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GATE 2015 Examination-1st February
EC: Electronics & Communications Engineering
Duration: 180 minutes

Maximum Marks: 100

1. To login, enter your Registration Number and password provided to you. Kindly go through the various symbols
used in the test and understand their meaning before you start the examination.
2. Once you login and after the start of the examination, you can view all the questions in the question paper, by
clicking on the View All Questions button in the screen
3. This question paper consists of 2 sections, General Aptitude (GA) for 15 marks and the subject specific GATE
paper for 85 marks. Both these sections are compulsory. The GA section consists of 10 questions. Question
numbers 1 to 5 are of 1-mark each, while question numbers 6 to 10 are of 2-mark each. The subject specific
GATE paper section consists of 55 questions, out of which question numbers 1 to 25 are of 1-mark each, while
question numbers 26 to 55 are of 2-mark each.
4. Depending upon the GATE paper, there may be useful common data that may be required for answering the
questions. If the paper has such useful data, the same can be viewed by clicking on the Useful Common Data
button that appears at the top, right hand side of the screen.
5. The computer allotted to you at the examination center runs specialized software that permits only one answer to
be selected for multiple-choice questions using a mouse and to enter a suitable number for the numerical answer
type questions using the virtual keyboard and mouse.
6. Your answers shall be updated and saved on a server periodically and also at the end of the examination. The
examination will stop automatically at the end of 180 minutes.
7. In each paper a candidate can answer a total of 65 questions carrying 100 marks.
8. The question paper may consist of questions of multiple choice type (MCQ) and numerical answer type.
9. 9. Multiple choice type questions will have four choices against A, B, C, D, out of which only ONE is the correct
answer. The candidate has to choose the correct answer by clicking on the bubble (⃝) placed before the choice.
10. 10. For numerical answer type questions, each question will have a numerical answer and there will not be any
choices. For these questions, the answer should be entered by using the virtual keyboard that appears on the
monitor and the mouse.
11. All questions that are not attempted will result in zero marks. However, wrong answers for multiple choice type
questions (MCQ) will result in NEGATIVE marks. For all MCQ questions a wrong answer will result in
deduction of 1/3 marks for a 1-mark question and 2/3 marks for a 2-mark question.
12. There is NO NEGATIVE MARKING for questions of NUMERICAL ANSWER TYPE
13. Non-programmable type Calculator is allowed. Charts, graph sheets, and mathematical tables are NOT allowed in
the Examination Hall. You must use the Scribble pad provided to you at the examination centre for all your rough
work. The Scribble Pad has to be returned at the end of the examination.
Declaration by the candidate:
―I have read and understood all the above instructions. I have also read and understood clearly the instructions given
on the admit card and shall follow the same. I also understand that in case I am found to violate any of these
instructions, my candidature is liable to be cancelled. I also confirm that at the start of the examination all the
computer hardware allotted to me are in proper working condition‖.

1

GATE-2015 SET-C(1st February )

GENERAL ABILITY

Q. 1 - Q. 5 carry one mark each
1.

Choose the correct verb to fill in the blank below :
Let us _____________
(A) introvert
(B) alternate
(C) atheist
Sol: (B)
2.

(D) altruist

If x > y > 1, which of the following must be true ?
(i) ln x > ln y

(ii) e x  e y

(A) (i) and (ii)
(B) (i) and (iii)
Sol: (A) Both (i) and (ii) are correct

(iii) y x  x y

(iv) cos x  cos y

(C) (iii) and (iv)

(D) (ii) and (iv)

3.

Find the missing sequence in the letter series below :
A, CD, GHI, ?, UVWXY
(A) LMN
(B) MNO
(C) MNOP
(D) NOPQ
Sol: (C)
4.
Choose the most suitable one word substitute for the following expression :
Connotation of a road or way
(A) Pertinacious
(B) Viaticum
(C) Clandestine
(D) Ravenous
Sol: (B)
5.

Choose the most appropriate word from the options given below to complete the following sentence.
If the athlete had wanted to come first in the race, he _________ several hours every day.
(A) should practise (B) should have practised
(C)practised (D) should be practising
Sol: (B)
Q. 6 - Q. 10 carry two marks each
6.

