Signals And Systems 2h2b5i
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CHAPTER 3 SIGNALS & SYSTEMS
YEAR 2012 MCQ 3.1
ONE MARK
If x [n] = (1/3) n − (1/2) n u [n], then the region of convergence (ROC) of its z -transform in the z -plane will be (B) 1 < z < 1 (A) 1 < z < 3 3 3 2 (C) 1 < z < 3 2
MCQ 3.2
(D) 1 < z 3 The unilateral Laplace transform of f (t) is 2 1 . The unilateral Laplace s +s+1 transform of tf (t) is (B) − 2 2s + 1 2 (A) − 2 s (s + s + 1) 2 (s + s + 1) (C) 2 s (D) 2 2s + 1 2 2 (s + s + 1) (s + s + 1) YEAR 2012
MCQ 3.3
Let y [n] denote the convolution of h [n] and g [n], where h [n] = (1/2) n u [n] and g [n] is a causal sequence. If y [0] = 1 and y [1] = 1/2, then g [1] equals (A) 0 (B) 1/2 (C) 1
MCQ 3.4
(D) 3/2
The Fourier transform of a signal h (t) is H (jω) = (2 cos ω) (sin 2ω) /ω . The value of h (0) is (A) 1/4 (B) 1/2 (C) 1
MCQ 3.5
TWO MARKS
(D) 2
The input x (t) and output y (t) of a system are related as y (t) = . The system is (A) time-invariant and stable
# x (τ) cos (3τ) dτ t
−3
(B) stable and not time-invariant (C) time-invariant and not stable (D) not time-invariant and not stable
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PAGE 116
SIGNALS & SYSTEMS
CHAP 3
YEAR 2011
ONE MARK 3
MCQ 3.6
MCQ 3.7
The Fourier series expansion f (t) = a 0 + / an cos nωt + bn sin nωt of n=1 the periodic signal shown below will contain the following nonzero
(A) a 0 and bn, n = 1, 3, 5, ...3
(B) a 0 and an, n = 1, 2, 3, ...3
(C) a 0 an and bn, n = 1, 2, 3, ...3
(D) a 0 and an n = 1, 3, 5, ...3
Given two continuous time signals x (t) = e−t and y (t) = e−2t which exist for t > 0 , the convolution z (t) = x (t) * y (t) is (A) e−t − e−2t (B) e−3t (C) e+t
(D) e−t + e−2t
YEAR 2011 MCQ 3.8
Let the Laplace transform of a function f (t) which exists for t > 0 be F1 (s) and the Laplace transform of its delayed version f (t − τ) be F2 (s). Let F1 * (s) be the complex conjugate of F1 (s) with the Laplace variable set s = σ + jω . F (s) F1 * (s) , then the inverse Laplace transform of G (s) is an ideal If G (s) = 2 F1 (s) 2 (A) impulse δ (t) (B) delayed impulse δ (t − τ) (C) step function u (t)
MCQ 3.9
TWO MARKS
(D) delayed step function u (t − τ)
The response h (t) of a linear time invariant system to an impulse δ (t), under initially relaxed condition is h (t) = e−t + e−2t . The response of this system for a unit step input u (t) is (A) u (t) + e−t + e−2t (B) (e−t + e−2t) u (t) (C) (1.5 − e−t − 0.5e−2t) u (t)
(D) e−t δ (t) + e−2t u (t)
YEAR 2010 MCQ 3.10
ONE MARK
For the system 2/ (s + 1), the approximate time taken for a step response to reach 98% of the final value is (A) 1 s (B) 2 s (C) 4 s
(D) 8 s
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CHAP 3
MCQ 3.11
SIGNALS & SYSTEMS
The period of the signal x (t) = 8 sin `0.8πt + π j is 4 (A) 0.4π s (B) 0.8π s (C) 1.25 s
MCQ 3.12
PAGE 117
(D) 2.5 s
The system represented by the input-output relationship y (t) =
5t
# x (τ) dτ, t > 0 −3
MCQ 3.13
(A) Linear and causal
(B) Linear but not causal
(C) Causal but not linear
(D) Neither liner nor causal
The second harmonic component of the periodic waveform given in the figure has an amplitude of
(A) 0
(B) 1
(C) 2/π
(D)
5
YEAR 2010 MCQ 3.14
x (t) is a positive rectangular pulse from t =− 1 to t =+ 1 with unit height as shown in the figure. The value of transform of x (t)} is.
MCQ 3.15
TWO MARKS
#- 3
3
X (ω) 2 dω " where X (ω) is the Fourier
(A) 2
(B) 2π
(C) 4
(D) 4π
Given the finite length input x [n] and the corresponding finite length output y [n] of an LTI system as shown below, the impulse response h [n] of the system is
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PAGE 118
SIGNALS & SYSTEMS
(A) h [n] = {1, 0, 0, 1} (C) h [n] = {1, 1, 1, 1} -
CHAP 3
(B) h [n] = {1, 0, 1} (D) h [n] = {1, 1, 1} -
Common Data Questions Q.6-7. Given f (t) and g (t) as show below
MCQ 3.16
g (t) can be expressed as
(C) g (t) = f`2t − 3 j 2
(B) g (t) = f` t − 3j 2 (D) g (t) = f` t − 3 j 2 2
The Laplace transform of g (t) is (A) 1 (e3s − e5s) s
(B) 1 (e - 5s − e - 3s) s
- 3s (C) e (1 − e - 2s) s
(D) 1 (e5s − e3s) s
(A) g (t) = f (2t − 3)
MCQ 3.17
YEAR 2009 MCQ 3.18
ONE MARK
A Linear Time Invariant system with an impulse response h (t) produces output y (t) when input x (t) is applied. When the input x (t − τ) is applied to a system with impulse response h (t − τ), the output will be (A) y (τ) (B) y (2 (t − τ)) (C) y (t − τ)
(D) y (t − 2τ)
YEAR 2009 MCQ 3.19
TWO MARKS
A cascade of three Linear Time Invariant systems is causal and unstable. From this, we conclude that (A) each system in the cascade is individually causal and unstable (B) at least on system is unstable and at least one system is causal (C) at least one system is causal and all systems are unstable GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SIGNALS & SYSTEMS
PAGE 119
(D) the majority are unstable and the majority are causal MCQ 3.20
The Fourier Series coefficients of a periodic signal x (t) expressed as x (t) = k3=- 3 ak e j2πkt/T are given by a- 2 = 2 − j1, a− 1 = 0.5 + j0.2 , a 0 = j2 , a 1 = 0.5 − j0.2 , a 2 = 2 + j1 and ak = 0 for k > 2 Which of the following is true ? (A) x (t) has finite energy because only finitely many coefficients are nonzero
/
(B) x (t) has zero average value because it is periodic (C) The imaginary part of x (t) is constant (D) The real part of x (t) is even MCQ 3.21
The z-transform of a signal x [n] is given by 4z - 3 + 3z - 1 + 2 − 6z2 + 2z3 It is applied to a system, with a transfer function H (z) = 3z - 1 − 2 Let the output be y [n]. Which of the following is true ? (A) y [n] is non causal with finite (B) y [n] is causal with infinite (C) y [n] = 0; n > 3 (D) Re [Y (z)] z = e =− Re [Y (z)] z = e Im [Y (z)] z = e = Im [Y (z)] z = e ; − π # θ < π ji
ji
- ji
- ji
YEAR 2008 MCQ 3.22
ONE MARK
The impulse response of a causal linear time-invariant system is given as h (t). Now consider the following two statements : Statement (I): Principle of superposition holds Statement (II): h (t) = 0 for t < 0 Which one of the following statements is correct ? (A) Statements (I) is correct and statement (II) is wrong (B) Statements (II) is correct and statement (I) is wrong (C) Both Statement (I) and Statement (II) are wrong (D) Both Statement (I) and Statement (II) are correct
MCQ 3.23
A signal e - αt sin (ωt) is the input to a real Linear Time Invariant system. Given K and φ are constants, the output of the system will be of the form Ke - βt sin (vt + φ) where (A) β need not be equal to α but v equal to ω (B) v need not be equal to ω but β equal to α (C) β equal to α and v equal to ω (D) β need not be equal to α and v need not be equal to ω GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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PAGE 120
SIGNALS & SYSTEMS
CHAP 3
YEAR 2008 MCQ 3.24
TWO MARKS
A system with x (t) and output y (t) is defined by the input-output relation : y (t) =
#- 3x (t) dτ - 2t
The system will be (A) Casual, time-invariant and unstable (B) Casual, time-invariant and stable (C) non-casual, time-invariant and unstable (D) non-casual, time-variant and unstable MCQ 3.25
A signal x (t) = sinc (αt) where α is a real constant ^sinc (x) = πx h is the input to a Linear Time Invariant system whose impulse response h (t) = sinc (βt), where β is a real constant. If min (α, β) denotes the minimum of α and β and similarly, max (α, β) denotes the maximum of α and β, and K is a constant, which one of the following statements is true about the output of the system ? (A) It will be of the form Ksinc (γt) where γ = min (α, β) sin (πx)
(B) It will be of the form Ksinc (γt) where γ = max (α, β) (C) It will be of the form Ksinc (αt) (D) It can not be a sinc type of signal MCQ 3.26
Let x (t) be a periodic signal with time period T , Let y (t) = x (t − t0) + x (t + t0) for some t0 . The Fourier Series coefficients of y (t) are denoted by bk . If bk = 0 for all odd k , then t0 can be equal to (B) T/4 (A) T/8 (C) T/2
MCQ 3.27
(D) 2T
H (z) is a transfer function of a real system. When a signal x [n] = (1 + j) n is the input to such a system, the output is zero. Further, the Region of convergence (ROC) of ^1 − 12 z - 1h H(z) is the entire Z-plane (except z = 0 ). It can then be inferred that H (z) can have a minimum of (A) one pole and one zero (B) one pole and two zeros (C) two poles and one zero D) two poles and two zeros
MCQ 3.28
z Given X (z) = with z > a , the residue of X (z) zn - 1 at z = a for 2 (z − a) n $ 0 will be (B) an (A) an - 1 (C) nan
(D) nan - 1
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CHAP 3
MCQ 3.29
MCQ 3.30
SIGNALS & SYSTEMS
PAGE 121
Let x (t) = rect^t − 12 h (where rect (x) = 1 for − 12 # x # 12 and zero otherwise. sin (πx) If sinc (x) = πx , then the FTof x (t) + x (− t) will be given by (B) 2 sinc` ω j (A) sinc` ω j 2π 2π (C) 2 sinc` ω j cos ` ω j (D) sinc` ω j sin ` ω j 2π 2 2π 2 Given a sequence x [n], to generate the sequence y [n] = x [3 − 4n], which one of the following procedures would be correct ? (A) First delay x (n) by 3 samples to generate z1 [n], then pick every 4th sample of z1 [n] to generate z2 [n], and than finally time reverse z2 [n] to obtain y [n]. (B) First advance x [n] by 3 samples to generate z1 [n], then pick every 4th sample of z1 [n] to generate z2 [n], and then finally time reverse z2 [n] to obtain y [n] (C) First pick every fourth sample of x [n] to generate v1 [n], time-reverse v1 [n] to obtain v2 [n], and finally advance v2 [n] by 3 samples to obtain y [n] (D) First pick every fourth sample of x [n] to generate v1 [n], time-reverse v1 [n] to obtain v2 [n], and finally delay v2 [n] by 3 samples to obtain y [n]
YEAR 2007 MCQ 3.31
ONE MARK
Let a signal a1 sin (ω1 t + φ) be applied to a stable linear time variant system. Let the corresponding steady state output be represented as a2 F (ω2 t + φ2). Then which of the following statement is true? (A) F is not necessarily a “Sine” or “Cosine” function but must be periodic with ω1 = ω2 . (B) F must be a “Sine” or “Cosine” function with a1 = a2 (C) F must be a “Sine” function with ω1 = ω2 and φ1 = φ2 (D) F must be a “Sine” or “Cosine” function with ω1 = ω2
MCQ 3.32
