Gate By Rk Kanodia Signals And Systems 5u454

GATE EC BY RK Kanodia

UNIT 5

ì e2 t, t < 0 30. y( t) = u( t) * h( t) , where h( t) = í -3t îe , t >0 1 5 1 (A) e -2 t u( - t - 1) + - e -3t u( -t) 2 6 3

Signal & System

(A)

5 3 1 13 -4 t sin t + cos t + e - t e , t ³ 0 34 34 6 61

(B)

5 3 13 -4 t 1 - t sin t + cos t e + e , t ³ 0 34 34 51 6

(B)

1 2t 5 1 e u( -t - 1) + - e-3t u( -t) 2 6 3

(C)

3 5 13 -4 t 1 - t sin t + cos t e + e , t ³ 0 34 34 51 6

(C)

1 2t 1 e + [5 - 3e 2 t - 2 e -3t ]u( t) 2 6

(D)

3 5 1 13 -4 t sin t + cos t + e -4 t e , t ³ 0 34 34 6 51

(D)

1 2t 1 e + [5 - 3e 2 t - 2 e -3t ]u( -t) 2 6

37.

d 2 y ( t) dy ( t) +6 + 8 y( t) = 2 x( t), dt 2 dt

Statement for Q.31-34:

y (0 - ) = -1,

The impulse response of LTI system is given. Determine the step response.

(A)

2 - t 5 -2 t 5 -4 t e - e + e , t ³0 3 2 6

31. h( t) = e - |t |

(B)

2 5 -2 t 5 -4 t + e + e , t ³ 0 3 2 6

(A) 2 + e t - e - t

(B) e t u( -t + 1) + 2 - e - t

(C) e t u( -t + 1) + [2 - e - t ]u( t)

(D) e t + [2 - e - t - e t ]u( t)

(C) 4 + 5( 3e -2 t + e -4 t ) , t ³ 0 (D) 4 - 5( 3e -2 t + e -4 t ), t ³ 0

32. h( t) = d( 2 ) ( t) (A) 1

(B) u( t)

(C) d( 3) ( t)

(D) d( t)

38.

d 2 y( t) 3dx( t) , + y( t) = dt 2 dt y (0 - ) = -1,

33. h( t) = u( t) - u( t - 4) (A) tu( t) + (1 - t) u( t - 4)

(B) tu( t) + (1 - t) u( t - 4)

(C) 1 + t

(D) (1 + t) u( t)

(A) sin t + 4 cos t - 3te -3t + t, t ³ 0 (B) 4 sin t - cos t - 3te - t , t ³ 0 (D) 4 sin t + cos t - 3te - t , t ³ 0

(A) u( t)

(B) t

(C) 1

(D) tu( t)

39. The raised cosine pulse x( t) is defined as ì 1 ï (cos wt + 1) , x ( t) = í 2 ïî 0,

Statement for Q.35-38: The system described by the differential equations has been specified with initial condition. Determine the output of the system and choose correct option. dy( t) + 10 y( t) = 2 x( t), y(0 - ) = 1, x( t) = u( t) dx

(A) 15 (1 + 4 e -10 t ) u( t)

(B) 15 (1 + 4 e -10 t )

(C) - 15 (1 + 4 e -10 t ) u( t)

(D) - 15 (1 + 4 e -10 t )

-

p p £t£ w w otherwise

The total energy of x ( t) is 3p 3p (B) (A) 4w 8w (C)

3p w

(D)

3p 2w

40. The sinusoidal signal x( t) = 4 cos (200 t + p 6) is

d 2 y( t) dy( t) dx( t) , 36. +5 + 4 y( t) = 2 dt dt dt dy( t) = 1, x( t) = sin t u( t) y (0 - ) = 0, dt 0 -

Page 252

dy( t) = 1, x( t) = 2 te- t u( t) dt 0 -

(C) sin t - 4 cos t + 3te -3t + t, t ³ 0

34. h( t) = y( t)

35.

dy( t) = 1, x( t) = e - t u( t) dt 0 -

ed through a square law device defined by the input output relation y ( t) = x 2 ( t). The DC component in the signal is (A) 3.46

(B) 4

(C) 2.83

(D) 8

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Continuous-Time Systems

41. The impulse response of a system is h( t) = d( t - 0.5). If two such systems are cascaded, the impulse response

Chap 5.1

46. The y( t) = x( t) * h( t) is y(t)

y(t)

of the overall system will be (A) 0.58( t - 0.25)

(B) d( t - 0.25)

(C) d( t - 1)

(D) 0.5 d( t - 1)

a

t

1+a

a

t

y(t)

h(t)

a 1

t

5

3

t

1+a 2

1+a

t

a

1-a

(C)

Fig P5.1.42

The output of the system is zero every where except for the

value of a is (A) 1

(B) 2

(C) 1 < t < 5

(D) 1 < t < 8

(C) 3

(D) 0

S1 : h1 ( t) = e - (1 - 2 j ) t u( t)

48. Consider the signal x( t) = d( t + 2) - d( t - 2).The value of E¥ for the signal y( t) =

S2 : h2 ( t) = e cos 2 t u( t) The stable system is (A) S1

(B) S2

(C) Both S1 and S2

(D) None

ò x( t) dt

is

(A) 4

(B) 2

(C) 1

(D) ¥

49. The

response of a system S to a complex input

x( t) = e j 5t is specified as y( t) = te j 5t . The system

44. The non-invertible system is (B) y( t) =

t

ò

(A) is definitely LTI x( t) dt

(B) is definitely not LTI

-¥

dx( t) dt

t

-¥

-t

(A) y( t) = x( t - 4)

t

47. If dy( t) dt contains only three discontinuities, the

(B) 0 < t < 8

43. Consider the impulse response of two LTI system

1

(D)

(A) 0 < t < 5

(C) y( t) =

1

(B)

y(t)

impulse response h( t) of the system.

1-a

(A)

42. Fig. P5.1.40 show the input x( t) to a LTI system and x(t)

1

(D) None of the above

(C) may be LTI (D) information is insufficient

45. A continuous-time linear system with input x( t) and output y( t) yields the following input-output pairs: x( t) = e j 2 t Û y( t) = e j 5t

(B) is definitely not LTI

If x1 ( t) = cos (2 t - 1), the corresponding y1 ( t) is (B) e- j cos (5 t - 1) (D) e j cos (5 t - 1)

Suppose that 0 £ t £1 elsewhere

(C) may be LTI (D) information is insufficient. 51. The auto-correlation of the signal x( t) = e - t u( t) is

Statement for Q.46–47:

ì 1, x( t) = í î 0,

response of a system S to a complex input

x( t) = e j8 t is specified as y( t) = cos 8 t. The system (A) is definitely LTI

x( t) = e - j 2 t Û y( t) = e - j 5t (A) cos (5 t - 1) (C) cos 5( t - 1)

50. The

and

ætö h( t) = xç ÷, where 0 < a £ 1. èaø

(A)

1 t 1 e u( -t) + e - t u( t) 2 2

(B)

et 1 - t + ( e - e t ) u( t) 2 2

(C)

1 -t 1 e u( -t) + e - t u( t) 2 2

(D)

1 t 1 e u( -t) - e - t u( t) 2 2

*********************

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GATE EC BY RK Kanodia

UNIT 5

Signal & System

11. (C)

SOLUTIONS

x(10t)

x(10t-5)

1

1. (A)

2p = 60 p T

2. (C) T1 =

Þ

T=

p 30

1

-0.5 -0.4

0.4 0.5

t

0.1

0.9

1

t

Fig S5.1.11

12. (D) Multiplication by 5 will bring contraction on

2p 2p æ 2p 2p ö s, T2 = s, LCM ç , ÷ = 2p 5 7 7 ø è 5

time scale. It may be checked by x(5 ´ 0.8) = x( 4).

3. (D) Not periodic because of t. 4. (D) Not periodic because least common multiple is

13. (A) Division by 5 will bring expansion on time scale. æ 20 ö It may be checked by y( t) = xç ÷ = x( 4). è 5 ø

infinite. 5. (C) y( t) is not periodic although sin t and 6 cos 2 pt are independently periodic. The fundamental frequency can’t be determined.

¥

ò | x ( t)|dt < ¥

=

-¥

¥

¥

-¥

0

-4 t -4 t ò e u( t) dt = ò e dt =

¥

7. (A) | x( t)| = 1, E¥ =

ò | x( t)| dt 2

1 4

1 T® ¥ 2T

T

5

-5

4

otherwise

=8+ =¥

-¥

ò | x( t)| dt 2

5

4

5

0

0

4

15. (D) E = 2 ò x 2 ( t) dt = 2 ò (1)1 dt + 2 ò (5 - t) 2 dt 2 26 = 3 3

16. (B) Let x1 ( t) = v( t) then y1 ( t) = u{v( t)}

So this is a power signal not a energy. P¥ = lim

-4

for - 5 < t < - 4 for 4 < t <5

E = ò (1) 2 dt + ò ( -1) 2 dt = 2

6. (C) This is energy signal because E¥ =

ì 1, ï 14. (C) y( t) = í -1, ï 0, î

Let x2 ( t) = kv( t) then y2 ( t) = u{kv( t)} ¹ ky1 ( t)

=1

(Not homogeneous not linear)

-T

8. (D) v( t) is sum of 3 unit step signal starting from, 1, 2, and 3, all signal ends at 4.

y1 ( t) = u{v( t)},

y2 ( t) = u{v( t - to)} = y1 ( t - to)

(Time invariant)

The response at any time depends only on the

9. (A) The function 1 does not describe the given pulse.

excitation at time t = to and not on any future value. (Causal)

It can be shown as follows : u(a-t)

u(t-b)

17. (C) y1 ( t) = v( t - 5) - v( 3 - t)

u(a-t) - u(t-b)

y2 ( t) = kv( t - 5) - kv( 3 - t) = ky1 ( t) t

a

t

b

t

(Homogeneous)

Let x1 ( t) = v( t) then y1 ( t) = v( t - 5) - v( 3 - t) Let x2 ( t) = 2 w( t) then y2 ( t) = w( t - 5) - w( 3 - t)

Fig S5.1.3.9

Let x3( t) = x( t) + w( t) 10. (B)

Then

r(t-4)

r(t-6)

r(t-4) - r(t-6)

2

2

2

y3( t) = v( t - 5) + w( t - 5) - v( 3 - t) - w( 3 - t)

= y1 ( t) + y2 ( t)

(Additive)

Since it is both homogeneous and additive, it is also linear.

4

6

8

t

4

6

8

Fig S5.1.10

Page 254

t

4

6

8

t

y1 ( t) = v( t - 5) - v( 3 - t) y2 ( t) = v( t - to - 5) - v( 3 - t + to) = y1 ( t - to)

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(Time invariant)

GATE EC BY RK Kanodia

Continuous-Time Systems

At

time,

t = 0, y(0) = x( -5) - x( 3).

Therefore

the

Chap 5.1

21. (C) All option are linear. So it is not required

response at time, t = 0 depends on the excitation at a

to check linearity.

later time t = 3.

d y1 ( t) - 8 y1 ( t) = v( t) dt d Let x2 ( t) = v( t - to) then t y2 ( t) - 8 y2 ( t) = v( t - to) dt

(Not causal)

If x( t) is bounded then x( t - 5) and x( 3 - t) are bounded and so is y( t).

(Stable)

ætö ætö 18. (D) y1 ( t) = vç ÷ , y2 ( t) = kvç ÷ = ky1 ( t) è2 ø è2 ø (Homogeneous)

Let x1 ( t) = v( t) then t

The first equation can be written as d ( t - to) y( t - to) - 8 y( t - to) = v( t - to) dt This equation is not satisfied if y2 ( t) = y1 ( t - to) therefore

x3 = v( t) + w( t) then ætö ætö y3( t) = vç ÷ + wç ÷ = y1 ( y) + y2 ( t) è2 ø è2 ø

(Additive)

Since it is both homogeneous and additive, it is also

y2 ( t) ¹ y1 ( t - to)

(Time Variant)

The system can be written as y( t) =

linear

x( l) y( l) dl + 8 ò dl l -¥ -¥ l t

t

ò

So the response at any time, t = to depends on the

ætö æt ö æ t - to ö y1 ( t) = vç ÷ , y2 ç - to ÷ ¹ y( t - to) = vç ÷ 2 2 è ø è ø è 2 ø

excitation at t £ to , and not on any future values.

(Time variant)

(Causal)

At time t = -2, y( -2) = x( -1), therefore, the response at

The Homogeneous solution to the differential equation

time t = -2, depends on the excitation at a later time,

is of the form y( t) = kt8 . If there is no excitation but the

t = -1.

zero excitation, response is not zero. The response will

(Not causal)

It x( t) is bounded then y( t) is bounded.

(Stable)

increases without bound as time increases. (Unstable)

19. (C) y1 ( t) = cos 2 pt v( t) y2 ( t) k cos 2 pt v( t) = ky1 ( t)

(Homogeneous)

22. (C) y1 ( t) =

x3( t) = v( t) + w( t)

t+ 3

ò v(l)dl

-¥

y3( t) = cos 2 pt [ v( t) + w( t)] = y1 ( t) + y2 ( t)

(Additive)

t+ 3

t+ 3

-¥

-¥

ò kv(l)dl = k ò v(l)dl = ky1 ( t)

y2 ( t) =

Since it is both homogeneous and additive. It is also linear.

x3( t) = v( t) + w( t)

y1 ( t) = cos 2 pt v( t)

y3( t) =

y2 ( t) = cos 2 pt ( t - to) ¹ y( t - to) = cos [2p( t - to)]v( t - to)

(Time Variant)

(Homogeneous)

t+ 3

t+ 3

t+ 3

-¥

-¥

-¥

ò [ v( l) + w( l)]dl =

ò v( l)dl +

ò w( l)dl

= y1 (t) + y2 (t)

(Additive)

The response at any time t = to depends only on the excitation at that time and not on the excitation at any

Since it is Homogeneous and additive, it is also linear.

later time.

(Causal)

y1 ( t) =

If x( t) is bounded then y( t) is bounded.

(Stable)

ò v(l)dl

-¥

y2 ( t) =

20. (C) y1 ( t) = |v( t)|, y2 ( t) = |kv( t)| = k y1 ( t)

t+ 3

t+ 3

t - to + 3

-¥

-¥

ò v(l - to )dl =

ò v(l)dl = y (t - t ) o

1

If k is negative k y1 ( t) ¹ ky1 ( t)

(Time invariant)

(Not Homogeneous Not linear). y1 ( t) = |v( t)|, y2 ( t) = | y( t - to)| = y1 ( t - to)

The response at any time, t = to , depends partially on the excitation at time to to < t < ( to + 3) which are in

(Time Invariant) The response at any time t = to depends only on the excitation at that time and not on the excitation at any later time. If x( t) is bounded then y( t) is bounded.

(Causal)

future. If x( t) is a constant k, then y ( t) =

(Not causal) t+ 3

t+ 3

-¥

-¥

ò kdl = k ò dl and as

t ® ¥ , y ( t) increases without bound.

(unstable)

(Stable) www.gatehelp.com

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GATE EC BY RK Kanodia

UNIT 5

Signal & System

æ pn ö æ pn ö æ pn p ö 14. x[ n] = cos ç + ÷ ÷ - sin ç ÷ + 3 cos ç 3ø è 2 ø è 8 ø è 4

11. x[ n + 2 ] y[ n - 2 ] x[n] 3

(A) periodic with period 16

2

(A)

(B) periodic with period 4

1 -6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

n

(C) periodic with period 2 (D) Not periodic

x[n] 3

15. x[ n] = 2 e

2

(B)

1 -6

-5

-4

-3

-2

-1

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

1

2

3

4

5

6

n

(A) periodic with 12p

(B) periodic with 12

(C) periodic with 11p

(D) Not periodic

16. The sinusoidal signal has fundamental period n

N = 10 samples. The smallest angular frequency, for which x[ n] is periodic, is 1 (A) rad/cycle 10

-1

(C)

æn ö j çç - p÷÷ è6 ø

-2 -3

(C) 5 rad/cycle -6

-5

-4

-3

-2

-1

1

2

3

4

5

6

n

-1

(D)

(B) 10 rad/cycle (D)

p rad/cycle 5

17. Let x[ n], - 5 £ n £ 3 and h[ n], 2 £ n £ 6 be two finite duration signals. The range of their convolution is

-2 -3

(A) -7 £ n £ 9

(B) -3 £ n £ 9

(C) 2 £ n £ 3

(D) -5 £ n £ 6

Statement for Q.12–15: A discrete-time signal is given. Determine the

Statement for Q.18–26:

period of signal and choose correct option.

x[ n] and h[ n] are given in the question. Compute the convolution y[ n] = x[ n] * h[ n]

pn æ pn 1 ö 12. x[ n] = cos + sin ç + ÷ 9 2ø è 7 (A) periodic with period N = 126 (B) periodic with period N = 32

Page 260

and choose correct

option. 18. x[ n] = {1, 2, 4}, h[ n] = {1, 1, 1, 1, 1} (A) {1, 3, 7, 7, 7, 6, 4} (B) {1, 3, 3, 7, 7, 6, 4}

(C) periodic with period N = 252

(C) {1, 2, 4}

(D) Not periodic

(D) {1, 3, 7}

æ nö æ pn ö 13. x[ n] = cos ç ÷ cos ç ÷ è8ø è 8 ø

19. x[ n] = {1, 2, 3, 4, 5}, h [ n] = {1} (A) {1, 3, 6, 10, 15}

(B) {1, 2, 3, 4, 5}

(A) Periodic with period 16p

(C) {1, 4, 9, 16, 20}

(D) {1, 4, 6, 8, 10}

(B) periodic with period 16( p + 1)

20. x[ n] = {1, 2, -1}, h [ n] = x [ n]

(C) periodic with period 8

(A) {1, 4, 1}

(B) {1, 4, 2, -4, 1}

(D) Not periodic

(C) {1, 2, -1}

(D) {2, 4, -2}

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Discrete-Time Systems

21. x[ n] = {1, -2, 3}, h [ n] = {0, 0, 1, 1, 1, 1} ­

Chap 5.2

25. x[ n] = {1, 4, -3, 6, 4}, h[ n] = {2, -4, 3} ­

­

­

(A) {2, 4, -19, 36, -25, 2, 12}

(A) {1, -2, 4, 1, 1, 1}

­

­

(B) {4, -19, 36, -25}

(B) {0, 0, 3}

­

­

(C) {1, 4, -3, 6, 4}

(C) {0, 0, 3, 1, 1, 1, 1}

­

­

(D) {1, 4, -3, 6, 4}

(D) {0, 0, 1, -1, 2, 2, 1, 3}

­

­

22. x[ n] = {0, 0, 1, 1, 1, 1}, h[ n] = {1, -2, 3} ­

­

(A) {0, 0, 1, -1, 2, 2, 1, 3}

ì 1, ï 26. x[ n] = í 2, ï0 î

n = -2, 0, 1 n = -1 elsewhere

h (n) = d[ n ] - d[ n - 1] + d[ n - 4]

­

(A) d[ n ] - 2 d[ n - 1 ] + 4 d[ n - 4 ] + d[ n - 5 ]

(B) {0, 0, 1, -1, 2, 2, 1, 3}

(B) d[ n + 2 ] + d[ n + 1 ] - d[ n ] + 2 d[ n - 3 ] + d[ n - 4 ] + d[ n - 5 ]

­

(C) d[ n + 2 ] - d[ n + 1 ] + d[ n ] + 2 d[ n - 3 ] - d[ n - 4 ] + 2 d[ n - 5 ]

(C) {1, -2, 3, 1, 1, 2, 1, 1}

(D) d[ n ] + 2 d[ n - 1 ] + 4 d[ n - 5 ] + d[ n - 5 ]

­

Statement for Q.27–30:

(D) {1, -2, 3, 1, 1, 1, 1} ­

In question y[ n] is the convolution of two signal.

