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EC-GATE 2018

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2

Section-I: General Ability

1. “By giving him the last ______of the cake, you will ensure lasting ____in our house today.”

The words that best fill the blanks in the above sentence are

(A) peas, piece (B) piece, peace (C) peace, piece (D) peace, peas

Key: (B)

2. “Even though there is a vast scope for its _____, tourism has remained a/an____ area.”

The words that best fill the blanks in the above sentence are

(A) improvement, neglected (B) rejection, approved

(C) fame, glum (D) interest, disinterested

Key: (A)

3. What is the value of 1 1 1 1

1 ....?4 16 64 256

(A) 2 (B) 7

4 (C)

3

2 (D)

4

3

Key: (D)

Exp: Given series is 1 1 1 1

1 ........ ... 14 16 64 256

Clearly, (1) is infinite Geometric series (∴ The first term a =1, common ratio (r)=1/4)

∴ The sum of infinite Geometric series

1 1 1 1 1 41 .......

14 16 64 256 31

4

(∵sum of infinite geometric series

a

1 r

)

4. If the number 715 ? 423 is divisible by 3 (? Denotes the missing digit in the thousands place), then

the smallest whole number in the place of ? is ______.

(A) 0 (B) 2 (C) 5 (D) 6

Key: (B)

Exp: We know that, if the sum of the digits is a multiple of 3 then the number is divisible by 3

∴ 7 1 5 x 4 2 3 22 x where 'x' is the missing digit

If x=0; then 22 is not multiple of 3

If x=2; then 24 is a multiple of 3

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If x=5; then 27 is a multiple of 3

If x=6; then 28 is not multiple of 3

∴ The minimum value of x=2.

5. A 1.5 m tall person is standing at a distance of 3 m from a lamp post. The light from the lamp at the

top of the post casts her shadow. The length of the shadow is twice the height. What is the height of

the lamp post in meters?

(A) 1.5 (B) 3 (C) 4.5 (D) 6

Key: (B)

Exp: B D ( Since lamp post as wellas the manarestanding vertical to the ground)

E E

le leso ABE CDE AA Similarity criterion

AB BE X 6X 3m.

CD DE 1.5 3

6. A coastal region with unparalleled beauty is home to many species of animals. It is dotted with

coral reefs and unspoilt white sandy beaches. It has remained inaccessible to tourists due to poor

connectivity and lack of accommodation. A company has spotted the opportunity and is planning to

develop a luxury resort with helicopter service to the nearest major city airport. Environmentalists

are upset that this would lead to the region becoming crowded and polluted like any other major

beach resorts.

Which one of the following statements can be logically inferred from the information given in the

above paragraph?

(A) The culture and tradition of the local people will be influenced by the tourists.

(B) The region will become crowded and polluted due to tourism.

(C) The coral reefs are on the decline and could soon vanish.

(D) Helicopter connectivity would lead to an increase in tourists coming to the region.

Key: (B)

7. A cab was involved in a hit and run accident at night. You are given the following data about the

cabs in the city and the accident.

(i) 85% of cabs in the city are green and the remaining cabs are blue.

(ii) A witness indentified the cab involved in the accident as blue.

(iii) It is known that a witness can correctly identify the cab colour only 80% of the time.

Which of the following is closest to the probability that the accident was caused by a blue cab?

(A) 12% (B) 15% (C) 41% (D) 80%

B D

C

E

A

1.5m

xLampheight

3m 3m

man height

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Key: (A)

Exp: P [ Green cab] = P[G] = 85%, P [Blue cab] =P[B] = 15%

Let A denote that a witness can correctly identify the cab colour as blue.

Given

A A

P 80%,P 20%B G

By Baye‟s theorem we have

P B P A BP B A

P B P A B P G P A G

15% 80%

15% 80% 85% 20%

0.41 41%

8. Leila aspires to buy a car Rs. 10,00,000 after 5 years. What is the minimum amount in Rupees that

she should deposit now in a bank which offers 10% annual rate of interest, if the interest was

compounded annually?

(A) 5,00,000 (B) 6,21,000 (C) 6,66,667 (D) 7,50,000

Key: (B)

Exp: Let Principal = x

Time period = 5 years

R=10% per year

After 5 years, Amount = P + Interest =10,000,00

Principal =x = 2

By compound interest, we have

510

Amount P 1100

121 121 11110 110 110 110 11010,00,000 P P

100 100 100 100 100 100000

10,000,00 1,00,000P 6,20,921 6,21,000

121 121 11

9. Two alloys A and B contain gold and copper in the ratios of 2:3 and 3:7 by mass, respectively.

Equal masses of alloys A and B are melted to make an alloy C. The ratio of gold to copper in alloy

C is _____.

(A) 5:10 (B) 7:13 (C) 6:11 (D) 9:13

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Key: (B)

Exp: Given

A B

2:3 3: 7

G C G C

Total parts in A =5 Total parts in B =10

The part of gold in A=2/5 Thepart of gold in B=3/10

Thepart of copper in A =3/5 Thepart of copper in B =7/10

We assume that, the equal quantities of A & B = L.C.M of 5 and 10 =10.

∴ The required ratio of gold & copper in C is

BB AA

G : C

2 3 3 7 4 3 : 6 7 7 :1310 10 : 10 105 10 5 10

10. The Cricket Board has long recognized John‟s potential as a leader of the team. However, his on-

field temper has always been a matter of concern for them since his junior days. While this

aggression has filled stadia with die-hard fans, it has taken a toll on his own batting. Until recently,

it appeared that he found it difficult to convert his aggression into big scores. Over the past three

seasons though, that picture of John has been replaced by a cerebral, calculative and successful

batsman-captain. After many years, it appears that the team has finally found a complete captain.

Which of the following statements can be logically inferred from the above paragraph?

(i) Even as a junior cricketer, John was considered a good captain.

(ii) Finding a complete captain is challenge.

(iii) Fans and the Cricket Board have differing views on what they want in a captain.

(iv) Over the past three seasons John has accumulated big scores.

(A) (i), (ii) and (iii) only (B) (iii) and (iv) only

(C) (ii) and (iv) only (D) (i), (ii), (iii) and (iv)

Key: (C)

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Section-II: Electronics & Communication Engineering

1. A binary source generates symbols X { 1,1} which are transmitted over a noisy channel. The

probability of transmitting X = 1 is 0.5. Input to the threshold detector is R = X + N. The probability

density function fN(n) of the noise N is shown below.

If the detection threshold is zero, then the probability of error (correct to two decimal places)

is____.

Key: (0.125)

Exp: Given

P x 1 P x 1 0.5, threshold is 0V

When x =1 is transmitted, error is going to take place if received voltage is less than 0V.

When x= -1 is transmitted, error is going is take place if received voltage is greater than 0V.

error x 1 x 1P P R 0 .P x 1 P R 0 .P x 0

Given that R= X+M, where M is a Random noise

error x 1 x 1P P x N 0 P x 1 P R 0 P x 1

P M 1 .P x 1 P M 1 P x 1

Using pdf figure,

Shaded portion represent the P M 1 and P M 1

e

1 1 1P 1 0.25 1 0.25 0.125

2 2 2 2

Nf (n)

0.5

22 n

Mf x

0.5

P M 1 P M 1

0.25 0.25

1 1 2 n2

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2. In the circuit shown below, the op-amp is ideal and Zener voltage of the diode is 2.5 volts. At the

input, unit step voltage is applied, i.e., IN (t) u(t) volts. Also, at t = 0, the voltage across each of

the capacitors is zero.

