8/9/2019 EC Forenoon New1

1/17

EC-GATE-2015

1

GATE 2015 – A Brief Analysis

(Based on student test experiences in the stream of EC on 31st

January, 2015 –Forenoon session)

Section wise analysis of the paper

Section Classification 1 Mark 2 Marks Total No ofQuestions

Engineering Mathematics 5 3 8

Networks 3 3 6

Electronic Devices 4 3 7

Analog Circuits 2 3 5

Digital Circuits 3 4 7

Signals and Systems 2 4 6

Control Systems 2 3 5

Communication2 4 6

Electromagnetics 2 3 5

Verbal Ability 3 3 6

Numerical Ability 2 2 4

30 35 65

Questions from the Paper

Aptitude

1.5/ 7

x

1log 3

−=

Find x.Key: (2.74)

Exp: 3

1 3 1 35 7 7x x x 2.747 5 5

− = = =

2. Frog(A) Croak (B) Roar (C) Hiss (D) Patter

Key: (A)Exp: Frogs make ‘croak’ sound.

3. Synonym of ‘Educe’(A) exert (B) educate (C) extract (D) extend

Key: (C)

8/9/2019 EC Forenoon New1

2/17

EC-GATE-2015

2

4. Principal presents chief guest with _____ as a token of gratitude.(A) Momento (B) Memento (C) Momentum (D) Moment

Key: (B)

5. If , and → are defined as

a ba ba b

−=+

a ba b

a b+=−

a b ab→ =

then find ( ) ( )66 6 66 6→ (A) 1− (B) 2− (C) 1 (D) 2

Key: (C)

Exp: 66 6 60 5

66 666 6 72 6

−= = =

+

6666 6 72 6

666 6 60 5

+= = =−

(66 6)→(66 6)5 6

16 5

= × =

6. Cube of side 3 unit is constructed using cube of side 1 unit. The ratio of no. of visiblefaces to non-visible faces is

(A) 1:3 (B) 1:4 (C) 1:2 (D) 2:3Key: (C)Exp:

Number of faces per cube = 6

Total number of cubes = 9×3 = 27 Total number of faces = 27×6 = 162 Total number of non visible faces = 162-54 = 108

Number of visible faces 54 1

Number of non visible faces 108 2= =

1 1 11

11

1 1 1

111

1 1 1

8/9/2019 EC Forenoon New1

3/17

EC-GATE-2015

3

7. Question based on rephrasing sentence.

8. Question based on paragraph conclusion.

9. Find the missing number ________

Key: (3)Exp: Middle number is the average of the numbers on both sides.

Average of 6 and 4 is 5Average of (7+4) and (2+1) is 7Average of (1+9+2) and (1+2+1) is 8Average of (4+1) and (2+3) is 5Therefore, Average of (3) and (3) is 3

Technical

1. Which of the following wave forms represent given function?f(x) = e -x (x2+x+1)(A) (B)

(C) (D) None

1 9 2 8 1 2 1

4 1 5 2 3

3 3?

7 4 7 2 1

6 5 4

( )f x

1.0

x

( )f x

1

( )f x

x

1

8/9/2019 EC Forenoon New1

4/17

EC-GATE-2015

4

Key: (B)Exp: f(1) = e -x (x2+x+1)

f(0) = 1f(0.5) = 1.067For positive values of x, function never goes negative.

2.

Which of the following represents output?

(A) (B)

(C) (D)

Key: (C)Exp: The circuit can be re drawn as

During positive pulse, both diodes are forward biased. So output pulse of +ve polarity isproduced. As polarity at ‘d’ and ‘c’ is given opposite to input terminals, hence +ve pulseis inverted. During negative pulse, both diodes are reverse biased. So, V 0 = 0V

inV

t

iV

t

oV

t

inV

t

inV

t

−b c

da

2D2D

1D

oV

+

−+

R+

−

+ +

−−

inV

c

d

a

b

R

8/9/2019 EC Forenoon New1

5/17

EC-GATE-2015

5

3. The result of convolution of the signal x(-t)* δ (-t-t 0) is(A) x(t-t 0) (B) x(t+t 0) (C) x(-t+t 0) (D) x(-t-t 0)

Key: (D)

Exp: ( )0 0x( t )* t t x( t ) * (t t )− δ − − = − δ +

( )( )( )

0

0

x t t t

x t t

= − +

= − −

4. What are values of V 1 and V 2?

(A) 5, 25 (B) 25, 30 (C) 5, 5 (D) 25, 25Key: (A)Exp: By nodal analysis

1

2 1

1

5 I I 2I 0

4I 5

5I A

4V 4I 5volts

V 4(5) V

20 V 25 volts

− + + + ==

=

= == += + =

5. In a lead-network, the feed forward path contain R parallel with C and transfer function is

given as ( ) s 2G ss 4

+=

+ . The value of RC is __________.