In the following question, the first and the last sentence of the passage are in order and numbered 1
and 6. The rest of the passage is split into 4 parts and numbered as 2, 3, 4, and 5. These 4 parts are
not arranged in proper order. Read the sentences and arrange them in a logical sequence to make a
passage and choose the correct sequence from the given options.
1. On Diwali, the family rises early in the morning.
2. The whole family, including the young and the old enjoy doing this.
3. Children let off fireworks later in the night with their friends.
4. At sunset, the lamps are lit and the family performs various rituals.
5. Father, mother, and children visit relatives and exchange gifts and sweets.
6. Houses look so pretty with lighted lamps all around.
(A) 2, 5, 3, 4
(B) 5, 2, 4, 3
(C) 3, 5, 4, 2
(D) 4, 5, 2, 3
Sol: (B)
2

7.

Ram and Shyam shared a secret and promised to each other that it would remain between them. Ram
expressed himself in one of the following ways as given in the choices below. Identify the correct
way as per standard English.
(A) It would remain between you and me (B) It would remain between I and you.
(C) It would remain between you and I.
(D) It would remain with me.
Sol: (A)
8.
Sol:

log tan1º  log tan 2º ..........  log tan 89º is.....

(B) 1/ 2

(A) 1
(C)

(C) 0

(D) –1

9.

Ms. X will be in Bagdogra from 01/05/2014 to 20/05/2015 and from 22/05/20014 to 31/05/2015. On
the morning of 21/05/2014, she will reach Kochi via Mumbai.
Which one of the statements below is logically valid and can be inferred from the above sentences?
(A) Ms. X will be in Kochi for one day, only in May.
(B) Ms. X will be in Kochi for only one day in May.
(C) Ms. X will be only in Kochi for one day in May.
(D) Only Ms. X will be in Kochi for one day in May.
Sol: (B)
10.

From a circular sheet of paper of radius 30 cm, a sector of 10% area is removed. If the remaining part
is used to make a conical surface, then the ratio of the radius and height of the cone is _________.
Sol: (2.064)
Area of circle =  r 2    30

2

Remaining area  0.9  30 

 810 .

2

if cone is made of some radius ‗r‘
Then lateral surface area  r
Now 810    r  where   30cm

810  r  30

r  27

h  171 so

r
 2.064
h

3

GATE-2015 SET-C(1st Feb Morning)

ELECTRONICS AND COMMUNICATION ENGINEERING-EC

Q.1- Q.25 carry one mark each:
1.

If the base width in a bipolar junction transistor is doubled, which one of the following statements
will TRUE ?
(A) Current gain will increase.
(B) Unity gain frequency will increase.
(C) Emitter-base junction capacitance will increase. (D) Early Voltage will increase.
Sol: (D) If base width is doubled then Early voltage will be achieved at higher value
Ref: Analog PANACEA Class Notes Page-114

2.

Consider the Bode plot shown in the figure. Assume that all the poles and zeros are real-valued.

The value of f H  f L (in Hz) is ___________.
Sol:

(8970)
Here

f L  30Hz by concept of decade & f H  9000Hz by concept of decade

So

B.W  f H  f L  8970 Hz

Ref: Q on basic definition of Bode plot. Many Questions did in classes on similar line.

4

3.

A coaxial cable is made of two brass conductors. The spacing between the conductors is filled with
Teflon  r  2.1, tan   0 . Which one of the following circuits can represent the lumped element
model of a small piece of this cable having length z ?

Sol: (B)
Since Loss tangent is zero between conductors hence loss will be zero in between medium.
Ref: Refer EMT Class Notes Page-102

5

4.

In the circuit shown, the voltage Vx (in Volts) is ___________.

Sol:

(8)

Applying KCL at point P
V V  0.25Vx
 5  0.5Vx   x  x
20
10



10 10  Vx   Vx  2Vx  0.5Vx



100  12.5 Vx 

Vx  8 volt

Ref: similar type of questions on page 9-11 in class notes of Network theory
10
5.
The phase margin (in degrees) of the system G  s  
is _________.
s  s  10 
Sol:

(84.29)
G s H s  

10
s  s  10 

For calculation of GCF G  s  H  s   1



 2 100  10

Approximately  = 1 rad/sec
Now phase

  90º  tan1  /10

at GCF,

  90º  tan1  0.1
  90º 5.71  95.71

PM =   180º  84.29º
Ref: On exact same concept many Questions did in class
6.
For the circuit shown in the figure, the Thevenin equivalent voltage (in Volts) across terminals a-b is
__________.

