EC-GATE-2014
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
GATE 2014 – A Brief Analysis
(Based on student test experiences in the stream of EC on 15th
February,
2014 - First Session)
1. Questions with numerical answers accounted for 40-50 % of the paper. Hence you
cannot guess your way through the paper. General Ability section also has such
questions.
2. In the papers of the past, there used to be some questions which could be solved by
eliminating the wrong options. But with an increase in the “numerical answer” type
questions, this method of problem solving takes a backseat.
3. With the high occurrence of “numerical answer” type questions, one has to be
thorough and careful with unit conversions so as to type in the correct answer.
Earlier, the four options more or less served as guidelines.
4. Among general ability questions, there was one question each from Data
Interpretation, Probability, Time and Speed (Relative Velocity), Series, logical
reasoning and critical reasoning.
5. Engineering Mathematics questions were of average difficulty level. There were
questions from Linear Algebra (2 questions), Differential Equation (1 question),
Complex Integration (1 question) and Vector Calculus.
6. There were some direct questions from a few topics while some of questions needed
concept of two topics to solve them. Such questions were tough to solve. Scoring
sections were Networks, Control Systems, Signal and Systems and Analog Circuits.
7. The sequence of questions was different for each candidate i.e. no two candidates
will have same sequence of questions.
8. There were no “Common Data” and “Linked” questions in this paper as well.
EC-GATE-2014
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
Section wise analysis of the paper
1 Mark 2 Marks Total No of Questions
Engineering Mathematics 3 3 6
Networks 3 3 6
Electronic Devices 2 3 5
Analog Circuits 3 3 6
Digital Circuits 3 3 6
Signals and Systems 3 4 7
Control Systems 3 4 7
Communications 3 3 6
Electromagnetics 3 3 6
Verbal Ability 3 2 5
Numerical Ability 2 3 5
31 34 65
Type of questions asked from each section
Engineering Mathematics
one question based on probability, one question on complex algebra,
two question based on matrix, one solution of differential equation
Networks
one question based on KCL, one question based on maximum power
transfer, and one question based on AC circuit and some basic question
Electronic Devices
one simple question based on concentration under forward biased and
one question to find transconductance and some formulae based
question
Analog Circuits
one question based on gain, one question based on MOSFET node vlg
for given circuit and threshold, one question based on op-amp offset vlg
and some questions based on simple analysis
Digital Circuits
one question based on essential prime implicants, one question on
simple combinational circuit, one question based on asynchronous
counter frequency analysis
Signals and Systems
one question based on fundamental period, one based on Z-transform,
one question based on Fourier series, and some simple bits
Control Systems
one question based on find the value of k for repeated poles, one
question based on GM, one question based on State space analysis
Communications
one question based on sampling frequency, one question on BSC, one
Random variable etc
Electromagnetics one question based on plane waves, one question transmission line etc
EC-GATE-2014
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
Questions from the paper
1. When the optical power incident on a photodiode is 10 wµ and the responsivity is
0.8 A / W, the photo current generated (in A)µ is________.
2. Find the fundamental period of the signal shown below.
2x[n] sin n = π
(A) periodic with 2
π (B) periodic with π
(C) periodic with 2
π (D) Not periodic
3. Ideal current buffer is having
(A) low input impedance and high output impedance
(B) High input impedance and high output impedance
(C) High input impedance and low output impedance
(D) low input impedance and low output impedance
4. The value of K for which both the poles will lie at same location______ for the given
open loop transfer function.
k
G(s)(s 1)(s 2)
=+ +
5. When practical voltage drop is 0.7 V for diode D1 (PN diode) and 0.3V for D2 (Schottky
diode), then
(A) Both diode ON
(B) Both diode OFF
(C) Diode D1 ON and D2 OFF
(D) Diode D1 OFF and D1 ON
1D
10V
1kΩ 20Ω
2D
EC-GATE-2014
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
6. The input frequency for the given counter is 1 MHz, the output frequency observed at Q4
is_______.
7. For the given combinational circuit, the expression for F is
(A) F xyz xyz= + (B) F xyz xyz= +
(C) F x y z xyz= + (D) F xyz xyz= +
8. For the given circuit, the output voltage Vo is
Z
Y
X
F
5Q
5Q
4Q
4Q
3Q
3Q
2Q
2Q
1Q
1Q
5J
5k
4J3J
2J1J
4k 3k2k 1k
1 1 1 1 1
1 1 1 1 1
Clock
−
+
1R
2R
1I
oV2I
EC-GATE-2014
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
(A) ( )1 1 2I R R− + (B) 2 1I R
(C) 2 1I R (D) ( )2 1 2I R R− +
9. Find the voltage observed at P, Q and R for NMOS transistor and threshold voltage is 1V.
(A) 5V, 4V, 3V (B) 5V, 5V, 5V
(C) 4V, 4V, 4V (D) 8V, 4V, 5V
5V
5V
5V
P
Q
R
5V
EC-GATE-2014
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
(Based on student test experiences in the stream of EC on 16th
February, 2014 – First Session)
1. Match the following.
P1 : Solution of Integration
P2 : Solution of transcendental equation
P3: Solution of linear equation
P4: Solution of differential equation
M1: Nweton Raphson method
M2: Runge kutta method
M3: Gauss Elimination method
M4: Simpson’s IIIrd rule.
(A) P1 – M4, P2 – M1, P3 – M3, P4 – M2
(B) P1 – M2, P2 – M1, P3 – M4, P4 – M3
(C) P1 – M1, P2 – M2, P3 – M3, P4 – M4
(D) P1 – M2, P2 – M1, P3 – M3, P4 – M3
Ans: A
2. For an All-pass system ( )1
1
z bH z
1 az
−
−
−=
− where ( )jH e 1− ω = for all .ω If
Re a 0≠ , ( )Img a 0≠ then b equals,
(A) a (B) a* (c) 1
a * (D)
1
a
Ans: B
( )1
1
Z b 1 bzH z
1 a z z a
−
−
− −= =
− −
1
pole a; zerob
∴ = =
‘a’ is complex in nature. j apole a e⇒ =
An all pass system for which ( )jH e 1− ω = for all ‘w’ must satisfy the following
condition:
If the pole lies at ‘a’ then the zero must present at 1
;a
j a1 1 1 1zero e
a a * b a *⇒ = = ⇒ =
b a *∴ = .
EC-GATE-2014
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
3. The maximum value of f(x) = ln (1 + x) –x (when x > 0 occurs at x equal to ………
Ans: f(x) = ln ( 1 + x) – x
( )1 1f x 1
1 x= −
+
For f(x) to be maximum or minimum f1(x) = 0
x 0⇒ =
Thus at x = 0 f(x) is maximum.
( )( )
1
2
1f x
1 x
−=
+ (Negative at x = 0 implies maximum).
4. The unbiased coin is tossed infinitely. The probability that fourth head appears at the 10th
toss is
(A) 0.063 (B) 0.082 (C) 0.072 (D) 0.16
Ans: B
Three heads can came at any place in first 9 tosses and 10th toss must be head.
th thP 4 head at 10 toss⇒
3 6
3
4th headthree head
1 1 19c .
