0 and d # Hv ! 0 (D) d $ Hv ! 0 and d # Hv = 0 The radiation resistance of a circular loop of one turn is 0.01 W. The radiation resistance of five turns of such a loop will be (B) 0.01 W (A) 0.002 W (C) 0.05 W (D) 0.25 W

The far field of an antenna varies with distance r as (A) 1 (B) 12 r r (C) 13 (D) 1 r r

9.125

ONE MARK

A transmission line of 50 W characteristic impedance is terminated with a 100 W resistance. The minimum impedance measured on the line is equal to (A) 0 W (B) 25 W (C) 50 W (D) 100 W A rectangular air filled waveguide has cross section of 4 cm #10 cm . The minimum frequency which can propagate in the waveguide is (A) 0.75 GHz (B) 2.0 GHz (C) 2.5 GHz (D) 3.0 GHz A parabolic dish antenna has a conical beam 2c wide, the directivity of the antenna is approximately (A) 20 dB (B) 30 dB (C) 40 dB (D) 50 dB

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

1997 9.126

9.127

Page 226

TWO MARKS

A very lossy, l/4 long, 50 W transmission line is open circuited at the load end. The input impedance measured at the other end of the line is approximately (A) 0 (B) 50 W (C) 3 (D) None of the above

(A) 0% (B) 4% (C) 20% (D) 100% 9.133

The skin depth at 10 MHz for a conductor is 1 cm. The phase velocity of an electromagnetic wave in the conductor at 1, 000 MHz is about (A) 6 # 106 m/ sec (B) 6 # 107 m/ sec (C) 3 # 108 m/ sec (D) 6 # 108 m/ sec 9.134

1996 9.128

ONE MARK

A lossless transmission line having 50 W characteristic impedance and length l/4 is short circuited at one end and connected to an ideal voltage source of 1 V at the other end. The current drawn from

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in the voltage source is (A) 0 (C) 3 9.129

9.130

9.131

A 1 km long microwave link uses two antennas each having 30 dB gain. If the power transmitted by one antenna is 1 W at 3 GHz, the power received by the other antenna is approximately (A) 98.6 mW (B) 76.8 mW (C) 63.4 mW (D) 55.2 mW Some unknown material has a conductivity of 106 mho/m and a permeability of 4p # 10-7 H/m . The skin depth for the material at 1 GHz is (A) 15.9 mm (B) 20.9 mm (C) 25.9 mm (D) 30.9 mm

(B) 0.02 A (D) none of these

The capacitance per unit length and the characteristic impedance of a lossless transmission line are C and Z 0 respectively. The velocity of a travelling wave on the transmission line is (A) Z 0 C (B) 1 Z0 C (C) Z 0 (D) C Z0 C A transverse electromagnetic wave with circular polarization is received by a dipole antenna. Due to polarization mismatch, the power transfer efficiency from the wave to the antenna is reduced to about (A) 50% (B) 35.3% (C) 25% (D) 0% A metal sphere with 1 m radius and a surface charge density of 10 Coulombs/m2 is enclosed in a cube of 10 m side. The total outward electric displacement normal to the surface of the cube is (A) 40p Coulombs (B) 10p Coulombs (C) 5p Coulombs (D) None of these 1996

9.132

9.135

The critical frequency of an ionospheric layer is 10 MHz. What is the maximum launching angle from the horizon for which 20 MHz wave will be reflected by the layer ? (A) 0c (B) 30c (C) 45c (D) 90c

TWO MARKS

A uniform plane wave in air is normally incident on infinitely thick slab. If the refractive index of the glass slab is 1.5, then the percentage of incident power that is reflected from the air-glass interface is

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 227

SOLUTION

Also, direction of propagation is v v avk = ax + az 2 x + z So, k = 2 Substituting it in equation (1), we get p # 10 ^x + z h Evi = Eo _avx - avz i e-j 3 2 2 4

9.1

Option (D) is correct. v around Stoke’s theorem states that the circulation a vector field A v over a closed path l is equal to the surface integral of the curl of A the open surface S bounded by l . vh : dsv v : dlv = ^d # A i.e., A

9.5

##

#

Here, line integral is taken across a closed path which is denoted by a small circle on the integral notation where as, the surface invh is taken over open surface bounded by the loop. tegral of ^d # A 9.2

Option (A) is correct. We obtain the reflection coefficient for parallel polarized wave (since, electric field is in the plane of wave propagation) as h cos qt - h1 cos qi Gz = 2 h2 cos qt + h1 cos qi ...(1) As we have already obtained

Option (D) is correct. Given, the vector field

qi = 45c, qt = 19.2c m 1 = h2 = = h0 e 4.5

Also, v = xavx + yavy + zavz A

so, v (Divergence of A v ) = 2Ax + 2Ay + 2Az d$A 2x 2y 2z = 1+1+1 = 3 9.3

Option (A) is correct. Given, the return loss of device as 20 dB i.e., G in dB =- 20 dB (loss) ^

or,

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h

20 log G =- 20

& G = 10-1 = 0.1 Therefore, the standing wave ration is given by 1+ G VSWR = 1- G

m = h0 1 = h0 1 e Substituting these in eq. (1) we get G z = cos 19.2c - 4.5 cos 45c cos 19.2c + 4.5 cos 45c =- 0.227 .- 0.23 Therefore, the reflected field has the magnitude given by Ero = T 11' Eio or Ero = G z Eio =- 0.23 Eio Hence, the expression of reflected electric field is p # 10 Evr =- 0.23 Eo _- avx - avz i e-j 3 k 2 (2)

Option (C) is correct. For the given incidence of plane wave, we have the transmitting angle qt = 19.2c From Snell’s law, we know n1 sin qi = n 2 sin qt

4

c m1 e1 sin qi = c m2 e2 sin qt ...(1) For the given interfaces, we have m1 = m2 = 1 e1 = 1, e2 = 4.5 So, from Eq. (1) sin qi = 4.5 sin 19.2 or, qi . 45c v Now, the component of Ei can be obtained as Evi = _Eox avx - Eoz avz i e-jbk (observed from the shown figure) Since, the angle qi = 45c so, Eox = Eoz = Eo 2 E v o v Therefore, Ei = _ax - avz i e-jbk 2 ...(1) Now, the wavelength of EM wave is l = 600 mm So,

b = 2p = p # 10 4 3 l

h1 =

and

= 1 + 0.1 = 1.1 = 1.22 1 - 0.1 0.9 9.4

h0 4.5

Again, we have the propagation vector of reflected wave as v v avk = ax - az 2 x z or, k = 2 Substituting it in Eq. (2), we get p # 10 x - z Evr =- 0.23 Eo _- avx - avz i e-j 3 b 2 l 2 jp # 10 ^x - z h V E m Evr = 0.23 o _avx + avz i e- 3 2 2 4

4

9.6

Option (C) is correct. Electric field of the propagating wave in free space is given as Ei = (8ax + 6ay + 5az ) e j (wt + 3x - 4y) V/m So, it is clear that wave is propagating in the direction (- 3ax + 4ay) . Since, the wave is incident on a perfectly conducting slab at x = 0 . So, the reflection coefficient will be equal to - 1. i.e. Er = (- 1) Ei =- 8ax - 6ay - 5az Again, the reflected wave will be as shown in figure. 0

0

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 228

l = (2m + 1) l 4 3 # 108 c = = 0.174 m f1 # 4 429 # 106 # 4 8 f2 = 1 GHz , l2 = c = 3 # 10 = 0.075 m 9 f2 # 4 1 # 10 # 4 Only option (C) is odd multiple of both l1 and l2 . (2m + 1) = 1.58 = 9 l1 (2m + 1) = 1.58 - 21 l2 f1 = 429 MHz,

i.e. the reflected wave will be in direction 3ax + 4ay . Thus, the electric field of the reflected wave will be. Ex = (- 8ax - 6ay - 5az ) e j (wt - 3x - 4y) V/m 9.7

Option (A) is correct. The field in circular polarization is found to be Es = E 0 (ay ! jaz ) e-jbx propagating in + ve x -direction. where, plus sign is used for left circular polarization and minus sign for right circular polarization. So, the given problem has left circular polarization. b = 25 = w c

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9.9

2pf c

9.11

Substituting above values, b = - j261

9.12

# H : dl

For r < a ,

#

cos 4 q sin qdqdfE

p/2 5 = 1 ;2p b- cos q lE = 1 # 2p :- 0 + 1 D 5 5 4p 4p 0 = 1 # 2p = 1 5 10 4p D = 1 = 10 10

9.10

D (in dB) = 10 log 10 = 10 dB

Option (C) is correct. Since

Z0 =

Z1 Z 2

100 = 50 # 200 This is quarter wave matching. The length would be odd multiple of l/4 .