Key: 0.5

Exp: Givens 2

G(s)s 4

+=+

Zero = 2 =1 1

RC=

τ

Pole = 4 =1 1

RC=

ατ τ

So, RC = 0.5

6. The polar plot of transfer function ( ) ( )10 s 1

G ss 10

+=

+ for 0 ≤ ω ≤ ∞ will be in

(A) First quadrant (B) Second quadrant(C) Third quadrant (D) Fourth quadrant

2V 5A

+

−

+

−

1V2I4Ω 4Ω

I4Ω

8/9/2019 EC Forenoon New1

6/17

EC-GATE-2015

6

Key: (A)

Exp: ( ) ( )10 s 1

G ss 10

+=

+

Put s j= ω

( ) ( )( )

10 j 1G j j 10

ω +ω = ω +

0,ω = M 1 0= <

,ω = ∞ M 10 0= <

So, zero is nearer to imaginary axis. Hence plot will move clockwise direction.

It is first quadrant.

7. Differential equation is given ( ) ( )2

2

d y 2dyy 0 ; y 0 y ' 0 1

dt dt+ + = = = .

The output y(t) is(A) (2+t)e -t (B) (1-2t)e -t (C) (2-t)e -t (D) (1+2t)e -t

Key: (D)Exp: Differential equation is (D 2+2D+1).y = 0

D2+2D+1 = 0 (D+1) 2 = 0 D = -1,-1

( ) ( )

( ) ( )( )

t1 2

t t2 1 2

solution is y t c c t e C.F

y ' t c e c c t e

−

− −

= + →

= + + −

y(0) = 1; ( )y' 0 1= gives c 1 = 1 and c 2+c1(-1) = 1 c2 = 2 y(t) = (1+2t)e -t

8. A complex function is given z = x+iy. Which of the following is not true?

(A) z is an analytic function (B) Residue of 2z 1

at z 1 isz 1 2

=−

(C) 2

c

z dz∫ value is 0 (D) None of aboveKey: (A)

jω

0ω= 10ω= ∞ σ

8/9/2019 EC Forenoon New1

7/17

EC-GATE-2015

7

Exp: ( )

x x

f z z

x iy

u x, v y

u 1and v 0

== −

= = − = =

y y

x y

u 0 and v 1

u v i.e., C R equations not satisfied

z is not analytic

= = −

≠ −

(B) z = 1 is a simple pole

( )2 2z 1 z 1z z z 1

Residue at z 1 is lim z 1 . limz 1 z 1 z 1 2→ →

= − = =

− − +

(C) Since z 2 is analytic everywhere

Using Cauchy’s integral theorem, 2

C

z dz 0=∫

9. Maximum power transfer to load is ___________W.

Key: 1.66

Exp: Th(rms ) Th (mq)4 2 j

V 2 2 45 ; V 2 42 2 j

×= = = +

Th

L th

Z 2 || 2j 1 j

R Z 2

= = +

= = Ω

Maximum power transfer to R L is2

2max L

2 2 45P I R 2 1.66W

2 1 j

°= × = × =+ +

10.

The value of R is 300 Ω. eqR is ____________ Ω.

2Ω

j2Ω LR4 0 °

eqR

R

R RR

R

R

R

RR

RR

RR

R

R

8/9/2019 EC Forenoon New1

8/17

EC-GATE-2015

8

Key: 100

Exp:

By bridge condition

eqRR 100

3= = Ω

11. If matrix

1

2

3

is eigen vector of the matrix

4 1 2

P 2 1

14 4 10

−

, then value of P is _______.

Key: 17

Exp:

( )

4 1 2 1

AX X P 2 1 2

14 4 10 3

12

P 7 2 12 136 3

= λ = λ −

λ

+ = λ λ = − − − − − λ

( )

( )

2 P 7 2

and 3 36 i.e., 12

Equation 2 gives P 7 24 P 17

λ = + − − − − −

λ = λ = + = =

12. Find the value of K so that point ( )1 j2− + lies on root locus of unity feedback system

whose open loop transfer function is given by( )

( )( )2K s 4

GHs 8 s 9

+=

+ −

Key: 25.5

Exp: By magnitude condition

( ) ( )s 1 2j

G s H s 1=− +

=

2R R R 2R

eqR

R R 2R

2R

RR 2R

2R

R R

eqR

8/9/2019 EC Forenoon New1

9/17

EC-GATE-2015

9

So,k 2 j 3

17 2j 2 2 j 4 2j

+=

+ + − +

20 8 53K

1325.5

=

=

So K value is = 25.5

13. If vector P is given by 3 2 2 2x y zˆ ˆ ˆx y a x y a x yza− − then P is

(A) solenoidal and irrotational(B) not solenoidal but irrotational(C) solenoidal but not irrotational(D) neither solenoidal non irrotational

Key: (C)

Exp: 3 2 2 2x y z2 2

P x ya x y a x yza

.P 3x y 2x y x y 02

= − −

= − − =

It is solenoidal.

( ) ( ) ( )

x y z

3 2 2 2

2 2 3x y z

a a a

Px y z

x y x y x yz

a x y a 2xyz a 2xy x 0

∂ ∂ ∂× =

∂ ∂ ∂− −

= − − − + − − ≠

So P is solenoidal but not irrotational.

14. Negative differential resistance is observed in PN junction diode if(A) Both P and N side are heavily doped.(B) Only P side heavily doped.(C) Only N side heavily doped(D) A intrinsic silicon is inserted between the junction of P and N.

