Signals & Systems Test - 1 - PDF Flipbook
Signals & Systems Test - 1
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GATE
EEE
Signals
&
Systems
Test-01Solutions
SIGNALS & SYSTEMS
1. Let x(t) be a continuous time periodic signal with fundamental
period T = 1 seconds Let { } be the complex Fourier series
coefficients of x(t), where k is integer valued. Consider the
following statement about x(3t):
I. The complex Fourier series coefficients of x(3t) are { }
where k is integer valued
II. The complex Fourier series coefficients of x(3t) are {3 }
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
Answer: (b)
Solution:
X(t) → aK, = 2π
X(at) → aK,
X(3t) → aK, 3 = 6π
So I and III are true
1
2. If a plot of signal x(t) is as shown in the Figure,
then the plot of the signal x (1 – t) will be
Answer: (a)
2
Solution:
x(1 + t) is the shifted version of x(t) towards left by one unit.
X(1 – t) is the mirror image of x(1 + t) along vertical axis.
3. Consider the following systems:
1. Y[K] = x[k] + a1x [k – 1] – b1y [k – 1] – b2y [k – 2]
2. Y[K] = x[k] + a1x [k – 1] – a2x [k – 2]
3. Y[K] = x[k +1] + a1x [k] + a2x [k – 1]
4. Y[K] = a1x [k] + a2x [k + 1] – b1y [k – 2]
Which of the systems given above represent recursive discrete
system?
a) 1 and 4
b) 1 and 2
c) 1, 2 and 3
d) 2,3 and 4
Answer: (a)
3
Solution:
System 2 & 3 are not recursive because it is not in the form
y(n – k) = ∑ ( − ).
4. The Fourier series of an odd periodic function contains only
a) odd harmonics
b) even harmonics
c) cosine terms
d) sine terms
Answer: (d)
Solution:
If a periodic function is odd, the d.c. term a0 = 0 and cosine
terms, which are even are absent. It contains only sine terms.
5. The lengths of two discrete time sequence x1(n) and x2(n) are 5
and 7, respectively. What is the maximum length of a sequence
x1(n) * x(n)?
a) 5
b) 6
c) 7
d) 11
Answer: (d)
Solution:
Maximum length of 1( )*x2(n) is
L = L1 + L2 – 1
Where
L1 is length of x1(n) & L2 is length of x2(n).
4
6. A square wave is defined by
X(t) = �− , ,0 0 0. If the initial conditions are zero and the input is e3t, the
output for t > 0 is
a) 3 − 2
b) 5
c) 3 + 2
d)
Answer: (a)
Solution:
Impulse response, h(t) = 2
So, H(s) = 1
( −2)
Input, x (t) = 3
LT[x(t)] = X(s) = 1
( −3)
Output y(t) = x(t)*h(t)
Taking Laplace transform Y(s) = X(s) H(s)
= 1 ∙ 1
( −2) ( −3)
= 1 ∙ 1
( −3) ( −2)
Taking inverse Laplace transform.
Y(t) = 3 − 2
Note:
3 is a real exponential.
9
14. Match List-I (Time Domain Property) with List-II
(Frequency Domain Property pertaining to Fourier
Representation Periodicity Properties) and select the correct
answer using the codes given below the lists:
List-I List-II
A. Continuous 1. Periodic
B. Discrete 2. Continuous
C. Periodic 3. Non-periodic
D. Non-periodic 4. Discrete
Codes:
ABC D
a) 3 4 1 2
b) 2 4 1 3
c) 2 1 4 3
d) 3 1 4 2
Answer: (d)
Solution:
Discrete in one domain, periodic in other domain. Continuous in
one domain, aperiodic in other domain.
15. The impulse response of a discrete system with a simple pole
shown in the below figure. The pole of the system must he
located on the
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a) Real axis at z = – 1
b) Real axis between z = 0 and z = 1
c) Imaginary axis at z = j
d) Imaginary axis between z = 0 and z = j
Answer: (b)
Solution:
The given impulse response has the form H (z) = − , | | > 0
⟹ h(n) = u(n), 0 < a < 1
So, the pole of the system must be located on the real axis
between z = 0 and z =1.
16. The input to a channel is a band pass signal. It is obtained by
linearly modulating a sinusoidal carrier with a single-tone
signal. The output of the channel due to this input is given by
y(t) = (1/100) cos (100t – 10–6) cos (106t – 1.56). The group
delay ( )and the phase delay ( ) ), in seconds, of the channel
are
a) = 10−6, = 1.56
b) = 1.56 , = 10−6
c) = 10−8, = 1.56 × 10−6
d) = 10−8, = 1.56
Answer: (c)
Solution:
According to the information given, the input to the channel can
be taken as x (t) = cos( ) cos( ),
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where cos( ) is the carrier and cos( ) is the single tone
modulating signal.