Ienclosed =

Io = (pa2) J

J (pr 2) Jr 2 = 2 a pa2

# H : dl

= Ienclosed 2 H # 2pr = Jr2 a H = Jr 2 2pa H \ r , for r < a

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# # 0

= Ienclosed

H # 2pr = (pa2) J H = Io 2pr H \ 1 , for r > a r

Fmax = 1 Favg = 1 F (q, f) dW 4p 2p 2p = 1 ; F (q, f) sin qdq dfE 4p 0 0

or,

Option ( ) is correct. For r > a , Ienclosed = (pa2) J

So,

p/2

6.283 # 1010 2 - (2.0942 + 2.6182) 10 4 # c 3 108 m #

b is imaginary so mode of operation is non-propagating. vp = 0

Option (A) is correct. The directivity is defined as D = Fmax Favg

# #

w 2 - (b 2 + b 2) x y c2

b =

Option (B) is correct. Let b " outer diameter a " inner diameter Characteristic impedance, m0 4p # 10-7 # 36p ln 2.4 = 100 W Z0 = ln b b l = b 1 l e0 er a 10-9 # 10.89

2p = 1 ; 4p 0

Option (D) is correct. Hz = 3 cos (2.094 # 102 x) cos (2.618 # 102 y) cos (6.283 # 1010 t - bz) bx = 2.094 # 102 by = 2.618 # 102 w = 6.283 # 1010 rad/s For the wave propagation,

8 f = 25 # c = 25 # 3 # 10 = 1.2 GHz 2p 2 # 3.14

&

l1 =

9.13

Option (B) is correct. Assuming the cross section of the wire on x -y plane as shown in figure.

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 229

Further if H z = 1.5utx + Auty + Buz Then from Boundary condition 10ut (3utx + 30uty) utx = (1.5utx + Auty + Butz ) xt + v y J =- 30utz =- Autz + Buty + 10uty Comparing we get A = 30 and B =- 10 So H z = 1.5utx + 30uty - 10utz A/m 9.18

Since, the hole is drilled along the length of wire. So, it can be assumed that the drilled portion carriers current density of - J . Now, for the wire without hole, magnetic field intensity at point P will be given as Hf1 (2pR) = J (pR2) Hf1 (2pR) = JR 2 Since, point o is at origin. So, in vector form H1 = J (xax + yay) 2 Again only due to the hole magnetic field intensity will be given as.

9.19

9.14

9.15

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9.17

From the expressions of Ev & Hv , we can write, b = 280 p 2p = 280 p & l = 1 or 140 l v E Wave impedance, Zw = E = p = 120 p 3 v er H again,

y = yl H2 = - J [(x - d) ax + yay] 2

Now

So, total magnetic field intensity = H1 + H2 = J dax 2 So, magnetic field inside the hole will depend only on ‘d ’.

or

Option (C) is correct. Power radiated from any source is constant.

Now

or

9.20

Option (C) is correct. We have d = 2 mm and f = 10 GHz Phase difference = 2p d = p ; 4 l

Option (A) is correct. TM11 is the lowest order mode of all the TMmn modes.

or or

Option (C) is correct. For a lossless network

(not lossless)

S12 = S21 = 0.9 90c (Reciprocal) 9.21

Option (A) is correct. From boundary condition Bn1 = Bn2 m1 Hx1 = m2 Hx2 Hx2 = Hx1 = 1.5 2 Hx2 = 1.5utx

f = 14 GHz 8 3 l = C = 3 # 10 9 = er f er 14 # 10 140 er 3 1 = 140 140 er er = 9 Ep = 120p = E p = 120p 3 9

S11 2 + S21 2 = 1 For the given scattering matrix S11 = 0.2 0c , S12 = 0.9 90c S21 = 0.9 90c , S22 = 0.1 90c Here, (0.2) 2 + (0.9) 2 ! 1 Reciprocity :

= l = 8d = 8 # 2 mm = 16 mm v = fl = 10 # 109 # 16 # 10-3 = 1.6 # 108 m/ sec

or

9.16

Option (D) is correct.

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(Hf2) (2pr) =- J (pr 2) Hf2 = - Jr 2 Again, if we take Ol at origin then in vector form H2 = - J (xlax + ylay) 2 where xl and yl denotes point ‘P ’ in the new co-ordinate system. Now the relation between two co-ordinate system will be. x = xl + d So,

Option (A) is correct. Since voltage maxima is observed at a distance of l/4 from the load and we know that the separation between one maxima and minima equals to l/4 so voltage minima will be observed at the load, Therefore load can not be complex it must be pure resistive. Now G = s-1 s+1 also RL = R 0 (since voltage maxima is formed at the load) s RL = 50 = 10 W 5

Option (D) is correct. For distortion less transmission line characteristics impedance Z0 = R G Attenuation constant a = So,

9.22

Option (C) is correct.

RG a = R = 0.1 = 0.002 50 Z0

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 230

tan bl = tan b 2p : l l = 3 4 l R ZL V S tan bl + jZo W 2 W = Z 0 = 60 W Z in = Zo S S Zo + jZL W ZL S tan bl W T X For length of l/8 transmission line Z + jZo tan bl Z in = Zo ; L Zo + jZL tan bl E

Intrinsic impedance of EM wave m m0 = = 120p = 60p h = e 2 4e0 Time average power density 2 1 = 1 Pav = 1 EH = 1 E = 2 2 h 2 # 60p 120p 9.23

So,

Option (C) is correct. v = xyatx + x 2 aty A v = dxatx + dyaty dl v = # (xyatx + x 2 aty) : (dxatx + dyaty) # Av : dl

= # (xydx + x 2 dy)

Zo = 30 W, ZL = 0 (short) tan bl = tan b 2p : l l = 1 8 l

C

C

Z in = jZo tan bl = 30j Circuit is shown below.

C

=

2/ 3

#1/

3

xdx +

1/ 3

#2/

3

3xdx +

#1

3

4 dy + 3

#3

1

1 dy 3

= 1 : 4 - 1 D + 3 :1 - 4 D + 4 [3 - 1] + 1 [1 - 3] = 1 2 3 3 2 3 3 3 3 9.24

Option (A) is correct.

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Reflection coefficient 60 + 3j - 60 t = ZL - Zo = = 60 + 3j + 60 ZL + Zo 1+ t VSWR = = 1 + 17 = 1.64 1- t 1 - 17 9.26

In the given problem

9.27

Reflection coefficient h - h1 t = 2 = 400p - 120p =- 1 h2 + h 1 2 40p + 120p t is negative So magnetic field component does not change its direction Direction of incident magnetic field atE # atH = atK atZ # atH = aty atH = atx ( + x direction) So, reflection magnetic field component Hr = t # 24 cos (3 # 108 + by) atx , y $ 0 h = 1 # 24 cos (3 # 108 + by) atx , y $ 0 2 # 120p

So, 9.25

8 b = w = 3 # 108 = 1 vC 3 # 10 Hr = 1 cos (3 # 108 + y) atx , y $ 0 10p

Option (B) is correct. For length of l/4 transmission line Z + jZo tan bl Z in = Zo ; L Zo + jZL tan bl E ZL = 30 W ,

Zo = 30 W, b = 2p , l = l 4 l

9.28

1 17

Option (D) is correct. Due to 1 A current wire in x - y plane, magnetic field be at origin will be in x direction. Due to 1 A current wire in y - z plane, magnetic field be at origin will be in z direction. Thus x and z component is non-zero at origin. Option (A) is correct. Rectangular and cylindrical waveguide doesn’t TEM modes and have cut off frequency. Coaxial cable TEM wave and doesn’t have cut off frequency. Option (B) is correct. We have V = 4# A By Stokes theorem

# A $ dl

=

## (4 # A) $ ds

...(1) ...(2)

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 From (1) and (2) we get

# A $ dl 9.29

=

Option (D) is correct.

