Download GATE 2016 – A Brief Analysis (Based on student test experiences in

EC-GATE-2016
GATE 2016 – A Brief Analysis
(Based on student test experiences in the stream of EC on 31st
January, 2016 – (Afternoon Session)
Section wise analysis of the paper
Section Classification
1 Mark
Engineering Mathematics
Networks
Electronic Devices
Analog Circuits
Digital Circuits
Signals and Systems
Control Systems
Communication
Electromagnetics
Verbal Ability
Numerical Ability
5
1
3
3
3
3
2
3
2
2
3
30
2 Marks
4
3
3
4
3
4
3
4
2
3
2
35
Total No of
Questions
9
4
6
7
6
7
5
7
4
5
5
65
Type of Questions asked from each section
Digital Circuits
Questions came from Boolean expression simplification,
microprocessor, logic gates, Sequential circuits.
Electronic Devices
Questions came from Solar cell, pn junction diode, MOS
capacitor, BJT
Analog Circuits
OP-amp, zener diode, diode, 555 timers, BJT, MOSFET
Signals and Systems
Questions came from Z-transforms, Sampling, Laplace
transform, Filters, DFT.
Networks
Questions came from Basics of networks, Two port networks,
Steady state Analysis.
Control Systems
Questions came from Root locus, R-H criterion, Block diagram
reduction, Time Domain analysis
Communications
Questions came from Digital communication, Information
theory.
Electromagnetics
Questions came from Wave guides, Polarization, Maxwells
equations
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
1
EC-GATE-2016
Questions from the paper
General Aptitude
1.
Key:
Exp:
If y=mx+c curve passes through (0,0) and (2,6) then m= _________.
3
y=mx+c passing through (0,0)  0  0  c  c  0
y=mx+c passing through (2,6)  6  2m m  3
2.
Key:
Exp:
It takes 10s, 15s for two trains moving in same direction, to completely pass a pole. Length of
first train is 120 m and other is 150m. The magnitude of the difference between speeds is m/s.
(A) 2
(B) 10
(C) 12
(D) 22
(A)
length
Speed 
 length  speed  time
time
120  10  s 1  s1  12
150  15  s 2  s 2  10
s1  s 2  2
3.
Key:
4.
Key:
Exp
Four undergraduates are staying is a room. They agreed that older enjoys the more space. Manu is
two months older than Sravan, who is one month younger than Trideep. Pavan is one month older
than Sravan. Who will enjoy more space in room.
(A) Manu
(B) Sravan
(C) Trideep
(D) Pavan
(A)
The area bounded by 3x+2y=14 and 2x-3y=5 in the first quadrant is
(A) 14.95
(B) 15.25
(C) 15.70
(D) 20.35
(B)
 14 
A  ,0
 3 
B   0, 7 
5 
C   , 0
2 
 5 
D   0,

3 

E   4,1
y
B
3x  2y  14
2x  3y  5
E
R
O
x
C
A
Required area is area of
 OAB – area of  CEA

1  14 
   7   15.25 sq.units
2 3 
D
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
2
EC-GATE-2016
Technical
1.
In below figure current ‘i’ is __________A.
1A

5
 8V
1
1
1
 8V
C1
1
Key:
Exp:
-1
Nodal equation at V
V 8 V V 8 V
 
 0
1
1
1
1
 4V  16
1A
5

8V
i1
a

1

1
v
 V  4V
By using KCL at node ‘a’.
84
1
 i1  0  i1  5A
1
KCL at b
4
  i1  i  0  4  5  i  0
1
 i  1A
0V
b

8V

i
1
Px
P1
2.
The charge density profile shown in figure. The resultant potential
distribution is best described by
0
Vx
(A)
(B)
a
b
V(x)
a
(D)
a
x
V(x)
b
0
x
b
x
V(x)
a
P1
0
(C)
b
0
x
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
3
EC-GATE-2016
Key:
(D)
Exp:
q
q
Electrical 0   N d x no   N d x po



x2 
x 

2w 0 

Potential (x)  0
P(x)
V(x)
P1
b
b
a
a
P1
P1
3.
Key:
Exp:
In the below circuit M1 is in saturation has
transconductance gm = 0.01 seimens, Ignoring
internal parasitic capacitance and assuming the
channel length modulation  to be zero the small
signal input pole frequency (kHz) is ______.
57.9
Cin  50PF 1  g m R   550PF
VDD
1K
VD
50pF
Vin
5k
M1
g m  0.015
R  1k
R  1k
1  g m R  11
R in  5k
f in 
4.
1
 57.9 kHz
2R in Cin
 z11 z12 
The Z – parameter matrix 
 for two port network shows
 z 21 z 22 
3
IP
OP
6
(A)
 2 2
 2 2 


