1 EC GATE-05-PAPER-03 General Aptitude Q. No. 5 Carry One Mark Each. An apple costs Rs. 0. An onion costs Rs. 8. Select the most suitable sentence w...

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General Aptitude Q. No. 1 – 5 Carry One Mark Each 1.

An apple costs Rs. 10. An onion costs Rs. 8. Select the most suitable sentence with respect to grammar and usage. (A) The price of an apple is greater than an onion. (B) The price of an apple is more than onion. (C) The price of an apple is greater than that of an onion. (D) Apples are more costlier than onions.

Key:

(C)

2.

The Buddha said, “Holding on to anger is like grasping a hot coal with the intent of throwing it at someone else; you are the one who gets burnt.” Select the word below which is closest in meaning to the word underlined above. (A) burning

(B) igniting

(C) clutching

(D) flinging

Key:

(C)

3.

M has a son Q and a daughter R. He has no other children. E is the mother of P and daughter-in-law of M. How is P related to M? (A) P is the son-in-law of M.

(B) P is the grandchild of M.

(C) P is the daughter-in law of M.

(D) P is the grandfather of M.

Key:

(B)

4.

The number that least fits this set: (324, 441, 97 and 64) is ________. (A) 324

(B) 441

(C) 97

Key:

(C)

Exp:

324 182 ; 441 212 ;64 82 but 97 x 2 for any positive integer

(D) 64

i.e. 97 is odd man out 5.

It takes 10 s and 15 s, respectively, for two trains travelling at different constant speeds to completely pass a telegraph post. The length of the first train is 120 m and that of the second train is 150 m. The magnitude of the difference in the speeds of the two trains (in m/s) is ____________. (A) 2.0

Key:

(A)

Exp:

Speed

(B) 10.0

(C) 12.0

(D) 22.0

length length speed time time

120 10 s 1 s1 12 150 15 s 2 s 2 10 s1 s 2 2 ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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Q. No. 6 – 10 Carry Two Marks Each 6.

The velocity V of a vehicle along a straight line is measured in m/s and plotted as shown with respect to time in seconds. At the end of the 7 seconds, how much will the odometer reading increase by (in m)?

(A) 0

(B) 3

(C) 4

(D) 5

Key:

(D)

7.

The overwhelming number of people infected with rabies in India has been flagged by the World Health Organization as a source of concern. It is estimated that inoculating 70% of pets and stray dogs against rabies can lead to a significant reduction in the number of people infected with rabies. Which of the following can be logically inferred from the above sentences? (A) The number of people in India infected with rabies is high. (B) The number of people in other parts of the world who are infected with rabies is low. (C) Rabies can be eradicated in India by vaccinating 70% of stray dogs (D) Stray dogs are the main sources of rabies worldwide. (A)

Key: 8.

Key: 9. Key:

A flat is shared by four first year undergraduate students. They agreed to allow the oldest of them to enjoy some extra space in the flat. Manu is two months older than Sravan, who is three months younger than Trideep. Pavan is one month older than Sravan. Who should occupy the extra space in the flat? (A) Manu (B) Sravan (C) Trideep (D) Pavan (C) Find the area bounded by the lines 3x+2y=14, 2x-3y=5 in the first quadrant. (A) 14.95 (B) 15.25 (C) 15.70 (D) 20.35 (B)

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2 22/12 2/14 7EC (2)

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14 A ,0 3 B 0, 7

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y

B

3x 2y 14

5 C , 0 2 5 D 0, 3 E 4,1

2x 3y 5

E R O

x C

A

Required area is area of OAB – area of CEA

10.

D

1 14 1 13 7 1 15.25 sq.units 2 3 2 6

A straight line is fit to a data set (ln x, y). This line intercepts the abscissa at ln x = 0.1 and has a slope of −0.02. What is the value of y at x = 5 from the fit?

Key:

(A) −0.030 (A)

(B) −0.014

Exp:

y a bx, where x l n x and

(C) 0.014

(D) 0.030

a 0.1, b 0.02 b a 0.002

a 0.02 x 0.002 0.02 x

at x 5, y 0.002 0.02 1.609 0.03018 0.030

Electronics and Communication Engineering Q. No. 1 – 25 Carry One Mark Each

1.

x ,

Consider a 2 × 2 square matrix A

Where x is unknown. If the eigenvalues of the matrix A are j and j , the x is equal to Key: Exp:

(A) j (B) j (D) Product of eigen values = det

(C)

(D)

j j 2 x 2 2 2 x x

2.

