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General Aptitude
Q. No. 1 – 5 Carry One Mark Each
1.
What is the adverb for the given word below?
Misogynous
(A) Misogynousness
(B) Misogynity
(C) Misogynously
(D) Misogynous
Answer: (C)
2.
Choose the appropriate word–phrase out of the four options given below, to complete the
following sentence
Dhoni, as well as the other ream members of Indian team _______________ present on the
occasion
(A) Were
(B) Was
(C) Has
(D) Have
Answer: (A)
3.
Ram and Ramesh appeared in an interview for two vacancies in the same department. The
probability of Ram’s selection is 1/6 and that of Ramesh is 1/8. What is the probability that
only one of them will be selected?
(A) 47/48
(B) 1/4
(C) 13/48
(D) 35/48
Answer:
(B)
Exp: P  Ram   1 ; p  Ramesh   1
6
8
7 1 5
p  only at   p  Ram   p  not ramesh   p  Ramesh   p  n 0  R am   1   
6 8 8 6
12 1


4
40
4.
Choose he word most similar in meaning to the given word:
Awkward
(A) Inept
(B) Graceful
(C) Suitable
Answer: (A)
5.
(D) Dreadful
An electric bus has onboard instruments that report the total electricity consumed since the
start of the trip as well as the total distance covered. During a single day of operation, the bus
travels on stretches M, N, O and P, in that order. The cumulative distances traveled and the
corresponding electricity consumption are shown in the Table below
Stretch
Cumulative distance (km)
Electricity used (kWh)
M
20
12
N
45
25
O
75
45
P
100
57
The stretch where the electricity consumption per km is minimum is
(A) M
(B) N
(C) O
(D) P
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Answer:
Exp:
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(B)
For M  12
N  25
O  45
P  57
20
45
75
6
 555
6
200
 57
Q. No. 6 – 10 Carry Two Marks Each
6.
Given below are two statements followed by two conclusions. Assuming these statements to
be true, decide which one logically follows.
Statements:
I.
All film stars are playback singers.
II. All film directors are film stars
Conclusions:
I.
All film directors are playback singers.
II. Some film stars are film directors.
(A) Only conclusion I follows
(C) Neither conclusion I nor II follows
Answer:
7.
(B) Only conclusion II follows
(D) Both conclusions I and II follow
(D)
Lamenting the gradual sidelining of the arts ill school curricula, a group of prominent artists
wrote to the Chief Minister last year, asking him to allocate rnore funds to support arts
education in schools. However, no such increase has been announced in this year’s Budget.
The artists expressed their deep anguish at their request not being approved, but many of them
remain optimistic about funding in the future
Which of the staternent(s) below is/are logically valid and can be inferred from the above
statements?
(i) The artists expected funding for the arts to increase this year
(ii) The Chief Minister was receptive to the idea of increasing funding for the arts
(iii) The Chief Minister is a prominent artist
(iv) Schools are giving less importance to arts education nowadays
(A) (iii) and (iv)
Answer:
8.
(B) (i) and (iv)
(C) (i), (ii) and (iv)
(D) (i) and (iii)
(C)
A tiger is 50 leaps of its own behind a deer. The tiger takes 5 leaps per minute to the deer’s 4.
If the tiger and the deer cover 8 metre and 5 metre per leap respectively. What distance in
metres will be tiger have to run before it catches the deer?
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Answer: 800
Exp: Tiger  1 leap  8 meter
Speed  5leap hr  40m min
Deer  1 leap  5meter
speed  4hr  20m min
Let at time ‘t’ the tiger catches the deer.
 Distance travelled by deer + initial distance between them
50  8  400m = distance covered by tiger.
 40  t  400  20t
400
t
 20 min
200
 total dis tance  400  20  t  800 ms
9.
If a 2  b2  c2  1, then ab  bc  ac lies in the interval
(A) [1, 2/3]
Answer: (B)
(B) [–1/2, 1]
(C) [–1, 1/2]
(D) [2, –4]
10.
In the following sentence certain parts are underlined and marked P, Q and R. One of the
parts may contain certain error or may not be acceptable in standard written communication.
Select the part containing an error. Choose D as your answer if there is no error.
The student corrected all the errors that the instructor marked on the answer book
P
Q
R
(A) P
(B) Q
(C) R
(D) No error
Answer: (B)
Exp: The is not required in ‘Q’
Electronics and Communication Engineering
Q. No. 1 – 25 Carry One Mark Each
1.
Let the signal f  t   0 outside the interval T1 ,T2 , where T1 and T2 are finite. Furthermore,
f  t    . The region of convergence (RoC) of the signal’s bilateral Laplace transform F(s)
is
1.
A parallel strip containing the j axis
2.
A parallel strip not containing the j axis
3.
4.
The entire s–plane
A half plane containing the j axis
(A) 1
Answer:
Exp:
(B) 2
(C) 3
(D) 4
(C)
For a finite duration time domain signal ROC is entire s-plane.
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K
. The
s  s  10 
gain K for the system to have a damping ratio of 0.25 is ____________________
Answer: 400
k
Exp: G(s)  2
s  10s
k
CLTF  2
s  10s  k
  0.25
2.
A unity negative feedback system has an open–loop transfer function G  s  
k  2n
n  k