2 2 2
=
9 8 7 1 1
0.0823 2 1 512 2
× ×= × × =
× ×
5. The phase function of an LTI system is given by
( ) ( )c cf 2 f f 2 fφ = − πα − − πβ
' 'α and ' 'β are contstants and fc is the center frequency of the system. The delay
introduced by system is
(A) α − β
α + β (B)
αβ
α + β (C) α (D) β
Ans: C
Delay introduced by system is given by group delay
( )g
d 1T f
df 2= − φ ×
π
= α [Remaining all terms are constants]
EC-GATE-2014
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
6. Let ( ) ( ) ( )x t cos 10 t cos 30 t= π + π be sampled at 20Hz and reconstructed using an ideal
low-pass filter with cutoff frequency of 20Hz. The frequency/frequencies present in the
reconstructed signal is/are,
(A) 5Hz and 15Hz only
(B) 10Hz and 15Hz only
(C) 5Hz, 10Hz and 15Hz only
(D) 5Hz only
Ans: A
( ) ( ) ( )x t cos 10 t cos 30 t= π + π
( ) ( ) ( ) ( ) ( )x t 10 10 30 30 = π δ ω − π + δ ω + π + δ ω − π + δ ω + π
By using a filter of cut off frequency 20Hz only 5Hz & 15Hz components are filtered out.
7. Analog voltage 0 to 8V is quantized into 16 bits which is encoded with 4 bits. The
maximum quantization error is _________
Ans: Maximum quantization error is step size
2
−
8 0 1
step size 0.5V16 2
−− = = =
Quantization error = 0.25 V
8. The output F is
(A) 1 2F W S S= ⊕ ⊕ (B)
1 1 2F WS S S W= +
(C) ( )11 1 2F WS W S S= + (D) 21 1 2F WS WS WS S= + +
Ans: A
0
1
0
1
1S
2S
F
W
20− 15− 5− 5 15 20( )f Hz→
( )X f
EC-GATE-2014
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.
9
9. The angle modulated wave is given by
( )c 1 1 2 2cos 2 f t sin f t sin 2 fπ + β 2π + β π
The maximum frequency deviation is given by (in Hz)
(A) 1 1 2 2f fβ + β
(B) 1 2β + β
(C) f1 + f2
(D) 2 2 2 2
1 1 2 2d dβ + β
Ans: A
Maximum frequency deviation
( )max
1 dphase deviation
2 dt=
π
Maximum phase deviation
1, 1 2 2 max
sin 2 f t sin 2 f t= β π + β π
Derivative of phase deviation
1 1 1 2 2 22 f cos2 f t 2 f cos 2 f t= β π π + β π π
⇒ Maximum frequency deviation
( )1 1 2 2
12 f 2 f
2= πβ + πβ
π
10. Which of the following in linear non-Homogeneous differential equation? [X and Y are
independent and dependent variables respectively]
(A) xdyxy e
dx
−+ =
(B) dy
xy 0dx
+ =
(C) ydyxy e
dx
−+ =
(D) ydye 0
dx
−+ =
Ans: A
Option (c) and (d) are non-linear.
Option (a) and (b) are linear.
(a) is non-homogeneous
(b) is homogeneous.
EC-GATE-2014
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.
10
11. Given z = xy [ln (xy)]. Then
(A) dy dz
x y 0dx dy
+ =
(B) dz dz
x ydx dy
=
(C) dz dz
y xdx dy
=
(D) dz dz
y x 0dx dy
+ =
Ans: B
( )dz 1
xy. .y ln xy .ydx xy
= +
( )y ln xy .y.= +
( )dz 1
xy. .x ln xy .xdy xy
= +
( )x ln xy .x= +
dz dz
x ydx dy
⇒ =
12. The input –3e2tu(t), where u(t) is the unit-step function, is applied to a system with transfer
function, is applied to a system with transfer function S 2
S 3
−
+ . If the initial value of output
is –2, then the value of the output at steady state is __________ .
Ans: ( )s 2
H s ;s 3
−=
+ ( ) ( )2tx t 3e u t= −
( ) ( )3 3
X s Y ss 2 s 3
− −= ⇒ =
− +∵
( ) ( )Lt
s 0
3sy SY s 0.
s 3→
−∞ = = =
+
13. The rectangular waveguide with dimension 5cm × 3cm is given. The cut off frequency for
TE21 mode is (in MHz) __________
Ans: For air filled rectangular waveguide,
2 2
21
c m nf
2 a b
= +
M = 2, n = 1
2 210
21
3 10 2 1f
2 5 3
× = × +
5546 MHz.=
EC-GATE-2014
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
14. A modulated signal is ( ) ( ) ( )y t m t cos 40000 t= π where the base band signal m(t) has
frequency components less than 5KHz only. The minimum required rate(in KHz) at
which y(t) should be sampled to recover m(t) is _________
Ans: . ( ) ( )x t x f↔
After modulation,
We can use band-pass sampling to find minimum sampling rate.
( )x
s min
2ff ;
k= h
h z
fK Integar part of
f f
=
−
FH = 25 KHz; fL = 15 KHz.
25
K Integer part of 210
∴ = =
( )s min
2 25Kf 25 KHz
2
×∴ = =
25− 20− 15− 15 20 25
( )in KHz→
5− 5 ( )f in KHz
EC-GATE-2014
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
15. Let X1, X2 and X3 be independent and identically distributed random variable with
uniform distribution between [0,1].
[ ]1 2 3P X X X+ ≤ is __________
Ans: 1 2 3 1 2 3x x x x x x 0+ ≤ ⇒ + − ≤
Let Z = x1 + x2 – x3
1 2 3P x x x z 0 .⇒ + ≤ = ≤
Pdf of z we need to determine.
It is the convolution of three pdf.
0 3 0
2
11
z 1pz 0 z 2 dz 0.1667
6 6−−
≤ = ⇒ = =∫
16. Consider a communication system an shown below
Two Binary symmetric channel are calculated as shown in figure above. i.e., Probability
of x = 1 is 1/3 and cross over probability in1/2.
Determine the value of H(Y1) + H(Y2).
Ans:
PX = 1 = 1/3
PX = 0 = 2/3.
To get H(y1) we should know probability of getting y1.
X BSC1Y
BSC
2Y
1
0
1
01− 0
**
*
1
1− 020
1⇒
1
1X
1 21 2
1 2 1 2
1 21 2
1 2
1 2
0
1y
2y
EC-GATE-2014
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
1
1 1 2P y 1 1 2 1 2
3 2 3= = × + × =
1P y 0 1 2.= =
Similarly
2P y 1 1 2= =
2P y 0 1 2= =
( ) ( )1 2 2 2
1 1H y H y log 2 log 2 2
2 2
+ = + ×
2 bits.=
17. The power spectral density of WSS random process is
( )( )610 3000 f f 3
S f0 otherwise
− − ≤× =
X(t) is DSB-SC modulated with carrier ( ) ( )c t cos 16000 t .= π The modulated signal is
passed through a Band pass filter with center frequency fc = 8000 Hz and bandwidth of 2
kHz.
