GATE 2023 SOLUTIONS SIGNALS and SYSTEMS (GATE EC SOLUTIONS SERIES)

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Citation preview

First Edition

Mechasoft

GATE 2023 EC conceptually empowered and error free chapter wise and year wise

S

olutions

to enhance solving skills

Signals and Systems GATE SOLUTIONS SERIES

Mechasoft Publishers

Rajeev Gupta

GATE -2023 EC conceptually empowered and error free

S

olutions

to enhance solving skills

Signals and Systems

Rajeev Gupta Mechasoft Publishers

Asst. Prof., Department of ECE MNNITAllahabad (INDIA)

Second Edition : 2022 MRP : 275.00 Copyright © : 2015,2022 Mechasoft Publishers No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording or otherwise or stored in a database or retrieval system without the prior written permission of the publisher. Mechasoft Publishers, Allahabad (INDIA)

E-mail : [email protected]

Dedicated to Promotion and Elevation of Engineering Education

and Way of Life as Gifted by

ALMIGHTY "of which We are an Integral Part"

M

e

h c

s a

t f o

Preface The prime objective of the Kindle version of ‘GATE , EC Signal and Systems , a sequel to GATE Solutions Series to Enhance Solving Skills' is to meet ever growing demand for awless, error free and succinct but conceptually empowered solutions to all GATE questions over the period 2003-2021. Step by step solutions together with theoretical explanations mark the strength of the text. Perusing the questions asked over the period of last 22 years , it is strongly felt that the concepts are repeated, not the questions. Keeping this in view , an adequate stress has been laid to build concepts while bringing out the solutions. This e-book is designed chapter wise and year wise to particularly supplement the texts for Electronics, Electrical and Instrumentation engineering. The key idea behind this text is to portray a ‘dialogue between a teacher and taught’ in the form of problems and solutions to unfold the intricacies. I do not sense any decit in believing that this title will, in many aspects, be different from the similar titles within the reach of students. This might fascinate readers to know that other published titles , GATE Solutions to enhance solving skills for EC, EE, Digital Electronics text book , Drill Series ; Anlaog , Digital Electronics, Control Systems, Communication Systems and Electrical Machines , are in wide circulation. Thousands of units of these titles have been shipped through www.ipkart.com , www.amazon.in and www.amazon.kindle.com .Over 500 reviews sought so far from renowned teachers , have been unarguably inspiring and render potential impetus to further develop allied texts. In particular, I wish to thank Shreyansh Mishra who typed the entire manuscript in expert fashion, not once but several times . I would also not forget to thank Mr. Triloki Nath Mishra, a sincere personality for his consistent persuasion to complete the task. Finally, I extend my appreciation to the family members, Shalini, Pranjali and Divyansh for their unwavering help and patience. The nal manuscript has been prepared with utmost care. However, going on line that there is always room for improvement in anything done , I would always welcome and greatly appreciate suggestions and corrections for further improvements.




































































Rajeev Gupta

Contents Preface

IV Questions

1.Continuous Time Signals

Solutions

1-6

7-12

13-20

21-30

3.Sampling Theorem and its applications

31-33

34-39

4.Discrete Systems

41-49

50-61

63-69

70-80

81

82

(Fourier series and Fourier transform ,Laplace transform )

2.Continuous Systems (causality , stability , impulse response , convolution, poles and zeros , frequency response , group/phase delay )

(causality , stability , impulse response , convolution ,poles and zeros , frequency response )

5.Discrete Time Signals (DTFT and z-transform )

6.GATE-2022 Signals and Systems

1

Fourier series and Fourier transform ,Laplace transform

Section 1

Questions

Continuous Time Signals

Chapter

Q1.The exponential Fourier series representation of a continuous time periodic signal x(t) is defined as ¥

x(t) =

åae

jkw0 t

k

k=–¥

where w0 is the fundamental angular frequency of x(t) and the coefficients of the series are ak . The following information is given about x(t) and ak . I. x(t) is real and even , having a fundamental period of 6 II. The average value of x(t) is 2 ìk , 1 £ k £ 3 III. ak = í k>3 î0 , The average power of the signal x(t) (rounded off to one decimal place) is ________ . ¥

Q2. X(w )is the Fourier transform of x(t) shown below . The value of

ò

X(w ) 2 dw

[GATE2021/2marks]

–¥

(rounded off to two decimal places )is ___________ .

[GATE2020/2marks]

x(t) 3 2 1 –2 –1 0 1 2 3 4

t

Q3. Let x(t) be a periodic function with period T = 10. The Fourier series coefficients for this ¥

series are denoted by ak , that is, x(t) = å a k e

jk

2p t T

k=– ¥

The same function x(t) can also be considered as a periodic function with period Tˊ= 40. Let bk be ¥

¥

the Fourier series coefficients when period is taken as Tˊ. If å | a k | = 16, then å | b k | is equal to k=– ¥

(a) 256

(b) 64

(c) 16

k=– ¥

(d) 4 [GATE2018/1mark]

02

Mechasoft

Q4. A periodic signal x(t) has a trigonometric Fourier series expansion ¥

x(t) = a 0 + å (a n cosnw 0 t + b n sin nw 0 t). n =1

If x(t) = – x( – t) = – x(t – π/ω0), we can conclude that (a) an are zero for all n and bn are zero for n even . (b) an are zero for all n and bn are zero for n odd . (c) an are zero for n even and bn are zero for n odd . (d) an are zero for n odd and bn are zero for n even .

[Set1/GATE2017/1mark]

Q5. Let x(t) be a continuous time periodic signal with fundamental period T = 1 second. Let {ak} be the complex Fourier series coefficients of x(t), where k is integer valued. Consider the following statements about x(3t): I. The complex Fourier series coefficients of x(3t) are {ak} where k is integer valued . II. The complex Fourier series coefficients of x(3t) are {3ak} where k is integer valued . III. The fundamental angular frequency of x(3t) is 6π rad/s. For the three statements above, which one of the following is correct? (a) only II and III are true (b) only I and III are true ( c) only III is true (d) only I is true [Set1/GATE2017/2marks] Q6. The Laplace transform of the causal periodic square wave of period T shown below, is f(t) 1 1 1 (b) F(s) = (a) F(s) = –sT/2 s 1 + e–sT/2 1+ e

(

1 (c) F(s) = s 1 – e–sT

(

(d) F(s) =

)

1 1 – e–sT

(

)

0

T/2

)

T 3T/2 2T

t

[Set1/GATE2016/2marks]

æ 2p ö t – cos(pt) is the input to an LTI system with the transfer function H(s)=es+e–s. Q7. A signal 2cos ç è 3 ÷ø

If Ck denotes the kth coefficient in the exponential Fourier series of the output signal then C3 is equal to (a) 0 (b) 1 ( c) 2 (d) [Set3/GATE2016/2marks] Q8. Let the signal f(t) = 0 outside the interval [T1, T2], where T1and T2 are finite. Furthermore, |f(t)| < ∞. The region of convergence (ROC) of the signal’s bilateral Laplace transform F(s) is (a) a parallel strip containing the jΩ axis.

(b) a parallel strip not containing the jΩ axis.

(c) the entire s-plane.

(d) a half plane containing the jΩ axis. ìï1 if a £ t £ b is ïî0 otherwise

[Set2/GATE2015/1mark]

Q9. The bilateral Laplace transform of a function f (t) = í

(a)

s

a-b s

(b)

e (a - b)

s

(c)

e

–as

-e s

– bs

(d)

e

s(a–b)

s [Set2/GATE2015/1mark]

Mechasoft

03 |ak|

Q10.The magnitude and phase of the complex Fourier series coefficient ak of a periodic signal x(t) are shown in the figure. Choose the correct statement from the four choices given. Notation : C is the set of complex number, R is the set of purely real numbers, and P is the set of purely imaginary numbers.

–5 –4 –3 –2 –1 0 1

(a) x(t) Î R

–5 –4 –3 –2 –1

3

3

2

2 1 k

2 3 4

Ðak 1 2 3 4

k

0

(b) x(t) Î P

–π ( c) x(t) Î(C – R) [Set2/GATE2015/1mark]

(d) the information given is not sufficient to draw any conclusion about x(t)

Q11. Let x(t) = α s(t) + s(–t) with s(t) = βe–4t u(t), where u(t) is unit step function. If the bilateral Laplace 16 transform of x(t) is X(s) = 2 - 4 < Re[s] < 4; then the value of β is ____________________. s - 16 [Set2/GATE2015/1mark] Q12. Consider the function g(t) = e–t sin (2πt) u(t) where u(t) is the unit step function. The area under g(t) is _________.

[Set3/GATE2015/1mark]

Q13.Consider the periodic square wave in the figure shown. x 1

0

1

2

3

t

4

-1 The ratio of the power in the 7th harmonic to the power in the 5th harmonic for this wave form is closest in value [Set2/GATE2014/1mark] to ____________. ¥

Q14. The value of the integral

2

ò sinc

(5t)dt is _________ .

[Set2/GATE2014/2marks]

–¥

Q15. Let g(t) = e

– pt 2

, and h(t) is a filter matched to g(t). If g(t) is applied as input to h(t), then the Fourier

transform of the output is (a)

e –π f

2

(b) e

–π f 2 /2

(c) e

–π f

(d) e

–2π f 2

[GATE2013/1mark]

04

Mechasoft

Q16. The Fourier transform of a signal h(t) is H(jω) = (2cosω) (sin 2ω) / ω. The value of h (0) is (a) 1/4

(b) 1/2

( c) 1

(d) 2 [GATE2012/2marks]

Q17.The trigonometric Fourier series of an even function does not have the (a) dc term

(b) cosine terms

( c) sine terms

(d) odd harmonic terms [GATE2011/1mark]

Q18.The trigonometric Fourier series for the waveform f(t) shown below contains f(t)

A

T

-T -

2 3T 4

-

T 2

-

T

T 0

T

3T

4

4

4

t

-2A

(a) only cosine terms and zero values for the dc components. (b) only cosine terms and a positive value for the dc components. (c) only cosine terms and a negative value for the dc components. (d) only sine terms and a negative value for the dc components.

[GATE2010/1mark]

3s + 1 ù –1 é Q19. Given f(t) = L ê 3 f (t) = 1 , then the value of k is 2 ú . If lim t ®¥ ë s + 4s + (k – 3)s û

(a) 1

(b) 2

(c) 3

(d) 4 [GATE2010/2marks]

Q20. The Fourier series of a real periodic function has only P: cosine terms if it is even Q: sine terms if it is even R: cosine terms if it is odd S: sine terms if it is odd Which of the above statements are correct? (a) P and S

(b) P and R

(c) Q and S

(d) Q and R

[GATE2009/1mark]

Mechasoft

05 ïì 1 for –1£ t £ +1 otherwise ïî 0

Q21.The signal x(t) is described by x(t) = í

Two of the angular frequencies at which its Fourier transform becomes zero, are (a) p, 2p

(b) 0.5p, 1.5p

( c) 0, p

(d) 2p, 2.5p [GATE2008/2marks]

Q22. Let x(t) « X (jw) be Fourier transform pair. The Fourier transform of the signal x(5t - 3) is 1 æ jw ö (a) e– j3w /5 X ç ÷ è 5ø 5

1 æ jw ö (b) ej3w /5 X ç ÷ è 5ø 5

1 æ jw ö (c) e– j3w X ç ÷ è 5ø 5

1 æ jw ö (d) ej3w X ç ÷ è 5ø 5

[GATE2006/1mark] Q23. Consider the function f (t) having Laplace transform F(s) = The final value of f (t) would be (a) 0

(b) 1

ω0 2

; Re[s] > 0

2

s +ω 0

(c) -1 £ f (¥) £ 1

(d) ¥ [GATE2006/2marks]

Q24.Choose the function f(t);– ¥ < t < ¥ for which a Fourier series cannot be defined (a) 3 sin (25t)

(b) 4 cos (20t + 3) + 2 sin (710t)

(c) exp(-|t|) sin(25t)

(d) 1

[GATE2005/1mark] Q25. The output y(t) of a linear time invariant system is related to its input x(t) by the following equations y(t) = 0.5x(t – t d + T) + x(t – t d ) + 0.5x(t – t d – T) .The filter transfer function H(w) is (a) (1 + cosωT)e

–jωt d

(b) (1 + 0.5cosωT)e

–jωt d

(c)(1 – cosωT)e

Q26. Match the following and choose the correct combination. Group 1

–jωt d

(d)(1 – 0.5cosωT)e

–jωt d

[GATE2005/2marks]

Group 2

E. Continuous and aperiodic signal

1. Fourier representation is continuous and aperiodic

F. Continuous and periodic signal

2. Fourier representation is discrete and aperiodic

G. Discrete and aperiodic signal

3. Fourier representation is continuous and periodic

H. Discrete and periodic signal

4. Fourier representation is discrete and periodic

(a) E-3, F-2, G-4, H-1

(b) E-1, F-3, G-2, H-4

[GATE2005/2marks] ( c) E-1, F-2, G-3, H-4 (d) E-2, F-1, G-4, H-3 Q27. For a signal x(t) the Fourier transform is X(f). Then ,the inverse Fourier transform of X(3f + 2) is

(a)

1 æ t ö j3πt xç ÷ e 2 è 2ø

(b)

1 æ t ö –j4πt/3 xç ÷ e 3 è 3ø

(c) 3x(3t)e

–j4πt

(d) x(3t + 2)

[GATE2005/2marks]

06

Mechasoft

Q28.The Fourier transform of a conjugate symmetric function is always (a) imaginary

(b) conjugate anti-symmetric

( c) real

(d) conjugate symmetric

[GATE2004/1mark]

Q29. Let x(t) and y(t) with Fourier transforms X(f) and Y(f) respectively be related as shown. Then ,Y(f) is x(t) y(t) 1 1 – jpf j2 pf (a) – X(f / 2)e (b) – X(f / 2)e 1 2 2 –2 0 (c) – X(f / 2)e

j2 pf

(d) – X(f / 2)e

– j2 pf

–2

0

2

t

t

–1 [GATE2004/2marks]

Q30. The Laplace transform of i(t) is given by I(s) = (a) 0

2 s(1 + s)

. At t ® ¥, the value of i(t) tends to

(b) 1

(c) 2

(d) ¥ [GATE2003/1mark] ¥

Q31. The Fourier series expansion of a real periodic signal with fundamental frequency f0 is gp (t) =

åce

j2 pf0 t

n

.

n=–¥

It is given that c3 = 3 + j5. Then, c-3 is (a) 5 + j3

(b) -3 -j5

(c) -5 +j3

(d) 3 -j5 [GATE2003/1mark]

Answer Key Q1 Q8 Q15 Q22 Q29

32 c d a b

Q2 Q9 Q16 Q23 Q30

58.64 c c c c

Q3 Q10 Q17 Q24 Q31

c a c c d

Q4 Q11 Q18 Q25

a –2 c a

Q5 Q12 Q19 Q26

b 0.155 d c

Q6 Q13 Q20 Q27

b 0.51 a b

Q7 Q14 Q21 Q28

b 0.2 a c

Section2

Solutions

Q1.x(t) is real and even Þ a – k = ak and average value of x(t) is 2 Þ a0 = 2 ìk , 1 £ k £ 3 ak = í Þ a –1 = a1 = 1, a –2 = a2 = 2 and a –3 = a3 = 3 k >3 î0 , 3

Use Parseval's theorem to get average power Pav =

k = –3

¥

Q2. Use Parseval's theorem to get 2

x(t)

1

(

0

(

)

)

3

¥ 2

X(w ) dw = 2p ´

–¥

0

28 = 58.64 3

1

ìï 4t3 üïù é1 4 ù 28 + 2t2 + t ýú = 2 ê + + 2 + 1ú = í 3 3 3 ë û 3 ú ïû 0þ îï

x(t)=2t+1

2

1

t+

é ìï t3 üï = 2 ê í + t2 + t ý + êë îï 3 ï –1 þ

–1

t)=

0

ò

–¥ 0

2

–1

)

2

x(t) dt = 2 ò (t + 1) dt + 2ò (2t + 1) dt = 2 ò t2 + 2t + 1 dt + 2ò 4t2 + 4t + 1 dt

–¥

(

)

¥

X(w ) dw = 2p ò x(t) dt

1 2

(

= a02 + 2 a12 + a22 + a32 = 22 + 2 12 + 22 + 32 = 32

1

x(

ò

0

2 k

2

ò –¥

¥

åa

–2 –1 0 1 2 3 4

t

Q3( c) The periodic function x(t) with time period T = 10, when considered as that with time period ¥

¥

Tˊ = 4 × 10 = 40, the values of Fourier series coefficients will not change, that is, å b k = å a k = 16 k– ¥

¥

Q4(a) In fourier expansion, x(t) = a 0 +

å (a

n

k– ¥

cos n w0 t + b n sin n w0 t)

n =1

x(t) = - x( - t) Þ x(t) has odd symmetry Þ a n = 0 for all n and x(t) = - x (t - p / w 0 ) = - x(t - T / 2); T = 2 p / w 0 implies that x(t) has half wave symmetry too. Half wave symmetry implies that bn = 0 for even n. Only bn for odd n, will survive. Q5(b) Fundamental period, T = 1sec and fundamental frequency w0 = ¥

x(t) =

å

¥

a k e jkw0 t

and

x(3 t) =

k =-¥

åa e

jk (3w0 )t

2p = 2p rad / sec. T

k

k =-¥

The fundamental frequency of x(3t), will be 3ω0 = 6π rad/sec but the Fourier series co-efficients( ak ) are same for both x(t) and x(3t). Q6(b)The periodic square waveformof period T, is expressible in terms of unit step functions as æ T ö æ 3T ö æ 5T ö f(t) = u(t) – u ç t– ÷ + u(t– T) – u ç t– ÷ø + u(t– 2 t) – u çè t– ÷ + ...... è 2 ø è 2 2 ø –

sT

–3

sT

–5

sT

1 e 2 e–sT e 2 e–2sT e 2 F(s) = – + – + – + ....... s s s s s s –sT sT –3 ù 1 1é 1 1 = ê1 – e 2 + e–sT – e 2 + ......ú = ´ = –sT –sT së û s 1 – æ –e 2 ö s æ 1 + e 2 ö çè ÷ø çè ÷ø

08

Mechasoft

æ 2p ö t – cos pt Q7(b) x(t)=2cos ç è 3 ÷ø

H(s)=es+e–s

y(t)=x(t+1)+x(t–1)

The output Y(s) = X(s) H(s) = es X(s) + e–s X(s) 2p 2p (t + 1) – cos p (t + 1) + 2cos (t – 1) – cos p (t – 1) 3 3 2p æ 2p ö The input x(t) = 2cos ç t – cos pt has two frequencies w1 = and w2 = p, è 3 ÷ø 3 2p 2p and corrensponding time periods T1 = = 3 and T2 = =2 w1 w2 y(t) = x(t + 1) + x(t – 1) = 2cos

x(t) and y(t) both will have same fundamental time period T0 = LCM (T1 ,T2 ) = 6 and corresponding frequency w0 =

2p p = T0 3

Note that C3 is the 3rd coefficient in the exponential series, that is ,coefficient of 3rd harmonic 3w0 = p. In order to find C3 , collect the terms of third harmonic in y(t) and that is 1 – cos p (t + 1) – cos p (t– 1) = – éëcos3w0 (t + 1) + cos3w0 (t – 1)ùû = – éëej3w0 (t +1) + e– j3w0 (t +1) + ej3w0 (t –1) + e– j3w0 (t –1) ùû 2 1 j3w0 = – éë e + e– j3w0 ej3w0 t + ej3w0 + e– j3w0 e– j3w0 t ùû 2 1 1 = – éë ejp + e– jp ej3w0 t + ejp + e– jp e– j3w0 t ùû = – ejp + e– jp éëej3w0 t + e– j3w0 t ùû 2 2 1 jp – jp C3 = – e + e = – cos p = 1 2

(

)

(

(

)

(

)

(

(

)

)

)

T2

Q8(c)The signal f(t) lying in interal [T1, T2] has bilateral Laplace transform F(s) = ò f (t)e

–st

dt

T1

Further , | f(t) |< ¥ and for s = s + jw to be in the ROC , it is required that f (t) e– st be absolutely T2

integrable , that is, I = ò | f(t) | e– st dt < ¥. T1

Note that I < ¥ for s < 0, s = 0 and s > 0 over the interval [T1 ,T2 ] and therefore , the ROC of F(s) includes entire s - plane . ¥

Q9( c) F(s) = ò f (t) e –¥

–st

b

dt = ò (1) e a

–st

dt =

1

b

ée –st ùû = a ( -s) ë

e

–as

-e

–bs

s

Q10(a) The observation of magnitude and phase of Fourier series coefficients, reveals that | ak |=| a – k | and Ðak = Ða –k , that is, ak is real and even . So , the periodic signal is real and even , that is, x(t) Î R.