23. x[ n] = {1, 1, 0, 1, 1}, h[ n] = {1, -2, - 3, 4} ­

­

(A) {1, -1, - 2, 4, 1, 1} ­

(B) {1, -1, - 2, 4, 1, 1} ­

(C) {1, -1, - 5, 2, 3, -5, 1, 4} ­

(D) {1, -1, - 5, 2, 3, -5, 1, 4} ­

24. x[ n] = {1, 2, 0, 2, 1}, h[ n] = x[ n] ­

(A) {1, 4, 4, 4, 10, 4, 4, 4, 1} ­

(B) {1, 4, 4, 4, 10, 4, 4, 4, 1} ­ (C) {1, 4, 4, 10, 4, 4, 4, 1} ­

(D) {1, 4, 4, 10, 4, 4, 4, 1} ­

Choose correct option for y[ n]. 27. y[ n] = ( -1) n * 2 n u[2 n + 2 ] (A)

4 6

(B)

4 u[ -n + 2 ] 6

(C)

8 ( -1) n u[ -n + 2 ] 3

(D)

8 ( -1) n 3

28. y[ n] =

1 u[ n] * u[ n + 2 ] 4n

æ1 1 ö (A) ç - n ÷u[ n] è3 4 ø

æ 1 12 ö (B) ç - n ÷u[ n + 2 ] è3 4 ø

æ 4 1 æ 1 ön ö ÷u[ n + 2 ] (C) ç ç 3 12 çè 4 ÷ø ÷ è ø

1 ö æ 16 (D) ç - n ÷u[ n + 2 ] è 3 4 ø

29. y[ n] = 3n u[ -n + 3] * u[ n - 2 ] ì 3n , ïï (A) í 2 ï 83 , ïî 2 ì 3n , ïï (C) í 2 ï 81 , ïî 2

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n £5 n³6 n £5 n³6

ì 3n , ï (B) í 83 ïî 2 , ì 3n , ïï (D) í 6 ï 81 , ïî 2

n £5 n³6

n £5 n³6 Page 261

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UNIT 5

30. y[ n] = u[ n + 3] * u[ n - 3]

37. x[ n] as shown in fig. P5.2.37

(A) ( n + 1) u[ n]

(B) nu[ n]

(C) ( n - 1) u[ n]

(D) u[ n]

31.

The

convolution

of

x[ n] = cos ( 2p n) u[ n]

5

x[n]

and

+

10

y[n]

Fig. P5.2.37

h[ n] = u[ n - 1] is f [ n]u[ n - 1]. The function f [ n] is n = 4 m + 1, 4 m + 2 ì 1, (A) í n = 4 m, 4 m + 3 î 0, ì 0, (B) í î 1,

Signal & System

(A) P, Q, R, S

(B) Q, R, S

(C) P, R, S

(D) P, Q, S

n = 4 m + 1, 4 m + 2 n = 4 m, 4 m + 3

ì 1, (C) í î 0,

n = 4 m + 1, 4 m + 3

ì 0, (D) í î 1,

n = 4 m + 1, 4 m + 3 n = 4 m, 4 m + 2

38. x[ n] as shown in fig. P5.2.38 x[n]

n = 4 m, 4 m + 2

+

y[n] +

Fig. P5.2.38

Statement for Q.32–38: Let P be linearity, Q be time invariance, R be causality and S be stability. In question discrete time

(A) P, Q, R, S

(B) P, Q, R

(C) P, Q

(D) Q, R, S

Statement for Q.39–41: and S2

Two discrete time systems S1

input x[ n] and output y[ n] relationship has been given. In the option properties of system has been given. Choose the option which match the properties for system.

fig. P5.2.39–41.

32. y[ n] = rect ( x[ n])

x[n]

(A) P, Q, R

(B) Q, R, S

(C) R, S, P

(D) S , P, Q

S2

S1

y[n]

Fig. P5.2.39–41.

39. Consider the following statements

33. y[ n] = nx[ n] (A) P, Q, R, S

(B) Q, R, S

(a) If S1 and S2 are linear, the S is linear

(C) P, R

(D) Q, S

(b) If S1 and S2 are nonlinear, then S is nonlinear

34. y[ n] =

(c) If S1 and S2 are causal, then S is causal

n +1

å

u[ m ]

(d) If S1 and S2 are time invariant, then S is time invariant

m = -¥

(A) P, Q, R, S

(B) R, S

(C) P, Q

(D) Q, R

35. y[ n] = x[ n] (A) Q, R, S

(B) R, S, P

(C) S, P, Q

(D) P, Q, R

36. x[ n] as shown in fig. P5.2.36 x[n]

2

y[n]

True statements are : (A) a, b, c

(B) b, c, d

(C) a, c, d

(D) All

40. Consider the following statements (a) If S1 and S2 are linear and time invariant, then interchanging their order does not change the system. (b) If S1 and S2 are linear and time varying, then interchanging their order does not change the system. True statement are

Fig. P5.2.36

Page 262

are

connected in cascade to form a new system as shown in

(A) P, Q, R, S

(B) Q, R, S

(C) P, Q

(D) R, S

(A) Both a and b

(B) Only a

(C) Only b

(D) None

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Discrete-Time Systems

41. Consider the statement

Chap 5.2

(D) Above all

(a) If S1 and S2 are noncausal, the S is non causal (b) If S1 and/or S2 are unstable, the S is unstable.

45. The system shown in fig. P5.2.45 is

(A) Both a and b

y[n]

+

1 4

x[n]

True statement are : (B) Only a

(C) Only b

D

y[n-2]

D

+

+

(D) None

1 4

+

-1 2

42. The following input output pairs have been

Fig. P5.2.45

observed during the operation of a time invariant (A) Stable and causal

system : x1 [ n] = {1, 0, 2}

S ¬¾ ®

y1 [ n] = {0, 1, 2}

(B) Stable but not causal

­

(C) Causal but unstable

­ x2 [ n] = {0, 0, 3}

¬¾® S

y2 [ n] = {0, 1, 0, 2}

­

­

x3[ n] = {0, 0, 0, 1}

¬¾® S

46. The impulse response of a LTI system is given as

y3[ n] = {1, 2, 1}

­

(D) unstable and not causal

n

æ 1ö h[ n] = ç - ÷ u[ n]. è 2ø

­

The conclusion regarding the linearity of the

The step response is

system is (A) System is linear

1 æç æ 1ö 2 -ç- ÷ ç 3è è 2ø

(C)

1 æç æ 1ö 2 + ç- ÷ 3 çè è 2ø

(B) System is not linear (C) One more observation is required.

n +1

(A)

(D) Conclusion cannot be drawn from observation.

n +1

ö ÷u[ n] ÷ ø

(B)

1 æç æ 1ö 2 -ç- ÷ ç 3è è 2ø

ö ÷u[ n] ÷ ø

(D)

1 æç æ 1ö 2 + ç- ÷ 3 çè è 2ø

n

n

ö ÷u[ n] ÷ ø ö ÷u[ n] ÷ ø

43. The following input output pair have been observed

47. The difference equation representation for a system

during the operation of a linear system:

is

x1 [ n] = { -1, 2, 1}

¬¾® S

y1 [ n] = {1, 2, - 1, 0, 1}

­ x2 [ n] = {1, - 1, - 1}

­ S ¬¾ ®

y2 [ n] = { - 1, 1, 0, 2}

­

­

x3[ n] = {0, 1, 1}

y[ n] -

S ¬¾ ®

­

(A)

3æ 1ö ç - ÷ u[ n] 2è 2ø

(C)

3æ 1ö ç ÷ u[ n] 2 è2 ø

(A) System is time-invariant

(A) y[ n] = x[ n] + 11 . y[ n - 1] (B) y[ n] = x[ n] -

1 ( y[ n - 1] + y[ n - 2 ]) 2

(D)

2æ1ö ç ÷ u[ n] 3è2 ø

n

y[ n] - 2 y[ n - 1] + y[ n - 2 ] = x[ n] - x[ n - 1] If y[ n] = 0 for n < 0 and x[ n] = d[ n], then y[2 ] will

(C) One more observation is required

44. The stable system is

2æ 1ö ç - ÷ u[ n] 3è 2 ø

48. The difference equation representation for a system is

(B) System is time variant

(D) Conclusion cannot be drawn from observation

n

(B)

n

The conclusion regarding the time invariance of the system is

y [ -1] = 3

The natural response of system is n

y3[ n] = {1, 2, 1}

­

1 y[ n - 1] = 2 x[ n], 2

be (A) 2

(B) -2

(C) -1

(D) 0

49. Consider a discrete-time system S whose response to a complex exponential input e jpn S : e jpn

(C) y[ n] = x[ n] - (15 . y[ n - 1] + 0.4 y[ n - 2 ]) www.gatehelp.com

2

Þ

e jp3n

2

is specified as

2

Page 263

GATE EC BY RK Kanodia

UNIT 5

24. (B) y[ n] = {1, 4, 4, 10, 4, 4, 4, 1}

Signal & System

28. (C) For n + 2 < 0

­

for

n + 2 ³0

or

n < - 2, y [ n] = 0

n ³ - 2, y[ n] =

or

n+2

1

å4k

4 1 1 , 3 12 4 n

=

k=0

1

2

0

2

1

1

1

2

0

2

1

2

2

4

0

4

2

Þ

æ 4 1 æ 1 ön ö ÷u[ n + 2 ] y[ n] = ç ç 3 12 çè 4 ÷ø ÷ è ø n -2

0

0

0

0

0

k=¥

0

2

2

4

0

4

2

1

1

2

0

2

1

for n - 2 ³ 4 or n ³ 6, y[ n] =

Fig. S5.2.24

25. (A) y[ n] = {2, 4, -19, 36, -25, 2, 12} ­

-3

6

4

2

2

8

-6

12

8

-4

-4

-16

12

-24

-16

3

3

12

ì 3n , ïï y[ n] = í 6 ï 81 , ïî 2

-9

18

=

k

81 , 2

n³6 n < 0,

y[ n] = 0 n -3

n > 0, y[ n] =

for n - 3 ³ - 3 or

å 1 = n + 1,

k = -3

y[ n] = ( n + 1) u[ n] 31. (A) For n - 1 < 0

12

For

Fig. S5.2.25

26. (B) x[ n] = {1, 2, 1, 1}, h[ n] = {1, -1, 0, 0, 1} ­

å3

n £5

30. (A) For n - 3 < - 3 or

4

3

k = -¥

Þ

1

3n 6

å 3k =

29. (D) For n - 2 £ 3 or n £ 5 , y[ n] =

Þ

n -1 ³ 0 ì 1, y[ n] = í î 0,

or

or

n <1 ,

y[ n] = 0

æp ö n ³ 1, y[ n] = å cos ç k ÷ k=0 è2 ø n -1

n = 4 m + 1, 4 m + 2 n = 4 m, 4 m + 3

­

32. (B) y1 [ n] = rect ( v[ n]) ,

y2 [ n] = rect ( kv[ n])

1

2

1

1

y2 [ n] ¹ k y1 [ n]

1

1

2

1

-1

y1 [ n] = rect ( v[ n]), y2 [ n] = rect ( v[ n - no ])

-1

-1

-2

-1

-1

0

0

0

0

0

0

0

0

0

0

(Not Homogeneous not linear)

y1 [ n - no ] = rect ( v[ n - no ]) = y2 [ n]

(Time Invariant)

At any discrete time n = no , the response depends only on the excitation at that discrete time.

(Causal)

No matter what values the excitation may have the response can only have the values zero or one.

1

1

2

1

(Stable)

1

Fig. S5.2.26

33. (C) y1 [ n] = nv[ n] ,

y[ n] = {1, 1, -1, 0, 0, 2, 1, 1}

ky1 [ n] = y2 [ n]

­ y[ n ] = d[ n + 2 ] + d[ n + 1 ] - d[ n ] + 2 d[ n - 3 ] + d[ n - 4 ] + d[ n - 5 ] ¥

å ( -1)

27. (D) y[ n] =

k

k=n -2

æ 1ö 2n ç - ÷ è 2ø = 1 1+ 2 Page 266

n -2

=

8 ( -1) n 3

2n -k = 2n

¥

å

k=n -2

y2 [ n] = nkv[ n]

æ 1ö ç- ÷ è 2ø

k

(Homogeneous)

Let x1 [ n] = v[ n]

then

y1 [ n] = nv[ n]

Let x2 [ n] = w[ n]

then

y2 [ n] = nw[ n]

Let x3[ n] = v[ n] + w[ n] then y3[ n] = n( v[ n] + w[ n]) = nv[ n] + nw[ n] = y1 [ n] + y2 [ n]

(Additive)

Since the system is homogeneous and additive, it is also linear. y1 [ n - no ] = ( n - no) v[ n - no ] ¹ yn [ n] = nv[ n - no ]

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Discrete-Time Systems

(Time variant)

Chap 5.2

If the excitation is bounded, the response is bounded.

At any discrete time, n = no the response depends only on the excitation at that same time.

(Causal)

(Stable).

If the excitation is a constant, the response is

37. (B) y1 [ n] = 10 v[ n] - 5,

unbounded as n approaches infinity.

y2 [ n] ¹ ky1 [ n]

n +1

34. (C) y1 [ n] =

å v[ m ] ,

y2 [ n ] =

m = -¥

(Unstable)

n +1

å

å kv[ m ]

n = -¥

y3[ n] =

y1 [ n - no ] = 10 v[ n - no ] - 5, = y2 [ n]

m = -¥

(Homogeneous)

v[ m ], y2 [ n] =

(Not Homogeneous so not linear)

y1 [ n] = 10 v[ n] - 5, y2 [ n] = 10 v[ n - no ] - 5

n +1

y2 [ n] = ky1 [ n] y1 [ n] =

y2 [ n] = 10 kv[ n] - 5

n +1

å w[ m ]

(Time Invariant)

At any discrete time, n = no the response depends only on the excitation at that discrete time and not on any future excitation.

n = -¥

(Causal)

If the excitation is bounded, the response is bounded.

n +1

å ( v[ n] + w[ m ])

(Stable).

m = -¥

=

n +1

n +1

m = -¥

m = -¥

å v[ m ] + å w[ n] = y [ n] + y [ n] 1

(Additive)

2

Since the system is homogeneous and additive it is also linear y1 [ n] =

n +1

å v[ n] , y2 [ n] =

m = -¥

y1 [ n - no ] =

y[ n] = x[ n] + x[ n - 1] + y[ n - 2 ], Then by induction ¥

y[ n] = x[ n - 1] + x[ n - 2 ] + K x[ n - k] + K = å x[ n - k] k=0

n +1

å v[ m - no ]

n +1

m = -¥

q = -¥

å v[ q - no ] = y2 [ n]

y1 [ n] =

n

å v[ m ] ,

n

m =n

m = -¥

n

å kv[ m ] = ky [ n]

y2 [ n ] =

m = -¥

-¥

å x[ m ] = å x[ m ]

Let m = n - k then y[ n] =

m = -¥

n -no +1

å v[ m ] =

38. (B) y[ n] = x[ n] + y[ n - 1], y[ n - 1] = x[ n - 1] + y[ n - 2 ]

1

m = -¥

(Time Invariant) At any discrete time, n = no , the response depends on the excitation at the next discrete time in future. (Anti causal)

(Homogeneous) y3[ n] =

n

¥

n

m = -¥

m = -¥

m = -¥

å (v[ m ] + w[ m ]) = å v[ m ] + å w[ m ]

= y1 [ n] + y2 [ n]

(Additive)

If the excitation is a constant, the response increases

System is Linear.

without bound.

y1 [ n] =

(Unstable)

¥

å v[ m ]

m = -¥

35. (A) y1 [ n] = v[ n] , y2 = kv[ n] = k v[ n] y1 [ n] = v[ n] , y2 [ n] = v[ n - no ] (Time Invariant)

At any discrete time n = no , the response depends only on the excitation at that time

o

m = -¥

y1 [ n] can be written as

ky1 [ n] = k v[ n] ¹ y2 [ n] (Not Homogeneous Not linear)

y1 [ n - no ] = v[ n - no ] = y2 [ n]

n

å v[ n - n ]

, y2 =

(Causal)

If the excitation is bounded, the response is bounded. (Stable). 36. (B) y[ n] = 2 x [ n]

y1 [ n - no ] =

n -no

n

m = -¥

q = -¥

å v[ m ] = å v[ q - n ] = y [ n] o

2

(Time Invariant) At any discrete time n = no the response depends only on the excitation at that discrete time and previous discrete time.

(Causal)

If the excitation is constant, the response increase without bound.

(Unstable)

2

Let x1 [ n] = v[ n]

Let x2 [ n] = kv[ n] then ky[ n] ¹ y2 [ n] Let x1 [ n] = v[ n]

39. (C) Only statement (b) is false. For example

then y1 [ n] = 2 v 2 [ n]

S1 : y[ n] = x[ n] + b, and S2 : y[ n] = x[ n] - b , where b ¹ 0

y2 [ n] = 2 k2 v 2 [ n]

S{x[ n]} = S2 {S1 {x[ n]}} = S2 {x[ n] + b} = x[ n]

(Not homogeneous Not linear) then

y1 [ n] = 2 v 2 [ n]

Hence S is linear.

Let x2 [ n] = v[ n - no ] then y2 [ n] = 2 v 2 [ n - no ]

40. (B) For example

y1 [ n - no ] = 2 v[ n - no ] = y2 [ n]

S1 : y[ n] = nx[ n]

(Time invariant)

At any discrete time, n = no , the response depends only on the excitation at that time.

(Causal)

and S2 : y[ n] = nx[ n + 1]

If x[ n] = d[ n] then S2 {S1 {d[ n]}} = S2 [0 ] = 0,

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CHAPTER

5.3 THE LAPLACE TRANSFORM

Statement for Q.1-12:

6. x( t) = u( t) - u( t - 2)

Determine the Laplace transform of given signal. 1. x( t) = u( t - 2) -2 s

(A)

-e s

(B)

e

-2 s

e -2 s - 1 s

(B)

1 - e -2 s s

(C)

2 s

(D)

-2 s

s 7. x( t) =

-2 s

e (C) 1+ s

(A)

(D) 0

(A)

1 s( s + 1) 2

(B)

s ( s + 1) 2

(C)

e- s s+1

(D)

e- s ( s + 1) 2

2. x( t) = u( t + 2) (A)

1 s

(B) -

(C)

e -2 s s

(D)

1 s

-e - 2 s s

3. x( t) = e -2 t u( t + 1) (A) (C)

1 s+2

(B)

e- ( s + 2 ) s+2

(D)

e- s s+2 -e - s s+2

4. x( t) = e 2 t u( -t + 2) (A)

e

2 (s - 2 )

-1 s -2 -2 ( s - 2 )

(C)

1-e s -2

8. x( t) = tu( t) * cos 2pt u( t) (A)

1 s( s 2 + 4 p2 )

(B)

2p s 2 ( s 2 + 4 p2 )

(C)

1 s ( s + 4 p2 )

(D)

s3 s + 4 p2

-3 s4

e s+2 -2 s

(D)

2

2

2

9. x( t) = t 3u( t) (A)

3 s4

(B)

(C)

6 s4

(D) -

-2 s

(B)

d { te - t u( t)} dt

e s -2

6 s4

10. x( t) = u( t - 1) * e -2 t u( t - 1)

5. x( t) = sin 5 t 5 (A) 2 s +5

s (B) 2 s +5

(A)

e -2 ( s + 1 ) 2s + 1

(B)

e -2 ( s + 1 ) s+1

5 (C) 2 s + 25

s (D) 2 s + 25

(C)

e- ( s + 2 ) s+2

(D)

e -2 ( s + 1 ) s+2

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UNIT 5

t

11. x( t) = ò e -3t cos 2 t dt

17. X ( s) =

0

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

(A)

-( s + 3) s(( s + 3) 2 + 4)

(B)

( s + 3) s(( s + 3) 2 + 4)

(A) ( e -2 t - 2 te - t ) u( t)

(B) ( e -2 t + 2 te - t ) u( t)

(C) ( e - t - 2 te -2 t ) u( t)

(D) ( e - t + 2 te -2 t ) u( t)

(C)

s( s + 3) ( s + 3) 2 + 4

(D)

-s( s + 3) ( s + 3) 2 + 4

18. X ( s) =

12. x( t) = t

( s + 2 s + 2) ( s 2 + 4 s + 2) 2

(C) ( 3e - t cos 3t - e - t sin 3t) u( t)

-( s + 2 s + 2) ( s 2 + 4 s + 2) 2 2

(D)

(D) ( 3e - t sin 3t + 3e- t cos 3t) u( t) 19. X ( s) =

Statement for Q.13–24: Determine the time signal x( t) corresponding to given X ( s) and choose correct option.

