Signals & Systems Test - 5

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GATE
EEE

Signals
&

Systems

Test-05Solutions

SIGNALS & SYSTEMS

1. The period of the function cos π/4 (t – 1) is
a) 1⁄8 s

b) 8s

c) 4s
d) 1⁄4 s

Answer: (b)

Solution:

f(t) = cos π (t − 1)
4

Period = 2π = 2π = 8S
ω π⁄4

2. Two linear time-invariant discrete time systems S1 and S2 are

cascaded as shown in the below figure. Each system is modelled

by a second order difference equation. The difference equation

of the overall cascaded system can be of the order of

a) 0,1, 2,3 or 4

b) either 2 or 4

c) 2

d) 4

Answer: (d)

Solution:

H(s) = H1(s) ∙ H2(s)
= (as2 +….) (ms2 +…) = (ps4 +…)

Where a, m and p are constants

1

Thus the overall cascaded system will have the difference
equation of the order of 4.
3. The Fourier series representations of a periodic current [2 +
6√2 cos ωt + √48 sin 2ωt]A The effective value of the current
is
a) [2 + 6 + √24]A
b) 8A
c) 6A
d) 2A
Answer: (b)
Solution:

Ieff = �Id2c + I12rms + I22rms = �(2)2 + �6√√22�2 + �√√428�2
= √4 + 36 + 24 = √64 = 8A

4. The input and output of a continuous time system are
respectively denoted by X (t) and y (t). Which of the following
descriptions corresponds to a causal system?
a) y(t) = x (t – 2) + x (t + 4)
b) y(t) = (t – a) x (t + 1)
c) y(t) = (t + a) x (t – 1)
d) y(t) = (t + 5) x (t + 5)
Answer: (c)
Solution:
For a causal system the output, y at any time, t0 depends only on
x(t) for t ≤ t0 and not on x(t) for t > t0.

2

Y(t) = (t + 4) x (t – 1) is a causal system because y at time t

depends upon previous value of x at (t – 1).

Note that the causality does not depend upon whether x(t) is

multiplied by a constant or function of time.

5. Given that F(s) is the one-sided Laplace transform of f(t), the

Laplace transform of ∫0t f(τ)dτ is

a) SF(S) – f (0)

b) 1 F(S)
S

c) ∫0S F(τ)dτ

d) 1 [F(S) − f(0)]
S

Answer: (c)

Solution:

If f(t)⟶F(s), then according to integration in time property

∫0t f(τ) dτ ⟶ F(S) and ∫∞t f(τ) dτ ⟶ F(S) + ∫−0∞− fS(τ)dτ.
s s

6. The Fourier transform of u(t) is

a) 1/jω

b) jω

c) 1/(1 + jω)

d) πδ(ω) + 1/jω

Answer: (d)

Solution:

U(t) �F�.T. 1 + πδ(ω)
jω

3

7. If the Fourier transform of f(t) is F(jω), then what is the Fourier
transform of f (– t)?
a) F (jω)
b) F (– jω)
c) – F(jω)
d) Complex conjugate of F(jω)
Answer: (b)
Solution:
This is the time reversal property of Fourier transform.

8. The Fourier series expansion of a real periodic signal with
fundamental frequency f0 is given by gP(t) = ∑∞n=−∞ Cnej2πf0t .
It is given that c3 = 3 + j5. Then C– 3 is
a) 5 + j3
b) – 3 – j5
c) – 5 + j3
d) 3 – j5
Answer: (d)
Solution:
Property of Fourier series (FS):
If a periodic signal gP(t) is real its exponential F > S coefficient
for the nth harmonic satisfies the relation cn = c−∗ n or c−n= cn∗ or
cn is conjugate even (*denotes complex conjugation)
∴ c−3 = c3∗ = 3 – j5

4

9. What is the input-output relation of the causal moving-average

system (discrete time)?

a) y[n] = 1 {x[n] + x[n − 1] + x[n − 2]}
3

b) y[n] = 1 {x[n − 1] + x[n] + x[n + 1]}
3

c) y[n] = 1 �x[n] + (x[n])2 + (x[n])1/2�
3

d) y[n] = 1 {x[n] + x[n + 1] + x[n + 2]}
3

Answer: (a)

Solution:

• System should be causal (b) and (d) are not causal.