10
2 k
10
k
0.5
k  (20) 2
 0.25
k  400
3.
A mod–n counter using a synchronous binary up–counter with synchronous clear input is
shown in the figure. The value of n is _____________________
4  Bit Binary
Counter
CLOCK
CLK
QA
QB
QC
QD
QA
QB
QC
QD
CLEAR
Answer: 6
Exp: To find the modulus of the counter, consider the status of the inputs (QB, QC) as 1.
So, QA QBQCQD  0110
So, it is a MOD-6 counter
4.
By performing cascading and/or summing/differencing operations using transfer function
blocks G1  s  and G 2  s  , one CANNOT realize a transfer function of the form
G1  s 
(A) G1  s  G 2  s 
(B)
 1

 G 2  s  
(C) G1  s  
 G1  s 

 1

 G 2  s  
(D) G1  s  
 G1  s 

Answer:
G2 s
(B)
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5.
The electric field of a uniform plane electromagnetic wave is



E   a x  j4a y  exp  j  2107 t  0.2z 
The polarization of the wave is
(A) Right handed circular
(B) Right handed elliptical
(C) Left handed circular
(D) Left handed elliptical
Answer: (D)
Exp:


E  a x  4 ja y e

j 2 107 t  0.2z

  2  107
  0.2
E z  cos t
E y  4cos     2   4sin t
So, it left hand elliptical polrization
6.
A piece of silicon is doped uniformly with phosphorous with a doping concentration of
1016 / cm3 . The expected value of mobility versus doping concentration for silicon assuming
full dopant ionization is shown below. The charge of an electron is 1.6 1019 C. The
conductivity  in S cm1  of the silicon sample at 300 K is _________________
Hole and Electron Mobility in Silicon at 300 K
1400
Electron
Mobility (cm2.V-1. s-1)
1200
Hole
1000
600
800
400
200
0
1.E  13 1.E  14 1.E  15
Answer:
Exp:
1.E  16
1.E  17 1.E  18

Doping concentration cm3
1.92
1.E  19 1.E  20

As per the graph mobility of electrons at the concentration 1016/cm3 is 1200
So,  n  1200
cm 2
Vs
cm2
Vs
 N  N D q n
 1016  1.6  1019  1200
 1.92 S cm 1
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In the figure shown, the output Y is required to be Y  AB  CD . The gates G1 and G2 must
be, respectively,
A
G1
B
G2
Y
C
D
(A) NOR, OR
Answer:
Exp:
(B) OR, NAND
(C) NAND, OR
(D) AND, NAND
(A)
Given expression is Y  AB  CD
The first term can be obtained by considering G1 as NOR gate, and second term  CD  is
obtained from another lower NOR-Gate. So, final expression can be implemented by
considering G2 as OR-Gate.
8.
In the bistable circuit shown, the ideal opamp has saturation level of  5V . The value of
R1  in k  that gives a hysteresis width of 500 mV is ______________.
R 2  20k
R1

vout
vin
Answer:



1
Exp:
V0
L
VI
VTL
VTH
L
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Hysteresis  VTH  VTL
R 
R 
 L  1   L  1 
 R2 
 R2 
 R   R 
500mV  ( 5)  1   5  1 
 20k   20k 
R
 1
2k
 R1  500  2  103  103
 1000   1k
9.
Two causal discrete–time signal x  n  and y  n  are related as y n    m0 x m
n
2
It the z–transform of y  n  is
Answer:
z  z  1
2
, the value of x  2 is __________
0
Exp: y  n  
n
 x  m
m0
According to accumulation property of z-transform
Y z 
X z
1z 
1
 X z 

z X z
2

z  z 1
 z 1
2z 2
2z 3

 z 1 1 z 1 
 x  n   2u  n 3 thus x  20
10.
1 if a  t  b
The bilateral Laplace transform of a function f  t   
0 otherwise
(A)
Answer:
Exp:
ab
s
(C)
(B)
1
0
Given f  t  
L f  t  


a

0

0
e2  a  b 
s
(C)
e as  e bs
s
(D)
es  a  b 
s
atb
otherwise
est f  t  dt
est f  t  


a
est f  t  dt 


b
e st f  t  dt
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
b
a
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e st dt  0
e st b 1  bs
 e  e  as 


s a
s

11.
e  as e  bs
s
The 2–port admittance matrix of the circuit shown is given by
0.3
(A) 
0.2
3.33
(C) 
 5
0.2 
0.3
15
(B) 
5
5 
3.33
5
5
15
10 
0.3 0.4 
(D) 

0.4 0.3 
10 
Answer:
Exp: Correct answer not given in options
 0.3  0.2
  0.2 0.3 