The time t, in milliseconds, at which the output voltage OUT crosses 10V is

(A) 2.5 (B) 5 (C) 7.5 (D) 10

Key: (C)

Exp: Given, INV t u t , also, t = 0 voltage across each capacitor is zero.

z

CC C

V 2.5V

C.V 1m ti ; V

t 1

Voltage across at second capacitor is 2.5,

By K.V.L, C1 0 0

C1

V 2.5 V 0, given V 10V

V 7.5

1t 7.5 7.5msec

1m

3. Consider the following amplitude modulated signal:

s(t) cos(2000 t) 4cos(2400 t) cos(2800 t)

The ratio (accurate to three decimal places) of the power of the message signal to the power of the

carrier signal is____.

Key: (*)

Exp: Given,

cos 2000 t 4cos 2400 t cos 2800 ts t

side band carrier sideband

4cos 2400 t 2cos 400 t cos 2400 t

t 0

1V

0V

1Nv t

1k

1 F

1 F

OUTv (t)

1Nv t

1k

1 F

1 F

OUTv (t)

EC-GATE 2018


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1

4 1 cos 400 t cos 2400 t2

Now at this point, some data is missing,

if we compare this with standard AM signal equation

c a

a

a

a

s t A 1 k m t cos 2400 t

if k 1 2, m t cos 400 t , PM 1 2

1if k 1, m t cos 400 t , PM 1 8

2

1if k 2, m t cos 400 t , PM 1 32

4

Like this infinite possibilities exist

Hence many answers are possible.

4. A p-n step junction diode with a contact potential of 0.65 V has a depletion width of 1 m at

equilibrium. The forward voltage (in volts, correct to two decimal places) at which this width

reduces to 0.6 m is____.

Key: (0.416)

Exp: We know that in pn-junction bi fW V V

1 f

2 f

1

2

Given, W 1 m ,V 0

W 0.6 m,V ?

W 5 0.65

W 3 0.65 x

25 250.65 x 0.65

9 9

x 0.416

5. The points P, Q, and R shown on the Smith chart (normalized impedance chart) in the following

figure represent:

(A) P: Open Circuit, Q: Short Circuit, R: Matched Load

(B) P: Open Circuit, Q: Matched Load, R: Short Circuit

(C) P: Short Circuit, Q: Matched Load, R: Open Circuit

(D) P: Short Circuit, Q: Open Circuit, R: Matched Load

Key: (C)

Exp: Constant resistance circle are given by

2 2

2r 1x y

1 r 1 r

P Q R

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r 1

center ,0 , radius ,1 r 1 r

r is normalized resistance constant reactance circle are given by

2

2

2

1 1x 1 y

x x

, x in normalized reactance

P is short circuit R in open circuit, Q is matched load.

At point P, normalized reactance is zero

At point Q, normalized reactance is zero

At point R, normalized reactance is zero

6. A lossy transmission line has resistance per unit length R = 0.05 / m. The line is distortionless and

characteristic impedance of 50 . The attenuation constant (in Np/m, correct to three decimal

places) of the line is_____.

Key: (0.001)

Exp: Given R 0.05 m.

Condition for distortion less transmission line

2

o

R LRC LG

G C

R50 50 R G

G

L Lz 50 50

C C

Alternation constant RG

0.05 0.05 0.05

0.00150 50 50

7. Three are two photolithography systems: one with light source of wavelength 1 156nm (System 1)

and another with light source of wavelength 2 325 nm(System 2). Both photolithography system

are otherwise identical. If the minimum feature sizes that can be realized System 1 and System 2 are

min1 min 2L and L respectively, the ratio min 1 min 2L / L (correct to two decimal places) is ____.

Key: (0.48)

Exp: 1 min

CD K LNA

CD is the minimum feature size

NA=Numerical aperture

min1 11 2 1 2

min 2 2

L 156k k and NA NA 0.48

L 325

Qr 5

r 1

R

r

xP

r 0

y

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8. Let x(t) be a periodic function with period T = 10. The Fourier series coefficients for this series are

denoted by ak, that is

2

jk tT

k

k

x(t) a e

The same function x(t) can also be considered as periodic function with period T 40. Let kb be

the Fourier series coefficients when period is taken as T . If kka 16,

then kk

b

is

equal to

(A) 256 (B) 64 (C) 16 (D) 4

Key: (C)

Exp: Only for case study let us consider the periodic waveform x(t) as a square wave.

2

jk tT

k

k

x t a e

We know ak is the exponential Fourier series coefficient and is given by

0

T

jk t

k

0

1a x t e dt

T

Since the waveform repeat itself, we can say if we evaluate ak using the period, as any integer

multiple of To like 2To, 3To…… the coefficient will not change, we can notice this in the above

integration.

So we can say,

if T=10, the E.F.S coefficient is ka , then

if T=40, the E.F.S coefficient is k kb a

Since, k k k k

k k

a b a b 16

For further more clarity, if we find the D.C. component (k=0) of x(t), then if we consider

T 10, T 20, T 40, etc, the value of oa remains same.

A

x(t)

T 10

T' 40

t

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9. Let M be a real 4 × 4 matrix. Consider the following statements:

S1: M has 4 linearly independent eigenvectors

S2: M has 4 distinct eigenvalues.

S3: M is non-singular (invertible)

Which one among the following is TRUE?

(A) S1 implies S2 (B) S1 implies S3

(C) S2 implies S1 (D) S3 implies S2

Key: (C)

Exp: We know that if a matrix A has „n‟ distinct eigen values then A has „n‟ linearly independent

eigenvectors.

S2 implies S1.

10. A discrete-time all-pass system has two of its poles at 0.25 0 and 2 30 . Which one of the

following statements about the system is TRUE?

(A) It has two more poles at 0.5 30 and 4 0 .

(B) It is stable only when the impulse response is two-sided.

(C) It has constant phase response over all frequencies.

(D) It has constant phase response over the entire z-plane.

Key: (B)

Exp: For the given all pass system the 2 poles are located at z 0.25 0 and z 2 30

If we are interested for stability of the given system, then the associated ROC must include the unit

circle as shown in the figure.

For the above highlighted ROC, with respect to the pole z 0.25 0 , the ROC is outside the circle.

So its impulse response will be right sided.

xP 2 30

x

P 0.25 0

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Similarly for the pole z 2 30 , the ROC is inside the circle of radius 2, so its impulse response is

left sided, will contain u n 1 term. So, overally the impulse response should be both sided for

the given stability.

So Option B is Correct.

By observing the pole pattern of an All Pass System, we can comment the location of zeros. But we

never can comment regarding rest of the poles, if present. Hence option A is wrong.

We know for an All Pass System, the magnitude response is constant for all frequency or in entire

z-plane but the phase response changes, Hence option C and option D are wrong.