Determine the output power is (watts) _______ .
Ans:
After modulation
3− 3
63000 10
−×( )x
S f
( )f in KHz
61500 10−×
11− 8− 5− 5 8 11 ( )f in KHz
9 8− 7−
1
7 8 9 ( )f in KHz
EC-GATE-2014
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
Output of BPF will have PSD as shown below.
The total power is area under shaded figure.
Total output power 3 6 3 612 2 10 1000 10 2 10 500 10
2
− − = × × × + × × × ×
[ ]2 2 0.5 5W= + =
18. A thin p-type silicon sample is uniformly illuminated with light which generates excess carriers. The recombination rate is directly proportional to
(A) The minority carrier mobility
(B) The majority carrier recombination life-time
(C) Majority carrier concentration
(D) Excess minority carrier concentration.
Ans: D
19. The maximum value of
( ) 3 2f x 2x 9x 12x 3= − + − in the interval [0, 3] is _________
Ans: ( )1 2f x 6x 18x 12= − +
( ) [ ]3 2f x 2x 9x 12x 3 in 0,3= − + −
For f(x) to be maximum and minimum,
( )1f x 0.=
26x 18x 12 0⇒ − + =
x 1or 2.⇒ =
f(0) = −3
f(1) = 2
f(2) = 1 f(3) = 6.
⇒ maximum value is 6 since the interval is closed interval.
9 8− 7− 7 8 9( )f in KHz
( )yS d
61500 10
−×
61000 10
−×
EC-GATE-2014
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
20. For the given op-amp circuit
Then Vo is _______ .
Ans:
2 02 1 2V VV V V
0R 2R 3R
−−+ + =
02 22 1
VV VV V 0
2 3 3− + + − =
o 21
V 11 VV
3 6= −
20 1
11 VV 3V 4V
2= − = −
21. At T = 300k, the hole mobility of a semiconductor p 500µ = cm2/V- S and KT
26mV.q
=
The hole diffusion constant Dp in the Cm2/S is _____________
Ans: 2
p 500 cm V s;µ = − KT
26 mV4
=
p 3
p
p
D26mV D 26 10 500−∴ = ⇒ = × ×
µ
Dp = 13 cm2/sec.
1VR
2R
−
+R
oV
3R
2V
1V 5V=R
2R
−
+R
oV
3R
2V 2V=
EC-GATE-2014
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
22. Consider a WSS Random process x(t) with power spectral density Sx(f). If another
Random process y(t) = x(2t – 1), then power spectral density of ( ) ( )yx t s f↔ in term of
Sx(f)
(A) ( ) ( )j f
y x 2
1s f e s f
2
− π=
(B) ( ) ( )y x
1S f S f 2
2=
(C) ( ) ( )j2 f
y x
1S f e S f 2
2
− π=
(D) ( )jnf
y x
1S (f) e S f 2
2=
Ans: B
( ) ( ) ( )x xx t R z S f→ ↔
( ) ( ) ( )x xx t 1 R z S f− → ↔
( ) ( ) ( )x x
1X 2t 1 R 2z S f 2
2− → ↔
( ) ( ) ( )y x
1y t S f S f 2
2⇒ → =
Shifting operation does not change PSD scaling operation change PSD
23. In MOSFET fabrication, the channel length is defined during the process of
(A) Isolation oxide growth
(B) Channel stop implantation
(C) Polysilicon Gate patterning
(D) Lithography step leading to the contact pads
Ans: C
24. Consider the following block diagram in the figure
The transfer function ( )( )
C sis,
R s
( )R s
1G
+
+
2G+ ( )C s
+
EC-GATE-2014
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.
17
(A) 1 2
1 2
G G
1 G G+ (B) G1G2 + G1 + 1 (C)
1 2 2G G G 1+ + (D) ( ) 1
1 2
GD
1 G G+
Ans: C
By drawing the signal flow graph for the given block diagram
No.of parallel paths are three.
P1 = G1G2
P2 = G2
P3 = 1.
( )( ) 1 2 3 1 2 2
C sP P P G G G 1
R s= + + = + +
25. Consider a transmission line setup as shown below
ZT is characteristic impedance of transmitter
ZO is characteristic impedance of transmission line
ZR is characteristic impedance of receiver.
Let x(t) be a signal from transmitter. Which of the following is true?
(A) Signal distortion will not take place if ZO = ZR, T OZ Z≠
(B) Signal distortion will not take place if ZT = Zo, O RZ Z≠
(C) Signal distortion will not take place given if T 0 RZ Z Z≠ ≠
(D) Signal distortion will never take place. The values of ZT, ZO and ZR will decide only
the power efficiency of the setup.
Ans: D
V+− TZ oZ RZ
TransmitterReceiver
( )R s 1
1G 2G
1
1
1 ( )C s
EC-GATE-2014
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.
18
26. For the MOSFET M1 shown in the figure, assume W
2,L
= VDD = 2V, x
2
n oc 100 A Vµ = µ
and Vth = 0.5V. The transistor M1 switches from saturation region to linear region when
Vin (in volts) is __________ .
Ans: Initially the transistor M1 is in saturation.
( )2
D o gs t
1 WI C V V
2 L
∴ = µ −
( )216
out
12 100 10 V
2
−= × × ×
4 2
D outI 10 V−= … (1)
From the given circuit,
DD D outV I 10K V 0.− + + =
4
out D2 V 10 I∴ − = … (2)
On substituting (2) in (1),
4 2out
out4
2 V10 V
10
−−=
2
out outV V 2 0.∴ + − =
outV 1V⇒ =
out in inV V V= −
inV 1.5V.∴ =
27. The z-Transform of the sequence x[n] is given by ( )( )
21
1x z ,
1 2z−=
− with the ROC z 2>
then x[2] is ___________
Ans: ( )( )
21
1x z
1 2z−=
−
Consider
( )( )
[ ] [ ]1
n
1 121
2zX z x n n2 u n
1 2z
−
−= ⇒ =
− [ ] [ ]
( )[ ]n 1
1
n 11x n x n 1 2 u n 1
2 2
++∴ = + = +
[ ] ( )33x 2 2 1 12.
2= × =
inV
1M
DDV
R 10k= Ω
outv
EC-GATE-2014
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.
19
28. Let h(t) denote the impulse response of a causal system with transfer 1
S 1+. Consider the
following three statements
S1: The system is stable
S2: ( )
( )h t 1
h t
+ is independent of t for t > 0
S3: A non-causal system with the same transfer function is stable.
For the above system,
(A) Only S1 and S2 are true
(B) Only S2 and S3 are true
(C) Only S1 and S3 are true
(D) S1,S2 and S3 are true.