Mechasoft

09

Q11. x(t) = α s(t) + s(–t) = α β e–4t u(t) + β e4t u(–t) X(s) =

ab s+4

-

b

=

(a –1) bs–4b

(a+1)

2

s-4

s –16

16

Compare with X(s) =

2

s - 16

: - 4 < Re[s] < 4 to get a = 1 and β = –2

é e j2 pt - e – j2 pt ù 1 –(1– j2 p )t –(1+ j2 p )t éëe u(t) = u(t) - e u(t) ùû Q12. g(t) = e sin 2 pt u(t) = e ê ú j2 j2 ë û –t

G(f ) =

–t

1 é

1

j2 êë (1 – j2 p ) + j2 pf

¥ 2p ù and area under g(t) is = 0.155 ò g(t)dt = G(0) = 2 ú (1 + j2 p ) + j2 pf û 1 + 4p –¥

1

-

Q13. The periodic square wave is an odd function. Fourier series coefficients a0 and an will be zero. é - cos nwt 1 ù 2 4 T/2 41 2p 2p bn = 1 - cos np ; w = = =p ò x(t)sin nwt dt = ò (1)sin nwt dt = 2 ê ú= T 0 20 n w n p T 2 0 ë û

[

]

ì4 ï for n = 1, 3, 5, 7 ¼.. b n = í np ïî 0 for n = 0, 2, 4, 6 ¼.. 2

th

th

Ratio of power in 7 harmonic to that in 5 harmonic =

b7

2

b5

=

(4 / 7 p )

2

(4 / 5p )

2

1 X(f ) = rect(f / 5) 5

Q14. Let x(t) = sinc (5t).Then, X(f)=(1/5) rect (f/5) ¥

()

Recall Parseval ' s relation ò x t

2

¥

¥

2

¥

2

2

dt = ò X (f ) df

–¥

–¥

2

2.5

= 0.51

1/5 2.5

f æ 1ö = 0.2 çè ÷ø df = 25 –2.5 –2.5 5

ò sinc 5t dt = ò X(f ) df = ò –¥

–¥

Q15(d) g(t) = e– pt

2

Matched Filter h(t)

–2.5

2.5

0

f

y(t)

Since , h(t) is a filter metched to g(t), h(t)is a time reversed and delayed version of g(t), that is, *

h(t) = g(T - t) Þ H(f ) = G (f ) e e

– pt 2

– j2 pfT

*

and Y(f ) = G (f ) H(f ) = G (f ) G(f ) e

– j2 pfT

FT ¬¾ ® e – pf Þ Y(f ) = e –2 pf |e – j2 pfT |Þ Fourier transformof output is Y(f) = e –2πf 2

2

2

= | G (f )| e 2

– j2 pfT

10

Mechasoft

Q16 (c) H( jw ) = Let x(t) =

(2 cos w ) (sin 2w ) w

2(e

=

jw

+e

– jw

) (sin 2w )

2w

=e

jw

sin 2w w

+e

– jw

sin 2w w x(t) = 0.5 rect (t/4) 1/2

1 4 sin 4pf sin (2w ) étù rect ê ú . Then, X(f ) = sinc (4f ) = 2 = 2 2 4pf w ë4û

FT x(t– 1) ¾¾ ® e– jw

sin (2w ) sin (2w ) FT , x(t + 1) ¾¾ ® e jw and h(t) = x(t – 1) + x(t + 1) –2 w w

t

2

0

Thus , h(0) = 1

h(t)=x(t–1)+x(t+1)

–1

x(t–1)

x(t+1)

1/2

1/2 t

3

0

–3

0

1 1/2 t

1

–1

–3

0

1

t

3

Q17( c) For an even function, trigonometric Fourier series contains only dc and cosine terms ,that is, the coefficients a0 and an exist while bn is always zero. The sine terms do not appear in the series. ¥

Q18 ( c) In Fourier series f (t ) = a 0 + å (a n cos nwt + b n sin nwt) n =1

dc component is a 0 =

1

T

T

0

ò f (t) dt =

1

T/4

ò A dt +

T – T/4

1

3T/4

T

T/4

T/4

(–2A )dt =

ò

3T/4

1 1 (At) + ( -2At) T T – T/4 T/4

=

A

–A= –

2

A 2

While dc component is negative, the waveform f(t) has an even symmetry and it will contain only cosine terms.

Q19 (d) lim f (t ) = lim sF (s) = lim t ®¥

s® 0

s® 0

s ´ (3s +1) 3

2

s + 4s + (k -3)s

=

1 1 and = 1 gives k = 4 k–3 k–3 ¥

(

Q20 (a)The Fourier series of a real periodic function is f (t ) = a 0 + å a n cos nwt + b n sin nwt n =1

)

If f(t) is even, Fourier series coefficients bn = 0. The series will contain only cosine terms. If f (t) is odd, Fourier series coefficients an = 0. The series will contain only sine terms. Statements P and S both are true.

Q21 (a) x(t) = rect (t / 2) and X(f ) = 2sinc (2f ) X(f )has zero crossing points coincident with f = ±1 / 2, ±1, ±3 / 2 and so on or w = ±p, ±2p, ±3p and so on.

–1

x(t)

X(f) 2

1

+1

t –1

–1/2

0

f 1/2

1

Mechasoft

11

FT FT 1 æ jw ö Q22 (a) x(t) ¬¾® X ( jw ). Use scaling property to get x(5t) ¬¾® X ç ÷ 5 è 5ø

ì æ 3 ö ü FT 1 æ jω ö –j3ω/5 Then , use time shifting property to get x í5 ç t - ÷ ý ¬¾® X ç ÷ e . 5 è 5ø î è 5ø þ w0 Q23 (c) F(s) = 2 Þ f (t) = sin w 0 t 2 s + w0 For any value of t, sinusoidal function f (t) alwayslies in the range – 1 £ f (t) £ +1.Thus, – 1 £ f( ¥ ) £ +1

Q24 ( c) Fourier series is defined for continues time periodic signals. Since , e

–t

sin 25t is not a periodic

signal , Fourier series can not be defined. Q25(a) y(t) = 0.5x (t - t d + T) + x (t - t d ) + 0.5 x (t - t d - T) Y(w ) = 0.5 X(w ) e Y(w ) X(w )

= H(w ) = e

– jw (t d –T)

– jw t d

+ X(w )e

– jw t d

+ 0.5X(w )e

(

– jw (t d + T)

=e

– jw t d

)

é1 + 0.5 e jwT + e – jwT ù = e –jωtd [1 + cosωT] ë û

Q26( c) The Fourier representation of different type of signals, is depicted below.

.

continuous aperiodic

continuous CFT

aperiodic

periodic FT DT

er uri Fo eries s

discrete

periodic

aperiodic

aperiodic periodic

Q27 (b) x(t) « X(f) ; x(at) «

discrete

Fourier series

periodic

1 æfö X ç ÷ and e– j2 pf0 t x(t) « X(f + f0 ) a è aø

1 æ tö 1 –j2π 2 t æ t ö x ç ÷ « X(3f ) and e 3 x ç ÷ « X(3f + 2) è 3ø 3 è 3ø 3

(

)

é1 + 0.5 e jwT + e – jwT ù X(w ) ë û

12

Mechasoft

Q28( c) A signal is said to be conjugate symmetric if its real part is even and imaginary part is odd, that is, x(t)=x*(–2).The signals and their corresponding Fourier spectra are depicted below . Signal Real

Imaginary

Fourier Spectrum

Even

Even

Odd

Odd

Odd

Odd

Even

Even

Real

Imaginary

Let x(t) = a(t) + jb(t) be a conjugate symmetric function where a(t) is even and b(t) is odd. Then, as depicted above, the conjugate symmetric function always provides real Fourier transform Q29 (b) As demonstrated below, y(t)= –x(2t+1) x(t)

x(2t)

1

1

–x(2t) 1

–1

–2 –1 0

1

2

t

–2 –1 0

–2

1

2

t

–x(2t+1)= y(t)

0

2

t

–2

–1

1 æfö x(t) ® X(f ) and – x {2 (t + 1)}® – X ç ÷ ej2πf 2 è 2ø 2 =2 Q30 ( c) lim i (t ) = lim sI (s) = lim s ´ t ®¥ s® 0 s® 0 s (s + 1)

Q31(d) For a real periodic signal, Fourier series coefficients have the following relation. *

Ck = C – k Þ C – k = Ck *

*

*

C –3 = C3 = (3 + j5) = 3 - j5

1

–1 0 –1

2

t

2

Continuous Systems

causality , stability , impulse response , convolution, poles and zeros , frequency response , group/phase delay

Chapter

Questions

Section 1

pö æ 2p ö æp t + 4cos ç t + ÷ where t is in seconds. Q1. Consider thesignal f (t) = 1 + 2cos (pt ) + 3sin ç è 3 ÷ø è2 4ø Its fundamental time period , in seconds ,is _________ .

[GATE2019/2marks]

Q2. Let the input be u and the output be y of a system, and the other parameters are real constants. Identify which among the following system is not a linear system: 3

(a)

d y dt

3

2

+ a1

d y dt

2

+ a2

dy dt

+ a 3 y = b 3u + b 2

du dt

2

+ b1

d u dt

2

(with initial rest conditions)

t

(b) y(t) = ò e a (t– t )bu( t ) d t 0

(c) y = au + b, b ≠ 0 (d) y = au

[GATE2018/1mark] Q3. Consider the following statements for continuous-time linear time invariant (LTI) systems. I. There is no bounded input bounded output (BIBO) stable system with a pole in the right half of the complex plane. II. There is no causal and BIBO stable system with a pole in the right half of the complex plane. Which one among the following is correct? (a) Both I and II are true (b) Both I and II are not true ( c) Only I is true (d) Only II is true [set1/GATE2017/1mark] Q4. A continuous time signal x(t) = 4 cos(200πt) + 8 cos(400πt) where t is in seconds, is the input to a linear time invariant (LTI) filter with the impulse response

ì 2 sin(300p t) ï , t¹0 h(t) = í pt ïî 600 , t=0 [set1/GATE2017/2marks]

Let y(t) be the output of this filter. The maximum value of |y(t)| is ________. t

Q5. Input x(t) and the output y (t) of a continuous-time system are related as y(t) = ò x(u)du. t–T The system is (a) linear and time-variant (b) linear and time-invariant (c) non-linear and time-variant (d) non-linear and time-invariant [set2/GATE2017/1mark]

14

Mechasoft

ì f ï1 - , f £ 20 Q6. Consider an LTI system with magnitude response H(f) = í 20 ï0, f > 20 î and phase response arg{H(f)} = – 2f. If the input to the system is x(t) = 8 cos (20 pt + ( p / 4)) + 16 sin (40 pt + ( p / 8)) + 24 cos (80 pt + ( p / 16)),

then ,the average power of the output signal y (t) is _________.

[set2/GATE2017/2marks]

Q7. The transfer function of a causal LTI system is H(s) = 1/s. If the input to the system is x(t) = [sin(t) /π t]u(t) where u(t) is a unit step function, the system output y(t) as t → ∞ is ______. [set2/GATE2017/2marks]

Q8.Consider the parallel combination of two LTI systems shown in the figure. h1(t) +

x(t)

y(t)

h2(t) The impulse responses of the systems are h1 (t) = 2d (t + 2) - 3d (t + 1) and h 2 (t) = d (t - 1). If the input x(t) is a unit step signal, then the energy of y(t) is ____________. [set2/GATE2017/2marks] Q9. The energy of the signal x(t) = Q10. If the signal x(t) = (a)

sin (t ) pt

sin(4pt) is _________ . 4pt

[set2/GATE2016/1mark]

sin (t ) sin (t ) * where (*) denotes convolution operation , then x(t) is equal to pt pt 2 æ sin (t )ö sin (2t ) 2sin (t ) (b) ( c) (d) ç 2pt pt è pt ÷ø

[set3/GATE2016/1mark] Q11. The waveform of periodic signal x(t) is shown in the figure. æ t - 1ö . A signal g(t) is defined by g(t) = x ç è 2 ÷ø The average power of g(t) is ________________.

x(t) 3 –2 –4 –3

1 –1

4 2 3

t

–3 [set1/GATE2015/1mark] Q12.The result of the convolution x(–t) * δ(–t –t0) is (a) x(t + t0)

(b) x(t – t0)

( c) x(–t + t0)

(d) x(–t –t0) [set1/GATE2015/1mark]

¥

Q13. The value of the integral

ò 12cos(2pt) –¥

sin(4pt) dt is _______________. 4pt

[set2/GATE2015/2marks]

Mechasoft

15

Q14.Input x(t) and output y(t) of an LTI system are related by the differential equation yʹʹ(t) – yʹ(t) – 6y(t) = x(t). If the system is neither causal nor stable, the impulse response h(t) of the system is

(a)

1 5

( c)

3t

e u( - t) +

1 5

e

–2t

u( - t)

1 1 (b) - e3t u( - t) + e –2t u( - t) 5 5 1 3t 1 –2t (d) - e u( - t) - e u(t) 5 5

1

1 –2t 3t e u( - t) - e u(t) 5 5

[set2/GATE2015/2marks]

Q15.A continuous, linear time – invariant filter has an impulse response h(t) described by ïì 3 for 0 £ t £3 otherwise ïî 0

h(t) = í

[set1/GATE2014/1mark]

When a constant input of value 5 is applied to this filter, the steady state output is _______________.

Q16.Let h(t) denote the impulse response of a casual system with transfer function

1 s +1

Consider the following three statements.

.

S1 : The system is stable. h(t + 1) S2 : is independent of t for t > 0 h(t) S3 : A non-casual system with the same transfer function is stable. (a) only S1 and S2 are true.

(b) only S2 and S3 are true.

(c) only S1 and S3 are true.

(d) S1, S2 and S3 are true. [set3/GATE2014/2marks] W Q17. A real valued signal x(t) limited to the frequency band f £ is passed through a linear time invariant 2

ì –j4πf W , f £ ïï e 2 system whose frequency response is H(f) = í W ï0 , f > ïî The output of the system is 2

(a) x (t + 4)

(b) x (t – 4)

(c) x (t + 2)

Q18.A stable linear time variant (LTI) system has a transfer function H(s) =

(d) x (t – 2) [set4/GATE2014/1mark] 1

. s +s –6 To make this system causal it needs to be cascaded with another LTI system having a transfer function H1(s). 2

A correct choice for H1(s) among the following options is (a) s + 3

(b) s – 2

(c) s – 6

(d) s + 1 [set4/GATE2014/2marks]

16

Mechasoft

Q19. A casual LTI system has zero initial conditions and impulse response h(t). Its input x(t) and output y(t) 2

are related through the linear constant-coefficient differential equation

d y(t) dt

t

Let another signal g(t) be defined as g(t) = a2 ò h(t ) dt + 0

2

+α

dh(t) + a h(t). dt

dy(t) dt

2

+ α y(t) = x(t).

[set4/GATE2014/2marks]

If G(s) is the Laplace transform of g(t), then the number of poles of G(s) is _____________________. Q20. The impulse response of a system is h(t) = t u(t). For an input u(t-1), the output is

(a)

t

2

(b)

t (t – 1)

2 t – 1) ( ( c) u(t – 1)

2

(d)

t –1

u(t – 1) 2 [GATE2013/1mark] Q21. For a periodic signal x(t) =30 sin100t + 10 cos300t + 6 sin(500t + p/4), the fundamental frequency is 2

u(t)

(a) 100 rad/sec

2

u(t – 1)

(b) 300rad/sec

2

(c) 500rad/sec

(d) 1500 rad/sec [GATE2013/1mark]

Q22. Which one of the following statements is NOT TRUE for a continuous time causal and stable LTI system? (a) All the poles of the system must lie on the left side of the jw axis. (b) Zeros of the system can lie anywhere in the s-plane. [GATE2013/1mark]

(c) All the poles must lie within |s| = 1 (d) All the roots of the characteristic equation must be located on the left side of the jw axis. t

Q23. The input x(t) and output y(t) of a system are related as y(t) = ò x(τ)cos(3τ)dτ. The system is –¥

(a) time invariant and stable

(b) stable but not time invariant

( c) time invariant but not stable

(d) neither time invariant nor stable [GATE2012/2marks] Q24. If the unit step response of a network is (1 - e-αt), then its unit impulse response is

(a) α e-αt

(b) α-1 e-αt

(c) (1 - α-1) e-αt

(d) (1-α) e-αt [GATE2011/1mark]

Q25. An input x(t) = exp (-2t) u(t) + d (t - 6) is applied to an LTI system with an impulse response h(t) = u(t). The output is (a) [1-exp (-2t)]u(t) + u (t + 6)

(b) [1-exp (-2t)] u(t) + u (t - 6)

( c) 0.5 [1-exp (-2t)]u(t) + u (t + 6)

(d) 0.5 [1-exp (-2t)] u(t) + u (t - 6) [GATE2011/2marks]

Mechasoft

17 2

Q26. A continuous time LTI system is described by

d y(t)

(a) (et - e3t ) u(t)

(b) (e-t - e-3t ) u(t)

+4

dy(t)

+ 3y(t) = 2

dx(t)

+ 4x(t) dt dt dt Assuming zero initial conditions, the response y(t) of the above system for the input x(t) = e-2t u(t) is 2

( c) (e-t + e-3t ) u(t)

(d) (et + e3t ) u(t) [GATE2010/2marks]

Q27. A function f(t) = sin2t + cos 2t has frequency components at (a) 0 and 1/2π Hz

(b) 0 and 1/π Hz

( c) 1/2π and 1/π Hz

(d) 0,1/2π and 1/π Hz

[GATE2009/2marks] ¥ Q28.Consider a system whose input x and output y are related as y(t) = ò x(t–τ) h (2τ)dτ –¥ h(t) where h(t) is shown in the graph. Which of the following four properties are possessed by the system? BIBO : Bounded input gives a bounded output Casual : The system is causal,

t

0

LP : The system is low pass. LTI : The system is linear and time invariant. (a) Casual, LP

(b) BIBO, LTI

( c) BIBO, Casual, LTI

(d) LP, LTI [GATE2009/2marks]

Q29. The input and output of a continuous time system are respectively denoted by x(t) and y(t). Which of the following description corresponds a causal system? (a) y(t) = x(t - 2) + x(t + 4)

(b) y(t) = (t - 4) x (t + 1)

( c) y(t) = (t + 4) x (t - 1)

(d) y(t) = (t + 5) x (t + 5)

[GATE2008/1mark]

Q30. The impulse response h(t) of a linear time invariant continuous time system is described by h(t) = exp (at) u(t) + exp (bt) u (-t) where u(-t) denotes the unit step function, and a and b are real constants. This system is stable if (a) a is positive and b is positive.

(b) a is negative and b is negative.

(c) a is positive and b is negative.

(d) a is negative and b is positive.