11. A traffic signal cycles from GREEN to YELLOW. YELLOW to RED and RED to GREEN. In each

cycle, GREEN is turned on for 70 seconds. YELLOW is turned on for 5 seconds and the RED is

turned on for 75 seconds. This traffic light has to be implemented using a finite state machine

(FSM). The only input to this FSM is a clock of 5 second period. The minimum number of flip-

flops required to implement this FSM is____.

Key: (5)

Exp: Green 70 seconds

Yellow 5 seconds

Red 75 seconds

So to complete one cycle i.e.,

Green yellowRedGreenyellowred…….

Then required time is 70+5+75=150 sec.

Since clock period is 5 sec; the number of clock needed for one cycle of 150 sec is 30, so we

can regard the FSM as MOD 30 counter.

n2 30 minn 5

so, minimum 5 flip flops are required.

12. Two identical nMOS transistor 1 2M and M are connected as shown below. The circuit is used as

amplifier with the input connected between G and S terminals and the output taken between D and

S terminals. Vbias and VD are so adjusted that both transistors are in saturation. The

transconductance of this combination is defined as Dm

GS

ig

v

while the output resistance is

DS0

D

vr ,

i

where iD is the current flowing into the drain of M2. Let m1g , gm2 be transconductances

and 01 02r , r be the output resistance of transistor M1 and M2, respectively.

Which of the following statements about estimates for gm and r0 is correct?

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(A) m m1 m2 02 0 01 02g g g .r and r r r .

(B) m m1 m2 0 01 02g g g and r r r .

(C) m m1 0 01 m2 02g g and r r .g .r .

(D) m m1 0 02g g and r r

Key: (C)

Exp: DSDm o

rms D

vig r

I

The given circuit is MOS cascode

om

i

ig

v

AC equivalent circuit of MOS cascade is

Writing a node equation for the node, we have

gs2 gs2

m2 gs2 m1 i

o1 o2

V Vg V g v ... 1

r r

Observe that the voltage 1, 2 gs2d s v

Equation (1) can be simplified as

m2 gs2 m i

o1 o2

1 1g v g v ... 2

r r

Since m2

o1 o2

m gs2 m1 i

1 1g ,

r r

g v g v ...(3)

Node equation at node d2 is

gs2

0 m2 gs2 m2 gs2

o2 o2

0 m2 gs2

v 1i g v g v

r r

i g v ... 4

DV

biasV

G

S

2M

1M

D

gs1 iV vm1 ig v o1r

gs2V

m2 gs2g v o2r

1 2d s

2dDi

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From equation (3) and (4)

0 Dm m1

i gs

i Ig g

v v

Similarly gsx

0

x D

vvR

I I

Voltage at the source node, which is gs2v can expressed in terms of xi is

gs2 x o1

x x m2 gs2 o2 x o1

v i r and ... 1

v i g v r i r ... 2

Substituting (1) in (2)

x x o1 o2 m2 o2 o1

o o1 o2 o1 o2 m2

v i r r g r r

R r r r r g

In above expression the last term will dominate, thus o m2 o2 o1R g .r r

13. Consider matrix 2 2

k 2kA

k k k

and vector

1

2

xx .

x

The number of distinct real values of k for

which the equation Ax = 0 has infinitely many solutions is____.

Key: (2)

Exp: Given that AX=0 has infinitely many solutions.

2 2

3 2

3 3 2 3 2

2

Then A 0

K 2K0

K K K

K 2K K K 0

K 2K 2K 0 K 2K 0

K K 2 0 K 0,0,2

Number of distinct values of K = 2. [i.e., 0 and 2].

gs2Vm2 gs2g v o2r

2s

xi

o1r

2v

xi

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14. Considered a binary channel code in which each codeword has a fixed length of 5 bits. The

Hamming distance between any pair of distance codewords in this code is at least 2. The maximum

number of codewords such a code can contain is ____.

Key: (16)

15. Taylor series expansion of f(x)

2tx 2

0e dt

around x = 0 has the form

2

0 1 2f (x) a a x a x .....

The coefficient a2 (correct to two decimal places) is equal to _____.

Key: (0)

Exp: 2x

t 2

0

f x e dt

2

2 2

x 2

x 2 x 2

'f x e

2x ''''f x e x.e f 0 02

By Taylor‟s series, of f(x) around x=0, we have

2x' ''f x f 0 x f 0 f 0 ........

2! ; where the coefficient of

2''f 0

x2!

But given the coefficient of 2

2x in f x a

2

2 2

''f 0a

2!

0''a 0 f 0 0 a 0.

2!

16. A good transimpedance amplifier has

(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

Key: (D)

Exp: Trans impedance amplifier contain, input of Norton's circuit and output of Thevenin circuit

A good transimpedance amplifier has low input impedance

sI sR iR

1I

m 1R I

oR

LR

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17. Consider 3 2

2 1 0p(s) s a s a s a with all real coefficients. It is known that its derivative p'(s) has

no real roots. The number of real roots of p(s) is

(A) 0 (B) 1 (C) 2 (D) 3

Key: (B)

Exp: Given p'(s) has no real roots.

Discriminant of p'(s) < 0.

p(s) will have exactly one real root.

18. Let 1 2 3 4X ,X , X and X be independent normal random variables with zero mean and unit variance.

The probability that X4 is the smallest among the four is _____.

Key: (0.25)

Exp: Since 1 2 3 4X ,X ,X and X are independent normal random variables with zero mean and unit

variance, then

1 2 3

4

P X is the smallest P X is the smallest P X is the smallest

1P X is the smallest 0.25.

4

19. The ABCD matrix for a two-port network

is defined by:

1 2

1 2

V VA B

I IC D

The parameter B for the given two-port network (in ohms, correct to two decimal places) is ____.

Key: (4.8)

Exp: The ABCD matrix equation is

1 2 2

1 2 2

V AV BI

I CV DI

If we make output port short i.e. 2V 0, then

1

2

VB

I

Making the output short, the network becomes

2 2

5

1I 2I

1V 2V

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Writing KVL on the input loop

1 1 2

1 1 2

V 2I 2I 0

V 2I 2I ... 1

Writing KCL at node X

1 2 2

1 2 2

1 2

2I I I 0

5

I 0.4I I

I 1.4I ... 2

Putting equation (2) in equation (1)

1 2 2

2 2

1

2

V 2 1.4I 2I

2.8I 2I

VB 4.8

I

20. In a p-n junction diode at equilibrium, which one of the following statements is NOT TRURE?

(A) The hole and electron diffusion current components are in the same direction.

(B) The hole and electron drift current components are in the same direction.

(C) ON an average, holes and electrons drift in opposite direction.

(D) ON an average, electrons drift and diffuse in the same direction.

Key: (D)

Exp: Since the concentration of holes on the p side is much greater than that in the n side, a very large

hole diffusion current tends to flow across the junction from the p to the n material. Hence an

electric field must build up across the junction in such a direction that a hole drift current will tend

to flow across the junction from n to p side in order to counterbalance the diffusion current.