For f z

Key:

1

sin z z2

, the residue of the pole at z = 0 is __________.

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1 z z2 sin z 1 z3 z5 z ..... ...... z2 z2 3! 5! z 3! 5! 1 Resideu coefficient of 1 z

Exp:

f z

3.

The probability of getting a “head” in a single toss of a biased coin is 0.3. The coin is tossed repeatedly till a “head” is obtained. If the tosses are independent, then the probability of getting “head” for the first time in the fifth toss is . 0.07 Required probability = TTTTH 0.7 0.7 0.7 0.7 0.3

Key: Exp:

0.07203 1

4.

The integral

dx

1 x

is equal to _________.

0

Key:

2

Exp:

1 x 2 1 1 x 2 1 x dx 0 1 1 1 2 2 0

1

1

1

1

1

1 2

2 1 x

1 1 2 0

2

0

5.

Consider the first order initial value problem

Key:

with exact solution y x x 2 ex . For x = 0.1, the percentage difference between the exact solution and the solution obtained using a single iteration of the second-order Runge-Kutta method with stepsize h = 0.1is 0.06 to 0.063

6.

Consider the signal x t cos 6t sin 8t , where t is in seconds. The Nyquist sampling rate (in

y' y 2 x x3 , y 0 1, 0 x

samples/second) for the signal y(t) = x(2t+5) is (A) 8

(B) 12

(C) 16

(D) 32

Key:

(C)

Exp:

Shifting doesn‟t effects the sampling rate due to scaling by a factor „2‟ spectral components are doubled. X(f )

4 3

3

4

Thus maximum frequency of Y(f ) 8.

Nyquist rate 8 2 16Hz ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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If the signal x t

(A)

GATE-2015-PAPER-03

sin t sin t with * denoting the convolution operation, then x(t) is equal to * t t

sin t

sin 2t

(B)

t

Key:

(A)

Exp:

sin t 1 sa t t

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2t

(C)

2sin t t

sin t (D) t

2

F sa t rect 2 1 F sa t rect 2

Convolution in time domain leads to multiplication in frequency domain 1 sin t rect rect X x t sa t t 2 2

8.

A discrete-time signal x n n 3 2 n 5 has z-transform X(z). If Y(z) = X(-z) is the ztransform of another signal y[n], then (A) y[n] = x[n]

(B) y[n] = x[-n]

Key:

(C)

Exp:

x n n 3 2 n 5

(C) y[n] = -x[n]

(D) y[n] = -x[-n]

x z z 3 2z 5

x z z 2 2 3

5

y z z 3 2z 5 y n n 3 2 n 5 y n x n 9.

In the RLC circuit shown in the figure, the input voltage is given by

i t 2cos 200t 4sin 500t The output voltage 0 t is (A) cos (200t) + 2sin (500t) (B) 2 cos (200t) + 4sin (500t) (C) sin (200t) + 2cos (500t) (D) 2 sin (200t) + 4cos (500t) Key:

(B) ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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Vi(t) = 2cos 200t + 4 sin 500t, since there are 2 frequency term output will also have 2 frequency term. If we take 4sin500t first i.e. W = 500 then on the output section, this parallel LC combination have ZLC , so it is open circuit and V0 = Vi

j200

j200

So w.r.t. 4sin500t output must be 4sin500t without any change in amplitude and phase, this is satisfied by only option B. 10.

The I-V characteristics of three types of diodes at the room temperature, made of semiconductors X, Y and Z, are shown in the figure. Assume that the diodes are uniformly doped and identical in all respects except their materials. If EgX, EgY and EgZ are the band gaps of X, Y and Z, respectively, then I

X

Y

Z

V

(A) EgX > EgY > EgZ (C) EgX < EgY < EgZ

(B) EgX = EgY = EgZ (D) no relationship among these band gaps exists

Key:

(C)

11.

The figure shows the band diagram of a Metal Oxide Semiconductor (MOS). The surface region of this MOS is in

Key: Exp:

(A) inversion (A)

(B) accumulation

(C) depletion

(D) flat band

The semiconductor used in the MOSFET is n-type. At the surface the intrinsic level is above EF as it is found at the distance of below EF, So, the surface is in inversion region. ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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12.

The figure shows the I-V characteristics of a solar cell illuminated uniformly with solar light of power 100 mW/cm2. The solar cell has an area of 3 cm2 and a fill factor of 0.7. The maximum efficiency (in %) of the device is ________.