Ans: A
( ) ( ) ( )t1H s h t e u t
s 1
−→ = ⇒ =+
∵ the pole is located at’−1’ the system is said to be stable.
1S is true.∴
( )
( )
( ) ( )( )
t 1
t
h t 1 e u t 1consider
h t e u t
− +
−
+ +→ =
( )( )
t
1 1
t
e u te e
e u t
−
− −
−
=
which is a constant.
( )( )
h t 1
h t
+∴ is independent of time.
2s is true∴
( ) ( ) ( )t1H s h t e u t
s 1
−→ = ⇒ = − −+
(for non-causal system)
∴ system is un-stable.
3S is False.∴
29. Let ( ) ( )1
1
1H z 1 pz ,
−−= − ( ) ( )
11
2H z 1 qz ,
−−= − ( ) ( )1 2H z rH z .+ The quantities p, q and r
are real numbers. Consider 1
p ,2
= 1
q ,4
= − r 1< If the zero of H(z) lies on the unit
circle then r = ___________
Ans: ( )( )1 1
1H z ;
1 pz−=
− ( )
( )2 1
1H z
1 qz−=
−
EC-GATE-2014
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.
20
( ) ( )1 211
1 rH z rH z
111 z1 z
42
−−
+ = + +−
1 1
1 1
1 11 z r 1 z
4 2
1 11 z 1 z
2 4
− −
− −
+ + −
=
− +
( )( )
( )1
1
1 1 1 1
1 r
4 21 r 1 z
1 rr r1 r z
4 2
1 1 1 11 z 1 z 1 z 1 z
2 4 2 4
−
− − −
−
+ ++
+ + − = =
− + − +
the zero lies on the unit circle,−∵
( )
1 r
1 r4 21 1 r
1 r 4 2
−
− = + ⇒ − + = ++
r 5 r 5
r2 2 2 4
− = ⇒ =
5
r 2.5.2
= =
30. The current through 5Ω resistor as shown in figure below is _________.
Ans: Apply the principle of superposition.
I = 0.5 Amp
31. The Donor and Acceptor impurities in an abrupt junction silicon diode are 1 × 1016 cm−3
and 5 × 1018 cm—3 respectively. Assume that the intrinsic carrier concentration in silicon
ni = 1.5 × 1010 cm—3 at 300k, KT
26mVq
= and the permittivity of silicon 12
si1.04 10−∈ = ×
F/cm. The built-in potential and depletion width of the diode under thermal equilibrium
onditions respectively are,
(A) 0.7V and 1×10−4 cm (B) 0.86V and 1×10−
4 cm
(C) 0.7V and 3.3×10−5 cm (D) 0.86V and 3.3×10−
5 cm
5V+−
5Ω
10Ω
5Ω1A
i
EC-GATE-2014
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.
21
Ans: D
Built in potential A Do 2
i
N NkTV ln
q n
=
( )
16 83
210
1 10 5 1026 10 ln
1.5 10
−
× × × = ×
×
0.86V.=
Depletion width
12
o
A D
2 V 1 1
q N N
∈= +
112 2
19 16 18
2 1.04 10 1 1
1.602 10 1 10 5 10
−
−
× × = +
× × ×
53.34 10 cm.−= ×
32. An ideal MOS capacitor has boron doping concentration of 1015
cm−3
in the substrate.
When a gate voltage is applied, a depletion region of width 0.5 mµ is formed with a
surface (channel) potential of 0.2V. Given that 14
o8.854 10−ε = × F/cm and the relative
permitivities of silicon and silicon dioxide are 12 and 4 respectively. The peak electric
field ( )in V mµ in the oxide region is ___________
Ans: s
2 0.2 VE 0.8m0.5
×= =
µ
sox s
ox
VE E 2.4m
ε= =
µε
33. The value of resistance R1 in the delta equivalent of the given star network is ________
(in ohm)
Ans: 1
5 3R 5 3 6
7.5
×= + + = Ω
2R3R
1R
5Ω 3Ω
7.5Ω
EC-GATE-2014
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.
22
34. In the given RC network the voltage equation is given by ( ) ( )t
o
1V Ri t i v dv.
c= + ∫
The voltage source is switched ‘ON” at t = 0, which of the following waveform represent
current i(t)
(A) (B)
(C) (D)
Ans: A
( ) ( ) ( )1
V s RI s I sCS
= +
( )( )
( )CS
I s V s .R s 1
=+
( )1
V sS
=
( )( )C
I sR s 1
⇒ =+
( ) t /RCI t 6ke−⇒ =
When ‘K’ is some constant
V
+
−
( )i t
( )i t
t
( )i t
t
( )i t
t
( )i t
t
EC-GATE-2014
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.
23
35. Given vector field x yˆ ˆF sin x cos y a cos xsin y a
→
= + .The magnitude of curl of F→
is
________
Ans: Zero
36. The given circuit represents
(A) SR Latch (B) JK Flip Flop
(C) Toggle (D) master slave D-flip flop
Ans: D
37. For the given circuit w and y are MSBs.
Determine F.
Ans: ( )F Wx xw .U.y= +
38. The line current density in the figure shown below in x
Ampˆ2 a
meter
XD Q
CLK Q
D Q
Q
W X Y Z
F
0
1
2
3
0
1
2
3
U
ConductionX
V
EC-GATE-2014
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.
24
The magnetic field intensity just above the surface of perfect conductor is
(A) ˆ2z Amp m−
(B) ˆ2z Amp m
(C) ˆ ˆx y Amp m+
(D) ˆ2 y Amp m
Ans: B
ˆH 2z Amp m.=
Apply Right hand thumb rule.
39. The slope of ID VS VGS curve of an n-channel MOSFET in linear regime is 10−3
1−Ω at
VDS = 0.1 V. For the same device neglecting channel length modulation, the slope of the
DI VS VGS curve ( )in A V under saturation region is approximately ________
Ans: In linear region,
[ ]2
DD GS T D
VI K V V V
2
= − −
3DD
GS
I10 KV
V
−∂= =
∂
In saturation region,
( ) ( )2
GS tD sat
1I k V V
2= −
( ) ( )GS tD sat
KI V V
2= −
D
GS
I K
V 2
∂=
∂
⇒ Slope of D S GSI V V curve under saturation is, 210
0.07 A V2
−
=
40. The steady state error of the system shown in the figure for a unit-step input is
__________ .
−
( )R s ( )E s
( )e t
( )c s1
s 2+k 4=
( )E t( )r t
+
2
S 4+
EC-GATE-2014
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.
25
Ans: The steady state error is defined only for unity feedback system. If non-unity feedback
system is given, we should represent the system with unity feedback. Then by considering
the open loop transfer function, the steady state error has to be evaluated.
The closed loop transfer function of given block diagram is,
( )( )
( )( ) ( )
C s G s
R s 1 G s H s=
+
Where ( )1
G ss 2
=+
and ( )2
H ss 4
=+
with k = 4
C.L.T.F is ( )( ) 2
C s 4s 16
R s s 6s 16
+=
+ +
The C.L.T.F is equivalent to the following unity feedback system is,
( ) ( )( )
4s 16Now G s H s
s s 2
+=
+
It is a type –1 system,
And given input is unit step.
