Signals and Systems - GATE_Electrical_By

YEAR 2010 ONE MARK

MCQ 3.1

For the system / (2 s+ , the approximate time taken for a step 1) response to reach 98% of the final value is

(A) 1 s (B) 2 s

MCQ 3.2

The period of the signal ( )x t 8sin 0.8 t π π4

= ` + j is

(A) 0.4π s (B) 0.8π s

MCQ 3.3

The system represented by the input-output relationship ( )

y t = ⁵^tx( )τ τd ,t>0

−3

#

(A) Linear and causal (B) Linear but not causal (C) Causal but not linear (D) Neither liner nor causal

NOTES

Common Data Questions Q.6-7.

Given ( )f t and ( )g t as show below

YEAR 2009 ONE MARK

MCQ 3.9

A Linear Time Invariant system with an impulse response ( )h t produces output ( )y t when input ( )x t is applied. When the input

( )

x t− is applied to a system with impulse response (τ h t− , the τ) output will be

(A) ( )y τ (B) y( (2 t−τ))

NOTES YEAR 2009 TWO MARKS

MCQ 3.10

A cascade of three Linear Time Invariant systems is causal and unstable. From this, we conclude that

(A) each system in the cascade is individually causal and unstable (B) at least on system is unstable and at least one system is causal (C) at least one system is causal and all systems are unstable (D) the majority are unstable and the majority are causal

MCQ 3.11

The Fourier Series coefficients of a periodic signal ( )x t expressed as ( )x t a ek j kt T^/

/

³_=-₃ 2^π are given by a-2= −2 j1, a−1=0.5+j0.2, a0= , j2 a1=0 5. −j0 2. , a2= + and a2 j1 k = for k0 > 2

Which of the following is true ?

(A) ( )x t has finite energy because only finitely many coefficients are non-zero

-YEAR 2008 ONE MARK

MCQ 3.13

The impulse response of a causal linear time-invariant system is given

NOTES

Page 122

minimum of α and β and similarly, max ( , )α β denotes the maximum of α and β, and K is a constant, which one of the following statements is true about the output of the system ?

(A) It will be of the form Ksinc( )γ where t γ=min( , )α β (B) It will be of the form Ksinc( )γ where t γ =max( , )α β (C) It will be of the form Ksinc(αt)

(D) It can not be a sinc type of signal

MCQ 3.17

Let x t be a periodic signal with time period T , Let ( )

( ) ( ) ( )

y t =x t−t0 +x t+ for some tt0 0. The Fourier Series coefficients of ( )y t are denoted by b_k. If b_k = for all odd k , then t0 0 can be equal to

(A) T 8 (B) / T 4/

MCQ 3.18

( )

H z is a transfer function of a real system. When a signal [ ]x n =(1+j)ⁿ is the input to such a system, the output is zero. Further, the Region of convergence (ROC) of ^{^}1−₂¹z^-¹^h H(z) is the entire Z-plane (except z= ). It can then be inferred that ( )0 H z can have a minimum of (A) one pole and one zero

(B) one pole and two zeros (C) two poles and one zero D) two poles and two zeros

MCQ 3.19

Given ( )

( )

X z z a

= 2

− with z > , the residue of ( )a X z z^{n 1}^- at z= a for n$ will be0

(A) a^{n 1}^- (B) aⁿ

(C) naⁿ (D) na^{n 1}

-MCQ 3.20

Let x t( )=rect^t− ₂¹h (where rect x( )= for 1 −₂¹ #x# ₂¹ and zero otherwise. If sinc( )x = ^sinπ⁽xπx⁾

, then the Fourier Transform of

NOTES

( ) ( )

x t + − will be given byx t (A) sinc 2

ωπ

` j (B) 2

sinc 2 ωπ

` j

2 2

sinc` ωπj `ωj (D) sin

2 2

sinc` ωπj `ωj

MCQ 3.21

Given a sequence [ ]x n , to generate the sequence [ ]y n =x[3−4n], which one of the following procedures would be correct ?

(A) First delay ( )x n by 3 samples to generate [ ]z n₁ , then pick every 4^th sample of [ ]z n1 to generate [ ]z n2 , and than finally time reverse

[ ]

z n₂ to obtain [ ]y n .

(B) First advance [ ]x n by 3 samples to generate [ ]z n₁ , then pick every 4^thsample of [ ]z n1 to generate [ ]z n2 , and then finally time reverse

[ ]

z n₂ to obtain [ ]y n

(C) First pick every fourth sample of [ ]x n to generate [ ]v n₁ , time-reverse [ ]v n1 to obtain [ ]v n2 , and finally advance [ ]v n2 by 3 samples to obtain [ ]y n

(D) First pick every fourth sample of [ ]x n to generate [ ]v n₁ , time-reverse [ ]v n₁ to obtain [ ]v n₂ , and finally delay [ ]v n₂ by 3 samples to obtain [ ]y n