[GATE2008/1mark]

Q31. A linear, time - invariant, causal continuous time system has a rational transfer function with simple poles at s = - 2 and s = - 4 and one simple zero at s = - 1. A unit step u(t) is applied at the input of the system. At steady state, the output has constant value of 1. The impulse response of this system is (a) [exp (- 2t) + exp (- 4t)] u(t)

(b) [- 4 exp (- 2t) - 12 exp (- 4t) - exp (- t)] u(t)

( c) [- 4exp (- 2t) + 12 exp (- 4t)] u(t)

(d) [- 0.5 exp (- 2t) + 1.5 exp (- 4t)] u(t) [GATE2008/2marks]

18

Mechasoft

Q32. Let x(t) be the input and y(t) be the output of a continuous time system. Match the system properties P1, P2 and P3 with system relations R1, R2, R3, R4 Properties

Relations

P1 : Linear but NOT time - invariant

R1 : y(t) = t2 x(t)

P2 : Time - invariant but NOT linear

R2 : y(t) = t |x(t)|

P3 : Linear and time - invariant

R3 : y(t) = |x(t)| R4 : y(t) = x (t - 5)

(a) (P1, R1), (P2, R3), (P3, R4)

(b) (P1, R2), (P2, R3), (P3, R4)

( c) (P1, R3), (P2, R1), (P3, R2)

(d) (P1, R1), (P2, R2), (P3, R3)

[GATE2008/2marks]

Statement for linked Answer Questions Q 33 and Q 34 The impulse response h(t) of linear time - invariant continuous time system is given by h(t) = exp (-2t) u(t), where u(t) denotes the unit step function. Q33. The frequency response H(w) is

(a)

1 1 + j2ω

(b)

sinω

1

( c)

ω

(d)

2 + jω

jω 2 + jω

Q34. The output of this system, to the sinusoidal input r(t) = 2 cos 2t for all time t, is (a) 0

(b) 2-0.25cos (2t - 0.125p)

(c) 2-0.5 cos (2t - 0.125p)

(d) 2-0.5 cos (2t - 0.25p)

Q35.If the Laplace transform of a signal Y(s) = (a) -1

(b) 0

[GATE2008/2marks]

1

, then its final value is s(s – 1) (c) 1

(d) unbounded [GATE2007/1mark]

-t

Q36. The 3-dB bandwidth of the low-pass signal e u(t), where u(t) is the unit step function, is

(a)

1 2π

Hz

(b)

1 2π

2 – 1 Hz

(c) ¥

(d) 1Hz [GATE2007/2marks]

Q37.The frequency response of a linear, time variant system is given by H(f) = The step response of the system is (a) 5 (1 - e-5t ) u(t)

(b) 5[1 – e

–t/5

]u(t)

( c)

1 2

(1 – e

–5t

)u(t)

5 1 + j10πf

.

1 – t/5 (d) (1 – e )u(t) 5 [GATE2007/2marks]

Mechasoft

19

Q38. The Dirac delta function d(t) is defined as ì1 , t = 0 ì1 , t = 0 (a) d (t) = í (b) d (t) = í and î0, otherwise î0, otherwise ì¥ , t = 0 ( c) d (t) = í î0, otherwise

ì¥ , t = 0 (d) d (t) = í and î0, otherwise

¥

ò d(t)dt = 1 –¥ ¥

ò d(t)dt = 1

[GATE2006/1mark]

–¥

Q39. In the system shown below, x(t) = sin t u(t). In steady-state, the response y(t) will be 1 (a) 1 sin(t – p / 4) (b) sin(t + p / 4) 2 2 1 x(t) s +1 1 –t (c) (d) sin t - cos t e sin t 2

y(t) [GATE2006/1mark]

Q40. A low- pass filter with frequency response H( jw ) = A(w )e– jf (w ) produces no phase distortions if (a) A(w ) = Cw2 , f(w ) = kw3

(b) A(w ) = C, f(w ) = kw

( c) A(w ) = Cw, f(w ) = kw2

(d) A(w ) = C, f(w ) = kw–1

[GATE2006/1mark] -3t

Q41. The unit step response of a system starting from rest is given by c(t) = 1-e for t ³ 0. The transfer function of the system is 1 (a) (b) 3 1 + 2s 3+ s

( c)

1 2+s

(d) 2s 1 + 2s

[GATE2006/2marks] Q42.The unit impulse response of a system is f(t) = e-t, t ³ 0. For this system the steady-state value of the output for unit step input is equal to (a) -1

(b) 0

(c) 1

(d) ¥

[GATE2006/2marks] Q43.The function x(t) is shown in the figure. Even and odd parts of a unit step function u(t) are respectively x(t) 1 1 (a) , x(t) 2 2 1

1 1 (b) – , x(t) 2 2 1

1

1

(d) – , – x(t) 2 2

( c) , – x(t) 2 2

[GATE2005/1mark]

Q44. Which of the following can be impulse response of a causal system? h(t) h(t) h(t) (a)

(b) t

h(t)

( c)

t

t

0 -1

1

(d) t

t [GATE2005/1mark]

20

Mechasoft

πö æ Q45. The power in the signal s(t) = 8cos ç 20πt – ÷ + 4sin(15πt) is è 2ø (a) 40 (b) 41 (c) 42

(d) 82 [GATE2005/1mark] Q46. Let x(t) be the input to a linear, time-invariant system. The required output is 4x(t-2) The transfer function of the system should be (a) 4ej4pf

(b) 2e-j8pf

(c) 4e-j4pf

(d) 2ej8pf [GATE2003/1mark]

Common data for Q47 & Q48 The system under consideration is an RC low-pass filter (RC-LPF) with R = 1kW and C = 1.0mF . Q47. Let H(f ) denote the frequency response of the RC-LPF. Let f1 be the highest frequency such that | H(f1 ) | ³ 0.95 for 0 £| f |£ f1.Then f1 (in Hz) is [GATE2003/2marks] H(0) (a) 324.8

(b) 163.9

(c) 52.2

(d) 104.4

Q48. Let tg(f ) be the group delay function of the given RC-LPF and f2 = 100 Hz. Then tg(f2) in ms, is (a) 0.717

(b) 7.17

(c) 71.7

(d) 4.505 [GATE2003/2marks]

Answer Key Q1 Q8 Q15 Q22 Q29 Q36 Q43

12 7 45 c c a a

Q2 Q9 Q16 Q23 Q30 Q37 Q44

c 0.25 a a d b d

Q3 Q10 Q17 Q24 Q31 Q38 Q45

d a d a c d a

Q4 Q11 Q18 Q25 Q32 Q39 Q46

8 2 b d a a c

Q5 Q12 Q19 Q26 Q33 Q40 Q47

b d 1 b c b c

Q6 Q13 Q20 Q27 Q34 Q41 Q48

8 3 c b d b a

Q7 Q14 Q21 Q28 Q35 Q42

0.5 b a b d c

Section2

GATE Solutions

æ 2p ö æp ö t + 4cos ç t + p ÷ 4ø è 3 ÷ø è2

Q1. The signal f (t) = 1 + 2cos ( p t) + 3sin ç

p , 3 2 2p 2p 2p and corresponding time periods T1 = = 2, T2 = = 3 , and T3 = = 4. w1 w2 w3 has 3 periodic functions with frequencies w1 = p , w 2 =

2p

and w 3 =

Then , f (t) has fundamental time period ,T0 = LCM(T1 ,T2 ,T3 ) = LCM [2,3, 4] = 12sec Q2( c) y = au + b represents a non linear system. It neither follows the law of additivity (supper position) nor the law of multiplicativity (homogeneity). System

Additivity test : u1

aλu + b ≠ λy. In fact, λy = λau + λb

System

Multiplicativity test : λu

y2 = au2 + b

y* = a(u1 + u2) + b ≠ y1 + y2 .In fact, y1 + y2 = a(u1 + u2) + 2b

System

u* = u1 + u2

System

u2

y1 = au1 + b

Q3(d) For a continuous time, linear, time invariant (LTI) and causal system, to be BIBO (bounded input bounded output) stable, all the poles must be in the left half of complex plane. A LTI system may be BIBO stable system even if its poles are in the right half of complex plane, but it is possible only if the system is non causal. Q4. x(t) = 4cos200pt + 8 cos 400 p t and X(f) = 2d (f - 100) + 2d (f + 100) + 4d (f - 200) + 4d (f + 200)

ì 2 sin(300p t) ï ;t ¹ 0 h(t) = í pt ïî600 ;t = 0 where

2 sin(300 p t) pt

= 600

h(t) 600

sin(300 p t) 300 p t

= 600 sinc(300 t)

h(t) = 2 ´ 300 sinc(300 t) and H(f) = 2P (f/ 300)

t −2/300

y(t) = x(t) * h(t) and Y(f) = X(f) H(f)

X(f) 4

H(f)

−150

0

1/300

2/300

Y(f)

2

2 −200 −100 0 100 200 f(Hz)

−1/300

4 150 f(Hz)

−100 0

100

Y(f) = 4d (f - 100) + 4d (f + 100) and y(t) = 8 cos 200 p t and the maximum value of y(t) is 8.

f(Hz)

22

Mechasoft t

Q5(b) The output y(t) = ò x(u) du ; x(t) is input .The integrator represents a linear system . t–T t–t 0

t

ˆ = ò x(u - t 0 )du = The time shift in input, that is x(t - t 0 ) gives output y(t) t–T

ò t–t 0 –T

x(u) du = y(t - t 0 )

The system is time invariant.

ì f ï1 - ; f £ 20 Q6. H(f) = í 20 ï0 ; f > 20 î

H(f)

arg{H(f)}

1

slope=1/20

-1/20=slope 10

1/2 −20 −10 0 10 20

arg{H(f)} = -2f

f(Hz)

−20

f(Hz)

–20 slope = −2

pö pö pö æ æ æ Input x(t) = 8 cos ç 20 pt + ÷ + 16 sin ç 40 pt + ÷ + 24 cos ç 80 pt + ÷ è è è 4ø 8ø 16 ø 144244 3 1442443 144244 3 x1 (t),10Hz

x 2 (t),20Hz

20

x 3 (t),40Hz

Since , |H(f)| = 0 for f ≥ 20Hz and system is linear, the response y(t) will be zero due to inputs x2(t) and x3(t). Only, due to x1(t), y(t) ≠ 0. 10 Y(f) = X1 (f) H(f) and H(f) f =10 = 1 = 0.5 and arg{H(f)} = -20 for f = 10Hz and 20 for f = -10. 20 y(t) = 0.5 ´ 8 cos(20 p t +

p 4

- 20) = 4 cos(20 p t +

Average power of output signal y(t) is Pav =

X1 (f)

4

p 4

- 20)

2

2

=8

arg{x(f)} π/4

4

−10 0

−10

0

10

f(Hz)

10

f(Hz)

−π/4 arg{Y(f)}

p - 20 4

Y (f) 2

−10 0

−10

0

10

f(Hz)

-

p + 20 4

10

f(Hz)

Mechasoft

23

é sin(t) ù u(t) and H(s) = 1 / s Q7. The input x(t) = ê ë pt úû LT ˆˆˆ † sin t u(t) ‡ˆˆ ˆ sin t u(t) t X(s) =

1 2

s +1 ¥

LT ˆˆˆ † ‡ˆˆ ˆò

s

; Re[s] > 0

1

–1

2

s +1

¥

ds = tan (s)

s

=

p 2

–1

- tan s

1 ép

1 ép ù –1 ù –1 - tan s and Y(s) = X(s) H(s) = - tan (s) ê ú ê úû p ë2 ps ë 2 û

lim y(t) = y(¥ ) = lim sY(s) = lim

t ®¥

s® 0

s® 0

1 ép

ù –1 - tan (s) = 0.5 ê úû p ë2

Q8. y(t) = x(t) * h1 (t) + x(t) * h 2 (t) ; x(t) = u(t) h1 (t) = 2d (t + 2) - 3d (t + 1) and h 2 (t) = d (t - 2) Since, convolutoin of any function with impulse, leaves the function itself . y(t) = 2 u(t + 2) - 3 u(t + 1) + u(t - 2) as depicted below. y(t)

y2(t) area=4

4

2

area=3

1 0

−1

2

t

−2

−2

0

−1

t

2

−1 2

Energy of y(t), E y = ò y (t) dt = total area = 4 + 3 = 7 ¥

¥

2 2 ò | x(t) | dt = ò | X(f ) | df

Q9. Recall Parseval's theorem for energy signal x(t) , that is, energy E =

–¥

–¥

sin (4pt ) 1 æfö FT x(t) = = sinc(4t) ¾¾ ® X(f ) = rect ç ÷ è 4ø 4pt 4 Area under X(f ) 2 vs f , that is, E = 0.25 J X(f)

X(f) 1/4

−2

1 16 = 0.25

area = 4 ´

1/16

0

2

f(Hz)

−2

2

0

2

f(Hz)

24

Mechasoft

ì æ töü sin p ç ÷ ï ï è pø ï 1 sin(t) sin(t) sin(t) 1 ï æ tö Q10(a) x(t) = * and let x1 (t) = = í ý = sinc ç ÷ è pø pt pt pt pï æ tö ï p pç ÷ ïî è p ø ïþ X1 (f ) =

1 sint p ´ rect (pf )}= rect (pf )and X(f ) = X1 (f )X1 (f ) = X1 (f ) = rect (pf ) Þ x(t) = x1 (t) = { p πt

X1 (f)

X1 (f)

1

X1 (f)

1

−1/2π

f

1/2π

0

−1/2π

1

=

× 1/2π

0

f

−1/2π

f

1/2π

0

Q11.The time shifting and time scaling operations in a periodic signal, do not change the average power of the signal. The average power Pav of g(t) is T/2

1

2

1

1 é 9t

3

1

ù Pav = ò x (t)dt = ò (–3t) dt = ê ú =2 T – T/2 3 –1 3ë 3 û –1 1

2

Q12 (d) The convolution of any function with impulse, is the shifted version of function itself, that is, x(t) * δ (t –t0) = x(t – t0) and x(–t) * δ (–t –t0) = x(–t–t0) .

é sin 4pt ù = x1 (t)x 2 (t) ë 4pt úû

Q13. Let x(t) = 12cos (2 pt )ê

where x1 (t) = 12 cos (2 pt ) and x 2 (t) =

sin 4 pt 4 pt

X2(t) 1

= sinc(4 t) –3/4

æfö X1 (f) = 6 {d (f– 1) + d (f + 1)} and X 2 (f ) = rect ç ÷ è 4ø 4 1

–2/4

–1/4

0

1/4

t 2/4

3/4

X(f ) = X1 (f ) * X 2 (f ) is plotted below noting that convolution of any function with impulse leaves the shifted version of function itself .

X2(f)

X1(f)

0

¥

ò 12 cos 2 pt –¥

X(0) = 3 1/4

6 –1

X(f)

* 1

f(Hz)

sin 4 pt 4 pt

1.5

= –2

0

+2

f(Hz)

f(Hz) –3

¥

dt = ò x(t) dt = X(f ) f =0 = X(0) = 1.5 + 1.5 = 3.0 –¥

–1

0

1

3

Mechasoft

25

→ [s2 Y(s) – s y(0) – yʹ(0)] – [s Y(s) – y(0)] – 6 Y(s) = X(s) Y(s) 1 1 1é 1 1 ù With zero initial conditions , H(s) = = 2 = = ê X(s) s - s - 6 (s - 3)(s + 2) 5 ë (s - 3) (s + 2) úû

Q14(b) yʹʹ(t) – yʹ(t) – 6y(t) = x(t)

Note (i) for a non causal LTI system , the ROC of the transfer function lies to the left of the left most pole. (ii) An LTI system is unstable if and only if the ROC of the transfer function does not include the jω axis.

The given system is non causal and unstable . The ROC should be the left of the left most pole so that it does jω not include jω axis . s-plane 1 1 3t –2t « - e u( - t) ; ROC Re[s] < 3 and « - e u( - t) ; ROC Re[s] < -2 –2 s-3 s+2 σ 3 1 3t 1 –2t ROC The impulse response h(t) = - e u( - t) + e u( - t) Re[s] < –2 5 5 h(t)

x(t) 3

Q15. 5 t

y(t) 3

0

é1

h (t) = 3[u(t) - u(t - 3)] Þ H(s) = 3 ê

ës

y(t) = x(t) * h(t) and Y(s) = X(s)H(s) =

15 s

2

e

t

–3s

ù 5 and x (t) = 5u(t) Þ X(s) = ú s û s

(1 – e ) –3s

2 ö üï 3s) 15 ì ( ï æ The steady state output y(¥ ) = lim y(t) = lim sY(s) = lim 1 – 1 – 3s + – .... í ç ÷ ý = 45 t ®¥ s® 0 s® 0 s 2! ø ïþ ïî è 1 Im Q16(a)Given H(s) = , impulse response of the causal system is s-plane s +1 h(t) = e-t u(t), that is, h(t) = 0 for t < 0 and ROC Re [s] > -1 Re 0 -1 The ROC includes jw axis and the system is stable. Statement S1 is true.

h(t + 1) h(t)

=

e

–(t +1)

e

–t

–1 – t

=

e e e

–t

=

1 e

is independent of t for t > 0 . Statement S 2 is true.

If the system with same transfer function is not causal, that is, 1 h(t) ¹ 0 for t < 0 and h(t) = -e-t u(-t). Then,H(s) = ; ROC : Re [s] < - 1 . s +1 The ROC does not include jw axis and the system is unstable. Statement S3 is false.

Q17(d) The output y(t) = x(t) * h(t) and Y(f) = X(f) H(f) = X(f)e– j4 pf Þ y(t) = x(t – 2)

Im

-1

0

s-plane Re

26

Mechasoft

Q18 (b) H(s) =

1

1

=

2

s +s-6

=

(s + 3) (s - 2)

1é 1

1 ù ê 5 ës - 2 s + 3ú û

Im

–3 +2 Re . Since, the given system is stable , ROC must include jw axis so that impulse response is -1 2t 1 –3t h(t) = e u ( - t) - e u(t) ; ROC : - 3 < Re[s] < 2 5 5 In order to make the stable system causal its ROC must be the right half plane to the right most pole but continue to include jω axis . Case (1) if cascaded with H1(s) = (s+3) the over all transfer function is H 0 (s) =

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

=

1

2t

(s - 2)

Þ h 0 (t) = - e u( - t) ROC : Re[s] < 2 (non causal)

Case (2) if cascaded with H1(s) = s-2 the overall transfer function is s-2

1

=

Þ h 0 (t) = e

–3t

u(t) ROC : Re[s] > 3 (causal ) (s + 3)(s - 2) (s + 3) 2 d y dy 2 2 2 Q19. +a + a y(t) = x(t) Þ s Y(s) - sy(0) - y '(0) + a[sY(s) - y(0)] + a Y(s) = X(s) 2 dt dt H 0 (s) =

2

2

Plug zero ICs to get s Y(s) + asY(s) + a Y(s) = X(s) and H(s) = 2

2

t

g(t) = a ò h( t ) dt+ 0

X(s)

=

1 2

2

(s + as + a )

a dh(t) + a h(t) and G(s) = H(s) + s H(s) + a H(s) dt s

2 2 é a 2 + s 2 + as ù 1 (s + as + a ) 1 = ´ = ú (s 2 + as + a 2 ) s s s ë û

G(s) = H(s) ê

Q20( c) The output y(t) = x(t) * h(t) where x(t) = u(t – 1) Þ X(s) = Y(s) = X(s) H(s) =

Y(s)

e

–s

s

3

Þ y(t) =

(t - 1)

2

has one pole at s = 0. e –s 1 and h(t) = tu(t) Þ H(s) = 2 s s

u(t - 1)

2

Q21(a) Let x(t) = 30sin100t + 10cos300t + 6sin(500 t + p / 4) = x1 (t) + x2 (t) + x3 (t) 2p 2p , x2 (t) =10cos300t has time period T2 = 100 300 2p and x3 (t) = 6sin (500t + p / 4 )has time period T3 = . 500 2p 2p Fundamental time period of x(t) is T = LCM(T1 ,T2 ,T3 ) = and angular frequency w = = 100 rad / sec 100 T Q22( c) For system to be causal and stable all the poles must lie on the left side of the jω axis. The zeroes of the causal and stable system can lie any where in s–plane .The causal and stable system is not required to have all the poles within |s| = 1. The stable system must have all the roots of the characteristics equation located on the left side of jω axis where x1 (t) = 30sin100t has time period T1 =

Mechasoft

27

Q23 (a)The system can be put in the form of diagram as shown below. x(t)

y(t)

ò (.) dt

cos3t Since, delayed input x(t) generates output y(t) delayed by same amount, the system is time invariant. if x(t) is bounded, that is,

t The system is stable. | x(t) |< ¥ , then| y(t) | £ ò |x( t )||cos3t| dt will also be bounded. –¥ Note that absolute value of integral is never greater than integral of absolute value of integrand.