2 2

5

1I

2I1V

X

22I

2

2I

5

22I

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21. The logic function f(X,Y) realized by the given circuit is

(A) NOR (B) AND (C) NAND (D) XOR

Key: (D)

Exp: If we focus the N-MOS section

f X.Y X Y XY X Y X Y

22. The Nyquist stability criterion and the Routh criterion both are powerful analysis tools for

determining the stability of feedback controllers. Identify which of the following statements is

FALSE:

(A) Both the criteria provide information relative to the stable gain range of the system.

(B) The general shape of the Nyquist plot is readily obtained from the Bode magnitude plot for all

minimum-phase systems.

(C) The Routh criterion is not applicable in the condition of transport lag, which can be readily

handled by the Nyquist criterion.

(D) The closed-loop frequency response for a unity feedback system cannot be obtained from the

Nyquist plot.

DDV

X

YY

X

f (X,Y)

X

Y

X

Y

XY XY

f X,Y

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Key: (B)

Exp: Analysis of option A

By observing the encirclement about the point -1+j0 in Nyquist plot we find the range of k for

which system will be stable similarly in Routh criterion by observing the sign change we find the

range of k for which system will be stable

Hence option A represents a correct statement.

Analysis of option B

Shape of Nyquist plot can‟t be commented readily form bode plot, first we have to find the

transfer function and by using this we have to draw the Nyquist plot, so readily we can‟t comment.

Hence option B represents a false statement.

Analysis of option C

For transportation log se Nyquist plot is suitable as its magnitude is 1 for all frequency and

phase is , so no approximation required but in Routh criterion we need to do approximation

first, so its not suitable.

Hence option C represents a correct statement.

Analysis of option D

With the help of Nyquist plot we cannot comment closed loop frequency response for this we need

M, N circles, Nichols chart etc.

Hence option D represents a correct statement.

23. A function F(A, B, C) defined by three Boolean variables A, B and C when expressed as sum of

products is given by

F A.B.C A.B.C A.B.C

Where, A, B, and C and the complements of the respective variables. The product of sums (POS)

form of the function F is

(A) F (A B C). A B C . A B C

(B) F A B C . A B C . A B C

(C) F (A B C).(A B C).(A B C). A B C .(A B C)

(D) f A B C . A B C .(A B C).(A B C).(A B C)

Key: (C)

Exp:

F A.B.C ABC A B C A B C m 0,2,4

M 1,3,5,6,7

A B C A B C A B C A B C A B C

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24. Let the input be u and the output be y of a system, and the other parameters are real constants.

Identify which among the following systems is not a linear system:

(A) 3 2 2

1 2 3 3 2 13 2 2

d y d y dy du d ua a a y b u b b

dt dt dt dt dt (with initial rest conditions)

(B) t

(t )

0y(t) a u( )d

(C) y = au + b, b 0

(D) y = au

Key: (C)

Exp: If we observe all the 4 options carefully, we can notice option C (i.e., y au b, b 0 ) is non

linear system due to the addition of non-zero constant b, since it does not satisfy homogeneity and

additivity as follows

1 2 1 2au b au b a u u b

a ku b k au b

Option A is linear differential equation with constant co-efficient, so it is linear.

Integrator system is generally linear, so option B represents a linear system.

y au, is a perfect linear system.

25. Let f(x,y) 2 2ax by

,xy

where a and b are constants. If

f f

x y

at x = 1 and y = 2, then the relation

between a and b is

(A) b

a4

(B) b

a2

(C) a 2b (D) a = 4b

Key: (D)

Exp: 2 2 2 2ax by ax by

Let f x,yxy xy xy

2 2

x 1,y 2

2 2

x 1,y 2

ax byf x, y

y x

f a by f ax b&

x y x y y x

f fGiven

x y

a by ax b

y x y x

EC-GATE 2018


21

a a2b b

2 4

a a2b b

2 4

2a a 4 3b

3a 12b a 4b.

26. In the circuit shown below, the (W/L) value for M2 is twice that for M1. The two nMOS transistors

are otherwise identical. The threshold voltage VT for both transistors is 1.0 V.

Note that VGS for M2 must be > 1.0 V.

Current through the nMOS transistors can be modeled as

2

DS ox GS T DS DS

W 1I C V V V V

L 2

for DS GS TV V V

2

DS ox GS T

WI C V V / 2

L

for DS GS TV V V

The voltage (in volts, accurate to two decimal places) at Vx is ________.

Key: (0.4266)

Exp: Let n oxn

C WK

2 L

2 1

n2 n1

1 GS1 T

s GS2 T x x

DS2 x GS2 T

W WGiven that 2

L L

So, K 2K

For M , V V 2 1 1V

For M , V V 2 V 1 1V V 1V

V 3.3 V V V

So, M1 will be in linear region and M2 will be in saturation region.

3.3V

2M

1M

xV

2.0V

EC-GATE 2018


22

D1 D2

22

n1 GS1 T DS1 DS1 n2 GS2 T

22

n1 x x n1 x

2 2 2

x x x x x x

2 2

x x x

I I

K 2 V V V V K V V

K 2 2 1 V V 2K 2 V 1

2V V 2 1 V 2V 2V 4V 2

23V 6V 2 0; V 2V 0

3

x

GS2 x T x

x

84

13V 1 1 V4 3

V 2 V V 1 V

1V 1 0.4226V

3

27. A junction is made between p- Si with doping density 15 3

A1N 10 cm and p Si with doping density 17 3

A2N 10 cm .

Given: Boltzmann constant k = 1.38 × 10–23 J.K–1, electronic charge q = 1.6 × 10–19 C.

Assign 100% acceptor ionization.

At room temperature (T = 300 K), the magnitude of the built-in potential (in volts, correct to two

decimal places) across this junction will be_____.

Key: (0.1192)

Exp: Setting JP = 0

TV dPE

P dx

T

121 2 1 T

2

dP dVdV V E

dx dx

PV V V V ln

P

Built in potential,

A2bi

A1

NkT 1.38 3V n n 100 V 0.1192V

q N 1.6 100

28. In the circuit shown below, a positive edge-triggered D Flip-Flop is used for sampling input data Din

using clock CK. The XOR gate output 3.3 volts for logic HIGH and 0 volts for logic LOW levels.

The data bit and clock periods are equal and the value of CKT / T 0.15, where the parameters

CKT and T are shown in the figure. Assume that the Flip-Flop and the XOR gate are ideal.

1V 2V

2P1P

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23

If the probability of input data bit (Din) transition in each clock period is 0.3, the average value (in

volts, accurate to two decimal places) of the voltage at node X, is_____.

Key: (0.8415)

Exp: If

P1: Probability of Output xV 1 ON state

P0: Probability of Output xV 0 OFF state

VX1: Voltages corresponding to logic 1 i.e. 3.3V

VX0: Voltage corresponding to logic i.e. 0 V

TON: The period when Output xV 1

TOFF: The period when Output xV 0

Then we can say the average value of xV is

1 0x 1 x ON 0 x OFFavg

V P V T P V T

The above relation is somewhat similar to the average probability of error calculation in digital

modulation Schemes.

Let us draw the wave form of xV and find the unknown parameter of the above equation.