 2 2 
(B) 

 2 2
 1 2
(C) 

 2 1 
 2 1
(D) 

 1 2 
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
4
EC-GATE-2016
Key:
Exp:
(A)
Since the given network is symmetric
and reciprocal Z11  Z22
Z12  Z21
 Z11 
V1
I1
V
 Z21  2
I1

I2  0
3
5.
Key:
Exp:
6.
output
input
3 6
2
36


6
I1
I2  0

We know V2  V1  Z21  2
Z
So  11
 Z21


V1
Z12   2 2 

Z22   2 2 

V2
3


IL  0
If a right handed circularly polarized wave is incident on a place perfect conductor, then the
reflected wave will be _______.
(A) Right handed circularly polarized
(B) left handed circularly polarized
o
(C) Elliptical with all angle 45
(D) horizontally polarized
(B)
If incident wave is right handed polarized then the reflected wave is left handed polarized.
The block diagram of a feedback system is shown in figure the overall closed loop gain of the
system is _________.
x


G2


y
G1
H1
Exp:


G2
G1

G1
1  G1H1
H1
G1G 2
Y

x 1  G1H1  G1G 2
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
5
EC-GATE-2016
1
dx
is equal to _______.
1 x
7.

Key:
2
Exp:

8.
f z 
Key:
Exp:
-1
0
1
0
1
1 x

d 0  2 1  x

02 2
0
sin  z 
, the residue of the pole at z = 0 is _______.
z 2
z  0 is a simple pole as
f z 
sin  z 
z 2


1 
z3 z5
z


 ......
2 
z 
3! 5!

 z  0    z  0   ......
1
 

z0
3!
5!
 Residue of f  z  at t  0 is  1
3
9.
Key:
Exp:
The first two rows in the routh table for characteristics of a closed loop control system are given
as
s3 1  2k  3
4
s 2 2k
The range of ‘k’ for which the system is stable is
(A) 2.0  k  0.5
(B) 0  k  0.5
(C) 0  k  8
(D) 0.5  k  
(D)
S3 1 2k  3
4
S2 2k
From the table we can find characteristic equation
s3  2ks2   2k  3 s  4  0
For stability  2k   2k  3  4
4k 2  6k  4  0
1

 k    k  2  0
2

1
So the conditions are k  and k  2 combiningly k > – 2
2
10.
Key:
A triangle in the xy plane is bounded by the straight lines 2x =3y, y = 0 and x = 3. The volume
above the triangle and under the plane x + y + z = 6 is _________.
10
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
6
EC-GATE-2016
Exp:
Volume =
 z dx dy
y
R
3

2
x
3
y 0
   6  x  y  dy dx
x 0
3


2

y
  6  x  y  z 
x 0
.dx
y0
3
xb
R
3

8  x3 
8
  2x 2      2  9    33   10 cubic units
9
3
27
 0

11.
B
2
x
3

 2x  1  4 2  
  6  x   3   2  9 x  dx
3
2
2 
   4x  x 2  x 2 dx

3
9 


2x  3y
x
O
A
x limits : 0 to 3
2
y limits: 0 to x
3
An analog baseband signal, band limited to 100MHz, is sampled at the nyquist rate. The samples
are quantized into four message symbols that occur independently with probabilities
P1  P4  0.125 and P2  P3 . The information rate  bits sec  of the message source is _____.
Key:
Exp:
213.9
Information rate = rH
H  bits symbol
r  symbols sec
If we assume each sample is mapped to symbol
Then r = 200 MHz
P1  P4  2P2  1
1 3
2P2  1  
4 4
3
P2 
8
3
P2  P3 
8
2
6
8
H  log 2 8  log 2    1.069
8
8
3
I.R  213.9 Mbps
12.
Faraday’s law of electromagnetic s induction is mathematically described by which one of the
followers.
(A)  .B  0
Key:

B
(C)   E  
t
(C)
(B)  .D  V
(D)   H   E 
D
t
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
7
EC-GATE-2016
13.
Key:
Exp:
The minimum number of 2 -input NAND gates required to implement a 2 – input XOR gate is
(A) 4
(B) 5
(C) 6
(D) 7
(A)
A
Y
AB  AB  A  B
B
14.
For the unity feedback control system shown in below figure the open loop transfer function G(s)
= 2
s  s  1
x t  et
G s
gt

The steady state error due to unit step i p is _______.
Key:
0.33
Exp:
For the unit feedback system G  S 
2

S S  1
For unit step input the steady state error is
1
ess 
1  kp
Where k p  limsG  s   lims
s 
 ess 
s 0
2
2
s  s  1
1
1
  0.33
1 2 3
Key:
The bit error probability of a memory less BSc is 10-5. If 105 bits are sent over this channel, then
the probability that not more than one bit will be in error is _________.
1
Exp:
Pebit  105
15.
n  105 bits
x  no. of bits  error 
P  x  1  P  x  0   P  x  1
5
5
105  5
105  5 1
5 10  0
5 10 1