Q24 (a) Unit step response s(t) = 1 – e

–a t

and unit impulse response h (t ) =

Q25 (d) y(t) = x(t) * h(t) Þ Y(s) = X(s)H(s) where X(s) =

Y(s) =

1 s(s + 2)

+

2

d y(t)

Q26 (b)

dt x(t) = e Y(s) =

–2t

2

+4

e

–6s

s d dt

=

1 é1

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

and H(s) =

–αt

1 s

d dt

2

x(t) + 4x(t) Þ s Y(s) + 4sY(s) + 3Y(s) = 2sX(s) + 4X(s)

s+2 =

2 (s + 1) (s + 3)

1 - cos 2t 2

and sinusoidal component equal to

1

=

1 (s + 1)

+ cos 2t =

1 2

-

+

1 (s + 3)

1 2

(

)

and y(t) = e –t –e –3t u(t)

cos 2t has dc component equal to

cos 2t of frequency f =

2 Q28(b)Since, h(t) ≠ 0 for t < 0, the system is not causal.

1

1 2

with f = 0Hz

Hz .

π

¥

y(t) = ò x(t – t )h(2 t )dt £ ò –¥

1- e – at )= αe ( dt

1

Q27(b) f (t) = sin t + cos 2t =

¥

–6s

d

–6s

1

-

y(t) + 3y(t) = 2

(2s + 4)

2

(s + 2)

+e

dt

s (t ) =

ù e + and y (t) = 0.5 éë1-e –2t ùû u(t) + u(t - 6) ê ú 2 ës s + 2 û s

u(t) and X(s) =

2

1

d

x(t – t ) h(2 t ) dt

–¥

The absolute value of integral is never greater than integral of absolute value of integrand. The bounded input will guarantee the bounded output. System is BIBO Stable.

28

Mechasoft

The convolution is valid only for linear time invariant (LTI) systems. The impulse response h(t) is of form sinw0t rect (t/T) and its Fourier transform H(f) = sinc (f + f0) + sinc (f - f0) as depicted below. The system is not low pass. In fact , it is band pass with centre frequency f0 . H(f)

0

-f0

f

f0

Q29 ( c) A system is said to be causal if its output at any time depends on only present and past values of input . y(t) = x(t - 2) + x(t + 4) and at t = 0 , y(0) = x(-2) + x(4) it is non causal system, output depends on future value. y(t) = (t - 4) x(t + 1) and at t = 0 , y(0) = -4x(1), it is non causal system, the output depends on future value. y(t) = (t + 4) x(t - 1) and at t = 0 , y(0) = 4x(-1) , it is causal system , the output depends on past value . y(t) = (t + 5) x(t + 5) and at t = 0 , y(0) = 5x(5) , it is non causal system , the output depends¥on future value. Q30(d) A system is said to be stable if its impulse response is absolutely integrable, that is ò h (t ) dt < ¥ –¥ ¥ ¥ 0 at bt at bt dt + ò e dt ò e u (t )+ e u (- t ) dt = ò e –¥ 0 –¥ ¥ 0 First term ò eat dt is finite only if α is negative and second term ò ebt dt is finite only if β is positive. 0 –¥

Q31( c) The transfer function of the system is H(s) =

Y(s) X(s)

=

K(s + 1) (s + 2)(s + 4)

The output y(t) in steady state is y(¥ ) = lim y(t) = lims Y(s) = lim t ®¥

Now, H(s) =

8(s + 1) (s + 2) (s + 4) 2

= 2

12 (s + 4)

-

4 (s + 2)

[

where X(s) =

K(s + 1)s

s(s + 2)(s + 4) – 4t –2t and h(t) = 12e u(t) – 4e u(t) s® 0

s® 0

]

Q32(a) R1: y(t) = t x(t) ; t a1x1 (t) + a 2 x 2 (t) = a1y1 (t) + a 2 y 2 (t) Þ linear 2

t x (t - t 0 ) ¹ y(t - t 0 ) Þ time variant R 2: y(t) = t| x (t)| ; t | a1x1 (t) + a 2 x 2 (t) | ¹ a1y1 (t) + a 2 y 2 (t) Þ non linear t | x (t - t 0 )| ¹ y(t - t 0 ) Þ time variant R 3: y (t) = | x (t)| ; | a1x1 ( t) + a 2 x 2 (t)| ¹ a1y1 (t) + a 2 y 2 (t) Þ non linear |x (t - t 0 )| = y(t - t 0 ) Þ time invaraint R 4: y(t) = x (t - 5) ; a1x1 (t - 5) + a 2 x 2 (t - 5) = a1y1 (t) + a 2 y 2 (t) Þ linear x( t - 5 - t 0 ) = y (t - t 0 ) Þ time invariant Note a1x1 (t) + a 2 x 2 (t) ¹ a1x1 (t) + a 2 x 2 (t)

1 s

=1Þ K = 8

Mechasoft

29

Q33( c) The impulse response is h(t) = e

–2t

Q34(d)The input r(t) = 2cos 2t has frequency w = 2rad / sec and H(w ) = 1

H(w ) w= 2 = H(2) =

1

=

2

2 2

w +4

and Ð H(w ) w= 2 = ÐH(2) = – tan–1

Output y(t) = 2 H(2) cos [2t – ÐH(2) ] =

Q35(d) Since, the signal Y(s) =

1 s(s - 1)

1

u(t) and its frequency response is H(w ) =

2 2 2

jω + 2

1 jw + 2

2 = – 45o = –0.25p rad 2

cos (2t – 0.25p) = 2–0.5 cos (2t – 0.25π)

has a pole in right half of s - plane , the final value theorem

is not applicable .In fact , the signal y(t) is of growing nature and its final value is unbounded . Y(s) =

1 s(s - 1)

=

-1

Q36(a) The signal e 1

|H(w )|=

2

w +1 Then,|H(w )|

Q37(b) H(f ) =

s –t

+

1

t

s -1

t

and y(t) = -1 + e where e grows without bound.

u(t) has frequency transfer function H(w ) =

1

. Let w b be 3 - dB band width .

w=w b

|H(ω)|

jw + 1

1 1/ 2

ωb=1 =

1 2 wb

= +1

1 2

2

gives w b + 1 = 2, w b = 1 rad / sec and f b =

1

ω

Hz.

2π

5 5 5 1 or H( jw ) = or H(s) = = 1 + j10pf 1 + j5w 1 + 5s s + (1 / 5)

Unit step resonse Y(s) =

Q38(d) As D ® 0,

1 D

1 5 5 = – Þ y(t) = 5 éë1 – e–t/5 ùû u(t) s(s + 1 / 5) s s + 1 / 5

® ¥ while the area under impulse remains unity .

d (t) : D ® 0

ïì¥ for t = 0 ïî0 otherwise

1

d (t) = í

D ¥

and area under d (t), that is, ò d (t)dt = 1 –¥

D

1 1 1 p and H(j w ) = ; H(j1) = and ÐH(j1) = – 45o = – rad s +1 jw + 1 4 2 1 æ pö sin ç t – ÷ In steady state , response y(t) = H( j1) sin {t + ÐH( j1)} = è 4ø 2

Q39(a) H(s) =

Q40(b)The output of a low pass filter depends on frequency response H(jω)=A(ω)ej Φ(w) It will not produce any phase distortion only if phase Φ(ω) varies linearly with ω, that is, Φ(ω) = kw.

t

30

Mechasoft

Q41(b) The unit step response c(t) = 1-e-3t and unit impulse response g(t) =

3

The system transfer function G(s) =

d c(t) dt

= 3e

–3t

s+3

Q42 ( c) The unit impulse response f (t) = e-t gives the transfer function F(s) =

1 s +1

The response for unit step input r(t) = u(t) or R(s) =1/s will be Y(s) = F(s)R(s) =

1 s (s + 1)

s

The steady state value of output, y(¥ ) = lt sY(s) = lt

=1 s(s + 1) Q43 (a) Let unit step function be expressed as sum of even and odd parts ,u(t) = even éë u(t) ùû + odd éë u(t) ùû s® 0

even éë u(t) ùû =

u(t) + u( - t) 2

and odd éë u(t) ùû =

=

s® 0

2

u(t) - u( - t) 2

u(–t) 1

u(t) 1

1 1

=

2

1

sgn (t) =

t

0

t

0

odd [u(t)]

even [u(t)] 1/2

x(t)

2

1/2 t

0 Q44 (d) A causal system has the impulse response, h (t) = 0 for t < 0. p æ ö Q45(a) The power Pav of signal s(t) = 8 cos ç 20 p t - ÷ + 4 sin15 p t = 8sin 20 p t + 4 sin15 p t is è ø 2 2 2 8 4 Pav = + = 40 2 2 Y(f ) Q46( c) The output y(t) = 4x (t – 2) gives Y(f ) = 4X(f )e– j4 pf and H(f ) = = 4e– j4πf X(f )

Q47( c) The product RC = 103 ´ 10–6 = 10–3 sec and H(f ) = H(f ) f =0 = H(0) = 1 and H(f ) =

1

(

1 + (2pfRC)

H(f1 ) 1 ³ 0.95 Þ H(0) H(0) 1 + 2 ´ 10–3 pf1

(

2

)

–1

1 + 2 ´ 10–3 pf

R = 1kW 2

)

C = 1mF

³ 0.95 Þ f1 £ 52.3Hz .

1 d æ 2p f ö and group delay function t g (f ) = 3 ÷ ø 10 2 p df

Q48(a) Phase response ÐH(f ) = - tan ç è

()

t g f2 =

1 2p

´

2p 10

3

´

1 2 2 4p f2 6

æ ö çè1 + 10 ÷ø

=

10

–3

2

–3

4

æ 4p ´ 10 ö çè1 + 106 ÷ø

t

1 / j2pfC 1 = R + (1 / j2pfC) 1 + j2pfRC

1

=

2

0 –1/2

= 0.717 ´ 10 sec = 0.717ms

é –1 2 pf ù êë- tan 103 úû

3

Sampling Theorem

Section 1

Questions

and its applications

Chapter

Q1. Consider a real valued base band signal x(t), band limited to 10kHz. The Nyquist rate for the signal tö æ y(t) = x(t)x ç1 + ÷ is è 2ø (a) 15kHz

(b) 30kHz

( c) 60kHz

Q2. A band limited low-pass signal x(t) of bandwidth 5 kHz is sampled at a sampling rate fs. The signal x(t) is reconstructed using the reconstruction filter H(f) whose magnitude response is shown below: The minimum sampling rate fs (in kHz) for perfect reconstruction of x(t) is _______.

(d)20kHz [GATE2021/1mark] |H(f)| K

–8 –6

0

6

8

f(kHz)

[GATE2018/2marks] Q3.The signal x(t) = sin(14000πt), where t is in seconds, is sampled at a rate of 9000 samples per second. The sampled signal is the input to an ideal low pass filter with frequency response H(f) as follows : H(f) =

ìï1, f £ 12kHz í ïî0, f > 12kHz.

What is the number of sinusoids in the output and their frequencies in kHz? (a) Number = 1, frequency = 7 (b) Number = 3, frequencies= 2,7,11 ( c) Number = 2, frequencies = 2, 7 (d) Number = 2, frequencies = 7, 11 [set2/GATE2017/2marks] Q4. A continuous time sinusoid of frequency 33 Hz is multiplied with a periodic Dirac impulse train of frequency 46Hz . The resulting signal is passed through an ideal low pass filter with a cut off frequency of 23Hz. The fundamental frequency (in Hz ) of the output , is _____________. [set1/GATE2016/1mark] Q5. A continuous time function x(t) is periodic with period T. The function is sampled uniformly with a sampling period Ts. In which one of the following cases is the sampled signal periodic ? (a) T = 2 Ts

(b) T = 1.2Ts

( c) Always

(d) Never [set1/GATE2016/1mark] Q6.Consider the signal x(t)=cos (6πt) +sin (8πt) where t is in seconds. The Nyquist sampling rate (in samples/seconds) for the signal y(t)=x(2t+5) is (a) 8 (b) 12 ( c)16 (d)32 [set3/GATE2016/1mark]

32

Mechasoft

pö æ Q7. The signal cos ç 10 pt + ÷ is ideally sampled at a sampling frequency of 15 Hz. è 4ø æ sin( pt) ö æ The sampled signal is passed through a filter with impulse response ç cos 40 pt è pt ÷ø çè The filter output is 15 æ sin( pt) ö pö æ 15 pö æ cos ç 10 pt + ÷ (a) cos ç 40 pt - ÷ (b) ç ÷ è 2 è pt ø 4ø è 2 4ø

( c)

pö æ cos ç 10 pt - ÷ è 2 4ø

15

(d)

15 æ sin( pt) ö

pö æ çè ÷ø cos çè 10pt - ÷ø 2 pt 2

pö

÷.

2ø

[set2/GATE2015/1mark]

æ sin( pt / 2 ö ¥ * å d (t - 10n) where ‘*’ denotes the è ( pt / 2) ÷ø n = – ¥

Q8. Consider a continuous time signal defined as x(t) = ç

convolution operation and t is in seconds. The Nyquist sampling rate (in samples/sec) for x(t) is __________. [set3/GATE2015/2marks] Q9. Let x (t) = cos (10πt) + cos (30πt) be sampled at 20 Hz and reconstructed using an ideal low- pass filter with cut- off frequency of 20 Hz. The frequency/ frequencies present in the reconstructed signal is/are (a) 5 Hz and 15 Hz only

(b) 10 Hz and 15 Hz only

( c) 5 Hz , 10 Hz and 15 Hz only

(d) 5 Hz only

[set3/GATE2014/1mark]

Q10. A band-limited signal with a maximum frequency of 5 kHz is to be sampled. According to the sampling theorem, the sampling frequency which is NOT valid, is (a) 5 kHz

(b) 12 kHz

(c) 15 kHz

Q11.The Nyquist sampling rate for the signal s(t) = (a) 400 Hz

sin(500 pt)

(b) 600 Hz 2

Q12. An LTI system having transfer function

sin(700π t)

× πt πt (c) 1200 Hz

(d) 20 kHz [GATE2013/1mark] is

(d) 1400 Hz [GATE2010/2marks]

s +1

and input x(t) = sin (t + 1) is in steady state. 2 s + 2s + 1 The output is sampled at a rate ws rad/s to obtain the final output {y(k)}. Which of the following is true? (a) y(.) is zero for all sampling frequencies ws (b) y(.) is nonzero for all sampling frequencies ws ( c) y(.) is nonzero for ws > 2, but zero for ws < 2 (d) y(.) is zero for ws > 2, but nonzero for ws < 2

[GATE2009/2marks]

Mechasoft

33

Q13. A signal m(t) with bandwidth 500 Hz is first multiplied by a signal g(t) where ¥

g(t) =

å (–1) d (t – 0.5 ´ 10 k ) k

–4

k=–¥

The resulting signal is then passed through an ideal low pass filter with bandwidth 1kHz. The output of the low pass filter would be (a) d(t)

(b) m(t)

(c) 0

(d) m(t)d(t) [GATE2006/2marks]

Q14.The minimum sampling frequency (in samples/sec) required to reconstruct the following signal from its samples without distortion, is 3

æ sin2π×1000t ö æ sin2π×1000t ö x (t ) = 5 ç + 7ç ÷ ÷ø è ø è πt πt (a) 2×103

2

(b) 4×103

(c) 6×103

(d) 8×103 [GATE2006/2marks]

Q15. Let x(t) = 2 cos (800pt) + cos (1400pt) be sampled with the rectangular pulse train shown in figure. The only spectral components (in kHz) present in the sampled signal in frequency range 2.5 kHz to 3.5 kHz are p(t) (a) 2.7, 3.4 (b) 3.3, 3.6

3

( c) 2.6, 2.7, 3.3, 3.4, 3.6 (d) 2.7, 3.3 T 0 T0 – 0 6 6

–T0

t

T0

T0 =10–3sec

Answer Key Q1 Q8 Q15

b Q2 0.4 Q9 d

13 a

Q3 Q10

b a

Q4 Q11

13 c

Q5 Q12

b a

Q6 Q13

c Q7 c Q14

a c

Section2

Solutions

t ö FT æ FT Q1(b) Let x(t) ¾¾ ® X(f ), then x ç1 + ÷ ¾¾ ® 2X(2f )ej4 pf where the factor 2ej4 pf is insignificant in è 2ø finding the largest frequency components in the spectrum of the signal y(t). The convolution of X(f) and X(2f) as demonstrated below, reveals that the largest frequency content in signal y(t) is 15kHz and the Nyquist rate of sampling is 2×15=30kHz . Y(f) X(f) X(2f)

=

* –10

0

10

f(kHz)

f(kHz)

–5 0 5

–15

0

15

Q2. Let x(t) ƒ X(f) ; | X(f) | = 0 for | f | > 5 kHz ¥ The signal x(t) sampled at rate fs has spectrum X d (f) = fs X(f) + fs å X(f - nfs ) n=– ¥ Note that Xδ(f) is periodic in fs and weighted by fs. n ¹0 As demonstrated below , for perfect reconstruction, fs ≥ 2 × 5kHz and ( fs –5) ≥ 8 or fs ≥ 13 kHz or fs,min = 13kHz. Xδ(f) Reconstruction filter pass band X(f) k X(0) fsX(0) .... f(kHz)

–5

0

–fs–5 –fs

5

–8

–6 –5

56

0

–fs+5

8 fs–5

fs

f(kHz)

.... f(kHz) fs+5

Q3(b) The frequency of input signal, fx = 7kHz and sampling rate, fs = 9kHz. The frequencies at the output of sampler, are fx, fs ± fx, 2fs ± fx, 3fs ± fx and so on, that is, 7kHz, 9 ± 7kHz(= 2kHz and 16kHz), 18 ± 7(= 11kHz and 25kHz), 27 ± 7(= 20kHz and 27kHz) and so on. X(f) When sampled signal is passed through low 0.5 pass filter with cut off frequency equal to –7 12kHz, the frequencies present at the output f(kHz) 0 +7 of low pass filter, will be 2kHz, 7kHz and 11kHz (number = 3). For more insight, x(t) = sin14000πt –0.5 X(f) = 0.5 δ(f – 7 ×103) – 0.5 δ(f + 7 × 103) ¥ The sampled signal Xδ(t) has spectrum X d (f) = fs X(f) + fs å X(f - nfs ) n=– ¥ n ¹0

pass band of low pass filter

Xδ(f)