6

Sol: (10)
Ref: Easy Question and many Qs did on this concept in PANACEA Network notes at Page no.80-85
7.

Sol:

The impulse response of an LTI system can be obtained by
(A) differentiating the unit ramp response (B) differentiating the unit step response
(C) integrating the unit ramp response
(D) integrating the unit step response
(B)

Ref: Page-89 Signal&sytem class notes of PANACEA Institute

8.

In the circuit shown in the figure, the BJT has a current gain () of 50. For an emitter-base voltage
VEB  600 mV , the emitter-collector voltage VEC (in Volts) is _________.

7

Sol: (2 )
It is case of PNP transistor
2.4
3  0.6  60 IB  IB 
mA
60
Since VEB  0.6 v, so it is in active region.
Now
Now

2.4  50
 2mA
60
3  VEC  0.5  2  VEC  2 volt
IC  I B 

Ref: Exact same concept many questions on page 233 of class notes of EDC


n

1
9.
The value of  n   is _________.
n 0  2 
Sol: (2 )
n

1
1
  u[n] 
1
2
1  Z1
2
n

d
1
2
1
n   u[n]  

 X[z]
dz 1  1 Z1  2z  12
2
2


n

1
X[0]=  n   =2
n 0  2 
Ref: Page- 254 Signal System Class notes of PANACEA Institute

8

10.

tan x 
 1
T
1
For A  
 , the determinant of A A is

tan
x
1



(A) sec2 x
Sol: (C)

(B) cos 4x

AT A1  AT A1  A 

(C) 1

(D) 0

1
=1
A

Ref: Mathematics Class Notes:

11.

Which one of the following processes is preferred to form the gate dielectric (SiO2) of MOSFETs ?
(A) Sputtering
(B) Molecular beam epitaxy (C) Wet oxidation
(D) Dry oxidation
Sol: (D) http://www.scientific.net/MSF.483-485.713 link for
Ref: Same type of Q was asked in GATE previous year papers. So discussed in class and above points were
given during assignment discussions
Dry oxidation and wet oxidation
1. Has a lower growth rate than wet oxidation
2. although the oxide film quality is better than the wet oxide film.
3. Therefore thin oxides such as screen oxide, pad oxide, and especially gate oxide normally use the dry
oxidation process.
4. Dry oxidation also results in a higher density oxide than that achieved by wet oxide and so it has a
higher breakdown voltage
5. Therefore the wet oxidation process has a significantly higher oxidation rate than the dry oxidation. It is
used to grow thick oxides such as masking oxide.
12.

If C is a circle of radius r with centre z0, in the complex z-plane and if n is a non-zero integer, then
dz
 C  z  z n 1 equals
0
nj
(D) 2n
2
It has zero value because it contains only poles and residue value at Z=Z0=0

(A) 2nj
Sol:

(B)

(B) 0

1

  z  z 
c

0

r 1

(C)

 1 r 2 ' 
z  z 0  2i 
  2i  0 = 0
  r  2 

Ref: Basic Q and was discussed in class

9

The contour on the x-y plane, where the partial derivative of x 2  y 2 with respect to y is equal to the

13.

partial derivative of 6y  4x with respect to x, is
(A) y = 2

(B) x = 2

(C) x  y  4

(D) x  y  0

z
w
 2y &
4
y
x

Sol: (A)


2y  4  y  2

14. Consider the function g  t   e t sin  2t  u  t  where u(t) is the unit step function. The area under g  t 
is ________
Sol: (0.1553)
Let g  t   e t sin 2t  u  t 
Let A is the area under curve g(t)


A

e

t

sin 2t u  t   G  0 



where

G s 

G s  







0

 st
t
 st
 g  t  e dt   e sin  2t  e dt

2

1  s    2 
2

2



G  0 

2
 0.1553
1  4 2

Ref: Exact same concept Q on Page-137 in class notes of Signal system
15.
The modulation scheme commonly used for transmission from GSM mobile terminals is
(A) 4-QAM
(B) 16-PSK
(C) Walsh-Hadamard orthogonal codes
(D) Gaussian Minimum Shift Keying (GMSK)
Sol: (D)
Reference: PANACEA Latest Technology Printed Notes PAGE-9

10

16.