##V $ ds

The transmission line are as shown below. Length of all line is l 4

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 231

2 2 Zi1 = Z01 = 100 = 200W ZL1 50

Reflection coefficient

2 2 Zi2 = Z02 = 100 = 200W ZL2 50

G=

Substituting values for h1 and h2 we have

ZL3 = Zi1 Zi2 = 200W 200W = 100W Zi = 9.30

Z02

ZL3

= 50 = 25W 100

er = 9

x a - y a y xm c 2 x + y2 x2 + y2 To convert in cylindrical substituting x = r cos f and y = r sin f

Now

9.31

9.32

+

mo eo mo eo

= 11+

er = 1 er 1+

9 9

since

=- 0.5 ...(1)

9.36

ax = cos far - sin faf ay = sin far + cos faf Bv = Bv0 af v Bv a Hv = B = 0 f m0 m0 Jv = 4# Hv = 0

mo eo er mo eo er

t =

2

Option (C) is correct. We have Bv = B0

and In (1) we have

h2 - h1 h2 + h1

Option (C) is correct. In single mode optical fibre, the frequency of limiting mode increases as radius decreases Hence r \ 1 f So. if radius is doubled, the frequency of propagating mode gets halved, while in option (D) it is increased by two times.

9.37

constant since H is constant

Option (C) is correct. The beam-width of Hertizian dipole is 180c and its half power beamwidth is 90c.

Option (D) is correct.

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Option (D) is correct. Maxwell equations 4- B = 0 4$ E = r/E 4# E =- B 4# Ht = D + J For static electric magnetic fields

8 l = c = 3 # 10 9 = 3 f 200 20 # 10 2 2 Gp = hp2 ` D j = 0.7 # p2 c 13 m = 30705.4 l 100

Gain

= 44.87 dB

4$ B = 0 9.38

4$ E = r/E

Option (A) is correct. g = b cos 30cx ! b sin 30cy

4# E = 0 S 4# H = J 9.33

= 2p 3 x ! 2p 1 y l 2 l 2 = p 3 x! py l l

Option (A) is correct. Cut-off Frequency is fc = c 2 For TE11 mode,

m 2 n 2 ` a j +`b j

3 # 1010 fc = 2 9.34

9.39

1 2 1 2 ` 4 j + ` 3 j = 6.25 GHz

For ZL = 0 , Zin = iZo tan (bl) The wavelength is 8 l = c = 3 # 109 = 0.1 m or 10 cm f 3 # 10 bl = 2p l = 2p # 1 = p 10 l 5 Thus Zin = iZo tan p 5 Thus Zin is inductive because Zo tan p is positive 5 Option (C) is correct. m We have h = e

Maxwell Equations

## 4# H $ ds = ## `J + 22Dt j .ds

Integral form

## `J + 22Dt j .ds

Stokes Theorem

s

s

ZL + iZo tan (bl) Zo + iZL tan (bl)

p 3 x! p y l l mE

Option (D) is correct. 4# H = J + 2D 2t

Option (D) is correct. Zin = Zo

9.35

E = ay E0 e j (wt - g) = ay E0 e j;wt - c

# H $ dl

=

s 9.40

Option (A) is correct. 2 wm p H sin ` 2px j sin (wt - bz) yt 2 `2j 0 a h This is TE mode and we know that mpy Ey \ sin ` mpx j cos ` a b j

E =

Thus m = 2 and n = 0 and mode is TE20 9.41

Option (C) is correct. The 2-port scattering parameter matrix is S11 S12 S == S21 S22 G

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 232

(ZL Z0) - Zo (50 50) - 50 = =- 1 (ZL Z0) + Zo (50 50) + 50 3 2 (ZL Zo) 2 (50 50) S12 = S21 = = =2 (ZL Zo) + Zo (50 50) + 50 3 (Z Z ) - Zo (50 50) - 50 S22 = L o = =- 1 (ZL Zo) + Zo (50 50) + 50 3

The Brewster angle is

S11 =

9.42

Option (D) is correct. The input impedance is

Zin1

tan 60c =

if l = l 4

2 2 = Zo1 = 50 = 25 ZL1 100

9.48

ZL = Zin1 Zin2 25 12.5 = 25 3 Zs =

(50) 2 = 300 25/3 9.49

For free space watts 9.44

Option (A) is correct. mo eo er mo eo er

+

mo eo mo eo

Pt = (1 - G2) Pi = 1 - 1 = 8 9 9 Pt = 8 Pi 9

Option (D) is correct. sin q = 1 = 1 er 2 p q = 45c = 4

Hy2

or

The configuration is shown below. Here A is point source.

m 2 n 2 ` a j +`b j

Since the mode is TE20, m = 2 and n = 0 8 fc = c m = 3 # 10 # 2 = 10 GHz 2 2 2 # 0.03 ho 377 = h' = = 400W 10 2 fc 2 1-c m 1 - c 10 10 m f 3 # 10 Option (B) is correct. Using the method of images, the configuration is as shown below

Now From geometry

AO = 1 m BO = 1 m

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 = pr2 = p # OB = p m 2

Thus area 9.51

Option (C) is correct. The cut-off frequency is fc = c 2

Here d = l, a = p, thus bd = 2p Array factor is

9.46

h2 - h1 = h2 + h1

Option (C) is correct. The cut-off frequency is fc = c 2

9.45

Prad = 10 Watts

or

2 2 2 = c 5 3 m + c 5 m = c 10 m H = ho ho ho 2 2 ho H E 2 h = = o c 10 m = 50 P = 2 ho ho 2ho 2

We have

Option (A) is correct. We have 10 log G = 10 dB or G = 10 Now gain G = Prad Pin or 10 = Prad 1W

= 1 + er = 1 - 4 =- 1 3 1 + er 1+ 4 The transmitted power is

9.50

Option (D) is correct. Hx2 +

, thus wave is left hand circularly polarized.

G=

G = ZS - Zo = 300 - 50 = 5 ZS + Zo 300 + 50 7 2

p 2

or

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9.43

Option (C) is correct. We have E = atxx sin (wt - bz) + aty sin (wt - bz + p/2) Here Ex = Ey and fx = 0, fy = p2 Phase difference is

2 2 Zin2 = Zo2 = 50 = 12.5 ZL2 200

Now

er2 = 3

or 9.47

2 Zin = Zo ; ZL

er2 er1 er2 1

tan qn =

Option (D) is correct.

bd cos y + a E 2 2p cos y + p = cos ; E = sin (p cos y) 2 = cos ;

m 2 m 2 ` a j +` b j

Since the mode is TE30 , m = 3 and n = 0 fc = c m 2 a or or

8 18 # 109 = 3 # 10 3 2 a a = 1 m = 5 cm 40 2

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 9.52

Option (C) is correct. We have E1 = 4ux + 3uy + 5uz Since for dielectric material at the boundary, tangential component of electric field are equal E21 = E1t = 3aty + 5atz at the boundary, normal component of displacement vector are equal i.e. Dn2 = Dn1 or e2 E2n = e1 E1n or 4eo E2n = 3eo 4atz or E2n = 3atx Thus E2 = E2t + E2a = 3atx + 3aty + 5atz

9.53

Option (C) is correct. w = 50, 000 and b =- 0.004 4 vP = w = 5 # 10 - 3 = 1.25 # 107 m/s b - 4 # 10

Phase Velocity is 9.55

9.61

9.62

9.63

Option (C) is correct. Refractive index of glass m = 1.5 Frequency f = 1014 Hz c = 3 # 108 m/sec

9.57

Option (D) is correct. ZZC

9.58

bd sin q + a l 2

p 2

The option (A) satisfy this equation. 9.59

4pU (q, f) = Prad GD (q, f) = 1 m # 3.98 = 3.98 mW 9.64

Option (D) is correct.

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The capacitance is - 12 -4 C = eo A = 8.85 # 10 - 3 # 10 = 8.85 # 10 - 13 d 10 The charge on capacitor is Q = CV = 8.85 # 10 - 13 = 4.427 # 10 - 13 Displacement current in one cycle Q I = = fQ = 4.427 # 10 - 13 # 3.6 # 109 = 1.59 mA T 9.65

Option (C) is correct. VL = ZO Vin Zin or VL = ZO Vin = 10 # 300 = 60 V Zin 50

9.66

Option (D) is correct.