(B) (2 e - t - e -2 t ) u( t)

(C) (2 e -2 t - e - t ) u( t)

(D) (2 e - t + e -2 t ) u( t)

(A) 2 d( t) + ( e -3t - e-2 t ) u( t)

(A) (2 e -2 t + 2 e - t sin 2 t - 2 e - t cos 2 t) u( t) (B) (2 e -2 t + 2 e - t cos 2 t - 2 e - t sin 2 t) u( t) (D) (2 e -2 t + 2 e - t sin 2 t - e - t cos 2 t) u( t)

(A) (2 e -2 t + e - t ) u( t)

2 s 2 + 10 s + 11 s2 + 5 s + 6

20. X ( s) =

(C) 2 d( t) + ( e

+ e ) u( t)

(D) 2 d( t) - ( e

-2 t

+ e -3t ) u( t)

15. X ( s) = -t

-3t

2s - 1 s2 + 2 s + 1 -t

(A) ( 3e - 2 te ) u( t) (B) ( 3e- t + 2 te- t ) u( t)

(B) ( e -2 t + 2 e -3t cos t - 6 e -3t sin t) u( t) (C) ( e -2 t + 2 e -3t cos t - 2 e -3t sin t) u( t) (D) (9 e -2 t - 6 e -3t cos t + 3e -3t sin t) u( t) 21. X ( s) =

16. X ( s) =

5s + 4 s 3 + 3s 2 + 2 s

2 s 2 + 11s + 16 + e -2 s ( s2 + 5 s + 6)

(A) 2 d( t) + ( 3e -2 t - 2 e-3t ) u( t - 2) (B) 2 d( t) + (2 e -2 t - e -3t + e -2 ( t - 2 ) + e -3( t - 2 ) ) u( t) (C) 2 d( t) + (2 e -2 t - e -3t ) u( t) + ( e -2 t - e -3t ) u( t - 2) (D) 2 d( t) + (2 e -2 t - e -3t ) u( t) + ( e -2 ( t - 2 ) - e -3( t - 2 ) ) u( t - 2) 22. X ( s) = s

(C) (2 e- t - 3te- t ) u( t) (D) (2 e- t + 3te- t ) u( t)

3s 2 + 10 s + 10 ( s + 2)( s 2 + 6 s + 10)

(A) ( e -2 t + 2 e -3t cos t + 2 e -3t sin t) u( t)

(B) 2 d( t) + ( e -2 t - e-3t ) u( t) -2 t

4 s 2 + 8 s + 10 ( s + 2)( s 2 + 2 s + 5)

(C) (2 e -2 t + 2 e - t cos 2 t - e - t sin 2 t) u( t)

s+3 13. X ( s) = 2 s + 3s + 2

14. X ( s) =

2

1 æ ö (B) ç 3e- t sin 3t - e- t cos 3t ÷u( t) 3 è ø

( s 2 + 4 s + 2) (B) 2 ( s + 2 s + 2) 2

2

(C)

3s + 2 s + 2 s + 10 1 æ -t ö (A) ç 3e cos 3t - e- t sin 3t ÷u( t) 3 è ø

d -t { e cos t u( t)} dt

-( s 2 + 4 s + 2) (A) 2 ( s + 2 s + 2) 2

d2 ds 2

æ 1 ö 1 çç 2 ÷÷ + 9 3 s + s + è ø

ö æ 2t t2 (A) çç e -3t + sin 3t + cos 3t ÷÷u( t) 3 9 ø è (B) ( e -3t + 2 t sin 3t + t 2 cos 3t) u( t)

(A) (2 + e - t + 3e -2 t ) u( t)

2t æ ö (C) ç e -3t + sin 3t + t 2 cos 3t ÷u( t) 3 è ø

(B) (2 + e - t - 3e -2 t ) u( t)

(D) ( e -3t + t 2 sin 3t + 2 t cos 3t) u( t)

(C) ( 3 + e - t - 3e -2 t ) u( t) (D) ( 3 + e - t + 3e -2 t ) u( t) Page 270

Signal & System

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The Laplace Transform

23. X ( s) = (A) e (C)

-0 .5t

Chap 5.3

Statement for Q.30–33:

1 (2 s + 1) 2 + 4

Given the transform pair

1 (B) e - t sin t u( t) 2

sin t u( t)

1 -0 .5t e sin t u( t) 4

x( t) u( t)

(D) e - t sin t u( t)

L ¬¾ ®

2s . s2 + 2

Determine the Laplace transform Y ( s) of the given time signal in question and choose correct option.

24. X ( s) = e

-2 s

1 d æ ç ds çè ( s + 1) 2

ö ÷÷ ø

30. y( t) = x( t - 2)

(A) - te - t u(1 - t)

(B) -te - t u( t - 1)

(C) -( t - 2) 2 e - ( t - 2 ) u( t - 2)

(D) te - t u( t - 1)

Statement for Q.25–29:

(A)

2 se -2 s s2 + 2

(B)

2 se 2 s s2 + 2

(C)

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

(D)

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

Given the transform pair below. Determine the 31. y( t) = x( t) *

time signal y( t) and choose correct option. cos 2t u( t)

L ¬¾ ®

X ( s).

dx( t) dt

(A)

4 s3 ( s 2 + 2) 2

(B)

4 ( s 2 + 2) 2

-4 s 3 ( s + 2) 2

(D)

4 ( s 2 + 2) 2

25. Y ( s) = ( s + 1) X ( s) (A) [cos 2 t - 2 sin 2 t ]u( t)

sin 2 t ö æ (B) ç cos 2 t + ÷u( t) 2 ø è

(C)

(C) [cos 2 t + 2 sin 2 t ]u( t)

sin 2 t ö æ (D) ç cos 2 t ÷u( t) 2 ø è

32. y( t) = e - t x( t) (A)

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

(B)

2( s + 1) s2 + 2 s + 2

(C)

2( s + 1) s2 + 2 s + 4

(D)

2( s + 1) s2 + 2 s

26. Y ( s) = X ( 3s) æ2 ö (A) cos ç t ÷u( t) è3 ø

1 æ2 ö (B) cos ç t ÷u( t) 3 è3 ø

(C) cos 6t u( t)

(D)

1 cos 6 t u( t) 3

33. y( t) = 2 tx( t) (A)

8 - 4 s2 ( s 2 + 2) 2

(B)

4 s2 - 8 ( s 2 + 2) 2

(C)

4 s2 s2 + 1

(D)

s2 s +1

27. Y ( s) = X ( s + 2) (A) cos 2( t - 2) u( t)

(B) e2 t cos 2 t u( t)

(C) cos 2( t + 2) u( t)

(D) e-2 t cos 2 t u( t)

2

2

Statement for Q.34–43: X ( s) 28. Y ( s) = 2 s (A) 4 cos 2 t u( t) 2

(C) t cos 2 t u( t)

29. Y ( s) =

Determine the bilateral laplace transform and (B)

1 - cos 2 t u( t) 4

cos 2 t (D) u( t) t2

d -3s [ e X ( s)] ds

(A) t cos 2( t - 3) u( t - 3)

(B) t cos 2( t - 3) u( t)

(C) -t cos 2( t - 3) u( t - 3)

(D) -t cos 2( t - 3) u( t)

choose correct option. 34. x( t) = e - t u( t + 2) (A)

e2 ( s + 1 ) , s+1

Re ( s) > -1

(B)

1 , 1+ s

Re ( s) < - 1

(C)

e2 ( s + 1 ) , s+1

Re ( s) < - 1

(D)

1 , 1+ s

Re ( s) > -1

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GATE EC BY RK Kanodia

UNIT 5

35. x( t) = u( -t + 3) 1 - e -3s , (A) s (B)

1-e (C) s (D)

-3s

Re ( s) > 3

(B)

e 3s , s-3

Re ( s) < 3

(C)

e 3( s -1 ) , s-3

Re ( s) > 3

(D)

e 3( s -1 ) , s-3

Re ( s) < 3

Re ( s) < 0

,

-e -3s , s

Re ( s) > 0

(A) e , Re( s) > 0 s

(B) e , Re ( s) < 0

s

(D) None of above

s

(C) e , all s 37. x( t) = sin t u( t) 1 , (A) (1 + s 2 )

Re ( s) > 0

-1 (C) , (1 + s 2 )

Re ( s) < 0

-1 , (1 + s 2 )

Re ( s) > 0

-

t 2

41. x( t) = cos 3t u( -t) * e - t u( t) -s , (A) ( s + 1)( s2 + 9) -s , ( s + 1)( s2 + 9)

-1 < Re ( s) < 0

(C)

s , ( s + 1)( s 2 + 9)

-1 < Re ( s) < 0

(D)

s , ( s + 1)( s 2 + 9)

Re ( s) > 0

42. x( t) = e t sin (2 t + 4) u( t + 2)

-t

38. x( t) = e u( t) + e u( t) + e u( -t) t

(A)

6 s2 + 2 s - 2 , (2 s + 1)( s 2 - 1)

Re ( s) < - 0.5

(B)

6 s2 + 2 s - 2 , (2 s + 1)( s 2 - 1)

-1 > Re ( s) > 1

(C)

1 1 1 , + + s + 0.5 s + 1 s - 1

-1 < Re ( s) < 1

(D)

1 1 1 , + s + 0.5 s + 1 s - 1

-0.5 < Re ( s) < 1

(A)

e 2 ( s -1 ) , ( s - 1) 2 + 4

Re ( s) > 1

(B)

e 2 ( s -1 ) , ( s - 1) 2 + 4

Re ( s) < 1

(C)

e( s - 2 ) , ( s - 1) 2 + 4

Re ( s) > 1

(D)

e( s - 2 ) , ( s - 1) 2 + 4

43. x( t) = e t (A)

1-s , s+1

Re ( s) > -1

(C)

s -1 , s+1

Re ( s) < - 1

s -1 , s+1

Re ( s) > -1

(1 - s) 1 1 , 0.5 < Re ( s) < 1 + + 2 ( s - 1) + 4 s + 1 s - 0.5

(D)

(B)

(1 - s) 1 1 , -1 < Re ( s) < 1 + + ( s - 1) 2 + 4 s + 1 s - 0.5

Statement for Q.44–49:

(C)

( s - 1) 1 1 , 0.5 < Re ( s) < 1 + + ( s - 1) 2 + 4 s + 1 s - 0.5

bilateral Laplace transform.

40. x( t) = e ( 3t + 6 ) u( t + 3) Page 272

Re ( s) < - 1

(A)

( s - 1) 1 1 , -1 < Re ( s) < 1 + + ( s - 1) 2 + 4 s + 1 s - 0.5

Re ( s) < 1

d -2 t [ e u( -t)] dt

1-s , s+1

(B)

t

39. x( t) = e t cos 2 t u( -t) + e - t u( t) + e 2 u( t)

(D)

Re ( s) > 0

(B)

Re ( s) < 0

1 , (1 + s 2 )

(D)

e 3s , s-3

Re ( s) < 0

36. y( t) = d( t + 1)

(B)

(A) Re ( s) > 0

-e -3s , s

Signal & System

Determine the corresponding time signal for given

44. X ( s) =

e 5s with ROC: Re ( s) < -2 s+2

(A) e -2 ( t + 5) u( t + 5) www.gatehelp.com

GATE EC BY RK Kanodia

UNIT 5

(A)

1 -t e sin t u( t) 2

(D)

58.

62. A stable system has input

(B) 2e- t cos t u( t)

(C) 2 e - t cos t u( t) +

is (A) d( t) - ( e -2 t cos t + e -2 t sin t) u( t)

1 -t e cos t u( t - 1) + 2 e - t sin t u( t - 1) 2

(B) d( t) - ( e -2 t cos t + e -2 t sin t) u( t - 2) (C) d( t) - ( e 2 t cos t + e 2 t sin t) u( t)

d 3 y( t) d 2 y( t) dy( t) + 4 +3 = x( t) 3 2 dt dt dt

(D) d( t) - ( e 2 t cos t + e 2 t sin t) u( t + 2)

All initial condition are zero, x( t) = 10 e -2 t

63. The relation ship between the input x( t) and output y( t) of a causal system is described by the differential

5 é5 ù (A) ê + 5 e - t - 5 e -2 t + e -3t ú u( t) 3 ë3 û

equation

5 é5 ù (B) ê - 5 e - t + 5 e -2 t + e -3t ú u( t) 3 ë3 û 5 5 u( t) - 5 u( t - 1) + 5 u( t - 2) + u( t - 3) 3 3

(D)

5 5 u( t) + 5 u( t - 1) - 5 u( t - 2) + u( t - 3) 3 3

dy( t) + 10 y( t) = 10 x( t) dt The impulse response of the system is

59. The transform function H ( s) of a causal system is

(A) -10 e -10 t u( -t + 10)

(B) 10 e -10 t u( t)

(C) 10 e -10 t u( -t + 10)

(D) -10 e -10 t u( t)

64. The relationship between the input x( t) and output y( t) of a causal system is defined as

2 s2 + 2 s - 2 H ( s) = s2 - 1

d 2 y( t) dy( t) dx( t) . - 2 y( t) = -4 x( t) + 5 2 dt dt dt

The impulse response is

The impulse response of system is

(A) 2d( t) - ( e- t + et ) u( -t) -t

(A) 3e - t u( t) + 2 e 2 t u( -t)

(B) 2d( t) - ( e + e ) u( t) t

-t

(B) ( 3e - t + 2 e 2 t ) u( t)

(C) 2d( t) + e u( t) - e u( -t) t

(C) 3e - t u( t) - 2 e 2 t u( -t)

(D) 2d( t) + ( e- t + et ) u( t)

(D) ( 3e - t - 2 e 2 t ) u( -t)

60. The transfer function H ( s) of a stable system is 2s - 1 H ( s) = 2 s + 2s + 1 The impulse response is (A) 2 u( -t + 1) - 3tu( -t + 1) (B) ( 3te- t - 2 e- t ) u( t) (C) 2 u( t + 1) - 3tu( t + 1) (D) (2 e- t - 3te- t ) u( t) 61. The transfer function H ( s) of a stable system is H ( s) =

s2 + 5 s - 9 ( s + 1)( s 2 - 2 s + 10)

The impulse response is (A) -e - t u( t) + ( e t sin 3t + 2 e t cos 3t) u( t) (B) -e - t u( t) - ( e t sin 3t + 2 e t cos 3t) u( -t) (C) -e - t u( t) - ( e t sin 3t + 2 e t cos 3t) u( t) (D) -e - t u( t) + ( e t sin 3t + 2 e t cos 3t) u( -t) Page 274

x( t) and output

y( t) = e -2 t cos t u( t). The impulse response of the system

1 -t e sin t u( t) 2

(C)

Signal & System

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*******

GATE EC BY RK Kanodia

The Laplace Transform

SOLUTIONS ¥

¥

0

2

1. (B) X ( s) = ò x( t) e - st dt = ò e - st dt =

Chap 5.3

r ( t) = e -2 t u( t)

L ¬¾ ®

v( t) = e -2 t u( t - 1)

e -2 s s

L ¬¾ ®

x( t) = q( t) * v( t)

¥

¥

¥

0

0

0

2. (A) X ( s) = ò x( t) e -3t dt = ò u( t + 2) -3t dt = ò e -3t dt =

1 s

L ¬¾ ®

3. (A) X ( s) = ò e -2 t e - st dt = 0

1 s+2

¥

¥

0

0

- 1 1 - e -2 ( 2 - s ) = 2-s s -2

p( t) dt

1 P ( s) p( t) dt + ò s -¥ s

L ¬¾ ®

Þ

X ( s) =

5 ( e j 5t - e - j 5t ) - st e dt = 2 2 25 j s + 0

q( t) =

d p( t) dt

x( t) = tq( t) 6. (B) X ( s) = ò e - st dt =

1 - e -2 s s

7. (B) p( t) = te - t u( t)

L ¬¾ ®

2

0

d p( t) dt

L ¬¾ ®

Þ P ( s) =

1 ( s + 1) 2

s ( s + 1) 2

X ( s) =

8. (A) p( t) = tu( t)

L ¬¾ ®

P ( s) =

1 s2

q( t) = cos 2 pt u( t)

L ¬¾ ®

Q( s) =

s s + 4 p2

x( t) = p( t) * q( t) X ( s) =

L ¬¾ ®

q( t) = - tp( t)

t n u( t)

X ( s) = P ( s)Q( s)

L ¬¾ ®

P ( s) =

Q( s) = X ( s) =

1 s2

d -2 P ( s) = 3 ds s

d 6 Q( s) = 4 ds s

n! sn + 1

L ¬¾ ®

10. (D) p( t) = u( t) q( t) = u( t - 1)

L ¬¾ ®

L ¬¾ ®

Q( s) = X ( s) = -

P ( s) =

s+1 ( s + 1) 2 + 1

s( s + 1) ( s + 1) 2 + 1 d Q( s) ds

-( s 2 + 4 s + 2) ( s 2 + 2 s + 2) 2

13. (B) X ( s) = A=

L ¬¾ ®

s+3 A B = + ( s + 3s + 2) s + 1 s + 2 2

s+3 s+3 = 2, B = = -1 s + 2 s = -1 s + 1 s = -2

x( t) = [2 e - t - e -2 t ]u( t) 14. (A) X ( s) = 2 -

1 1 1 =2 + ( s + 2) ( s + 3) ( s + 2) ( s + 3)

x( t) = 2 d( t) + ( e-3t - e-3t ) u( t) 2s - 1 A B = + s 2 + 2 s + 1 ( s + 1) ( s + 1) 2

B = (2 s - 2) s = -1 = -3, A = 2 x( t) = x( t) = [2 e - t - 3te - t ]u( t) 16. (B) X ( s) =

5s + 4 A B C = + + 2 s + 3s + 2 s s s + 1 s + 2 3

A = sX ( s) s = 0 = 2, B = ( s + 1) X ( s) s = -1 = 1, C = ( s + 2) X ( s) s = -2 = -3 x( t) = [2 + e - t - 3e -2 t ]u( t)

L ¬¾ ®

L ¬¾ ®

L ¬¾ ®

15. (C) X ( s) =

2

9. (C) p( t) = tu( t)