The circuit shown consists of J-K flip-flops, each with an active low asynchronous reset R d input .
The counter corresponding to this circuit is

Sol:

(A) a modulo-5 binary up counter
(C) a modulo-5 binary down counter
(A)

(B) a modulo-6 binary down counter
(D) a modulo-6 binary up counter

Since 101  state becomes 000
here it is MOD–5 counter (UP)
Ref: Exact same Question on page 144 in class notes of Digital Electronics

17.

In the circuit shown, diodes D1, D2 and D3 are ideal, and the inputs E1, E2 and E3 are ―0 V‖ for logic
‗0‘ and ―10 V‖ for logic ‗1‘. What logic gate does the circuit represent ?

(A) 3-input OR gate
(B) 3-input NOR gate
(C) 3-input AND gate
(D) 3-input XOR gate
Sol:(C) It is +ve logic AND gate here 10V is nothing but Vcc

11

Ref: Exact same question on Page-42 in class notes of Digital Electronics

18.

In the circuit shown, assume that diodes D1 and D2 are ideal. In the steady state condition, the
average voltage Vab (in Volts) across the 0.5 F capacitor is ________.

Sol:

(100)

It is voltage doubler

Ref: EDC PANACEA Class Notes Page-166

19.

A message signal m  t   Am sin  2fm t  is used to modulate the phase of a carrier Ac cos  2fc t  to
get the modulated signal y  t   Ac cos  2fc t  m  t   . The bandwidth of y(t)
(A) depends on Am but not on fm

(B) depends on fm but not on Am
12

(C) depends on both Am and fm
Sol:

(C)

(D) does not depend on Am or fm

B.W of PM  2  f  fm  where f  k P A m f m

B.W  2  kP Am fm  fm 
Ref: Formula Available on Page-125 in class notes of Communication system

20.

Which one of the following 8085 microprocessor programs correctly calculates the product of two 8bit numbers stored in registers B and C ?
MVI A, 00H
MVI A, 00H
JNZ LOOP
CMP C
(A)
(B) LOOP DCR B
CMP C
LOOP DCR B
JNZ LOOP
HLT
HLT

MVI A, 00H
LOOP ADD C
DCR B
(C)
JNZ LOOP
HLT

Sol:
21.

Sol:

MVI A, 00H

(D)

ADD C
JNZ LOOP
LOOP INR B
HLT

(C)
In the circuit shown using an ideal opamp, the 3-dB cut-off frequency (in Hz) is _______.

(159.24)

13

It is the case of LPF here its value should be f H 

1
2 RC

1
103

 159.24 Hz
2104 107 2
Ref: Basic Q of LPF in OPAMP
22.
At very high frequencies, the peak output voltage V0 (in Volts) is ________.
fH 

Sol: (0.5)

23.

At high frequency capacitor becomes short circuited Here v0 

vi
2

The transfer function of a first-order controller is given as

K s  a 
sb
where K, a and b are positive real numbers. The condition for this controller to act as a phase lead
compensator is
(A) a  b
(B) a  b
(C) K  ab
(D) K  ab
Sol: (A)
Ref: Class notes of Control system
GC s  

14

24.

Consider a four-point moving average filter defined by the equation y  n   i 0 i x  n  i  .
3

The condition on the filter coefficients that results in a null at zero frequency is

Sol:

(A) 1   2  0;  0  3

(B) 1   2  1;  0   3

(C)  0  3  0; 1   2

(D) 1   2  0;  0  3

(A)

y  n    i 0 i x  n  i    0 x  n   1 x  n  1   2 x  n  2  3 x  n  3
3

Y[z]  X(z) 0  z 11  z 2 2  z 23  here z  e jw
Ref: Asked same type of Q many times in IES&GATE exam. Discussed in class.
25.
The directivity of an antenna array can be increased by adding more antenna elements, as a larger
number of elements
(A) improves the radiation efficiency
(B) increases the effective area of the antenna
(C) results in a better impedance matching (D) allows more power to be transmitted by the antenna
Sol: (B)
26.