Form the given logic circuit we can say that the Output of the ex or Din and previous state of

the D flip flop, because the elements are given as ideal.

inD

CK

D Q

DFlip-Flop

CLK

X

CK

inD

CKT

T T T

T T T T

CLKT

inD

3.3V

CLK

T

T

CLK

T1

T

x

in

When there

V is a transition

of D

EC-GATE 2018


24

1 ox 1 x ON 1 x OFFavg

CLK CLK

V P V T P V T

T T0.3 3.3 1 0.7 0

T T

0.3 3.3 1 0.15 0.7 0

0.3 3.3 0.85 0.8415 V.

29. The circuit shown in the figure is used to provide regulated voltage (5V) across the 1k resistor.

Assume that the Zener diode has a constant reverse breakdown voltage for a current range, starting

from a minimum required Zener current, minZI 2mA to its maximum allowable current. The input

voltage VI may vary by 5% from its nominal value of 6V. The resistance of the diode in the

breakdown region is negligible.

The value of R and the minimum required power dissipation rating of the diode, respectively, are

(A) 186 and 10 mW (B) 100 and 40 mW

(C) 100 and10mW (D) 186 and 40mW

Key: (B)

Exp: zV 6V 5% 6V 0.3V 5.7V to 6.3V

L

Ls min Z min

I zs

zI min

s min

max

max

s max

Lz max s max

zz min z max

5VI 5mA

1k

I I I 5mA

V VI ;

R

V VI 7mA

R

5.7 5So, R 100

7

6.3 5I 13mA

100

I I I 13mA

P V I 5 8mW 40mW

R

1kIV

5V

R

1k1V5V

I2I

zI

EC-GATE 2018


25

30. For the circuit given in the figure, the magnitude of the loop current (in amperes, correct to three

decimal places) 0.5 second after closing the switch is _____.

Key: (0.316)

Exp:

We can redraw the above circuit as

t /

L L L Li t i i 0 i e ; t 0

At t 0 , switch is in open state, so the loop is not closed hence, Li 0 0A

At t switch is closed and inductor is short Circuited

Time constant tn

L

R

1

1H1

1V

2

1 V

L

1i A

2

2

1nR 2

2

1H1 V

t 0

Li t

assume switchingis happening at t 0

1

1H1

1V

t 0

EC-GATE 2018


26

tn

L 1,

R 2

So we have L L

1 1i 0 0 i

2 2

t /

L L L L

2t

2t

i t i i 0 i e ; t 0

1 10 e ; t 0

2 2

11 e ; t 0

2

2 0.5 1

L

1 1i t 0.5 1 e 1 e

2 2

11 0.368 0.5 0.632 0.316amp

2

31. Consider a white Gaussian noise process N(t) with two-sided power spectral density SN(f) = 0.5

W/Hz as input to a filter with impulse response 0.52t / 2e

(where t is in seconds) resulting in output

Y(t). The power in Y(t) in watts is

(A) 0.11 (B) 0.22 (C) 0.33 (D) 0.44

Key: (B)

Exp: Given, MS t 0.5 w / Hz.

2

y M

2

y

y

y

2 2

S t H jt .S t

S t 0.5 H t

Power in y t area under s t

s t dt

0.5 H t dt 0.5 H t dt

Applying parsevals theorem,

2

2

2 2

2t

t2

H t dt h t dt

Power in y t 0.5 0.5e dt 0.5 0.25 e dt

2t

12

21 1

0.5 0.25 2 . e dt12

2a.2

0.22

Gaussian function

Area 1

2

t 2h t 0.5.e

y t

y

PSD

S f

N t

M

PSD

S f

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27

32. The position of a particle y(t) is described by the differential equation:

2

2

d y dy 5y.

dt dt 4

The initial conditions are y(0) = 1 and t 0

dy0.

dt

The position (accurate to two decimal places) of the particle at t is _____.

Key: (-0.21)

Exp: Given, 2

2

d y dy 5y;

dt dt 4

where y(t) represents the position of a particle

2

2

d y dy 5y 0 1

dt dt 4

The auxiliary equation of the D.E is

2m m 5 4 0

1 1 4 1 5 4 1 i 2 1m i complex

2 1 2 2

Roots are complex

The required solution of equation (1) is

1t

21 2

t 0

1

t 2 t 2

1 2 1 2

2 1

12 1

y t e C cost +C sint 2

dyGiven y 0 1& 0

dt

From 2 ;if y 0 1; then C 1

dyif 0at t 0;

dt

dy 1then e C sin t C cos t C cos t C sin t e

dt 2

10 C C

2

C 1C C 1

2 2

The position of the particle at t is

t 2

t

2 2

1y e cos t sin t

2

e 1 e 0.21

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28

33. A random variable X takes value –0.5 and 0.5 with probabilities 1 3

and ,4 4

respectively. The noisy

observation of X is Y = X + Z, where Z has uniform probability density over the interval (–1, 1). X

and Z are independent. If the MAP rule based detector outputs X̂ as

0.5, Y

X̂0.5, Y ,

Then the value of (accurate to two decimal places) is _____.

Key: (-0.75)

Exp: Given, 1

P x 0.5 , T x 0.3 3 4 , y x m4

Noise is with uniform density function as shown

pdf of y when x 0.5 is transmitted

pdf of y when x 0.5 is transmitted

In a general rule, threshold shift towards that side where probability is less

Using MAP rule,

T

x 0.5

T 0.5

x 0.5

T x 0.5

P y x P x 0.5

P y x P x 0.5

1 1z

zf g

1 2

1.5 0.5y

y

y/x 0.5f 0.5 p x 0.5 1 4 1 2

x1

0.5xg 1.5y

x 0.5f y y p x 0.5 3 4 1 2

EC-GATE 2018


29

Equality can be used to find threshold

T T

T T

T T

1 1 1 31.5 x x 1.5 x

2 4 2 4

1.5 x 4.5 3x

3 4x x 3 4 0.75

34. The state equation and the output equation of a control system are given below:

4 1.5 2x x u,

4 0 0

y 1.5 0.625 x.

The transfer function representation of the system is

(A) 2

3s 5

s 4s 6

(B)

2

3s 1.875

s 4s 6

(C)

2

4s 1.5

s 4s 6

(D)

2

6s 5

s 4s 6

Key: (A)

Exp: 1 2

2 2 2 1

4 1.5 2A , B , C 1.5 0.625

4 0 0

1

1

2 2

2 22 2

T F C sI A B

s 4 1.5sI A

4 s

s 1.51sI A

4 s 4s s 4 6

s 1.5

s 4s 6 s 4s 6

4 s 4

s 4s 6 s 4s 6

1 1

2 22 1

2

22 1

1

2s I A B s I A

0

2s

s 4s 6

8

s 4s 6

C s I A B

2

1 2

22 1

2s

s 4s 61.5 0.625

8

s 4s 6

EC-GATE 2018


30

2 2 2

1 1

3s 5 3s 5

s 4s 6 s 4s 6 s 4s 6

So the closed loop transfer function is 2

3s 5T s

s 4s 6

35. The figure below shows the Bode magnitude and phase plots of a stable transfer function

0

3 2

2 1 0

nG(s)

s d s d s d

Consider the negative unity feedback configuration with gain k in the feedforward path. The closed

loop is stable for k < k0. The maximum value of k0 is _____.