• (c) is not moving average.

So, moving average = 1 [x(n + x(n − 1) + x(n − 2)]
3

10. A system has a phase response given by ϕ(ω) where ω is the

angular frequency. The phase delay and group delay at ω = ω0

are respectively given by

a) − ∅(ωω00), − d∅d(ωω)�ω = ω0
b) ∅(ω0), − d2d∅ω(2ω)�ω = ω0
c) ∅(ωω00), − d∅d(ωω)�ω = ω0
d) ω0∅(ω0), ∫−ω∞0 ∅(λ)dλ

Answer: (d)

5

Solution:

For an LTI system, if the phase response is ϕ(ω), then at f = f0

or ω = ω0

Phase delay, tp= − ∅(ωω00), − 1 ∅(f0)
2π f0

and group delay tg = −dd∅ω(ω)�atω = ω0= − 1 −dd∅f(f)�atf = f0
2π

11. If �2S72S++339S7� is the Laplace transform of f (t) then f (0+)is

a) Zero

b) 97/33

c) 27

d) Infinity

Answer: (c)

Solution:

f(0+) = sl→im∞ SF(s)

= sl→im∞ S �2S72s++3937s�

= sl→im∞ 27 + 97 = 27
s
33
1 + s

12. The region of convergence of z-transform of the sequence

�56�n u(n) − �56�n u(−n − 1) must be

a) |Z| < 5
6

b) |Z| > 6
5

c) 5 < |Z| < 6
6 5

6

d) 6 < |Z| < ∞
5

Answer: (c)

Solution:
Let x(n) = �65�nu(n) − �56�nu (– n – 1)

The given function x(n) = x1(n) + x2(n)

Where, X1(n) = �56�nu(n) �Z�T �z − (z5/6)�;|Z| > 5
6

X2(n) = − �56�nu (– n – 1) Z��T �z − (z6/5)�;|Z| < 6
5

∴ X(z) = �z − (z5/6)� + �z − (z6/5)�

The overlapping region 5 < |Z| < 6 is the region of convergence.
6 5

13. What is the Laplace transform of cosω0t?

a) ω0
S2 + ω02

b) S2 S
+ ω02

c) ω0
(S + ω0)2

d) (S S
+ ω0)2

Answer: (b)

14. The network function, H(s) is equal to

a) Y(S)
X(S)

b) X(S)
Y(S)

c) X(S) Y(S)

d) 1
X(S)Y(S)

7

Answer: (a)

15. The frequency response H(Ω) of a system for impulse

sequence response h[n] = δ[n] + δ [n – 1]

a) H (Ω) = 2cos �Ω2 � ∠ − Ω
2

b) H (Ω) = 2cos Ω∠ − Ω

c) H (Ω) = 2cos Ω ∠ − Ω
2

d) H (Ω) = 2∠ − Ω
2

Answer: (a)

Solution:

Given, h[n] = δ [n] + δ [n – 1]

Taking Fourier transform

H(Ω) = 1 + e – jΩ

⇒ H(Ω) = 2e−jΩ⁄2 �ejΩ⁄2 +2e−jΩ⁄2�

⇒ H(Ω) = 2 cos �Ω2 � jΩ

e2

⇒ H(Ω) = 2 cos �Ω2 � ∠ − Ω
2

16. Let x (t) = cos (10 πt) + cos (30 πt) 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) 5Hz,10Hz and 15 Hz only

d) 5 Hz only

8

Answer: (a)

Solution:

fm1 = 5Hz, fs = 20 HZ, fm2 = 15 Hz

Frequency at output of sampler,

fm1 ± nfs ⇒ 5 ± n(20)
fm2 ± nfs ⇒ 5 ± n(20)

n = 1: 15, 25, 5, 35

When these frequencies are applied to L.P.F with fc = 20 Hz, we

will retain 5 Hz and 15 Hz only.