Where ( ) ( )lim
p s 0k G s H s→=
( )
sslimps 0
A 1 1e 0
4s 161 k 11
s s 2→
= = = =++ + ∞++
41. The state equation of a second order linear system is given by
x(t) = Ax(t), x(0) = xo
For ( )t
o t
1 ex , x t
1 e
−
−
= = − −
and
for ( )t 2t
o t 2t
0 e ex , x t
1 e 2e
− −
− −
− = =
− +
Wheno
3x
5
=
, x(t) is
(A) t 2 t
t 2t
8e 11e
8e 22e
−
− −
− +
− (B)
t 2t
t 2t
11e 8e
11e 16e
− −
− −
− − +
(C) t 2t
t 2 t
3e 5e
3e 10e
− −
− −
− − +
(D) t 2t
t 2t
5e 3e
5e 6e
− −
− −
− − +
( )R s +
− ( )4s 16
s s 2
+
+( )C s
EC-GATE-2014
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.
26
Ans: B
Apply linearity property.
3 1 0
a h5 1 1
= + −
3 = a
5 = − a + b
⇒ b = 8.
( )t t 2t
t t 2 t
e e ex t 3 8
e e 2e
− −
− − −
−⇒ = +
− − +
t 2t
t 2t
11e 8e
11e 16e
−
− −
−=
− +
42. Consider the building block called ‘Networks N’ shown in figure. Let C 100 F= µ
and R 10k= Ω
Two such blocks are connected in cascade as shown in the figure.
The transfer function ( )( )
3
1
V s
V s of the cascaded network is
(A) S
S 1+ (B)
2
2
S
1 3S S+ + (c)
2S
S 1
+ (D)
S
S 2+
+
−
( )1V s Network NNetwork N
+
−
( )3V s
( )1V s
C
+
−
( )2V S
−
+
R
EC-GATE-2014
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.
27
Ans: B
Apply mesh analysis to determine the current ( )2I s
( ) ( ) ( )1 2 1
1R I s RI s V s R
Sc
+ − = ×
( ) ( )
( ) ( )
1 2
2
2 1
1 1RI s 2R I s 0 R
SC SC
1 12R R R I s RV s
SC SC
− + + = × +
+ + − =
( ) ( )3 2V s I s R;=∵
( )( ) ( )( )
2 2 23
2 2 2
1
V s R S C
V s 2RSC 1 1 RSC R S C=
+ + −
( )( )
2 2 2
2 2 2
S R C
2RSC 1 1 RSC R S C=
+ + −
2 2 2
2 2 2
S R C
1 3SRC S R C=
+ +
R 10k & C 100 F; RC 1= Ω = µ =∵
( )( )
23
2
1
V s S
V s S 3S 1=
+ +
43. Advice is
(A) verb (B) Noun (C) Adjective (D) Verb and Noun.
Ans: A
Advice is noun. Advise is verb
44. The next number in series
81, 54, 36, 24, _________
Ans: 2
81 54 27 183
− = × =
2
54 36 18 123
− = × =
2
36 24 12 83
− = × =
( )1V s
1
SC
−
+
( )1I s R ( )2I s R
+
−
1
SC
( )3V s
EC-GATE-2014
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.
28
24 x 8− =
x 16∴ =
45. A man row at a speed of 8km/hr in still water. It takes thrice as long to cover a distance
when he moves upstream as compared to downstream. The speed of stream is ________ .
Ans:
Let the distance be x.
Let the stream speed be V km/hr.
x x
3.8 v 8 v
⇒ =− +
8 V 24 3V⇒ + = −
4V = 16
V 4km hr.⇒ =
46. India is a country of rich heritage and culture. Which of the following statement support
best the above statement?
(a) India has 28 states and 7 union territories.
(b) India has a population of 1 billion
(c) India has many different languages and dialects
(d) India has many places to visit.
Ans: C
47. For the given condition P < M, which of the following is surely represent above
condition?
(A) N < M > O < P
(B) P = A < B < M
(C) O < P > N < M
(D) None
Ans: B
I . M > B
& B > A i.e., M > A
II. P = A i.e., M P> or P M<
48. The next element in series is 7G 11K 13M
(A) 17 P (B) 17 Q (C) 15 Q (D) 17 S
Ans : B
EC-GATE-2014
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.
29
(Based on student test experiences in the stream of EC on 16th
February, 2014 - Second Session)
1. ( ) ( ) ( )
( ) ( ) ( )
( )
n
n
x n 0.5 u n
If y n x n x n
Then y n _________∞
=−∞
=
= ⊗
=∑
Ans: 4
( ) ( ) ( )
( )( )
j
2j
y n x n x n
1y e
1 0.5e
ω
− ω
= ⊗
=−
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( )
n
n
n n
n n 0
n n
2n 0 n 0
n
1y n n 1 a u n n 1 u n
2
So n 1 a u n n 1 a
11 1 12n
12 2 1 11 22
1 22 y n 4
1 4
∞ ∞
=−∞ =
∞ ∞
= =
∞
=−∞
= + = +
+ = +
= + = +
−−
= + = =
∑ ∑
∑ ∑
∑
Alternatively
It can also be solved in frequency domain.
( ) ( ) ( )
( ) ( )
( )
( ) ( )
( ) ( )
( ) ( )
( )
2
2
j
j n
n
n
n
2
jn
y n x n x n
Y x
1Y
1 0.5e
Y y n e
Substituting 0 in both side of above equation
Y 0 y n
y n Y 0
1y n 4
1 0.5e0
− ω
∞− ω
=−∞
∞
=−∞
∞
=−∞
∞
− ω=−∞
= ⊗
ω = ω
ω =
−
ω =
ω =
=
⇒ =
⇒ = =
− ω =
∑
∑
∑
∑
2. The series n 0
1
n!
∞
=
∑ converges to ______
(A) 2 n2 (B) 2 (C) 2 (D) e
EC-GATE-2014
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.
30
Ans: D
n 0
2 3x
1 1 1 11 ................ e
n! 1! 2! 3!
x x xe 1 ...............
1! 2! 3!
∞
=
= + + + ∞ =
= + + + ∞
∑
∵
3. The solution of D.E is 2
2
dx dx2 x 0,dtdt
+ + = where a, b are constants ?
(A) tae− (B) t tae bte− −+ (C) t tae bte−+ (D) 2tae−
Ans: B
( )
( )
2
t t
D 2D 1 x 0
D 1, 1,
x t ae b te− −
+ + =
= − −
= +
4.