12.
The value of x for which all the eigen–values of the matrix given below are real is
10
x

 4
5 j 4 
20
2 
2
 10 
(B) 5 – j
(A) 5 + j
Answer:
Exp:
(C) 1 – 5j
(D) 1+ 5 j
(B)
10 5  J 4 

20
2 
Let A  x

 4
2 10 
Given that all eigen values of A are real.
 A is Hermitian
 
 A   A ie. A
T
A
4  10 5 j 4 
 10 x
5 j 20 2    x 20
2   x  5 j

 
 4
2 10   4
2 10 
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

The signal cos 10t   is ideally sampled at a sampling frequency of 15 Hz. The sampled
4

 sin  t  


signal is passed through a filter with impulse response 
 cos  40t   . The filter
2

 t 
output is
15  sin  t  

15



(A)
(B)
cos  40t  

 cos 10t  
2
4
2  t 
4


15  sin  t  

15



(C)
(D)
cos 10t  

 cos  40t  
2
4
2  t 
2


Answer:
(A)


Exp: Given signal is x  t  cos 10t 


4

and it can be inserted at the end result.
4
1
L
 If x1  t  cos 10t 
 X1  f     f 5     f 5  
2
Neglect the phase-shift
Given filter impulse response is,

 sin t 

h t 
 cos  40t  
2
 t 

  sin c t  sin  40t 
 H  f   rect f *

1
  f  20     f  20  
2j 
1
 rect  f  20   rect  f  20  
2j 
X1  f  repeats with a value f o 15Hz and each impulse value is
15
2
Thus the sampled signal spectrum and the spectrum of the filter are as follows:
X s f 
20 15 10 5
5
15 / 2
10
15 20
f  H(f 
H f 
20
20
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 Xs  f  H  f  
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15
  f  20     f  20  
4j 
15
sin  40 t   recovered signal
2
15


=
cos  40 t  
2
2


Insert the neglected phase shift
4
15
   15



x r  t   cos  40 t     cos  40 t  
2
2 4 2
4


xr  t  
14.
A sinusoidal signal of amplitude A is quantized by a uniform quantizer Assume that the signal
utilizes all the representation levels of the quantizer. If the signal to quantization noise ratio is
31.8 dB, the number of levels in the quantizer is ____________________
Answer: 32
Exp: Signal power  A 2 2
2A
Quantization step size,  
L
2
Quantization noise power 
12
4A 2 A 2


12L2 3L2
3
 Signal to quantiation noise ratio  L2
2
Given signal to quantization noise ratio  31.8dB or 1513.56
3
 L2  1513.56
2
 L  31.76
 32
15.
3
3
ak
The magnitude and phase of the complex Fourier
series coefficients a k of a periodic signal x(t) are
shown in the figure. Choose the correct statement
form the four chices given. Notation: C is the set of
complex numbers, R is the set of purely real numbers,
and P is the set purely imaginary numbers.
(A) x  t   R
5 4 3 2 1
0 1
ak
(B) x  t   P
5 4 3 2 1
(C) x  t    C  R 
(D) The information given is not sufficient to draw
2
2
k
2 3 4
1 2 3 4
k
0

any conclusion about x(t)
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Answer: (A)
Exp: a k   only changes the sign of the magnitude a k . Since the magnitude spectrum a k
is even the corresponding time-domain signal is real.
16.
dy 1  cos 2y
is

dx 1  cos 2x
(B) tan x  cot y  c  c is a cons tan t 
The general solution of the differential equation
(A) tan y  cot x  c  c is a cons tan t 
(C) tan y  cot x  c  c is a cons tan t 
Answer: (D)
Exp:
(D) tan x  coty  c  c is a cons tan t 
dy 1cos 2y

dx 1  cos 2x
dy
dx
Variable - Separable


1 cos 2y 1  cos 2x
dy
dx


2
2sin y 2 cos 2 x
Given
  cos ec 2 ydy   sec 2 x dx
 cot y  tan x  k
  tan x  cot y  k
 tan x  cot y  c where c   k
17.
An n–type silicon sample is uniformly illuminated with light which generates 1020 electron
hole pairs per cm3 per second. The minority carrier lifetime in the sample is 1s . Int eh
steady state, the hole concentration in the sample is approximately 10x , where x is an integer.
The value of x is __________________
Answer: 14
Exp: The concentration of hole-electron pair in 1 sec  1020 106  1014 cm3
So, the power of 10 is 14.
x  14
18.
If the circuit shown has to function as a clamping circuit, which one of the following
conditions should be satisfied for sinusoidal signal of period T?
C