X ( s) =

L ¬¾ ®

2

1 s( s + 4 p2 )

x( t) = - tq( t)

( s + 3) s[( s + 3) 2 + 4 ]

12. (A) p( t) = e - t cos t u( t)

¥

s+3 ( s + 3) 2 + 4

0

ò

5. (C) X ( s) = ò

Þ

X ( s) = Q( s) V ( s)

2 (2 - s )

0

x( t) =

e- ( s + 2 ) s2

e X ( s) = s+2

Þ

t

4. (C) X ( s) = ò x( t) e - st dt = ò e 2 t u( -t + 2) e - st dt e

V ( s) =

-2 ( s + 1 )

-¥

= ò e t ( 2 - s ) dt =

1 s+2

L 11. (B) p( t) = e -3t cos 2 t u( t) ¬¾ ® P ( s) =

¥

2

R( s) =

P ( s) =

Q( s) =

e- s s2

1 s

17. (C) X ( s) = =

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

A B C + + ( s + 2) ( s + 1) ( s + 1) 2

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Page 275

GATE EC BY RK Kanodia

UNIT 5

A = ( s + 2) X ( s) s = -2 = 1, C = ( s + 1) 2 X ( s) A + B =1 x( t) = [ e

-2 t

Þ

B =0

æsö L 23. (C) P ç ÷ ¬¾ ® ap( at) èaø 1 1 -t L ¬¾ ® e sin 2 t u( t) ( s + 1) 2 + 4 2

= -2

s = -1

-t

- te ]u( t) 3s + 2 3( s + 1) 1 = 2 2 ( s + 1) 2 + 32 s + 2 s + 10 ( s + 1) + 3

18. (A) X ( s) =

2

1 é ù x( t) = ê3e - t cos 3t - e - t sin 3t ú u( t) 3 ë û

Q( s) =

Þ

A = ( s + 2) X ( s) s = -2 = 2 Þ

Þ

C = -2

Þ

x( t) = [2 e -2 t + 2 e - t cos 2 t - e - t sin 2 t ]u( t)

A = ( s + 2) X ( s) s = -2 = 1, A + B = 3 x( t) = [ e

-2 t

+ 2e

-3t

21. (D) X ( s) =

Þ

Þ

-3t

sin t ]u( t)

( s + 2)(2 s 2 + 11s + 16) =2 ( s2 + 5 s + 6) s = -2

B=

( s + 3)(2 s + 11s + 16) = -1 ( s2 + 5 s + 6) s = -3

R( s) = sQ( s)

L ¬¾ ®

e-2 t x( t)

X ( s) s

L ¬¾ ®

p( t) =

ò x( t) dt

-¥ t

ò cos 2 t u( t) dt =

P ( s) s

1 sin 3t u( t) 3

sin 2 t 1 - cos 2 t dt = u( t), 2 4 0

ò

L ¬¾ ®

31. (A) p( t) =

q( t) = - p( t)

32. (A) e - t x ( t)

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L ¬¾ ®

d x( t) dt

y( t) = x( t) * p( t)

é2 t ù x( t) = v( t) + r ( t) = ê sin 3t u( t) + t 2 cos 3t u( t) + e -3t ú u( t) ë3 û

p( t) = x( t - 3)

= -t cos 2( t - 3) u( t - 3).

t2 sin 3t u( t) 3

2t sin 3t u( t) + t 2 cos 3t u( t) 3 1 L V ( s) = ¬¾ ® v( t) = e -3t u( t) s+3

L ¬¾ ®

= cos 2( t - 3) u( t - 3) d L Q( s) = P ( s) ¬¾ ® ds

30. (A) x( t - 2)

d q( t) - q(0 - ) dt

sin 2 t 2

t

29. (C) P ( s) = e -3s X ( s)

q( t) = ( -1) 2 t 2 p( t) =

r ( t) =

t

L ¬¾ ®

-¥

=

Page 276

L ¬¾ ®

L ¬¾ ®

x( t) = 2 d( t) + [2 e -2 t - e-3t ]u( t) + [ e -2 ( t - 2 ) - e -3( t - 2 ) ]u( t - 2)

L ¬¾ ®

ax( at)

1 æ2 ö cos ç t ÷ u( t) 3 è3 ø

L ¬¾ ®

28. (B) P ( s) =

2

d2 P ( s) ds 2

L ¬¾ ®

2 s 2 + 11s + 16 + e -2 s ( s2 + 5 s + 6)

A=

Q( s) =

dx( t) + x( t) dt

L ¬¾ ®

C = -6

cos t - 6 e

1 s2 + 9

x( t) = q( t - 2)

y( t) = ( -2 sin 2 t + cos 2 t ) u( t)

27. (D) X ( s + 2)

B =2

A B e -2 s e -2 s =2 + + + ( s + 2) ( s + 3) ( s + 2) ( s + 3)

22. (C) P ( s) =

L ¬¾ ®

x( t) = - ( t - 2) e ( t - 2 ) u( t - 2)

X ( 3s)

A B( s + 3) C = + + ( s + 2) ( s + 3) 2 + 1 ( s + 3) 2 + 1

q( t) = -tp( t) = -t 2 e - t u( t)

L ¬¾ ®

æsö 26. (B) X ç ÷ èaø

3s 2 + 10 s + 10 20. (B) X ( s) = ( s + 2)( s 2 + 6 s + 10)

10 A + 6 B + 2 C = 10

d P ( s) ds

p( t) = te - t u( t)

L ¬¾ ®

25. (A) sX ( s) + X ( s)

B =2

5 A + 2 B + 2 C = 10

1 ( s + 1) 2

X ( s) = e -2 sQ( s)

A B( s + 1) C = + + 2 2 ( s + 2) ( s + 1) + 2 ( s + 1) 2 + 2 2 A + B=4

1 -0 .5t e sin t u( t) 4

L ¬¾ ®

x( t)

24. (C) P ( s) =

4 s 2 + 8 s + 10 ( s + 2)( s 2 + 2 s + 5)

19. (C) X ( s) =

Signal & System

e -2 s X ( s), Y ( s) =

L ¬¾ ®

L ¬¾ ®

L ¬¾ ®

2 se -2 s s2 + 2

P ( s) = sX ( s)

Y ( s) = P ( s) X ( s) = s( X ( s)) 2 X ( s + 1) =

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

GATE EC BY RK Kanodia

The Laplace Transform

33. (B) 2 tx( t)

L ¬¾ ®

¥

ò x( t) e

34. (A) X ( s) =

-2

- st

d 4 s2 - 8 X ( s) = 2 ds ( s + 2) 2

Chap 5.3

42. (A) x( t) = e -2 e t + 2 sin (2 t + 4) u( t + 2) p( t + 2) X ( s) =

dt

L ¬¾ ®

e 2 s P ( s),

e 2 s e -2 , ( s - 1) 2 + 4

Re ( s) > 1

-¥ ¥

¥

-2

-2

= ò e - t e - st dt = ò e - t ( s + 1 ) dt =

e2 ( s + 1 ) , Re( s) > - 1 s+1

¥

-3s

-e 35. (B) X ( s) = ò u( -t + 3) e - st dt = ò e - st dt = s -¥ -¥ 36. (C) Y ( s) =

3

¥

ò d( t + 1) e

- st

dt = e s ,

43. (A) p( t) = e -2 t u ( -t) q( t) = Re ( s) < 0

( e - e ) - st e dt 2j 0

37. (B) X ( s) = ò ¥

X ( s) = e 5s P ( s)

¥

1 1 1 , e t ( j - s ) dt e - t ( j + s ) dt = = 2 j ò0 2 j ò0 1 + s2 ¥

-

¥

t 2

38. (D) X ( s) = ò e e - st + ò e - t e - st dt + 0

=

0

Þ

Re ( s) > 0

òee

t - st

dt

-¥

òe

-¥

¥

X ( s) = -

0.5 < Re ( s) < 1

q( t) = p( t + 3) X ( s) =

e 3( s -1 ) , s-3

¬¾® L

L ¬¾ ®

Re ( s) > 3

41. (B) p( t) * q( t)

d2 P ( s) ds 2

1 R( s) s

u( -( t + 5))

p( t) = e 3t u( t)

L ¬¾ ®

x( t) = t 2 e 3t u( t)

Þ

L ¬¾ ®

Re ( s) > -1, Re ( s) < 0 -1 < Re ( s) < 0

P ( s)Q( s)

p( t) = u( t)

L ¬¾ ®

r ( t) = -tq( t) = -tu( t - 3) v( t) =

t

ò r( t) dt

-¥

t

3

e 3s s-3

L ¬¾ ®

q( t) = p( t - 3) = u( t - 3)

v( t) = -ò tdt = -

X ( s) =

1 s

L ¬¾ ®

L ¬¾ ®

1 v( s) s

1 2 ( t - 9) 2

L ¬¾ ®

x( t) = -

t

1 2 ò ( t - 9) 2 -¥

9 é -1 ù x( t) = ê ( t 3 - 27) + ( t - 3) ú u( t - 3) 6 2 ë û

48. (B) X ( s) =

-s æ 1 ö X ( s) = 2 ç ÷ s + 9 è s + 1ø Þ

x( t) = p( t + 5)

47. (C) Right-sided, P ( s) =

Þ

1 P ( s) = s-3 Q( s) = e 3s P ( s) =

-

V ( s) =

40. (C) x( t) = e -3e -3( t + 3) u( t + 3) p( t) = e u( t)

-2 ( t + 5)

Q( s) = e-3s P ( s) d R( s) = Q( s) ds

s -1 1 1 + + ( s - 1) 2 + 4 ( s + 1) s - 0.5

3t

L ¬¾ ®

t 2

( e - e ) - st e dt + ò e - t e- st dt + ò e e - st dt 2j 0 0

Re ( s) < 1, Re ( s) > -1, Re ( s) > 0.5 Therefore

p( t) = -e -2 t u( -t)

x( t) = -u( -t) + u ( -t + 1) + d( t + 2) ¥

- jt

jt

t

1-s 1+ s

46. (D) Left-sided

-0.5 < Re ( s) < 1

39. (A) X ( s) =

x( t) = - e

X ( s) =

Re ( s) > -0.5, Re ( s) > -1, Re ( s) < 1

0

X ( s) = Q( s - 1) =

45. (A) Right-sided 1 L P ( s) = ¬¾ ® ( s - 3)

0

1 1 1 + s + 0.5 s + 1 s - 1

Þ

L ¬¾ ®

-1 , Re ( s) < - 2 s+2

Q( s) = sP ( s)

44. (C) Left-sided 1 L P ( s) = ¬¾ ® s+2

- jt

jt

x( t) = e t q( t)

L ¬¾ ®

P ( s) =

Re ( s) < - 1 thus left-sided .

All s

-¥

¥

d p( t) dt

L ¬¾ ®

-s - 4 -3 2 = + s 2 + 3s + 2 ( s + 1) s + 2

Left-sided, x( t) = 3e - t u( -t) - 2 e -2 t u( -t) 49. (A) X ( s) = Left-sided, www.gatehelp.com

5 1 ( s + 1) ( s + 1) 2

x( t) = -5 u( -t) + te - t u( -t) Page 277

GATE EC BY RK Kanodia

UNIT 5

50. (D) x(0 + ) = lim sX ( s) = x ®¥

h( t) = (2 e - t - 3te - t ) u( t).

s =0 s + 5s - 2 2

61. (A) H ( s) =

s2 + 2 s 51. (A) x(0 + ) = lim sX ( s) = 2 =1 s ®¥ s + 2s - 3 52. (D) x(0 + ) = lim sX ( s) = s ®¥

53. (A) x( ¥) = lim sX ( s) = s ®0

e

-2 s

h( t) = - e - t u( t) + (2 e t cos 3t + et sin 3t) u( -t)

(6s + s ) =0 s2 + 2 s - 2 3

s ®0

2

62. (A) X ( s) =

2 s3 + 3s =0 s2 + 5 s + 1

H ( s) = =1 -

1 , s+1

Y ( s) =

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

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

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

h( t) = d( t) - ( e-2 t cos t + e-2 t sin t) u( t)

e -3s (2 s2 + 1) 1 = s2 + 5 s + 4 4

63. (B) sY ( s) + 10 Y ( s) = 10 X ( s) H ( s) =

-

56. (C) sY ( s) - y(0 ) + 10 Y ( s) = 10( s)

Þ

1 y(0 ) = 1, X ( s) = s 10 1 1 Y ( s) = + = s( s + 1) ( s + 1) s -

Þ

-1 2( s - 1) 3 + + ( s + 1) ( s - 1) 2 + 32 ( s - 1) 2 + 32

System is stable

s+2 54. (C) x( ¥) = lim sX ( s) = 2 =2 s ®0 s + 3s + 1 55. (B) x( ¥) = lim sX ( s) =

Signal & System

Y ( s) 10 = X ( s) s + 10

h( t) = 10 e-10 t u( t)

64. (B) Y ( s)( s 2 - s - 2) = X ( s)(5 s - 4) H ( s) =

y( t) = u( t)

Y ( s) 5s - 4 3 2 = = + X ( s) s 2 - s - 2 s + 1 s - 2

h( t) = 3e - t u( t) + 2 e 2 t u( t).

57. (C) s Y ( s) - 2 s + 2 sY ( s) - 2 + 5 Y ( s) = 1 2

( s 2 + 2 s + 5) Y ( s) = 3 + 2 s 2s + 3 2( s + 1) 1 Y ( s) = 2 = + s + 2 s + 5 ( s + 1) 2 + 2 2 ( s + 1) 2 + 2 2 Þ

y( t) = 2 e - t cos t u( t) +

1 -t e sin t u( t) 2

58. (B) s 3Y ( s) + 4 s 2 Y ( s) + 3sY ( s) = Y ( s) =

***********

10 ( s + 2)

10 A B C D = + + + s( s + 1)( s + 2)( s + 3) s ( s + 1) ( s + 2) s + 3

5 A = sY ( s) s = 0 = , B = ( s + 1) Y ( s) s = -1 = -5, 3 C = ( s + 2) Y ( s) s = -2 = 5, D = ( s + 3) Y ( s) s = 0 = Þ

5 é5 ù y( t) = ê - 5 e - t + 5 e -2 t + e -3t ú u( t) 3 3 ë û

59. (D) For a causal system H ( s) = 2 + Þ

h( t) = 0

for t < 0

1 1 + s + 1 s -1

h( t) = 2d( t) + ( e- t + e t ) u( t)

60. (D) H ( s) = Page 278

5 3

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CHAPTER

5.4 THE Z-TRANSFORM

Statement forQ.1-12: Determine the z-transform and choose correct option. 1. x[ n] = d[ n - k] , k > 0 k

(A) z , k

(C) z ,

z >0 z ¹0

(C) (B) z

-k

,

z >0

(D) z

-k

,

z ¹0

(A) z - k , z ¹ 0

(B) z k , z ¹ 0

(C) z - k , all z

(D) z k , all z

(C)

z , |z|< 1 1 - z -1

(B) (D)

1 , |z|< 1 1 - z -1 z , |z|> 1 1 - z -1

(A)

z - 0.25 , z > 0.25 z 4 ( z - 0.25)

(B)

z - 0.25 , z >0 z 4 ( z - 0.25)

(C)

z 5 - 0.25 5 , z < 0.25 z 3( z - 0.25)

(D)

z 5 - 0.25 5 , all z z 4 ( z - 0.25)

5

(B)

-5 z 2 3 , <|z|< (2 z - 3)( 3z - 2) 3 2

(C)

5z 2 2 , <|z |< (2 z - 3) ( 3z - 2) 3 3

(D)

5z 3 2 , - <(2 z - 3)( 3z - 2) 2 3

(A)

4z 1 , |z|> 4z - 1 4

(B)

4z 1 , |z|< 4z - 1 4

(C)

1 1 , |z|> 1 - 4z 4

(D)

1 1 , |z|< 1 - 4z 4

n

æ1ö æ1ö 8. x[ n] = ç ÷ u[ n] + ç ÷ u[ -n - 1] è2 ø è4ø (A)

(B)

4

æ1ö 5. x[ n] = ç ÷ u[ -n] è4ø

3 , |z|< 3 3-z

-5 z 3 2 , - <(2 z - 3)( 3z - 2) 2 3

n

5

(D)

(A)

n

5

z , |z|< 3 3-z

|n|

æ1ö 4. x[ n] = ç ÷ ( u[ n] - u[ n - 5 ]) è4ø 5

3 , |z|> 3 3-z

(B)

æ2 ö 7. x[ n] = ç ÷ è 3ø

2. x[ n] = d[ n + k] , k > 0

3. x[ n] = u[ n] 1 (A) , |z|> 1 1 - z -1

6. x[ n] = 3n u[ -n - 1] z (A) , |z|> 3 3-z

(C)

1 1 , 1 -1 1 -1 1- z 1- z 2 4

1 1 <|z|< 4 2

1 1 + , 1 -1 1 -1 1- z 1- z 4 2

1 1 <|z|< 4 2

1 1 1 , |z|> 1 -1 1 -1 2 1- z 1- z 2 4

(D) None of the above www.gatehelp.com

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GATE EC BY RK Kanodia

UNIT 5

33. X ( z) =

Signal & System

37. Consider three different signal

1 1 , |z|< -2 1 - 4z 4

n é æ1ö ù x1 [ n] = ê2 n - ç ÷ ú u[ n] è 2 ø úû êë

¥

(A) - å 2 2 ( k + 1 ) d[ -n - 2( k + 1)] k=0 ¥

(B) - å 2 2 ( k + 1 ) d[ -n + 2( k + 1)]

x2 [ n] = -2 n u[ -n - 1] +

1 u[ -n - 1] 2n

x3[ n] = - 2 n u[ -n - 1] -

1 u[ n] 2n

k=0 ¥

(C) - å 2 2 ( k + 1 ) d[ n + 2( k + 1)] k=0 ¥

(D) - å 2 2 ( k + 1 ) d[ n - 2( k + 1)]

Fig. P.5.4.37 shows the three different region.

k=0

Choose the correct option for the ROC of signal

34. X ( z) = ln (1 + z -1 ) , |z|> 0 ( -1) (A) k

k -1

Im

( -1) (B) k

d[ n - 1]

( -1) k (C) d[ n - 1] k

k -1

R1 z - plane

d[ n + 1]

R2 R3 2

( -1) k (D) d[ n + 1] k

Re 1 2

35. If z-transform is given by X ( z) = cos ( z -3), |z|> 0,

Fig. P5.4.37

R1

R2

R3

(A)

x1 [ n]

x2 [ n ]

x3[ n]

(B)

x2 [ n ]

x3[ n]

x1 [ n]

(C)

x1 [ n]

x3[ n]

x2 [ n ]

(D)

x3[ n]

x2 [ n ]

x1 [ n]

The value of x[12 ] is (A) -

1 24

(B)

(C) -

1 6

(D)

1 24 1 6

36. X ( z) of a system is specified by a pole zero pattern in fig. P.5.4.36.

38. Given 7 -1 z 6 X ( z) = 1 -1 öæ 1 -1 ö æ ç 1 - z ÷ç 1 + z ÷ 2 3 è øè ø

Im

1+

z - plane

1 3

Re 2

For three different ROC consider there different Fig. P.5.4.36

solution of signal x[ n] :

Consider three different solution of x[ n]

(a) |z|>

n é æ1ö ù x1 [ n] = ê2 n - ç ÷ ú u[ n] è 3 ø úû êë

n é 1 1 æ -1 ö ù , x[ n] = ê n -1 - ç ÷ ú u[ n] 2 è 3 ø úû êë2

(b) |z|<

é -1 æ -1 ö 1 , x[ n] = ê n -1 + ç ÷ 3 è 3 ø êë2

1 u[ n] 3n 1 x3[ n] = - 2 n u[ n - 1] + n u[ -n - 1] 3 Correct solution is x2 [ n] = - 2 n u[ n - 1] -