17. The output y[n] of a discrete time LTI system is related to the
input x[n]as given below: y[n] = ∑k∞= 0 x[k]. Which one of the
following correctly relates the z-transforms of the input and

output, denoted by X(z) and Y(z), respectively?

a) Y(Z) = (1 – z –1) X(Z)

b) Y(Z) = z –1 X(Z)

c) Y(Z) = X(Z)
1 – Z−1

d) Y(Z) = dX(Z)
dz

Answer: (c)

Solution:

Y[n] = ∑∞K = 0 x[k] = ∑∞K = −∞ x[k]u[k]

Y[z] = ∑∞K = −∞ u[n]Z−n = ∑K∞= 0(Z−1)n
X[z]

⇒ Y[z] = 1 1
X[z] − z−1

⇒ y[z] = X[z]
1 − z−1

9

18. Unit step response of the system described by the equation

Y(n) + y (n – 1) = x(n) is

a) Z2
(Z + 1)(Z − 1)

b) (Z + Z − 1)
1)(Z

c) Z+ 1
(Z − 1)

d) Z(Z − 1)
(Z + 1)

Answer: (a)

Solution:

Given y(n) + y (n – 1) = x(n)

Taking Z-transform

Y(z) + Z – 1y(z) = X(z)

∴ Y(Z) = (1 1
X(Z) + Z−1)

And x(n) = u(n)

∴ x(Z) = (1 1
+ Z−1)

Hence Y(z) = (1 1 X(Z)
+ Z−1)

⇒ Y(Z) = (1 1 × (1 1 = (Z − Z2 + 1)
+ Z−1) + Z−1) 1)(Z

⇒ (Z + Z2 − 1)
1)(Z

19. Consider two infinite duration input sequences {x1[n], X2[n]}.

When will the Region of Convergence [ROC] of z-transform of

their superposition i.e., {x1[n], X2[n]}be entire z- plane except

Possibly at z = 0 or z = ∞?

10

a) When their linear combination is of finite duration

b) When they are left sided sequences

c) When they are right sided sequences

d) When their linear combination is causal

Answer: (a)

Solution:

x1(n) + x2(n)

For finite duration, the ROC is entire Z-plane except Z = 0 and

Z=∞

Example: δ (n – 1) + δ(n + 1) ⟷ Z – 1 + Z = 1 + Z
Z

Discontinuity occurs at z = 0 and Z = ∞. Hence ROC = entire z-

plane except Z = 0 and Z = ∞

20. The Laplace transform of a unit ramp function starting at t = a,

is

a) 1
(S + a)2

b) e−as
(S + a)2

c) e−as
S2

d) a
S2

Answer: (c)

Solution:

11

Unit ramp function starting at t = 0 is given by r(t) = t u(t)

Unit ramp function starting at t = a (shown in Fig. 1) is given by

X(t) = r(t – a) = (t – a) u(t – a)

Remember the LT pair:

r(t) = t u(t) ⟶ 1/(s2), σ > 0

using the time shift property,

x(t) = r(t – a)

= (t – a) u (t – a)

X(s) = e−as , σ > 0
S2

21. Laplace transform of e-at f(t) is

a) F(s)e-at

b) F (s – a)

c) F (s + a)

d) F(s) + a
s

Answer: (c)

22. Inverse Fourier transforms of signal X(jω) = 2 πδ(ω) + πδ(ω –

4π) + πδ(ω + 4π)

a) 1 + cos 4πt

b) π (1 – cos4πt)

c) 2π (1 – cos4πt)

d) 2π (1 + cos4πt)

Answer: (a)

Solution:
X(jω) = 2 δ(ω) + δ (ω – 4 ) + δ (ω + 4 )

12

1 �F�.T.2 δ(ω)
cos ω0t F��.T. [δ (ω – ω0) + δ (ω + ω0)]
23. Match each of the items A, B and C with an appropriate item

from 1, 2, 3, 4 and 5.