v
(A) RC<<T
Answer: (D)
(B) RC = 0.35T
R
(C) RC  T
(D) RC >> T
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19.
In a source free region in vacuum, if the electrostatic potential   2x 2  y2  cz2 , the value of
constant c must be __________________________
Answer:
Exp:
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3
  2x 2  y 2  cz 2
E    4xa x  2ya y  2cza z
.E  0
4  2  2c  0
c3
20.
In an 8085 microprocessor, which one of the following instructions changes the content of the
accumulator?
(A) MOV B, M
Answer:
Exp:
(B) PCHL
(C) RNZ
(D) SBI BEH
(D)
Generally arithmetic or logical instructions update the data of accumulator and flags. So, in
the given option only SBT BE H is arithmetic instruction.
SBI BE H  Add the content of accumulator with immediate data BE H and store the result
in accumulator.
21.
The voltage  VC  across the capacitor (in Volts) in the network shown is _____________
80V
40V
Vc
100V, 50 Hz
Answer:
Exp:
100
V  VR2   Vc  VL 
2
(100) 2  (80) 2   Vc  40 
2
 Vc  40   (180)(20)
 Vc  40    2  90  20
2
Vc  40  60
Vc  60  40
Vc  60  40
Vc  100V
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22.
az  b
. If f  z1   f  z 2  for all z1  z 2 ,a  2,b  4 and c  5, then d should be
cz  d
equal to __________.
Let f  z  
Answer:
Exp:
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10
f z 
az  b
if f  z1   f  z 2  , for z1  z 2
cz  d
a  2, b  4, c 5
f z 
2z  4
5z  d
f  z1   f  z 2  
2z1  4 2z 2  4

5z1  d 5z 2  d
 10 z1 z 2  20 z 2  2dz1  4d 10 z1 z 2  20 z1  2dz 2  4d
 20  z 2  z1   2d  z 2  z1 
 d 10
23.
In the circuit shown the average value of the voltage Vab (in Volts) in steady state condition
is _____________.
1k
5 sin  5000t 
b
1F

a
1mH
2k

Vab
5V
Answer:
24.
5
For the signal flow graph shown in the figure, the value of
C s
R s
is
H3
1 X1
R s 
C s 
X3 1 X 4
G1 X 2 G 2
H1
G 3 X5 G 4
H 2
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(A)
G1G 2 G 3G 4
1  G1G 2 H1  G 3G 4 H 2  G 2G 3H3  G1G 2G 3G 4 H1H 2
(B)
G1G 2 G 3G 4
1  G1G 2 H1  G 3G 4 H 2  G 2 G 3H3  G1G 2G 3G 4 H1H 2
(C)
1
1  G1G 2 H1  G 3G 4 H 2  G 2 G 3H3  G1G 2G 3G 4 H1H 2
(D)
1
1  G1G 2 H1  G 3G 4 H 2  G 2G 3H3  G1G 2G 3G 4 H1H 2
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Answer: (B)
Exp: Using mason gain formula we get it directly.
25.
In the circuit shown, V0  V0A for switch SW in position A and V0  V0B for SW in position
V
B. Assume that the opamp is ideal. The value of 0B is ____________________
V0A
1 k
1 k
5V

B
V0
A 1 k
SW

1 k
1 k
1V
Answer:
Exp:
1.5
 1   1k 
VOB  5 
  1

 1k   1k 
 6V
 1k  1  1k 
VOA  5 
  1 
  4V
 1k  2  1k 

1 

 V  V 
2 

VOB
 1.5
VOA
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Q. No. 26 – 55 carry Two Marks Each
26.
Let X 0,1 and Y 0,1 be two independent binary random variables. If P (X = 0) = p and
P(Y = 0) = q, then P  X  Y  1 is equal to
(A) pq  1  p 1  q 
(B) pq
(C) p 1  q 
(D) 1  pq
Answer:
Exp:
(D)
P x  0  P  P x  1  1  p
P y  0  q  P y  1  1  q
Let Z  X  Y
X Y Z
0 0 0
0 1 1
1 0 1
1 1 2
From above table,
P X  Y  Z  P  Z  B
P Z  1  P X  0 and Y  1  P X  1 and Y  0  P X  1 and Y  1
 1  P X  0 and Y  0
 1  pq
27.
An LC tank circuit consists of an ideal capacitor C connected in parallel with a coil of
inductance L having an internal resistance R. The resonant frequency of the tank circuit is
(A)
(C)
Answer:
Exp:
1
2 LC
1
2 LC
1
L
R 2C
(B)
1
C
1 R2
L
2 LC
(D)
1
C
1 R2
L
2 LC
(D)
C
Y  Yc  YLR
Y  jC 
 R  jL 
1
 jC  2
 jL  R 
 R  2 L2 
Placing Imaginary part to zero we get option (D).
L
R
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28.
Xn n 
n 
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is an independent and identically distributed (i,i,d,) random process with X n
1 or  1.Yn n 
n 
equally likely to be
is another random process obtained as
Yn  Xn  0.5Xn 1 . The autocorrelation function of Yn n  denoted by R Y  k  is
n 
R Y k 
(A)
R Y k 
(B)
1
0.5
k
3 2 1 0 1 2 3
R Y k 
(C)
 