Output of the channel is given as
Y (t) = 1 cos(100 − 10−6) cos(106 − 1.56)
100
Where signal frequency =100 r/s and
Carrier frequency = =106r/s
Rearranging y(t),
Y(t) =1100 cos(100( − 10−8)) cos(106( − 1.56 × 10−6))
Group delay or signal delay = = 10−8sec
Phase delay or carrier delay = = 1.56× 10−6 sec.
17. Which one of the following is the inverse z-transform of X(Z)
= ( −2) ( −3), | |< 2?
a) [2 − 3 ] (− − 1)
b) [3 − 2 ] (− − 1)
c) [2 − 3 ] ( + 1)
d) [2 − 3 ] ( )
Answer: (a)
Solution:
X(Z) = ( −2) ( −3), | |< 2
= −
−3 −2
⟹ x(n) = −3 u(– n – 1)+ 2 u(– n – 1)
⟹ x(n) = (2 − 3 ) u(– n – 1)
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18. Let X (t) is the input and y (t) is the output of a continuous
time system. Match the system properties P1, P2 and P3 with
System relations R1, R2, R3, R4.
Properties
P1: Linear but NOT time-invariant
P2: Time-invariant but NOT linear
P3: Linear and time-invariant
Relations
R1: Y (t) = t2 X (t)
R2: y (t) = t| ( )|
R3: y (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)
Answer: (a)
Solution:
R1: y(t) = 2x(t) is linear Multiplication of x(t) by a function of
time t2 keeps the system linear but makes the system time
varying or not time invariant.
R1: y(t) = 2x(t) is linear but not time-invariant. (matches with
P1)
R4: y(t) = x (t – 5) is linear and time invariant, as time shifting
of input is a linear operation. (matches with P3)
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R3:Y(t) = | ( )| = x(t) for x(t) > 0
= – x(t) for x(t) < 0
Modulus operation is a nonlinear, time invariant operation as
shown below
X(t) → y(t) = | ( )|
1(t) → 1(t) = | 1( )|
2(t) → 2(t) = | 2( )|
α 1( ) + β 2(t) → output = | 1( ) + β 2(t) |…..(1)
However, α 1( ) + β 2(t) = | 1( ) + β 2(t) |…..(2)
As (1) and (2) give different expressions, the system is not linear
X(t) → y(t) = | ( )|
X(t – t0)→ output = | ( − 0)|…….(3)
X(t – t0) = | ( − 0)|…….(4)
As (3) and (4) give the same expression, the system is time
invariant.
⸫y(t) = | ( )|is nonlinear and time invariant.
(R3 matches with P2)
R2: y(t) = t| ( )| is nonlinear and time varying.
Correct matching:
(Pl, R1), (P2, R3), (P3, R4)
⸫Option (a) is correct.
19. The z-transform of – u (– n – 1) is
a) with | | >1
−1
b) with | | 1
b) |Z| < 1
c) (Real part of z) > 0
d) (Real part of z) < 0
Answer: (a)
Solution:
U (n) = unit step function or unit step sequence
U (n) ↔ , | | >1
−1
ROC is outside the circle of radius = 1, with its center at the
origin in the z-plane.
21. Consider a two-sided discrete-time signal (neither left sided,
nor right sided). The region of convergence (ROC) of the z-
transform of the sequence is
1. All region of z-plane outside a unit circle (in z-plane)
2. All region of z-plane inside a unit circle (in z-plane)
3. Ring in z-plane
Which of the above is/are correct?
a) 1 only
b) 2 only
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c) 3 only
d) 1 and 3
Answer: (c)
Solution:
From the properties of ROC of z-transform, for a two sided
sequences the ROC of its z-transform is in the form of circular
strip or annular strip i.e., in the form of ring.
22. If α, β, γ are the roots of equations x3 + px2 + qx + p, Then the
value of tan-1 + tan-1 β + tan-1 γ is
a) n ⁄2
b) n
c) 2n
d) n ⁄4
Answer: (b)
Solution:
tan−1 + tan−1 + tan−1 = tan−1 �1− (+ + + − + )�
Given equation is 3 + 2 + qx + p = 0
α + β + γ = – p, αβγ = – p
+ + =q
tan−1 + tan−1 + tan−1 = tan−1 �1−− (+ ) �
= tan−1(0)
= nπ.