Key:

21

Exp:

Efficiency

13.

The diodes D1 and D2 in the figure are ideal and the capacitors are identical. The product RC is very large compared to the time period of the ac voltage. Assuming that the diodes do not breakdown in the reverse bias, the output voltage V0 (in volt) at the steady state is

Key:

0

Exp:

Vo 0volts

FF.Voc Isc 0.7 0.5 180 100% = 21%. Pin (100 3)

10V

Vo 0V

10V

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14.

Consider the circuit shown in the figure. Assuming VBE1 = VEB2 = 0.7 volt, the value of the dc voltage VC2 (in volt) is

Key:

0.5

Exp:

VE2 2.5 VBE1 1.8V VB2 VE2 0.7 1.1V 0.1 105 A 10k 50 105 A

I B2 I C2

VC2 5 104 103 0.5V

15.

In the astable multivibrator circuit shown in the figure, the frequency of oscillation (in kHz) at the output pin 3 is .

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Key:

5.64

Exp:

f

16.

In an 8085 microprocessor, the contents of the accumulator and the carry flag are A7 (in hex) and 0, respectively. If the instruction RLC is executed, then the contents of the accumulator (in hex) and the carry flag, respectively, will be

1.44 1.44 5.64kHz R A 2R B C [2200 24700] 0.022 106

(A) 4E and 0 Key:

(D)

Exp:

Accumulator

(B) 4E and 1

(C) 4F and 0

(D) 4F and 1

CY

0

1

0

1

0

0

1

1

1

RLCRotate left accumulator content without carry

Accumulator

CY 1

17.

0

1

0

0

1

1

1

4FH

1

The logic functionality realized by the circuit shown below is

(A) OR

(B) XOR

(C) NAND

(D) AND

Key:

(D)

Exp:

All the transistor are n-mos we know when input to gate is 0 n mos behave as open circuit. Input to Gate is 1 n mos is short circuit.

A

we can redraw the circuit as

B

T1 Y

B

T2

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to get the functionality get find its truth table A 0 0 1 1

B 0 1 0 1

T1 OFF ON OFF ON

T2 ON OFF ON OFF

Y 0 A0 0 A 1

So, Y satisfy AND gate table.

18.

The minimum number of 2-input NAND gates required to implement a 2-input XOR gate is (A) 4

Key:

(B) 5

(C) 6

(D) 7

(A)

Exp: A

Y

AB AB A B

B

19.

The block diagram of a feedback control system is shown in the figure. The overall closed-loop gain G of the system is

(A) G

G1G 2 1 G1H1

(B) G

G1G 2 1 G1G 2 G1H1

(C) G

G1G 2 1 G1G 2 H1

(D) G

G1G 2 1 G1G 2 G1G 2 H1

Key:

(B)

Exp:

G2

G1

G1 1 G1H1

H1 ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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G1G 2 Y x 1 G1H1 G1G 2

20.

For the unity feedback control system shown in the figure, the open-loop transfer function G(s) is given as G s

2 s s 1

The steady state error ess due to a unit step input is (A) 0

(B) 0.5

Key:

(A)

Exp:

For unit step input ess

(C) 1.0

(D) ∞

1 . 1 kp

2 s 0 s(s 1)

k p lim G(s) lim s 0

So, ess

1 0 1

21.

For a superheterodyne receiver, the intermediate frequency is 15 MHz and the local oscillator frequency is 3.5 GHz. If the frequency of the received signal is greater than the local oscillator frequency, then the image frequency (in MHz) is .

Key:

3485

Exp:

f image = ?

fS = f L0 (fSi - f L0 ) fSi = Image freq

f L0 = Signal freq fS - f L0 fSi + f L0

fSi 2f L0 fS fS f L0 f1F

fSi 2f L0 f1F f L0 f L0 f1F 3500 15 =3485 mHz ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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22.

An analog baseband signal, band limited to 100 Hz, is sampled at the Nyquist rate. The samples are quantized into four message symbols that occur independently with probabilities p1 = p4 = 0.125 and p2 = p3. The information rate (bits/sec) of the message source is

Key:

360 to 363

23.