9.67

Option (A) is correct. Ravg = 1 Re [E # H*] 2

p p m = cos ` sin q + j 4 2

Option (C) is correct. From the diagram, VSWR is s = Vmax = 4 = 4 Vmin 1 When minima is at load ZO = s.ZL or ZL = Zo = 50 = 12.5W s 4

9.60

4pU (q, f) Prad

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Option (A) is correct. The array factor is

Here b = 2p , d = l and a = 90c l 4 2p l sin q + Thus A = cos c l 4 2

GD (q, f) =

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Zo2 = ZOC .ZSC 2 = Zo = 50 # 50 = 50 ZOC 100 + j150 2 + 3j 50 (2 - 3j) = = 7.69 - 11.54j 13

A = cos b

Option (A) is correct.

Prad = Pin Here we have Prad = Pin = 1 mW and 10 log GD (q, f) = 6 dB or GD (q, f) = 3.98 Thus the total power radiated by antenna is

-6 lg = a = 3 # 10 = 2 # 10 - 6 m 1.5 m

Option (D) is correct.

Option (D) is correct. We know that vp > c > vg .

For lossless antenna

8 l = c = 3 # 10 = 3 # 10 - 6 14 f 10 wavelength in glass is

9.56

Option (C) is correct. The given figure represent constant reactance circle.

We have

Option (C) is correct. Since antenna is installed at conducting ground, 2 2 2 50 Rrad = 80p2 ` dl j = 80p2 c = 4p W m 3 5 l 0.5 # 10

9.54

Page 233

Option (A) is correct. The reflection coefficient is G = ZL - ZO = 12.5 - 50 =- 0.6 ZL + ZO 125. + 50

Thus 9.68

E # H* = (atx + jaty) e jkz - jwt # k (- jatx + aty) e-jkz + jwt wm = atz ; k - (- j) (j) k E = 0 wm wm Ravg = 1 Re [E # H*] = 0 2

Option (A) is correct. Suppose at point P impedance is Z = r + j (- 1) If we move in constant resistance circle from point P in clockwise direction by an angle 45c, the reactance magnitude increase. Let us consider a point Q at 45c from point P in clockwise direction. It’s impedance is Z1 = r - 0.5j

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 234

or Z1 = Z + 0.5j Thus movement on constant r - circle by an +45c in CW direction is the addition of inductance in series with Z . 9.69

or

Option (D) is correct. We have

VSWR

or

G

Thus

G

Now

G -2 3

or

1- G = Emax = 5 = Emin 1+ G =2 3 =- 2 3 h - h1 = 2 h2 + h1 h - 120p = 2 h2 + 120p

Thus 9.74

or Thus or

=

2=

1- G 1+ G

1

+

0

0

#2 ydyutz + #3 zdzuzt 0

2 0 y2 +z G 2 2 2 3

Option (D) is correct. m e

h = Reflection coefficient t =

h2 - h1 h2 + h1

Substituting values for h1 and h2 we have t = er = 4

G =1 3

mo eo er mo eo er

+

m0 eo mo eo

= 11+

er = 1 er 1+

4 4

since

= - 1 = 0.333+180c 3

Pref = G2= 1 Pinc 9 Pref = Pinc 9

9.76

9.77

Thus 4# H has unit of current density J that is A/m2 Option (B) is correct.

or

xdxutx +

=- 1 [22 - 12 + 02 - 22 + 02 - 32] = 5 2 9.75

4# H = 2D + J 2t

Thus

2

2 2

Option (C) is correct. By Maxwells equations

We know that

#1

Y

#XE.dl

=-= x 2

i.e. 11.11% of incident power is reflected.

9.73

Option (C) is correct. We have E = xux + yuy + zuz dl = utx dx + uty dy + utz dz VXY =-

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9.72

E2 = E2t + E2n =- 3uy + uz + 1.2ux

Option (D) is correct. The VSWR

9.71

e1 E1n = e2 E2n 1.5eo 2ux = 2.5eo E2n E2n = 3 ux = 1.2ux 2.5

h2 = 24p

or 9.70

or or

d \ 1 f d2 = d1 d2 = 25 d2 =

f1 f2 1 4 1 # 25 = 12.5 cm 4

Option (C) is correct. We have E1 = 2ux - 3uy + 1uz E1t = - 3uy + uy and E1n = 2ux Since for dielectric material at the boundary, tangential component of electric field are equal (x = 0 plane) E1t =- 3uy + uy = E2t E1n = 2ux At the boundary the for normal component of electric field are D1n = D2n

Option (B) is correct. We have E (z, t) = 10 cos (2p # 107 t - 0.1pz) where w = 2p # 107 t b = 0.1p 7 Phase Velocity u = w = 2p # 10 = 2 # 108 m/s b 0.1p Option (A) is correct. The fig of transmission line is as shown below. [Z + jZo tan bl] We know that Zin = Zo L [Zo + jZL tan bl] For line 1, l = l and b = 2p , ZL1 = 100W l 2

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 [ZL + jZo tan p] = ZL = 100W [Zo + jZL tan p] For line 2, l = l and b = 2p , ZL2 = 0 (short circuit) l 8 [0 + jZo tan p4 ] Thus = jZo = j50W Zin2 = Zo [Zo + 0] Y = 1 + 1 = 1 + 1 = 0.01 - j0.02 Zin1 Zin2 100 j50 Thus

Zin1 = Zo

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 235 9.84

Option (A) is correct. 1.7 # 10 - 4 tan a = s = we 2p # 3 # 109 # 78eo

Loss tangent

-4 9 = 1.7 # 10 9# 9 # 10 = 1.3 # 10 - 5 3 # 10 # 39 9.85

Option (D) is correct. The flux density is s = eE = e0 er E = 80 # 8.854 # 10 - 12 # 2 s = 1.41 # 10 - 9 C/m 2

or 9.86

9.78

9.79

P \ 12 r P1 = r22 Thus P2 r12 3 dB decrease $ Strength is halved P1 = 2 Thus P2

Option (A) is correct. 8 u = c = 3 # 10 = 1.5 # 108 2 e0 In rectangular waveguide the dominant mode is TE10 and fC = v ` m j2 + ` n j2 2 a b 8 1 2 + 0 2 = 1.5 # 108 = 2.5 GHz = 1.5 # 10 ` 0.03 j ` b j 2 0.06

Option (D) is correct. Normalized array factor = 2 cos

y 2

Option (B) is correct.

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y = bd sin q cos f + d q = 90c,

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d =

Now

2 s, f = 45c, d = 180c bd sin q cos f + d y = 2 cos ; 2 cos E 2 2

= 2 cos 8 2p 2 s cos 45c + 180 B l. 2 2 = 2 cos 8 ps + 90cB = 2 sin ` ps j l l

9.80

Option (D) is correct. VSWR

Substituting values we have

or Distance to move 9.87

9.88

s = 1+G 1-G

where G varies from 0 to 1

9.82

Option (B) is correct. Reactance increases if we move along clockwise direction in the constant resistance circle.

9.89

Option (A) is correct. In wave guide vp > c > vg and in vacuum vp = c = vg where vp $ Phase velocity c $ Velocity of light vg $ Group velocity

f 2 1-c c m f

9.90

Option (C) is correct. We have

Option (B) is correct.

VC

When wave propagate in waveguide fc < f $ VP > VC 9.83

Option (C) is correct. A transmission line is distortion less if LG = RC d2 Ex = c2 d2 Ex dz2 dt2 This equation shows that x component of electric fields Ex is traveling in z direction because there is change in z direction.