Key: (0.1)

Exp: Basically here we are supposed to find the value of 0K K for which system will be marginally

stable.

We know for marginal stable, pc gc , if we notice both the magnitude and phase plot

carefully we can say, the frequency of which the phase is –180° in the phase plot the corresponding

gain is 20dB in the magnitude plot.

So for marginal stable, at this frequency, the gain plot should touch 0 db line, so this to happens,

if we select K 0.1 20dB, then each point in the gain plot will be subtracted by + 20dB,

including the point where gain is + 20dB, it becomes 0 dB & the system becomes marginal stable.

k G s

G j

36 dB

20 dB

180

G j

270

0

0 dB

EC-GATE 2018


31

Since K=0.1, it is positive phase plot remains unaffected so no change in pc but

gc changed

so that we have pc gc .

So if pc gcK 0.1, which is the condition for marginal stable.

if K 0.1, system stable

K 0.1, unstable

K 0.1, marginal stable

36. The contour C given blow is on the complex plane z x jy, where j 1.

The value of the integral 2

C

1 dz

j z 1 is ____.

Key: (2)

Exp: 2

1 1Let f z

z 1 z 1 z 1

The singular points of f(z) are z = -1,1 both lies inside „C‟.

By Cauchy‟s integral formula,

1 2

2

C C

C C

1 dz 1 dz

j z 1 j z 1 z 1

1 1 1 1 1dz dz

j 2 z 1 2 z 1

1 1 12 j 1 2 j 1

j 2 2

(Since the closed curve „C1‟ is clock wise & 2'C ' is anti clock wise)

1 2 j

j j 2j j

37. A band limited low –pass signal x(t) of bandwidth

5 kHz is sampled at a sampling rate fs. The signal

x(t) is reconstructed using the reconstruction filter

H(f) whose magnitude response is shown below:

The minimum sampling rate fs( in kHz) for perfect

reconstruction of x(t) is ____.

f (kHz)

K

H(f )

8 6 6 80

1C

111

2C

C

y

x1 1

EC-GATE 2018


32

Key: (13)

Exp: It is given that x t is a low pass signal, and is band limited to 5 kHz, let its spectrum X f is as

follows:

If we sample x t ideally, the sampled signal is s

n

X f X f nfs ,

where fs is the sampling

frequency.

If we think to sample it on critical sampling rate

s mf 2f 10kHz, Then the spectrum of the sampled signal will be

If we pass now sX f through the non ideal filter (shown in dotted line), at the output of the filter

along with the original signal, we get some undesired frequency of 5 to 8 kHz.

However, if we select sf 13kHz, then the filter will capture only the desired part, as follows, (this

idea we can get by observing the spectrum when we have selected fs = 10 kHz).

f (kHz)

sx f

8 5013 5 6 1318 6 8 18

sf 13kHz

X f

f (kHz)5 5

sf 13kHz

f (kHz)

sX (f )

10 5015 5 6 8 10 158 6

EC-GATE 2018


33

11V

1R1R

1R 1R

1R 1R

2R

2R

3R3R

B

DC

A

I

3R

38. Consider the network shown below with 1 2 3R 1 , R 2 and R 3 . The network is connected

to a constant voltage source of 11V.

The magnitude of the current (in amperes, accurate to two decimal places) through the source is

____.

Key: (8)

Exp: By observing the network we can say, it holds a symmetry

and hence we can say Node A and Node B are

equipotential. Similarly Node C and Node D are also

equipotential.

When 2 nodes are equipotential, the element connected

across these 2 nodes (R2 in this case) can be replaced by

either short circuit, or open circuit or any other value of

resistor without affecting the network.

If we replace the R2 by short circuit, then the network can

be redrawn as

eq 1 1 1 3 1 3 1 1

3 31 1 1 11

1 2 3

eq

R R R R R R R R R

R RR R R RR

2 2 2 2 2 2

1 31 R , R , R values are given

2 2

3/4 3 111 1

4/2 8 8

11 11I 8A

R 11/ 8

11V

1R1R

1R 1R

1R 1R

2R

2R

3R3R

11V

1R 1R

1R 1R3R 3R

1R 1R

I eqR

EC-GATE 2018


34

39. A dc current of 26 A flows through the circuit shown. The diode in the circuit is forward biased

and it has an ideality factor of one. At the quiescent point, the diode has a junction capacitance of

0.5 nF. Its neutral region resistances can be neglected. Assume that the room temperature thermal

equivalent voltage is 26 mV.

For 62 10 rad/s, the amplitude of the small-signal component of diode current (in A, correct

to one decimal place) is _____.

Key: (6.4)

Exp: 3T

d 6

D

9

d

nV 1 26mr 10 1k

I 26 10

C 0.5 10 F

3

d

6 3

d dd d 3 3

d d

6 6

33

6

jX j10

C

r .X j10 10Z r X

r X 10 10 j 1 j

5m 5 10 5 10I

110 0.6 0.5j0.10.1 10

1 j1 j

6.4 10 39.8 I 6.4 A

40. Let X[K] k 1,0 k 7 be 8-point DFT of a sequence x[n],

Where N 1 j2 nk/N.

n 0X[k] x[n] e

The value (correct to two decimal places) of 3

n 0x[2n]

is _______.

100

V

5sin( t)mV

dC

A.C equivalent

dr

EC-GATE 2018


35

Key: (3)

Exp: It is given that

X k k 1, 0 k 7 for 8 point DFT of x n

So X k 1, 2, 3, 4, 5, 6, 7, 8

We are asked to find

3

n 0

x 2n x 0 x 2 x 4 x 6

For N-point DFT we know that

N 1

n 0

N 1n

n 0

X 0 x nCentralordinate

Propertyof DFTNX x n 1

2

If we put N=8 in the above 2 equations

8 1

n 0

8 1n

n 0

3

n 0

X 0 x n x 0 x 1 x 2 ....x 7

X 4 x n 1 x 0 x 1 x 2 x 3 ....x 6 x 7

X 0 X 4 2 x 0 x 2 x 3 x 6

1x 0 x 2 x 4 x 6 X 0 X 4

2

1x 2n 1 5 3

2

41. For the circuit given in the figure, the voltage VC (in volts) across the capacitor is

(A) 1.25 2sin(5t 0.25 ) (B) 1.25 2sin(5t 0.125 )

(C) 2.5 2sin(5t 0.25 ) (D) 2.5 5sin(5t 0.125 )

5sin(5t)V

1 F

CV

100 k

100 k

EC-GATE 2018


36

Key: (C)

Exp: Using sine function as reference phasor, if we transform the given time domain network into its

equivalent phasor domain, it will be as follows

Here R 100k , C 1 F and 5rad / sec

Using the voltage division rule for VC we have

C

3 6

C

1/ j C 1V 5 0 5 0

2R 1/ j C 1 j2 RC

15 0

1 j 2 5 100 10 1 10

15 0

1 j

1 545 5 0 45

2 2

5 245 2.5 2 0.25

2 2

V t 2.5 2 sin 5t 0.25

42. An op-amp based circuit is implemented as shown below.

In the above circuit, assume the op-amp to be ideal. The voltage (in volts, correct to one decimal

place) at node A, connected to the negative input of the op-amp as indicated in the figure is____.