1  10

 10
 10  1  10 
0 
1 
105
5
105  5 1
5 10 1

 10  1  10 
1 
 1  very close 
 105  1 


 105 
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
8
EC-GATE-2016
16.
For the circuit shown in figure R1  R 2  R 3  1, L  1H, C  1F.
If input Vin  cos 106 t  , then the overall voltage gain of the circuit is _________.
R1
R3
L
C




Vout
R2
 V
in
Key:
Exp:
-1

R  
1

A1  1  1   1  6
2
6 
jL   10 10 

R3
1
1 1
A2  



1
R 2  XC
11 2
1 6
10 106
V
The overall voltage gain A v  out  A1  A 2
Vin
Vout
 1
 2     1
Vin
 2
17.
Key:
Exp:
The forward path transfer function and the feedback path transfer function of single loop negative
feedback control system is given as
k s  2
G s  2
and H  s   1
s  2s  2
If the variable perimeter k is real positive, then the location of the breakaway point on the root
locus diagram of the system is _______.
-3.414
To find break point, from characteristic equation we need to arrange k as function of s, then the
dk
 0 gives break point.
root of
ds
Characteristic equation is given by
s2  2s  2  k  2k  0
 k  s  2     s 2  2s  2 
 s 2  2s  2 
 k  

 s2 

  2
d
d 2

 s  2 ds s  2s  2  s  2s  ds s  2 
dk

 

ds
 s  2 2


Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
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EC-GATE-2016
  s  2  2s  2    s 2  2s  2  
dk
 

ds
 s  2 2


dk

0
ds
 2s 2  2s  4s  4  s 2  2s  2  0

j
1
2
 s 2  4s  2  0
 s  0.58 and  3.414
To find the valid break point we need to find that lies on root locus

– 3.414 lies on root locus

So break point – 3.414.

18.
j
Current  I 
The I  V characteristics of a solar cell is as
shown in figure, when it is illuminated
uniformly with solar light of power 100
mW/cm2. The solar cell has an area of 3 cm2 Isc  180mA
and fill factor is 0.7. The maximum efficiency
of the cell is _________%.
Key:
21
Exp:
Efficiency 
FF.Voc Isc 0.7  0.5 180

100%
Pin
(100  3)
Voc  0.5V
(0,0)
Voltage  V 
= 21%.
19.
Consider the circuit shown below, Assuming VBE1  VEB2  0.7V . The value of D.C voltage
Vc2 (V)is ____ .
Vcc  2.5V
1  100
Q1
Q2
10 k
2  50
Vc2
1V
1k
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
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EC-GATE-2016
Key:
Exp:
0.5
VE2  2.5  VBE1  1.8V
VB2  VE2  0.7  1.1V
0.1
 105 A
10k
 50  105 A
I B2 
I C2
VC2  5  104  103  0.5V
20.
The I-V characteristics of three types of diodes at room temperature made of semiconductors
x, y, z are shown in the figure. Assume that the diodes are uniformly doped and identical in all
respects except their materials. If E gx , E gy , E gz are the band gaps of x,y,z respectively then
(A) E gx  E gy  E gt
I
(B) E gx  E gy  E g2
x
z
y
(C) E gx  E gy  E gz
(D) No relation
V
Key: (C)
21.
The direct form structure of an FIR filter is shown in figure
unity
delay
x(n)
unity
delay
5
5


Key:
Exp:
The filter can be used to approximate a
(A) LPF
(B) HPF
(C)
(C) BPF

y[n]
(D) BSF
y  n   5x  n   5x  n  2 
 H  e j   5 1  e 2 j 
At   0, H  e j   0
At   ; H  e j   0
Thus the filter is band pass filter
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
11
EC-GATE-2016
22.
Two diodes D1 and D2 shown in figure are Ideal. The two capacitors are identical and the value
of RC is very large than time period of ac signal. If the diodes don’t breakdown in reverse bias,
the value of output voltage V0 (volts) in steady state is __________
D1

10sin t
C

R
V0
C

Key:
Exp:
0
Vo  0volts
D2

10V


Vo  0V

10V

23.