4500 –25

–7

–16 –20

–11

2 –2

11 7

20 16

25

f(kHz) –4500

Mechasoft

35

Q4. Multiplying sinusoidal signal of frequency fx = 33Hz with impulse train of frequency 46Hz, is equivalent to sampling the sinusoidal signal with sampling frequency fs = 46 Hz and the frequency components generated in doing so, are nfs ± fx ; n = 0, 1, 2, 3, ........ =33, 46 ± 33, 92 ± 33, ....... = 13, 33, 79, 59, 125 Hz and so on. On passing these frequency components through low pass filter with cut off frequency of 23 Hz, the fundamental frequency of output will be 13Hz. Q5(b) The function x(t) is periodic with period T and the sampled signal x(nTs ) will be periodic T is rational, that is, T = 1.2Ts . Ts Q6( c) Signal x(t)=cos(6πt) +sin (8πt) has the largest frequency component equal to 4Hz. Then, the signal y(t)=x(2t+5) will have the largest frequency component equal to 8Hz. The time compression by a factor of 2, results is frequencies of sinusoidal components increased by a factor of 2 while time delay does not change the frequencies. The Nyquist rate of sampling for signal y(t) is fs=2×8=16 samples/sec . if

Q7(a) The signal is x(t) = cos(10πt + π/4) is ideally sampled at a sampling frequency fs = 15 Hz. The spectrum of the signal x1(t) = cos(10π t) is X 1 (f ) = 0.5 éd ë (f - 5 ) + d (f + 5 )ùû as demonstrated below. The initial phase π/4 of x(t) is ignored for the time being to be accounted later in the final filter output. ¥

(

The ideally sampled signal X1d (f ) = fs X1 (f ) + fs å X1 f - nfs n = –¥ n ¹0

15

=

15 2

|X1(f)| 1/2

15

ëéd (f + 5) + d (f - 5)ûù + 2 ëéd (f + 5 - 15) + d (f - 5 - 15) 2 + d (f + 5 + 15) + d (f - 5 + 15)ùû + L =

)

–5

0

+5

f(Hz)

éëd (f + 5) + d (f - 5) + d (f + 10) + d (f - 10) + d (f + 20)+d (f - 20) + Lùû X1d(f)

15 / 2

–20

–15

–10

–5

0

5

10

15

The impulse response of filter h(t) =

pö æ cos ç 40 pt - ÷ = sinc(t) sin 40 pt = h1 (t) ´ h 2 (t) , è pt 2ø

sin pt

where h1(t) = sinc(t) and h2(t) = sin 40πt H1 (f ) = rect (f ) and H 2 (f ) =

1 j2

[d (f - 20)- d (f + 20)]

fs

20

f(Hz)

36

Mechasoft

H(f) = H1(f) * H2(f) ; * denotes convolution. Recall replication property, that is, the convolution of any function with impulse leaves the function itself. The frequency response of filter H(f) is illustrated below. |H(f)| |H (f)| 2

|H1(f)|

1

*

–0.5 0 0.5

1/ 2

1/ 2

= f(Hz)

–20

f(Hz)

20

–20.5 –19.5 –20

19.5

0

-1/ 2

20 20.5

f(Hz)

-1/ 2

On passing the sampled signal [X1δ(f)] through filter [H(f)], the spectral components δ(f ± 20) will only be available at the output as demonstrated below. The filter output Y (f ) = X1d (f ) H(f ) = and y(t) =

15 2

sin 40 pt =

15 j4

[d (f - 20)- d (f + 20)]

pö æ cos ç 40 pt - ÷ . è 2 2ø

15

Taking into the account the initial phase p/4, the final filter output

15 / 4

–20 0

p ö 15 + ÷ = cos (40πt - π 4). 2 2 4ø 2 ¥ sin( pt / 2) * å d (t - 10n) = x1 (t) * x 2 (t) Q8. Let x(t) = n=– ¥ ( pt / 2) ¥ sin( p t/ 2) where x1 (t) = = sinc(0.5t) and x 2 (t) = å d (t–10n) ( p t/ 2) n=–¥ ¥ 1 ¥ n æ ö X1 (f) = 2 rect(2 f) and X 2 (f) = å d ç f - ÷ = 0.1 å d (f - 0.1n) n=– ¥ 10 n = – ¥ è 10 ø =

15

æ è

|Y(f)|

cos ç 40 pt -

p

X1(f)

20

-15 / 4

X2(f)

2

0.1

–0.25

0

0.25

f(Hz)

–0.4 –0.3 –0.2 –0.1

0

0.1 0.2

f(Hz)

0.3 0.4

X(f)

x(t) = x1(t) * x2(t) and X(f) = X1(f) X2(f) 0.2

The largest frequency in the spectrum of x(t) is 0.2 Hz. The Nyquist sampling rate is 2 ´ 0.2 = 0.4 Hz. –0.2 –0.1

0

0.1 0.2

f(Hz)

f(Hz)

Mechasoft

37

Q9(a) The signal x(t) has two frequency components f1 = 5 Hz and f2 = 15 Hz. The sampling frequency fs = 20 Hz. The sampled signal will have the frequency components f1 , f2 , fs ± f1 , fs ± f2 , 2 fs ± f1 , 2fs ± f2 and so on. These frequency components are 5 Hz , 15 Hz , 20 ± 5 = 15 Hz and 25 Hz, 20 ± 15 = 5 Hz and 35 Hz, 40 ± 5 = 35 Hz and 45 Hz , 40 ± 15 = 25 Hz and 55 Hz and so on.When these components are passed through ideal low pass filter with cut off frequency of 20 Hz, only components with frequencies 5 Hz and 15 Hz will be within pass band and be present in reconstructed signal. Q10(a) According to the sampling theorem, the sampling frequency fs ≥ 2fx ; fx = maximum frequency in band limited signal. For fx = 5kHz, fs ≥ 10 kHz and therefore , fs = 5kHz is not valid. Q11( c) s(t) =

sin 500pt sin 700pt æ f ö æ f ö ´ = 35 ´ 104 sinc(500t) sinc(700t) and S(f) = rect ç * rect ç è 500 ÷ø è 700 ÷ø pt pt rect(f/700)

rect(f/500)

S(f)

* –250

0

f(Hz)

250

= –350

0

350

f(Hz)

0

–600

600

f(Hz)

The largest frequency in the spectrum S(f) is 600Hz and Nyquist sampling rate = 2 × 600Hz = 1200Hz. Q12(a) The transfer function H(s) =

Y(s) s2 + 1 1 – w2 = 2 and H( jw ) = X(s) s + 2s + 1 1 – w2 + j2w

For sinusoidal input x(t) = sin(t + 1), the output y(t) = H( j1) sin(t + 1 + ÐH( j1) Since, H( jw ) w=1 = H( j1) = 0, y(.) is zero for all sampling frequencis ws . ¥

Q13( c)The signal g (t ) =

å ( -1)

k

d (t - 0.5 ´ 10 -4 k) represents impulse train of alternate

k = -¥

polarity +1 for k = 0, ± 2, ± 4.... and -1 for k = ± 1, ± 3, ± 5,.............with periodicity 0.05 m sec, as shown below.

g(t) 1 ...... –0.2

0.05

–0.05

–0.15

0

–0.1

0.15 0.2

0.1

...... t(ms)

–1 ¥

Let g1 (t) = å d (t - 10

–4

k)

g1(t)

k=– ¥

1 ...... –0.3

–0.2

–0.1

0

0.1

0.2

0.3

...... t(ms)

38

Mechasoft –4

–4

ù and G(f) = G1 (f) é1 - e – j p´10 f ù ë û ë û +¥ ¥ 4 4 4 4 – jp´10 –4 f ù é where G1 (f ) = 10 å d (f - 10 k )and G (f ) = 10 å d (f - 10 k )ù é1 - e û êë k = – ¥ úû ë k =-¥

Then , g(t) = g1 (t) – g1 é t– 0.5 ´ 10

Let X(f ) = 1 - e

– jp´10 –4 f

–4

–4

= 1 - cos p ´ 10 f + jsin p ´ 10 f

4

then, for f = k ´ 10 and k = 0, X(f ) = 1 - 1 = 0 4

for f = ± k ´ 10 and k is odd, X(f ) = 1 - ( -1) = 2 4

for f = ± k ´ 10 and k is even, X(f ) = 1 - 1 = 0 4 4 ì 0 for k = 0, ± 2, ± 4,.,.,. or f = 0, ±2 ´ 10 , ±4 ´ 10 Hz,.,.,. ï ï ¥ Thus, G(f ) = í 4 4 4 4 ï2 ´ 10 k =å– ¥ d (f - 10 k ) for k = ±1, ± 3, ± 5,.,.,.or f = ± 10 , ±3 ´ 10 Hz,.,.,. ï î

G( f ) 2×10

–50

–30

–10

M(0)

4

0

10

30

50

f(kHz)

–0.5

M(f)

0

0.5

f(kHz)

The output of multiplier y(t) = m(t) ´ g(t) and Y(f) = M(f) *G(f) ; * denotes convolution. Noting that convolution of any waveform with impulse leaves the wave form itself, Y(f) is sketched below. Form Y(f), it is obvious that no portion of spectrum of y(t) falls within pass band of low pass filter with bandwidth 1 kHz. Thus , the output of low pass filter is 0. Y(f ) pass band of LPF

–30 –30.5 –29.5

–10 –10.5

–1 –9.5

2×104×M(0)

0

f (kHz) 10 1 9.5 10.5

30 29.5

30.5

Mechasoft

39 3

æ sin 2p ´ 1000t ö æ sin 2p ´ 1000t ö Q14(c)The signal x(t) = 5 ç ÷ø + 7 çè ÷ø è pt pt =

2

1 7 é1 1 é3 3 3 ù 3 ù sin 2 p ´ 10 t - sin 2 p ´ 3 ´ 10 t + 2 2 úû p t êë 2 - 2 cos 2p ´ 2 ´ 10 t úû p t êë 4 4 5

3 3

has the largest frequency = 3 ´ 103 Hz. 3 Minimum (Nyquist) sampling frequency = 2 ´ 3 ´ 10 = 6 ´ 103 samples/sec Q15(d) The spectral components present in x(t) = 2cos(800p t) + cos(1400p t), are 0.4 kHz and 0.7 kHz. The pulse train p(t) has fundamental period T0 = 10−3 sec and fundamental period f0 = 1 kHz. Use exponential form of Fourier series to get ¥

p(t) =

åC

n

n=–¥

= =

1 T0

e

j2 pnt /T0

1 where Cn = T0

T0 /6

ò – T0 /6

3 ´ e– j2 pnt /T0 dt =

T0 /2

ò

p(t)e– j2 pnt /T0 dt

– T0 /2

T0 / 6 T0 3 ´ ´ éëe– j2 pnt /T0 ùû –T0 / 6 T0 – j2pn

3 é e– jpn/3 – ejpn/3 ù 3 ´ ú = np sin(np / 3) pn êë – j2 û

Now, it is easy to identify from C n =

3 æ np ö sin ç ÷ that all harmonics except integral multiples of è 3ø np

3 ( i.e. 3, 6, 9 and so on), are present in p(t). The spectral components that are present in sampled signal s(t) = x(t) p(t) are 1 ± 0.4 and 1 ± 0.7 or 0.6, 1.4, 0.3 and 1.7 kHz 2 ± 0.4 and 2 ± 0.7 or 1.6, 2.4, 1.3 and 2.7 kHz 4 ± 0.4 and 4 ± 0.7 or 3.6, 4.4, 3.3 and 4.7 kHz and so on. Out of these, the spectral components present in the range 2.5 kHz to 3.5 kHz, are 2.7 and 3.3 kHz.

4

Discrete Systems

causality , stability , impulse response , convolution ,poles and zeros , frequency response

Chapter

Questions

Section 1 Q1. For a unit step input u [n ], a discrete - time LTI system produces an output signal æ æ 1ön ö 2d [n + 1]+ d [n ]+ d [n – 1]. Let y [n ] be the output of the system for an input ç ç ÷ u [n ]÷ . è è 2ø ø The value of y [0] is ___

[GATE2021/2marks]

Q2. The output y[n] of a discrete time system for an input x[n] is y[n] = max x[k] . – ¥£ k £ n The unit impulse response of the system is (a) 0 for all n (b) 1 for all n ( c) unit step signal u[ n ] (d) unit impulse signal δ [n] [GATE2020/1mark] Q3.Which one of the following pole -zero plots corresponds to the transfer function of an LTI system characterized by the input-output difference equation given below? 3

k

y[n] = å (–1) x[n – k] k =0 Im 1 3rdorder pole

(a)

1

–1

Re

Im 1

(b)

1

–1

Im

Im

( c)

–1

1

rd

3 order pole 1

–1

Re

–1

–1

1

3rdorder pole

Re

(d)

4rthorder pole 1

–1

Re

–1 [GATE2020/1mark]

42

Mechasoft

K(z – a) where K and α are real z + 0.5 numbers. The value of α (rounded off to one decimal place ) with |α| < 1 , for which the magnitude response of the system is constant over all frequencies , is ______. [GATE2020/2marks] æ 1ö Q5. Let H(z) be the z-transfrom of real-valued discrete-time signal h[n]. If P(z) =H(z)H ç ÷ has a zero at è zø 1 1 z = + j , and P(z) has a total of four zeros, which one of the following plots represents all the zeros 2 2 correctly?

Q4. The transfer function of a stable discrete- time LTI system is H(z) =

(a)

(b)

Im z- plane

Im

2

z- plane 2

|z|=1

0.5

|z|=1 0.5

–2

0.5

Re

2

–0.5

–2

–0.5

0.5

2

Re

–2 –2 Im

( c)

(d)

z- plane

2

2

z- plane

1

|z|=1 –2

Im

|z|=1

0.5 0.5 –0.5 –0.5

0.5

Re 2

–2

0.5 1

–1 –0.5

Re 2

–1 –2

–2 [GATE2019/1mark]

Mechasoft

43

Q6. It is desired to find a three- tap causal filter which gives zero signal as an output to an input of the form æ jpn ö æ jpn ö x [n ]= c1 exp ç + c 2 exp ç , ÷ è 2 ø è 2 ÷ø where c1and c2 are arbitrary real numbers. The desired three-tap filter is given by h[0] = 1, h[1] = a , h[2] = b and h [n] = 0 for n < 0 or n > 2 What are the values of the filter taps a and b if the output is y [n] =0 for all n, when x [n] is as given above? n =0

x[n]

(a) a = 1 , b = 1

¯ h[n] = {1, a, b}

(b) a = 0 , b = –1

y[n] = 0

(c) a = –1 , b = 1

(d) a = 0 , b = 1 [GATE2019/2marks]

Q7. Let h[n] be a length-7 discrete-time finite impulse response filter, given by h[0] = 4 , h[1] = 3, h[2] = 2, h[3] = 1, h[–1] = –3, h[–2] = –2, h[–3] = –1, and h[n] is zero for |n| ≥ 4. A length-3 finite impulse response approximation g[n] of h[n] has to be obtained p

jw jw 2 such that E(h,g) = ò | H (e ) - G (e )| dw –p

is minimized, where H(ejɷ) and G (ejɷ) are the discrete-time Fourier transforms of h[n] and g[n], respectively. For the filter that minimizes E (h,g), the value of 10g[–1]+g[1], rounded off to 2 decimal places, is________________. [GATE2019/2marks] Q8. A discrete - time all - pass system has two of its poles at 0.25Ð0° and 2 Ð30°. Which one of the following statements about the system is TRUE? (a) It has two more poles at 0.5Ð30° and 4 Ð 0° . (b) It is stable only when the impulse response is two - sided. (c) It has constant phase response over all frequencies. [GATE2018/1mark] (d) It has constant phase response over the entire z - plane. Q9. Consider a single input single output discrete-time system with x[n] as input and y[n] as output,

ì

where the two are related as y[n] = í

n x[n] ,

for 0 £ n £ 10

otherwise. îx[n] - x[n - 1], Which one of the following statements is true about the system? (a) It is causal and stable (b) It is causal but not stable (c) It is not causal but stable (d) It is neither causal nor stable [set1/GATE2017/1mark] Q10.Two discrete-time signals x[n] and h[n] are both non-zero for n = 0, 1, 2 and are zero otherwise. It is given that x[0] = 1, x[1] = 2, x[2] = 1, h[0] = 1. Let y[n] be the linear convolution of x[n] and h[n]. Given that y[1] = 3 and y[2] = 4, the value of the expression (10y[3] + y[4]) is _________. [set1/GATE2017/2marks]

Q11. An LTI system with unit sample response h[n] = 5δ[n]– 7δ[n – 1]+ 7δ[n – 3]– 5δ[n – 4] is a (a) low – pass filter (b) high – pass filter (c) band – pass filter (d) band– stop filter [set2/GATE2017/1mark]

44

Mechasoft n

é x[n] ù é1 1 ù é1 ù Q12. A sequence x[n] is specified as ê ú= ê ú ê ú ; n ³ 2. ë x[n – 1] û ë1 0 û ë0 û The initial conditions are x[0] = 1, x[1] = 1 and x[n] = 0 for n < 0. The value of x[12] is _________ .

[Set1/GATE2016/2marks] Q13. The direct form structure of an FIR (finite impulse response) filter is shown below . This can be used to approximate a unit unit x[n] delay

(a) low pass filter (b) high pass filter ( c) band pass filter (d) band -stop filter

delay

5

5 +

y[n]

[Set3/GATE2016/2marks] Q14. The pole-zero diagram of a causal and stable discrete-time system is shown in the figure. The zero at the origin has multiplicity 4. The impulse response of the system is h[n]. If h[0] = 1, we can conclude Im (z)

(a) h[n] is real for all n (b) h[n] is purely imaginary for all n

0.5

( c) h[n] is real for only even n

4

(d) h[n] is purely imaginary for only odd n

0.5

–0.5

Re (z)

–0.5

[Set1/GATE2015/2marks] Q15. For the discrete-time system shown in the figure, the poles of the system transfer function are located at x[n]

(a) 2, 3

(b)

1 2

( c)

1 1 , 2 3

(d) 2,

,3

+

+

-1

5

6

6

y[n]

1 3

Z–1

Z–1

[Set1/GATE2015/2marks]

Mechasoft

45

1 ù éa ù éA ù é1 1 2p j –1 –2 ê Q16.Two sequences [a, b, c] and [A, B, C] are related as, ê B ú = 1 W3 W3 ú êb ú where W3 = e 3 ú êC ú ê –2 –4 ê ú ë û ëê1 W3 W3 úû ë c û

ép ù é1 1 If another sequence [p, q, r] is derived as, êq ú = ê1 W1 3 êr ú ê 2 ë û êë1 W3

1 ù é1 2 W3 4 W3

0 0 ù é A / 3ù ú ê0 W 2 0 ú ê B / 3 ú 3 úê ú ú 4 ê úû êë0 0 W3 úû ë C / 3 û

then, the relationship between the sequence [p, q, r] and [a, b, c] is (a) [p, q, r] = [b, a, c]

(b) [p, q, r] = [b, c, a]

(c) [p, q, r] = [c, a, b]

(d) [p, q, r] = [c, b, a] [Set1/GATE2015/2marks]

n =¥

{ }n= – ¥ is an independent and identically distributed random process with X equally likely to be +1 n =¥ or –1. {Yn } is another random process obtained as Y = X + 0.5 X . The autocorrelation function of n=– ¥ n =¥ [Set2/GATE2015/2marks] {Yn }n= – ¥ denoted by R [k],is Q 17. X n

n

n

n

n–1

y

RY(k)

0.5

(a) –3 –2 –1

0

1

2

0.25

0.5

0.5

–3 –2 –1

0

1

0

0.25

0.25 2

3

k

1

2

3

2

3

k

1.25

RY(k)

( c) –3 –2 –1

0.5

(b)

k

3

1.25

RY(k)

1.25

RY(k)

1

0.25

(d) –3 –2 –1

0

1

k

n

¥ æ 1ö Q18. The value of å n ç ÷ is __________ . n =0 è 2 ø

[Set3/GATE2015/1mark] 3

Q19.Consider a four point moving average filter defined by the equation y[n] = å ai x[n–i]. i=0

The condition on the filter coefficients that results in a null at zero frequency is (a) α1 = α2 = 0; α0 = –α3

(b) α1 = α2= 1; α0 = –α3

(c) α0 = α3 = 0; α1 = α2

(d) α1 = α2= 0; α0 = α3 [Set3/GATE2015/1mark]

46

Mechasoft

Q20. Two sequences x1[n] and x2[n] have the same energy. Suppose x1[n] = α 0.5n u[n], where α is a positive real

number and u[n] is the step sequence. Assume x 2 [n] = Then the value of α is _________________.