5 0

1

j C

CV

R

R

1k

1V

A

15V

31k

15V

o

EC-GATE 2018


37

Key: (0.5)

Exp: Given f0

sat sat

0 sat 0

R 31kV 31

R 1k

V 15Vand V 15V

As V V V 15V

Applying KVL at point A

A A

A A

A A

V 1 V 150

1 31

31V 31 V 15 0

1632V 16 V 0.5V

32

43. A solar cell of area 1.0 cm2, operating at 1.0 sun intensity, has a short circuit current of 20 mA, and

an open circuit voltage of 0.65 V. Assuming room temperature operation and thermal equivalent

voltage of 26 mV, the open circuit voltage (in volts, correct to two decimal places) at 0.2 sun

intensity is ____.

Key: (0.608)

Exp: 2

L oc TA 1.0cm ; I 20mA; V 0.65; V 26mV

oc

T

oc1

T oc2

Toc2

T

2

2

V2L LVL oc T L S

s

2 2

L

V

V V 0.65 10LV 26mV

LV

oc

oc

T

I JJ 20mA/cm V V n 1 ; J J

e 1A J

'J 0.2 20mA/cm 4mA/cm

J e 1; n 1 5 7.2 10

e 1e 1J 'e 1

V23.39 V 26m 23.39 0.608

V

44. The cutoff frequency of TE01 mode of an air filled rectangular waveguide having inner dimensions

a cm × b cm (a > b) is twice that of the dominant TE10 mode. When the waveguide is operated at

frequency which is 25% higher than the cutoff frequency of the dominant mode, the guide

wavelength is found to be 4 cm. The value of b (in cm, correct to two decimal places) is ____.

Key: (0.75)

Exp: Cut off frequency for 01

CTE

2b

Cut off frequency for 10

CTE

2a

R 1k

AV

15V

fR 31k

15V

oV

1V

EC-GATE 2018


38

It is given that C C

22b 2a

a

2b

Given waveguide frequency 5 C

f4 2a

(25% higher than the cut off frequency of dominant mode)

c

5f f

4

01guide

2

c

01 01guide

2

01

10

10

f1

f

54cm

341

5

3 42.4cm

5

2.4 1.25 3

a 1.5cm2

b 0.75cm

45. The distance (in meters) a wave has to propagate in a medium having a skin depth of 0.1m so that

the amplitude of the wave attenuates by 20 dB, is

(A) 0.12 (B) 0.23 (C) 0.46 (D) 2.3

Key: (B)

Exp: 1

Skin depth 0.1mf

Attenuation constant1

f 100.1

Amplitude of wave varies as ze

z 1e

10

1z n 10z 2.3

10

2.3z 0.23m

10

46. The logic gates shown in the digital circuit below use strong pull-down nMOS transistors for LOW

logic levels at the outputs. When the pull-downs are off, high-value resistors set the output logic

EC-GATE 2018


39

levels to HIGH (i.e., the pull-ups are weak). Note that some nodes are intentionally shorted to

implement “wired logic”. Such shorted nodes will be HIGH only if the outputs of all the gates

whose outputs are shorted are HIGH.

The numbers of distinct values of 3 2 1 0X X X X (out of the 16 possible values) that give Y = 1 ____ .

Key: (8)

Exp:

If we evaluate F by referring the above logic circuit

0 0F X X G 0

So the node F is always 0. Irrespective any possible combinations of 0 1 2 3X ,X ,X ,X

Since already one of the input to the OR gate is logic 0 (fixed) and we are expecting Y=1, so

we must need the other input of the or gate is logic 1 for which 3X 1.

So for Y=1. We need the fixed value of 3X 1, with this 0 1 2X ,X ,X can take any possible

combinations,

So the number of possible combination of 0 1 2X ,X ,X is 32 8, for each of the combination

3X 1.

Hence there are 8 combinations of 0 1 2 3X ,X ,X ,X for which output Y=1

47. A 2 × 2 ROM array is built with the help of diodes as shown in the circuit below. Here W0 and W1

are signals that select the word lines and B0 and B1 are signals that are output of the sense amps

based on the stored data corresponding to the bit lines during the read operation. During the read

operation, the selected word line goes high and the other word line is in a high impedance state.

0X

1X

2X

3X

Y

0X

1X

2X

3X

0X0X G

G

1 2X X

31 2G X X .X

F

1

Y 1

3X

0X

0 fixed

EC-GATE 2018


40

As per the implementation shown in the circuit diagram above, what are the bits corresponding to

Dij (where i = 0 or 1 and j = 0 or 1) stored in the ROM?

(A) 1 0

0 1

(B) 0 1

1 0

(C) 1 0

1 0

(D) 1 1

0 0

Key: (A)

Exp: Form the given statement; it is clear that, if we want to read the stored information in the ROM

word line value should be logic 1.

So in all the 4 options, the top row of matrix indicate if 0W 1, what are the value of

0 1B and B similarly the bottom row indicates if 1W 1. what are the value of 0 1B & B .

If we make 0W 1, it‟s corresponding diode will be ON. And DDV will be available on 0B ,

since the diode corresponding to 1W is in off state 1W 1now , by default 1B 0.

So when 0 0 1W 1, then B 1and B 0.

Similarly when 1 0 1W 1, then B 0 and B 1

So the matrix is

0 1

0

1

B B

W 1 1 0

W 1 0 1

or

0 1

0

1

B B

W 1 0

W 0 1

48. A curve passed through the point (x = 1, y = 0) and satisfies the differential equation 2 2dy x y y

.dx 2y x

The equation that describes the curve is

(A) 2

2

yln 1 x 1

x

(B)

2

2

1 yln 1 x 1

2 x

(C) y

ln 1 x 1x

(D)

1 yln 1 x 1

2 x

W0

W1

VDD

B0 B1

Sense amps

0 1

00 010

10 111

B B

D DW

D DW

Bits stored in the ROM Array

EC-GATE 2018


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Key: (A)

Exp: Given D.E is

2 2

2

22

2

dy x y y

dx 2y x

dy 1 1 xy

dx x 2 2y

dy 1 1 xy y 1

dx x 2 2

dy dt dy 1 dtLet y t 2y y

dx dx dx 2 dx

2

2

1 1 x2 dx2x 2

2

1 dt 1 1 xFrom 1 t

2 dx x 2 2

dt 1 12 t x 2 ; which is linear D.E.

dx x 2

1 1 eI.F e P 2 & Q x I.F

x 2 x

x x2

2 2

xx

2

x2 x

2

The solution of eq 2 is

e et. x dx

x x

et. e c

x

ey e c 3

x

Given that eq(3) passes through the point (1,0).