Page 282

n

ù ú u[ -n + 1] úû n

(c)

1 1 1 æ -1 ö <|z |< , x[ n] = - n -1 u[ -n - 1] - ç ÷ u[ n] 3 2 2 è 3 ø

Correct solutions are

(A) x1 [ n]

(B) x2 [ n]

(A) (a) and (b)

(B) (a) and (c)

(C) x3[ n]

(D) All three

(C) (b) and (c)

(D) (a), (b), (c)

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The z-Transform

39. X ( z) has poles at z = 1 2 and z = -1. If

x [1] = 1

Chap 5.4

44. The transfer function of a causal system is given as

x [ -1] = 1, and the ROC includes the point z = 3 4. The

H ( z) =

time signal x[ n] is (A)

1 u[ n] - ( -1) n u[ -n - 1] 2 n -1

The impulse response is (A) ( 3n + ( -1) n 2 n + 1 ) u[ n]

1 (B) n u[ n] - ( -1) n u[ -n - 1] 2 (C) (D)

1 2 n -1

(B) ( 3n + 1 + 2 ( -2) n ) u[ n] (C) ( 3n -1 + ( -1) n 2 n + 1 ) u[ n]

u[ n] + u[ -n + 1]

(D) ( 3n -1 - ( -2) n + 1 ) u[ n]

1 u[ n] + u[ -n + 1] 2n

45. A causal system has input

40. x[ n] is right-sided, X ( z) has a signal pole, and x[0 ] = 2, x[2 ] = 1 2. x[ n] is u[ -n] (A) n -1 2 (C)

u[ n] (B) n -1 2

u[ -n] 2n +1

as 3 -1 z 2

(1 - 2 z -1 ) (1 +

1 1 d[ n - 1] - d[ n - 2 ] and output 4 8

y[ n] = d[ n] -

3 d[ n - 1] . 4

(A)

n n 1 é æ -1 ö æ1ö ù 5 2 ÷ ç ÷ ú u[ n] ê ç 3 êë è 2 ø è 4 ø úû

(B)

1 é æ1ö æ -1 ö ÷ ê5ç ÷ + 2ç 3 êë è 2 ø è 4 ø

(C)

1 3

(D)

1 é æ1ö æ1ö ê5ç ÷ + 2ç ÷ 3 êë è 2 ø è4ø

41. The z-transform function of a stable system is given

2-

x[ n] = d[ n] +

The impulse response of this system is

u[ -n] 2n +1

(D)

H ( z) =

1 -1 z ) 2

n

n

æ1ö (A) 2 n u[ -n + 1] - ç ÷ u[ n] è2 ø

ù ú u[ n] úû

n

ù ú u[ n] úû

46. A causal system has input x[ n] = ( -3) n u[ n] and

n

æ -1 ö (B) 2 n u[ -n - 1] + ç ÷ u[ n] è 2 ø

output n é æ1ö ù y[ n] = ê4(2) n - ç ÷ ú u[ n]. è 2 ø úû êë

n

æ -1 ö (C) -2 n u[ -n - 1] + ç ÷ u[ n] è 2 ø

The impulse response of this system is

n

æ1ö (D) 2 n u[ n] - ç ÷ u[ n] è2 ø Let

n

n é æ 1 ön æ -1 ö ù 5 2 ç ÷ ç ÷ ê ú u[ n] è 4 ø úû êë è 2 ø n

The impuse response h[ n] is

42.

5z2 z2 - z - 6

x[ n] = d[ n - 2 ] + d[ n + 2 ].

The

unilateral

z-transform is (A) z -2

(B) z 2

(C) -2 z -2

(D) 2 z 2

n é æ 1 ön æ1ö ù (A) ê7ç ÷ - 10ç ÷ ú u[ n] è 2 ø úû êë è 2 ø

n é æ1ö ù (B) ê7(2 n ) - 10ç ÷ ú u[ n] è 2 ø úû êë

é æ 1 ö2 ù (C) ê10ç ÷ - 7(2) n ú u[ n] êë è 2 ø úû

n é æ1ö ù (D) ê10 (2 n ) - 7ç ÷ ú u[ n] è 2 ø úû êë

47. A system has impulse response h[ n] =

43. The unilateral z-transform of signal x[ n] = u[ n + 4 ] is (A) 1 + z + z 2 + 3z + z 4

(B)

1 1-z

(C) 1 + z -1 + z -2 + z -3 + z -4

(D)

1 1 - z -1

1 u[ n] 2n

The output y[ n] to the input x[ n] is given by y[ n] = 2 d[ n - 4 ]. The input x[ n] is (A) 2 d[ -n - 4 ] - d[ -n - 5 ]

(B) 2 d[ n + 4 ] - d[ n + 5 ]

(C) 2 d[ -n + 4 ] - d[ -n + 5 ]

(D) 2 d[ n - 4 ] - d[ n - 5 ]

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GATE EC BY RK Kanodia

UNIT 5

48. A system is described by the difference equation y[ n] -

1 2

n -2

53. The z-transform of a signal x[ n] is given by

1 y[ n - 1] = 2 x[ n - 1] 2

X ( z) =

The impulse response of the system is -1 1 (B) n - 2 u[ n + 1] (A) n - 2 u[ n - 1] 2 2 (C)

Signal & System

u[ n - 2 ]

(D)

If X ( z) converges on the unit circle, x[ n] is 1

(A) -

-1 u[ n - 2 ] 2n -2

(B) 49. A system is described by the difference equation (C)

y[ n] = x[ n] - x[ n - 2 ] + x[ n - 4 ] - x[ n - 6 ]

3n -1 8 1

3n -1 8 1 3n -1 8

The impulse response of system is (D) -

(A) d[ n] - 2 d[ n + 2 ] + 4 d[ n + 4 ] - 6 d[ n + 6 ] (B) d[ n] + 2 d[ n - 2 ] - 4 d[ n - 4 ] + 6 d[ n - 6 ]

u[ n] -

3n + 3 u[ -n - 1] 8

u[ n] -

3n + 3 u[ -n] 8

1 n -1

3

3n + 3 u[ -n - 1] 8

u[ n] -

8

u[ n] -

3n + 3 u[ -n] 8

54. The transfer function of a system is given as

(C) d[ n] - d[ n - 2 ] + d[ n - 4 ] - d[ n - 6 ] (D) d[ n] - d[ n + 2 ] + d[ n + 4 ] - d[ n + 6 ]

H ( z) =

50. The impulse response of a system is given by h[ n] =

3 u[ n - 1]. 4n

4 z -1 1 -1 ö æ ç1 - z ÷ 4 è ø

2

, |z|>

1 4

The h[ n] is

The difference equation representation for this system is (A) 4 y[ n] - y[ n - 1] = 3 x[ n - 1]

(A) Stable

(B) Causal

(C) Stable and Causal

(D) None of the above

55. The transfer function of a system is given as

(B) 4 y[ n] - y[ n + 1] = 3 x[ n + 1]

1ö æ 2ç z + ÷ 2ø è . H ( z) = 1 öæ 1ö æ ç z - ÷ç z - ÷ 2 øè 3ø è

(C) 4 y[ n] + y[ n - 1] = - 3 x[ n - 1] (D) 4 y[ n] + y[ n + 1] = 3 x[ n + 1]

Consider the two statements

51. The impulse response of a system is given by

Statement(i) : System is causal and stable.

h[ n] = d[ n] - d[ n - 5 ] The difference equation representation for this

Statement(ii) : Inverse system is causal and stable.

system is

The correct option is

(A) y[ n] = x[ n] - x[ n - 5 ]

(B) y[ n] = x[ n] - x[ n + 5 ]

(A) (i) is true

(C) y[ n] = x[ n] + 5 x[ n - 5 ]

(D) y[ n] = x[ n] - 5 x[ n + 5 ]

(B) (ii) is true

52. The transfer function of a system is given by H ( z) =

z( 3z - 2) 1 z2 - z 4

(C) Both (i) and (ii) are true (D) Both are false 56. The impulse response of a system is given by n

(A) Causal and Stable (B) Causal, Stable and minimum phase (C) Minimum phase (D) None of the above

n

æ -1 ö æ -1 ö h[ n] = 10ç ÷ u[ n] - 9ç ÷ u[ n] è 2 ø è 4 ø

The system is

Page 284

3 10 -1 1z + z -2 3

For this system two statement are Statement (i): System is causal and stable Statement (ii): Inverse system is causal and stable.

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The z-Transform

Chap 5.4

62. The impulse response of the system shown in fig.

The correct option is (A) (i) is true

(B) (ii) is true

(C) Both are true

(D) Both are false

P5.4.62 is X(z)

Y(z) z-1

+

z-1

57. The system

z-1

y[ n] = cy[ n - 1] - 0.12 y[ n - 2 ] + x[ n - 1] + x[ n - 2 ]

Fig. P5.4.62

is stable if (A) c < 112 .

(B) c > 112 .

(C) |c|< 112 .

(D) |c|> 112 .

æn ö çç - 2 ÷÷ ø

(A) 2 è 2 (B)

58. Consider the following three systems

y2 [ n] = x[ n] - 0.1 x[ n - 1]

(C) 2 è 2 (D)

y3[ n] = 0.5 y[ n - 1] + 0.4 x[ n] - 0.3 x[ n - 1]

(B) y2 [ n] and y3[ n]

(C) y3[ n] and y1 [ n]

(D) all

1 d[ n] 2

2n 1 [1 + ( -1) n ] u [ n] - d [ n] 2 2

H ( z) =

z z2 + z + 1

is shown in fig. P5.4.63. This system diagram is a X(z)

59. The z-transform of a causal system is given as X ( z) =

(1 + ( -1) n ) u[ n] -

63. The system diagram for the transfer function

The equivalent system are (A) y1 [ n] and y2 [ n]

1 d[ n] 2

2n 1 (1 + ( -1) n ) u[ n] + d[ n] 2 2 æn ö çç - 2 ÷÷ ø

y1 [ n] = 0.2 y[ n - 1] + x[ n] - 0.3 x[ n - 1] + 0.02 x[ n - 2 ]

(1 + ( -1) n ) u[ n] +

. z -1 2 - 15 -1 . z + 0.5 z -2 1 - 15

Y(z)

+

+

+

The x[0 ] is

z-1

z-1

(A) -15 .

(B) 2

(C) 1.5

(D) 0

Fig. P5.4.63

(A) Correct solution 60. The z-transform of a anti causal system is X ( z) =

12 - 21z 3 - 7 z + 12 z 2

The value of x[0 ] is 7 (A) 4

(B) Not correct solution (C) Correct and unique solution (D) Correct but not unique solution

(B) 0

(C) 4

*****************

(D) Does not exist

61. Given the z-transforms X ( z) =

z( 8 z - 7) 4z2 - 7z + 3

The limit of x[ ¥ ] is (A) 1

(B) 2

(C) ¥

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The z-Transform

z 2 - 3z 1 - 3z -1 = 3 3 z 2 + z -1 1 + z -1 - z -2 2 2 2 1 1 , ROC : <|z|< 2 = 1 + 2 z -1 1 - 1 z -1 2 2 1 Þ x[ n] = -2(2) n u[ -n - 1] - n u[ n] 2

Chap 5.4

ìæ 1 ö 2 ï , n even and n ³ 0 = í çè 4 ÷ø ï n odd î 0 , n

24. (A) X ( z) =

ì2-n , =í î 0,

n even and n ³ 0 n odd ¥

¥

k=0

k=0

33. (C) X ( z) = -4 z 2 å (2 z) 2 k = - å 2 2 ( k + 1 ) z 2 ( k + 1 )

25. (A) x[ n] is right sided 1 49 47 z - z -1 4 32 32 = + X ( z) = 1 - 16 z -1 1 + 4 z -1 1 - 4 z -1 47 n ù é 49 Þ x[ n] = ê ( -4) n + 4 ú u[ n] 32 ë 32 û

Þ

¥

x[ n] = - å 2 2 ( k + 1 ) d[ n + 2 ( k + 1)] k=0

34. (A) ln (1 + a) =

( -1) k -1 ( a) k k k =1 ¥

å

( -1) k -1 ( z -1 ) k k k =1 ¥

X ( z) = å

26. (C) x[ n] is right sided

( -1) k -1 d[ n - 1] k k =1 ¥

x [ n] = å

1 -1 ö 2 æ X ( z) = ç 2 + + ÷z -1 1+ z 1 - z -1 ø è

Þ

Þ

35. (B) cos a = å

x[ n] = 2 d[ n + 2 ] + (( -1) n - 1) u[ n + 2 ]

( -1) k 2 k a k = 0 (2 k) ! ¥

27. (A) d[ n] + 2 d[ n - 6 ] + 4 d[ n - 8 ]

( -1) k -3k 2 k (z ) k = 0 (2 k) ! ¥

X ( z) = å 10

1

å k d[ n - k]

28. (B) x[ n] is right sided, x[ n] =

Þ

k=5

x [ n] =

( -1) k

¥

å (2 k) ! d [ n - 6 k]

k=0

n = 12

29. (D) x[ n] is right sided signal X ( z) = 1 + 3z Þ

-1

+ 3z

-2

+z

x[ n] = d[ n] + 3d[ n - 1] + 3d[ n - 2 ] + d[ n - 3]

12 - 6 k = 0, k = 2,

36. (D) All gives the same z transform with different

30. (A) x[ n] = d[ n + 6 ] + d[ n + 2 ] + 3d[ n] + 2 d [ n - 3] + d[ n - 4 ] 31. (B) X ( z) = 1 + z +

z2 z3 + + ......... 2 ! 3! 1 1 1 ......... +1 + + + 2 z 2!z 3! z3

d[ n + 2 ] d[ n + 3] x[ n] = d[ n] + d[ n + 1] + + ...... 2! 3! d[ n - 2 ] d[ n - 3] +d[ n] + d[ n - 1] + + ......... 2! 3! 1 x [ n] = d [ n] + n! 32. (A) X ( z) = 1 + Þ

Þ

( -1) 2 1 x[12 ] = = 4! 24

-3

¥

k

-2

-2

æz z + çç 4 è 4

æ1ö x [ n] = å ç ÷ d[ n - 2 k] k=0 è 4 ø

2

ö æ1 ö ÷÷ = å ç z -2 ÷ k=0 è 4 ø ø ¥

k

ROC. So all are the solution. 37. (C) x1 [ n] is right-sided signal z1 > 2 , z1 >

1 gives z1 > 2 2

x2 [ n] is left-sided signal 1 1 z 2 < 2, z 2 < gives z 2 < 2 2 x3[ n] is double sided signal 1 1 z 3 > and z 3 < 2 gives < z3 < 2 2 2 38. (B) X ( z) =

2 -1 , + 1 -1 1 1- z 1 + z -1 2 z n

2 æ -1 ö u[ n] - ç ÷ u[ n] 2n è 3 ø

|z|>

1 (Right-sided) Þ 2

x[ n] =

|z|<

1 (Left-sided) Þ 3

é -2 æ -1 ön ù x[ n] = ê n + ç ÷ ú u[ -n - 1] è 3 ø úû êë 2

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UNIT 5

n

1 1 2 æ -1 ö <|z|< (Two-sided) x[ n] = - n u[ -n - 1] - ç ÷ u[ n] 3 2 2 è 3 ø So (b) is wrong. 3 39. (A) Since the ROC includes the z = , ROC is 4 1 <|z|< 1, 2 A B X ( z) = + 1 -1 1 + z -1 1- z 2 A Þ x[ n] = n u[ n] × B ( -1) n u[ -n - 1] 2 A 1= Þ A =2 , 2

Signal & System

-2 5 Y ( z) 3 3 , H ( z) = = + X ( z ) 1 - 1 z -1 1 + 1 z -1 4 2 Þ

h[ n] =

n n 1 é æ -1 ö æ1ö ù 5 2 ç ÷ ç ÷ ê ú u[ n] 3 êë è 2 ø è 4 ø úû

Y ( z) =

H ( z) =

4 1 3 = -1 1 1 1 - 2z æ ö 1 - z -1 (1 - 2 z -1 )ç 1 - z -1 ÷ 2 2 è ø 10 Y ( z) -7 = + X ( z ) 1 - 2 z -1 1 - 1 z -1 2

x[ -1] = 1 = ( -1) B( -1) Þ B = 1 1 Þ x[ n] = n -1 u[ n] - ( -1) n u[ -n - 1] 2

Þ

40. (B) x[ n] = n u[ n] , x[0 ] = 2 = C

47. (D) H ( z) =

x[2 ] =

1 = 2 p2 2

Þ

p=

X ( z) =

n

41. (B) H ( z) =

Þ

1 1 + -1 1 1 - 2z 1 + z -1 2

n

æ -1 ö h[ n] = (2) u[ -n - 1] + ç ÷ u [ n] è 2 ø ¥

-n

Y ( z) = 2 z -4 - z -5 H ( z)

x[ n] = 2[ d - 4 ] - d[ n - 5 ]

H ( z) =

Þ

n

å x[ n]z

= d[ n - 2 ]z - n = z -2

Y ( z) 2 z -1 = z -1 X ( z) 12

æ1ö h[ n] = 2ç ÷ è2 ø

49. (C) H ( z) =

n =0

+

43. (D) X ( z) =

¥

å x[ n]z

n =0

-n

=

¥

åz

n =0

Þ -n

1 = 1 - z -1

3 2 44. (B) H ( z) = + -1 1 - 3z 1 + 2 z -1 h[ n] is causal so ROC is |z|> 3, Þ

h[ n] = [ 3n + 1 + 2 ( -2) n ]u[ n]

45. (A) X ( z) = 1 +

1 , Y ( z ) = 2 z -4 1 -1 1- z 2

é z -1 ù -1 48. (A) Y ( z) ê1 ú = 2 z X ( z) 2 ë û

h[ n] is stable, so ROC includes |z|= 1 1 ROC : <|z|< 2 , 2

42. (A) X + ( z) =

n é æ1ö ù h[ n] = ê10(2) n - 7ç ÷ ú u( n) è 2 ø úû êë

1 , 2

æ1ö x[ n] = 2ç ÷ u( n) è2 ø

Page 288

1 1 + 3z -1

46. (D) X ( z) =

z -1 z -2 3z -1 , Y ( z) = 1 4 8 4

n -1

u[ n - 1] Y ( z) = (1 - z -2 + z -4 - z -6 ) X ( z)

h[ n] = d[ n] - d[ n - 2 ] + d[ n - 4 ] - d[ n - 6 ]

50. (A) h[ n] =

3æ1ö ç ÷ 4è4ø

n -1

u[ n - 1]

3 -1 z Y ( z) 4 H ( z) = = X ( z ) 1 - 1 z -1 4 1 3 Y ( z ) - z -1 Y ( z) = z -1 X ( z) 4 4 1 3 Þ y[ n] - y[ n - 1] = x[ n - 1] 4 4

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The z-Transform

51. (A) H ( z) = Þ

Y ( z) = 1 - z -5 X ( z)

So y1 and y2 are equivalent. 59. (B) Causal signal x[0 ] = lim X ( z) = 2

y[ n] = x[ n] - x[ n - 5 ]

52. (D) Zero at : z = 0,

Chap 5.4

z ®¥

1± 2 2 , poles at z = 2 3

60. (C) Anti causal signal, x[0 ] = lim X ( z) = 4 z ®¥

(i) Not all poles are inside |z|= 1, the system is not causal and stable.