0.5
0.25
(D)
2 1 0 1 2
Answer:
Exp:
R Y k 
0.5
 
0.5
3 2 1 0 1 2 3
1.25
0.25
1.25

k
1.25
0.25 0.25
k
3 2 1 0 1 2 3
k
(B)
R Y  k   R y  n, n  k 
 E  Y  n  .Y  n  k  
Y  n   x  n   0.5x  n  1
R y  k   E  x  n   0.5x  n  1  x  n  k   0.5x  n  k  1  
 E  x  n  .x  n  k   x  n  0.5 x  n  k  1  0.5 x  n  1 .x  n  k 
 0.25x  n  1 x  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  
 R X  k   0.5R x  k  1  0.5R x  k  1  0.25R x  k 
R y  k   1.25R x  k   0.5R x  k  1  0.5R x  k  1
R x  k   E  x  n  .x  n  k  
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if k  0,
R x  0   E  x 2  n  
1
1
2
 12.   1 
2
2
1
if k  0,
R x  k   E  x  n   .E  x  n  k  
0

 E  x  n    0



 E  x  n  k    0 



R y  0   1.25R x  0   0.5R x  1  0.5R x 1
 1.25
R y 1  1.25R x 1  0.5R x  0   0.5R x  2 
 0.5
R y  1  1.25R x  1  0.5R x  2   0.5R x  0 
 0.5
R y  k  for k other than 0, 1 and – 1 = 0
 Ry k
1.25
0.5
1
0.5
1
k
29.
In a MOS capacitor with an oxide layer thickness of 10 nm, the maximum depletion layer
thickness is 100 nm. The permittivities of the semiconductor and the oxide layer are s and ox
respectively. Assuming s / ox  3, the ratio of the maximum capacitance to the minimum
capacitance of this MOS capacitor is _____________________
Answer: 4.33
Exp:
ox
t ox
 X
Cmax
 
 100 1 

 1  d max  ox   1 
  4.33


Cmin
t ox
s 
10 3 

ox
s


t ox X d max
ox
s

t ox X d max
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30.
Let the random variable X represent the number of times a fair coin needs to be tossed till two
consecutive heads appear for the first time. The expectation of X is _____________________
Answer: 1.5
Exp: Let x be a random variable which denotes number of tosses to get two heads.
1 1

2 2
1 1 1
P  x 3  THH   
2 2 2
1 1 1 1
P  x  4   TTHH    
2 2 2 2
.........................................................
P  x  2   HH 
1 1
1 1 1
1 1 1 1
E  x   2     3       4       ..........
2 2
2 2 2
2 2 2 2
1
1
1
 2  2  3  3  4  4  ............
2
2
2
1  1
1
1

  2.  3. 2  4 3  .............
2  2
2
2


1 
1
1
 
1  2.  3 2  ......  1

2 
2
2
 
2

1  1 
1
3
 1  1   41 
2  2 
2
2

31.
In the circuit shown, the Norton equivalent resistance
______________.
2
a


2
4I
 in  
across terminals a–b is
4
I
b
Answer: 1.333
Exp: Nodal @ ‘a’
Vth  4I Vth Vth


1  0
2
2
4
Also Vth  4I
 Vth 
4
3
4
1
R th   1  1.333
3
3
Vth
a

2


4I
2
Vth
4
1A
I

b
R th 
Vth
1
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32.
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The figure shows a binary counter with synchronous clear input. With the decoding logic
shown, the counter works as a
Binary
Counter Q
3
Q2
CLK
Q1
Q0
CLEAR
(A) mod–2 counter
Answer:
Exp:
(B) mod–4 counter
(C) mod–5 counter
(D) mod–6 counter
(B)
Q3
0
0
0
0
0
Q2
0
0
0
0
1
Q1 Q0
0
0
0
1
1
0
1 1
0
0
Once the output of Ex-NOR gate is 0 then counter will be RESET. So, Ex-NOR-gate will
produce logic 0 for Q3 = 0, Q2 = 1. So, the counter will show the sequence like:
0 1  2  3
So, it is MOD-4 counter.
33.
In the ac equivalent circuit shown, the two BJTs are bbiased in active region and have
identical parameters with   . The open circuit small signal voltage gain is approximately
________________________
vout
vin
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Answer:
Exp:
www.gateforum.com
-1
Vout
Vin
When Base and collector is shorted, it act as a diode.
So V0  0.7V.
Gain ,
34.
V0
0.7V

 1
Vin
0.7V
The state variable representation of a system is given as
0 1 
1 
x  
x; x  0    

0  1
0
y  0
1 x
The response y(t) is
(B) 1  et
(A) sin(t)
(C) 1  cos  t 
(D) 0
Answer: (D)
  AX
Exp: X
X(s)   sI  A  X(0)
1
1
 s 1  1 
X(s)  
  
0 s  1 0 
1 
X(s)   s 
 0 
1 
xt   
0
1 
y(t)  [0 1]    0
0
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35.
Consider the differential equation
response x  t  for t  0 is
(A) 2  e0.2t
Answer:
Exp:
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dx
 10  0.2x with initial condition x  0   1 . The
dt
(C) 50  49e0.2t
(B) 2  e0.2t
(D) 50  49e0.2t
(C)
dx
100.2x
x  0  1
dt
dx