16
23. What does the following integral evaluate to?
∫0 ⁄2 sin6θ. dθ
a) 5 ⁄16
b) 5 ⁄8
c) 0
d) 5 ⁄32
Answer: (d)
∫0 ⁄2 = ( −1) ∙ ( −3) ∙ ( −5) ∙∙∙ 3 ∙ 1 ∙
( −2) −4 4 2 2
Where n is odd
= ( −1) ∙ ( −3) ∙ ( −5) ∙∙∙ 3 ∙ 1 ∙ 1
( −2) −4 4 2
∫0 ⁄2 6
Here n = even
∫0 ⁄2 = 5 ∙ 3 ∙ 1 ∙
6 4 2 2
= 53 2 .
24. The Fourier transform F {e – t u(t)} is equal to 1+ 1 2 .
Therefore, F �1+ 1 2 � is
a) u(f)
b) − u(f)
c) u (–f)
d) u(–f)
Answer: (c)
17
Solution:
− u (t) → 1/(1 + j2πf)
Use the duality property:
If g (t) → G(f)
G (t) → g(– f )
⸫1/(1 + j2πt) → (− )
25. If φ = 2x2y – xz3, then the Laplacian of φ is
a) 4yz – 6xy
b) 4z – 6xy
c) 4y – 6xz
d) 2xy – 6yz
Answer: (c)
Solution:
Laplacian of ∅ = ∇2∅ = 2∅ + 2∅ + 2∅
2 2 2
∅ = 2 2 − 3
2∅ = 4Y
2
2∅ = 0
2
2 ∅2= – 6xyz
⸫ 2∅ + 2∅ + 2∅ = 4Y – 6xz
2 2 2
26. A signal x(n) = sin ( 0n + ∅) is the input to a linear time-
invariant system having a frequency response H ( ). If the
output of the system is AX (n – n0), then the most general form
of ∠H ( ) will be
18
a) − 0 0+ β for any arbitrary real β
b) − 0 0+ 2πk for any arbitrary integer k
c) 0 0+ 2πk for any arbitrary integer k
d) − 0 0 + ∅
Answer: (b)
Solution:
Let x(n) X( )
→
Let the output, y(n) = AX (n – n0)
According to Time shifting property
Y� � = − X( )
The frequency response is given by
H� � = � �
� �
⸫H( ) = A −
�H( )�= A, for A > 0
= – A, for A < 0
∠ � � = − 0 + ∠A
= − 0 + kπ,
K = ±1, ±3, ±5, … for A < 0
= − 0 + 2kπ,
K = 0, ±1, ±2, …… for A > 0
If A > 0, ∠ � � = − 0 + 2πk
For k = 0, ±1, ±2, …….
For x(n) = sin( 0 + ∅), at = 0,
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∠ � 0� = − 0 0 + 2 , for any arbitrary integer k (most
general form).
The answer can also be directly by writing the output,
y(n) = Ax(n – no)for the input, x(n) = sin( 0 + ∅).
Y(n) = A sin[ 0( − 0) + ∅]
= A sin( 0 + ∅ − 0 0)
The added phase angle to the input angle is (− 0 0) or
(− 0 0 + 2 ) for any arbitrary integer, K.
This should be equal to ∠ � � at ω = 0
27. The value of x at which y has minimum for y = x2 – 3x + 1
a) – 3/2
b) 3/2
c) 0
d) None of these
Answer: (b)
Solution:
Y = 2 −3x + 1
= 2x – 3 = 0 ⟹ x = 3
2
2 = 2 > 0
2
So, at x = 32, y has minimum value.
28. (cos 5θ - I sin 5θ) is same as
a) cos 10θ + sin 10θ
b) cos 25θ - I sin 25θ
c) (cos θ + i sin θ)-10
20
d) (cos θ - i sin θ) -10
Answer: (c)
29. Statement (I): Aliasing occurs when the sampling frequency
is less than twice the maximum frequency in the signal.
Statement (II): Aliasing is a reversible process.
a) Both Statement (I) and Statement (II) are individually true
and Statement (II) is the correct explanation of Statement (I).
b) Both Statement (I) and Statement (II) are individually true
but Statement (II) is not the correct explanation of Statement
(I).
c) Statement (I) is true but Statement (II) is false.
d) Statement (I) is false but Statement (II) is true.
Answer: (c)
Solution:
Aliasing is an irreversible process. Once aliasing has occurred
then signal cannot be recovered back.
30. Match List-I (Discrete Time Signal) with List-II
(Transform) and select the correct answer using the codes
given below the lists:
List-I List-II
A. Unit step function 1. 1
B. Unit impulse function
C. sin , t = 0, T, 2T 2. −cos
D. cos , t = 0, T, 2T 2−2 cos +1
3.
−1
4. −sin
2−2 cos +1
21
Codes:
A BCD
a) 2 4 1 3
b) 3 1 4 2
c) 2 1 4 3
d) 3 4 1 2
Answer: (b)
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