ì 1.5 for n = 0,1 í otherwise î 0

[set3/GATE2015/2marks]

Q21. A discrete time signal x[n] = sin(p2n) , n being an integer, is (a) periodic with period p

(b) periodic with period p2

(c) periodic with period p/2

(d) not periodic

[set1/GATE2014/1mark]

Q22. An FIR system is described by the system function H(z) = 1 + (a) maximum phase

(b) minimum phase

( c) mixed phase

(d) zero phase

7 2

z

–1

+

3 2

z

–2

is

[set2/GATE2014/1mark]

Q23. Let x[n] = x[-n]. Let X(z) be the z-transform of x[n]. If 0.5 + j 0.25 is a zero of X(z), which one of the following must also be a zero of X(z). (a) 0.5 – j 0.25

(b) 1/(0.5 + j 0.25)

( c) 1/(0.5 – j 0.25)

(d) 2 + j 4 [set2/GATE2014/1mark]

ìï n for 0£ n £10 Q24.Consider a discrete-time signal x[n] = í ïî0 otherwise

If y[n] is the convolution of x[n] with itself, the value of y[4]is _____________________. [set2/GATE2014/2marks] Q25. The input-output relationship of a causal stable LTI system is given as y[n] =αy[n – 1] + β x[n] . If the ¥

impulse response h[n] of this system satisfies the condition å h[n] = 2 , the relationship between α and β is n=0

(a) α = 1 -β/2

(b) α = 1 + β/2

Q26.For an all-pass system H(z) =

(z

–1

(c) α = 2 β

[set2/GATE2014/2marks] – b) –1

(1 – az ) If Re (a) ¹ 0, Im (a) ¹ 0, then b equals

(a) a

(d) α = -2β

( ) = 1 for all w.

where H e

(b) a*

–jω

( c) 1/a*

(d) 1/a [set3/GATE2014/1mark]

Q27. Let H1 (z) = (1 – pz–1 ) –1 , H2 (z) = (1 – q z–1 ) –1 and H(z) = H1 (z) + rH2 (z).The quantities p,q, r are real numbers.Consider p = 1 / 2, q = –1 / 4, r < 1. If the zero of H(z) lies on the unit circle , then r = ________ .

[set3/GATE2014/2marks]

Mechasoft

47

Q28. The sequence x[n] = 0.5n u[n], where u[n] is the unit step sequence, is convolved with itself to obtain y[n]. ¥

Then, å

is ____________________________.

y[n]

[set4/GATE2014/1mark]

n=– ¥

Q29. 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

[GATE2012/2marks]

(b) 1/2

(c) 1

(d) 3/2

Q30. A system is defined by its impulse response h[n] = 2nu[n - 2]. The system is (a) stable and causal

(b) causal but not stable

( c) stable but not causal

(d) unstable and non-causal

[GATE2011/1mark] Q31. Two system H1(z) and H2(z) are connected in cascade as shown below. The overall output y(n) is the same as the input x(n) with a one unit delay. The transfer function of the second system H2(z) is –1

x(n)

(a)

1 – 0.6z –1

H1 (z) = –1

–1 –1

(b)

(1 - 0.4z )

–1

z (1 – 0.6z ) –1

y(n)

H2(z)

–1

(1 - 0.6z ) –1

( c)

–1

z (1 – 0.4z )

1 – 0.4z

(d)

–1

–1

–1 –1

z (1 – 0.6z ) [GATE2011/2marks] Q32.Two discrete time systems with impulse response h1 [n] = d [n - 1] and h2[n] = d [n - 2] are connected in z (1 – 0.4z )

(1 – 0.4z )

(1 – 0.6z )

cascade. The overall impulse response of the cascaded system is (a) d [n - 1] + d [n - 2]

(b) d [n - 4]

(c) d [n - 3]

(d) d [n - 1] d [n - 2] [GATE2010/1mark] 2–

Q33.The transfer function of a discrete time LTI system is given by H(z) = 1–

Consider the following statements.

3 4

z

3 4

–1

z

–1

1 –2 + z 8

S1: The system is stable and casual for ROC: |z| > 1/2 S2: The system is stable but not casual for ROC: |z| < 1/4 S3: The system is neither stable nor casual for ROC: 1/4 < |z|< 1/2 Which one of the following statements is valid? (a) Both S1 and S2 are true

(b) Both S2 and S3 are true

( c) Both S1 and S3 are true

(d) S1, S2 and S3 are all true [GATE2010/2marks]

48

Mechasoft

Q34.A system with transfer function H(z) has impulse response h(.) defined as h(2) = 1, h(3) = -1 and h(k) = 0 otherwise. Consider the following statements. S1: H(z) is a lowpass filter. S2: H(z) is an FIR filter. Which of the following is correct? (a) Only S2 is true.

(b) Both S1 and S2 are false.

(c) Both S1 and S2 are true, and S2 is a reason for S1. (d) Both S1 and S2 are true, but S2 is not a reason for S1.

[GATE2009/2marks]

Q35.A discrete time linear shift invariant system has an impulse response h[n] with h[0] = 1, h[1] = -1, h[2] = 2,and zero otherwise. The system is given an input sequence x[n] with x[0] = x[2] = 1, and zero otherwise. The number of non zero samples in the output sequence y[n], and the value of y[2] are respectively (a) 5, 2

(b) 6, 2

(c) 6, 1

(d) 5, 3 [GATE2008/2marks]

æ 5pn ö Q36. A system with input x[n] and output y[n] is given as y[n] = ç sin ÷ x[n] . The system is è 6 ø (a) linear, stable and invertible (b) non-linear, stable and non-invertible

( c) linear, stable and non-invertible

(d) linear, unstable and invertible

[GATE2006/2marks] Q37. A signal x(n) = sin (w0n + f) is the input to a linear time - invariant system having a frequency response

H(ejw ). If the output of the system is A x(n - n0), then the most general form of ÐH (ejw ) will be (a) –n0w0 + b for any arbitrary real k

(b) –n0w0 + 2pk for any arbitrary integer k

( c) n0w0 + 2pk for any arbitrary integer k

(d) –n0w0 f

[GATE2005/2marks] Q38.The impulse response h[n] of a linear time invariant system is given by h[n]=u[n+3]+u[n–2]–2u[n–7]

where u[n] is the unit step sequence. The above system is (a) stable but not causal

(b) stable and causal

(c) causal but unstable

(d) unstable and not causal

[GATE2004/1mark]

Q39. A causal LTI system described by equation 2y[n] =αy[n – 2] – 2x[n] + βx[n – 1] is stable only if (a) | a |= 2,| b | < 2

(b) | a |> 2,| b | > 2

( c) | a |< 2,any value of b

(d) | b |< 2,any value of a

[GATE2004/2marks]

Mechasoft

49

ì –2 2 n = 1, –1 ï Q40.The impulse response h[n] of a linear time invariant system is given as h[n] = í 4 2 n = 2, –2 ï 0 otherwise If the input to the above system is the sequence ejpn/4, then the output is î (a) 4 2 ejpn/4

(d) –4ejpn/4

( c) 4ejpn/4

(b) 4 2 e– jpn/4

[GATE2004/2marks] Q41. A sequence x[n] with the z-transform X(z) = z4 + z2 - 2z + 2 - 3z-4 is applied as an input to a linear, time ìï 1, n=0 ïî 0, otherwise

invariant system with the impulse response h[n] = 2d[n - 3] where δ[n] = í The output at n = 4 is (a) -6

(b) zero

(c) 2

(d) -4 [GATE2003/1mark]

Q42. Let P be linearity, Q be time invariance, R be causality and S be stability. A discrete time system has the

ì x[n] , n ³ 1 ï , n=0 input-output relationship y[n] = í 0 ïîx[n + 1] , n £ –1 where x(n) is the input and y(n) is the output. The above system has the properties (a) P, S but not Q, R

(b) P, Q, S but not R

( c) P, Q, R, S

(d) Q, R, S but not P

[GATE2003/2marks]

Answer Key Q1 Q8 Q15 Q22 Q29 Q36

0 b c c a c

Q2 Q9 Q16 Q23 Q30 Q37

c a c b b b

Q3 Q10 Q17 Q24 Q31 Q38

a 31 b 10 b a

Q4 Q11 Q18 Q25 Q32 Q39

–2 c 2 a c c

Q5 Q12 Q19 Q26 Q33 Q40

d 233 a b c d

Q6 Q13 Q20 Q27 Q34 Q41

d c 1.5 –0.5 a b

Q7 Q14 Q21 Q28 Q35 Q42

–27 a d 4 d b

Section2

GATE Solutions

Q1. x1[n]=u[n] H(z) =

y1[n]=2δ[n+1]+δ[n]+δ[n–1]

H(z)

Y1 (z) 2z + 1 + z–1 = = 2z – 1 – z–2 and h[n] = 2d[n + 1] – d[n] – d[n – 2] X1 (z) 1 / 1 – z–1

(

)

n

n

æ 1ö æ 1ö y[n] = ç ÷ u[n] * h[n] = ç ÷ u[n] * [2d[n + 1] – d[n] – d[n – 2]] è 2ø è 2ø æ 1ö = 2ç ÷ è 2ø æ 1ö y[0] = 2 ç ÷ è 2ø

n +1

0 +1

n

æ 1ö æ 1ö u[n + 1] – ç ÷ u[n] – ç ÷ è 2ø è 2ø

n –2

u[n – 2]

0

æ 1ö –ç ÷ –0= 0 è 2ø

ì1 for k = 0 Q2( c) The response y[n] for unit impulse, that is, x[n] = d[n] is y[n] = max d (k) ; d (k) = í – ¥£ k £ n î0 for k ¹ 0 For n < 0 , y[n] = max d (k) = max {0 ,0 ,0.....,0} = 0 – ¥£ k £ –1

For n = 0 , y[n] = max d (k) = max {0 ,0 ,0.....,1} = 1 – ¥£ k £ 0

For n > 0 , y[n] = max d (k ) = max {0 ,0 ,0.....,1,0,0,0....} = 1 – ¥£ k £ n

ì0 for – ¥ £ n < 0 y[n] = í = u[n] , that is , the unit impulse response of the system is unit step signal. î1 for 0 £ n £ ¥

Q3(a) In order to find the pole - zero plot, the transfer function H(z) is determined as demonstrted below. 3

k

y[n] = å (–1) x[n – k] = x[n] – x[n–1] + x[n– 2] – x[n– 3] and Y(z) = X(z) – z–1X(z) + z–2 X(z) – z–3 X(z) k =0 2 Y(z) z3 – z2 + z –1 z (z –1) + (z –1) (z –1) (z + j) (z– j) = 1 – z–1 + z–2 – z–3 = = = X(z) z3 z3 z3 The transfer function H(z) has3 poles at origin and 3 zeros at z = 1, ± j .

H(z) =

Q4. A stable discrete - time LTI system contributes constant magnitude response over all frequencies if the poles and the zeros occur in conjugate reciprocal pairs. Such a system is also referred to as all pass system. K (z – a ) 1 H(z) = has a pole at z = – 0.5 , then the zero must be located at z = a = – = –2 z + 0.5 0.5 so as to form conjugate reciprocal pair . Q5(d) H(z) is the z-transfrom of a real valued signal h[n] and P(z)=H(z)(z –1). This implies, if there is a zero at z = 0.5 +j0.5 , there will be 3 more zeros at Im 1 2 (1 - j) –1 unit 1 z* = 0.5 - j0.5 , = (0.5 + j0.5) = = 1- j circle z (1 + j) (1 - j) 1/2

*

æ 1ö and ç ÷ = 1 + j è zø The complex zeros always appear in conjugate pair.

–1/2 –1

1/2

Re

Mechasoft

51

Q6(d) x[n] = c1e – jpn/2 + c 2 jpn/2 has w 0 = p / 2 h[0] = 1, h[1] = a , h[2] = b and h[n] = 0 for n < 0 or n > 2 Þ H(w ) = 1 + a e – jw + b e – j2w The zero output , that is , y[n] = 0 for all n when x[n] is applied , implies H (w 0 )

w0 =

p 2

=0Þ

(1 + ae – jp /2 + b e jp ) = 0 Þ 1 - ja - b = 0 Þ a = 0 and b = 1

p

3

2

jw jw Q7(d) E(h, g) = ò H (e ) - G (e ) dw = 2p å

h[n] - g[n]

2

n = –3

–p

where g[n] is length - 3 approximation of h[n] such that p

2

E(h, g) = ò H(e jw ) - G (e jw ) dw is minimal or

2

h[n] - g[n] is minimal

n = –3

–p 3

å

3

å

h[n] - g[n]

2

2

2

= h[ -3] - g[ -3] + h[ -2] - g[ -2] +

2

h[ -1] - g[ -1] + h[0] - g[0]

2

n = –3 2

+ h[1] - g[1] +

h[2] - g[2]

2

+ h [ 3] - g[ 3]

2

where g[ -3] = g[ -2] = g[2] = g[3] = 0 3

2

2

2

2

h[n] - g[n] = ( -1) 2 + ( -2) 2 + -3 - g[ -1] + 4 - g[0] + 3 - g[1] + 2 2 + 12

Thus , å n = –3

2

2

= 10 + -3 - g[ -1] + 4 - g[0] + 3 - g[1]

2

gives minimal value equal to 10 for g[ -1] = -3, g[0] = 4 and g[1] = 3 Then, 10g[ -1] + g[1] = 10 ´ ( -3) + 3 = –27

Q8(b) Each pole of an all pass system H(z) is paired with a conjugate reciprocal zero. If a pole is at z = ak then a zero is at z = (1 / ak* ), that is, a pole at ak = rejθ is paired with a zero at (1/ak*) = (1/r)ejθ. If h[n] is real, then ak = ak*. Two poles at 0.25 Ð0o and 2Ð 30o will be paired with two zeros at 4Ð0o and 0.5Ð30o. An all pass system has the gain |H(ejω)| = A where A is a Im real constant but the phase Ð H(ejω) is not constant. 2Ð30o × The system has two poles, one inside and one Unit circle outside the unit circle. For stability ROC must include 0.5Ð30o 30o unit circle. Thus, it is stable only when the impulse × Re response h[n] is two sided. 0.25 4 Q9(a) y[n] =

ìn x[n] , for 0 £ n £ 10 í îx[n] - x[n - 1] , otherwise

[

]

[

]

= u[n] - u[n - 11] nx[n] + [u[n - 11] + u( - n - 1)] x[n] - x[n - 1]

For any n = n0, the output y[n] depends on present input x[n] at n = n0 and past input n = n0 − 1. The system is causal . If input x[n] is bounded for all n then, the output is also bounded (finite). The system is stable. y[n]=

[ 4244 14 3] u[n]–u[n–11]

n x[n] 123

finite

finite ,n max =10

144424443

[14243]

finite

finite

+ {u[n–11] + u(–n–1)} x[n]–x[n–1]

52

Mechasoft

Q10. x[n] = {x1 , x 2 , x 3 } = {1, 2,1} and h[n] = {h1 , h 2 , h 3 } = {1, h 2 , h 3 } ­

­

­

­

y[n] = x[n] * h[n]; y[1] = 3 and y[2] = 4; * denotes linear convolution

h[n]

x[n]

y[n]

1

h2

h3

1

1

h2

h3

2

2

2h2

2h3

1

1

h2

h3

Þ y[n] = {1,(2+h2), (1+2h2+h3),(h2+2h3),h3}

y[1] = 2 + h 2 = 3 Þ h 2 = 1 and y[2] = 1 + 2 h 2 + h 3 = 4 Þ h 3 = 1 y[n] = {1, 3, 4, 3,1} and 10y[3] + y[4] = 10 ´ 3 + 1 = 31 ­

Im

Q11( c) Unit sample response h[n] = 5d[n] - 7d[n - 1] + 7d[n - 3] - 5d[n - 4] –1

H(z) = 5 - 7 z + 7z 2

=

–3

- 5z

2

–4

=

1 z

4

4

2

5(z - 1)(z + 1) - 7 z(z - 1) 4

3

[5 z - 7 z + 7 z - 5] 2

=

1

–1

Re

2

(z - 1)(5 z - 7 z + 5) 4

z z Zero at z = −1, pulls down the frequency response at high frequencies and the zero at z = +1, pulls down the frequency response at low frequencies. To sum up, H(z) acts as a band pass filter. n

é x[n] ù é1 1 ù é1 ù Q12. ê ú=ê ú ê ú ; for n £ 2 while x[0] = x[1] = 1 and x[n] = 0 for n < 0 ë x[n – 1]û ë1 0 û ë0 û 2

é x[2]ù é1 1 ù é1 ù é 2 ù For n = 2, ê ú=ê ú ê ú = ê ú Þ x[2] = 2 = x[0] + x[1] ë x[1] û ë1 0 û ë0 û ë1 û 3

é x[3] ù é1 1 ù é1 ù é 3 ù For n = 3, ê ú=ê ú ê ú = ê ú Þ x[3] = 3 = x[2] + x[1] ë x[2]û ë1 0 û ë0 û ë 2 û Thus , in general , for n ³ 2 , x[n] = x[n – 1] + x[n – 2] gives x[4] = x[3] + x[2] = 5 Similarly, x[5] = 8, x[6] = 13, x[7] = 21, x[8] = 34, x[9] = 55, x[10] = 89, x[11] = 144 and x[12] = 233

Q13( c) y[n] = 5x[n] – 5x[n – 2] Þ H(z) =

(

Y(z) = 5 1 – z–2 X(z)

(

)

z–1

x[n]

)

H(ejw ) = 5(1 – e– j2w ) = 5e– jw ejw – e– jw = j10 ´ e– jw sin w

z–1

5

5 -

jw

H(e ) = 10sin w is magnitude response of band pass filter.

