CWT–04
A
Booklet No.:
Booklet Series:
03092014
Signal System
Electrical Engineering
Student Name:
Roll Number:
Duration: 90 Minutes
PAPER
MAXIMUM
MARKS: 60
INSTRUCTIONS
1.
IMMEDIATELY AFTER THE COMMENCEMENT OF THE EXAMINATION, YOU SHOULD CHECK
THAT THIS TEST BOOKLET DOES NOT HAVE ANY UNPRINTED OR TORN OR MISSING PAGES
OR ITEMS ETC. IF SO, GET IT REPLACED BY A COMPLETE TEST BOOKLET.
2.
This Test Booklet contains 30 questions. Each question comprises four responses (answers).
You will select the response which you want to mark on the Answer Sheet. In case you feel that
there is more than one correct response, mark the response which you consider the best. In any
case, choose ONLY ONE response for each item.
3.
You have to mark all your response ONLY on the separate Answer Sheet provided.
4.
All items carry equal marks.
5.
Before you proceed to mark in the Answer Sheet the response to various items in the Test
Booklet, you have to fill in some particulars in the Answer Sheet as per instructions.
6.
Each questions 2 marks and 2/3 negative mark is assigned for the wrong answer.
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Signal System
(1.)
The period of signal x t   10sin5t  4cos 7t
(a.) x1 t  and x 2 t 
is
(b.) x 2 t  and x 3 t 
(a.)
12
35
(b.)
4
35
(c.) x 3 t  and x1 t 
(d.) x1 t  and x 3 t 
Ans: (b)
(c.) 2
(5.)
(d.) Not Periodic
(a.) z k z  0
Ans: (c)
(2.)
(b.) z k , z  0
The signal x t   e 4t u t  is a
(c.) z k all z
1
(a.) Power signal with P 
4
(d.) z k , all z
(b.) Power signal with P  0
(c.) Energy signal with E  
Ans: (d)
1
4
(6.)
(a.)  n   2 n  6  4  n  8
Ans: (c)
(b.)  n   2 n  6  4  n  8
Common Data for Questions 3 and 4:
(c.)  n   2 n  6  4 n  8
Consider 3 continuous time periodic signal
(d.)  n   2 n  6  4 n  8
whose Fourier series representation one as
follows
Ans: (a)
(7.)
x1 t  
100
k
1
  3 
e
 2 
j K
t
 50 
x 3 t  
100

cos k e
 2 
j K
t
 50 

signal
corresponding
to

K 0
 2 
j sin
K 100

k  j K  50 t
e
2
(b.)   2
The even signals are

(c.)   2
Exp: For a signal to be even its Fourier
series coefficients must be even this is true
only for n2 t 
 n  2  K  1
 n  2  K  1
(d.) None
Ans: (c)
(c.) x1 t  and x 3 t 
Ans: (a)
2 K 1
K 0
(b.) x 2 t  and x 3 t 
(d.) x1 t  only
2 K 1
K 0
(a.) x 2 t  only
2
time
(a.)   22 K 1  n  2  K  1
K 100
100
The
1
1
is
z 
2
1 4z
4
K 0
x 2 t  
(4.)
The time signal corresponds to
1  2z 6  4z 8 z  0 is
(d.) Energy signal with E   0
(3.)
The Z transform of  n  k , k  0 is
1
(8.)
The time signal corresponding to e z  e z
z  0 is
(a.)
1
n
(b.)  n  
1
n
The real value of signals are
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Signal System

 2 n  k 
(c.)
K 0

 2 n  k 
(d.)
K 0
Ans: (b)
(12.)
g t  can be expressed as
(a.) g t   f  2t  3
Common Data for Question 9 and 10:
Consider a waveform voltage should in
t

(b.) g t   f   3 
2

figure
(c.)
3

g t   f  2t  
2

 t 3
(d.) g t   f   
2 2
Ans: (d)
(9.)
(13.)
The DC component of V is
(a.) 0.4
The Laplace transform of g t  is
(a.)
1 3s
 e  e 5s 
s
(b.)
1 5s
e  e 3s 
s
(c.)
e 3s
1  e 2s 
s
(d.)
1 5s
e  e 3s 
s
(b.) 0.2
(c.) 0.8
(d.) 0.1
Ans: (b)
(10.)
The amplitude of fundamental component of
V is
(a.) 1.2 V
Ans: (c)
(b.) 2.4 V
(14.)
(c.) 2 V
Fourier
continuous
time
signals
for t  0, the contribution z t   x t  .y t  is
Ans: (a)
The
two
x t   e t u t  and y t   e 2t u t  which exit
(d.) 1 V
(11.)
Given
series
for
the
(a.) e t  e 2t
function
f  x   sin2 x
(b.) e 3t
(a.) sin x  sin 2x
(c.) e t
(b.) 1  cos 2x
(d.) e t  e 2t
(c.) sin 2x  cos 2x
Ans: (a)
(d.) 0.5  0.5 cos 2x
(15.)
For
a
periodic
signal V t   30sin100t
Ans: (d)