The frequency spectrum of a signal is shown in the figure. If this is ideally sampled at intervals of 1 ms, then the frequency spectrum of the sampled signal will be
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PAGE 122
SIGNALS & SYSTEMS
CHAP 3
YEAR 2007 MCQ 3.33
TWO MARKS
A signal x (t) is given by 1, − T/4 < t # 3T/4 x (t) = *− 1, 3T/4 < t # 7T/4 − x (t + T) Which among the following gives the fundamental fourier term of x (t) ? (B) π cos ` πt + π j (A) 4 cos ` πt − π j π T 4 4 2T 4 (C) 4 sin ` πt − π j (D) π sin ` πt + π j π T 4 4 2T 4
Statement for Linked Answer Question 34 and 35 : MCQ 3.34
A signal is processed by a causal filter with transfer function G (s) For a distortion free output signal wave form, G (s) must (A) provides zero phase shift for all frequency (B) provides constant phase shift for all frequency GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SIGNALS & SYSTEMS
PAGE 123
(C) provides linear phase shift that is proportional to frequency (D) provides a phase shift that is inversely proportional to frequency MCQ 3.35
G (z) = αz - 1 + βz - 3 is a low digital filter with a phase characteristics same as that of the above question if (B) α =− β (A) α = β (C) α = β(1/3)
MCQ 3.36
(D) α = β(- 1/3)
Consider the discrete-time system shown in the figure where the impulse response of G (z) is g (0) = 0, g (1) = g (2) = 1, g (3) = g (4) = g = 0
This system is stable for range of values of K (A) [− 1, 12 ] (B) [− 1, 1] (C) [− 12 , 1] MCQ 3.37
If u (t), r (t) denote the unit step and unit ramp functions respectively and u (t) * r (t) their convolution, then the function u (t + 1) * r (t − 2) is given by (B) 12 (t − 1) u (t − 2) (A) 12 (t − 1) u (t − 1) (C)
MCQ 3.38
(D) [− 12 , 2]
1 2
(t − 1) 2 u (t − 1)
(D) None of the above
X (z) = 1 − 3z - 1, Y (z) = 1 + 2z - 2 are Z transforms of two signals x [n], y [n] respectively. A linear time invariant system has the impulse response h [n] defined by these two signals as h [n] = x [n − 1] * y [n] where * denotes discrete time convolution. Then the output of the system for the input δ [n − 1] (A) has Z-transform z - 1 X (z) Y (z) (B) equals δ [n − 2] − 3δ [n − 3] + 2δ [n − 4] − 6δ [n − 5] (C) has Z-transform 1 − 3z - 1 + 2z - 2 − 6z - 3 (D) does not satisfy any of the above three
YEAR 2006 MCQ 3.39
ONE MARK
The following is true (A) A finite signal is always bounded (B) A bounded signal always possesses finite energy (C) A bounded signal is always zero outside the interval [− t0, t0] for some t0 (D) A bounded signal is always finite GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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PAGE 124
MCQ 3.40
SIGNALS & SYSTEMS
CHAP 3
x (t) is a real valued function of a real variable with period T . Its trigonometric Fourier Series expansion contains no of frequency ω = 2π (2k) /T; k = 1, 2g Also, no sine are present. Then x (t) satisfies the equation (A) x (t) =− x (t − T) (B) x (t) = x (T − t) =− x (− t) (C) x (t) = x (T − t) =− x (t − T/2) (D) x (t) = x (t − T) = x (t − T/2)
MCQ 3.41
A discrete real all system has a pole at z = 2+30% : it, therefore (A) also has a pole at 12 +30% (B) has a constant phase response over the z -plane: arg H (z) = constant constant (C) is stable only if it is anti-causal (D) has a constant phase response over the unit circle: arg H (eiΩ) = constant YEAR 2006
MCQ 3.42
TWO MARKS
x [n] = 0; n < − 1, n > 0, x [− 1] =− 1, x [0] = 2 is the input and y [n] = 0; n < − 1, n > 2, y [− 1] =− 1 = y [1], y [0] = 3, y [2] =− 2 is the output of a discrete-time LTI system. The system impulse response h [n] will be (A) h [n] = 0; n < 0, n > 2, h [0] = 1, h [1] = h [2] =− 1 (B) h [n] = 0; n < − 1, n > 1, h [− 1] = 1, h [0] = h [1] = 2 (C) h [n] = 0; n < 0, n > 3, h [0] =− 1, h [1] = 2, h [2] = 1 (D) h [n] = 0; n < − 2, n > 1, h [− 2] = h [1] = h [− 1] =− h [0] = 3
MCQ 3.43
n The discrete-time signal x [n] X (z) = n3= 0 3 z2n , where 2+n denotes a transform-pair relationship, is orthogonal to the signal n (A) y1 [n] ) Y1 (z) = n3= 0 ` 2 j z - n 3
/
/ (B) y2 [n] ) Y2 (z) = /n3= 0 (5n − n) z - (2n + 1) (C) y3 [n] ) Y3 (z) = /n3=- 3 2 - n z - n (D) y4 [n] ) Y4 (z) = 2z - 4 + 3z - 2 + 1 MCQ 3.44
A continuous-time system is described by y (t) = e - x (t) , where y (t) is the output and x (t) is the input. y (t) is bounded (A) only when x (t) is bounded (B) only when x (t) is non-negative GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SIGNALS & SYSTEMS
PAGE 125
(C) only for t # 0 if x (t) is bounded for t $ 0 (D) even when x (t) is not bounded MCQ 3.45
The running integration, given by y (t) =
#- 3 x (t') dt' t
(A) has no finite singularities in its double sided Laplace Transform Y (s) (B) produces a bounded output for every causal bounded input (C) produces a bounded output for every anticausal bounded input (D) has no finite zeroes in its double sided Laplace Transform Y (s)
YEAR 2005 MCQ 3.46
For the triangular wave from shown in the figure, the RMS value of the voltage is equal to
(A)
1 6
(C) 1 3 MCQ 3.47
MCQ 3.48
TWO MARKS
(B) (D)
1 3 2 3
2 The Laplace transform of a function f (t) is F (s) = 5s 2+ 23s + 6 as s (s + 2s + 2) t " 3, f (t) approaches (A) 3 (B) 5 (D) 3 (C) 17 2
The Fourier series for the function f (x) = sin2 x is (A) sin x + sin 2x (B) 1 − cos 2x (C) sin 2x + cos 2x (D) 0.5 − 0.5 cos 2x
MCQ 3.49
If u (t) is the unit step and δ (t) is the unit impulse function, the inverse z -transform of F (z) = z +1 1 for k > 0 is (A) (− 1) k δ (k)
(B) δ (k) − (− 1) k
(C) (− 1) k u (k)
(D) u (k) − (− 1) k
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PAGE 126
SIGNALS & SYSTEMS
CHAP 3
YEAR 2004 MCQ 3.50
MCQ 3.51
TWO MARKS
The rms value of the periodic waveform given in figure is
(A) 2 6 A
(B) 6 2 A
(C)
(D) 1.5 A
4/3 A
The rms value of the resultant current in a wire which carries a dc current of 10 A and a sinusoidal alternating current of peak value 20 is (A) 14.1 A (B) 17.3 A (C) 22.4 A
(D) 30.0 A
YEAR 2002 MCQ 3.52
ONE MARK
Fourier Series for the waveform, f (t) shown in Figure is
(A) 82 8sin (πt) + 1 sin (3πt) + 1 sin (5πt) + .....B 9 25 π (B) 82 8sin (πt) − 1 cos (3πt) + 1 sin (5πt) + .......B 9 25 π (C) 82 8cos (πt) + 1 cos (3πt) + 1 cos (5πt) + .....B 9 25 π (D) 82 8cos (πt) − 1 sin (3πt) + 1 sin (5πt) + .......B 9 25 π MCQ 3.53
Let s (t) be the step response of a linear system with zero initial conditions; then the response of this system to an an input u (t) is t t (A) s (t − τ) u (τ) dτ (B) d ; s (t − τ) u (τ) dτ E dt 0 0
#
#
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CHAP 3
SIGNALS & SYSTEMS
(C) MCQ 3.54
#0 s (t − τ); #0 u (τ1) dτ1Edτ t
t
(D)
#0 [s (t − τ)] 2 u (τ) dτ 1
Let Y (s) be the Laplace transformation of the function y (t), then the final value of the function is (B) LimY (s) (A) LimY (s) s"0
s"3
(C) Lim sY (s)
(D) Lim sY (s)
s"0
MCQ 3.55
PAGE 127
s"3
What is the rms value of the voltage waveform shown in Figure ?
(A) (200/π) V
(B) (100/π) V
(C) 200 V
(D) 100 V
YEAR 2001 MCQ 3.56
ONE MARK
Given the relationship between the input u (t) and the output y (t) to be y (t) =
#0 (2 + t − τ) e- 3(t - τ)u (τ) dτ , t
The transfer function Y (s) /U (s) is - 2s (A) 2e s+3 (C) 2s + 5 s+3
s+2 (s + 3) 2 (D) 2s + 72 (s + 3) (B)
Common data Questions Q.57-58* Consider the voltage waveform v as shown in figure
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PAGE 128
MCQ 3.57
MCQ 3.58
SIGNALS & SYSTEMS
CHAP 3
The DC component of v is (A) 0.4
(B) 0.2
(C) 0.8
(D) 0.1
The amplitude of fundamental component of v is (A) 1.20 V (B) 2.40 V (C) 2 V
(D) 1 V
***********
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CHAP 3
SIGNALS & SYSTEMS
PAGE 129
SOLUTION SOL 3.1
Option (C) is correct. n n x [n] = b 1 l − b 1 l u [n] 3 2 n −n n = b 1 l u [n] + b 1 l u [− n − 1] − b 1 l u (n) 3 3 2 Taking z -transform X 6z @ = − =
3
/ n =− 3 3
/
n =− 3 3
1 n −n b 3 l z u [ n] +
1 −n b 2 l z u [ n] = n
3
3
/ n =− 3 3
/ b 13 l z
n=0
14 42 4 43 I
m
m=1
n
−n
n=0
/ b 31z l + / b 13 z l n
1 −n −n b 3 l z u [ − n − 1]
−
1 44 2 44 3 II
3
+
−1
/ n =− 3
1 −n −n b3l z −
/ b 21z l
n
n=0
3
/ b 12 l z n
−n
n=0
Taking m =− n
14 42 4 43 III
1 < 1 or z > 1 3 3z 1 Series II converges if z < 1 or z < 3 3 Series III converges if 1 < 1 or z > 1 2 2z Region of convergence of X (z) will be intersection of above three So, ROC : 1 < z < 3 2 Series I converges if
SOL 3.2
Option (D) is correct. Using s -domain differentiation property of Laplace transform. If
f (t)
SOL 3.3
F (s)
dF (s) ds 2s + 1 L [tf (t)] = − d ; 2 1 = ds s + s + 1E (s2 + s + 1) 2 tf (t)
So,
L
L
−
Option (A) is correct. Convolution sum is defined as y [n] = h [n] * g [n] = For causal sequence,
y [n] =
3
/ h [n] g [n − k]
k =− 3
3
/ h [n] g [n − k]
k=0
y [n] = h [n] g [n] + h [n] g [n − 1] + h [n] g [n − 2] + ..... GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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PAGE 130
SIGNALS & SYSTEMS
CHAP 3
For n = 0 ,
y [0] = h [0] g [0] + h [1] g [− 1] + ........... = h [ 0] g [ 0 ] g [− 1] = g [− 2] = ....0 ...(i) = h [ 0] g [ 0 ]
For n = 1,
y [1] = h [1] g [1] + h [1] g [0] + h [1] g [− 1] + .... = h [ 1] g [ 1 ] + h [ 1 ] g [ 0 ] 1 h [1] = b 1 l = 1 2 2
1 = 1 g [1] + 1 g [0] 2 2 2 1 = g [1] + g [0] g [1] = 1 − g [0] y [0] 1 g [0] = = =1 h [0] 1
From equation (i),
g [1] = 1 − 1 = 0
So, SOL 3.4
Option (C) is correct. (2 cos ω) (sin 2ω) H (jω) = = sin 3ω + sin ω ω ω ω We know that inverse Fourier transform of sin c function is a rectangular function.
So, inverse Fourier transform of H (jω) h (t) = h1 (t) + h2 (t) h (0) = h1 (0) + h2 (0) = 1 + 1 = 1 2 2 SOL 3.5
Option (D) is correct. y (t) =
# x (τ) cos (3τ) dτ t
−3
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CHAP 3
SIGNALS & SYSTEMS
PAGE 131
Time invariance : Let, x (t) = δ (t) y (t) =
# δ (t) cos (3τ) dτ t
−3
= u (t) cos (0) = u (t)
For a delayed input (t − t 0) output is y (t, t 0) = Delayed output
# δ (t − t ) cos (3τ) dτ t
0
−3
= u (t) cos (3t 0)
y (t − t 0) = u (t − t 0) y (t, t 0) ! y (t − t 0) System is not time invariant. Stability : Consider a bounded input x (t) = cos 3t y (t) =
#
t
−3
cos2 3t =
1 − cos 6t = 1 2 2 −3
#
t
# 1dt − 12 # cos 6t dt t
t
−3
−3
As t " 3, y (t) " 3 (unbounded) System is not stable. SOL 3.6
Option (D) is correct.