+

y[n]

Mechasoft

53

5(z + 1)(z – 1) has two zeros , one at z = 1 that pulls down the frequency response at low z2 frequencies and another at z = –1 that pulls down the frequency response at high frequencies. Im To sum up, it acts as bpf . |H(ejω)| Also, H(z) =

–1

1

10

Re

Q14(a) The transfer function H(z) =

Kz

ω

π

0 4

æ 1 1ö æ 1 1ö æ 1 1ö æ 1 1ö çè z - + j ÷ø çè z - - j ÷ø çè z + + j ÷ø çè z + - j ÷ø 2 2 2 2 2 2 2 2

= 1+

K 1 4

z

–4

The initial value theorem, h(0) = lim H(z) = K = 1 z®¥

Use long division to get H(z) = h[n] = d[n] -

1 4

d[n - 4] +

Q15( c) y[n] = x[n] + Y(z) = X(z) + H(z) =

Y(z) X(z)

5 6

6

16

–1

-

1 + 4z

1 6

1

5

z

–1

+

1

6 6 1 and z = 2 3

= 1-

1 64

1 4

z

Y(z)z

–4

1

+

16

z

–8

x[n]

z

–2

1

64

z

–12

+

+ .............

y[n]

+

–2

-

=

1

-

d[n - 12] + ........... Þ h[n] is real for all n.

y[n - 2]

6

1 1-

–4

d[n - 8] -

y[n - 1] -

y(z)z

=

has poles at z =

5

1

1

z

2

1

5

6

6

æ 1 ö æ 1ö çè z - ÷ø çè z - ÷ø 2 3

y[n–2] z–1

y[n–1] z–1

é1 W32 W34 ù éA ù 1 ù é1 0 0 ù éA/3ù ép ù é1 1 1ê 3 6ú Q16( c) êq ú = ê1 W31 W32 ú ê0 W32 0 ú ê B/3ú = 1 W3 W3 ê B ú ê ú úê ú êr ú ê ú 3 2 4 4 ê 4 8 ê ú êë1 W3 W3 úû ë C û ë û êë1 W3 W3 úû êë0 0 W3 úû ëC/3û é1 W32 W34 ù é1 1 é1 + W32 + W34 1 + W3 + W32 1 ù éa ù 1 + 1 + 1 ù éa ù 1ê 1ê 3 6ú ê –1 –2 ú ê ú 3 6 2 4 2ú = 1 W3 W3 1 W3 W3 b = 1 + W4 + W3 1 + W3 + W3 1 + W3 + W3 êb ú ê ú ê ú ê ú 3 3 4 8 –2 –4 ê ú 4 8 3 6 2 4 ê ú êë1 W3 W3 úû êë1 W3 W3 úû ë c û êë1 + W3 + W3 1 + W3 + W3 1 + W3 + W3 úû ë c û Fit the values j

W3 = e

2p 3

=-

1 2

+j

3 2

=

4 3 W3 , W3

=

6 W3

= 1,

2 W3

=-

1 2

-j

3 2

=

8 W3

ép ù é0 0 3 ù é a ù é c ù 1 ê ú toget q = ê3 0 0ú êb ú = êa ú ê r ú 3 ê0 3 0ú ê c ú êb ú ë û ë ûë û ë û

54

Mechasoft

Q17(b) An independent and identically distributed random process Xn has auto correlation RX(k) = RX(n, n + k) = E[Xn Xn+k] and auto correlation of Yn = Xn + 0.5 Xn–1 is

[

] [

R Y (k) = E Yn Yn + k = E (X n + 0.5X n–1 )(X n + k + 0.5X n + k–1 )

]

[ ] = E [X n X n + k ]+ 0.5E [X n X n + k–1 ]+ 0.5E [X n–1 X n + k ]+ 0.25E [X n–1 X n + k–1 ] = E X n X n + k + 0.5X n X n + k–1 + 0.5X n–1 X n + k + 0.25X n–1 X n + k–1

RY(k) 1.25

= R X (k) + 0.5R X (k - 1) + 0.5R X (k + 1) + 0.25R X (k)

0.5

= 1.25R X (k) + 0.5R X (k - 1) + 0.5R X (k + 1)

0.5

R Y (0) = 1.25R X (0) + 0.5R X ( -1) + 0.5R X (1)

–3 –2 –1

R Y (1) = R Y ( -1) = 1.25R X (1) + 0.5R X (0) + 0.5R X (2)

0

1

2

3

k

R Y (2) = R Y (–2) = 1.25R X (2) + 0.5R X (1) + 0.5R X (3) and so on 2

2

where R X (0) = E[X n X n + 0 ] = E éëX n ùû = (1) ´

1 2

2

+ ( -1) ´

1 2

=1

X n is equally likely to be ± 1 and R X (k) k ¹0 = 0

Fit these values to get RY(0) = 1.25 , RY(1) = RY(–1) = 0.5 and R Y (k) |k| >1 = 0 n 2 3 2 üï 1ì 1 ï æ 1ö æ 1ö æ 1ö æ 1ö æ 1ö = 0.(1) + 1 + 2 + 3 + ..... = 1 + 2 ´ + 3 ´ ....¥ ý í ÷ ç ÷ ç ÷ ç ÷ ç ÷ è 2ø è 2ø è 2ø è 2ø è 2ø 2ï 2 ïþ î ì 1 ü ï ï 1´ 1ï 1 ï 2 = í + 2ý= 2 1 2ï æ 1ö ï 1– 1– ïî 2 çè 2 ÷ø ïþ

¥

Q18.

å nç

n =0

Note that sum to infinity of AGP with initial term ' a ' , common ratio ' r ' : | r |< 1 and common difference ' d ' is S¥ =

a 1– r

+

dr (1 – r)

2

2

. In variables ,AGP lookslike a, (a + d)r , (a + 2d)r ,....., {a + (n – 1) d}r

n–1

3

Q19(a) y[n] = å a i x[n - i] = a 0 x[n] + a1 x[n - 1] + a 2 x[n - 2] + a 3 x[n - 3] i=0

–1

–2

–3

Y(z) = a 0 X(z) + a1 z X(z) + a 2 z X(z) + a 3 z X(z) H(z) =

Y(z) X(z)

= a 0 + a1 z

–1

+ a2 z

–2

+ a3 z

–3

jw

and H(e ) = a 0 + a1 e

– jw

+ a2 e

– j2 w

+ a3 e

– j3w

Seeking the condtion on filter coefficients resulting in null at zero frequency , one gets j0

H(e ) = a 0 + (a1 + a 2 + a 3 )e

j0

= 0 Þ a 0 + a1 + a 2 + a 3 = 0 Þ α1 = α 2 = 0 and α 0 = –α 3

+ .....¥.

Mechasoft

55 ¥

n

æ 1ö u[n] è 2 ÷ø

¥

2

2

Q20. The energy of x1 [n] is E1 = å x1[n] = å a ç n=– ¥

æ 1ö = åa ç ÷ è 2ø n =0 ¥

2n

n=– ¥ 2

2

4a é 1 1 ù a = a 1+ + êë 4 16 + .........úû = 1 = 3 where a is a positive real number . 1-

2

2

4

¥

1

2

2

The energy of x 2 [n] is E 2 = å x 2 [n] = å ( 1.5) = 1.5 + 1.5 = 3 n=– ¥

E1 = E 2 gives

4a

n =0

2

3

= 3 and a = 1.5 2

2

Q21(d) The fundamental angular frequency of x[n] = sin[ p n] is w 0 = p = N=

2m p

2 pm N

.

,m is an integer and N is an irrational number ,signal is not periodic .

Q22( c) H(z) = 1 +

7 2

z

–1

+

3 2

2

z

–2

=

2z + 7z + 3 2z

2

=

(2z + 1) (z + 3) 2z

2

For minimum phase system H(z) must have all zeros inside unit circle and for maximum phase system H(z) must have all zeros outside unit circle. 1 Out of two zeros of H(z) located at z = - and - 3 , one lies inside the unit circle and another outside . 2 The system is mixed phase . æ 1ö Z Z ® X(z), then x[– n] ¾¾ ® X ç ÷ , time reversal property ; Q23(b) x[n] ¾¾ è zø if a = 0.5 + j0.25 is a zero of X(z) : z – a = 0, then

1 1 = will also be zero of X(z) : z–1 – a = 0. a 0.5 + j0.25

Note that x[n] = x[–n] and z transform of x(–n) is obtainable by replacing z by z–1.

Q24.x[n] = n = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and y[n] = x[n] * x[n]

x[n] x[n] 0 1 2 3 4 . . 10

0

1

2

3

4

5

6

7

8

9

10

0 0 0 0 0

0 1 2 3 4

0 2 4 6 8

0 3 6 9 12

0 4 8 12 16

0 5 10 15 20

0 6 12 18 24

0 7 14 21 28

0 8 16 24 32

0 9 18 27 36

0 10 20 30 40

y[4]=0+3+4+3+0=10

56

Mechasoft –1

Q25(a) y[n] = a y[n - 1] + b x[n] Þ Y(z) = a Z Y(z) + b X(z) Þ H(z) =

Y(z) X(z)

=

b 1- az

–1

Since , the system is causal and stable ,H(z) must have all the poles inside the unit circle of z - plane n

and ROC must include unit circle. Therefore,h[n] must be right sided sequence , that is,h[n] = ba u[n] ¥

¥

n

å h[n] = å b (a ) u[n] = 2 Þ

b 1- a

= 2 or a = 1 -

β

2 1 z– z–1 – b 1 – bz b ; a is complex. Q26(b) H(z) = = = – b 1 – a z–1 z – a z–a In an all pass system , the poles and zeros occur in conjugate reciprocal pairs. 1 The pole located at z = a and zero at z = . Then , b = a* . b 1 1 –1 –1 –1 –1 Q27. H(z) = H1 (z) + r H 2 (z) = (1 - p z ) + r (1 - q z ) where p = , q = - and |r| 1 and 0.5r – 0.25 = – (1 + r) gives r = – 0.5 : | r | < 1 n

¥ æ 1ö Q28. x[n] = ç ÷ u[n] and y[n] = x[n] * x[n] = å x[k] x[n - k] è 2ø k=– ¥

x[k]

x[–k] 1

æ 1ö çè ÷ø æ 1 ö 2 2 ç ÷ æ 1ö3 è 2ø ç ÷ è 2ø ..... ∞ 0

1

2

3

+∞.....

k

2æ 1ö 3 1 æ 1 ö æç ö÷ çè 2 ÷ø çè ÷ø è 2 ø 2

–3

–2

Note that x(k)=0 for k < 0 and x[n–k] = 0 for k >n.

–1

x[n–k]

1

1 +∞ .....

0

k

0

n

k

Mechasoft

57 k

æ 1ö æ 1ö y[n] = å ç ÷ ç ÷ k =0 è 2 ø è 2 ø n

n–k

n

n

æ 1ö n æ 1ö = ç ÷ å (1) = ç ÷ (n + 1)u[n] è 2 ø k =0 è 2ø

n

n

0

1

2

3

¥ æ 1ö æ 1ö æ 1ö æ 1ö æ 1ö æ 1ö å y[n] = å ç ÷ (n + 1)u[n] = å (n + 1)ç ÷ = 1 ´ ç ÷ + 2 ´ ç ÷ + 3 ´ ç ÷ + 4 ´ ç ÷ + ....¥ è 2ø è 2ø è 2ø è 2ø è 2ø n=– ¥ n=– ¥ è 2 ø n =0 ¥

¥

æ 1ö è 2 ÷ø

0

1´ ç =

1–

æ 1ö è 2 ÷ø

1´ ç +

1

æ 1ö çè1 – ÷ø 2

2

2

= 2+2 = 4

0

1 æ 1ö In present referecnce , sum to infinity of AGP is S¥ = + 2 ; a = 1 ´ ç ÷ = 1, d = 1 and r = è 2ø 1 – r (1 – r ) 2 k k ¥ ¥ æ 1ö ¥ æ 1ö Q29(a) y[n] = h[n] * g[n] = å h[k] g[n - k] = å ç ÷ u[k] g[n - k] = å ç ÷ g[n - k] k=– ¥ k=– ¥ è 2 ø k =0 è 2 ø a

k

dr

2

1 æ 1ö æ 1ö g[ - k] = g[0] + g(–1) + ç ÷ g(–2) + .....¥ = 1 Þ g[0] = 1 ç ÷ è 2ø k =0 è 2 ø 2 ¥

y[0] = 1 Þ å

Since , g[n] is a causal sequence , g[–1] = g[–2] = .... = 0 k

2

1 1 1 1 1 1 1 æ 1ö æ 1ö y[1] = å ç ÷ g[1 – k] = Þ g[1] + g[0] + ç ÷ g[–1] + .... = Þ g[1] + g[0] = Þ g[1] = – = 0 è 2ø k =0 è 2 ø 2 2 2 2 2 2 2 ¥

¥

Q30(b) For a system to be stable, its impulse response must be absolutely summable, that is, å h[n] < ¥ n=– ¥ ¥

¥

¥

n

n

2

3

but å h[n] = å 2 u[n - 2] = å 2 = 2 + 2 + ............. = ¥. The system is unstable. n=– ¥

n– ¥

n =2

For a system to be causal , h[n]=0 for n< 0 . Since , h[n] = 2n u[n-2] is defined only for n ≥ 2, the system is causal. Y(z) –1 –1 Q31(b) The ouput y[n] = x[n – 1] Þ Y(z) = z X(z) and H(z) = =z X(z) H(z) = H1 (z) H 2 (z) Þ z

–1

(1-0.4z –1)´ H (z) Þ H = (1 - 0.6z ) 2

–1

Q32( c) h1[n] = d[n – 1] Þ H1 (z) = z –1

H(z) = H1 (z) H 2 (z) = z . z

–2

=z

–1

–3

z 2 (z)

=

(1 – 0.6z ) (1 – 0.4z )

–1

and h 2 [n] = d[n – 2] Þ H 2 (z) = z

and h[n] = δ[n – 3]

–1

–1

–2

58

Mechasoft 3

2Q33( c) H (z) = 1For ROC z >

1

3

z

4

z

4

–1

+

–1

=

1

z

8

–2

1-

1 1 2

+ z

–1

1-

1 1 4

has poles at z = z

–1

1 2

and z =

1 4

, ROC includes unit circle and all poles lie within unit circle , the system is both stable and causal .

2

The statement S1 is true .

Im

1/2

1/ 2

Re

1

1/ 4

1 4

1/4

ROC : | z |
2

The ROC : z
|b| ( c) |z| < |a| (d) |a| < |z| < |b| [set1/GATE2016/1mark] Q8. Consider x(n) = 6d [n + 2] + 3d[n + 1] + 8d[n] + 7d[n–1] + 4d[n– 2]. p

1 X(ejw ) sin2 (2w )dw is equal to _____ . p –òp [set1/GATE2016/2marks] 2s + 6 Q9. A continuous time filter with transfer function H(s) = 2 is converted to a discrete time filter with s + 6s + 8 2 2z – 0.5032z transfer function G(z) = 2 so that the impulse response of the continuous time filter, sampled 2z – 0.5032z + k at 2 Hz ,is identical at the sampling instants to the impulse response of the discrete time filter . The value of k is______________. [set2/GATE2016/2marks] If X(ejw ) is the discrete time Fourier transform of x[n], then

Q10. The discrete Fourier transform (DFT) of the 4 - point sequence x[n] = {x[0], x[1],x[2],x[3]} = {3, 2,3, 4} is X[k] = {X[0],X[1],X[2],X[3]} = {12 , 2 j ,0, – 2 j} . X1[8] is ___ . X1[11]

If X1[k] is DFT of the 12 - point sequence x1[n] = {3,0,0, 2,0,0,3,0,0, 4,0,0}, the value of

[set2/GATE2016/2marks] Q11. A discrete time signal x[n]= δ[n–3] + 2δ[n–5] has z-transform X(z). If Y(z) =X(–z) is the z-transform of another signal y[n] , then (a) y[n] = x[n] (b) y[n] = x[–n] ( c) y[n] = –x[n] (d) y[n] = –x[–n] [set3/GATE2016/1mark] Q12. Let the ROC (region of convergence ) of the z-transform of a discrete time signal be represented by the shaded region in the z-plane. If the signal x[n]=2|n| , – ∞ < n< +∞ , then the ROC of its z-transform is represented by Im Im unit circle unit circle

(a)

0.5 2

Re

(b)

0.5 2

Re

Mechasoft

65 Im

0.5

( c)

Im

unit circle

unit circle 2

0.5

(d)

Re

Re

2

[set3/GATE2016/2marks] Q13. A continuous time speech signal xa (t) is sampled at a rate 8 kHz and the samples are subsequently grouped in blocks , each of size N .The DFT of each block is to be computed in real time using the radix -2 decimation- in- frequency FFT algorithm . If the processor performs all operations sequentially, and takes 20µs for computing each complex multiplication (including multiplications by 1 and –1 ) and the time required for addition /subtraction is negligible , then the maximum value of N is _________. [set3/GATE2016/2marks] n

Q14.Two causal discrete-time signals x[n] and y[n] are related as y[n] = å x[m]. If the z-transform of y[n] is m=0 2 , [set2/GATE2015/1mark] the value of x[2] is _____________. 2 z(z - 1) Q15.Consider two real sequences with time-origin marked by the bold value, x1[n] = {1, 2, 3, 0} and x2[n] = {1, 3, 2, 1}.Let X1(k) and X2(k) be 4-point DFTs of x1[n] and x2[n], respectively . Another sequence x3[n] is derived by taking 4-point inverse DFT of X1(k)X2(k).The value of x3[2] is _______________. [set2/GATE2015/2marks] æ pn ö % = 1 + cos ç ÷ be a periodic signal with period 16. Its DFS coefficients are defined by Q16.Let x[n] è 8ø ak =

1

æ

15

% å x[n]exp çè - j

16 n =0

p 8

ö ø

kn÷ for all k. The value of the coefficient α31 is ________________.

[set3/GATE2015/2marks] Q17.Suppose x[n] is an absolutely summable discrete-time signal. Its z-transform is a rational function with two poles and two zeroes. The poles are at z = ±2j. Which one of the following statements is TRUE for the signal x[n]? (a) It is a finite duration signal

(b) It is a causal signal

( c) It is non-causal signal

(d) It is a periodic signal

[set3/GATE2015/2marks] Q18.A realization of a stable discrete time system is shown in figure. If the system is excited by a unit step x[n]

sequence input x[n], the response y[n] is n

n

æ 1ö æ 2ö (a) 4 ç - ÷ u[n] - 5 ç - ÷ u[n] è 3ø è 3ø æ 1ö

n

æ 2ö

n

(c)5 ç ÷ u[n] - 5 ç ÷ u[n] è 3ø è 3ø

n

n

æ 2ö æ 1ö (b)5 ç - ÷ u[n] - 3 ç - ÷ u[n] è 3ø è 3ø æ 2ö

n

+

æ 1ö

Z–1 +

1

+

y[n]

Z–1

n

(d)5 ç ÷ u[n] - 5 ç ÷ u[n] è 3ø è 3ø

–5/3

–2/9

5/3

[set3/GATE2015/2marks]

66

Mechasoft n

n

æ 1ö æ 1ö Q19. Let x[n] = ç – ÷ u(n) – ç – ÷ u(–n – 1).The region of convergence (ROC) of z-transform of x[n] è 9ø è 3ø (a) is z >

1

(b) is z
z >

3

1

(d) does not exist

9

[set1/GATE2014/2marks]

æ πn ö Q20.Consider a discrete time periodic signal x[n] = sin ç ÷ è 5ø

.

Let ak be the complex Fourier series coefficients of x[n]. The coefficients {ak} are non-zero when k = Bm ± 1, where m is any integer. The value of B is __________________.

Q21.The z-transform of the sequence x[n] is given by X(z) = Then, x[2] is __________.

[set1/GATE2014/2marks] 1 –1 2

(1 – 2z )

n

æ 2ö FT ® Q22.A Fourier transform pair is given by ç ÷ u[n + 3] ¬¾¾ è 3ø

Ae

,with the region of convergence|z|>2.

[set3/GATE2014/2marks]

j6πf

æ 2 ö –j2πf e è 3 ÷ø

1– ç

where u[n] denotes the unit step sequence. The values of A is _________________. [set4/GATE2014/1mark] Q23.The N-point DFT X of a sequence x[n], 0 £ n £ N - 1 is X[k] =

1

N–1

N

n=0

Denote this relation as X = DFT (x).

–j

å x[n]e

2π nk N

, 0 £ k £ N – 1.

For N = 4, which one of the following sequence satisfies DFT (DFT(x)) = x? (a) x = [ 1 2 3 4]

(b) x = [ 1 2 3 2]

(c) x = [ 1 3 2 2]

(d) x = [1 2 2 3] [set4/GATE2014/2marks]

Q24.The DFT of a vector [a b c d] is the vector [a b g d]. Consider the product

éa êd [p q r s] = [a b c d] ê êc êëb

b c dù b cú

a

ú ú c d a úû

d a

b

[GATE2013/2marks]

The DFT of the vector [ p q r s] is a scaled version of (a) [a2 b2 g2 d2]

(b) [Öa Öb Ög Öd]

( c) [a+b b+d d+g g+a]

(d) [a b g d]

n

Q25. x[n] = (1 / 3) – (1 / 2) u[n], then the region of convergence (ROC) of its z-transform in the z-plane will be |n|

(a)

1 3

< z e

– 0.0005

–5

[GATE2008/2marks]

Q34.A 5-point sequence x[n] is given as x[-3] = 1, x[-2] =1, x[-1] = 0, x[0] = 5 and x[1] = 1. π

Let X(ejw ) denoted the discrete-time Fourier transform of x[n]. The value of ò X(e jω )dω is –π

(a) 5

(b) 10p

(c) 16p

Q35.The z- transform X(z) of a sequence x[n] is given by X[z] =

(d) 5 + j10p [GATE2008/2marks]

0.5 –1

1 – 2z It is given that the region of convergence of X(z) includes the unit circle. The value of x[0] is

(a) -0.5

(b) 0

( c) 0.25

(d) .05

[GATE2007/2marks] 1 2 Q36.If the region of convergence of x1[n] + x2[n] is 2 which is zero . 2 ROC does not exist and x[n] will not have z - transform . However , the right sided sequence

2n u[n] and left sided sequence 2– n u[–n – 1] do have z - transforms individually.