The characteristic equation of an LTI system is given by F s   s5  2s4  3s3  6s2  4s  8  0 . The

Sol:

number of roots that lie strictly in the left half s-plane is ___________.
(2)

s5 1

3

4

s4 2

6

8

s3 0

0

0

s3 8

12

0

s2 3

8

0

0

0

0

0

1

s 32
s

0

8

Auxilary equation:F s   2s4  6s2  8  0  s  1, 2j
There is only 1 sign change so only 1 root in RHS, Here two roots are on Imaginary axis so there will
be only two roots on LHS

15

27.

A 200 m long transmission line having parameters shown in the figure is terminated into a load RL.
The line is connected to a 400 V source having source resistance RS through a switch, which is
closed at t = 0. The transient response of the circuit at the input of the line (z = 0) is also drawn in the
figure. The value of RL (in ) is __________.

Sol: (30)
c
v
 2 108 so in 2 micro second wave will return back to source.
r

vo (at t  2 s)  v  at t  0 1  k1  k1 k 2 
(at t = 2  sec) so 62.5  100(1  k1  k1  0.5)

150  50
 0.5
150  50
R  50
So here k 1 =-0.25= L
where RL=30 ohm
R L  50
k2 

Ref: Refer page-130of class notes of EMT at PANACEA Institute for this type of concepts

16

28.

 n 
Let x  n   1  cos   be a periodic signal with period 16. Its DFS coefficients are defined by
 8 
ak 

1 15
  
x  n  exp   j kn  for all k. The value of the coefficient a31 is _______

16 n 0
 8 

Sol: (0.5)
2

a 31  a 1 

1 15
 
1
   1 15 

x
n
exp
   j kn    1  exp  j n   16  0.5

16 n 0
 8  32 n 0 
 8   32

17

29.

A three bit pseudo random number generator is shown. Initially the value of output Y  Y2 Y1 Y0 is
set to 111. The value of output Y after three clock cycles is

Sol:

(A) 000
(D)

(B) 001

Initially

Q2  1

Q1  1 &

D2  0

Now

D2  0
D2  1

after 3 clock pulses

&

D1  0
Q2  1

D0  1

Q1  1

Q0  1
D0  1

D1  0

after 2 clock cycle Q2  0
Now

D0  Q1

D1  1

after 1 clock cycle Q2  0

(D) 100

Q0  1

D1  Q 2

D 2  Q1  Q0

Initially

(C) 010

Q1  0

Q0  1

D0  0
Q1  0

Q0  0

Ref: Exact many same concept questions on page 168 of class notes of Digital Electronics

18

30.

An NPN BJT having reverse saturation current IS  10 15 A is biased in the forward active region with
VBE  700 mV . The thermal voltage  VT  is 25 mV and the current gain () may vary from 50 to

150 due to manufacturing variations. The maximum emitter current (in A) is __________.
Sol: (1475)
V
IC  I0 (e
 1)  1016  (e700/25  1)  1.44626 mA
VT
But For Emitter current to Maximum Beta Value should be less……If Beta is taken as 150 then answer will
be 1455.9µA but here Beta will be taken as 50 so answer will be 1475 µA
Ref: Exact same concept Q on page 205 of class notes of EDC

31.

In the circuit shown, assume that the diodes D1 and D2 are ideal. The average value of voltage Vab (in
Volts), across terminals ‗a‘ and ‗b‘ is ________.

Sol: (5)
At high frequency capacitor will be short circuited in +ve half of i/p D1 will ON & D2 will be OFF
so circuit will be :

19

v 
v0   i 
3
In –ve half D1 will be OFF & D2 will be ON.