A discrete sequence x  n     n  3  2  n  5 has a z-transform X(z); if Y(z) = X(-z) is the ztransform for another sequence y[n], then,
(A) y[n]  x[n]
(B) y[n]  x[ n]
(C) y[n]   x[n]
(D) y[n]   x[ n]
Key:
(B)
Exp:
X(z)  z3  2z5
 Y(z)  X(z)  z3  2z5
On applying inverse z-transform,
y[n]  [n 3]  2[n 5]
Consider
x[ n]  [ n  3]  2[ n  5]
 [n  3]  2[n  5]
 y[n]  x[ n]
24.
A continuous time signal x(t) =cos 6t  sin 8t, the Nyquist sampling rate (samples/sec) of
y(t)=x(2t+5) is
(A) 8
(B) 12
(C) 16
(D) 20
Key:
(C)
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
12
EC-GATE-2016
Exp:
X(f )
Shifting doesn’t effects the sampling rate due to scaling
by a factor ‘2’ spectral components are doubled.
Thus maximum frequency of Y(f )  8.
4
 Nyquist rate  8  2  16Hz
3
3
4
2
t  cos t is passed through an LTI system with impulse transfer
3
function H  s   es  es . If ck represents kth coefficient of exponential Fourier series of output
25.
An input signal x  t   2cos
Key:
then the value of c3 is
(A) 0
(C)
Exp:
H  j  2cos 
(B) 1
(C) 2
(D) 3
 2 
 2 
For 2cos  t  , output  2k1 cos  t 
 3 
 3 
 2 
 2cos 
k1 
 2cos    1
2
 3 

3
For cos t, output  k 2 cos t
k2 
2  cos 
 2
  
2
 output  2cos t  2 cos t
3
26.
xt 
(A)
Key:
Exp:
sin t sin t
*
,
t
t
sin t
t
 c3  2
If * is the convolution then the value of x  t  is
(B)
sin 2t
2 t
(C)
2sin t
t
(D)
sin 2 t
 t
2
(A)
sin t 1
 sa  t 
t 
 
F
 sa  t  
  rect  
2
1
 
F
sa  t  
 rect  

2
Convolution in time domain leads to multiplication in frequency domain
1
sin t
 
 
 rect   rect    X    x  t   sa  t  

t
2
2
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
13
EC-GATE-2016
27.
Figure I and II shows MOS capacitors of unit area. The capacitor in figure I has insulator X (with
thickness t1  1m and r1  4 ) and Y (thickness t 2  3nm and r2  20 ). The capacitor in figure II
has only insulator material X of thickness tq . Find tq  _____________ nm.
Metal
t2
2 , Y
t1
1 , X
Metal
1 , X
Figure II
Figure I
Key:
Exp:
28.
t eq
1.6
E1 E 2
.
t1 t 2
E1E 2
4  20
CI 


 2.5
E1 E 2 E1t 2  E 2 t1 (4  2)  (20  1)

t1 t 2
E
4
CII  1 
 2.5  t Eq  1.6nm
t Eq
t Eq
The injected electron concentration profile in the base region of npn BJT, biased in the active
region is linear as shown in figure. If the area of the emitter – base junction is 0.001 cm2,
n  800 cm2 V  s
in the base region, the collector current Ic  mA  at room temperature is __________
 VT  26mV, q  1.6 1019 c  .
IB
n
IE
Key:
Exp:
p
n
IC
 0.5 m 
6.65
Ic  qADn
1014 cm3
 1014  0 
dn
dn
 1.6  1019  0.001  800  26  103 
 qAn V 
4 
dx
dx
 0.5  10 
Ic  6.65mA
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
14
EC-GATE-2016
29.
Frequency of oscillation at pin 3 is ___________kHz..
4
8
R A  2.2k
VCC
7
R B  4.7k
6
555
2
Out
3
1
C  0.022F
Key:
5.64
Exp:
f
30.
The I-V characteristics of diodes D1 and D2 are given in figure. The simply voltage is varied from
0 to 100V. The breakdown occurs at
1.44
1.44

 5.64kHz
 R A  2R B  C [2200  24700]  0.022 106
ID
D1
80V 70V
V
0 to100V
D2
D1
(A)
Key:
Exp:
D2
D1 only
(B) D2 only
(C)
Both D1 and D2
(D) None of D1 and D2
(D)
When two diodes are connected in series, the effective breakdown voltages becomes equal to the
sum of their individual breakdown voltage.
Vb  Vb1  Vb2  80  70  150V
Since the applied voltage has a maximum of 100V (100<Vb), No diodes will be in breakdown
region.
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
15
EC-GATE-2016
31.
The surface region of the MOSFET is in
(A) Accumulation
(B) Inversion
(C) Depletion
(D) None of the above
EC
B
B
Key:
Exp:
EF
Ei
EV
(B)
The semiconductor used in the MOSFET is n-type. At the surface the intrinsic level is above EF
as it is found at the distance of below EF, So, the surface is in inversion region.
Disclaimer – This paper analysis and questions have been collated based on the memory of some students who appeared in the
paper and should be considered only as guidelines. GATEFORUM does not take any responsibility for the correctness of the
same.
16