A binary baseband digital communication system employs the signal

1 , 0 t TS p t TS 0, otherwise for transmission of bits. The graphical representation of the matched filter output y(t) for this signal will be (A)

(B)

(C)

(D)

Key:

(C)

Exp:

P(t) h(t) y(t)

h(t) = P(t S -t)

1

1 TS t

0

0

TS

2TS

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If a right-handed circularly polarized wave is incident normally on a plane perfect conductor, then the reflected wave will be (A) right-handed circularly polarized

(B) left-handed circularly polarized 0

(C) elliptically polarized with a tilt angle of 45

(D) horizontally polarized

Key:

(B)

Exp:

If incident wave is right handed polarized then the reflected wave is left handed polarized.

25.

Faraday‟s law of electromagnetic induction is mathematically described by which one of the following equations?

(A) .B 0

(B) .D V

B (C) E t

Key:

(C)

26.

Q. No. 26 – 55 carry Two Marks Each The particular solution of the initial value problem given below is

D (D) H E t

d2 y dy dy 12 36y 0 with y 0 3 and 36 2 dx dx dx x 0 6x (A) 3 18x e

Key:

(A)

Exp:

D2 12D 36 0

6x (B) 3 25x e

(C)

3 20x e6x

6x (D) 3 12x e

D 6, 6 C.F C1 C 2 x e 6x y C1 C 2 x e 6x y 0 3 3 C1 dy C 2 e 6x C1 C 2 x e 6x 6 dx dy 36 dx x 0 36 C 2 3 0 .1 6 C 2 18 y 3 18x e 6x

27.

If the vectors e 1 = (1, 0, 2), e 2 = (0, 1, 0) and e 3 = (−2, 0, 1) form an orthogonal basis of the threedimensional real space R3, then the vector u 4,3, 3 R 3 can be expressed as (A) u e1 3e2

2 5

11 e3 5

(B) u e1 3e2

2 5

11 e3 5

2 5

11 e3 5

(D) u e1 3e2

2 5

11 e3 5

(C) u e1 3e2

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Key:

(D)

Exp:

From option (D),

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2 11 e1 3e 2 e3 3 5 2 11 1,0, 2 3 0,1,0 2,0,1 5 5 4,3, 3

u

28.

A triangle in the xy-plane is bounded by the straight lines 2x = 3y, y = 0 and x = 3. The volume above the triangle and under the plane x + y + z = 6 is

Key:

10

Exp:

Volume =

y

z dx dy

2x 3y

R

3

B

2 x 3 y 0

6 x y dy dx

x 0

2

x

y2 3 6 x y .dx z y 0 x 0 3

2x 1 4 6 x 3 2 9 x

3

2

3

dx

x

O

A

2 2 2 2 4x x x dx 3 9 3

8 x3 8 2x 2 2 9 33 10 cubic units 9 3 27 0

29.

The values of the integral

xb

R

x limits : 0 to 3 2 y limits: 0 to x 3

1 ez dz along a closed contour c in anti-clockwise direction for 2j c z 2

(i) The point z0 = 2 inside the contour c, and (ii) The point z0 = 2 outside the contour c, Respectively, are (A) (i) 2.72, (ii) 0

(B) (i) 7.39, (ii) 0

Key:

(B)

Exp:

(i)

1 e2 1 2jf 2 e2 7.39 2j z 2 2j

(ii)

0

30.

(C) (i) 0, (ii) 2.72

(D) (i) 0, (ii) 7.39

2 t cos t is the input to an LTI system with the transfer function 3

A signal 2 cos

H s es es . If Ck denotes the kth coefficient in the exponential Fourier series of the output signal, then C3 is equal to (A) 0

(B) 1

(C) 2

(D) 3

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|EC| Key:

(B)

Exp:

output 2cos

GATE-2015-PAPER-03 2 t 2cos t 0 3 3

nd

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rd

2 harmonic 3 harmonic j 3t

In E.F.S 3rd harmonic 2cos t e 3 e

j 3t 3

j 3t

The coefficient of e 3 is1So C3 1. 31.

The ROC (region of convergence) of the z-transform of a discrete-time signal is represented by the shaded region in the z-plane. If the signal x n 2.0 , n , then the ROC of its z-transform is n

represented by

(A)

(B)

(C)

(D)

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Key:

(D)

Exp:

x n 2 , n

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n

x n 2n u n 2 n u n 1 2n u n ; ROC : z 2 2 n u n 1 ; ROC : z 0.5 Thus combined ROC does not exist for x[n] 32.