f (t) = a 0 +
3
/ (an cos ωt + bn sin nωt)
n=1
The given function f (t) is an even function, therefore bn = 0 f (t) is a non zero average value function, so it will have a non-zero value of a 0 T/2 a 0 = 1 # f (t) dt (average value of f (t)) ^T/2h 0 • an is zero for all even values of n and non zero for odd n T an = 2 # f (t) cos (nωt) d (ωt) T 0
• •
So, Fourier expansion of f (t) will have a 0 and an , n = 1, 3, 5f3 SOL 3.7
Option (A) is correct. x (t) = e−t Laplace transformation GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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SIGNALS & SYSTEMS
X (s) =
CHAP 3
1 s+1
y (t) = e−2t Y (s) = 1 s+2 Convolution in time domain is equivalent to multiplication in frequency domain. z (t) = x (t) ) y (t) Z (s) = X (s) Y (s) = b 1 lb 1 l s+1 s+2 By partial fraction and taking inverse Laplace transformation, we get Z (s) = 1 − 1 s+1 s+2 z (t) = e−t − e−2t SOL 3.8
Option (D) is correct. f (t)
L
F1 (s)
L
e−sτ F1 (s) = F2 (s) F (s) F 1)(s) e−sτ F1 (s) F 1)(s) G (s) = 2 = F1 (s) 2 F1 (s) 2 e−sE F1 (s) 2 ) 2 = "a F1 (s) F 1 (s) = F1 (s) F1 (s) 2
f (t − τ)
= e−sτ Taking inverse Laplace transform g (t) = L − 1 [e−sτ] = δ (t − τ) SOL 3.9
Option (C) is correct. h (t) = e−t + e−2t Laplace transform of h (t) i.e. the transfer function H (s) = 1 + 1 s+1 s+2 For unit step input r (t) = μ (t) or R (s) = 1 s Output, Y (s) = R (s) H (s) = 1 : 1 + 1 D s s+1 s+2 By partial fraction Y (s) = 3 − 1 − b 1 l 1 2s s + 1 s+2 2 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SIGNALS & SYSTEMS
PAGE 133
Taking inverse Laplace e−2t u (t) y (t) = 3 u (t) − e−t u (t) − 2 2 = u (t) 61.5 − e−t − 0.5e−2t@ SOL 3.10
Option (C) is correct. System is given as 2 (s + 1) R (s) = 1 s
H (s) = Step input Output
Y (s) = H (s) R (s) =
2 1 =2− 2 b s (s + 1) l s (s + 1)
Taking inverse Laplace transform y (t) = (2 − 2e− t) u (t) Final value of y (t), yss (t) = lim y (t) = 2 t"3
Let time taken for step response to reach 98% of its final value is ts . So, 2 − 2e− ts = 2 # 0.98 0.02 = e− ts ts = ln 50 = 3.91 sec. SOL 3.11
Option (D) is correct. Period of x (t), T = 2π = 2 π = 2.5 sec 0.8 π ω
SOL 3.12
Option (B) is correct. Input output relationship y (t) =
#- 3x (τ) dτ, 5t
t>0
Causality : y (t) depends on x (5t), t > 0 system is non-causal. For example t = 2 y (2) depends on x (10) (future value of input) Linearity : Output is integration of input which is a linear function, so system is linear.
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PAGE 134
SOL 3.13
SIGNALS & SYSTEMS
CHAP 3
Option (A) is correct. Fourier series of given function x (t) = A0 +
3
/ an cos nω0 t + bn sin nω0 t n=1
So,
a x (t) =− x (t) odd function A0 = 0 an = 0 T bn = 2 x (t) sin nω0 t dt T 0
#
T /2 T = 2= (1) sin nω0 t dt + (− 1) sin nω0 t dt G T 0 T /2 T /2 T = 2 =c cos nω0 t m − c cos nω0 t m G − nω0 0 − nω0 T/2 T = 2 6(1 − cos nπ) + (cos 2nπ − cos nπ)@ nω0 T = 2 61 − (− 1) n @ nπ 4 , n odd bn = * nπ 0 , n even So only odd harmonic will be present in x (t) For second harmonic component (n = 2) amplitude is zero.
#
SOL 3.14
Option (D) is correct. By parsval’s theorem 1 3 X (ω) 2 dω = 2π - 3
#
#- 3
3
SOL 3.15
#
#- 3 x2 (t) dt 3
X (ω) 2 dω = 2π # 2 = 4π
Option (C) is correct. Given sequences x [n] = {1, − 1}, 0 # n # 1 y [n] = {1, 0, 0, 0, − 1}, 0 # n # 4 If impulse response is h [n] then y [ n] = h [ n] * x [ n] Length of convolution (y [n]) is 0 to 4, x [n] is of length 0 to 1 so length of h [n] will be 0 to 3. Let h [n] = {a, b, c, d} GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SIGNALS & SYSTEMS
PAGE 135
Convolution
y [n] = {a, − a + b, − b + c, − c + d, − d} By comparing a =1 −a + b = 0 & b = a = 1
So,
SOL 3.16
−b + c = 0 & c = b = 1 −c + d = 0 & d = c = 1 h [n] = {1, 1, 1, 1} -
Option (D) is correct. We can observe that if we scale f (t) by a factor of 1 and then shift, we will 2 get g (t). First scale f (t) by a factor of 1 2 g1 (t) = f (t/2)
Shift g1 (t) by 3,
g (t) = g1 (t − 3) = f` t − 3 j 2
g (t) = f` t − 3 j 2 2 SOL 3.17
Option (C) is correct. g (t) can be expressed as GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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PAGE 136
SIGNALS & SYSTEMS
CHAP 3
g (t) = u (t − 3) − u (t − 5) By shifting property we can write Laplace transform of g (t) - 3s G (s) = 1 e - 3s − 1 e - 5s = e (1 − e - 2s) s s s
SOL 3.18
Option (D) is correct. L Let x (t) X (s) L y (t) Y (s) L h (t) H (s) So output of the system is given as Y (s) = X (s) H (s) Now for input
x (t − τ)
L
h (t − τ)
L
e - sτ X (s) e
− sτ
(shifting property)
H (s)
Y' (s) = e - sτ X (s) $ e - τs H (s)
So now output is
= e - 2sτ X (s) H (s) = e - 2sτ Y (s) y' (t) = y (t − 2τ) SOL 3.19
Option (B) is correct. Let three LTI systems having response H1 (z), H2 (z) and H 3 (z) are Cascaded as showing below
Assume H1 (z) = z2 + z1 + 1 (non-causal) H2 (z) = z3 + z2 + 1 (non-causal) Overall response of the system H (z) = H1 (z) H2 (z) H3 (z) H (z) = (z2 + z1 + 1) (z3 + z2 + 1) H3 (z) To make H (z) causal we have to take H3 (z) also causal. Let
H3 (z) = z - 6 + z - 4 + 1 = (z2 + z1 + 1) (z3 + z2 + 1) (z - 6 + z - 4 + 1) H (z) " causal
Similarly to make H (z) unstable atleast one of the system should be unstable. SOL 3.20
Option (C) is correct. Given signal x (t) =
3
/ak e j2πkt/T k =- 3
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CHAP 3
SIGNALS & SYSTEMS
PAGE 137
Let ω0 is the fundamental frequency of signal x (t) x (t) =
3
a 2π = ω0 T
/ak e jkω t 0
k =- 3
x (t) = a - 2 e - j2ω t + a - 1 e - jω t + a0 + a1 e jω t + a2 e j2ω t 0
0
0
0
= (2 − j) e - 2jω t + (0.5 + 0.2j) e - jω t + 2j + 0
0
+ (0.5 − 0.2) e jω t + (2 + j) e j2ω t 0
0
= 2 6e - j2ω t + e j2ω t @ + j 6e j2ω t − e - j2ω t @ + 0
0
0
0
0.5 6e jω t + e - jω t @ − 0.2j 6e+ jω t − e - jω t @ + 2j = 2 (2 cos 2ω0 t) + j (2j sin 2ω0 t) + 0.5 (2 cos ω0 t) − 0.2j (2j sin ω0 t) + 2j = 6 4 cos 2ω0 t − 2 sin 2ω0 t + cos ω0 t + 0.4 sin ω0 t @ + 2j 0
0
0
0
Im [x (t)] = 2 (constant) SOL 3.21
Option (A) is correct. Z-transform of x [n] is X (z) = 4z - 3 + 3z - 1 + 2 − 6z2 + 2z3 Transfer function of the system H (z) = 3z - 1 − 2 Output Y (z) = H (z) X (z) Y (z) = (3z - 1 − 2) (4z - 3 + 3z - 1 + 2 − 6z2 + 2z3) = 12z -4 + 9z -2 + 6z -1 − 18z + 6z2 − 8z -3 − 6z -1 − 4 + 12z2 − 4z3 = 12z - 4 − 8z - 3 + 9z - 2 − 4 − 18z + 18z2 − 4z3 Or sequence y [n] is y [n] = 12δ [n − 4] − 8δ [n − 3] + 9δ [n − 2] − 4δ [n] − 18δ [n + 1] + 18δ [n + 2] − 4δ [n + 3] y [n] = Y 0, n < 0 So y [n] is non-causal with finite .
SOL 3.22
Option (D) is correct. Since the given system is LTI, So principal of Superposition holds due to linearity. For causal system h (t) = 0 , t < 0 Both statement are correct.
SOL 3.23
Option (C) is correct. For an LTI system output is a constant multiplicative of input with same frequency. GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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PAGE 138
SIGNALS & SYSTEMS
Here
CHAP 3
input g (t) = e - αt sin (ωt)
output y (t) = Ke - βt sin (vt + φ) Output will be in form of Ke - αt sin (ωt + φ) So \= β, v = ω SOL 3.24
Option (D) is correct. Input-output relation y (t) =
#- 3x (τ) dτ - 2t
Causality : Since y (t) depends on x (− 2t), So it is non-causal. Time-variance : y (t) =
#- 3x (τ − τ0) dτ =Y y (t − τ0) - 2t
So this is time-variant. Stability : Output y (t) is unbounded for an bounded input. For example Let
x (τ) = e - τ (bounded) y (t) =
SOL 3.25
- τ - 2t
#- 3e- τ dτ = 8 −e 1 B- 3 $ Unbounded - 2t
Option (A) is correct. Output y (t) of the given system is y (t) = x (t) ) h (t) Or Y (jω) = X (jω) H (jω) Given that, x (t) = sinc (αt) and h (t) = sinc (βt) Fourier transform of x (t) and h (t) are X (jω) = F [x (t)] = π rect` ω j, − α < ω < α α 2α H (jω) = F [h (t)] = π rect` ω j, − β < ω < β β 2β 2 Y (jω) = π rect` ω j rect` ω j αβ 2α 2β
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CHAP 3
SIGNALS & SYSTEMS
Y (jω) = K rect ` ω j 2γ
So,
γ = min (α, β) y (t) = K sinc (γt)
Where And SOL 3.26
PAGE 139
Option (B) is correct. Let ak is the Fourier series coefficient of signal x (t) Given y (t) = x (t − t0) + x (t + t0) Fourier series coefficient of y (t) bk = e - jkωt ak + e jkωt ak bk = 2ak cos kωt0 bk = 0 (for all odd k ) kωt0 = π , k " odd 2 k 2π t0 = π 2 T For k = 1, t0 = T 4 0
SOL 3.27
Option ( ) is correct.