In the circuit current through ( )2R is _____________ mA ,
Ans: 2.8
By source Transformation theorem By K.V.L
28
20 10kI 8 0 I 2.8mA10k
− + = ⇒ = =
5. A unilateral Laplace transform of ( ) 2
1f t
s s 1=
+ +
If ( ) ( ) ( )g t t.f t , A Unilateral Laplace Transform of g t _____ ?= =
(A)
( )2
2
s
s s 1
−
+ + (B)
( )
( )2
2
2s 1
s s 1
− +
+ + (C)
( )2
2
s
s s 1+ + (D)
( )2
2
2s 1
s s 1
+
+ +
Ans: D
10mA
1R 3R
2R
4R
2K
3K
1K
4K2mA
20v
2K
3K
21K R= 4K
8v+−
I
+−
EC-GATE-2014
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.
31
By Laplace Transform Property we know that
( ) ( ) ( )
( )
( ) ( )
( ) ( )
nnn
n
2
2
22 2
dL t f t 1 F s
ds
d 1L tf t 1. .
ds s s 1
0. s s 1 2s 1 2s 11
s s 1 s s 1
= −
= − + +
+ + − + + = − =+ + + +
6. In the magnitude Bode plot, which of the following exhibits at high frequency
of 4th order all pole system ?
(A) dB80dec
− (B) dB40dec
− (C) dB40dec
(D) dB80dec
Ans: A
In a transfer function if all are poles if we plot the BODE diagram then on each and every corner frequency we have to introduce a line of slope
dB20dec
− and hence on the 4th corner frequency the slope of line will
become dB80dec
− and will continue upto infinity.
7. In the second order unit feedback system, what is the natural frequency
(rad/sec) ? (A) 16 (B) 4 (C) 2 (D) 1 Ans: B
2n
2 2 2 2
n n
2n n
w4TF s 4s 4 s 2 w w
w 4 w 2
= =+ + + ξ +
= ⇒ =
8. The cut –off wavelength ( )c mλ µ of light that can be used for intrinsic
excitation of semiconductor has Band gap gE 1.1ev,= is ___________ ?
Ans: 1.127
( )( )c
g
1.24 1.24m 1.127
1.4E 1.1evλ µ = = =
9.
( )4
s s 4+( )u s ( )u s
+
−
OpAmp
−
+
12+
12v−
2C
iv
2v−
2R
1C
ov
1R
EC-GATE-2014
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.
32
Which of the following represent above circuit? (A) Band Pass Filter (B) VCO (C) AM(Amplitude Modulation) (D) Monostable Multivibrator Ans: D
10. If C=0, the o/p expression for circuit is
(A) Y AB AB= + (B) Y A B= + (C) Y A B= + (D) Y AB=
Ans: A
( )( ) ( )
( ) ( )
y AB A B .1 AB. A B
A B A B AB AB
= + + = +
= + + = +
11.
What is eR
R in the above circuit ?
Ans: 2.618
We know that in a infinite ladder network if all resistance are comprises of
same value R then the equivalent resistance is ( )1 5 R
2
+ . Then the above
given network can be redrawn as R series with R equivalent as follows:
A
B
A
B
C 0=
1
y
2R R R
R R R
eR
R
eqReR
( )1 5 R
2
+
R
EC-GATE-2014
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.
33
e
e
R R 1.618R 2.618R
R2.618
R
= + =
=
12. Which of the following represents the expression of output Y ?
(A) Y ABC ACD= + (B) Y ABC ABD= +
(C) Y ABC ACD= + (D) Y ABD ABC= +
Ans: C
( )
1 3 6y ABC.I ABCI ABCI
y ABC.D ABCD ABC.1
ACD B B ABC
ABC ACD
= + +
= + +
= + +
= +
13. In sample-and-Hold circuit, the value of hold capacitor increased is (A) Drop rate decreased and Acquisition time decreased (B) Drop rate decreased and Acquisition time increased (C) Drop rate increased and Acquisition time decreased (D) Drop rate increased and Acquisition time increased Ans: B
In a capacitor drop rate is given as dv
dt.
→ We know that in a capacitor
dvi C
dt
dv 1
dt C
=
⇒ ∝
From the above relation it is clear that if capacitor value increases than the drop rate decreases because of inverse relation.
→ We know that in a capacitor
Q CV i t
C Vti
t C
= = ×
×⇒ =
⇒ ∝
From the above relation it is clear that if capacitor value increases acquisition time also increases because of proportionality relation.
14. ˆ ˆ ˆF z ax x ay y az,= + + where s represents the portion of the sphere 2 2 2x y z 1+ + =
If s
z 0, F.ds is _________≥ ∇ ×∫
?
0
1
2
3
4
5
6
7
I
I
I
I
I
I
I
I
0
D
0
D
0
0
1
0
8 x 1
MUX
2S 1S 0S
A B C
y
EC-GATE-2014
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.
34
Ans: π2
( )s
s s s
ˆ ˆ ˆax ay az
ˆ ˆ ˆF ax ay azx y z
z x y
ˆ ˆ ˆ ˆF.ds ax ay az . az dx dy dxdy 2−
∂ ∂ ∂∇ × = = + +
∂ ∂ ∂
∇ × = + + = = π∫ ∫ ∫
15. For Antenna radiating in free space, the E-field at a distance of 1 km is found
to be v12m .m
Given 0 120η = πΩ magnitude of avgP density due to this
antenna at a distance of 2 km from the antenna ___________ ( )2Wn .m
Ans: 47.7
In far-field zone
( )
2 1 12 1 22
1 2
2 26
2
avg 20
E r r1 12mv mE E E . E 6mv m
r E r 2r
E 6m 36 10 wP density
2 2 120 240 m
−
∝ ⇒ = ⇒ = = ⇒ =
×= = =
η × π π
16.
Column A Column B
(1) Point Electro-Magnetic source
(P) Highly directional
(2) Dish antenna (Q) End-fire array
(3) Yagi-uda antenna (R) Isotropic
(A) 1 – P, 2-Q, 3-R (B) 1-R, 2-P, 3-Q (C) 1-Q, 2-P, 3-R (D) 1-R, 2-Q,
3-P
Ans: B
17. Represents about Circuit is (A) Voltage controlled voltage source (B) Voltage controlled current source (C) Current controlled current source (D) Current controlled voltage source
Ans: C
Ri Ai
EC-GATE-2014
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.
35
In the above circuit in the output side there is a dependent current source which is controlled by the input current and hence it is a current controlled current source.
18. What happens if Emitter Resistance in a Common-Emitter Voltage Amplifier is
not by-passed? (A) Reduce both voltage gain and i/p impedance (B) Reduce voltage gain and increase i/p impedance (C) Increase voltage gain and reduce i/p impedance (D) Increase both voltage gain and i/p impedance
Ans: B
19. ( )n j6 f
F.T
j2 f
2 eFourier Transform u n 3 A
231 e
3
+ π
− π
→+ ← −
What is the value of A _______? Ans: 3.375
( )
( )
n
n 3
2F u n 3
3
2F u n 3
3
+
= +
= = +
( )
( )
3 n 3
3
j6 f
j2 f
2 2F u n 3
3 3
2 1e .