Option (C) is correct. Phase velocity VP =

p 2

t j ) e j (wt - kz) E = (0.5xt + ye

9.91

Ex = 0.5e j (wt - kz) jp 2

Ey = e e

Option (A) is correct. In a wave guide dominant gives lowest cut-off frequency and hence the highest cut-off wavelength. Option (A) is correct. or or

j (wt - kz)

Ex p = 0.5e- 2 Ey Ex Since ! 1, it is elliptically polarized. Ey

= 7071 - 5000 = 2071 m

We have

Thus s varies from 1 to 3. 9.81

2 2 = r22 5 r2 = 5 2 kM = 7071 m

9.92

Ic = Id sE = jw d E s = 2pfeo er

-2 9 s = 2s = 9 # 10 # 2 # 10 2p # eo er 4peo er 4

or

f =

or

f = 45 # 106 = 45 MHz

Option (B) is correct.

w = 2pf and e = er e0

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

or or Now

VSWR = 1 + G 1-G 3 = 1+G 1-G

Page 236

9.99

G = 0.5 Pr = G2 = 0.25 Pi

Option (A) is correct. We have

Distance between elements

or

Because of end fire y =0 q = 60c 0 = 2p # l cos 60c + d = p # 1 + d 2 2 l 4 d =- p 4

9.96

9.97

ZOC .ZSC = 100 # 25 = 10 # 5 = 50W

BW \

1 (Diameter)

Option (C) is correct. The fig is as shown below :

As per snell law sin qt = 1 sin qi er sin 30c = 1 or sin 45c er 1 2 = 1 1 er 2

l1 = l2

e2 e1

Option (D) is correct. 2 l ` 2 jd = l

8 l = c = 3 # 10 9 = 3 m f 40 4 # 10

2 3 ` 40 # 2 j d = (2.4)

or 9.102

d =

80 # (2.4) 2 . 150 m 3

Option (C) is correct. We know that for a monopole its electric field varies inversely with r 2 while its potential varies inversely with r . Similarly for a dipole its electric field varies inversely as r 3 and potential varies inversely as r 2 . In the given expression both the a _ r1 + r1 i are present, so this potential is due to both monopole & dipole. -1

9.103

9.104

As diameter increases Bandwidth decreases. 9.98

9.101

Option (C) is correct. As the impedance of perfect conductor is zero, electric field is minimum and magnetic field is maximum at the boundary. Option (B) is correct.

l \ 1 e

we get

Option (B) is correct. Zo =

For air

b = 2p = w me l l = 2p w me

Phase Velocity

Thus

Thus

m 2 n 2 ` a j +`b j

Option (B) is correct.

Option (C) is correct. The array factor is

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9.95

9.100

or

y = bd cos q + d d =l 4

vp 2

= 15 GHz

l = 492 m

25 W.

where

fc =

For rectangular waveguide dominant mode is TE01 8 v Thus fc = p = 3 # 10- 2 = 15 # 109 2a 2 # 10 vp = 3 # 108

and height of antenna = 124 m . l 4 It is a quarter wave monopole antenna and radiation resistance is 9.94

Option (C) is correct. Cutoff frequency

Thus 25% of incident power is reflected. 9.93

er = 2

or

9.105

-2

Option (D) is correct. In TE mode Ez = 0 , at all points within the wave guide. It implies that electric field vector is always perpendicular to the waveguide axis. This is not possible in semi-infine parallel plate wave guide. Option (A) is correct. Option (C) is correct. A scalar wave equation must satisfy following relation

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 2 2 E - m 22 2 E = 0 ...(1) 2t 2 2z 2 Where m = w (Velocity) b Basically w is the multiply factor of t and b is multiply factor of z or x or y . In option (A) E = 50e j (wt - 3z) m =w=w 3 b We can see that equations in option (C) does not satisfy equation (1)

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 9.106

Option (B) is correct. We know that distance between two adjacent voltage maxima is equal to l/2 , where l is wavelength. l = 27.5 - 12.5 2 Frequency

9.107

9.108

9.110

Option (D) is correct. Power received by antenna -4 PR = PT 2 # (apeture) = 251 # 500 # 102 = 100 mW 4pr 4 # p # (100) Option (C) is correct. Electrical path length = bl Where b = 2p , l = 50 cm l We know that l =u =1# 1 au= f f LC 1 1 = 6 # -6 25 # 10 10 # 10 # 40 # 10-12

9.115

1 LC

Option (D) is correct. In a lossless dielectric (s = 0) median, impedance is given by mr m0 mr m = 120p # h = 0c = er e0 er e = 120p # 2 = 188.4 W 8

d\ 1 \ l f so depth increases with increasing in wavelength. 9.116

or

#l H $ dl

=

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Click to Buy www.nodia.co.in along a line that makes an angle f with x -axis as shown below. 9.117

(ampere's law)

9.118

9.119

Option (A) is correct.

9.113

Option (B) is correct.

9.114

Power Re ceived Polynting vector of incident wave A =W P =

r = a + ib = 0.1p + j0.2p b = 2p = 0.2p l

2 P = E h0 = 120p is intrinsic impedance h0

of space

Based on continuity equation

So

A -6

= 2 # 10 E2 c h0 m

Option (B) is correct. Propagation constant here

Option (C) is correct. We have

#s Jds

4# H = J then it is modified to 4# H = J + 2D 2t 9.112

Option (D) is correct. Radiation resistance of a circular loop is given as Rr = 8 hp3 :ND2 S D 3 l 2 Rx \ N N " no. of turns So, Rr 2 = N 2 # Rr 1 = (5) 2 # 0.01 = 0.25 W

Aperture Area

Applying curl theorem =

Option (C) is correct. v v Hv = 1 4 A m # v is auxiliary potential function. where A So 4: H = 4: (4 # A) = 0 4# H = 4# (4 # A) = Y 0

#s Jds

#s (4 # H) $ ds

...(1)

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Option (A) is correct. This equation is based on ampere’s law as we can see = I enclosed

Option (A) is correct. Given j (wt + bz) v a x + e0 e j (wt + bz) avy E (z, t) = Eo e

...(2) E (z) = avx E1 (z) + avy E2 (z) Comparing (1) and (2) we can see that E1 (z) and E2 (z) are in space quadrature but in time phase, their sum E will be linearly polarized

Option (D) is correct. Impedance is written as

#l H $ dl

Option (C) is correct. The depth of penetration or skin depth is defined as – 1 d= pfms

Generalizing

jwm h = s + jwe Copper is good conductor i.e. s >> we jwm wm So h = = 45c s s Impedance will be complex with an inductive component. 9.111

l = 2 = 10 m 0.2

l = 2 # 15 = 30 cm 10 u = C = 3 # 10 = 1 GHz 30 l

7 = 5 # 10 6 = 2 m 25 # 10 Electric path length = 2p # 50 # 10-2 = p radian 5 2 9.109

Page 237

=

-6

2 # 10 # 120 # 3.14 (20 # 10-3) 2 -6 3.14 = 1.884 m2 = 2 # 10 # 12 -# 400 # 10 6

9.120

Option (B) is correct. Maximum usable frequency

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

fo sin Ae fm = 8MHz = sin 60c

Page 238

fm =

9.121

9.122

8 = 16 MHz 3 3 c 2 m

or

Option (D) is correct. When a moving circuit is put in a time varying magnetic field educed emf have two components. One for time variation of B and other turn motion of circuit in B .

f 2 = 1000 MHz is 4pf 2 V = ms

Option (A) is correct.

V

Option (B) is correct. Z in min = Z 0 S where S = standing wave ratio 1 + GL S = 1 - GL

= 9.128

ZL = 0 (Short circuited) Z 0 = 50 W 0 + j50 tan p/2 =3 Zin = 50 = 50 + j0 tan p/2G

and

9.129

Option (A) is correct. The cutoff frequency is given by ml m 2+ n 2 fc = 2 a a k a2k Here a < b , so minimum cut off frequency will be for mode TE 01 a ml = c 2 *

c = 3 # 108

Option (B) is correct. For lossless transmission line, we have Velocity V =w= 1 b LC Characteristics impedance for a lossless transmission line Z0 = L C From eqn. (1) and (2) 1 = 1 V = C (Z 0 C ) Z 0 C

9.130

Option (C) is correct.

9.131

Option (A) is correct.

9.132

Option (C) is correct.