SOL 3.28
Option (D) is correct. Given that
z , z >a (z − a) 2 at z = a is = d (z − a) 2 X (z) zn - 1 z = a dz z = d (z − a) 2 zn - 1 dz (z − a) 2 z=a n-1 n d z = = nz z = a = nan - 1 dz z = a
X (z) =
Residue of X (z) zn - 1
SOL 3.29
0
Option (C) is correct. Given signal
So,
x (t) = rect `t − 1 j 2 1, − 1 # t − 1 # 1 or 0 # t # 1 2 2 2 x (t) = * 0, elsewhere
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PAGE 140
SIGNALS & SYSTEMS
CHAP 3
1, − 1 # − t − 1 # 1 or − 1 # t # 0 2 2 2 x (− t) = * 0, elsewhere F [x (t) + x (− t)] = =
#- 3 x (t) e- jωt dt + #- 3 x (− t) e- jωt dt 3
3
#0 (1) e- jωt dt + #- 1 (1) e- jωt dt 0
1
- jω t - jω t = ; e E + ; e E = 1 (1 − e - jω) + 1 (e jω − 1) − jω 0 − jω - 1 jω jω 1
0
- j ω /2 j ω /2 =e (e jω/2 − e - jω/2) + e (e jω/2 − e - jω/2) jω jω
(e jω/2 − e - jω/2) (e - jω/2 + e jω/2) jω = 2 sin ` ω j $ 2 cos ` ω j = 2 cos ω sinc` ω j ω 2 2 2 2π
=
SOL 3.30
Option (B) is correct. In option (A) z1 [n] = x [n − 3] z2 [n] = z1 [4n] = x [4n − 3] y [n] = z2 [− n] = x [− 4n − 3] = Y x [3 − 4n] In option (B) z1 [n] = x [n + 3] z2 [n] = z1 [4n] = x [4n + 3] y [n] = z2 [− n] = x [− 4n + 3] In option (C) v1 [n] = x [4n] v2 [n] = v1 [− n] = x [− 4n] y [n] = v2 [n + 3] = x [− 4 (n + 3)] = Y x [3 − 4n] In option (D) v1 [n] = x [4n] v2 [n] = v1 [− n] = x [− 4n] y [n] = v2 [n − 3] = x [− 4 (n − 3)] = Y x [3 − 4n]
SOL 3.31
Option ( ) is correct. The spectrum of sampled signal s (jω) contains replicas of U (jω) at frequencies ! nfs . Where n = 0, 1, 2....... 1 fs = 1 = = 1 kHz Ts 1 m sec GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SIGNALS & SYSTEMS
PAGE 141
SOL 3.32
Option (D) is correct. For an LTI system input and output have identical wave shape (i.e. frequency of input-output is same) within a multiplicative constant (i.e. Amplitude response is constant) So F must be a sine or cosine wave with ω1 = ω2
SOL 3.33
Option (C) is correct. Given signal has the following wave-form
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PAGE 142
SIGNALS & SYSTEMS
CHAP 3
Function x(t) is periodic with period 2T and given that x (t) =− x (t + T) (Half-wave symmetric) So we can obtain the fourier series representation of given function. SOL 3.34
Option (C) is correct. Output is said to be distortion less if the input and output have identical wave shapes within a multiplicative constant. A delayed output that retains input waveform is also considered distortion less. Thus for distortion less output, input-output relationship is given as y (t) = Kg (t − td ) Taking Fourier transform. Y (ω) = KG (ω) e - jωt = G (ω) H (ω) H (ω) & transfer function of the system d
So, H (ω) = Ke - jωt Amplitude response H (ω) = K Phase response, θn (ω) =− ωtd For distortion less output, phase response should be proportional to frequency. d
SOL 3.35
Option (A) is correct. G (z) z = e = αe− jω + βe− 3jω for linear phase characteristic α = β . jω
SOL 3.36
Option (A) is correct. System response is given as G (z) H (z) = 1 − KG (z) g [n] = δ [n − 1] + δ [n − 2] G (z) = z - 1 + z - 2 (z - 1 + z - 2) = 2 z+1 -1 -2 z − Kz − K 1 − K (z + z ) For system to be stable poles should lie inside unit circle. So
H (z) =
z #1 z = K! K2 + 4K # 2 − K
K2 + 4K # 1 K ! 2
K2 + 4K # 2
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CHAP 3
SOL 3.37
SIGNALS & SYSTEMS
PAGE 143
Option (C) is correct. Given Convolution is, h (t) = u (t + 1) ) r (t − 2) Taking Laplace transform on both sides, H (s) = L [h (t)] = L [u (t + 1)] ) L [r (t − 2)] We know that, L [u (t)] = 1/s L [u (t + 1)] = es c 12 m s L [r (t)] = 1/s2 L r (t − 2) = e - 2s c 12 m s s 1 H (s) = ;e ` jE;e - 2s c 12 mE s s
and
So
(Time-shifting property)
(Time-shifting property)
H (s) = e - s c 13 m s Taking inverse Laplace transform h (t) = 1 (t − 1) 2 u (t − 1) 2 SOL 3.38
Option (C) is correct. Impulse response of given LTI system. h [ n ] = x [ n − 1] ) y [ n ] Taking z -transform on both sides. H (z) = z - 1 X (z) Y (z) We have X (z) = 1 − 3z - 1 and Y (z) = 1 + 2z - 2 So
a x [n − 1]
Z
z - 1 x (z)
H (z) = z - 1 (1 − 3z - 1) (1 + 2z - 2) Output of the system for input u [n] = δ [n − 1] is , y (z) = H (z) U (z)
U [n]
Z
So Y (z) = z - 1 (1 − 3z - 1) (1 + 2z - 2) z - 1 = z - 2 (1 − 3z - 1 + 2z - 2 − 6z - 3) = z - 2 − 3z - 3 + 2z - 4 − 6z - 5 Taking inverse z-transform on both sides we have output. y [n] = δ [n − 2] − 3δ [n − 3] + 2δ [n − 4] − 6δ [n − 5] SOL 3.39
Option (B) is correct. A bounded signal always possesses some finite energy. E =
t0
#- t
g (t) 2 dt < 3
0
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U (z) = z - 1
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PAGE 144
SOL 3.40
SIGNALS & SYSTEMS
CHAP 3
Option (C) is correct. Trigonometric Fourier series is given as 3
/an cos nω0 t + bn sin nω0 t
x (t) = A0 +
n=1
Since there are no sine , so bn = 0 bn = 2 T0
#0
= 2= T0
T0
#0
x (t) sin nω0 t dt
T0 /2
x (τ) sin nω0 τ dτ +
#T /2 x (t) sin nω0 t dt G T
0
Where τ = T − t & dτ =− dt T0 /2
= 2; T0
#T
= 2; T0
#T /2 x (T − t) sin n` 2Tπ T − t j dt ++ #T /2 x (t) sin nω0 t dt E
= 2; T0
x (T − t) sin nω0 (T − t) (− dt)+
0
#T /2 x (t) sin nω0 t dt E T
0
TO
T
0
O
T0
T0
#T /2 x (T − t) sin (2nπ − nω0) dt+ #T /2 x (t) sin nω0 t dt E 0
= 2 ;− T0
0
T0
T0
0
0
#T /2 x (T − t) sin (nω0 t) dt + + #T /2 x (t) sin nω0 t dt E
bn = 0 if x (t) = x (T − t) From half wave symmetry we know that if x (t) =− x`t ! T j 2 Then Fourier series of x (t) contains only odd harmonics. SOL 3.41
Option (C) is correct. Z -transform of a discrete all system is given as −1 ) H (z) = z − z−0 1 1 − z0 z It has a pole at z 0 and a zero at 1/z) 0. Given system has a pole at z = 2+30% = 2
( 3 + j) = ( 3 + j) 2
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CHAP 3
SIGNALS & SYSTEMS
PAGE 145
system is stable if z < 1 and for this it is anti-causal. SOL 3.42
Option (A) is correct. According to given data input and output Sequences are x [n] = {− 1, 2}, − 1 # n # 0 y [n] = {− 1, 3, − 1, − 2}, − 1 # n # 2 If impulse response of system is h [n] then output y [n] = h [ n] ) x [n] Since length of convolution (y [n]) is − 1 to 2, x [n] is of length − 1 to 0 so length of h [n] is 0 to 2. Let h [n] = {a, b, c} Convolution
y [n] = {− a, 2a − b, 2b − c, 2c} So, a=1
y [n] = {− 1, 3, − 1, − 2} -
2a − b = 3 & b =− 1 2a − c =− 1 & c =− 1 Impulse response h [n] = "1, − 1, − 1, SOL 3.43
Option ( ) is correct.
SOL 3.44
Option (D) is correct. Output y (t) = e - x (t) If x (t) is unbounded, x (t) " 3 y (t) = e - x (t) " 0 (bounded) So y (t) is bounded even when x (t) is not bounded.
SOL 3.45
Option (B) is correct. Given
y (t) =
t
# x (t') dt' −3
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PAGE 146
SIGNALS & SYSTEMS
CHAP 3
Laplace transform of y (t) X (s) , has a singularity at s = 0 Y (s) = s For a causal bounded input, y (t)=
t
# x (t') dt'
is always bounded.
−3
SOL 3.46
Option (A) is correct. RMS value is given by 1 T
Vrms = Where
#0
T
V2 (t) dt
2 T `T j t, 0 # t # 2
V (t) = * 0, So
1 T
#0
T
V 2 (t) dt = 1 = T
SOL 3.47
#0
T /2
2t 2 ` T j dt +
#T/2 (0) dt G T
T /2
T /2
3 t2 dt = 43 ; t E T 3 0 3 = 43 # T = 1 6 24 T = 1 V 6
#0
= 1 $ 42 T T
Vrms
T
Option (A) is correct. By final value theorem lim f (t) = lim s F (s) = lim s
t"3
s"0
s"0
(5s2 + 23s + 6) s (s2 + 2s + 2)
= 6 =3 2 SOL 3.48
Option (D) is correct. f (x) = sin2 x = 1 − cos 2x 2 = 0.5 − 0.5 cos 2x f (x) = A0 +
3
/an cos nω0 x + bn sin nω0 x n=1
f (x) = sin2 x is an even function so bn = 0 A0 = 0.5 − 0.5, n = 1 an = ) 0 , otherwise ω0 = 2π = 2π = 2 T0 T GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SOL 3.49
SIGNALS & SYSTEMS
Option (B) is correct. F (z) =
Z-transform
1 = 1− z = 1− 1 z+1 z+1 1 + z− 1
f (k) = δ (k) − (− 1) k Z 1 (− 1) k 1 + z− 1
so, Thus
SOL 3.50
PAGE 147
Option (A) is correct. Root mean square value is given as
#
1 T I2 (t) dt T 0 − 12 t, 0 # t < T 2 From the graph, I (t) = * ` T j 6, T /2 < t # T Irms =
So
1 T
#0
T
I2 dt = 1 = T
#0
T /2
− 12t 2 ` T j dt +
#T/2 (6) 2 dt G T
T /2
t3 = 1 e 144 + 36 6 t @TT/2 o ; 2 T T 3 E0 T3 + 36 T = 1 [6T + 18T] = 24 = 1 ; 144 c ` 2 jE T T T2 24 m Irms = SOL 3.51
24 = 2 6 A
Option (B) is correct. Total current in wire I = 10 + 20 sin ωt Irms =
SOL 3.52
(10) 2 +
(20) 2 = 17.32 A 2
Option (C) is correct. Fourier series representation is given as f (t) = A0 +
3
/an cos nω0 t + bn sin nω0 t n=1
From the wave form we can write fundamental period T = 2 sec Z 4 T ]]`T j t, − 2 # t # 0 f (t) = [ ]]−` 4 j t, 0 # t # T T 2 \ f (t) = f (− t), f (t) is an even function So, bn = 0 A0 = 1 f (t) dt T
#
T
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SIGNALS & SYSTEMS
= 1= T
#- T/2 `T4 j tdt + #0 0
CHAP 3
T /2
4 `− T j tdt G
T /2
0
2 2 = 1 e 4 ;t E − 4 ;t E o T T 2 - T /2 T 2 0 2 2 = 1 ; 4 c T m − 4 c T mE = 0 T T 8 T 8 an = 2 f (t) cos nω0 t dt T
#
T
0 4 t cos nω t + T/2 − 4 t cos nω tdt = 2= 0 0 G ` ` Tj T - T /2 T j 0 By solving the integration 8 , n is odd 2 2 an = * n π 0, n is even So, f (t) = 82 8cos πt + 1 cos (3πt) + 1 cos (5πt) + ....B 9 25 π
#
SOL 3.53
#
Option (A) is correct. Response for any input u (t) is given as y (t) = u (t) ) h (t) y (t) =
h (t) " impulse response
#- 3 u (τ) h (t − τ) dτ 3
Impulse response h (t)and step response s (t) of a system is related as h (t) = d [s (t)] dt 3 3 So u (τ) d s [t − τ] dτ = d u (τ) s (t − τ) dτ y (t) = dt dt - 3 -3
#
SOL 3.54
#
Option (B) is correct. Final value theorem states that lim y (t) lim Y (s)
t"3
SOL 3.55
s"3
Option (D) is correct. 1 V2 (t) dt Vrms = T0 T
#
0
here T0 = π 1 T0
#TV2 (t) dt = π1 = #0
π /3
(100) 2 dt +
0
2π /3
#π/3
(− 100) 2 dt +
π
#2π/3 (100) 2 dt G
= 1 810 4 ` π j + 10 4 ` π j + 10 4 ` π jB = 10 4 V π 3 3 3 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SIGNALS & SYSTEMS
Vrms = SOL 3.56
PAGE 149
10 4 = 100 V
Option (D) is correct. Let h (t) is the impulse response of system y (t) = u (t) ) h (t) y (t) = =
#0 u (τ) h (t − τ) dτ t
#0 (2 + t − τ) e- 3(t - τ)u (τ) dτ t
So h (t) = (t + 2) e - 3t u (t), t > 0 Transfer function Y (s) 1 H (s) = = + 2 2 (s + 3) U (s) (s + 3) (2s + 7) = 1 + 2s +26 = (s + 3) (s + 3) 2 SOL 3.57
Option (B) is correct. Fourier series representation is given as v (t) = A0 +
3
/an cos nω0 t + bn sin nω0 t n=1
period of given wave form T = 5 ms DC component of v is A0 = 1 v (t) dt T
#
T
3
5
= 1 > # 1dt + # − 1dtH 5 0 3 1 = 63 − 5 + 3@ = 1 5 5 SOL 3.58
Option (A) is correct. Coefficient, an = 2 v (t) cos nω0 t dt T
#
T
3
5
= 2 > # (1) cos nwt dt + # (− 1) cos nwt dtH 5 0 3
sin nωt sin nωt = 2 f : nω D − : nω D p 5 0 3 2 π 2 π ω= = 5 T an = 1 6sin 3nω − sin 5nω + sin 3nω@ nπ 3
Put
5