1

2 xx 1

2

2 2 2x x 1

2 2 2

C e from 3

From 3 ;

y ee e

x

y 1 y y1 e 1 e n 1 x 1.

x e x x

49. A uniform plane wave traveling in free space and having the electric field

8

x zˆ ˆE 2a a cos 6 3 10 t 2 (x 2 z) V / m

E

is incident on a dielectric medium (relative permittivity > 1, relative permeability = 1) as shown in

the figure and there is no reflected wave.

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The relative permittivity (correct to two decimal places of the dielectric medium is ____.

Key: (2)

Exp: 811 zGiven E 2 a a cos 3 10 t 2 x 2z V m

The wave is parallel to the x, z plane

Since, there is no reflection,

Angle of incidence in Brewster angle is

1 12

i B 2

1

tan tan

i

2 2

tan 2

2 2

50. For a unity feedback control system with the forward path transfer function

K

G(s)s(s 2)

The peak resonant magnitude Mr of the closed-loop frequency response is 2. The corresponding

value of the gain K (correct to two decimal places) is _____.

Key: (15.02)

Exp: The O.L.T.F of the unity –ve feed back system is

KG s

s s 2

The C.L.T.F is

2

G s KT s

1 G s s 2s K

By comparing with standard 2nd order equation

x

r

Dielectric

Medium( 1)

z

Free space

i

i2

x

z

Free space

1

EC-GATE 2018


43

2 2

ns n

n

2 2

n n n

2

s 2 , we have

K and

2 2 1 1

1K

Let us obtain for the resonant peak

r2

2 2

2

2

1M

2 1

1 12 4

2 1 1

116

1

2 2

2

2

2

2

2

16 1 1

let x 16x 1 x 1

16x 16x 1 0

16x 16x 1 0, by solving this we get

x 0.067

1 1K 15.02

0.067

51. A four-variable Boolean function is realized using 4 × 1 multiplexers as shown in the figure.

The minimized expression for F(U,V,W,X) is

(A) UV UV W (B) UV UV WX WX

(C) UV UV W (D) UV UV WX WX

F(U,V,W,X)

CCV4 1

MUX

4 1

MUX

W XU V

0I

1I

2I

3I

0I

1I

2I

3I1S 0S 1S 0S

EC-GATE 2018


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Key: (C)

Exp:

Output of the first MUX

Y UV.0 UV.1 U V.1 UV.0 UV U V

Output of second MUX

F W X Y WXY WX.0 WX.0

W Y X X WY YW

UV U V W

Although option C and option D are representing same function, but we can say the minimized

expression is option C.

52. Let c cc(t) A cos(2 f t)and mm(t) cos(2 f t). It is given that c mf 5 f . The signal c(t) + m (t) is

applied to the input of a non-linear device, whose output 0v (t) is related to the input

2

i 0 i iv (t) as v (t) av (t) bv (t), where a and b are positive constants. The output of the non-linear

device is passed through an ideal band-pass filter with center frequency fc and bandwidth 3fm, to

produce an amplitude modulated (AM) wave. If it is desired to have the sideband power of the AM

wave to the half of the carrier power, then a/b is

(A) 0.25 (B) 0.5 (C) 1 (D) 2

Key: (D)

Exp: Given

1

C c m

c m

c t A cos 2 f t , m t cos 2 f t

f 5f

„a‟ and „b‟ are constants

0I

1I

2I

3I1S 2S

0I

1I

2I

3I0

Y

YY0

1

1

0

ccV F U,V,W,X

U V W X

0

4 1

MUX

4 1

MUX

m t

iv t

c t

2

i iav t bv t 0v t

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2

0 c c c c m

2 2

c cm c c c m

c m c

v t a A cos 2 f t cos 2 fmt b A cos 2 f t cos 2 f t

bA bA b ba cos 2 f t a A cos 2 f t cos 4 f t cos 4 f t

2 2 2 2

2b A cos 2 f t cos 2 f t .

After passing through BPF of center frequency fc and bandwidth m3f ,

c c c m ca A cos 2 f t 2b A cos 2 f t cos 2 f t

y tcarrier side bands

Carrier power 2 2

ca A,

2 side band power =

2 2

cb A

It is given that sideband power 1

carrier power 2

2 2 2

c

2 2 2

c

b A 1 b 1 a2

a A 2 a 4 b

2

53. The input 4 sinc (2t) is fed to a Hilbert transformer to obtain y(t), as shown in the figure below:

Here sinc(x) = sin( x)

.x

The value (accurate to two decimal places) of

2y(t) dt

is____.

Key: (8)

Exp: By Parseval‟s theorem we can say

2 21

y t dt Y d2

The impulse response of the Hilbert transform is 1

h t ,t

and Fourier transform

H jsgn( )

If x t ksinc(at)

kthen X rect

a 2a

Hilbert

Transform4sinc(2t) y(t)

X

k/a

-a a

j

j

EC-GATE 2018


46

4sinc 2t

1

y t 4sin c2t Y 2rect . jsgn x t h t X .H*t 4

2 2 2

Finally,

1 1 1y t dt Y d area of Y 4 4 8

2 2 2

54. Red (R), Green (G) and Blue (B) Light Emitting Diodes (LEDs) were fabricated using p-n junctions

of three different inorganic semiconductors having different band-gaps. The built-in voltage of red,

green and blue diodes are R G BV , V and V , respectively. Assume donor and acceptor doping to be

the same (NA and ND, respectively) in the p and n sides of all the three diodes. Which one of the

following relationships about the built-in voltages in TRUE?

(A) R G BV V V (B) R G BV V V

(C) R G BV V V (D) R G BV V V

Key: (B)

Exp: Built in potential of pn diode

2

A Dbi

i

2

bi A D i

N .NV kT n

n

V kT n N N 2kT n n ... 1

Where in can be defined as

24sinc2t y t 1

h tt

2

-2 2

2

Y 4

-2 2

j2

j2

2

2

Y

j

j

Y

2

-2 2

*

EC-GATE 2018


47

g

g

E

2 kTi C V

E

2kTi

n N .N e

n k ' e ... 2

So substitute equation (2) in (1)

bi A D g

bi g g

V kT n N N 2kT n k ' E

V M E ;where E is band gap

All diodes A DN and N are same, so that M same

bi g g

hcV E ; E

We know that g Red g green g blue

Red green blue

R G B

E E E because

V V V

55. Let 2r x y z and

3 3z xy yz y 1. Assume that x and y are independent variables. At

(x,y,z) = (2, –1, 1 ), the value (correct to two decimal places) of r

x

is ____.

Key: (4.50)

Exp: Given, 2r x y z... i

3 3

2

2

2

2

2, 1,1

and z xy yz y 1... ii

r z2x ... 1 x, y are independent

x x

from i

z z3z y y 0 form ii

x x

z3z y y

x

z y2

x 3z y

r yform 1 & 2 ; 2x

x 3z y

r 12 2

x 3 1 1

14 4.50

2

EC-GATE 2018


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EC-GATE 2018 - d1zttcfb64t0un.cloudfront.netd1zttcfb64t0un.cloudfront.net/gatepapers/GATE2018/Gateforum_EC... · CD DE 1.5 3 ''? ... In the circuit shown below, the op-amp is ideal