A. a1 d2y + a2Y dy + a3y = a4
dX2 dx

B. a1 d3y + a2Y = a3
dX3

C. a1 d2y + a2X dy + a3x2y = 0
dX2 dx

1. Nonlinear differential equation.

2. Linear differential equation with constant coefficients.

3. Linear homogeneous differential equation.

4. Non - linear homogeneous differential equation.

5. Nonlinear first order differential equation.

a) A – 1, B – 2, C – 4.

b) A – 2, B – 5, C – 1.

c) A – 1, B – 3, C – 4.

d) A – 3, B – 4, C – 2.

Answer: (a)

Solution:

(A) a1 d2y + a2Y dy + a3y = a4⟶
dX2 dx

(1) Nonlinear Differential equation

(because of the product term y ddyx)

(B) a1 d3y + a2Y = a3 ⟶
dX3

13

(2) Linear Differential equation with constant coefficients

(C) a1 d2y + a2X dy + a3x2y = 0 ⟶
dX2 dx

(4) Non-Linear Homogenous Differential equation

(because RHS = 0)

The correct matching is A – 1, B – 2, C – 4.

24. A real - valued signal x(t) limited to the frequency band |f| ≤

W is passed through a linear time invariant system whose
2

e−j4πf, |f| ≤ W
frequency response is H(f) = � |f| >
2 The output of
0, W

2

the system is

a) X(t + 4)

b) X (t – 4)

c) X(t + 2)

d) X (t – 2)

Answer: (d)

Solution:

H(f) = Y(f) = e−j4πf
X(f)

Y(f) = e−j4πf X(f)

Using time shifting Y(t) = x (t – 2).

25. The function f(t) has the Fourier transform g(ω). The Fourier

transform of g(t) is

a) f (ω)/2π

b) f (–ω)/2π

14

c) 2π f (–ω)

d) None of these

Answer: (c)

Solution:

f(t) F��.T.g (ω)
g(t) �F�.T.2 f (– ω)

26. The process of imitating one system with another so that the

imitating systems accepts the same data, executes same

programming and achieves same results as the imitated system

is known as

a) Simulation

b) Modification

c) Translation

d) Emulation

Answer: (d)

27. The purpose of Design for Test (OFT) process in ASIC design

flow is

a) To capture functional errors

b) To capture manufacturing defects

c) To capture timing violations

d) For radiation mitigation

Answer: (b)

15

28. The region of the z plane for which |z − a| = 1(Re a≠ 0) is
|z + a|

a) x-axis

b) y-axis

c) The straight-line z = |a|

d) None of the above

Answer: (b)

Solution:

�ZZ − aa� = 1
+

|Z − a| = |z + a|

|x + jy − a| = |x + jy + a| (∵ Z = x + jy)

(x – a)2 + Y2 = (x + a)2 + y2

X2 + a2 – 2ax + y2 = x2 + a2 + 2ax + y2

4ax = 0
∴x=0

Hence equation will give y-axis.

29. Unit step response of the system described by difference

equation y(n) + y (n – 1) = x(n) is

a) Z2
(Z + 1)(Z − 1)

b) (Z + Z − 1)
1)(Z

c) (Z + 1)
(Z − 1)

d) Z(Z − 1)
(Z + 1)

Answer: (a)

16

Solution:

y(n) + y (n – 1) = x(n)

Taking z-transform of above equation y(Z) + Z – 1Y(z) = X(Z)

⇒ Y(Z) = 1 = Z
X(Z) 1 + Z−1 Z+1

Or Y(Z) = �Z Z 1�X(Z)
+

But given x(n) = u(n)

Or X(Z) = 1 1 = Z
− Z−1 Z−1

So Y(Z) = �Z Z 1� �Z Z 1�
+ −

= (Z + Z2 − 1)
1)(Z

30. A function c (t) satisfies the differential equation Ċ (t) + c(t) =

δ(t). For zero initial condition c(t) can be represented by
a) ∈−t
b) ∈t
c) ∈t u(t)
d) ∈−t u(t)

Where u(t) is a unit step function

Answer: (d)

Solution:

dc + c(t) = δ(t)
dt

sC(s) + C(s) = 1

C(s) = S 1 1
+

C(t) = ϵ−t u(t)

17

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