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SIGNALS & SYSTEMS
CHAP 3
= 1 ;2 sin b 3n 2π l − sin b 5n 2π lE nπ 5 5 = 1 ;2 sin b 6πn l − sin ^2nπhE nπ 5 = 2 sin b 6πn l nπ 5 Coefficient, bn = 2 v (t) sin nω0 t dt T
#
T
3
5
= 2 > # (1) sin nwt dt + # (− 1) sin nwt dtH 5 0 3
cos nωt cos nωt = 2 f 9− nω C − 9− nω C p 5 0 3 2 π 2 π put ω= = 5 T bn = 1 6− cos 3nω + 1 + cos 5nω − cos 3nω@ nπ = 1 6− 2 cos 3nω + 1 + cos 5nω@ nπ = 1 ;− 2 cos b 3n 2π l + 1 + cos b 5n 2π lE nπ 5 5 = 1 ;− 2 cos b 6πn l + 1 + 1E nπ 5 = 2 ;1 − cos b 6πn lE nπ 5 Amplitude of fundamental component of v is 3
5
v f = a12 + b12 a1 = 2 sin b 6π l, b1 = 2 b1 − cos 6π l π 5 π 5 vf = 2 π
sin2 6π + b1 − cos 6π l 5 5
2
= 1.20 Volt
***********
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YEAR 2012 MCQ 3.1
ONE MARK
If x [n] = (1/3) n − (1/2) n u [n], then the region of convergence (ROC) of its z -transform in the z -plane will be (B) 1 < z < 1 (A) 1 < z < 3 3 3 2 (C) 1 < z < 3 2
MCQ 3.2
(D) 1 < z 3 The unilateral Laplace transform of f (t) is 2 1 . The unilateral Laplace s +s+1 transform of tf (t) is (B) − 2 2s + 1 2 (A) − 2 s (s + s + 1) 2 (s + s + 1) (C) 2 s (D) 2 2s + 1 2 2 (s + s + 1) (s + s + 1) YEAR 2012
MCQ 3.3
Let y [n] denote the convolution of h [n] and g [n], where h [n] = (1/2) n u [n] and g [n] is a causal sequence. If y [0] = 1 and y [1] = 1/2, then g [1] equals (A) 0 (B) 1/2 (C) 1
MCQ 3.4
(D) 3/2
The Fourier transform of a signal h (t) is H (jω) = (2 cos ω) (sin 2ω) /ω . The value of h (0) is (A) 1/4 (B) 1/2 (C) 1
MCQ 3.5
TWO MARKS
(D) 2
The input x (t) and output y (t) of a system are related as y (t) = . The system is (A) time-invariant and stable
# x (τ) cos (3τ) dτ t
−3
(B) stable and not time-invariant (C) time-invariant and not stable (D) not time-invariant and not stable
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PAGE 116
SIGNALS & SYSTEMS
CHAP 3
YEAR 2011
ONE MARK 3
MCQ 3.6
MCQ 3.7
The Fourier series expansion f (t) = a 0 + / an cos nωt + bn sin nωt of n=1 the periodic signal shown below will contain the following nonzero
(A) a 0 and bn, n = 1, 3, 5, ...3
(B) a 0 and an, n = 1, 2, 3, ...3
(C) a 0 an and bn, n = 1, 2, 3, ...3
(D) a 0 and an n = 1, 3, 5, ...3
Given two continuous time signals x (t) = e−t and y (t) = e−2t which exist for t > 0 , the convolution z (t) = x (t) * y (t) is (A) e−t − e−2t (B) e−3t (C) e+t
(D) e−t + e−2t
YEAR 2011 MCQ 3.8
Let the Laplace transform of a function f (t) which exists for t > 0 be F1 (s) and the Laplace transform of its delayed version f (t − τ) be F2 (s). Let F1 * (s) be the complex conjugate of F1 (s) with the Laplace variable set s = σ + jω . F (s) F1 * (s) , then the inverse Laplace transform of G (s) is an ideal If G (s) = 2 F1 (s) 2 (A) impulse δ (t) (B) delayed impulse δ (t − τ) (C) step function u (t)
MCQ 3.9
TWO MARKS
(D) delayed step function u (t − τ)
The response h (t) of a linear time invariant system to an impulse δ (t), under initially relaxed condition is h (t) = e−t + e−2t . The response of this system for a unit step input u (t) is (A) u (t) + e−t + e−2t (B) (e−t + e−2t) u (t) (C) (1.5 − e−t − 0.5e−2t) u (t)
(D) e−t δ (t) + e−2t u (t)
YEAR 2010 MCQ 3.10
ONE MARK
For the system 2/ (s + 1), the approximate time taken for a step response to reach 98% of the final value is (A) 1 s (B) 2 s (C) 4 s
(D) 8 s
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CHAP 3
MCQ 3.11
SIGNALS & SYSTEMS
The period of the signal x (t) = 8 sin `0.8πt + π j is 4 (A) 0.4π s (B) 0.8π s (C) 1.25 s
MCQ 3.12
PAGE 117
(D) 2.5 s
The system represented by the input-output relationship y (t) =
5t
# x (τ) dτ, t > 0 −3
MCQ 3.13
(A) Linear and causal
(B) Linear but not causal
(C) Causal but not linear
(D) Neither liner nor causal
The second harmonic component of the periodic waveform given in the figure has an amplitude of
(A) 0
(B) 1
(C) 2/π
(D)
5
YEAR 2010 MCQ 3.14
x (t) is a positive rectangular pulse from t =− 1 to t =+ 1 with unit height as shown in the figure. The value of transform of x (t)} is.
MCQ 3.15
TWO MARKS
#- 3
3
X (ω) 2 dω " where X (ω) is the Fourier
(A) 2
(B) 2π
(C) 4
(D) 4π
Given the finite length input x [n] and the corresponding finite length output y [n] of an LTI system as shown below, the impulse response h [n] of the system is
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SIGNALS & SYSTEMS
(A) h [n] = {1, 0, 0, 1} (C) h [n] = {1, 1, 1, 1} -
CHAP 3
(B) h [n] = {1, 0, 1} (D) h [n] = {1, 1, 1} -
Common Data Questions Q.6-7. Given f (t) and g (t) as show below
MCQ 3.16
g (t) can be expressed as
(C) g (t) = f`2t − 3 j 2
(B) g (t) = f` t − 3j 2 (D) g (t) = f` t − 3 j 2 2
The Laplace transform of g (t) is (A) 1 (e3s − e5s) s
(B) 1 (e - 5s − e - 3s) s
- 3s (C) e (1 − e - 2s) s
(D) 1 (e5s − e3s) s
(A) g (t) = f (2t − 3)
MCQ 3.17
YEAR 2009 MCQ 3.18
ONE MARK
A Linear Time Invariant system with an impulse response h (t) produces output y (t) when input x (t) is applied. When the input x (t − τ) is applied to a system with impulse response h (t − τ), the output will be (A) y (τ) (B) y (2 (t − τ)) (C) y (t − τ)
(D) y (t − 2τ)
YEAR 2009 MCQ 3.19
TWO MARKS
A cascade of three Linear Time Invariant systems is causal and unstable. From this, we conclude that (A) each system in the cascade is individually causal and unstable (B) at least on system is unstable and at least one system is causal (C) at least one system is causal and all systems are unstable GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SIGNALS & SYSTEMS
PAGE 119
(D) the majority are unstable and the majority are causal MCQ 3.20
The Fourier Series coefficients of a periodic signal x (t) expressed as x (t) = k3=- 3 ak e j2πkt/T are given by a- 2 = 2 − j1, a− 1 = 0.5 + j0.2 , a 0 = j2 , a 1 = 0.5 − j0.2 , a 2 = 2 + j1 and ak = 0 for k > 2 Which of the following is true ? (A) x (t) has finite energy because only finitely many coefficients are nonzero
/
(B) x (t) has zero average value because it is periodic (C) The imaginary part of x (t) is constant (D) The real part of x (t) is even MCQ 3.21
The z-transform of a signal x [n] is given by 4z - 3 + 3z - 1 + 2 − 6z2 + 2z3 It is applied to a system, with a transfer function H (z) = 3z - 1 − 2 Let the output be y [n]. Which of the following is true ? (A) y [n] is non causal with finite (B) y [n] is causal with infinite (C) y [n] = 0; n > 3 (D) Re [Y (z)] z = e =− Re [Y (z)] z = e Im [Y (z)] z = e = Im [Y (z)] z = e ; − π # θ < π ji
ji
- ji
- ji
YEAR 2008 MCQ 3.22
ONE MARK
The impulse response of a causal linear time-invariant system is given as h (t). Now consider the following two statements : Statement (I): Principle of superposition holds Statement (II): h (t) = 0 for t < 0 Which one of the following statements is correct ? (A) Statements (I) is correct and statement (II) is wrong (B) Statements (II) is correct and statement (I) is wrong (C) Both Statement (I) and Statement (II) are wrong (D) Both Statement (I) and Statement (II) are correct
MCQ 3.23
A signal e - αt sin (ωt) is the input to a real Linear Time Invariant system. Given K and φ are constants, the output of the system will be of the form Ke - βt sin (vt + φ) where (A) β need not be equal to α but v equal to ω (B) v need not be equal to ω but β equal to α (C) β equal to α and v equal to ω (D) β need not be equal to α and v need not be equal to ω GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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SIGNALS & SYSTEMS
CHAP 3
YEAR 2008 MCQ 3.24
TWO MARKS
A system with x (t) and output y (t) is defined by the input-output relation : y (t) =
#- 3x (t) dτ - 2t
The system will be (A) Casual, time-invariant and unstable (B) Casual, time-invariant and stable (C) non-casual, time-invariant and unstable (D) non-casual, time-variant and unstable MCQ 3.25
A signal x (t) = sinc (αt) where α is a real constant ^sinc (x) = πx h is the input to a Linear Time Invariant system whose impulse response h (t) = sinc (βt), where β is a real constant. If min (α, β) denotes the minimum of α and β and similarly, max (α, β) denotes the maximum of α and β, and K is a constant, which one of the following statements is true about the output of the system ? (A) It will be of the form Ksinc (γt) where γ = min (α, β) sin (πx)
(B) It will be of the form Ksinc (γt) where γ = max (α, β) (C) It will be of the form Ksinc (αt) (D) It can not be a sinc type of signal MCQ 3.26
Let x (t) be a periodic signal with time period T , Let y (t) = x (t − t0) + x (t + t0) for some t0 . The Fourier Series coefficients of y (t) are denoted by bk . If bk = 0 for all odd k , then t0 can be equal to (B) T/4 (A) T/8 (C) T/2
MCQ 3.27
(D) 2T
H (z) is a transfer function of a real system. When a signal x [n] = (1 + j) n is the input to such a system, the output is zero. Further, the Region of convergence (ROC) of ^1 − 12 z - 1h H(z) is the entire Z-plane (except z = 0 ). It can then be inferred that H (z) can have a minimum of (A) one pole and one zero (B) one pole and two zeros (C) two poles and one zero D) two poles and two zeros
MCQ 3.28
z Given X (z) = with z > a , the residue of X (z) zn - 1 at z = a for 2 (z − a) n $ 0 will be (B) an (A) an - 1 (C) nan
(D) nan - 1
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CHAP 3
MCQ 3.29
MCQ 3.30
SIGNALS & SYSTEMS
PAGE 121
Let x (t) = rect^t − 12 h (where rect (x) = 1 for − 12 # x # 12 and zero otherwise. sin (πx) If sinc (x) = πx , then the FTof x (t) + x (− t) will be given by (B) 2 sinc` ω j (A) sinc` ω j 2π 2π (C) 2 sinc` ω j cos ` ω j (D) sinc` ω j sin ` ω j 2π 2 2π 2 Given a sequence x [n], to generate the sequence y [n] = x [3 − 4n], which one of the following procedures would be correct ? (A) First delay x (n) by 3 samples to generate z1 [n], then pick every 4th sample of z1 [n] to generate z2 [n], and than finally time reverse z2 [n] to obtain y [n]. (B) First advance x [n] by 3 samples to generate z1 [n], then pick every 4th sample of z1 [n] to generate z2 [n], and then finally time reverse z2 [n] to obtain y [n] (C) First pick every fourth sample of x [n] to generate v1 [n], time-reverse v1 [n] to obtain v2 [n], and finally advance v2 [n] by 3 samples to obtain y [n] (D) First pick every fourth sample of x [n] to generate v1 [n], time-reverse v1 [n] to obtain v2 [n], and finally delay v2 [n] by 3 samples to obtain y [n]
YEAR 2007 MCQ 3.31
ONE MARK