So

v 
v0   i 
2

Vm Vm 5Vm


3 2
6
Average value of output = 5 volt

Average value of o/p 
Vm  6

Here

Ref: Same type of Q was asked in GATE test series which is based upon FWR
32.
A realization of a stable discrete time system is shown in the figure. If the system is excited by a unit
step sequence input x[n], the response y[n] is

n

n

 1
 2
(A) 4    u  n   5    u  n 
 3
 3
n

n

1
2
(C) 5   u  n   5   u  n 
 3
 3
Sol: (C)
Refer : Basic Q
33.
Consider the differential equation

n

n

 2
 1
(B) 5    u  n   3    u  n 
 3
 3
n

n

2
1
(D) 5   u  n   5   u  n 
3
 3

d2x  t 
dx  t 
3
 2x  t   0 .
2
dt
dt

Given x  0  20 and x 1  10/e , where e = 2.718, the value of x(2) is ___________
Sol:

(0.855)
d 2 x 3dx

 2x  0
dt 2
dt

20

10
e
x  20

x  0   20, x 1 
t  0;

x  C1e  t  C 2 e 2t

20  C1  C2

— (1)

10
 C1e1  C2e2
e
Here C1=4.181 &C2=15.819

x  2  C1e2  C2e4 =0.855
34.

The variance of the random variable X with probability density function f  x  

Sol:

(6)
f x 

1
x
xe
2

E  X 

1
x
x  x e dx = 0 as it is an odd function.

2 

E X 2  

1
x
x e is ______.
2







1
x
x 2  x e dx   x 3e x dx ( It is an even function) = 3!=6

2 
0

x2  E  X 2    E[X]  12
Ref: This is direct Q from GATE Test series.
2

35.

the electric field profile in the depletion region of a p-n junction in equilibrium is shown in the
figure. Which one of the following statements is NOT TRUE ?

(A) The left side of the junction is n-type and the right side is p-type
(B) Both the n-type and p-type depletion regions are uniformly doped
(C) The potential difference across the depletion region is 700 mV
(D) If the p-type region has a doping concentration of 1015 cm–3, then the doping concentration in the
n-type region will be 1016 cm3
Sol:

(C) Here width of depletion layer in P-side is 0.1 m and width of depletion layer in N-side is 1 m.
Here

xpo Na  x no nd 

Means values of N a & N d  are such that
So

0.11015  11016

N a & N d  values are designed correctly

21

Here contact potential V0 is approximately
10 v/cm
E 
V0   0   W 
1.1104 cm  5.5 104 volt  0.55 mv
2
2
 
Ref: Refer page-79 of class notes of EDC for conclusion of all above concepts

36.

The position control of a DC servo-motor is given in the figure. The values of the parameters are
K T  1N-m/A, R a  1 , L a  0.1H , J  5kg-m 2 , B  1N-m/  rad/sec  and Kb  1V/  rad/ sec  . The

steady-state position response (in radians) due to unit impulse disturbance torque Td is ________.

Sol:
37.

(-0.5)
The Newton-Raphson method is used to solve the equation f  x   x3  5x 2  6x  8  0 . Taking the

Sol:

initial guess as x = 5, the solution obtained at the end of the first iteration is _______.
(4.29)

Sol:

x n 1  x n 

f  x n

f '  x n

Here x n  5

 x  5x  6x  8
 5
 3x  10x  6 
3

x n 1

2

2

22

putting x = 5

x n 1  4.29

Refer: Maths class notes

38.

The ABCD parameters of the following 2-port network are

3.5  j2 20.5 
(A) 

 20.5 3.5  j2

3.5  j2 30.5 
(B) 
3.5  j2
 0.5

2  j0
 10
(C) 

2  j0 10 
Sol: (B)
z11  7  j4
z12  z 21  2
2 
7  j4
7  j4
 2

 z  
A

z11 7  j4

z 21
2

7  j4 0.5 
(D) 

 30.5 7  j4
z 22  7  j4

z = 91
B

z 91

z 21 2

C

1
 0.5
2

D

7  j4
2

Ref: Exact same concept question on page 117 of class notes of Network theory

23

39.

A coaxial capacitor of inner radius 1 mm and outer radius 5 mm has a capacitance per unit length of
172 pF/m. If the ratio of outer radius to inner radius is doubled, the capacitance per unit length (in
pF/m) is ________.

Sol: (120.22)
Capacitance of coaxial cable is C 
If b/a is doubled then C ' 



C n  2b/a 

C' n  b/a 
C' 

2 0
n  b/a 

a &b Inner diameter & outer diameter

2 0
n  2b/a 



172 n10

C'
n 5

172  n 5
 120.22 pF
n10

Ref: Exact same concept Q on Page-110 in class notes of Electromagnetic Theory

24

40.