Option (B) is correct. Option (A) is correct. For any transmission line we can write input impedance Z + jZ 0 tanh lg Zin = Z 0 ; L Z 0 + jZL tanh lg E R jZ tanh lg V S1 + 0 W Z0 ZL W= so Zin = Z 0 lim S Z " 3S Z 0 j tanh lg W S ZL + j tanh lg W T X Option (A) is correct. We know that skin depth is given by 1 s = = 1 # 10-2 m pf1 ms

...(1)

...(2)

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 Reflected power

Here given ZL = 3 (open circuited at load end) So

L

9.127

Z 0 = Charateristic impedance of line ZL = Load impedance b = 2p l = length l bl = 2p l = p 2 l 4

Thus infinite impedance, and current will be zero.

m = 0, n = 1

9.126

Option (A) is correct. Input impedance of a lossless transmission line is given by Z + jZ 0 tan bl Zin = Z 0 ; L Z 0 + jZL tan bl E

so

1+1 3 =2 S = 1 13 50 Z in min = = 25 W 2

9.125

4 # p # 1000 # 10 # p - 6 106 m/ sec # 10-3

so here

GL = ZL - Z 0 = 100 - 50 = 50 = 1 ZL + Z 0 100 + 50 150 3

1 (10 # 10-12) 3 # 108 = 0.75 GHz = 2 # 2 # 10 # 10-2

6

where

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8 fc = 3 # 10 2#2

-3

ms = 10 in above equation p

Put

GL = reflection coefficient

9.124

f 1 = 10 MHz

Now phase velocity at another frequency

Far field \ 1 r 9.123

1 = 10-2 6 p # 10 # 10 # m s -3 ms = 10 p

or

9.133

Er = GEi G = Reflection coefficient h - h1 G = 2 = 1.5 - 1 = 1 h 2 + h1 1.5 + 1 5 Er = 1 # Ei 5 Er = 20% Ei

Ei " Incident power

Option (B) is correct. We have maximum usable frequency formulae as f0 fm = sin Ae

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

20 # 106 = 10 # 10 sin Ae sin Ae = 1 2

Page 239

6

Ae = 30c 9.134

Option (C) is correct.

9.135

Option (A) is correct. Skin depth

d=

1 pfms

Putting the given value d= = 15.9 mm

1 3.14 # 1 # 109 # 4p # 10-7 # 106

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

UNIT 10

(C) 10.8

GENERAL APTITUDE

YEAR 2013 10.1

10.2

Page 240

ONE MARK

Choose the grammatically CORRECT sentence: (A) Two and two add four (B) Two and two become four (C) Two and two are four (D) Two and two make four Statement: You can always give me a ring whenever you need. Which one of the following is the best inference from the above statement? (A) Because I have a nice caller tune. (B) Because I have a better telephone facility (C) Because a friend in need is a friend indeed

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10.9

10.10

10.4

10.5

10.1

In the summer of 2012, in New Delhi, the mean temperature of Monday to Wednesday was 41°C and of Tuesday to Thursday was 43cC . If the temperature on Thursday was 15% higher than that of Monday, then the temperature in cC on Thursday was (A) 40 (B) 43 (C) 46 (D) 49

10.11

They were requested not to quarrel with others. Which one of the following options is the closest in meaning to the word quarrel? (A) make out (B) call out (C) dig out (D) fall out

10.7

10.14

Find the sum to n of the series 10 + 84 + 734 + ... 9 ^9n - 1h 9 ^9n + 1h (B) (A) +1 +1 8 10

Statement: There were different streams of freedom movements in colonial India carried out by the moderates, liberals, radicals, socialists, and so on. Which one of the following is the best inference from the above statement? (A) The emergence of nationalism in colonial India led to our Independence (B) Nationalism in India emerged in the context of colonialism (C) Nationalism in India is homogeneous (D) Nationalism in India is heterogeneous The set of values of p for which the roots of the equation 3x2 + 2x + p ^p - 1h = 0 are of opposite sign is (A) ^- 3, 0h (B) ^0, 1h (C) ^1, 3h (D) ^0, 3h What is the chance that a leap year, selected at random, will contain 53 Sundays? (A) 2/7 (B) 3/7 (C) 1/7 (D) 5/7 ONE MARK

If (1.001) 1259 = 3.52 and (1.001) 2062 = 7.85, then (1.001) 3321 (B) 4.33 (A) 2.23 (C) 11.37 (D) 27.64 Choose the most appropriate alternate from the options given below to complete the following sentence : If the tired soldier wanted to lie down, he..................the mattress out on the balcony. (A) should take (B) shall take (C) should have taken (D) will have taken Choose the most appropriate word from the options given below to complete the following sentence : Give the seriousness of the situation that he had to face, his........ was impressive. (A) beggary (B) nomenclature (C) jealousy (D) nonchalance Which one of the following options is the closest in meaning to the

word given below ? Latitude (A) Eligibility (C) Coercion

TWO MARKS

A car travels 8 km in the first quarter of an hour, 6 km in the second quarter and 16 km in the third quarter. The average speed of the car in km per hour over the entire journey is (A) 30 (B) 36 (C) 40 (D) 24

9 ^9n - 1h + n2 8

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Option (D) is correct. They were requested not to quarrel with others. Quarrel has a similar meaning to ‘fall out’ YEAR 2013

10.6

10.12

10.13

Complete the sentence: Dare .................. mistakes. (A) commit (B) to commit (C) committed (D) committing

(D)

2012

(D) Because you need not pay towards the telephone bills when you give me a ring 10.3

9 ^9n - 1h +n 8

10.15

(B) Freedom (D) Meticulousness

One of the parts (A, B, C, D) in the sentence given below contains an ERROR. Which one of the following is INCORRECT ? I requested that he should be given the driving test today instead of tomorrow. (A) requested that (B) should be given (C) the driving test (D) instead of tomorrow

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

2012 10.16

10.17

10.18

10.19

TWO MARKS

One of the legacies of the Roman legions was discipline. In the legious, military law prevailed and discipline was brutal. Discipline on the battlefield kept units obedient, intact and fighting, even when the odds and conditions were against them. Which one of the following statements best sums up the meaning of the above age ? (A) Through regimentation was the main reason for the efficiency of the Roman legions even in adverse circumstances. (B) The legions were treated inhumanly as if the men were animals (C) Disciplines was the armies inheritance from their seniors (D) The harsh discipline to which the legions were subjected to led to the odds and conditions being against them. Raju has 14 currency notes in his pocket consisting of only Rs. 20 notes and Rs. 10 notes. The total money values of the notes is Rs. 230. The number of Rs. 10 notes that Raju has is (A) 5 (B) 6 (C) 9 (D) 10 There are eight bags of rice looking alike, seven of which have equal weight and one is slightly heavier. The weighing balance is of unlimited capacity. Using this balance, the minimum number of weighings required to identify the heavier bag is (A) 2 (B) 3 (C) 4 (D) 8

Category

Amount (Rs.)

Food

4000

Clothing

1200

Rent

2000

Savings

1500

Other Expenses

1800

10.23

10.24

The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relations in the original pair : Gladiator : Arena (A) dancer : stage (B) commuter : train (C) teacher : classroom (D) lawyer : courtroom Choose the most appropriate word from the options given below to complete the following sentence : Under ethical guidelines recently adopted by the Indian Medical Association, human genes are to be manipulated only to correct diseases for which...................treatments are unsatisfactory. (A) similar (B) most (C) uncommon (D) available Choose the word from the from the options given below that is most opposite in meaning to the given word : Frequency (A) periodicity (B) rarity

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Click to Buy www.nodia.co.in (C) gradualness 10.25

(D) persistency

Choose the most appropriate word from the options given below to complete the following sentence : It was her view that the country’s had been ............. by foreign techno-crafts, so that to invite them to come back would be counter-productive. (A) identified (B) ascertained (C) exacerbated (D) analysed 2011

10.26

TWO MARKS

The fuel consumed by a motor cycle during a journey while travelling at various speed is indicated in the graph below.

A and B are friends. They decide to meet between 1 PM and 2 PM on a given day. There is a conditions that whoever arrives first will not wait for the other for more than 15 minutes. The probability that they will meet on that days is (A) 1/4 (B) 1/16 (C) 7/16 (D) 9/16 2011

10.21

10.22

The data given in the following table summarizes the monthly budget of an average household.