Let a signal a1 sin (ω1 t + φ) be applied to a stable linear time variant system. Let the corresponding steady state output be represented as a2 F (ω2 t + φ2). Then which of the following statement is true? (A) F is not necessarily a “Sine” or “Cosine” function but must be periodic with ω1 = ω2 . (B) F must be a “Sine” or “Cosine” function with a1 = a2 (C) F must be a “Sine” function with ω1 = ω2 and φ1 = φ2 (D) F must be a “Sine” or “Cosine” function with ω1 = ω2
MCQ 3.32
The frequency spectrum of a signal is shown in the figure. If this is ideally sampled at intervals of 1 ms, then the frequency spectrum of the sampled signal will be
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SIGNALS & SYSTEMS
CHAP 3
YEAR 2007 MCQ 3.33
TWO MARKS
A signal x (t) is given by 1, − T/4 < t # 3T/4 x (t) = *− 1, 3T/4 < t # 7T/4 − x (t + T) Which among the following gives the fundamental fourier term of x (t) ? (B) π cos ` πt + π j (A) 4 cos ` πt − π j π T 4 4 2T 4 (C) 4 sin ` πt − π j (D) π sin ` πt + π j π T 4 4 2T 4
Statement for Linked Answer Question 34 and 35 : MCQ 3.34
A signal is processed by a causal filter with transfer function G (s) For a distortion free output signal wave form, G (s) must (A) provides zero phase shift for all frequency (B) provides constant phase shift for all frequency GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SIGNALS & SYSTEMS
PAGE 123
(C) provides linear phase shift that is proportional to frequency (D) provides a phase shift that is inversely proportional to frequency MCQ 3.35
G (z) = αz - 1 + βz - 3 is a low digital filter with a phase characteristics same as that of the above question if (B) α =− β (A) α = β (C) α = β(1/3)
MCQ 3.36
(D) α = β(- 1/3)
Consider the discrete-time system shown in the figure where the impulse response of G (z) is g (0) = 0, g (1) = g (2) = 1, g (3) = g (4) = g = 0
This system is stable for range of values of K (A) [− 1, 12 ] (B) [− 1, 1] (C) [− 12 , 1] MCQ 3.37
If u (t), r (t) denote the unit step and unit ramp functions respectively and u (t) * r (t) their convolution, then the function u (t + 1) * r (t − 2) is given by (B) 12 (t − 1) u (t − 2) (A) 12 (t − 1) u (t − 1) (C)
MCQ 3.38
(D) [− 12 , 2]
1 2
(t − 1) 2 u (t − 1)
(D) None of the above
X (z) = 1 − 3z - 1, Y (z) = 1 + 2z - 2 are Z transforms of two signals x [n], y [n] respectively. A linear time invariant system has the impulse response h [n] defined by these two signals as h [n] = x [n − 1] * y [n] where * denotes discrete time convolution. Then the output of the system for the input δ [n − 1] (A) has Z-transform z - 1 X (z) Y (z) (B) equals δ [n − 2] − 3δ [n − 3] + 2δ [n − 4] − 6δ [n − 5] (C) has Z-transform 1 − 3z - 1 + 2z - 2 − 6z - 3 (D) does not satisfy any of the above three
YEAR 2006 MCQ 3.39
ONE MARK
The following is true (A) A finite signal is always bounded (B) A bounded signal always possesses finite energy (C) A bounded signal is always zero outside the interval [− t0, t0] for some t0 (D) A bounded signal is always finite GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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PAGE 124
MCQ 3.40
SIGNALS & SYSTEMS
CHAP 3
x (t) is a real valued function of a real variable with period T . Its trigonometric Fourier Series expansion contains no of frequency ω = 2π (2k) /T; k = 1, 2g Also, no sine are present. Then x (t) satisfies the equation (A) x (t) =− x (t − T) (B) x (t) = x (T − t) =− x (− t) (C) x (t) = x (T − t) =− x (t − T/2) (D) x (t) = x (t − T) = x (t − T/2)
MCQ 3.41
A discrete real all system has a pole at z = 2+30% : it, therefore (A) also has a pole at 12 +30% (B) has a constant phase response over the z -plane: arg H (z) = constant constant (C) is stable only if it is anti-causal (D) has a constant phase response over the unit circle: arg H (eiΩ) = constant YEAR 2006
MCQ 3.42
TWO MARKS
x [n] = 0; n < − 1, n > 0, x [− 1] =− 1, x [0] = 2 is the input and y [n] = 0; n < − 1, n > 2, y [− 1] =− 1 = y [1], y [0] = 3, y [2] =− 2 is the output of a discrete-time LTI system. The system impulse response h [n] will be (A) h [n] = 0; n < 0, n > 2, h [0] = 1, h [1] = h [2] =− 1 (B) h [n] = 0; n < − 1, n > 1, h [− 1] = 1, h [0] = h [1] = 2 (C) h [n] = 0; n < 0, n > 3, h [0] =− 1, h [1] = 2, h [2] = 1 (D) h [n] = 0; n < − 2, n > 1, h [− 2] = h [1] = h [− 1] =− h [0] = 3
MCQ 3.43
n The discrete-time signal x [n] X (z) = n3= 0 3 z2n , where 2+n denotes a transform-pair relationship, is orthogonal to the signal n (A) y1 [n] ) Y1 (z) = n3= 0 ` 2 j z - n 3
/
/ (B) y2 [n] ) Y2 (z) = /n3= 0 (5n − n) z - (2n + 1) (C) y3 [n] ) Y3 (z) = /n3=- 3 2 - n z - n (D) y4 [n] ) Y4 (z) = 2z - 4 + 3z - 2 + 1 MCQ 3.44
A continuous-time system is described by y (t) = e - x (t) , where y (t) is the output and x (t) is the input. y (t) is bounded (A) only when x (t) is bounded (B) only when x (t) is non-negative GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SIGNALS & SYSTEMS
PAGE 125
(C) only for t # 0 if x (t) is bounded for t $ 0 (D) even when x (t) is not bounded MCQ 3.45
The running integration, given by y (t) =
#- 3 x (t') dt' t
(A) has no finite singularities in its double sided Laplace Transform Y (s) (B) produces a bounded output for every causal bounded input (C) produces a bounded output for every anticausal bounded input (D) has no finite zeroes in its double sided Laplace Transform Y (s)
YEAR 2005 MCQ 3.46
For the triangular wave from shown in the figure, the RMS value of the voltage is equal to
(A)
1 6
(C) 1 3 MCQ 3.47
MCQ 3.48
TWO MARKS
(B) (D)
1 3 2 3
2 The Laplace transform of a function f (t) is F (s) = 5s 2+ 23s + 6 as s (s + 2s + 2) t " 3, f (t) approaches (A) 3 (B) 5 (D) 3 (C) 17 2
The Fourier series for the function f (x) = sin2 x is (A) sin x + sin 2x (B) 1 − cos 2x (C) sin 2x + cos 2x (D) 0.5 − 0.5 cos 2x
MCQ 3.49
If u (t) is the unit step and δ (t) is the unit impulse function, the inverse z -transform of F (z) = z +1 1 for k > 0 is (A) (− 1) k δ (k)
(B) δ (k) − (− 1) k
(C) (− 1) k u (k)
(D) u (k) − (− 1) k
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PAGE 126
SIGNALS & SYSTEMS
CHAP 3
YEAR 2004 MCQ 3.50
MCQ 3.51
TWO MARKS
The rms value of the periodic waveform given in figure is
(A) 2 6 A
(B) 6 2 A
(C)
(D) 1.5 A
4/3 A
The rms value of the resultant current in a wire which carries a dc current of 10 A and a sinusoidal alternating current of peak value 20 is (A) 14.1 A (B) 17.3 A (C) 22.4 A
(D) 30.0 A
YEAR 2002 MCQ 3.52
ONE MARK
Fourier Series for the waveform, f (t) shown in Figure is
(A) 82 8sin (πt) + 1 sin (3πt) + 1 sin (5πt) + .....B 9 25 π (B) 82 8sin (πt) − 1 cos (3πt) + 1 sin (5πt) + .......B 9 25 π (C) 82 8cos (πt) + 1 cos (3πt) + 1 cos (5πt) + .....B 9 25 π (D) 82 8cos (πt) − 1 sin (3πt) + 1 sin (5πt) + .......B 9 25 π MCQ 3.53
Let s (t) be the step response of a linear system with zero initial conditions; then the response of this system to an an input u (t) is t t (A) s (t − τ) u (τ) dτ (B) d ; s (t − τ) u (τ) dτ E dt 0 0
#
#
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CHAP 3
SIGNALS & SYSTEMS
(C) MCQ 3.54
#0 s (t − τ); #0 u (τ1) dτ1Edτ t
t
(D)
#0 [s (t − τ)] 2 u (τ) dτ 1
Let Y (s) be the Laplace transformation of the function y (t), then the final value of the function is (B) LimY (s) (A) LimY (s) s"0
s"3
(C) Lim sY (s)
(D) Lim sY (s)
s"0
MCQ 3.55
PAGE 127
s"3
What is the rms value of the voltage waveform shown in Figure ?
(A) (200/π) V
(B) (100/π) V
(C) 200 V
(D) 100 V
YEAR 2001 MCQ 3.56
ONE MARK
Given the relationship between the input u (t) and the output y (t) to be y (t) =
#0 (2 + t − τ) e- 3(t - τ)u (τ) dτ , t
The transfer function Y (s) /U (s) is - 2s (A) 2e s+3 (C) 2s + 5 s+3
s+2 (s + 3) 2 (D) 2s + 72 (s + 3) (B)
Common data Questions Q.57-58* Consider the voltage waveform v as shown in figure
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MCQ 3.57
MCQ 3.58
SIGNALS & SYSTEMS
CHAP 3
The DC component of v is (A) 0.4
(B) 0.2
(C) 0.8
(D) 0.1
The amplitude of fundamental component of v is (A) 1.20 V (B) 2.40 V (C) 2 V
(D) 1 V
***********
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CHAP 3
SIGNALS & SYSTEMS
PAGE 129
SOLUTION SOL 3.1
Option (C) is correct. n n x [n] = b 1 l − b 1 l u [n] 3 2 n −n n = b 1 l u [n] + b 1 l u [− n − 1] − b 1 l u (n) 3 3 2 Taking z -transform X 6z @ = − =
3
/ n =− 3 3
/
n =− 3 3
1 n −n b 3 l z u [ n] +
1 −n b 2 l z u [ n] = n
3
3
/ n =− 3 3
/ b 13 l z
n=0
14 42 4 43 I
m
m=1
n
−n
n=0
/ b 31z l + / b 13 z l n
1 −n −n b 3 l z u [ − n − 1]
−
1 44 2 44 3 II
3
+
−1
/ n =− 3
1 −n −n b3l z −
/ b 21z l
n
n=0
3
/ b 12 l z n
−n
n=0
Taking m =− n
14 42 4 43 III
1 < 1 or z > 1 3 3z 1 Series II converges if z < 1 or z < 3 3 Series III converges if 1 < 1 or z > 1 2 2z Region of convergence of X (z) will be intersection of above three So, ROC : 1 < z < 3 2 Series I converges if
SOL 3.2
Option (D) is correct. Using s -domain differentiation property of Laplace transform. If
f (t)
SOL 3.3
F (s)
dF (s) ds 2s + 1 L [tf (t)] = − d ; 2 1 = ds s + s + 1E (s2 + s + 1) 2 tf (t)
So,
L
L
−
Option (A) is correct. Convolution sum is defined as y [n] = h [n] * g [n] = For causal sequence,
y [n] =
3
/ h [n] g [n − k]
k =− 3
3
/ h [n] g [n − k]
k=0
y [n] = h [n] g [n] + h [n] g [n − 1] + h [n] g [n − 2] + ..... GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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PAGE 130
SIGNALS & SYSTEMS
CHAP 3
For n = 0 ,
y [0] = h [0] g [0] + h [1] g [− 1] + ........... = h [ 0] g [ 0 ] g [− 1] = g [− 2] = ....0 ...(i) = h [ 0] g [ 0 ]
For n = 1,
y [1] = h [1] g [1] + h [1] g [0] + h [1] g [− 1] + .... = h [ 1] g [ 1 ] + h [ 1 ] g [ 0 ] 1 h [1] = b 1 l = 1 2 2
1 = 1 g [1] + 1 g [0] 2 2 2 1 = g [1] + g [0] g [1] = 1 − g [0] y [0] 1 g [0] = = =1 h [0] 1
From equation (i),
g [1] = 1 − 1 = 0
So, SOL 3.4
Option (C) is correct. (2 cos ω) (sin 2ω) H (jω) = = sin 3ω + sin ω ω ω ω We know that inverse Fourier transform of sin c function is a rectangular function.
So, inverse Fourier transform of H (jω) h (t) = h1 (t) + h2 (t) h (0) = h1 (0) + h2 (0) = 1 + 1 = 1 2 2 SOL 3.5
Option (D) is correct. y (t) =
# x (τ) cos (3τ) dτ t
−3
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CHAP 3
SIGNALS & SYSTEMS
PAGE 131
Time invariance : Let, x (t) = δ (t) y (t) =
# δ (t) cos (3τ) dτ t
−3
= u (t) cos (0) = u (t)
For a delayed input (t − t 0) output is y (t, t 0) = Delayed output
# δ (t − t ) cos (3τ) dτ t
0
−3
= u (t) cos (3t 0)
y (t − t 0) = u (t − t 0) y (t, t 0) ! y (t − t 0) System is not time invariant. Stability : Consider a bounded input x (t) = cos 3t y (t) =
#
t
−3
cos2 3t =
1 − cos 6t = 1 2 2 −3
#
t
# 1dt − 12 # cos 6t dt t
t
−3
−3
As t " 3, y (t) " 3 (unbounded) System is not stable. SOL 3.6
Option (D) is correct.