  0.2  x 10
dt
Auxiliary equation is m  0.2  0
m  0.2
Given D.E
 0.2  t
Complementary solution x c  C e
xp 
1
10 e0t
10 et 
D   0.2 
0.2
 50 e0t  50
x  x c  x p  C e 0.2 t  50
Given x  0  1  C  501  C 49
x  50 49 e
36.
0.2  t
For the voltage regulator circuit shown, the input voltage
 Vin  is 20V  20%
and the
regulated output volage  Vout  is 10V. Assume the opamp to be ideal. For a load R L drawing
200 mA, the maximum power dissipation in Q1 (in Watts) is _____________.
Answer:
Exp:
2.8056
PQ1 (max)  VCE(max)  Icmax
Q1
Vin
VCE(max)   24  10  V  14V
VVrefref
4V

R1
RL

Ic max  (200  0.4)mA
IE  Ic  200mA  0.4mA
 200.4mA
Vout
….(i)
40


mA 
 I R 2  I R1 
10


R 2 10k
Put values in Equation (1), we get
PQ1 (max)  14  200.4  103 Watt
 2.8056 Watt
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37.
Input x(t) and output y(t) of an LTI system are related by the differential equation
y" t   y'  t   6y  t   x  t  . If the system is neither causal nor stable, the impulse response
h(t) of the system is
(A)
1 3t
1
e u  t   e2t u  t 
5
5
1
1
(B)  e3t u   t   e2t u  t 
5
5
(C)
1 3t
1
e u   t   e2t u  t 
5
5
1
1
(D)  e3t u   t   e2t u  t 
5
5
Answer:
Exp:
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(B)
The given differential equation is,
y" t   y '  t   6y  t   x  t 
On applying Laplace transform on both sides,
s2 y  s   sy  0   y  0   s y  s   y  0   6y s   x s 
To calculate the transfer function all initial conditions are taken as ‘0’.
  s 2 s  6  y  s   x  s 
H s 
1
1
1 1
1 




 s2 s6   s3 s  2  5  s3 s  2 
It is given that h(t) is non-casual and un-stable.
To satisfy both the conditions ROC should be left of the left most pole.
Using the following standard pair
1

  e at u   t  ;   a
sa
1

  eat u   t  ;   a
s a
1 1
1 


5  s 3 s  2 
1
  e3t u   t   e 2t u   t  
5
1
1
 e 3t u   t   e 2t u   t 
5
5
H s 
So option (B) is correct.
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38.
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The diode in the circuit given below has VON  0.7V but is ideal otherwise. The current (in
mA) in the 4k resistor is __________.
3k
2 k
D 1k
1mA
6k
4k
Answer:
39.
0.6
N0
is passed through an
2
LTI filter whose impulse response h(t) is shown in the figure. The variance of the filtered
noise at t = 4 is
A zero mean white Gaussian noise having power spectral density
ht
A
0
2
1
3
t
A 
3 2
3
(B) A 2 N 0
A N0
2
4
Answer: (A)
Exp: Let N(t) be the noise at the output of filter.
(C) A 2 N0
(A)


Variance of N  t   E N2  t   E  N  1 
(D)
1 2
A N0
2
2
Since the input noise is zero mean,
Output noise mean is also zero.


E  N  t    E  W  t   .  h  t  dt 


 

E  W  t   0

W  t  is white noise

 var  N  t    E N 2  t 

 R N  0
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Since R N     h    * h    * R    
Since R N    
No
.   
2
R N      h    * h     .
R N   
R N  0 
No
2
No
2
No
2

 h  k .h    k  dk


N
 h  k  dk  2 3A 
2
o
2

3
 .A 2 .N o
2
40.
Assuming that the opamp in the circuit shown below is ideal, the output voltage V0 (in volts)
is ___________________
2k
12V
1k

Vo

12V
1V
Answer:
3
Exp:
2k
12V
1k

Vo

12V
1V
V0  1 1  0

2k
1k
V0  3V
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41.
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A 1–to–8 demultiplexer with data input Din , address inputs S0 ,S1 ,S2 (with S0 as the LSB)
and Y0 to Y7 as the eight demultiplexed output, is to be designed using two 2–to–4 decoders
(with enable input E and address input A0 and A1 ) as shown in the figure Din , S0 ,S1 and S2
are to be connected to P, Q, R and S, but not necessarily in this order. The respective input
connections to P, Q, R and S terminals should be
P
Q
R
S
1E
1Yo
2to4 1Y
1
Decoder
1A 0
1Y2
1A1
1Y3
Yo
Y1
Y2
Y3
2Yo
2to4 2Y
1
Decoder
2A 0
2Y2
2A1 7
2Y3
Y4
Y5
Y6
Y7