231 e
3
− +
−
π
− π
= +
= −
By comparing with the relation mentioned in question
32
A 3.3753
−
= =
20. The Impedance parameters [z] of network are given below:
(A) 6 24
12 9
(B) 9 6
6 24
(C) 9 8
8 24
(D) 42 6
6 60
Ans: B
+
−
+
−
2V1V
30Ω
10Ω 60Ω
EC-GATE-2014
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.
36
1 1 1
1 1 1
1
30 10 30y
30 60 30
0.133 0.0333 9 6Z y
0.0333 0.05 6 24
− − −
− − −
−
+ −=
− +
− = ⇒ = =
−
21.
In the circuit shown in figure above ( ) ( )oV t in volts for t is ______→ ∞ .
Ans: 31.25
Since we have to evaluate ( )oV t in steady state the inductor will behave as a
short circuit and hence A xV 5i=
By Nodal Analysis at Node A
( ) ( )
A xx
x xx
xx
o x
V 2i10 i 0
5
5i 2i10 i 0
5
8i 5010 i
5 8
50 250V t 5i t 5 31.25V
8 8
−− + + =
−− + + =
= ⇒ =
= = × = =
22. ( ) ( ) ( )3 2 2 2ˆ ˆ ˆE 2y 3yz x 6xy 3xz y 6xyz z= − − − − + is E-field in source free region.
Valid expression for potential (V) is
(A) 3 2xy yz− (B) 3 22xy xyz− (C) 3 22y xyz+ (D)
3 22xy 3xyz−
Ans: D We know that E V.= − ⇒∇
From option (D) we can get E-field.
2H+
− x2i
5
( )10u t
xi
5
+
−
( )oV t
A
+
− x2i
5
( )10u t
xi
5+
−
( )oV t5Ω
5H
EC-GATE-2014
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.
37
23. In Double-side bond (DSB) full carrier AM Transmission, if modulation index doubled, the ratio of total side bond power to carrier power increased by a factor _______
Ans: 4
In AM 2 2
T c c cP P P P 12 2
µ µ = + = +
Ratio of total sideband power to total carrier power =
2 2
e
2
1 1
2 2
1
2
2 1
P2 2
P
P
P 1
P 4
P 4P
µ µ=
µ=
µ
=
=
24. The characteristic equation of unity feedback system
( )1 KG s 0;+ = ( )G s has poles,
one pole at origin, and two poles at -1. If damping factor is 0.5, distance from origin to A (OA = 0.5) what is the value of K in root locus meet at A _______ ? Ans: 0.375
If we know the coordinate of point A of the given root locus then by
magnitude condition ( ) ( )G s H s 1= we can open the value of K at A. So, we
have to open the coordinate of A first. In the question it is given that damping factor 0.5ξ= and length of OA=5 then
in the right angle triangle OY
cosOA
OYcos60
0.5
1OY
4
θ =
⇒ =
⇒ =
1− 2
3
−1
3
− 00
LE 0.5= A
As we know 1cos cos−θ = ξ ⇒ θ = ξ
EC-GATE-2014
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.
38
AY AY 3
sin sin60 AYOA 0.5 4
θ = ⇒ = ⇒ =
So the coordinate of point A is 1 3
j4 4
+
Substituting the above value of A in the transfer function and equating it to 1 ( by magnitude condition) we obtain
( )
2
1 3S j
4 4
2
k1
s s 1
1 3 9 3K . 0.5 0.75 0.375
16 16 16 16
= − +
=+
= + + = × =
Alternatively
K value at A can also be obtained by taking the phasors of poles joining point A.
In both case answer is same. Can be verified
25. If ( ) ( ) ( )2
2
d y dyy 0 y ' 0 1, D.E is 4 4y x 0
dxdx= = + + = , then
( )y x is _______ ?x 1=
Ans: 0.541
Apply Laplace Transform to D.E
( ) ( ) ( ) ( ) ( ) ( )
( )
( )( )
( )( )
( )
( )
2
2
2 2
2
2x 2x
2 2 2
s y s sy 0 y ' 0 4 s y s y 0 4y s 0
y s s 4s 4 s 1 4 0
s 5 s 2 3y s
s 4s 4 s 2
1 3y s
s 2 s 2
y x e 3x e
y 1 e 3e 4e 0.541
− −
− − −
− − + − + =
+ + − − − =
+ + += =
+ + +
= ++ +
= +
= + = =
26. Let STA 1234 H, (Store Content of the accumulator into the address 1234)
let starting address is IFFE, the instruction fetched and executed sequentially,
then what are the 15 8A A Pins−
(A) 1F 1F 20 12 (H) (B) 1F 1F 1F 1F 20 (H)
(C) 1F 1F 12 20 (H) (D) 1F 1F 20 1F 20 (H)
0.5ξ = A
1− 2
3
−1
3
− 00yθ
EC-GATE-2014
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.
39
Ans: (A)
Starting address is
15 0A A
first1F F E STA
fetched
1F F F
Next execut
34
20
ed
0 0 12
12 3 4=
=
So, ANS = 1F 1F 20 12 (H)
27. Let ( )( )
2
p 2
Ps 3Ps 2T F of G s ,
s 3 P s 2 p
+ −=
+ + + − P is positive real number, the
maximum value of P in PG (S) remains stable is _____ ?
Ans: -3
We can choose the value of P upto a extent of marginal stable not beyond that so for the above transfer function we have to find value of P that will drive the system to marginal stable.
( )2C.E s 3 P s 2 P
3 P 0 P 3
= + + + −
+ = ⇒ = −
Note that in this case system will produce oscillatory output. 28. If propagation carry delay = 12ns
Propagation sum delay = 15ns
What is the worst case delay in 16-bit ripple carry adder _______ ( )ns
0A
0B
0S
0FA
1A
1B
1S
1FA
15A
15B
15S
15FA0C 1
C 15C
A
EC-GATE-2014
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.
40
Ans: 195
In a n bit ripple carry adder the worst case delay is given by
( ) ( ) ( ) ( )( )( )
pd carry pd carry pd sumworst delay n 1 t Max t , t
16 1 12 15
180 15 195
= − +
= − +
= + =
29. A LTI system has initially relaxed, if o/p and i/p relation
( ) ( )2
2
2
d y dyy t x t
dtdt+ α + α = and
( ) ( )( )
( ) ( ) ( )t
2
0
dh tg t h t h t , L g t G s ,
dt= α + + α =∫
No. of Poles of G(s) have _____ ? Ans: -1
Apply Laplace Transform on both sides
( )( ) ( )
( )( )
( )
22
2
2 2
dy td yL y t L x t
dtdt
y s 1H s
x s s s
+ α + α =
= =+ α + α
Similarly apply Laplace transform to equation (2)
( )( )
( ) ( )
( )( )
( )
2
2 22
2 2
2 2
H sG s sH s H s
s
H s s sH s s
s s
1 s s 1G s
s ss s
α= + + α
+ α + α α = + + α =
+ α + α= = =
+ α + α
30. A LTI system has ( ) 2
1H s ,
s s 6=
+ − if connected another cascaded system
( )1H s , to get overall system stable.
