gate 2014 – a brief analysisd1zttcfb64t0un.cloudfront.net/images/gate 2014 paper...signals and...

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.

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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

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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

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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

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(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

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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 *∴ = .

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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]

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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

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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.

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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.=

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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

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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

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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

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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

−×

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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=

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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

+

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(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

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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

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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−=

−

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( ) ( )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

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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

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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

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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

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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+

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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

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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

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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

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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

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29


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

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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

+−

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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+

−

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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

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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

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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%)

−

−

−

−

−

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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

∴ = × × + + + +

= × ×

=

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