The approximate percentages of the monthly budget NOT spent on savings is (A) 10% (B) 14% (C) 81% (D) 86% 10.20

Page 241

ONE MARK

There are two candidates P and Q in an election. During the campaign, 40% of voter promised to vote for P , and rest for Q . However, on the day of election 15% of the voters went back on their promise to vote for P and instead voted for Q . 25% of the voter went back on their promise to vote for Q and instead voted for P . Suppose, P lost by 2 votes, then what was the total number of voters ? (A) 100 (B) 110 (C) 90 (D) 95

The distance covered during four laps of the journey are listed in the table below Lap

Distance (km)

Average speed (km/hour)

P

15

15

Q

75

45

R

40

75

S 10 10 From the given data, we can conclude that the fuel consumed per kilometre was least during the lap

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) P (C) R 10.27

10.28

10.29

(B) Q (D) S

Page 242 10.34

The horse has played a little known but very important role in the field of medicine. Horses were injected with toxins of disease until their blood build up immunities. Then a serum was made from their blood. Serums to fight with diphteria and tetanus were developed this way. It can be inferred from the age, that horses were (A) given immunity to diseases (B) generally quite immune to diseases (C) given medicines to fight toxins (D) given diphtheria and tetanus serums The sum of n of the series 4 + 44 + 444 + ........ (A) (4/81) [10n + 1 - 9n - 1] (B) (4/81) [10n - 1 - 9n - 1] (C) (4/81) [10n + 1 - 9n - 10] (D) (4/81) [10n - 9n - 10]

10.35

10.36

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in

10.30

10.37

(B) - 1 (D) 2

Three friends R, S and T shared toffee from a bowl. R took 1/3 rd of the toffees, but returned four to the bowl. S took 1/4 th of what was left but returned three toffees to the bowl. T took half of the remainder but returned two back into the bowl. If the bowl had 17 toffees left, how many toffees were originally there in the bowl ? (A) 38 (B) 31 (C) 48 (D) 41

10.38

10.39

2010 10.31

10.32

10.33

ONE MARK

Which of the following options is the closest in meaning to the word below ? Circuitous (A) Cyclic (B) Indirect (C) Confusing (D) Crooked The question below consist of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair. Unemployed : Worker (A) Fallow : Land (B) Unaware : Sleeper (C) Wit : Jester (D) Renovated : House Choose the most appropriate word from the options given below to complete the following sentence : If we manage to ........ our natural resources, we would leave a better planet for our children. (A) unhold (B) restrain (C) cherish (D) conserve

25 persons are in a room 15 of them play hockey, 17 of them play football and 10 of them play hockey and football. Then the number of persons playing neither hockey nor football is (A) 2 (B) 17 (C) 13 (D) 3 2010

Given that f (y) = y /y, and q is any non-zero real number, the value of f (q) - f (- q) is

(A) 0 (C) 1

Choose the most appropriate word from the options given below to complete the following sentence : His rather casual remarks on politics..................his lack of seriousness about the subject. (A) masked (B) belied (C) betrayed (D) suppressed

10.40

TWO MARKS

Modern warfare has changed from large scale clashes of armies to suppression of civilian populations. Chemical agents that do their work silently appear to be suited to such warfare ; and regretfully, their exist people in military establishments who think that chemical agents are useful fools for their cause. Which of the following statements best sums up the meaning of the above age ? (A) Modern warfare has resulted in civil strife. (B) Chemical agents are useful in modern warfare. (C) Use of chemical agents in ware fare would be undesirable. (D) People in military establishments like to use chemical agents in war. If 137 + 276 = 435 how much is 731 + 672 ? (A) 534 (B) 1403 (C) 1623 (D) 1531 5 skilled workers can build a wall in 20 days; 8 semi-skilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall ? (A) 20 days (B) 18 days (C) 16 days (D) 15 days Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4 how much distinct 4 digit numbers greater than 3000 can be formed ? (A) 50 (B) 51 (C) 52 (D) 54 Hari (H), Gita (G), Irfan (I) and Saira (S) are siblings (i.e. brothers

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 and sisters.) All were born on 1st January. The age difference between any two successive siblings (that is born one after another) is less than 3 years. Given the following facts : 1. Hari’s age + Gita’s age > Irfan’s age + Saira’s age. 2. The age difference between Gita and Saira is 1 year. However, Gita is not the oldest and Saira is not the youngest. 3. There are no twins. In what order were they born (oldest first) ? (A) HSIG (B) SGHI (C) IGSH (D) IHSG ***********

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 243

SOLUTIONS 10.1

10.2

10.3

S2 =

10.8

Option (D) is correct Two and two make four Option (C) is correct. You can always given me a ring whenever you need. Because a friend is need is a friend indeed Option (C) is correct. Let the temperature on Monday, Tuesday, Wednesday and Thursday be respectively as TM , TTU , TW , TTH So, from the given data we have TH + TTU + TW = 41 ....(1) 3 TTU + TW + TTH = 43 ....(2) and 3 also, as the temperature on Thursday was 15% higher than that of Monday ....(3) i.e. TTH = 1.15 TM solving eq (1), (2) and (3), we obtain

10.9

S3 9 ^93 - 1h = + 32 = 828 8 Option (D) is correct. Nationalism in India is heterogeneous Option (B) is correct. Given, the quadratic equation 3x2 + 2x + P ^P - 1h = 0 It will have the roots with opposite sign if

P ^P - 1h < 0 So it can be possible only when P < 0 and P - 1 > 0 or P > 0 and P - 1 < 0 The 1 st condition tends to no solution for P .

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TTH = 46cC 10.4

10.5

10.6

Option (B) is correct. Dare to commit mistakes

Hence, from the second condition, we obtain

Option (D) is correct. They were requested not to quarrel with others. Quarrel has a similar meaning to ‘fall out’ Option (C) is correct. Given, the distance travelled by the car in each quarter intervals as Distance

0

<1 i.e., P is in the range ^0, 1h 10.10

Time Duration

8 km

1 4

hr

6 km

1 4

hr

16 km

1 4

hr

Therefore, the total time taken = 1 + 1 + 1 + 3 hr 4 4 4 4 Hence,

10.7

Total distance travelled = 8 + 6 + 16 = 30 km average speed = Total distance travelled Total time taken = 30 = 40 km/hr 3/4

10.11

Option (D) is correct. It will be easy to check the options for given series. From the given series.

Option (A) is correct. In a leap year, there are 366 days So, 52 weeks will have 52 saturdays and for remaining two days ^366 - 52 # 7 = 2h. We can have the following combinations Saturday, Sunday Sunday, Monday Monday, Tuesday Tuesday, Wednesday Wednesday, Thursday Thursday, Friday Friday, Saturday Out of these seven possibilities, only two consist a saturday. Therefore, the probability of saturday is given as P =2 7 Option (D) is correct. Let 1.001 = x So in given data :

10 + 84 + 734 + ...... We get Sum of 1 term = S1 = 10 Sum of 2 = S2 = 10 + 84 = 94 and sum of 3 = S 3 = 10 + 84 + 734 = 828 Checking all the options one by one, we observe that only (D) option satisfies as 9 ^9n - 1h + n2 Sn = 8 9 ^92 - 1h so, S1 + 22 = 10 8

9 ^9 - 1h + 22 = 94 8

Again

10.12

Option (C) is correct.

10.13

Option (D) is correct.

10.14

Option (B) is correct.

x1259 = 3.52 x2062 = 7.85 x3321 = x1259 + 2062 = x1259 x2062 = 3.52 # 7.85 = 27.64

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 10.15

Option (B) is correct.

10.16

Option (A) is correct.

10.17

Option (A) is correct. Let no. of notes of Rs.20 be x and no. of notes of Rs. 10 be y . Then from the given data.

Page 244

= 9000 # 100% = 86% 10500 10.20

Option (S) is correct. The graphical representation of their arriving time so that they met is given as below in the figure by shaded region.

x + y = 14 20x + 10y = 230 Solving the above two equations we get x = 9, y = 5 So, the no. of notes of Rs. 10 is 5. 10.18

Option (A) is correct. We will categorize the 8 bags in three groups as : (i) A1 A2 A 3 , (ii) B1 B2 B 3 , (iii) C1 C2 Weighting will be done as bellow : 1st weighting " A1 A2 A 3 will be on one side of balance and B1 B2 B 3 on the other. It may have three results as described in the following cases. Case 1 : A1 A 2 A 3 = B1 B 2 B 3

So, the area of shaded region is given by Area of 4PQRS - (Area of TEFQ + Area of TGSH ) = 60 # 60 - 2 b 1 # 45 # 45 l 2

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in This results out that either C1 or C2 will heavier for which we will have to perform weighting again. 2 nd weighting " C1 is kept on the one side and C2 on the other. if then C1 is heavier. C1 > C 2 then C2 is heavier. C1 < C 2 Case 2 : A1 A 2 A 3 > B1 B 2 B 3 it means one of the A1 A2 A 3 will be heavier So we will perform next weighting as: 2 nd weighting " A1 is kept on one side of the balance and A2 on the other. it means A 3 will be heavier if A1 = A 2 then A1 will be heavier A1 > A 2 then A2 will be heavier A1 < A 2 Case 3 : A1 A 2 A 3 < B 1 B 2 B 3 This time one of the B1 B2 B 3 will be heavier, So again as the above case weighting will be done. 2 nd weighting " B1 is kept one side and B2 on the other if B 3 will be heavier B1 = B 2 B1 > B 2 B1 will be heavier B1 < B 2 B2 will be heavier So, as described above, in all the three cases weighting is done only two times to give out the result so minimum no. of weighting required = 2. 10.19