(A) s 6+ (B) s 2− (C) s 1− (D) s 3−
Ans: B
( )( ) ( )
( ) ( )
( ) ( )
1
1H s
s 3 s 2
So H s s 2
1H s H s overall system , which is a stable system
s 3
=+ −
= −
= =+
( )H s ( )1H s
EC-GATE-2014
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.
41
31.
( ) ( ) ( )y t A sin t = ω ω + θ ω ;
( )A 0.25,ω = What is the frequency ω ?
(A) 1
3RC (B)
2
3RC (C)
1
RC (D)
2
RC
Ans: B
By Nodal analysis
( ) 2 2 2
2 2 2
x 1 0 x x0
1 2R
Cj Cj
1 Cj 1 0x Cj
R 2 R
2 3RCj 1 0x
2R R
2x
2 3RCj
x 1y
2 2 3RCj
1 1 2A 9 R C 12
4 3RC4 9R C
− ∠+ + =
ω ω
ω < + ω + =
+ ω < =
=+ ω
= =+ ω
ω = = ⇒ ω = = ω =+ ω
32. What is the state transition matrix for ( )( )
1 1
22
x t x0 1
x0 0x t
=
(A) 1 1
0 t
(B) t 1
0 1
(C) 1 t
0 1
(D) t t
0 t
Ans: C
( )
( )
11
11
2
21
t L sI A
s 1 s 11SI A
0 s 0 ss
1 11 ts st L0 11
0s
−−
−−
−
φ = −
− − = =
φ = =
33. The two BJT’s have some collector current, and have
1 2A 0.2 m 0.2 m, A 300 m 300 m= µ × µ = µ × µ , where 10 3xi 1.5 10 / cm= × and
Kq26mv.
T= What is
1BE BE2V V _____ in(mV)−
iV sin t= ω
R
C
C
C
( )y t
+
−
x+
−
EC-GATE-2014
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.
42
Ans: 381
BE
T
BE2 BE1
T T
V
nV
c o
C2 C1
V V
nV nV
o2 o1
I I e
I I
I e I e
=
=
=
BE1 BE2
T
1
V V
nV o2
01
o2BE BE2 T
o1
Ie
I
IV V nV n
I
300 30026m n
0.2 0.2
2.6 14.162mv 381mv
−
=
− =
×=
×
= × =
34.
cE = Bottom energy level in conduction bond
vE = top energy level in valence bond
E = Fermi-energy level
(A) (B) (C) (D) Ans: C
35. Which of the following options is the closest in meaning to the word underlined in the sentence below?
In a democracy, everybody has the freedom to disagree with the government.
(A) Dissent (B) descent (C) decent (D) decadent
Ans: A
cE
FE
VE
cE
FE
VE
cE
FE
VE
cE
FE
cE
FE
vE
N typeSemi conductor
−
V
EC-GATE-2014
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.
43
36. While receiving the award the scientist said “I Feel vindicated”. The meaning of the word vindicated is closed to
(A) Punished (B) substantiated (C) appreciated
Ans: B
37. In the sequence of 12 consecutive odd numbers the sum of first 5 numbers is 425 then the sum of last 5 numbers in sequence is _________.
Ans: 495
8th observation is 7×2=14 more than 1st observation
9th observation is 14 more than 2nd observation
10th observation is 14 more than 3rd observation
11th observation 14 more than 4th observation
12th observation 14 more than 5th observation
Total 14×5=70
Sum of the first five numbers =425
Sum of last five numbers =495
38. Let n mf (x, y) x y P.= = If x is doubled and y is halved the new value of f is
(A) n m2 P− (B) m n2 P− (C) 2(n m)P− (D) 2(m n)P−
Ans: A
m
n n m1P ' 2 X y
2
=
n m n m
n m
2 X Y
2 P
−
−
=
=
39. After discussion, Tom said to me, “please revert” He expect me to
(A) Retract (B) get back to him
(C) move in reverse (D) retreat
Ans: B
40. If KCLFTSB stands for best of luck and SHSWDG stands for good wishes, which of following indicates “ace the Exam’
(A) MCHTX (B) MXHTC (C) XMHCT (D) XMHTC
Ans: B
KCLFTSB: BST-Best, F-Of, LCK-Luck (Reverse order)
SHSWDG: GD-Good, WSHS-Wishes (Reverse order)
Similarly “ace the Exam’- C-Ace, T-The, XM-Exam
EC-GATE-2014
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.
44
41. A firm producing air purifiers sold 200 units in 2012. The following Pie chart represents the share of raw material, labour cost, transportation, energy, plant and machinery cost in total manufacturing cost of firm 2012. The expenditure of labour in 2012 is Rs. 4, 50,000. The raw material expenses in 2013 are increased by 30%. All other expenses increased by 20%. What is the percentage increase in total cost for company in 2013?
Ans: 22
2012 2013
Transport (10%) 300,000 360,000
Labour (15%) 450,000 540,000
Raw material (20%) 750,000 780,000
Energy (25%) 750,000 900,000
Plant and Machinery (30%) 900,000 1,080,000
Total 3,000,000 3,660,000
Percentage increase in total cost =22%
42. Industrial consumption of power doubled from 2000-2001 to 2010-2011. Assuming it to be uniform over the year find the annual rate of increase in percentage
(A) 5.6 (B) 7.2 (C) 10 (D) 12.2
Ans: B
n 10r r
A P 1 n 10 years A 2P 1 2100 100
r 7.2
= + = = ⇒ + =
=
43. Find the next term in the sequence 13M, 17Q, 19S, _____.
(A) 21W (B) 21V (C) 23 W (D) 23 V
Ans: C
44. A five digit is formed using the digits 1, 3, 5, 7 & 9 without repeating any one of them. What is the sum of all such possible five digit numbers?
(A) 6666660 (B) 6666600 (C) 6666666 (D) 6666606
Ans: B
The digit in unit place is selected in 4! Ways
The digit in tens place is selected in 4! Ways
(a)(b)
(c)
(d)
(e)
a transport (10%)
b labour (15%)
c raw material (20%)
d energy (25%)
e plant and machinery (30%)
−
−
−
−
−
EC-GATE-2014
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.
45
The digit in hundreds place is selected in 4! Ways
The digit in thousands place is selected in 4! Ways
The digit in ten thousands place is selected in 4! Ways
Sum of all values for 1
( )0 1 2 3 44! 1 10 10 10 10 10
4! 11111 1
× × + + + +
= × ×
Similarly for ‘3’ 4 ! (11111) 3× ×
Similarly for ‘5’ 4 ! (11111) 5× ×
Similarly for ‘7’ 4 ! (11111) 7× ×
Similarly for ‘9’ 4 ! (11111) 9× ×
( )sum of all such numbers 4! (11111) 1 3 5 7 9
24 (11111) 25
6666600
∴ = × × + + + +
= × ×
=
EC-GATE-2014
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.
46
Top Related