3 61. (A) The function has poles at z = 1, . Thus final 4

(ii) Not are poles and zero are inside |z|= 1, the system

value theorem applies.

is not minimum phase.

7 ) 4 =1 lim x( n) = lim( z - 1) X ( z) = ( z - 1) n ®¥ z ®1 3ö æ ( z - 1) ç z - ÷ 4ø è z(2 z -

3 27 8 8 53. (A) X ( z) = + 1 -1 1 - 3z -1 1- z 3 -

62. (C) [2 Y ( z) + X ( z)] z -2 = Y ( z)

Since X ( z) converges on |z|= 1. So ROC must include this circle. 1 ROC : <|z|< 3, 3

H ( z) =

Þ

3n + 3 x[ n] = - n -1 u[ n] u[ -n - 1] 3 8 8 1

=-

z -2 1 - 2 z -2

1 1 1 4 4 h[ n] = - + + 2 1 - 2 z -1 1 + 2 z -1 1 1 d[ n] + {( 2 ) n + ( - 2 ) n } u[ n] 2 4

n

æ1ö 54. (C) h[ n] = 16 nç ÷ u[ n]. So system is both stable and è4ø

63. (D) Y ( z) = X ( z) z -1 - { Y ( z) z -1 + Y ( z) z -2 }

causal. ROC includes z = 1.

Y ( z) z -1 z = = 2 -1 -2 X ( z) 1 + z + z z + z +1

1 1 , 2 3 1 Pole of inverse system at : z = 2

So this is a solution but not unique. Many other correct

55. (C) Pole of system at : z = -

diagrams can be drawn.

For this system and inverse system all poles are inside

***********

|z|= 1. So both system are both causal and stable. 56. (A) H ( z) =

=

10 9 1 -1 1 1+ z 1 + z -1 4 2

1 - 2 z -1 1 -1 öæ 1 -1 ö æ ç 1 + z ÷ç 1 + z ÷ 2 4 è øè ø

Pole of this system are inside |z|= 1. So the system is stable and causal. For the inverse system not all pole are inside |z|= 1. So inverse system is not stable and causal. 57. (C) |a2|= 0.12 < 1, a1 =|-c|< 1 + 0.12, |c|< 112 . 58. (A) Y1 ( z) = 1 - 0.1z -1 , Y2 ( z) = 1 - 0.1z -1 Y3( z) =

0.4 - 0.3z -1 1 - 0.5 z -1

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CHAPTER

5.6 THE DISCRETE-TIME FOURIER TRANSFORM

-n

Statement for Q.1–9: Determine the discrete-time Fourier Transform for the given signal and choose correct option. |n|£ 2

ì 1, 1. x[ n] = í î 0, sin 5W (A) sin W (C)

otherwise (B)

sin 2.5W sin W

sin 4W sin W

(C)

(

3 4

e - jW

)

4

(B)

1 - 43 e - jW 3 4

e - jW

1+

3 4

)

(

3 4

e jW

)

(B)

2 e jW 2 - e - jW

(C)

e jW 2 - e jW

(D)

2 e jW 2 - e jW

(A) 2 e - j 2 W

(B) 2 e j 2 W

(C) 1

(D) None of the above

(A) pd(W) -

1 1 + e - jW

(B)

1 1 - e - jW

(C) pd(W) +

1 1 - e - jW

(D)

1 1 + e - jW

4

1 - 43 e jW

4

(D) None of the above

e jW

8. x[ n] = {-2, - 1,

3. x[ n] = u[ n - 2 ] - u[ n - 6 ] (A) e 3 jW + e 3 jW + e 4 jW + e 5 jW (C) e -2 jW + e -3 jW + e -4 jW + e -5 jW

(B) (D)

e -2 jW(1 - e 3 jW) 1 - e jW e

-2 jW

-3 jW

(1 - e 1 - e - jW

)

0, 1, 2} ­

(A) 2 j(2 sin 2W + sin W)

(B) 2(2 cos 2W - cos W)

(C) -2 j(2 sin 2W + sin W)

(D) -2(2 cos 2W - cos W)

æp ö 9. x[ n] = sin ç n ÷ è2 ø

4. x[ n] = a|n| , |a|< 1

(A) p( d[W - p 2 ] - d[W + p 2 ])

(A)

1-a 1 - 2 a sin W + a 2

(B)

(C)

1 - a2 1 - 2 ja sin W + a 2

(D) None of the above

2

Page 300

e jW 2 - e - jW

7. x[ n] = u[ n]

n

(

(A)

6. x[ n] = 2 d[ 4 - 2 n]

(D) None of the above

æ 3ö 2. x[ n] = ç ÷ u[ n - 4 ] è4ø (A)

æ1ö 5. x[ n] = ç ÷ u[ -n - 1] è2 ø

1-a 1 - 2 a cos W + a 2 2

(B)

j ( d[W + p 2 ] - d[W - p 2 ]) 2

(C) 2 p( d[W - p 2 ] - d[W + p 2 ]) (D) jp( d[W + p 2 ] - d[W - p 2 ])

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The Discrete-Time Fourier Transform

16. X ( e jW) = j 4 sin 4W - 1

Statement for Q.10–21: Determine

the

signal

having

the

Fourier

(C) d[ n + 4 ] - d[ n - 4 ] - d[ n]

1 , |a|< 1 (1 - ae - jW) 2

(A) ( n - 1) a n u[ n]

(D) None of the above

(B) ( n + 1) a n u[ n]

n

(D) None of the above

(C) na u[ n]

(A) (d[ n + 2 ] + 2 d[ n] + d[ n - 2 ]) (C) -4(d[ n + 2 ] + d[ n] + d[ n - 2 ]) (d[ n + 2 ] + d[ n] + d[ n - 2 ])

(C)

ì 2 j, 0 <W £ p 12. X ( e jW) = í î - 2 j, - p < W £ 0 4 æ pn ö (A) sin 2 ç ÷ pn è 2 ø (C)

2-n 5

2 + e - jW + 6

n +1 æ ö ç 1 + æç -2 ö÷ ÷u[ n] ç è 3 ø ÷ø è n +1

ö ÷u[ n] ÷ ø

n +1 æ ö ç ( -1) n + æç 2 ö÷ ÷u[ n] ç è 3 ø ÷ø è

(D) None of the above 4 æ pn ö (B) sin 2 ç ÷ pn è 2 ø

8 æ pn ö sin 2 ç ÷ pn è 2 ø

5 2-n

-e

- j2 W

æ æ -2 ö (B) 2 - n ç 1 - ç ÷ ç è 3 ø è

(B) 2(d[ n + 2 ] + 2 d[ n] + d[ n - 2 ])

1 2

17. X ( e jW) = (A)

11. X ( e jW) = 8 cos 2 w

(D)

(A) 4 pd[ n + 4 ] - 4 pd[ n - 4 ] - 2 pd[ n] (B) 2 d[ n + 4 ] - 2 d[ n - 4 ] - d[ n]

transform given in question. 10. X ( e jW) =

Chap 5.6

(D) -

18. X ( e jW) =

2+ - 18 e - j 2 W

e - jW + 14 e - jW + 1 1 4

(A) 2 - n + 1 [1 + ( -2) - n ]u[ n]

8 æ pn ö sin 2 ç ÷ pn è 2 ø

(B) 2 - n [1 + ( -2) - n ]u[ n] (C) 2 - n + 1 [( -1) n + 2 - n ]u[ n]

ì ï 1, jW 13. X ( e ) = í ï 0, î (A)

(B)

3p p £|W|< 4 4 p 3p 0 £|W|< , £|W|£ p 4 4

(D) 2 - n [( -1) n + 2 - n ]u[ n] 19. X ( e jW) =

2æ æ 3pn ö æ pn ö ö ç sin ç ÷ - sin ç ÷÷ nè è 4 ø è 4 øø

(A) 2 n -1 [1 + ( -1) n ]u[ n]

1 æ æ 3pn ö æ pn ö ö ç sin ç ÷ - sin ç ÷÷ pn è è 4 ø è 4 øø

(C) 21 - n [1 - ( -1) n ]u[ n]

(B) 21 - n [1 + ( -1) n ]u[ n] (D) 2 n -1 [1 - ( -1) n ]u[ n]

2æ æ 3pn ö æ pn ö ö (C) ç cos ç ÷ + cos ç ÷÷ nè è 4 ø è 4 øø (D)

20. X ( e jW) =

1 æ æ 3pn ö æ pn ö ö ç cos ç ÷ + cos ç ÷÷ pn è è 4 øø è 4 ø

14. X ( e jW) = e

-

æ2 (A) ç ç9 è

jW 2

for

-p £ W £ p

(A) pd[ n - 1 2 ] (C)

2 e - jW 1 - 14 e - j 2 W

( -1) n + 1 p(n - 12 )

(B) 2pd[ n - 1 2 ] (D) None of the above

1 - 13 e - jW 1 - 14 e - jW - 18 e -2 jW

n

7 æ 1ö æ1ö ç ÷ + ç- ÷ 2 9 è 4ø è ø

n

ö ÷u[ n] ÷ ø

æ 2 æ 1 ön 7 æ 1 ön ö (B) ç ç - ÷ + ç ÷ ÷u[ n] ç9 è 2 ø 9 è 4 ø ÷ø è æ 2 æ 1 ön 7 æ 1 ön ö (C) ç ç - ÷ - ç ÷ ÷u[ n] ç9 è 2 ø 9 è 4 ø ÷ø è

15. X ( e jW) = cos 2W + j sin 2W (A) 2 pd[ n + 2 ]

(B) d[ n + 2 ]

(C) 0

(D) None of the above

æ2 (D) ç ç9 è

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n

7 æ 1ö æ1ö ç ÷ - ç- ÷ 2 9 è 4ø è ø

n

ö ÷u[ n] ÷ ø Page 301

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UNIT 5

n

n

æ 1ö æ1ö x[ n] = ç - ÷ u[ n] + ç ÷ u[ n] = 2 - n [( -1) n + 2 - n ]u[ n] 2 è ø è4ø

Signal & System

¥

å x[ n] , This condition is satisfied only

27. (D) X ( e j 0 ) =

n = -¥

if the samples of the signal add up to zero. This is true 2 e - jW 2 2 19. (C) X ( e ) = = 1 - j2 W 1 - jW 1 - jW 1- e 1- e 1+ e 4 2 2

for signal (b) and (h).

jW

n

28. (A) X ( e j 0 ) =

n

æ1ö æ 1ö x[ n] = 2ç ÷ u[ n] - 2ç - ÷ u[ n] è2 ø è 2ø 1 = n -1 [1 - ( -1) n ]u[ n] = 21 - n [1 - ( -1) n ]u[ n] 2

n = -¥

29. (A) y[ n] = x[ n + 2 ] Y ( e ) is real and even. Y ( e jW) = e j 2 W X ( e jW)

Thus arg{X ( e jW)} = -2W 30. (C)

=

p

ò X (e

jW

) dW = 2 px[0 ] = 4 p

-p

¥

å ( -1)

31. (A) X ( e jp) =

2æ1ö 7æ 1ö x[ n] = ç ÷ u[ n] + ç - ÷ u[ n] 9 è2 ø 9è 4ø

e j2 W

X ( e jW) = e - j 2 WY ( e jW),

Þ

Since Y ( e jW) is real. This imply arg{ Y ( e jW)} = 0

n

21. (C) X ( e jW) =

is an even signal. Therefore

jW

1 - jW 1- e jW 3 20. (A) X ( e ) = 1 1 1 - e - jW - e -2 jW 4 8 2 7 9 9 = + 1 - jW 1 - jW 1- e 1+ e 2 4 n

¥

å x[ n] = 6

n

x[ n] = 2

n = -¥

DTFT 32. (C) Ev{x[ n]} ¬ ¾¾ ® Re{X ( e jW)}

( b - a) e jW - ( a + b) e jW + ab

Ev{x[ n] =

- jW

( b - a) e 1 -1 = + 1 - ( a + b) e - jW + abe- j 2 W 1 - be - jW 1 - ae - jW

( x[ n] + x[ -n]) 2

1ü 1 1 ì 1 , 1, 0, 0, 1, 2, 1, 0, 0, 1, , 0, - ý = í - , 0, 2 2 2 2 î þ ­

x[ n] = bn u[ n] + a n u[ -n - 1] . 22. (D) The signal must be read and odd. Only signal ( h)

33. (D)

23. (A) The signal must be real and even. Only signal (c) and (e) are real and even. 24. (A) Y ( e jW) = e jaW X ( e jW),

y[ n] = x[ n + a ]

real.). Therefore x [ n + a ] is even and x [ n] has to be symmetric about a.This is true for signal (a), (c), (e), (f) and (g).

ò X (e

2

n = -¥

DTFT ¬ ¾¾ ®

j

) dW = 2 px[0 ] ,

35. (A) Y ( e jW) = e - j 4 W X ( e jW) |n - 4|

æ 3ö y[ n] = x[ n - 4 ] = ( n - 4) ç ÷ è4ø

imaginary. Thus y[ n] = 0.

x[0 ] = 0 is for signal (c), (f), (g) and (h).

37. (D) X 2 ( e jW) = X ( e j 2 W)

is always periodic with period 2p.

X ( e j 2 W)

Therefore all signals satisfy the condition.

DTFT ¬ ¾¾ ®

ì x[ n] , n even x 2 [ n] = í otherwise î 0,

|n| ì 2æ 3 ö jn ï ç ÷ , n even y[ n] = í è4ø ï0 , otherwise î

Page 306

d X ( e jW) dW

36. (C) Since x[ n] is real and odd, X ( e jW) is purely

-p

26. (D) X ( e jW)

34. (C) nx[ n]

¥

= 2 p å | x[ n]| = 28 p

¥ 2 ½dX ( e jW)½ ½ |n|2 x[ n] = 316 p ò- p½ dW ½½ = 2 pnå = -¥

If Y ( e ) is real, then y[ n] is real and even (if x[ n] is

25. (D)

2

p

jW

jW

jW

-p

is real and odd.

p

p

ò | X ( e )|

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The Discrete-Time Fourier Transform

38. (B) Y ( e jW) = X ( e jW ) * X ( e j ( W- p 2 ) ) y[ n] = 2 px[ n] x1 [ n], x1 [ n] = e y[ n] = 2 pn2 e jpn

Þ

39. (C) Y ( e jW) =

æ 3ö ç ÷ è4ø

2

jpn 2

This is possible only if b = -a.

x[ n] , 44. (A) For x[ n] = d[ n], X ( e jW) = 1,

2|n|

h[ n] =

d X ( e jW) dW

= |n|

æ 3ö y[ n] = - jnx[ n] = - jn2 ç ÷ è4ø

Þ

41. (C) For a real signal x[ n]

)

3=

2

dW =

å | x[ n]|

2

n = -¥

x[0 ] = ± 1,

å | x[ n]|

2

= ( x [0 ]) + 2 But

x[0 ] = 0,

Hence

n

n

n =0

=

-p

n

æ1ö h2 [ n] = ç ÷ u[ n] è 3ø

¬ ¾¾® DTFT

1 2 - jW 1- e 3 d j dW

æ ç 1 ç çç 1 - 2 e - jW è 3

2 - jW ö e ÷ ÷= 3 ÷÷ 1 - 2 e - jW ø 3

2 - jW e Y ( e jW ) 3 47. (B) H ( e ) = = X ( e jW) 1 - 2 e - jW 3 2 2 æ ö Þ ç 1 - e - jW ÷ Y ( e jW) = e- jW X ( e jW) 3 3 è ø jW

x[0 ] = 1

1 æ1ö ç ÷ 1 - jW è 4 ø 1- e 4

DTFT ¬ ¾¾ ®

n

1 - jW æ ö n e ÷ d ç 1 æ1ö DTFT 4 ç ÷= nç ÷ u[ n] ¬ ¾¾® j 2 dW ç 1 - 1 e - jW ÷ æ è2 ø ç ÷ ç 1 - 1 e - jW ö÷ è 4 ø è 4 ø 1

e jWn dW

2

æ1ö 42. (C) ç ÷ u[ n] è4ø

¥

- jW

2 - jW e 2 e - jW H( e ) = 3 = - jW 2 1 - e - jW 3 - 2 e 3

x[ n] = d[ n] + d[ n + 1] - d[ n + 2 ]

å næçè 2 ö÷ø

p

òe

jW

n =- ¥

-¥

-1

¥

DTFT ¬ ¾¾ ®

n

æ2 ö nç ÷ u[ n] è 3ø

Using Parseval’s relation jW

1 2p

Y ( e jW) , X ( e jW)

n

x[ n] = 2od{ x[ n] } = d[ n + 1] - d[ n + 2 ] For n < 0

ò | X ( e )|

1 , 1 - jW 1- e 3

æ2 ö ç ÷ u[ n] è 3ø

Since x[ n] = 0 for n > 0,

¥

-p

46. (D) H ( e jW) =

Therefore od{x[ n]} = F -1 { jIm{X ( e jW)}} 1 = ( d[ n + 1] - d[ n - 1] - d[ n + 2 ] + d[ n - 2 ]) 2 x[ n] - x[ -n] Od{ x[ n]} = 2

1 2p

) e jWn dW =

1 sin p( n - 1) e jW ( n -1 ) dW = 2 p -òp p( n - 1)

H 2 ( e jW) =

jIm{X ( e jW)}= j sin W - j sin 2W, 1 jW = e - e - j W - e 2 jW + e -2 jW 2

(

jW

p

jIm{X ( e jW)}

DTFT ¬ ¾¾ ®

p

ò Y (e

-12 + 5 e - jW 1 -2 = + 1 - jW 1 12 - 7 e - jW + e - j 2 W 1- e 1 - e - jW 3 4

y[ n] = x[ n] + x[ -n] = 0

od{x[ n]}

1 2p

dX ( e jW) =0 dW

45. (B) H ( e jW) = H1 ( e jW) + H 2 ( e jW)

40. (B) Y ( e jW) = X ( e jW ) + X ( e - jW) Þ

Chap 5.6

¥

å x[ n] = X ( e

j0

)=

n = -¥

y[ n] -

Þ

3 y[ n] - 2 y[ n - 1] = 2 x[ n - 1] .