A universal logic gate can implement any Boolean function by connecting sufficient number of them
appropriately. Three gates are shown.

Which one of the following statements is TRUE?
(A) Gate 1 is a universal gate.
(B) Gate 2 is a universal gate.
(C) Gate 3 is a universal gate.
(D) None of the gates shown is a universal gate.
Sol: (C)
If we put Y=0 then Gate-3 is behaving like an Inverter and once inverter can be designed by this gate then It
is easy to design AND & OR gate by this.
41.

Consider the 3 m long lossless air-filled transmission line shown in the figure. It has a characteristic

impedance of 120 , is terminated by a short circuit, and is excited with a frequency of 37.5 MHz. What is
the nature of the input impedance (Zin) ?

(A) Open
Sol: (D)

(B) Short

(C) Inductive

(D) Capacitive

2
3
l 
 so tan l  ve

4
Ref: Exact same concept Q on Page-107 in PANACEA class notes of Electromagnetic Theory
Z  jZo tan l

42.

Here l 

Two sequence x1  n  and x 2  n  have the same energy. Suppose x1  n   0.5n u n  , where  is a
positive real number and u[n] is the unit step sequence. Assume

 1.5 for n  0,1
x2 n  
otherwise.

 0

25

Then the value of  is ________


Sol:

Energy of x1  n      0.25 
2

n 0

n

 2 1
4

 2
1  0.25 3

Energy of x 2  n   1.5 1   2.5 1 = 3
2

4 2
3

 3
   1.5
3
2
Ref: General Qs based upon definition of energy of the signal
43.
A random binary wave y(t) is given by
So

yt 



 X p  t  nT   

n 

n

where p(t) = u(t) – u(t – T), u(t) is the unit step function and  is an independent random variable
with uniform distribution in [0, T]. the sequence

Xn 

consists of independent and identically

distributed binary valued random variables with PXn  1  PXn  1  0.5 or each n.


 3T 
 3T  
The value of the autocorrelation R yy    E  y  t  y  t    equals _________.
4 
 4 


Sol: (1)
44.
A network is described by the state model as
x 1  2x1  x 2  3u
x 2  4x 2  u
y  3x1  2x 2

 Y s  
The transfer function H  s   
 is
 U s  

(A)

11s  35
 s  2  s  4 

(B)

11s  35
 s  2  s  4 

(C)

11s  38
 s  2  s  4 

(D)

11s  38
 s  2  s  4 

Sol: (A)
45.
Consider a continuous-time signal defined as
 sin  t/2   
x  t   
 *    t  10n 
  t/2   n 

Sol:

where ‗*‘ denotes the convolution operation and t is in seconds. The Nyquist sampling rate (in
samples/sec) for x(t) is ___________
0.4 Hz ( Impulse train has frequency 0.1 Hz and Sa function has frequency of 0.25 Hz but
convolution in time domain means multiplication in frequency domain. So resultant frequency of
message x(t) will be 0.1 Hz only.

26

Ref: Exact same concept on page 170 of class notes of Communication theory

46.

Suppose x[n] is an absolutely summable discrete-time signal. Its z-transform is a rational function
with two poles and two zeroes. The poles are at z  2j . Which one of the following statements is

TRUE for the signal x[n] ?
(A) It is a finite duration signal.
(B) It is a causal signal.
(C) It is a non-causal signal.
(D) It is a periodic signal.
Sol: (C) If pole are on unit circle either as real or complex imaginary it would be oscillatory.
It can‘t be causal because ROC must be right of the right most pole but it is stable so best option is Non
causal only
47. A fair die with faces {1, 2, 3, 4, 5, 6} is thrown repeatedly till ‗3‘ is observed for the first time.
Let X denote the number of times the die is thrown. The expected value of X is _______
Sol: (6)
1
5 1
5 5 1
E  x   1  2    3     ..........
6
6 6
6 6 6
5
1 5
5 5 1
5 5 5 1
E  x   1   2     3      ..........
6
6 6
6 6 6
6 6 6 6

By using E[X]-5/6 E[X]

5 5 5 5 5 5
1
E  x   1        .......... 
6
5
6 6 6 6 6 6
1
6

27

48.