Option (D) is correct. Total budget = 4000 + 1200 + 2000 + 1500 + 1800

= 1575

10.21

Now Voter for P 40 - 6 = 34 Also, 25% changed form Q to P (out of 60%) 25 Changed voter from Q to P 60 = 15 100 # Now Voter for P 34 + 15 = 49 Thus P P got 49 votes and Q got 51 votes, and P lost by 2 votes, which is given. Therefore 100 voter is true value. 10.22

Option (A) is correct. A gladiator performs in an arena. Commutators use trains. Lawyers

For more GATE Resources, Mock Test and Study material the community http://www.facebook.com/gateec2014 performs, but do not entertain like a gladiator. Similarly, teachers educate. Only dancers performs on a stage. 10.23

10.24

= 10, 500 The amount spent on saving = 1500 So, the amount not spent on saving = 10, 500 - 1500 = 9000 So, percentage of the amount

So, the required probability = 1575 = 7 3600 16 Option (A) is correct. Let us assume total voters are 100. Thus 40 voter (i.e. 40 %) promised to vote for P and 60 (rest 60 % ) promised to vote fore Q. Now, 15% changed from P to Q (15 % out of 40) 15 Changed voter from P to Q 40 = 6 100 #

10.25

Option (D) is correct. Available is appropriate because manipulation of genes will be done when other treatments are not useful. Option (B) is correct. Periodicity is almost similar to frequency. Gradualness means something happening with time. Persistency is endurance. Rarity is opposite to frequency. Option (C) is correct. The sentence implies that technocrats are counterproductive

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(negative). Only (C) can bring the same meaning. 10.26

10.27

10.28

Option (B) is correct. Since fuel consumption/litre is asked and not total fuel consumed, only average speed is relevant. Maximum efficiency comes at 45 km/ hr, So least fuel consumer per litre in lap Q

Page 245 10.33

10.34

10.35

Option (B) is correct. Option B fits the sentence, as they built up immunities which helped humans create serums from their blood. Option (C) is correct. 4 + 44 + 444 + .............. 4 (1 + 11 + 111 + .......) = 4 (9 + 99 + 999 + ............) 9

10.37

= 4 [10 (1 + 10 + 102 + 103) - n] 9 n = 4 :10 # 10 - 1 - nD 9 10 - 1 = 4 610n + 1 - 10 - 9n@ 81 10.29

Option (D) is correct. y y -y f (- y) = =- f (y) y f (q) - f (- q) = 2f (q) = 2

Option (C) is correct. Betrayed means reveal unintentionally that is most appropriate. Option (D) is correct. Number of people who play hockey n (A) = 15 Number of people who play football n (B) = 17 Persons who play both hockey and football n (A + B) = 10 Persons who play either hockey or football or both : n (A , B) = n (A) + n (B) - n (A + B) = 15 + 17 - 10 = 22 Thus people who play neither hockey nor football = 25 - 22 = 3

10.36

= 4 [(10 - 1) + (100 - 1) + ........] 9

Option (D) is correct. Here conserve is most appropriate word.

Option (D) is correct. Option (C) is correct. Since 7 + 6 = 13 but unit digit is 5 so base may be 8 as 5 is the remainder when 13 is divided by 8. Let us check.

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f (y) =

Now or 10.30

Option (C) is correct. Let total no of toffees be x . The following table shows the all calculations. Friend

Bowl Status

R

= x -4 3

= 2x + 4 3

S

= 1 :2x + 4D - 3 4 3 = x +1-3 = x -2 6 6

= 2x + 4 - x + 2 3 6 = x +6 2

T

= 1 a x + 6k - 2 2 2 = x +1 4

= x +6-x -1 2 4 = x +5 4

Now, or

x + 5 = 17 4 x = 17 - 5 = 12 4 x = 12 # 4 = 48

10.31

10.32

Option (B) is correct. Circuitous means round about or not direct. Indirect is closest in meaning to this circuitous (A) Cyclic : Recurring in nature (B) Indirect : Not direct (C) Confusing : lacking clarity of meaning (D) Crooked : set at an angle; not straight Option (B) is correct. A worker may by unemployed. Like in same relation a sleeper may be unaware.

Click to Buy www.nodia.co.in 137 8 276 8 435 10.38

Thus here base is 8. Now

731 8 672 8 1623

Option (D) is correct. Let W be the total work. =W 20 Per day work of one skill worker = W =W 5 # 20 100 Similarly per day work of 1 semi-skilled workers = W = W 8 # 25 200 Similarly per day work of one semi-skill worker = W = W 10 # 30 300 Thus total per day work of 2 skilled, 6 semi-skilled and 5 unskilled workers is = 2W + 6W + 5W = 12W + 18W + 10W = W 100 200 300 600 15 Therefore time to complete the work is 15 days. Per day work of 5 skilled workers

10.39

Option (B) is correct. As the number must be greater than 3000, it must be start with 3 or 4. Thus we have two case: Case (1) If left most digit is 3 an other three digits are any of 2, 2, 3, 3, 4, 4, 4, 4. (1) Using 2, 2, 3 we have 3223, 3232, 3322 i.e. 3! = 3 no. 2! (2) Using 2, 2, 4 we have 3224, 3242, 3422 i.e. 3! = 3 no. 2! (3) Using 2, 3, 3 we have 3233, 3323, 3332 i.e. 3! = 3 no. 2! (4) Using 2, 3, 4 we have 3! = 6 no. (5) Using 2, 4, 4 we have 3244, 3424, 3442 i.e. 3! = 3 no. 2! (6) Using 3, 3, 4 we have 3334, 3343, 3433 i.e. 3! = 3 no. 2!

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 246

(7) Using 3, 4, 4 we have 3344, 3434, 3443 i.e. 3! = 3 no. 2! (8) Using 4, 4, 4 we have 3444 i.e. 3! = 1 no. 3! Total 4 digit numbers in this case is 1 + 3 + 3 + 3 + 6 + 3 + 3 + 3 + 1 = 25 Case 2 : If left most is 4 and other three digits are any of 2, 2, 3, 3, 3, 4, 4, 4. (1) Using 2, 2, 3 we have 4223, 4232, 4322 i.e. . 3! = 3 no 2! (2) Using 2, 2, 4 we have 4224, 4242, 4422 i.e. . 3! = 3 no 2! (3) Using 2, 3, 3 we have 4233, 4323, 4332 i.e. . 3! = 3 no 2! (4) Using 2, 3, 4 we have i.e. . 3! = 6 no

(5) Using 2, 4, 4 we have 4244, 4424, 4442 i.e. . 3! = 3 no 2! (6) Using 3, 3, 3 we have 4333 i.e 3! = 1. no. 3! (7) Using 3, 3, 4 we have 4334, 4343, 4433 i.e. . 3! = 3 no 2!

GATE Electronics & Communication by RK Kanodia Now in 3 Volume Purchase Online at maximum discount from online store and get POSTAL and Online Test Series Free visit www.nodia.co.in (8) Using 3, 4, 4 we have 4344, 4434, 4443 i.e. . 3! = 3 no 2! (9) Using 4, 4, 4 we have 4444 i.e. 3! = 1. no 3! Total 4 digit numbers in 2nd case = 3 + 3 + 3 + 6 + 3 + 3 + 1 + 3 + 1 = 26 Thus total 4 digit numbers using case (1) and case (2) is = 25 + 26 = 51 10.40

Option (B) is correct. Let H , G , S and I be ages of Hari, Gita, Saira and Irfan respectively. Now from statement (1) we have H + G > I + S Form statement (2) we get that G - S = 1 or S - G = 1 As G can’t be oldest and S can’t be youngest thus either GS or SG possible. From statement (3) we get that there are no twins (A) HSIG : There is I between S and G which is not possible (B) SGHI : SG order is also here and S > G > H > I and G + H > S + I which is possible. (C) IGSH : This gives I > G and S > H and adding these both inequalities we have I + S > H + G which is not possible. (D) IHSG : This gives I > H and S > G and adding these both inequalities we have I + S > H + G which is not possible.

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