10cos 300t  6sin  500t  
4

Common Data for Question 12 and 13:
The fundamental frequency in rad/se is
Given f t  and g t  as shown below
(a.) 100
(b.) 300
(c.) 500
3
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Signal System
(d.) 1500
(a.)
Ans: (a)
n
(16.)
n
1
1
If x n        u n  , then the region
3
 2
(b.)
of convergence (ROC) of its 2-transform in
(c.)
the 2-done will be

n 
n 


n 
n 


n 
n 


n 
n 
(d.)
(b.)
1
1
 z 
3
2
Ans: (d)
1
 3 3
2
 x n    x n 
e
which of the following relation must be true
(a.)
Ans: (c)
(b.)
of
e
x 0 n  is the add part of a signal x n , then
1
(d.)
z
3
Which
e
 x n   2  x n   0
1
 3 3
2
(19.)
e
 x n   2  x n 
(a.)
(c.)
(17.)

 x n    x n   0
the
following
represents
y n   2 x  n   4
(c.)


n 
n 



n 
n 
n 
 x n    x n 
 x n    x n   2  x n 

 x n   0
n 
(a.)
(d.)
0
0


n 
n 
 x n    x n 
e
Ans: (c)
(20.)
A differential equation for a LTI system with
specified input and initial conditions is
d 2y t 
dy t 
dx t 
2
 5 y t  
dt 2
dt
dt
y  0   2
(b.)
dy t 
 0, x t   u t 
dt t 0
(a.) 2e t cos t u t 
(b.)
1 t
e sin t u t 
2
(c.) 2 e t cos t u t  
(c.) Opposite of (b)
(d.) Opposite of (a)
(d.)
1 t
e sin t u t 
2
1 t
e cos t u t  1  2 e t sin t u t  1
2
Ans: (c)
Ans: (b)
(18.)
If a signal and its even part is x n  and
Common Data for Questions 21, 22 & 23:
xe n  respectively then which one of the
Consider
following relation must be true.
4
the
cos 2t u t  LT
transform
pair
X s 
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
given
Signal System
(21.)
The time signal corresponding to X s  2 is
(b.)
a
1  a e  j
(c.)
1
1  a e  j
(a.) cos 2 t  2 u t 
(b.) e 2t cos 2t u t 
1
cos 6t u t 
3
(c.)
(d.) None
Ans: (a)
(d.) cos 2 t  2 u t 
Ans: (b)
(22.)
Common Data for
The time signal corresponding to
X s 
is
s2
Questions 26, 27, 28 & 29:
Consider the signal X  j  shown below
(a.) 4cos 2t u t 
whose inverse Fourier transform is x t 
1  cos 2t
(b.)
u t 
4
(c.) t 2 cos 2t u t 
cos 2t
u t 
t2
(d.)
Ans: (b)
(23.)
The

time
signal
corresponding
to
(26.)
(b.) 2 
(b.) t cos 2 t  3 u t 
(c.) t cos 2 t  3 u t  3
(d.) 
Ans: (a)
(d.) t cos 2 t  3 u t 

(27.)
Ans: (c)
Fourier
transform
of
j  2
is
X  j  
2
  5 j  4
(a.)
16
3
(a.)
 2 e 4t  e t  u t 
(b.)
16
3
(b.)
 2e
4 t
(c.)
32
3
(c.)
 2e
t
 e t  u t 
 2e 4t  u t 
The

Fourier
 a  t  m  , a
m
m 0
(a.)
1
1  a e  j
 x t 
2
dt is
(d.) None
Ans: (b)

Ans: (b)
5
The value of

(d.) None
(25.)
1
2
(c.)
inverse
is
(a.) 1
(a.) t cos 2 t  3 u t  3
The
 x t  dt

d
e 3s X s  is
ds 
(24.)
The value of
(28.)
transform
of
Determine the value
 x t  e
3 j t
dt

signal
(a.) 2
 1 is
(b.) 3
(c.) 0
(d.) 1
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Signal System
Ans: (a)
(29.)
The impulse response of this system is
The value of x  0 is
(a.)
(a.)
4

n
n
1   1
1 
(b.)
5     2    u n 
3   2 
 4  
(b.) 8
(c.) 0
(c.)
Ans: (a)
A causal system has input
x n    n  
1
1
 n  1   n  2
4
8
Output y n    n  
6
1  1
 1
5    2   
3   2 
 4
n
(d.) 4
(30.)
n
n
1  1
1 
5

2
  
   u n 
3   2 
 4  
n

 u n 

(d.) None
Ans: (b)
3
 n  1
4
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P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Signal System
Answer Key
(1.)
(9.)
(17.)
(25.)
(2.)
(10.)
(18.)
(26.)
(3.)
(11.)
(19.)
(27.)
(4.)
(12.)
(20.)
(28.)
(5.)
(13.)
(21.)
(29.)
(6.)
(14.)
(22.)
(30.)
(7.)
(15.)
(23.)
(8.)
(16.)
(24.)
7
(b)
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m