Assume that the circuit in the figure has reached the steady state before time t = 0 when the 3 resistor suddenly burns out, resulting in an open circuit. The current i(t) (in ampere) at t = 0+ is ______.

Key:

1

Exp:

At t 0, the circuit is on steady state i.e. the capacitor is open circuited so the circuit will be

V3F V2 V3 V3 4V V2F V3 6V at t = 0+ when is open circuited, the capacitors will have an ideal voltage source of values 4V and 6V so the circuit will be 2A

3F 2

1 2

12V 2F

So the current through 2 resistor at t = 0+ should be

3

4 1A 22

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2

4V

1

4V

1

2

12 12

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2

2

6V

6V

Burn out

at t 0

33.

In the figure shown, the current i (in ampere) is ______.

Key:

-1

Exp:

Nodal equation at V

1A

V 8 V V 8 V 0 1 1 1 1

8V

i1

a 1

4V 16 V 4V

1

v b 8V

By using KCL at node „a‟. 1

5

84 i1 0 i1 5A 1

0V

i

1

KCL at b

4 i1 i 0 4 5 i 0 1 i 1A

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|EC| 34.

GATE-2015-PAPER-03 z11 z12 for the two-port network shown in z 21 z 22

The z-parameter matrix

2 2 2 2

2 2 2 2

(A)

(B)

Key:

(A)

Exp:

Since the given network is symmetric

Z12 Z21 Z11 Z21

V1 I1

I2 0

V2 I1

I2 0

9 3 6 9

(D)

3

output

input

3 6 2 3 6

6 I1

We know V2 V1 Z21 2

Z11 So Z21

9 3 6 9

(C)

and reciprocal Z11 Z22

35.

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Z12 2 2 Z22 2 2

V1

V2

3

IL 0

A continuous-time speech signal xa(t) is sampled at a rate of 8 kHz and the samples are subsequently grouped in blocks, each of size N. The DFT of each block is to be computed in real time using the radix2 decimation-in-frequency FFT algorithm. If the processor performs all operations sequentially, and takes 20 s for computing each complex multiplication (including multiplications by 1 and −1) and the time required for addition/subtraction is negligible, then the maximum value of N is _____

Key:

4096

36.

The direct form structure of an FIR (finite impulse response) filter is shown in the figure. The filter can be used to approximate a

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(A) low-pass filter

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(B) high-pass filter

Key:

(C)

Exp:

y n 5x n 5x n 2

(C) band-pass filter

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(D) band-stop filter

H e j 5 1 e 2 j At 0, H e j 0 At ; H e j 0

Thus the filter is band pass filter 37.

The injected excess electron concentration profile in the base region of an npn BJT, biased in the active region, is linear, as shown in the figure. If the area of the emitter-base junction is 0.001 cm2,

n 800cm2 V s in the base region and depletion layer widths are negligible, then the collector

current Ic (in mA) at room temperature is __________.

(Given: thermal voltage VT = 26 mV at room temperature, electronic charge q 1.6 10 Key: Exp:

19

C)

6.65 Ic qAD n

1014 0 dn dn 1.6 1019 0.001 800 26 103 qA n Vt 4 dx dx 0.5 10

Ic 6.65mA ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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|EC| 38.

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Figures I and II show two MOS capacitors of unit area. The capacitor in Figure I has insulator materials X (of thickness t1 = 1 nm and dielectric constant 1 4 ) and Y (of thickness t2 = 3 nm and dielectric constant 2 20 ). The capacitor in Figure II has only insulator material X of thickness tEq. If the capacitors are of equal capacitance, then the value of tEq (in nm) is ___________.

Key:

1.6

Exp:

E1 E 2 . t1 t 2 E1E 2 4 20 CI 2.5 E1 E 2 E1t 2 E 2 t1 (4 2) (20 1) t1 t2

CII

39.

E1 4 2.5 t Eq 1.6nm t Eq t Eq

The I-V characteristics of the zener diodes D1 and D2 are shown in Figure I. These diodes are used in the circuit given in Figure II. If the supply voltage is varied from 0 to 100 V, then breakdown occurs in

(A) D1 only Key:

(B) D2 only

(C) both D1and D2

(D) none of D1 and D2

(A)

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For the circuit shown in the figure, R1 R 2 R 3 1, L 1H and C 1F. If the input

Vin cos 106 t , then the overall voltage gain Vout Vin of the circuit is ______.