YEAR 2007 ONE MARK

MCQ 3.22

The frequency spectrum of a signal is shown in the figure. If this is ideally sampled at intervals of 1 ms, then the frequency spectrum of the sampled signal will be

NOTES

Page 125

YEAR 2007 TWO MARKS

MCQ 3.24

Which among the following gives the fundamental fourier term of ( )

Statement for Linked Answer Question 25 and 26 :

MCQ 3.25

A signal is processed by a causal filter with transfer function ( )G s For a distortion free output signal wave form, ( )G s must

(A) provides zero phase shift for all frequency (B) provides constant phase shift for all frequency

(D) provides a phase shift that is inversely proportional to frequency

MCQ 3.26

( )

G z =αz^-¹+βz^-³ is a low pass digital filter with a phase characteristics same as that of the above question if

(A) α= (B) β α=−β

MCQ 3.27

Consider the discrete-time system shown in the figure where the impulse response of ( )G z is (0)g =0, (1)g =g(2)=1, (3)g =g(4)=g=0

NOTES

This system is stable for range of values of K

(A) [−1, ₂¹] (B) [−1, 1]

MCQ 3.28

If ( ), ( )u t r t denote the unit step and unit ramp functions respectively and ( ) * ( )u t r t their convolution, then the function (u t+1) * (r t− 2) is given by

(A) (₂¹ t−1) (u t− (B) 1) ₂¹(t−1) (u t−2) (C) (₂¹ t−1)²u t( − 1) (D) None of the above

MCQ 3.29

( ) 1 3 , ( ) 1 2

X z = − z^-¹ Y z = + z^-² are Z transforms of two signals [ ], [ ]

x n y n respectively. A linear time invariant system has the impulse response [ ]h n defined by these two signals as [ ]h n =x n[ −1] * [ ]y n where * denotes discrete time convolution. Then the output of the system for the input [δ −n 1]

(A) has Z-transform z X z Y z^-¹ ( ) ( )

(B) equals [δ n− −2] 3δ[n− +3] 2δ[n− −4] 6δ[n−5] (C) has Z-transform 1−3z^-¹+2z^-²−6z^-³

(D) does not satisfy any of the above three

YEAR 2006 ONE MARK

MCQ 3.30

The following is true

(A) A finite signal is always bounded

(B) A bounded signal always possesses finite energy

(D) A bounded signal is always finite

NOTES

Page 129

(A) 61 (B)

3 1

3 2

MCQ 3.38

The Laplace transform of a function ( )f t is ( )

( 2 2)

F s s s s

s s

5 23 6

= 2

+ +

+ + as , ( )

t"3 f t approaches

(A) 3 (B) 5

MCQ 3.39

The Fourier series for the function ( )f x =sin²x is (A) sinx+sin2x (B) 1−cos x2 (C) sin2x+cos2x (D) 0 5. −0 5. cos x2

MCQ 3.40

If ( )u t is the unit step and ( )δ is the unit impulse function, the t inverse z -transform of ( )F z = _z₊¹₁ for k> is0

(A) (−1)^kδ( )k (B) δ( )k − −( 1)^k (C) (−1)^ku k( ) (D) u k( )− −( 1)^k

YEAR 2004 TWO MARKS

MCQ 3.41

The rms value of the periodic waveform given in figure is

(A) 2 6 A (B) 6 2 A

NOTES

MCQ 3.45

Let ( )Y s be the Laplace transformation of the function ( )y t , then the final value of the function is

(A) LimY s( )

What is the rms value of the voltage waveform shown in Figure ?

(A) (200/ )π V (B) (100/ )π V

YEAR 2001 ONE MARK

MCQ 3.47

Common data Questions Q.48-49*

Consider the voltage waveform v as shown in figure

NOTES

Page 132

MCQ 3.48

The DC component of v is

(A) 0.4 (B) 0.2

MCQ 3.49

The amplitude of fundamental component of v is

(A) 1.20 V (B) 2.40 V

***********

NOTES

Output is integration of input which is a linear function, so system is linear.

Hence (B) is correct option.

SOL 3.4

Fourier series of given function ( )

For second harmonic component (n= amplitude is zero.2) Hence option (A) is correct.

SOL 3.5

By parsval’s theorem ( ) Hence option (D) is correct.

NOTES

Hence option (C) is correct.

SOL 3.7

NOTES Hence (D) is correct option.

SOL 3.10

Let three LTI systems having response H z H z₁( ), ₂( ) and H z3( ) are Cascaded as showing below

Assume H z₁( ) = + + (non-causal)z² z¹ 1 ( )

H z₂ = + + (non-causal)z³ z² 1 Overall response of the system

( )

Similarly to make ( )H z unstable atleast one of the system should be unstable.

Hence (B) is correct option.

SOL 3.11

NOTES Hence (C) is correct option.

SOL 3.12

Z-transform of [ ]x n is ( )

X z =4z^-³+3z^-¹+ −2 6z²+2z³ Transfer function of the system

( )

Hence (A) is correct option.

SOL 3.13

Since the given system is LTI, So principal of Superposition holds due to linearity.