Q13. The sampling frequency fs =8kHz and sampling period Ts=125µs. The block size is N and time required for N samples is TN=N Ts=125N µsec. N The number of complex multiplications required in radix-2 algorithm for N-point DFT is log 2 N. 2 Time required for each complex multiplication , t1 = 20m sec. æN ö Time required for finding N - point DFT is T = ç log2 N÷ ´ t1 è2 ø Since , the computation should be performed before start of next block ,T £ TN . æN ö n 12 çè log2 N÷ø ´ 20 £ 125N Þ log2 2 £ 12.5 Þ nmax = 12 and Nmax = 2 = 4096 2 X(z)

n

Q14. y[n] = å x[m] and Y(z) = m=0

2 z(z - 1)

2

=

zX(z) (z - 1)

Þ X(z) =

1- z

2z

–1

=

z X(z)

x[n]

(z - 1)

–3 –1

(1 - z )

2

x[n] = 2u[n – 3] Þ x[2] = 0

0

1

2

3

4

n

74

Mechasoft

Q15. x1[n] = {1, 2, 3, 0} and x 2 [n] = {1, 3, 2,1} ­

­

DFT

[

]

x1[n] Ä x 2 [n] ¾¾¾ ® X1 (k) X 2 (k) and x 3 [n] = IDFT X1 (k)X 2 (k) ; Ä denotes circular convolution

é 9 ù é x 3 [0]ù ê 8 ú ê x [1] ú = ê ú º ê 3 ú Þ x 3 [2] = 11 ú ê ú 2 1 0 2 ú ê ú ê11ú ê x 3 [2]ú 3 2 1 úû êë1 úû êë14 úû êë x 3 [3]úû

é1 ê2 x 3 [n] = x1[n] Ä x 2 [n] = ê ê3 êë0

0 3 2 ù é1 ù 1 0 3ú ê3ú

1 j2 pn/16 1 – j2 pn/16 1 1 æ pn ö = 1+ e + e Þ a 0 = 1 , a1 = , a –1 = and N = 16 ÷ è 8ø 2 2 2 2

Q16. x[n] = 1 + cos ç

Use periodicity property a k = a k + N = a k + 2N to get a –1 = a –1+ 2´16 Þ a 31 = a –1 =

1 2

Im

Q17( c) Since ,x[n] is absolutely summable , its ROC must include unit circle |z|=1. Therefore , ROC lies inside the outer most pole located at z = ± j2 as demonstrated. x[n] is left sided, non causal signal .

j2 ROC 1

Re

2

unit circle –j2 Q18( c) Use Mason's gain formula to get Y(z)

G D +G D =H(z)= 1 1 2 2 = D X(z)

–1 5 –2 5 z + z – z –1 (1–z –1 ) 3 3 = 3 2 –1 2 –2 1–z –1 + z –2 1–z + z 9 X(z) 9

–

where x[n]=u[n] and X(z)=

–

Y(z)=

5 3

z

5

1

1

Y(z)

–1

5

æ 1 –1ö æ 2 –1ö æ 1 –1ö çè1– z ÷ø çè1– z ÷ø çè1– z ÷ø 3 3 3 3

1

z

(1–z )

=

2

–5/3 –1

1 –2/9

–1

Note that the ROC : |z|>

z–1

1

–

5

æ 2 –1ö çè1– z ÷ø 3

n

5/3 n

2 æ 1ö æ 2ö u[n]–5 ç ÷ u[n] ; ROC= z > ÷ è 3ø è 3ø 3

Þ y[n]=5 ç

includes unit circle and therefore, the system is stable .

Mechasoft

75 n

Im

n

æ –1ö æ –1ö Q19( c)Let x[n] = ç ÷ u[n] – ç ÷ u[–n – 1] = x1[n] – x2 [n] è 9ø è 3ø

–1/9

Re

n

æ –1ö where x1[n] = ç ÷ u[n] is right sided sequence è 9ø

ROC : |z|>1/9

n

æ –1ö and x2 [n] = ç ÷ u[–n – 1] is left sided sequence. è 3ø X1 (z) =

Im

1 z 1 = ; ROC1 : | z |> , outside the outermost pole . 9 æ 1 –1 ö z + 1 1– ç – z ÷ è 9 ø 9 ¥

å x [n] z

X2 (z) =

n

–1

–n

2

n=–¥

¥ –3z k æ 1ö = å ç – ÷ z– n = å (–3z ) = è ø 3 1 – (–3z) n=–¥ k =1

–1/3

Re

ROC : |z| | z |> 3 9

–1/3 –1/9

ROC :1/3> |z|>1/9

1 jnp /5 – jnp /5 1 –1 (e –e ) Þ a1 = , a –1 = , k = ±1 and N = 10 j2 j2 j2 The coefficients are non zero when k = Bm ± 1 = Nm ± 1 ; m is any integer Þ B=10 1 n 2 ; ROC : z > 2 Þ x[n] = (n + 1)(2) u[n] Þ x[2] = (2 + 1)(2) (1) = 12 Q21. X(z) = –1 2 (1 - 2z )

Q20. x[n] = sin [np / 5] =

z n Note (n + 1)a u[n] ¾®

1 –1 2

(1 - a z )

n

æ 2ö æ 2ö Q22. Let y[n] = ç ÷ u [n + 3] = ç ÷ è 3ø è 3ø

; ROC : | z | > a n +3–3

–3

æ 2ö æ 2ö u [n + 3] = ç ÷ ç ÷ è 3ø è 3ø

n +3

n

æ 2ö u[n] è 3 ÷ø

where x[n] = ç FT

FT

jw

Note x[n] ¾¾ ® X(e ) and x[n + 3] ¾¾ ®e n

æ 2ö FT ® çè ÷ø u[n] ¾¾ 3 æ 3ö y[n] ¾¾ ®ç ÷ è 2ø

1 1-

3

2 3

e

– j2 pf

j6 pf

FT

1-

e 2 3

e

– j2 pf

æ 2ö and ç ÷ è 3ø

j3w

jw

X(e )

n +3

j6 pf

FT

u[n + 3] ¾¾ ® 13

27 æ 3ö ÞA=ç ÷ = = 3.375 è 2ø 8

e 2 3

e

3

æ 3ö u[n + 3] = ç ÷ x[n + 3], è 2ø

– j2 pf

Re

76

Mechasoft

Q23(b)The N -point DFT of a sequence x[n] isX(k) =

1

N–1

kn

å x(n) WN where WN = e

– j2 p / N

N n =0 For N=4 , let x [n] = [a b c d] while W4= e–j2π/4= e–jπ/2 = cos (π/2) –jsin (π/2) = –j. 1 WN x N Use linear transformation matrix method for N- point DFT, to get X N = N 1 1 ù éa ù é1 1 1 2 3 1 ê1 W4 W4 W4 ú êb ú ê ú For N = 4 , DFT (x) = X 4 = 2 4 6 ê ú 4 ê1 W4 W4 W4 ú c

=

é1 1 ê1 DFT (DFT (x)) = ê 4 ê1 êë1

ê ú êë1 W43 W46 W49 úû ëêd ûú é1 1 1 1 ù éa ù é a+b+c+d ù ê ú ê ú 1 1 - j -1 j 1 êa - jb - c + jd ú b ê1 -1 1 -1ú ê c ú = ê a - b + c - d ú 2ê ú ê ú 2ê ú êë1 j -1 - júû êëd úû êëa + jb - c - jd úû 1 1 1 ù é a + b + c + d ù éa ù - j -1 j ú êa - jb - c + jd ú êd ú ú ê a - b + c - d ú = êc ú -1 1 -1 úê ú ê ú j -1 - júû êëa + jb - c - jd úû êëb úû

éa ù éa ù êd ú ê b ú DFT (DFT(x)) = x Þ ê ú = ê ú Þ b = d ; option (b) satisfies this. êc ú êc ú ëêb ûú ëêd ûú Q24(a) Given vector x[n] = [ a b c d ] and X(k) = [ a b g d ]

éa êb Inspecting the product [ a b c d ] ê êc êëd ép ù éa êq ú êb ê r ú = x[n]Ä x[n] = ê c ê ú ê êë s úû êëd

d c bù a d cú

ú = [ p q r s ] transpires that ú c b a úû

b a d

d c b ù éa ù a d c ú êb ú

ú ê c ú where Ä denotes circular convolution úê ú c b a úû êëd úû

b a d

2

DFT of the vector [ p q r s ] = DFT (x[n]Ä x[n])= X(k).X(k) = X (k) = éëα

2

β

2

γ

2

δ

2

ùû

Mechasoft

77

æ 1ö Q25( c) x[n] = ç ÷ è 3ø

|n|

n

æ 1ö æ 1ö u[n] = ç ÷ ÷ è 2ø è 3ø

–n

-ç

æ 1ö where x1[n] = ç ÷ è 3ø

–n

n

n

æ 1ö æ 1ö u[n] - ç ÷ u[n] = x1[n] + x 2 [n] – x 3 [n] ÷ è 3ø è 2ø

u[ - n - 1] + ç

n

n

æ 1ö æ 1ö u[ - n - 1] is left sided while x 2 [n] = ç ÷ u[n] and x 3 [n] = ç ÷ u[n] è 3ø è 2ø

are right sided sequences. X1 (z) =

–1

æ1 ö 1 – ç z÷ è3 ø

X 2 (z) = 1– X 3 (z) = 1–

1 1 3 1 1 2

–1

–z

=

z–3

z

= z

–1

z–

z

; ROC 2 = | z |>

1

, outside the outermost pole

3

3

z

= –1

1

; ROC1 = | z |< 3, inside the innermost pole

z–

1

; ROC3 = | z |>

1 2

, outside the outermost pole

2

ROC of x[n] is ROC x = ROC1 Ç ROC 2 Ç ROC3 =| z |< 3 Ç | z |>

1 3

Ç | z |>

1 2

=

1

e–0.05 .

Q33( c) x[n] = 5e–0.05n u[n] Þ X(z) =

1

2p kr N

| X(k) |

2

N

The time constant RC = 200 ´ 10 ´ 10 ´ 10 = 2sec

X(s) =

j

å x( r) X(k) e

Q32(b) The sampling frequency fs = 10Hz and sampling period t s = 3

2p kn N

1 fs

= 0.1 sec.

C +

10µF 200kΩ

R x(t) –

to sampler

Mechasoft

79 1 ¥

ìï üï jw ò X(e jw ) e jwn dw where X(e ) is DTFT of x[n] = í1,1,0,5,1ý ïî 2p – ¥ ­ ï þ ¥ ¥ ¥ 1 1 Plug n=0 to get x[0] = ò X(e jw ) e jw´ 0 dw = ò X(e jw ) dw and ò X (e jw )dw = 2 p x[0] = 2 p ´ 5 = 10π 2p – ¥ 2p – ¥ –¥

Q34(b) x[n] =

Im

0.5 0.5z Q35(b) X(z) = = ; ROC : | z |< 2 –1 1 – 2z z–2 Given that ROC includes unit circle and the pole is located at z = 2, the sequence x[n] is obviously left sided.

unit circle

1

Re

2

n

x[n] = –0.5 (2) u [–n – 1]and x[0] = 0

ROC :|z|>2

Q36(d) The ROC of x1 [n] –x2 [n] is same as that of x1 [n] +x2 [n] and it is 1/3 < |z|< 2/3. n

æ 5ö æ 6ö Q37( c) x[n] = ç ÷ u[n] - ç ÷ u[ - n - 1] and X(z) = è 6ø è 5ø 1 424 3 14243

right sided

1-

1 5

z

–1

6 123

left sided

-1 6 –1 1- z 5 123

; ROC :

5

6

5 ROC : |z| < 6 6 5 n 2n n é1ù æ 1ö 2 æ 1ö 2 Q38(d) x[n] = ê ú u[n] and y[n] = x [n] = ç ÷ u [n] = ç ÷ u[n] è ø è 4ø 2 ë2û ¥

jw

Y(e ) = å y[n] e

– jw n

n=– ¥ j0

n

n

¥ æ 1ö æ 1ö – jw n – jw n = å ç ÷ u[n] e = åç ÷ e n=– ¥ è 4 ø n =0 è 4 ø ¥

n

n

1/ 4 4 æ 1 ö – jn ´0 ¥ æ 1 ö e = åç ÷ = = ç ÷ n =0 è 4 ø n =0 è 4 ø 1 – (1 / 4) 3 ¥

Y(e ) = å

Q39(a) y[0] = x[–1] = 1 , y[2] = x

y[6] = x

é6 ù êë 2 – 1úû = x[2] =

é2 ù é4 ù – 1 = x[0] = 2 , y[4] = x êë 2 úû êë 2 – 1úû = x[1] = 1 ,

1 1 é –2 ù and y[–2] = x êë 2 – 1úû = x[–2] = 2 2

ì ïx y[n] = í ï0 î

x[n] 2

2

1 1/ 2 –2 –1

0

én ù êë 2 – 1úû for n even for n odd

1

1 1/ 2

1/ 2 1

2

n

n –2

–1

0

1

2

3

4

5

6

80

Mechasoft

Q40( c) Let g[n] = y[2n] , then g[0] = y[0] = 1 , g[–1] = y[–2] = 1 / 2, g[1] = y[2] = 2 , g[2] = y[4] = 1 and g[3] = y[6] = 1 / 2

g[n] 2

1

1 1/ 2

1/ 2

–1

0

1

2

3

n

¥

( )

– jw n jw – jw – j2 w – j3w G e jw = å g[n] e = g[–1] e + g[0] + g[1] e + g[2] e + g[3] e

=

n=– ¥

1 2

=e

e

jw

– jw

Q41( d) H(z) =

+ 1 + 2e

– jw

+e

– j2 w

(

+

1 2

e

– j3w

=e

– jw

1 – j2w ù é 1 j2w jw – jw êë 2 e + e + 2 + e + 2 e úû

)

é e j2w + e – j2w 2 e jw + e – jw ù –jω + + 2ú = e [cos2ω + 2cosω + 2] ê 2 2 êë úû

z 1 = has a pole located at z = 0.2 and ROC : | z |< 0.2 z– 0.2 1 – 0.2 z–1

The sequence is , obviously left sided , that is , h[n] = – (0.2)n u[–n – 1].

6

GATE-2022

Chapter

Signals and Systems

Section-1

Questions t is (1 + t2 )2 p (c) e–|w| 2j

Q1. The Fourier transform X( jw ) of the signal x(t) = (a)

p w e–|w| 2j

p (b) w e–|w| 2

p (d) e–|w| 2

Q2. The frequency response H(f) of a linear time-invariant system has magnitude as shown in the figure. Statement I : The system is necessarily a pure delay system for inputs which are band limited to – α≤f≤α. Statement II : For any wide-sense stationary input process with power spectral density SX(f), the output power spectral density SY(f), obeys SY(f) =SX(f) for – α≤f≤α. |H(f)| Which one of the following combinations is true ? (a) Statement I is correct , Statement II is correct. 1 (b) Statement I is correct , Statement II is incorrect. (c) Statement I is incorrect , Statement II is correct. f(Hz) (a) Statement I is incorrect , Statement II is incorrect. α –α 0 Q3. Let x1(t)=e–t u(t) and x2t = u(t)–u(t–2), where u(.) denotes the unit step function. If y(t) denotes the convolution of x1(t) and x2(t), then lim y(t) = ______(rounded off to one decimal place). t ®¥

Q4. The outputs of four systems S1,S2,S3,S4 corresponding to the input signal sin(t), for all time t, are shown in the figure. Based on the given information, which of the four systems is/are definitely NOT LTI (linear and time-invariant)? (a) S1 (b) S2 (c) S3 (d) S4

sin (t)

S1

sin (–t)

sin (t)

S3

sin (2t)

sin (t)

sin (t)

sin (t+1)

S2

sin2 (t)

S4

Q5. For a vector x = [x[0], x[1],....., x[7 ], the 8 - point discrete Fourier transform (DFT)is denoted by 7

æ 2p ö X = DFT(x) = [X[0], X[1],....., X[7]] , where X[k] = å x[n]exp ç – j nk ÷ . è ø 8 n =0 Here, j = –1. If x = [1,0,0,0, 2,0,0,0] and y = DFT (DFT (x)) , then the value of y[0]is__________ (rounded off to one decimal place).

Answer Key Q1

(a)

Q2

(c)

Q3

(0)

Q4

(c,d)

Q5

(8)

Solutions

Section 2 FT Q1(a) e–|t / t| ¾¾ ®

2t 2 d æ 2 ö – j4w FT FT gives e–|t| ¾¾ ® and te–|t| ¾¾ ®j ç ÷= 1 + w2 t2 1 + w2 dw è 1 + w2 ø 1 + w2 2

(

FT

)

FT

Use Duality theorem that states if x(t) ¾¾® X(w ), then X(t) ¾¾® 2px(– w ) to get

π –j4t t –|ω| FT FT ¾¾ ® 2π (–ω)e–|–ω| Þ ¾¾ ® ωe j2 (1+t2 )2 (1+t2 ) 2

Q2( c) A typical pure delay system has transfer function G(jw )=e–jwt whose magnitude | G(jw ) | = 1 is independent of ω.Statement (I) is incorrect. Output psd SY (f) =|H(f)|2 SX (f) where |H(f)|2 =1 for –α ≤ f ≤ α. Thus, SY (f) = SX (f) for –α ≤ f ≤ α.Statement (II) is correct. 1 1 and x2 (t) = u(t) – u(t– 2) Þ X2 (s) = (1 – e–2s ) s +1 s –2s 1– e y(t) = x1 (t) * x2 (t) Þ Y(s) = X1 (s)X2 (s) = s(s + 1)

–t Q3. x1 (t) = e u(t) Þ X1 (s) =

ì1 – e–2s ü lim y(t) = limsY(s) = lim í ý=0 t ®¥ s® 0 s® 0 î (s + 1) þ

Q4(c,d) For input sint , S1 generates output –sint and S2 generates sin(t+θ) where θ = 1rad. Both S1 and S2 produce sinusoid outputs of frequency same as input frequency. Obviously , S1 and S2 are LTI systems. Both systems S3 and S4 generate output sinusoids of frequency twice the frequency of input sinusoid. Obviously , S3 and S4 are definitely not LTI systems. Q5. On taking DFT of DFT of x[n],one gets scaled and inverted sequence back with x[0] remaining at 0th index. DFT of DFT x[n] = { Nx[0], Nx[N – 1], Nx[N – 2] ,..,.., Nx[2] , Nx[1] } For x = [x[0], x[1], x[2],..,.., x[7]] = [1,0,0,0, 2,0,0,0] ; N = 8 y = DFT(DFT (x)) = [8 x[0],8 x[8 – 1],8 x[8 – 2],..,..,8 x[2],8 x[1]] = [8,0,0,0,16,0,0,0] Þ y[0] = 8

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Rajeev Gupta is Assistant Professor in Department of ECE Motilal Nehru National Institute of Technology Allahabad, India. He has been awarded M.Tech. from IIT BHU and Ph.D. from MNNIT Allahabad. He has teaching and research experience of more than 18 years. His current research interest is in the field of RF and Microwave,Antenna and Wave Propagation, Signal and Image processing.

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