Key:

-1

Exp:

R 1 A1 1 1 1 6 2 6 j L 10 10

A2

R3 R 2 XC 1

1 1 10 106

1 1 11 2

6

The overall voltage gain A v

Vout A1 A2 Vin

Vout 1 2 1 Vin 2 41.

In the circuit shown in the figure, the channel length modulation of all transistors is non-zero

0 .

Also, all transistors operate in saturation and have negligible body effect. The ac small

signal voltage gain V0 Vin of the circuit is

(A) gm1 r01 || r02 || r03

1 || r02 || r03 g m2

(C) g m1 r01 || Key:

1 || r03 g m3

1 || r03 || r02 g m3

(B) g m1 r01 ||

(D) g m1 r01 ||

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In the circuit shown in the figure, transistor M1 is in saturation and has transconductance gm = 0.01 siemens. Ignoring internal parasitic capacitances and assuming the channel length modulation to be zero, the small signal input pole frequency (in kHz) is .

Key:

57.9

Exp:

Cin 50PF 1 g m R 550PF g m 0.015 R 1k R 1k 1 g m R 11 R in 5k f in

43.

Key:

1 57.9 kHz 2R in Cin

Following is the K-map of a Boolean function of five variables P, Q, R, S and X. The minimum sumof-product (SOP) expression for the function is

(A) P QS X P QS X Q R S X QRSX

(B) Q SX Q S X

(C) Q SX Q S X

(D) QS QS

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44.

For the circuit shown in the figure, the delays of NOR gates, multiplexers and inverters are 2 ns, 1.5 ns and 1 ns, respectively. If all the inputs P, Q, R, S and T are applied at the same time instant, the maximum propagation delay (in ns) of the circuit is .

Key: Exp:

(6) Case (i) When T = 0 Ttotal = delay of NOR + delay of 1st MUX + delay of 2nd MUX= 2+1.5+1.5 = 5ns Case (ii) When T = 1 Ttotal = delay of 1st NOT-gate + delay of 1st MUX + delay of 2nd NOR-gate + delay of 2nd MUX = 1+1.5+2+1.5 = 6 ns So, the maximum delay = 6 ns.

45.

For the circuit shown in the figure, the delay of the bubbled NAND gate is 2 ns and that of the counter is assumed to be zero. If the clock (Clk) frequency is 1 GHz, then the counter behaves as a

(A) mod-5 counter

(B) mod-6 counter

(C) mod-7 counter

Key:

(D)

Exp:

The time period of clock is 1nsec and that of nand gate is 2n sec.

(D) mod-8 counter

The time period of clock 1n sec and that of nand gate is 2n.sec. ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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If the nand gate would have 0 delay then the counter must be MOD 6 because on 6 th CP it will be rejected. But as the delay is 2nsec, the effect of 6th CP will be observed on reset in after 2nsec, but in this additional 2n.sec, 2 more clock cycle would passed in total 8 cycles are needed to rest the counter so MOD8 counter.

46.

The first two rows in the Routh table for the characteristic equation of a certain closed-loop control system are given as

The range of K for which the system is stable is (A) −2.0 < K < 0.5 Key:

(D)

Exp:

S3 S2

(B) 0 < K < 0.5

(C) 0 K

(D) 0.5 K

1 2k 3 2k 4

From the table we can find characteristic equation s3 2ks2 2k 3 s 4 0

For stability 2k 2k 3 4 4k 2 6k 4 0 1 k k 2 0 2

So the conditions are k 47.

1 and k 2 combiningly k > – 2 2

A second-order linear time-invariant system is described by the following state equations

d x1 t 2x1 t 3u t dt d x2 t x2 t u t dt where x1(t) and x2(t) are the two state variables and u(t) denotes the input. If the output c(t) = x1(t), then the system is (A) controllable but not observable (B) observable but not controllable (C) both controllable and observable (D) neither controllable nor observable Key:

(A)

Exp:

The set of equation of the system are ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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dx1 t dt

dx 2 t

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2x1 t 3u t x i t 2x1 t 0x 2 t 3u t

x 2 t u t x 2 t 0 x1 t x 2 t u t

dt c t x1 t 0x 2 t

we can frame the state space of the system as

x 1 2 0 x 1 3 4 0 1 x 2 1 x2 x y 1 0 1 x2

2 0 A matrix is 0 1 3 B matrix is 1 C matrix is 1 0

for controllability determinant of B AB 0

3 6 1 1 3 6 3 0 so controllable

C 0 Ca

for observability determinant of

1 0 2 0 0 so not observable final controllable but not observable 48.