For the system shown in the figure, S  2.75 lies on the root locus if K is _________.

Sol: (0.3)
If It lies on Root Locus then 1+G(s)H(s)=0
49.

A vector field D  22 a   z a z exists inside a cylindrical region enclosed by the surfaces   1 ,

z  0 and z = 5. Let S be the surface bounding this cylindrical region. The surface integral of this

  D.ds  is __________.
  D.ds    .DdV

field on S
Sol: (78.5 )

S

S

V

.D 

1 

A    A z   (6  1) A   0

 
z
2 5 1

 .D dV     .D  dddz =25π
V

50.

0 0 0

An SR latch is implemented using TTL gates as shown in the figure. The set and reset pulse inputs
are provided using the push-button switches. It is observed that the circuit fails to work as desired.
The SR latch can be made functional by changing

(A) NOR gates to NAND gates
(C) NOR gates to NAND gates and inverters to buffers

(B) inverters to buffers
(D) 5 V to ground

Sol: (A )
Ref: Exact same concept on page 113-114 of class notes of Digital Electronics

28

51.

 sin  t / 5  

The complex envelope of the bandpass signal x  t    2 
 sin  t   , centered about
4
 t / 5  
f

1
Hz , is
2

 sin  t / 5  j 4
(A) 
e
 t / 5 
(C)

 sin  t / 5  j 4
2
e
 t / 5 

 sin  t / 5   j 4
(B) 
e
 t / 5 
(D)

 sin  t / 5   j 4
2
e
 t / 5 

Sol: (C)

xp  t  



x(t)  jx(t)

 sin  t / 5   
 sin  t / 5  



  2
 sin  t    2 j 
 cos  t  
4
4

 t / 5  
 t / 5 
 sin  t / 5  


 j 2
 exp j  t  
4

 t / 5 
 sin  t / 5  


 2
 exp  t  
4

 t / 5 
29

 sin  t / 5  


xc  t   2 
 exp  t   exp(t)
4

 t / 5 

Ref: Exact same concept on page 82 in class notes of Communication theory and many Qs did on this
concept

52.

In the circuit shown, assume that the opamp is ideal. If the gain  vo / vin  is –12, the value of R(in
k) is __________.

Sol: (1)
vin  0 0  Vx
0  Vx Vx  V0 Vx

&


10k
10
10
10
R
Ref: Exact same Q on page 212 in class notes of Analog electronics

30

53.

In the circuit shown, both the enhancement mode NMOS transistors have the following
characteristics:

k n  n Cox  W / L  1mA/V2 :

VTN  1V . Assume that the channel length

modulation parameter  is zero and body is shorted to source. The minimum supply voltage VDD (in
volts) needed to ensure that transistor M1 operates in saturation mode of operation is ________.

Sol: (3)
Lower transistor is in saturation and VGS-VT=1 V and ID=1 mA
For Upper transistor: I D  1(VDD  1  1) 2  1  VDD  3V
Ref: Exact same concept many questions on page 298-303 in class notes of EDC
54.

In the circuit shown, the current I flowing through the 50  resistor will be zero if the value of
capacitor C (in F) is __________.

31

Sol : (20)
Just Use KCL method and solve it
55.
The current in an enhancement mode NMOS transistor biased in saturation mode was measured to be
1 mA at a drain-source voltage of 5 V. When the drain-source voltage was increased to 6 V while
keeping gate-source voltage same, the drain current increased to 1.02 mA. Assume that drain to
source saturation voltage is much smaller than the applied drain-source voltage. The channel length
modulation parameter  (in V–1) is __________.
Sol : (0.022)

ID  k  VGS  VT  1  VDS  if MOSFET is in saturation
2

1  k  VGS  VT  1  5 

----- (1)

1.02  k  VGS  VT  1  6 

----- (2)

2

2

1.02 1  6

1
1  5

1.02  5.10  1  6 

0.9  0.02

  0.022 v 1

Note: we have taken utmost care to give exact solution but still if You have any doubt in any problem
you are free to send mail at [email protected] or can do whats App at “9911382221”
Or can call at “9811382221”
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32

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