f (t) = a 0 +
3
/ (an cos ωt + bn sin nωt)
n=1
The given function f (t) is an even function, therefore bn = 0 f (t) is a non zero average value function, so it will have a non-zero value of a 0 T/2 a 0 = 1 # f (t) dt (average value of f (t)) ^T/2h 0 • an is zero for all even values of n and non zero for odd n T an = 2 # f (t) cos (nωt) d (ωt) T 0
• •
So, Fourier expansion of f (t) will have a 0 and an , n = 1, 3, 5f3 SOL 3.7
Option (A) is correct. x (t) = e−t Laplace transformation GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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SIGNALS & SYSTEMS
X (s) =
CHAP 3
1 s+1
y (t) = e−2t Y (s) = 1 s+2 Convolution in time domain is equivalent to multiplication in frequency domain. z (t) = x (t) ) y (t) Z (s) = X (s) Y (s) = b 1 lb 1 l s+1 s+2 By partial fraction and taking inverse Laplace transformation, we get Z (s) = 1 − 1 s+1 s+2 z (t) = e−t − e−2t SOL 3.8
Option (D) is correct. f (t)
L
F1 (s)
L
e−sτ F1 (s) = F2 (s) F (s) F 1)(s) e−sτ F1 (s) F 1)(s) G (s) = 2 = F1 (s) 2 F1 (s) 2 e−sE F1 (s) 2 ) 2 = "a F1 (s) F 1 (s) = F1 (s) F1 (s) 2
f (t − τ)
= e−sτ Taking inverse Laplace transform g (t) = L − 1 [e−sτ] = δ (t − τ) SOL 3.9
Option (C) is correct. h (t) = e−t + e−2t Laplace transform of h (t) i.e. the transfer function H (s) = 1 + 1 s+1 s+2 For unit step input r (t) = μ (t) or R (s) = 1 s Output, Y (s) = R (s) H (s) = 1 : 1 + 1 D s s+1 s+2 By partial fraction Y (s) = 3 − 1 − b 1 l 1 2s s + 1 s+2 2 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SIGNALS & SYSTEMS
PAGE 133
Taking inverse Laplace e−2t u (t) y (t) = 3 u (t) − e−t u (t) − 2 2 = u (t) 61.5 − e−t − 0.5e−2t@ SOL 3.10
Option (C) is correct. System is given as 2 (s + 1) R (s) = 1 s
H (s) = Step input Output
Y (s) = H (s) R (s) =
2 1 =2− 2 b s (s + 1) l s (s + 1)
Taking inverse Laplace transform y (t) = (2 − 2e− t) u (t) Final value of y (t), yss (t) = lim y (t) = 2 t"3
Let time taken for step response to reach 98% of its final value is ts . So, 2 − 2e− ts = 2 # 0.98 0.02 = e− ts ts = ln 50 = 3.91 sec. SOL 3.11
Option (D) is correct. Period of x (t), T = 2π = 2 π = 2.5 sec 0.8 π ω
SOL 3.12
Option (B) is correct. Input output relationship y (t) =
#- 3x (τ) dτ, 5t
t>0
Causality : y (t) depends on x (5t), t > 0 system is non-causal. For example t = 2 y (2) depends on x (10) (future value of input) Linearity : Output is integration of input which is a linear function, so system is linear.
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PAGE 134
SOL 3.13
SIGNALS & SYSTEMS
CHAP 3
Option (A) is correct. Fourier series of given function x (t) = A0 +
3
/ an cos nω0 t + bn sin nω0 t n=1
So,
a x (t) =− x (t) odd function A0 = 0 an = 0 T bn = 2 x (t) sin nω0 t dt T 0
#
T /2 T = 2= (1) sin nω0 t dt + (− 1) sin nω0 t dt G T 0 T /2 T /2 T = 2 =c cos nω0 t m − c cos nω0 t m G − nω0 0 − nω0 T/2 T = 2 6(1 − cos nπ) + (cos 2nπ − cos nπ)@ nω0 T = 2 61 − (− 1) n @ nπ 4 , n odd bn = * nπ 0 , n even So only odd harmonic will be present in x (t) For second harmonic component (n = 2) amplitude is zero.
#
SOL 3.14
Option (D) is correct. By parsval’s theorem 1 3 X (ω) 2 dω = 2π - 3
#
#- 3
3
SOL 3.15
#
#- 3 x2 (t) dt 3
X (ω) 2 dω = 2π # 2 = 4π
Option (C) is correct. Given sequences x [n] = {1, − 1}, 0 # n # 1 y [n] = {1, 0, 0, 0, − 1}, 0 # n # 4 If impulse response is h [n] then y [ n] = h [ n] * x [ n] Length of convolution (y [n]) is 0 to 4, x [n] is of length 0 to 1 so length of h [n] will be 0 to 3. Let h [n] = {a, b, c, d} GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SIGNALS & SYSTEMS
PAGE 135
Convolution
y [n] = {a, − a + b, − b + c, − c + d, − d} By comparing a =1 −a + b = 0 & b = a = 1
So,
SOL 3.16
−b + c = 0 & c = b = 1 −c + d = 0 & d = c = 1 h [n] = {1, 1, 1, 1} -
Option (D) is correct. We can observe that if we scale f (t) by a factor of 1 and then shift, we will 2 get g (t). First scale f (t) by a factor of 1 2 g1 (t) = f (t/2)
Shift g1 (t) by 3,
g (t) = g1 (t − 3) = f` t − 3 j 2
g (t) = f` t − 3 j 2 2 SOL 3.17
Option (C) is correct. g (t) can be expressed as GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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SIGNALS & SYSTEMS
CHAP 3
g (t) = u (t − 3) − u (t − 5) By shifting property we can write Laplace transform of g (t) - 3s G (s) = 1 e - 3s − 1 e - 5s = e (1 − e - 2s) s s s
SOL 3.18
Option (D) is correct. L Let x (t) X (s) L y (t) Y (s) L h (t) H (s) So output of the system is given as Y (s) = X (s) H (s) Now for input
x (t − τ)
L
h (t − τ)
L
e - sτ X (s) e
− sτ
(shifting property)
H (s)
Y' (s) = e - sτ X (s) $ e - τs H (s)
So now output is
= e - 2sτ X (s) H (s) = e - 2sτ Y (s) y' (t) = y (t − 2τ) SOL 3.19
Option (B) is correct. Let three LTI systems having response H1 (z), H2 (z) and H 3 (z) are Cascaded as showing below
Assume H1 (z) = z2 + z1 + 1 (non-causal) H2 (z) = z3 + z2 + 1 (non-causal) Overall response of the system H (z) = H1 (z) H2 (z) H3 (z) H (z) = (z2 + z1 + 1) (z3 + z2 + 1) H3 (z) To make H (z) causal we have to take H3 (z) also causal. Let
H3 (z) = z - 6 + z - 4 + 1 = (z2 + z1 + 1) (z3 + z2 + 1) (z - 6 + z - 4 + 1) H (z) " causal
Similarly to make H (z) unstable atleast one of the system should be unstable. SOL 3.20
Option (C) is correct. Given signal x (t) =
3
/ak e j2πkt/T k =- 3
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CHAP 3
SIGNALS & SYSTEMS
PAGE 137
Let ω0 is the fundamental frequency of signal x (t) x (t) =
3
a 2π = ω0 T
/ak e jkω t 0
k =- 3
x (t) = a - 2 e - j2ω t + a - 1 e - jω t + a0 + a1 e jω t + a2 e j2ω t 0
0
0
0
= (2 − j) e - 2jω t + (0.5 + 0.2j) e - jω t + 2j + 0
0
+ (0.5 − 0.2) e jω t + (2 + j) e j2ω t 0
0
= 2 6e - j2ω t + e j2ω t @ + j 6e j2ω t − e - j2ω t @ + 0
0
0
0
0.5 6e jω t + e - jω t @ − 0.2j 6e+ jω t − e - jω t @ + 2j = 2 (2 cos 2ω0 t) + j (2j sin 2ω0 t) + 0.5 (2 cos ω0 t) − 0.2j (2j sin ω0 t) + 2j = 6 4 cos 2ω0 t − 2 sin 2ω0 t + cos ω0 t + 0.4 sin ω0 t @ + 2j 0
0
0
0
Im [x (t)] = 2 (constant) SOL 3.21
Option (A) is correct. Z-transform of x [n] is X (z) = 4z - 3 + 3z - 1 + 2 − 6z2 + 2z3 Transfer function of the system H (z) = 3z - 1 − 2 Output Y (z) = H (z) X (z) Y (z) = (3z - 1 − 2) (4z - 3 + 3z - 1 + 2 − 6z2 + 2z3) = 12z -4 + 9z -2 + 6z -1 − 18z + 6z2 − 8z -3 − 6z -1 − 4 + 12z2 − 4z3 = 12z - 4 − 8z - 3 + 9z - 2 − 4 − 18z + 18z2 − 4z3 Or sequence y [n] is y [n] = 12δ [n − 4] − 8δ [n − 3] + 9δ [n − 2] − 4δ [n] − 18δ [n + 1] + 18δ [n + 2] − 4δ [n + 3] y [n] = Y 0, n < 0 So y [n] is non-causal with finite .
SOL 3.22
Option (D) is correct. Since the given system is LTI, So principal of Superposition holds due to linearity. For causal system h (t) = 0 , t < 0 Both statement are correct.
SOL 3.23
Option (C) is correct. For an LTI system output is a constant multiplicative of input with same frequency. GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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SIGNALS & SYSTEMS
Here
CHAP 3
input g (t) = e - αt sin (ωt)
output y (t) = Ke - βt sin (vt + φ) Output will be in form of Ke - αt sin (ωt + φ) So \= β, v = ω SOL 3.24
Option (D) is correct. Input-output relation y (t) =
#- 3x (τ) dτ - 2t
Causality : Since y (t) depends on x (− 2t), So it is non-causal. Time-variance : y (t) =
#- 3x (τ − τ0) dτ =Y y (t − τ0) - 2t
So this is time-variant. Stability : Output y (t) is unbounded for an bounded input. For example Let
x (τ) = e - τ (bounded) y (t) =
SOL 3.25
- τ - 2t
#- 3e- τ dτ = 8 −e 1 B- 3 $ Unbounded - 2t
Option (A) is correct. Output y (t) of the given system is y (t) = x (t) ) h (t) Or Y (jω) = X (jω) H (jω) Given that, x (t) = sinc (αt) and h (t) = sinc (βt) Fourier transform of x (t) and h (t) are X (jω) = F [x (t)] = π rect` ω j, − α < ω < α α 2α H (jω) = F [h (t)] = π rect` ω j, − β < ω < β β 2β 2 Y (jω) = π rect` ω j rect` ω j αβ 2α 2β
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CHAP 3
SIGNALS & SYSTEMS
Y (jω) = K rect ` ω j 2γ
So,
γ = min (α, β) y (t) = K sinc (γt)
Where And SOL 3.26
PAGE 139
Option (B) is correct. Let ak is the Fourier series coefficient of signal x (t) Given y (t) = x (t − t0) + x (t + t0) Fourier series coefficient of y (t) bk = e - jkωt ak + e jkωt ak bk = 2ak cos kωt0 bk = 0 (for all odd k ) kωt0 = π , k " odd 2 k 2π t0 = π 2 T For k = 1, t0 = T 4 0
SOL 3.27
Option ( ) is correct.