4 9

|

2 2 y[ n - 1] = x[ n - 1] 3 3

Þ

*********

|

43. (A) For all system H ( e jW) = 1 for all W H ( e jW) =

- jW

b+ e , b + e - jW = 1 - a e - jW 1 - ae- jW

|

| |

|

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CHAPTER

5.7 THE CONTINUOUS-TIME FOURIER SERIES

x(t)

Statement for Q.1-5:

A

Determine the Fourier series coefficient for given periodic signal x( t). -2p 3

1. x( t) as shown in fig. P5.7.1

Fig. P5.7.3

10

-5

0

5

10

(C) - j

t

Fig. P5.7.1

(A) 1

æp ö (B) cosç k ÷ è2 ø

æp ö (C) sinç k ÷ è2 ø

(D) 2

A æç - j ççè e 2 pk ç è

æ 4 pk ö ÷÷ 3 ø

(B) j

ö - 1÷ ÷ ø

(D)

A æç - j ççè e 2 pk ç è

æ 4 pk ö ÷÷ 3 ø

- A æç - j ççè e 2 pk ç è

æ 4 pk ö ÷÷ 3 ø

ö - 1÷ ÷ ø

ö - 1÷ ÷ ø

4. x( t) as shown in fig. P5.7.4 x(t) A 1 -1 -A

2. x( t) as shown in fig. P5.7.2

Fig. P5.7.4

x(t) A

-T

0

T 4

T 2

T

t

A (A) (1 - ( - 1 ) k ) kp A (C) (1 - ( - 1 ) k ) jkp

A (1 + ( - 1 ) k ) kp A (D) (1 + ( - 1 ) k ) jkp

(B)

5. x( t) = sin 2 t

Fig. P5.7.2

(A)

A æp ö sin ç k ÷ jpk è2 ø

(B)

A æp ö cos ç k ÷ jpk è2 ø

(A) -

1 1 1 d[ k - 1] + d[ k] - d[ k + 1] 4 2 4

(C)

A æp ö sin ç k ÷ pk è2 ø

(D)

A æp ö cos ç k ÷ pk è2 ø

(B) -

1 1 1 d [ k - 2 ] + d[ k] - d[ k + 2 ] 4 2 4

(C) -

1 1 d[ k - 1] + d[ k] - d[ k + 1] 2 2

(D) -

1 1 d[ k - 2 ] + d[ k] - d[ k + 2 ] 2 2

3. x( t) as shown in fig. P5.7.3

Page 308

t

4p

2p

æ 4 pk ö ö A æç - j ççè 3 ÷÷ø (A) - 1÷ e ÷ 2 pk ç è ø

x(t)

-10

-4p 3

0

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The Continuous-Time Fourier Series

Statement for Q.6-11:

(A)

sin 9pt sin pt

(B)

(C)

sin 18 pt 2 sin pt

(D) None of the above

In the question, the FS coefficient of time-domain signal have been given. Determine the corresponding time domain signal and choose correct option. 7. X [ k] = jd[ k - 1] - jd[ k + 1] + d[ k + 3] + d[ k - 3], wo = 2 p (A) 2(cos 3pt - sin pt)

(B) -2(cos 3pt - sin pt)

(C) 2(cos 6 pt - sin 2 pt)

(D) -2(cos 6 pt - sin 2 pt)

Chap 5.7

sin 9pt p sin pt

11. X [ k] As depicted in fig. P5.7.11, wo = p X [k] 3 2

|k|

æ -1 ö 8. X [ k] = ç ÷ , wo = 1 3 ø 4è (A) 5 + 3 sin t

5 (B) 4 + 3 sin t

5 (C) 4 + 3 cos t

4 (D) 5 + 3 cos t

1 k -6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

Fig. P5.7.11

(A) 3 cos 3pt + 2 cos 2 pt + cos pt (B) 3 sin 3pt + 2 sin 2 pt + sin pt

9. X [ k] as shown in fig. P5.7.9 , wo = p |{X [k]}|

(C) 6 sin 3pt + 4 sin 2 pt + 2 sin pt

2

(D) 6 cos 3pt + 4 cos 2 pt + 2 cos pt

1 k -6

-5

-4

-3

-2

-1

1

0

2

3

4

5

6

Statement for Q.12-16:

Ð{X [k]} p 4

Consider a continuous time periodic signal x( t) k

-p 4

with

coefficients

(C) e

-1

1

0

Determine

the

Fourier

series

2

3

4

5

- to

X [ k] + e to X [ -k]

æ 2p ö (B) 2 sin ç kto ÷ X [ k] T è ø (D) e

- to

X [ -k] + e to X [ k]

(A)

X [ k] + X [ - k] 2

(B)

X [ k] - X [ - k] 2

6

(C)

X [ k] + X *[ - k] 2

(D)

X [ k] + X *[ - k] 2

Ð{X [k]} 8p 6p 4p 2p

14. y( t) =Re{x( t)} k

-2p -4p -6p -8p

series

13. y( t) = Ev{x( t)}

k -2

Fourier

æ 2p ö (A) 2 cos ç kto ÷ X [ k] T è ø

1 -3

and

12. y( t) = x( t - to ) + x ( t - to )

|{X [k]}|

-4

T

choose correct option.

10. X [ k] As shown in fig. P5.7.10 , wo = 2 p

-5

X [ k].

period

coefficient of the signal y( t) given in question and

Fig. P5.7.9

pö pö æ æ (A) 6 cos ç 2 pt + ÷ - 3 cos ç 3pt - ÷ 4 4 è ø è ø pö pö æ æ (B) 4 cos ç 4 pt - ÷ - 2 cos ç 3pt + ÷ 4ø 4ø è è pö pö æ æ (C) 2 cos ç 2 pt + ÷ - 2 cos ç 3pt - ÷ 4ø 4ø è è pö pö æ æ (D) 4 cos ç 4 pt + ÷ + 2 cos ç 3pt - ÷ 4 4 è ø è ø

-6

fundamental

(A)

X [ k] + X [ - k] 2

(B)

X [ k] - X [ - k] 2

(C)

X [ k] + X *[ - k] 2

(D)

X [ k] + X *[ - k] 2

Fig. P5.7.10

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UNIT 5

d 2 x( t) dt 2

15. y( t) =

2

æ 2 pk ö (A) ç ÷ X [ k] è T ø 2

æ 2 pk ö (C) jç ÷ X [ k] è T ø

Signal & System

Statement for Q.20-21: Let x1 ( t) and x2 ( t) be continuous time periodic

2

æ 2 pk ö (B) - ç ÷ X [ k] è T ø

signal with fundamental frequency w1 and w2 , Fourier series coefficients X1 [ k] and X 2 [ k] respectively. Given

2

æ 2 pk ö (D) - jç ÷ X [ k] è T ø

that

x2 ( t) = x1 ( t - 1) + x1 (1 - t)

20. The relation between w1 and w2 is 16. y( t) = x( 4 t - 1) 8p (A) X [ k] T (C) e

æ 8 pö - jk çç ÷÷ èTø

(A) w2 =

4p (B) X [ k] T

X [ k]

(D) e

æ 8 pö jk çç ÷÷ èTø

w1 2

(B) w2 = w12

(C) w2 = w1

X [ k]

(D) w2 = w1

21. The Fourier coefficient X 2 [ k] will be

17. Consider a continuous-time signal

(A) ( X1 [ k] - jX1 [ -k]) e - jw 1 k

x( t) = 4 cos 100 pt sin 1000 pt 1 with fundamental period T = . The nonzero FS 50 coefficient for this function are

(B) ( X1 [ -k] - jX1 [ k]) e - jw 1 k

(A) X[ -4 ], X[ 4 ], X[ -7 ], X[7 ]

(C) ( X1 [ k] + jX1 [ -k]) e - jw 1 k (D) None of the above Statement for Q.22-23:

(B) X[ -1], X[1], X[ -10 ], X[10 ]

Consider three continuous-time periodic signals

(C) X[ -3], X[ 3], X[ -4 ], X[ 4 ]

whose Fourier series representation are as follows.

(D) X[ -9 ], X[9 ], X[ -11], X[11]

2p

k

100 æ 1 ö - jk t x1 ( t) = å ç ÷ e 50 k=0 è 3 ø

18. A real valued continuous-time signal x( t) has a fundamental period T = 8. The nonzero Fourier series

x2 ( t) =

coefficients for x( t) are

100

å cos kp e

- jk

2p t 50

k = -100

x3( t) =

X [1] = X [ -1] = 4, X [ 3] = X *[ -3] = 4 j

100

å

k = -100

2p

æ kp ö - jk 50 t j sinç ÷e è 2 ø

The signal x( t) would be æp ö æ 3p ö (A) 4 cos ç t ÷ + 4 j sin ç t÷ è4 ø è 4 ø

22. The even signals are (A) x2 ( t) only

(B) x2 ( t) and x3( t)

æp ö æ 3p ö (B) 4 cos ç t ÷ - 4 j cos ç t ÷ è4 ø è 4 ø

(C) x1 ( t) and x3( t)

(D) x1 ( t) only

pö æp ö æ 3p (C) 8 cos ç t ÷ + 8 cos ç t + ÷ 2ø è4 ø è 4

23. The real valued signals are (A) x1 ( t) and x2 ( t)

(B) x2 ( t) and x3( t)

(D) None of the above

(C) x3( t) and x1 ( t)

(D) x1 ( t) and x3( t)

19. The continuous-time periodic signal is given as æ 2p ö æ 5p ö x( t) = 4 + cos ç t ÷ + 6 sin ç t÷ è 3 ø è 3 ø The nonzero Fourier coefficients are (A) X [0 ], X [ -1], X [1], X [ -5 ], X [5 ]

24. Suppose the periodic signal x( t) has fundamental period T and Fourier coefficients X [ k]. Let Y [ k] be the Fourier coefficient of y( t) where Fourier coefficient X [ k] will be (A)

TY [ k] ,k¹0 j2 pk

(B)

TY [ k] j2pk

(C)

TY [ k] ,k¹0 jk

(D)

TY [ k] jk

(B) X [0 ], X [ -2 ], X [2 ], X [ -5 ], X [5 ] (C) X [0 ], X [ -4 ], X [ 4 ], X [ -10 ], X [10 ] (D) None of the above Page 310

y( t) = dx( t) dt . The

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The Continuous-Time Fourier Series

25. Suppose we have given the following information

The signal will be æp ö æp ö (A) 4 cos ç t ÷ - 2 sin ç t ÷ è4 ø è4 ø

about a signal x( t) : 1. x( t) is real and odd.

æp ö æp ö (B) 2 cos ç t ÷ + 4 sin ç t ÷ è4 ø è4 ø

2. x( t) is periodic with T = 2 3. Fourier coefficients X [ k] = 0 for |k|> 1

æp ö æp ö (C) 2 cos ç t ÷ + 2 sin ç t ÷ è4 ø è4 ø

2

4.

Chap 5.7

2 1 x( t) dt = 1 ò 20

(D) None of the above

The signal, that satisfy these condition, is (A)

2 sin pt and unique

(B)

2 sin pt but not unique

Statement for Q.29-31: Consider the following three continuous-time

(C) 2 sin pt and unique

signals with a fundamental period of T = 1

(D) 2 sin pt but not unique

x( t) = cos 2pt , y( t) = sin 2pt , z( t) = x( t) y( t)

26. Consider a continuous-time LTI system whose

29. The Fourier series coefficient X [ k] of x( t) are

frequency response is

(A)

1 2

( d[ k + 1] + d[ k - 1])

(B)

1 2

( d[ k + 1] - d[ k - 1])

(C)

1 2

( d[ k - 1] - d[ k + 1])

H ( jw) =

¥

ò h( t) e

- jwt

dt =

-¥

sin 4 w w

(D) None of the above

The input to this system is a periodic signal ì 2, 0 £ t £ 4 x( t) = í î -2, 4 £ t £ 8

30. The Fourier series coefficient of y( t), Y [ k] will be

with period T = 8. The output y( t) will be æ pt ö (A) 1 + sin ç ÷ è4ø

æ pt ö (B) 1 + cos ç ÷ è4ø

æ pt ö æ pt ö (C) 1 + sin ç ÷ + cos ç ÷ è4ø è4ø

(D) 0

2

2

(A)

j 2

( d[ k + 1] + d[ k + 1])

(B)

j 2

( d[ k + 1] - d[ k - 1])

(C)

j 2

( d[ k - 1] - d[ k + 1])

(D)

1 2j

( d[ k + 1] + d[ k + 1])

31. The Fourier series coefficient of z( t) , Z [ k] will be

27. Consider a continuous-time ideal low filter having the frequency response ì 1, |w|£ 80 H ( jw) = í î 0, |w|> 80

(A)

1 4j

( d[ k - 2 ] - d[ k + 2 ])

(B)

1 2j

( d[ k - 2 ] - d[ k + 2 ])

(C)

1 2j

d[ k + 2 ] - d[ k - 2 ])

(D) None of the above

When the input to this filter is a signal x( t) with fundamental frequency wo = 12 and Fourier series

32. Consider a periodic signal x( t) whose Fourier series coefficients are ì 2, ï X [ k] = í æ 1 ö|k| ï jç 2 ÷ , î è ø

S coefficients X [ k], it is found that x( t) ¬¾ ® y( t) = x( t).

The largest value of|k, | for which X [ k] is nonzero, is (A) 6

(B) 80

(C) 7

(D) 12

28.

A

continuous-time

periodic

otherwise

Consider the statements signal

has

a

fundamental period T = 8. The nonzero Fourier series coefficients are as, X [1] = X [ -1] = j , X [5 ] = X [ -5 ] = 2, *

k =0

1. x( t) is real.

2. x( t) is even

3.

dx( t) is even dt

The true statements are (A) 1 and 2

(B) only 2

(C) only 1

(D) 1 and 3

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UNIT 5

Statement for Q.33-36:

(A)

A 2A 1 1 + (sin t - sin 3t + sin 5 t....) 2 p 3 5

(B)

A 2A 1 1 + (cos t - cos 2 t + cos 3t....) 2 p 2 3

(C)

A 2A 1 1 + (cos t - cos 3t + cos 5 t....) 2 p 3 5

(D)

A 2A 1 1 + (sin t + cos t + sin 3t + cos 3t ....) 2 p 3 3

A waveform for one peroid is depicted in figure in question. Determine the trigonometric Fourier series and choose correct option. 33. x(t) 1 -p p

Signal & System

t

36. -1

x(t)

Fig. P5.7.33

2

2 1 1 1 (A) (cos t + cos 2 t + cos 3t + cos 4 t +....) p 2 3 4 2 1 1 1 (sin t - sin 2 t + sin 3t - sin 4 t +....) p 2 3 4

(C)

2 1 1 1 (sin t + cos t - sin 2 t - cos 2 t + sin 3t +....) p 2 2 3

Fig. P5.7.36

(A)

A

1 12 1 1 + (cos pt + cos 3pt + cos 5 pt +....) 2 p2 9 25

(B) 3 + (C)

x(t)

p

*****

Fig. P5.7.34

(A)

A 4A æ 1 1 ö + ç sin t + sin 2 t + sin 3t +.... ÷ 2 p è 2 3 ø

(B)

A 4A æ 1 1 ö + ç cos t + cos 3t + cos 5 t +.... ÷ 2 p è 3 5 ø

(C)

4A æ 1 1 ö ç sin t + sin 3t + sin 5 t + .... ÷ p è 3 5 ø

(D)

4A æ 1 1 ö ç cos t + cos 2 t + cos 3t +.... ÷ p è 2 3 ø

35. x(t) A

-p 2

p 2

p

t

Fig. P5.7.35

Page 312

12 1 1 (sin pt - sin 3pt + sin 5 pt -....) 2 p 9 25

t

-A

-p

12 1 1 (cos pt + cos 3pt + cos 5 pt +....) p2 9 25

1 12 1 1 + (sin pt - sin 3pt + sin 5 pt -....) 2 p2 9 25

(D) 3 +

-p

t

-1

2 1 1 1 (D) (sin t + cos t + sin 3t + cos 3t + sin 5 t + ....) p 3 3 5 34.

1

-1

(B)

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The Continuous-Time Fourier Series

SOLUTIONS 1 T

T 2 - jkw t ò Ad( t) e o dt =

-T 2

- jkw ot

-T 2

1 dt = T

ò Ae

- jkw ot

- jkt ò x( t) e dt = 0

dt

1 2p

Ae- jkt dt =

0

jA é êe 2 pk ê ë

ù - 1ú úû

æ 4 pk ö ÷÷ j çç è 3 ø

å X [ k]e

k = -¥

å X [ k]e

=

sin 9 pt sin pt

jpkt

1 T

X1 [ k] =

ò x( t - t ) e o

-k

e-

T

j kw o to

T

ò x( t) e

- jkw ot

dt

T

Y [ k] = X1 [ k] + X 2 [ k] = e - jkw oto X [ k] + e jkw oto X [ k] = 2 cos ( wo kto ) X [ k] 13. (A) Ev{ x( t)} =

2p =2 , p

x( t) + x( - t) , 2

The FS coefficients of x( t) are 1 1 X1 [ k] = ò x( -t) e - jkw ot dt = ò x( t) e jkw ot dt = X [ -k] TT T T Therefore, the FS X [ k] + X [ -k] Y [ k] = 2

j 2 pt

æ -1 ö jkt = åç ÷ e + k = -¥ è 3 ø

-1 - jt e 4 1 = 3 + = 1 1 1 + e - jt 1 + e j t 5 + 3cos t 3 3

dt =

The FS coefficients of x( t - to ) + x( t + to ) are

2

-1

- jkw ot

X 2 [ k] = e jkw oto X [ k]

ö - jkpt ò0 Ae dt ÷÷ ø 1

= je j 2 p t - je -

series coefficients X1 [ k] of x( t - to ) are

+ e j6 pt + e -

k

æ -1 ö jkt ÷ e ç å k=0 è 3 ø ¥

coefficients

of

Ev{ x( t)}

are

j6 pt

14. (C) Re{ x( t)} =

jkt

- j 2 pk ( t -1 )

k = -4

Similarly, the FS coefficients of x( t + to ) are

ö -1 2 jt ÷÷ = ( e - 2 + e -2 jt ) 4 ø

j 2 pkt

4

åe

e jpkt =

= e - jkw oto X [ k]

= - 2 sin 2 pt + 2 cos 6 pt 8. (D) x( t) =

p

12. (A) x( t - to ) is also periodic with T. The Fourier

k = -¥

¥

j

e j ( 3) pt + 2 e 4 e j (4 )pt

= 6 cos 3pt + 4 cos 2 pt + 2 cos pt

-1 1 1 d[ k - 1] + d[ k] - d[ k + 1] 4 2 4 ¥

4

= 3e j ( -3) pt + 2 e j ( -2 ) pt + e j ( -1 ) pt + e j (1 ) pt + 2 e j ( 2 ) pt + 3e j ( 3) pt

4p 0 < 3 4p < t < 2p 3

The fundamental period of sin 2 ( t) is p and wo =

å X [ k]e

jp

k = -¥

A æ 1 - e jkp e - jkp -1 ö A ÷÷ = (1 - ( -1) k ) = çç + 2 è jkp - jkp ø jkp

7. (C) x( t) =

- j 2 pk

¥

x( t) =

1 0 1 1æ X [ k] = ò x( t) e - jkt dt = çç ò - Ae - jkpt dt + 2 -1 2 è -1

X [ k] =

-

11. (D) X [ k] =|k|, - 3 £ k £ 3

4p 3

ò

4

k = -4

ì - A, - 1 < t < 0 2p 4. (C) T = 2, wo = = p, x( t) = í 0< t<1 2 î A,

æ e jt - e - jt 5. (A) sin 2 t = çç è 2j

p

-T 4

ì ï A, 2p 3. (B) T = 2p , wo = = 1, x( t) = í 2p ï0, î 2p

j

e j ( -4 ) pt + e 4 e j ( -3) pt + e

åe

x( t) =

T 4

4

1 2p

p 4

= 4 cos ( 4 pt + p 4) + 2 cos( 3pt - p 4)

é e - jkw o t ù 4 A æ pk ö ê - jkw ú - T = pk sinç 2 ÷ è ø o û ë

X [ k] =

- j

= 2( e - j ( 4 pt + p 4 ) + e j ( 4 pt + p 4 ) ) + ( e - j ( 3pt - p 4 ) + e j ( 3pt - p 4 ) )

A , T

T

A = T

jpkt

10. (A) X [ k] = e - j 2 pk , -4 £ k £ 4

T 2

ò x( t) e

å X [ k]e

k = -¥

A = 10 , T = 5, X [ k] = 2 1 2. (C) X [ k] = T

¥

9. (D) x( t) = = 2e

1. (D) X [ k] =

Chap 5.7

x ( t) + x *( t) , 2

The FS coefficient of x *( t) is 1 X1 [ k] = ò x *( t) e - jkw ot dt = X1*[ -k] TT X1*[ k] =

1 T

ò x( t) e

jkw ot

dt = X [ -k]

T

X1 [ k] = X *[ -k] Y [ k] = www.gatehelp.com

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UNIT 5

æ 10 p ö æ 4p ö 13. X [ k] = cos ç k ÷ + 2 j sin ç k÷ è 19 ø è 19 ø

18. y[ n] = x[ n] - x[ n - N 2 ] , (assume that N is even) (A) (1 - ( -1) k + 1 ) X [2 k]

(B) (1 - ( -1) k ) X [ k] (D) (1 - ( -1) k\ ) X [2 k]

(A)

19 ( d[ n + 5 ] + d[ n - 5 ]) + 19(d[ n + 2 ] - d[ n - 2 ]),|n|£ 9 2

(C) (1 - ( -1) k + 1 ) X [ k]

(B)

1 ( d[ n + 5 ] + d[ n - 5 ]) + (d[ n + 2 ] - d[ n - 2 ]),|n|£ 9 2

19. y[ n] = x *[ -n]

9 (C) ( d[ n + 5 ] + d[ n - 5 ]) + 9(d[ n + 2 ] - d[ n - 2 ]),|n|£ 9 2 (D)

1 ( d[ n + 5 ] + d[ n - 5 ]) + (d[ n + 2 ] - d[ n - 2 ]),|n|£ 9 2

æ pk ö 14. X [ k] = cos ç ÷ è 21 ø

(A) - X *[ k ]

(B) - X *[ -k ]

(C) X *[ k ]

(D) X *[ -k ]

20. y[ n] = ( -1) n x[ n], (assume that N is even) N ù é (A) X êk 2 úû ë

N ù é (B) X êk + 2 úû ë N é ù (D) X êk + - 1ú 2 ë û

(A)

21 (d[ n + 4 ] + d[ n - 4 ]),|n|£ 10 2

N é ù (C) X êk + 1ú 2 ë û

(B)

1 (d[ n + 4 ] + d[ n - 4 ]),|n|£ 10 2

Statement for Q.21-23:

(C)

21 (d[ n + 4 ] - d[ n - 4 ]),|n|£ 10 2

(D)

1 (d[ n + 4 ] - d[ n - 4 ]),|n|£ 10 2

Consider a discrete-time periodic signal ì 1, 0 £ n £ 7 x[ n] = í î 0, 8 £ n £ 9 with period N = 10. Also y[ n] = x[ n] - x[ n - 1 ]

Statement for Q.15-20: Consider a periodic signal x[ n] with period N and

21. The fundamental period of y[ n] is

FS coefficients X [ k]. Determine the FS coefficients Y [ k]

(A) 9

(B) 10

of the signal y[ n] given in question.