(A) S2 ,Din ,S0 ,S1
Answer:
Exp:
(B) S1 ,Din ,S0 ,S2
(C) Din ,S0 ,S1 ,S2
(D) Din ,S2 ,S0 ,S1
(D)
We need to implement 1 : 8 DEMUX
Y0
Select lines of DEMUX should be mapped
to address lines of decoder. So, LSB of
DEMUX should be connected to LSB of
address lines of decoder.
R  S0
S  S1
Y1
Din
Input to both the decoder should be same so
P  Din
NOT gate along with OR gate in case to
select one decoder at a time so Q  Sa .
Y7
S2
S1
S0
P  Din
Q  S2
R  S0
S  S1
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42.
The value of the integral
Answer:
Exp:



12cos  2t 
sin  4t 
4t
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dt is ______________________
3



12 cos 2t
sin 4  t
dt
4 t
12  2 cos 2t sin 4t
dt
4 0
t
 sin 2 t dt 
3   sin 6t dt

0 t   sin A cos B  sin  A  B  sin  A B  
  0
t

3 
6sin 6t
sin 2t 
   et
dt   et
dt 
0
0

t
t

3   sin 6t 
 sin 2t 
 L 
  L
 with s 0
   t 
 t 

3   6
ds 
  s s 2  362


s

2
ds  with s 0
2
s  4

2
3
1
1
 s 
 s 
  6.
tan 1    2.
tan 1  
 
6
2
 6 
 2  s


 with s  0

 1
1  s 
1
1  s  
 tan   tan  6   tan     tan  2  
 
 

3 


   tan 1 0   tan 10 
 2
2


3


43.
3 
 
3
0  0      3

 2
2 

A function of Boolean variables X, Y and Z is expressed in terms of the min–terms as
F  X,Y, Z   1,2,5,6,7 
Which one of the product of sums given below is equal to the function F  X,Y, Z  ?
(A)  X  Y  Z. X  Y  Z. X  Y  Z 
(B)
(C)
(D)
 X  Y  Z . X  Y  Z . X  Y  Z 
 X  Y  Z  .  X  Y  Z  .  X  Y  Z .  X  Y  Z .  X  Y  Z 
 X  Y  Z  .  X  Y  Z  .  X  Y  Z .  X  Y  Z .  X  Y  Z 
Answer: (B)
Exp: Given minterm is : F  X,Y, Z    1,2,5,6,7 
So, maxterm is :F  X,Y, Z   M  0,3, 4 
POS   X  Y  Z   X  Y  Z  X  Y  Z 
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44.
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The transfer function of a mass–spring damper system is given by
1
G s 
2
Ms  Bs  K
The frequency response data for the system are given in the following table.
 in rad/s
G  j in dB
arg  G  j  in deg
–18.5
–18.5
–18.4
–16
–11.4
–21.5
–32.8
–45.3
0.01
0.1
0.2
1
2
3
5
10
–0.2
–1.3
–2.6
–16.9
–89.4
–151
–167
–174.5
The unit step response of the system approaches a steady state value of ________________
Answer: 0.12
45.
Two half–wave dipole antennas placed as shown in the figure are excited with sinusoidally
varying currents of frequency 3 MHz and phase shift of  / 2 between them (the element at
the origin leads in phase). If the maximum radiated E–field at the point P in the x–y plane
occurs at an azimuthal angle of 60o. the distance d (in meters) between the antennas is
_______________.
z
O
y
d
60
Answer: 50
Exp:  = δ+βdcosθ
OP d
x
For maximum field,  = 0
3  108
3  106
 100 m

δ+βdcosθ = 0
 2
 
d cos 60  0
2 
 2
1

d
2 100
2
d  50m
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46.
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An air–filled rectangular waveguide of internal dimensions a cm  bcm  a  b  has a cutoff
frequency of 6 GHz for the dominant TE10 mode. For the same waveguide, if the cutoff
frequency of the TM11 mode is 15 GHz, the cutoff frequency of the TE 01 mode in GHz is
_____________.
Answer:
13.7
Exp:
TE10
TE 01
f c  6GHz
2
m n
fc 
   
2   a   b 
1
a
2
3  108 1
.
2
b
f c  13.7 GHz
fc 
1
40
TM11
15  109 

47.
3  108
2
1
1
 2
2
a
b
1
 91.65
b
Consider two real sequences with time–origin marked by the bold value
x1  n   1,2,3,0 , x 2 n   1,3,2,1
Let X1  k  and X2  k  be 4–point DFTs of x1  n  and x 2  n  , respectively
Another sequence x 3  n  is derived by taking 4–point inverse DFT of x 3  k   x1  k  x 2  k  .
The value of x 3  2 is __________
Answer:
Exp:
11
x1  n   1, 2,3, 0 , x 2  n   1,3, 2,1
X3  k   X1  k  X 2  k 
Based on the properties of DFT,
x1  n   x 2  n   X1  k  X2  k   x 3 n 
Circular convolution between two 4-point signals is as follows:
1 0 3 2 
2 1 0 3