For causal system ( )h t = , 0 t<0 Both statement are correct.

Hence (D) is correct option.

SOL 3.14

For an LTI system output is a constant multiplicative of input with same frequency.

NOTES Hence (A) is correct option.

SOL 3.17

Let ak is the Fourier series coefficient of signal ( )x t Given y t ( ) =x t( −t₀)+x t( +t₀)

Fourier series coefficient of ( )y t b_k =e^-^{jk t}^ω⁰a_k+e^{jk t}^ω⁰a_k

Hence (B) is correct option.

SOL 3.18

NOTES

Page 143

SOL 3.23

For an LTI system input and output have identical wave shape (i.e.

frequency of input-output is same) within a multiplicative constant (i.e. Amplitude response is constant)

So F must be a sine or cosine wave with ω₁=ω₂ Hence (D) is correct option.

SOL 3.24

Given signal has the following wave-form

Function x(t) is periodic with period T2 and given that ( )

x t =−x t( +T) (Half-wave symmetric)

So we can obtain the fourier series representation of given function.

Hence (C) is correct option.

SOL 3.25

Output is said to be distortion less if the input and output have

NOTES

identical wave shapes within a multiplicative constant. A delayed output that retains input waveform is also considered distortion less.

Thus for distortion less output, input-output relationship is given as ( )

y t =Kg t( −t_d) Taking Fourier transform.

( )

Y ω =KG( )ω e^-^{j t}^ω^d ( )

Y ω =G( )ω H( )ω ( )

H ω & transfer function of the system So, H( )ω Ke= ^-^{j t}^ω^d

Amplitude response H( )ω =K Phase response θ ω n( ) =−ωtd

For distortion less output, phase response should be proportional to frequency.

Hence (C) is correct option.

SOL 3.26

( )

G z z=e^j^ω =αe⁻^j^ω+βe⁻³^j^ω for linear phase characteristic α= .β

Hence (A) is correct option.

SOL 3.27

System response is given as ( )

For system to be stable poles should lie inside unit circle.

z #1

NOTES Hence (A) is correct option.

SOL 3.28

Given Convolution is, ( )

h t =u t( +1))r t( −2) Taking Laplace transform on both sides,

( )

Taking inverse Laplace transform ( )

h t (t ) u t( ) 2

1 1 ² 1

= − −

Hence (C) is correct option.

SOL 3.29

Impulse response of given LTI system.

[ ]

h n =x n[ −1 )] y n[ ] Taking z -transform on both sides.

( )

NOTES

-Taking inverse z-transform on both sides we have output.

[ ]

y n =δ[n− −2] 3δ[n− +3] 2δ[n− −4] 6δ[n−5] Hence (C) is correct option.

SOL 3.30

A bounded signal always possesses some finite energy.

E g t( ) dt<

-Hence (B) is correct option.

SOL 3.31

Trigonometric Fourier series is given as ( )

NOTES

Page 148

Since length of convolution ( [ ])y n is 1− to 2, [ ]x n is of length 1− to0 so length of [ ]h n is 0 to 2.

Let h n [ ] ={ , , }a b c -Convolution

y n [ ]

-{ a a,2 b b,2 c c,2 }

= − − −

[ ] y n

-{ 1, 3, 1, 2}

= − − −

So, a=1

a b b

2 − =3 & =−1

a c c

2 − =−1 & =−1 Impulse response

-[ ] , 1, 1 h n ="1 − − , Hence (A) is correct option.

SOL 3.34

Option () is correct

SOL 3.35

Output y t ( ) =e^-^{x t}^{( )} If ( )x t is unbounded, ( )x t " 3

( )

y t =e^-^{x t}^{( )} "0 (bounded) So ( )y t is bounded even when ( )x t is not bounded.

Hence (D) is correct option.

SOL 3.36

Given y t ( ) x t dt( )^' ^'

= t

−3

#

Laplace transform of ( )y t ( )

Y s = ( ) s

X s , has a singularity at s=0

NOTES

For a causal bounded input, ( )y t x t dt( )^' ^'

−3

#

is always bounded.

Hence (B) is correct option

SOL 3.37

RMS value is given by

V_rms ( ) Hence (A) is correct option

SOL 3.38

By final value theorem ( ) Hence (A) is correct option

SOL 3.39

NOTES

Page 151

SOL 3.42

Total current in wire

I =10+20sin tω Irms ( ) ( )

10 2

2 20 2

= + =17.32 A

Hence (B) is correct option

SOL 3.43

Fourier series representation is given as ( )

From the wave form we can write fundamental period T=2 sec ( )

By solving the integration

a_n ,

Hence (C) is correct option

NOTES

-Hence (A) is correct option

SOL 3.45

Final value theorem states that ( )

lim y t

t "3 lim Y s( )

s "3

Hence (B) is correct option.

SOL 3.46

Hence (D) is correct option.

SOL 3.47

NOTES

Amplitude of fundamental component of v is vf a12 b Hence (A) is correct option.

***********

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