(C) 11

(D) None of the above

15. y[ n] = x[ n - no ] (A) e

æ 2 pö ÷÷ n ok j çç è Nø

(B) e

X [ k]

æ 2 pö ÷÷ n ok - j çç è Nø

22. The FS coefficients of y[ n] are (A)

æ 8 pö j çç ÷÷ k ö 1 æç 1-e è 5ø ÷ ÷ 10 ç è ø

(B)

æ 8 pö - j çç ÷÷ k ö 1 æç 1-e è 5ø ÷ ÷ 10 ç è ø

(C)

1 10

æ 4 pö ÷k ö æ jç ç 1 - e çè 5 ÷ø ÷ ç ÷ è ø

(D)

1 10

X [ k]

(D) -kX [ k]

(C) kX [ k] 16. y[ n] = x[ n] - x[ n - 2 ] æ 4p ö (A) sin ç k ÷ X [ k] è N ø

æ 4p ö (B) cos ç k ÷ X [ k] è N ø

æ 4 pö ÷÷ k ö æ - j çç (C) ç 1 - e è N ø ÷ X [ k] ç ÷ è ø

æ 4 pö ÷÷ k ö æ j çç (D) ç 1 - e è N ø ÷ X [ k] ç ÷ è ø

23. The FS coefficients of x[ n] are æ pk ö

(A) -

17. y[ n] = x[ n] + x[ n + N 2 ] , (assume that N is even) æN ö (A) 2 X [2 k - 1], for 0 £ k £ ç - 1÷ è 2 ø (B) 2 X [2 k - 1], for 0 £ k £

Page 318

Signal & System

N 2

(C) 2 X [2 k],

æN ö for 0 £ k £ ç - 1÷ è 2 ø

(D) 2 X [2 k],

for 0 £ k £

N 2

j - j ççè 10 ÷÷ø æ pk ö e cosec ç ÷ Y [ k], k ¹ 0 2 è 10 ø æ pk ö

(B)

j - j ççè 10 ÷÷ø æ pk ö e cosec ç ÷ Y [ k], k ¹ 0 2 è 10 ø æ pk ö

(C) -

1 - j ççè 10 ÷÷ø æ pk ö e sec ç ÷ Y [ k] 2 è 10 ø æ pk ö

(D)

1 - j ççè 10 ÷÷ø æ pk ö e sec ç ÷ Y [ k] 2 è 10 ø

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The Discrete-Time Fourier Series

Statement for Q.24-27:

29. A real and odd periodic signal x[ n] has fundamental

Consider a discrete-time signal with Fourier

period N = 7 and FS coefficients X [ k]. Given that X [15 ] = j,

representation. x[ n]

¬ ¾ ¾¾®

X [ k]

Determine the corresponding signal y[ n] and choose correct option.

æp ö (B) 2 cos ç n ÷ x[ n] è5 ø

æp ö (C) 2 sin ç n ÷ x[ n] è2 ø

æp ö (D) 2 cos ç n ÷ x[ n] è2 ø

(A) 0, j, 2 j, 3 j

(B) 1, 1, 2, 3

(C) 1, -1, -2, -3

(D) 0, - j , -2 j, -3 j

(B)

4.

1 ( x[ n + 2 ] + x[ n - 2 ]) 2

9

1 10

å X [ k]

The signal x[ n] is æ p ö (A) 5 cos ç n÷ è 10 ø

æ p ö (B) 5 sin ç n÷ è 10 ø

æp ö (C) 10 cos ç n ÷ è5 ø

æp ö (D) 10 sin ç n ÷ è5 ø

31. Each of two sequence x[ n] and y[ n] has a period N = 4. The FS coefficient are

(B) j2 p( x[ n]) 2

The FS coefficient Z [ k] for the signal z[ n] = x[ n] y[ n] will be

x [ n] - x[ -n] 2

(B)

x [ n] + x[ -n] 2p

(D)

(A) 6

(B) 6|k|

(C) 6|k|

(D) e

æ 11p ö sin ç n÷ 20 ø è x[ n] = æ p ö sin ç n÷ è 20 ø

1. x[ n] is periodic with N = 6 5

å x[ n] = 2

n =0

n

with a fundamental period N = 20. The Fourier

x[ n] = 1

n -2

4. x[ n] has the minimum power per period among the set of signals satisfying the preceding three condition. The sequence would be.. ì 1 1 1 1 1 ü (A) í ... , , , , , ...ý (B) î 2 6 2 6 2 þ ­

ì 1 1 1 1 1 ü (C) í ... , , , , , ...ý î 3 6 3 6 3 þ ­

p j k 2

32. Consider a discrete-time periodic signal

28. Consider a sequence x[ n] with following facts :

7

1 1 X [1] = X [2 ] = 1 and 2 2

Y [0 ], Y [1], Y [2 ], Y [ 3] = 1

(D) 2 p( x[ n]) 2

x [ n] - x[ -n] 2p

å ( -1)

= 50

X [0 ] = X [ 3] =

27. Y [ k] = Re{ X [ k]} x [ n] + x[ -n] (A) 2

3.

2

n =0

2

(C) ( x[ n]) 2

2.

of

30. Consider a signal x[ n] with following facts

26. Y [ k] = X [ k] * X [ k]

(C)

values

3. X[11] = 5

1 ( x[ n + 10 ] + x[ n + 10 ]) (D) None of the above 2

( x[ n]) 2p

The

2. The period of x[ n] is N = 10

æp ö (A) 2 sin ç n ÷ x[ n] è5 ø

(A)

X [17 ] = 3 j.

1. x[ n] is a real and even signal

24. Y [ k ] = X [ k - 5 ] + X [ k + 5 ]

æ pk ö 25. Y [ k] = cos ç ÷ X [ k] è 5 ø 1 (A) ( x[ n + 5 ] + x[ n + 5 ]) 2

X [16 ] = 2 j,

X [0 ], X [ -1], X [ -2 ], and X [ -3] will be

p DTFS ; 10

In question the FS coefficient Y [ k] is given.

(C)

Chap 5.8

series coefficients of this function are 1 (A) ( u[ k + 5 ] - u[ k - 6 ]), |k|£ 10 20 (B)

1 1 1 ü ì í ...0, 1, , , , ...ý 2 3 4 þ î ­

1 ( u[ k + 5 ] - u[ k - 5 ]), |k|£ 10 20

(C) ( u[ k + 5 ] - u[ k + 6 ]), |k|£ 10 (D) ( u[ k + 5 ] - u[ k - 6 ]), |k|£ 10

(D) {...0, 1, 2, 3, 4, ...} ­

************

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Page 319

GATE EC BY RK Kanodia

The Discrete-Time Fourier Series

11. (D) N = 7, W o = x[ n] =

3

å X [ k]e

2p , 7

æ 2 pö ÷÷ kn j çç è 7 ø

17. (C) Note that y[ n] = x[ n] + x [ n + N 2 ] has a period

= 2e

æ 2 pö ÷÷ n j ( -1 ) çç è 7 ø

- 1 + 2e

of N 2 and N has been assumed to be even,

æ 2 pö ÷÷ n j (1 ) çç è 7 ø

Y [ k] =

n = -3

æ 2p ö = 4 cos ç n÷ - 1 è 7 ø

6

æ pö - j çç ÷÷ k è6 ø

e

æ pö j çç ÷÷ kn è6 ø

e =

9p j ( n -1 ) 6

p j ( n -1 ) ö æ ÷ ç 1-e 6 ÷ ç ø è

13. (A) N = 19, W o =

ö ÷ sin æç 3p ( n - 1) ö÷ ÷ ø = è 4 ø æ p ö sin ç ( n - 1) ÷ è 12 ø

æ 4 pö ÷÷ kn - j çç è Nø

n =0

19. (C) y[ n] = x *[ -n] 1 Y [ k] = N

2p 19

N -1

å x [ -n]e *

æ 2 pö ÷÷ kn - j çç è Nø

= X *[ k]

n =0

20. (A) With N even

æ 10 p ö æ 10 p ö X [ k] = cos ç k ÷ + 2 j sin ç k÷ 19 è ø è 19 ø =

( x[ n] + x[ n + N 2 ]) e

k even ì 0, =í 2 X [ k ], k odd î

k = -6

æ ç 1-e ç è

å

æ 2 pö N ÷÷ k ö æ - j çç Y [ k] = ç 1 - e è N ø 2 ÷ X [ k] = (1 - e - jpk ) X [ k] ç ÷ è ø

æ pö j çç ÷÷ k ( n -1 ) è6 ø

6

åe

=

k = -6

p j ( -4 ) ( n -1 ) 6

N 2 -1

18. (B) y[ n] = x[ n] - x [ n - N 2 ]

æ pö

åe

2 N

= 2 X [2 k] for 0 £ k £ ( N 2 - 1)

- j çç ÷÷ k p 12. (C) N = 12, W o = , X [ k] = e è 6 ø 6

x[ n] =

Chap 5.8

y[ n] = ( -1) n x[ n] = e jpn x[ n] = e

2p 2p 2p 2p - j ( 5) kö k - j(2 ) kö æ - j ( -2 ) 19 1 æ - j ( -5) 19 k 19 ÷ 19 ÷ çe çe + e + + e ÷ ç ÷ 2 çè ø è ø

Y [ k] =

By inspection 19 x[ n] = ( d[ n + 5 ] + d[ n - 5 ]) + 19 ( d[ n + 2 ] - d [ n - 2 ]), 2 Where |n|£ 9

=

1 N

1 N

N -1

åe

æ 2 pö N ÷÷ j çç è Nø 2

x[ n]e

æ 2 pö N ÷÷ j çç è Nø 2

x[ n]

æ 2 pö ÷÷ kn - j çç è Nø

n =0

N -1

å x[ n]e

æ 2 pö æ Nö ÷÷ n çç k - ÷÷ - j çç 2 ø è Nø è

= X [k - N 2]

n =0

21. (B) y[ n] is shown is fig. S5.8.21. It has fundamental y[n]

14. (A) N = 21, W o =

2p 21

1 9 2p - j ( -4 ) k 21

2p - j(4 ) k 21

ö æ 8 p ö 1 æç ÷ X [ k] = cos ç k÷ = ç e +e ÷ 21 2 è ø è ø 1 Since X [ k] = å x[ n]e- jkWon , By inspection N n=N

-1

15. (B) Y [ k] = æ 2 pö

=

1 - j ççè N ÷÷ø kn o e N

1 N

N -1

å x[ n - n ]e o

n =0

å x[ n]e

æ 2 pö ÷÷ kn - j çç è Nø

æ 2 pö ÷÷ kn o - j çç è Nø

22. (B) Y [ k] = =

1 æç 1-e 10 ç è

16. (C) Y [ k] = X [ k] - e

æ X [ k] = ç 1 - e ç è

5

7

8

n 10 11

9

å y[ n]e

æ 2 pö ÷÷ kn - j çç è 10 ø

n =0

ö 1 ÷= ÷ 10 ø

æ ç1 - e ç è

æ 8 pö - j çç ÷÷ k è 5ø

ö ÷ ÷ ø

23. (A) y [ n] = x [ n] - x [ n - 1]

X [ k] æ 4 pö ÷÷ k - j çç è Nø

1 10

æ 2 pö ÷÷ k8 - j çç è 10 ø

n =0

æ 2 pö ÷÷ 2 k - j çç è Nø

4

period of 10.

æ 2 pö ÷÷ kn - j çç è Nø

=e

3

Fig. S5.8.21

ì 21 ï , n = ±4 x[ n] = í 2 ïî 0, otherwise n Î { -10, - 9, ......9. 10} N -1

2

1

ö ÷ X [ k] ÷ ø

Y [ k] = X [ k] - e

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æ 2 pö ÷÷ k - j çç è 10 ø

X [ k]

Þ

X [ k] =

Y [ k] 1-e

æ pö - j çç ÷÷ k è 5ø

Page 321

GATE EC BY RK Kanodia

UNIT 5

Þ

e

X [ k] = e

=

-j e 2

æ p ö - j çç k ÷÷ è 10 ø

æ pö j çç ÷÷ k è 10 ø

Y [ k]

-e

æ pö - j çç ÷÷ k è 10 ø

æ p cosec ç è 10

24. (D) W o = Þ

æ pö j çç ÷÷ k è 10 ø

=

æ pö j çç ÷÷ k è 10 ø

29. (D) Since the FS coefficient repeat every N. Thus

Y [ k] æ pk ö 2 j sin ç ÷ è 10 ø e

Signal & System

X [1] = X [15 ], X [2 ] = X [16 ], X [ 3] = X [17 ] The signal real and odd, the FS coefficient X [ k] will be purely imaginary and odd. Therefore X[0 ] = 0

ö k ÷ Y [ k] ø

X [ -1] = - X [1], X [ -2 ] = - X [2 ], X [ -3] = - X [ 3] Therefore (D) is correct option.

p , Y [ k] = X [ k - 5 ] + X [ k + 5 ] 10

30. (C) Since N = 10, X [11] = X [1] = 5

p j ( -5) n ö æ j ( 5) p n æp ö y[ n] = çç e 10 + e 10 ÷÷ x[ n] = 2 cos ç n ÷ x[ n] è2 ø è ø

Since x[ n] is real and even X [ k] is also real and even. Therefore X [1] = X [ -1] = 5. Using Parseval’s relation

æ çe æp ö 25. (B) Y [ k] = cos ç k ÷ X [ k] = ç è5 ø ç è p

p j k 5

+e 2

p -j k 5

ö ÷ ÷ X [ k] ÷ ø

2

2

2

X [ -1] + X [1] + X [0 ] + 2

X [0 ] +

8

å X [ k]

2

27. (A) Y [ k] =Re{ X [ k]} Þ y[ n] =Ev{ x[ n]} =

x[ n] = å X [ k]e 2

x[ n] + x[ -n] 2

æ = 5ç e ç è

æ 2 pö ÷÷ kn j çç è Nø

=

2p , 6

åe

k = -1

1 Þ X [0 ] = , 3

Þ

1 6

åe

n =0

3

Z[ k ] = X [ 0 ] Y [ k ] + X [ 1 ] Y [ k - 1 ] + X [ 2 ] Y [ k - 2 ] + X [ 3 ] Y [ k - 3 ]

32. (A) N = 20 We know that

1 1 x[ n] = , X [ 3] = 6 6

P = å X [ k] ,

|n|£ 5 ì 1, í î 0, 5 <|n|£ 10

The value of P is minimized by choosing X [1] = X [2 ] = X [ 4 ] = X [5 ] = 0 Therefore x[ n] = X [0 ] + X [ 3]e

p 10

¬ ¾ ¾¾®

ö ÷÷ 3n ø

=

1 1 1 1 + ( -1) n = + ( -1) n 3 6 3 6

Using duality æ 11p ö sin ç n÷ p DTFS ; è 20 ø ¬ ¾ 10 ¾¾ ® æ p ö sin ç n÷ è 20 ø

ì 1 1 1 1 1 ü x[ n] = í ... , , , , , ...ý î 2 6 2 6 2 þ

æ 11p ö sin ç k÷ è 20 ø æ p ö sin ç k÷ è 20 ø

|k|£ 5 1 ì 1, í 20 î 0, 5 <|k|£ 10

*********

­

Page 322

DTFS ;

2

k=0

æ 2p çç è 6

å X [ l ]Y [ k - l ]

Thus Z [ k] = 6, for all k.

By Parseval’s relation, the average power in x[ n] is 5

DTFS ¬ ¾¾ ®

Since Y [ k] is 1 for all values of k.

å ( -1) n x[ n] = 1

æ 2 pö ÷÷ ( 3) k j çç è 6 ø

ö ÷ = 10 cos æç p n ö÷ ÷ è5 ø ø

= Y [ k] + 2 Y [ k - 1] + 2 Y [ k - 2 ] + Y [ k - 3]

1 x[ n] = 3

n =2 5

æ 2 pö ÷÷ n j çç è 10 ø

l= 0

Þ

7

From fact 3,

æ 2 pö ÷÷ kn j çç è 10 ø

Z [ k] = å X [ l ]Y [ k - l ]

Þ

å x[ n] = 2

n =0

+e

å X [ k]e

k =< N>

n =0

Þ

æ 2p ö - j çç ÷÷ n è 10 ø

8

31. (A) z[ n] = x[ n] y[ n]

5

1 6

= 50

Therefore X [ k] = 0 for k = 0, 2, 3, ..... 8.

26. (C) Y [ k] = X [ k] * X [ k] Þ y[ n] = x[ n] x[ n] = ( x[ n])

æ 2 pö ÷÷ ( 0 ) k j çç è 6 ø

2

2

k = -1

=0

N

5

8

å X [ k]

8

å X [ k]

k=2

1 y[ n] = ( x[ n - 2 ] + x[ n + 2 ]) 2

From fact 2,

= 50 =

k=2

p

28. (A) N = 6, W o =

2

N

j ( -2 ) kö 1 æ j(2 ) k 10 ÷ = çç e 10 + e ÷ X [ k] 2è ø

Þ

å X [ k]

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