SOL 3.28
Option (D) is correct. Given that
z , z >a (z − a) 2 at z = a is = d (z − a) 2 X (z) zn - 1 z = a dz z = d (z − a) 2 zn - 1 dz (z − a) 2 z=a n-1 n d z = = nz z = a = nan - 1 dz z = a
X (z) =
Residue of X (z) zn - 1
SOL 3.29
0
Option (C) is correct. Given signal
So,
x (t) = rect `t − 1 j 2 1, − 1 # t − 1 # 1 or 0 # t # 1 2 2 2 x (t) = * 0, elsewhere
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PAGE 140
SIGNALS & SYSTEMS
CHAP 3
1, − 1 # − t − 1 # 1 or − 1 # t # 0 2 2 2 x (− t) = * 0, elsewhere F [x (t) + x (− t)] = =
#- 3 x (t) e- jωt dt + #- 3 x (− t) e- jωt dt 3
3
#0 (1) e- jωt dt + #- 1 (1) e- jωt dt 0
1
- jω t - jω t = ; e E + ; e E = 1 (1 − e - jω) + 1 (e jω − 1) − jω 0 − jω - 1 jω jω 1
0
- j ω /2 j ω /2 =e (e jω/2 − e - jω/2) + e (e jω/2 − e - jω/2) jω jω
(e jω/2 − e - jω/2) (e - jω/2 + e jω/2) jω = 2 sin ` ω j $ 2 cos ` ω j = 2 cos ω sinc` ω j ω 2 2 2 2π
=
SOL 3.30
Option (B) is correct. In option (A) z1 [n] = x [n − 3] z2 [n] = z1 [4n] = x [4n − 3] y [n] = z2 [− n] = x [− 4n − 3] = Y x [3 − 4n] In option (B) z1 [n] = x [n + 3] z2 [n] = z1 [4n] = x [4n + 3] y [n] = z2 [− n] = x [− 4n + 3] In option (C) v1 [n] = x [4n] v2 [n] = v1 [− n] = x [− 4n] y [n] = v2 [n + 3] = x [− 4 (n + 3)] = Y x [3 − 4n] In option (D) v1 [n] = x [4n] v2 [n] = v1 [− n] = x [− 4n] y [n] = v2 [n − 3] = x [− 4 (n − 3)] = Y x [3 − 4n]
SOL 3.31
Option ( ) is correct. The spectrum of sampled signal s (jω) contains replicas of U (jω) at frequencies ! nfs . Where n = 0, 1, 2....... 1 fs = 1 = = 1 kHz Ts 1 m sec GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SIGNALS & SYSTEMS
PAGE 141
SOL 3.32
Option (D) is correct. For an LTI system input and output have identical wave shape (i.e. frequency of input-output is same) within a multiplicative constant (i.e. Amplitude response is constant) So F must be a sine or cosine wave with ω1 = ω2
SOL 3.33
Option (C) is correct. Given signal has the following wave-form
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PAGE 142
SIGNALS & SYSTEMS
CHAP 3
Function x(t) is periodic with period 2T and given that x (t) =− x (t + T) (Half-wave symmetric) So we can obtain the fourier series representation of given function. SOL 3.34
Option (C) is correct. Output is said to be distortion less if the input and output have identical wave shapes within a multiplicative constant. A delayed output that retains input waveform is also considered distortion less. Thus for distortion less output, input-output relationship is given as y (t) = Kg (t − td ) Taking Fourier transform. Y (ω) = KG (ω) e - jωt = G (ω) H (ω) H (ω) & transfer function of the system d
So, H (ω) = Ke - jωt Amplitude response H (ω) = K Phase response, θn (ω) =− ωtd For distortion less output, phase response should be proportional to frequency. d
SOL 3.35
Option (A) is correct. G (z) z = e = αe− jω + βe− 3jω for linear phase characteristic α = β . jω
SOL 3.36
Option (A) is correct. System response is given as G (z) H (z) = 1 − KG (z) g [n] = δ [n − 1] + δ [n − 2] G (z) = z - 1 + z - 2 (z - 1 + z - 2) = 2 z+1 -1 -2 z − Kz − K 1 − K (z + z ) For system to be stable poles should lie inside unit circle. So
H (z) =
z #1 z = K! K2 + 4K # 2 − K
K2 + 4K # 1 K ! 2
K2 + 4K # 2
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CHAP 3
SOL 3.37
SIGNALS & SYSTEMS
PAGE 143
Option (C) is correct. Given Convolution is, h (t) = u (t + 1) ) r (t − 2) Taking Laplace transform on both sides, H (s) = L [h (t)] = L [u (t + 1)] ) L [r (t − 2)] We know that, L [u (t)] = 1/s L [u (t + 1)] = es c 12 m s L [r (t)] = 1/s2 L r (t − 2) = e - 2s c 12 m s s 1 H (s) = ;e ` jE;e - 2s c 12 mE s s
and
So
(Time-shifting property)
(Time-shifting property)
H (s) = e - s c 13 m s Taking inverse Laplace transform h (t) = 1 (t − 1) 2 u (t − 1) 2 SOL 3.38
Option (C) is correct. Impulse response of given LTI system. h [ n ] = x [ n − 1] ) y [ n ] Taking z -transform on both sides. H (z) = z - 1 X (z) Y (z) We have X (z) = 1 − 3z - 1 and Y (z) = 1 + 2z - 2 So
a x [n − 1]
Z
z - 1 x (z)
H (z) = z - 1 (1 − 3z - 1) (1 + 2z - 2) Output of the system for input u [n] = δ [n − 1] is , y (z) = H (z) U (z)
U [n]
Z
So Y (z) = z - 1 (1 − 3z - 1) (1 + 2z - 2) z - 1 = z - 2 (1 − 3z - 1 + 2z - 2 − 6z - 3) = z - 2 − 3z - 3 + 2z - 4 − 6z - 5 Taking inverse z-transform on both sides we have output. y [n] = δ [n − 2] − 3δ [n − 3] + 2δ [n − 4] − 6δ [n − 5] SOL 3.39
Option (B) is correct. A bounded signal always possesses some finite energy. E =
t0
#- t
g (t) 2 dt < 3
0
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PAGE 144
SOL 3.40
SIGNALS & SYSTEMS
CHAP 3
Option (C) is correct. Trigonometric Fourier series is given as 3
/an cos nω0 t + bn sin nω0 t
x (t) = A0 +
n=1
Since there are no sine , so bn = 0 bn = 2 T0
#0
= 2= T0
T0
#0
x (t) sin nω0 t dt
T0 /2
x (τ) sin nω0 τ dτ +
#T /2 x (t) sin nω0 t dt G T
0
Where τ = T − t & dτ =− dt T0 /2
= 2; T0
#T
= 2; T0
#T /2 x (T − t) sin n` 2Tπ T − t j dt ++ #T /2 x (t) sin nω0 t dt E
= 2; T0
x (T − t) sin nω0 (T − t) (− dt)+
0
#T /2 x (t) sin nω0 t dt E T
0
TO
T
0
O
T0
T0
#T /2 x (T − t) sin (2nπ − nω0) dt+ #T /2 x (t) sin nω0 t dt E 0
= 2 ;− T0
0
T0
T0
0
0
#T /2 x (T − t) sin (nω0 t) dt + + #T /2 x (t) sin nω0 t dt E
bn = 0 if x (t) = x (T − t) From half wave symmetry we know that if x (t) =− x`t ! T j 2 Then Fourier series of x (t) contains only odd harmonics. SOL 3.41
Option (C) is correct. Z -transform of a discrete all system is given as −1 ) H (z) = z − z−0 1 1 − z0 z It has a pole at z 0 and a zero at 1/z) 0. Given system has a pole at z = 2+30% = 2
( 3 + j) = ( 3 + j) 2
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CHAP 3
SIGNALS & SYSTEMS
PAGE 145
system is stable if z < 1 and for this it is anti-causal. SOL 3.42
Option (A) is correct. According to given data input and output Sequences are x [n] = {− 1, 2}, − 1 # n # 0 y [n] = {− 1, 3, − 1, − 2}, − 1 # n # 2 If impulse response of system is h [n] then output y [n] = h [ n] ) x [n] Since length of convolution (y [n]) is − 1 to 2, x [n] is of length − 1 to 0 so length of h [n] is 0 to 2. Let h [n] = {a, b, c} Convolution
y [n] = {− a, 2a − b, 2b − c, 2c} So, a=1
y [n] = {− 1, 3, − 1, − 2} -
2a − b = 3 & b =− 1 2a − c =− 1 & c =− 1 Impulse response h [n] = "1, − 1, − 1, SOL 3.43
Option ( ) is correct.
SOL 3.44
Option (D) is correct. Output y (t) = e - x (t) If x (t) is unbounded, x (t) " 3 y (t) = e - x (t) " 0 (bounded) So y (t) is bounded even when x (t) is not bounded.
SOL 3.45
Option (B) is correct. Given
y (t) =
t
# x (t') dt' −3
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PAGE 146
SIGNALS & SYSTEMS
CHAP 3
Laplace transform of y (t) X (s) , has a singularity at s = 0 Y (s) = s For a causal bounded input, y (t)=
t
# x (t') dt'
is always bounded.
−3
SOL 3.46
Option (A) is correct. RMS value is given by 1 T
Vrms = Where
#0
T
V2 (t) dt
2 T `T j t, 0 # t # 2
V (t) = * 0, So
1 T
#0
T
V 2 (t) dt = 1 = T
SOL 3.47
#0
T /2
2t 2 ` T j dt +
#T/2 (0) dt G T
T /2
T /2
3 t2 dt = 43 ; t E T 3 0 3 = 43 # T = 1 6 24 T = 1 V 6
#0
= 1 $ 42 T T
Vrms
T
Option (A) is correct. By final value theorem lim f (t) = lim s F (s) = lim s
t"3
s"0
s"0
(5s2 + 23s + 6) s (s2 + 2s + 2)
= 6 =3 2 SOL 3.48
Option (D) is correct. f (x) = sin2 x = 1 − cos 2x 2 = 0.5 − 0.5 cos 2x f (x) = A0 +
3
/an cos nω0 x + bn sin nω0 x n=1
f (x) = sin2 x is an even function so bn = 0 A0 = 0.5 − 0.5, n = 1 an = ) 0 , otherwise ω0 = 2π = 2π = 2 T0 T GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SOL 3.49
SIGNALS & SYSTEMS
Option (B) is correct. F (z) =
Z-transform
1 = 1− z = 1− 1 z+1 z+1 1 + z− 1
f (k) = δ (k) − (− 1) k Z 1 (− 1) k 1 + z− 1
so, Thus
SOL 3.50
PAGE 147
Option (A) is correct. Root mean square value is given as
#
1 T I2 (t) dt T 0 − 12 t, 0 # t < T 2 From the graph, I (t) = * ` T j 6, T /2 < t # T Irms =
So
1 T
#0
T
I2 dt = 1 = T
#0
T /2
− 12t 2 ` T j dt +
#T/2 (6) 2 dt G T
T /2
t3 = 1 e 144 + 36 6 t @TT/2 o ; 2 T T 3 E0 T3 + 36 T = 1 [6T + 18T] = 24 = 1 ; 144 c ` 2 jE T T T2 24 m Irms = SOL 3.51
24 = 2 6 A
Option (B) is correct. Total current in wire I = 10 + 20 sin ωt Irms =
SOL 3.52
(10) 2 +
(20) 2 = 17.32 A 2
Option (C) is correct. Fourier series representation is given as f (t) = A0 +
3
/an cos nω0 t + bn sin nω0 t n=1
From the wave form we can write fundamental period T = 2 sec Z 4 T ]]`T j t, − 2 # t # 0 f (t) = [ ]]−` 4 j t, 0 # t # T T 2 \ f (t) = f (− t), f (t) is an even function So, bn = 0 A0 = 1 f (t) dt T
#
T
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PAGE 148
SIGNALS & SYSTEMS
= 1= T
#- T/2 `T4 j tdt + #0 0
CHAP 3
T /2
4 `− T j tdt G
T /2
0
2 2 = 1 e 4 ;t E − 4 ;t E o T T 2 - T /2 T 2 0 2 2 = 1 ; 4 c T m − 4 c T mE = 0 T T 8 T 8 an = 2 f (t) cos nω0 t dt T
#
T
0 4 t cos nω t + T/2 − 4 t cos nω tdt = 2= 0 0 G ` ` Tj T - T /2 T j 0 By solving the integration 8 , n is odd 2 2 an = * n π 0, n is even So, f (t) = 82 8cos πt + 1 cos (3πt) + 1 cos (5πt) + ....B 9 25 π
#
SOL 3.53
#
Option (A) is correct. Response for any input u (t) is given as y (t) = u (t) ) h (t) y (t) =
h (t) " impulse response
#- 3 u (τ) h (t − τ) dτ 3
Impulse response h (t)and step response s (t) of a system is related as h (t) = d [s (t)] dt 3 3 So u (τ) d s [t − τ] dτ = d u (τ) s (t − τ) dτ y (t) = dt dt - 3 -3
#
SOL 3.54
#
Option (B) is correct. Final value theorem states that lim y (t) lim Y (s)
t"3
SOL 3.55
s"3
Option (D) is correct. 1 V2 (t) dt Vrms = T0 T
#
0
here T0 = π 1 T0
#TV2 (t) dt = π1 = #0
π /3
(100) 2 dt +
0
2π /3
#π/3
(− 100) 2 dt +
π
#2π/3 (100) 2 dt G
= 1 810 4 ` π j + 10 4 ` π j + 10 4 ` π jB = 10 4 V π 3 3 3 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY
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CHAP 3
SIGNALS & SYSTEMS
Vrms = SOL 3.56
PAGE 149
10 4 = 100 V
Option (D) is correct. Let h (t) is the impulse response of system y (t) = u (t) ) h (t) y (t) = =
#0 u (τ) h (t − τ) dτ t
#0 (2 + t − τ) e- 3(t - τ)u (τ) dτ t
So h (t) = (t + 2) e - 3t u (t), t > 0 Transfer function Y (s) 1 H (s) = = + 2 2 (s + 3) U (s) (s + 3) (2s + 7) = 1 + 2s +26 = (s + 3) (s + 3) 2 SOL 3.57
Option (B) is correct. Fourier series representation is given as v (t) = A0 +
3
/an cos nω0 t + bn sin nω0 t n=1
period of given wave form T = 5 ms DC component of v is A0 = 1 v (t) dt T
#
T
3
5
= 1 > # 1dt + # − 1dtH 5 0 3 1 = 63 − 5 + 3@ = 1 5 5 SOL 3.58
Option (A) is correct. Coefficient, an = 2 v (t) cos nω0 t dt T
#
T
3
5
= 2 > # (1) cos nwt dt + # (− 1) cos nwt dtH 5 0 3
sin nωt sin nωt = 2 f : nω D − : nω D p 5 0 3 2 π 2 π ω= = 5 T an = 1 6sin 3nω − sin 5nω + sin 3nω@ nπ 3
Put
5
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PAGE 150
SIGNALS & SYSTEMS
CHAP 3
= 1 ;2 sin b 3n 2π l − sin b 5n 2π lE nπ 5 5 = 1 ;2 sin b 6πn l − sin ^2nπhE nπ 5 = 2 sin b 6πn l nπ 5 Coefficient, bn = 2 v (t) sin nω0 t dt T
#
T
3
5
= 2 > # (1) sin nwt dt + # (− 1) sin nwt dtH 5 0 3
cos nωt cos nωt = 2 f 9− nω C − 9− nω C p 5 0 3 2 π 2 π put ω= = 5 T bn = 1 6− cos 3nω + 1 + cos 5nω − cos 3nω@ nπ = 1 6− 2 cos 3nω + 1 + cos 5nω@ nπ = 1 ;− 2 cos b 3n 2π l + 1 + cos b 5n 2π lE nπ 5 5 = 1 ;− 2 cos b 6πn l + 1 + 1E nπ 5 = 2 ;1 − cos b 6πn lE nπ 5 Amplitude of fundamental component of v is 3
5
v f = a12 + b12 a1 = 2 sin b 6π l, b1 = 2 b1 − cos 6π l π 5 π 5 vf = 2 π
sin2 6π + b1 − cos 6π l 5 5
2
= 1.20 Volt
***********
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