The forward-path transfer function and the feedback-path transfer function of a single loop negative feedback control system are given as

G s

K s 2 s 2 2s 2

and H s 1,

Respectively, If the variable parameter K is real positive, then the location of the breakaway point on the root locus diagram of the system is . Key:

-3.414

Exp:

To find break point, from characteristic equation we need to arrange k as function of s, then the root of dk 0 gives break point. ds Characteristic equation is given by ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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s2 2s 2 k 2k 0 k s 2 s 2 2s 2 s 2 2s 2 k s2 2 d d 2 s 2 ds s 2s 2 s 2s ds s 2 dk ds s 2 2

s 2 2s 2 s 2 2s 2 dk ds s 2 2 dk 0 ds

2s2 2s 4s 4 s 2 2s 2 0 s 2 4s 2 0 s 0.58 and 3.414

j 1

2

j

49.

To find the valid break point we need to find that lies on root locus

– 3.414 lies on root locus

So break point – 3.414.

A wide sense stationary random process X(t) passes through the LTI system shown in the figure. If the

autocorrelation function of X(t) is R X , then the autocorrelation function R Y of the output Y(t) is equal to

Key:

(A) 2R X R X T0 R X T0

(B) 2R X R X T0 R X T0

(C) 2R X 2R X 2T0

(D) 2R X 2R X 2T0

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26 2626/ 12626 /14 7EC

|EC| Exp:

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R YY (τ) = E[Y(t)Y(t - τ)] E[{X(t)-X(t - T0 )}{X(t-τ)-X(t -τ T0 )}] = R X (τ) - R X (-τ-T0 )-R X (-T+T0 )+R X (-τ)

R YY (τ) = 2R X (τ) - R X (τ + T0 )-R X (τ-T0 ) 50.

A voice-grade AWGN (additive white Gaussian noise) telephone channel has a bandwidth of 4.0 kHz and two-sided noise power spectral density

2.5 105 Watt per Hz. If information at the rate 2

of 52 kbps is to be transmitted over this channel with arbitrarily small bit error rate, then the minimum bit-energy Eb (in mJ/bit) necessary is . Key:

31.5

Exp:

Information rate

R C

Channel capacity

For Error free transmission

S C = B log 2 1 N B = 4KHz

S = E b / Tb E b R b

N=

η .2B = ηB 2

S R b B log 2 1 N

ER R b B log 2 1+ b b ηB E b 31.5mJ/bit E bmin 31.5mJ/bit 51.

The bit error probability of a memoryless binary symmetric channel is 10−5. If 105 bits are sent over this channel, then the probability that not more than one bit will be in error is _________.

Key:

0.7 to 0.75 ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test Series Leaders in GATE Preparations 65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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52.

Consider an air-filled rectangular waveguide with dimensions a = 2.286 cm and b = 1.016 cm. At 10 GHz operating frequency, the value of the propagation constant (per meter) of the corresponding propagating mode is _________.

Key:

157

53.

Consider an air-filled rectangular waveguide with dimensions a = 2.286 cm and b = 1.016 cm. The increasing order of the cut-off frequencies for different modes is

Key:

(A) TE01 < TE10 < TE11 < TE20

(B) TE20 < TE11 < TE10 < TE01

(C) TE10 < TE20 < TE01 < TE11

(D) TE10 < TE11 < TE20 < TE01

(C) 2

Exp:

2

C m n fc 2 a b C 1 C for TE 01 f c 98.4 2 2 1.016 10 2 for TE10 fc

C 1 C 43.74 2 2 2.286 10 2

C 2 C 2 C fc 87.49 2 2 2.286 10 2

For TE11 f c 107.7 For TE20

TE10 < TE20 < TE01 < TE11 54.

A radar operating at 5 GHz uses a common antenna for transmission and reception. The antenna has a gain of 150 and is aligned for maximum directional radiation and reception to a target 1 km away having radar cross-section of 3 m2. If it transmits 100 kW, then the received power in W is

Key:

__________. 0.012

55.

Consider the charge profile shown in the figure. The resultant potential distribution is best described by

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GATE-2015-PAPER-03

(A)

(B)

(C)

(D)

Key:

(D)

Exp:

q q Electrical 0 N d x no N d x po

Potential (x) 0

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x2 x 2w 0

P(x) V(x)

P1

b

a

b

a

P1 P1

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