3 2 1 0


0 3 2 1 
 x 3  2 11
1   9 
3  8 
   
 2  11
   
1  14 
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Let x  t   a s  t   s  t  with s  t   e4t u  t  , where u  t  is unit step function. If the
bilateral Lapalce transform of x(t) is
16
 4  Re s  4
s  16
Then the value of  is ___________________________
X s 
Answer:
Exp:
2
-2
x  t   s  t   s   t  & s  t   e 4t u  t 
x  t   e 4t u  t   e 4t u   t 

s4

L
e 4t u   t  

s4
L
e 4t u  t  

B


s 4 s4
   s4  s 4 
16

;  4   4
 2
2
s 16

 s 16
 X s 
On solving the numerator
2
49.
Consider a binary, digital communication system which used pulses g  t  and  g  t  for
transmitting bits over an AWGN channel. If the receiver uses a matched filter, which one of
the following pulses will give the minimum probability of bit error?
(A) gt
(B)
1
1
0
(C)
gt
1
0
t
gt
(D)
1
t
1
t
gt
1
1
0
1
t
0
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Answer:
Exp:
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(A)
Optimum receiver for AWGN channel is given by matched filter.
In case of matched filter receiver,
 2E 
Probability of error  Q 
 N 
u 

 Probability of error is minimum for which E is maximum.
Now looking at options
Energy in option  A   12  1
Energy in option (C) and (D) is same = 1/3
1 2

2
Energy in option (B)  2   2t  dt 
 0

1 2 2 
 2  4t dt 
 0



12
 t3 
 2.4  
 3 0
1 3
Thus option (A) is correct answer.
50.
The electric field of a plane wave propagating in a lossless non–magnetic medium is given by
the following expression


E  z, t   a x 5cos  2109 t  z   a y 3cos  2 109 t  z  
2

The type of the polarization is
(A) Right Hand Circular.
(B) Left Hand Elliptical
(C) Right Hand Elliptical
(D) Linear
Answer:
Exp:
(B)
E x  5cos  t  z 


E y  3cos  t  z  
2



2
But the wave is propagating along negative z-direction
So, Left hand elliptical (LED).
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51.
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The energy band diagram and electron density profile n(x) in a semiconductor are shown in
 qx  3

 cm
the figure. Assume that n  x   1015 e kT 
, with   0.1V / cm and x expressed in cm.
kT
D kT
. The electron current density  in A / cm2 
 0.026V, Dn  36cm2s 1 , and 
q

q
EeV
lognx
at x = 0 is
Gvien
Slope0.1eV/cm
Ec
Ev
x0
(A) 4.4 102
x0
x
(B) 2.2 102
(D) 2.2 102
Answer:
Exp:
(B) OR (D)
dn  x 
J n,diff  qDn
dx
(C) 0
x
 qax 


n  x   1015 e kT 
Given :
d  n  x  
dx
 qax 
 1015 
d  n  x 
dx
… (1)
qa  kT 
e
kT
 1015
x 0
0.1
1
0.026
 3.846  1015 cm 4
J n ,diff  1.6  1019  36  3.846  1015
 2.2  102 A cm2
52.
A dc voltage of 10V is applied across an n–type silicon bar having a rectangular cross–section
and a length of 1cm as shown in figure. The donor doping concentration N D and the mobility
of electrons n are 1016 cm3 and 1000cm2 V1s 1 , respectively. The average time  in s 
taken by the electrons to move from one end of the bar to other end is _______________.
10V
n  Si
1cm
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Answer:
Exp:
100
V 10
    10 V m
d 1
vd    1000  10  104 cm s
vd 
53.
L
L
1  100
T
 4
 100 s
T
Vd 10  102
In the circuit shown, the initial voltages across the capacitors C1 and C2 are 1V and 3V,
respectively. The switch is closed at time t = 0. The total energy dissipated (in Joules) in the
resistor R until steady state is reached is ________________________
t0


Answer:
Exp:
R10

C1 3F

C2 1F
1.5
 3 1
  
s s
I(s) 
1 3

10   
3s 3s 

2
2
I(s) 

4
4 


10s   10  s  
3

 30 
1
i(t)  e
5
4
 t
30
;t0
20
1
s
1
3s


I(s)


3
s
1
s

E R   i 2 (t)10dt
0
4
 10   t
    e 15 dt
 25  0


10 e 
 .

25 4 
15  0

4
t
15
10 15

25 4
 1.5 J
0
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54.
The output of a standard second–order system for a unit step input is given as
2 t


yt 1
e cos  3t   . The transfer function of the system is
6
3

2
(A)
Answer:
Exp:
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s  2 s 
3

(B)
1
s  2s  1
2
(C)
3
s  2s  3
2
(D)
4
s  2s  4
2
(D)
Here n  1
1  2 
3
2
1
2
n  2

55.
If C denotes the counterclockwise unit circle, the value of the contour integral
1
 C Rez dz is _____________.
2j 
Answer:
Exp:
0.5
1
2J
 Re  z  dz
where C is z 1
e
Put z  e j  d  je j d
2
1
Re  e j  je j d
2j 0
1 2
cos  . j  cos   jsin   d
2j 0
2
j  2 2

cos

d


0 cos  sin  d
2j  0
j
1

   0 
2j
2

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any form without the written permission. Discuss this questions paper at www.gatementor.com.
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