Key: (A)

15.

Zeros of multiplicity 4.

j0.5

0.54

j0.5−

0.5−

×

×

×

×

8/9/2019 EC Forenoon New1

10/17

EC-GATE-2015

10

Figure above shows pole zero location in z-plane of a system. The impulse response ofthe system is h[n]. If h[0] = 1, then(A) h[n] is real for all ‘n’(B) h[n] is imaginary for all ‘n’(C) h[n] is real for even ‘n’(D) h[n] is imaginary for odd ‘n’.

16. 3-input majority gate represents as M(a,b,c) ab bc ca .= + +

The function ( )f M(a,b,c),M(a,b, c),c is realized by(A) 3-input XOR gate (B) 3-input XNOR gate(C) 3-input OR gate (D) 3-input AND gate

Key: (A) and (B)

Exp: ( ) ( )

( ) ( ) ( )

( ) ( ) ( )( ) ( )

M a,b,c ab bc ac m 3,5,6,7

M a,b,c m 0,1,2,4 X let say for simplicity

M a,b,c ab bc ac m 2,4,6,7 Y letc m 1,3,5,7 z let

= + + =

= =

= + + = == =

∑∑

∑∑

( ) ( )( ]

( )( ) ( )( ) ( ) ( )( )( ) ( )

( ) ( ) ( )

( )

f M a,b, c ,M a,b,c , c

f x, y,z

xy yz zx

m 0,1,2, 4 m 2,4,6,7 m 2,4,6,7 m 1,3,5,7

m 1,3,5,7 m 0,1, 2, 4

m 2,4 m 7 m 1

AND operater is like intersectionm 1,2,4,7OR operator is like union

A B C A B C s

= =

= + + = +

+ = + +

=

= =

∑ ∑ ∑ ∑∑ ∑

∑ ∑ ∑

∑

( )( ) ( )

tandardresult

Both options A and B are correct

17. The open loop transfer function isk

G(s)s(s 1)(s 3)

=+ +

for unity negative feedback

system.The value of gain k (>0), at which root locus crosses the imaginary axis is ______.

Key: 12

Exp: ( )( )k

C.E is1 0s s 1 s 3+ =+ + 3 2s 4s 3s k 0+ + + =

By using Routh Table, s 1 row should be zero. For poles to be on imaginary axis

12 k 0

4−

= ; k should be 12.

8/9/2019 EC Forenoon New1

11/17

EC-GATE-2015

11

18. For figure shown below, the values of currents are

2 1 TI 8mA, I 4mA, V (thermal voltage)= = ato27 K is 26 mV. What is the value of

1 2V V− ato50 C ? _______mV

Key: -19.2

Exp: BE T2V V

2 SI I e η=

( )

2

BE T1

1 2

T

1 2

BE 2

V V1 S

BE 1

V VV1

2

T

V V27.8m

1 2

V V

I I e

V V

Ie

I

50 273V at 50 C 27.8mV

11,600

1e

2V V 19.2mV

η

−η

−

=

==

=

+° = =

=

− = −

19. Propagation delay of each gate is 10nsec in below circuit. Initially (at t

8/9/2019 EC Forenoon New1

12/17

EC-GATE-2015

12

As per information given on question the waveforms of A, B, C are as follows

→ The logic to solve this question is first obtain X, Y waveform and using this obtain Z.→ To obtain X, initially assume delay of NOT gate is 0 and draw its waveform and

finally shift it by 10nsec to obtain actual X. Similar procedure to be followed forobtaining Y and Z i.e., first draw waveform with 0 delay and at the end shift by theamount of delay given in question.

A

B

Ct 0=

B

B

t 0=

t 0=

t 10 nsec=( )B

withdelay

( )X B with 0 delay→ =

Y B.A→ =

A

( )B

withdelay

( )Y B

withdelay

=

( )Y AB

withoutdelay

=

t 0=

t 10=

t 10 nsec=

t 20n sec=

8/9/2019 EC Forenoon New1

13/17

EC-GATE-2015

13

( )z AB C→ =

Clearly we can say that output is high during 10 nsec to 30 nsec i.e. a duration of 20nsec

Even though procedure seems to be lengthy but it is pretty systematic and we don’thave to think much so that we may stuck somewhere.

20. For the given system

2

x (t ) m(t )cos(2000 t )

y(t) x (t) 10x(t)

= π

= +

Bandpass characteristic are shown below

Message signal m(t) spectrum is as below

When output is 10x(t), the value of w is strictly less than _________ .

x(t)y(t)

Amplifier B.P.F o u t p u t

w−

m(t)

wf

1700− 70 0− 70 0 1700f ( in H z)

AB

C

( )Z

withoutdelay

( )Z

withdelay

t 20=

t 20=t 10=t 0=

t 0= 10 20

20 30100

8/9/2019 EC Forenoon New1

14/17

EC-GATE-2015

14

Key: 350Exp:

When output is 10 x(t)It must be 1200 w 2w− > … (1)2400 2w 1700− > … (2)From (1) & (2)w 400< …. (3)w

8/9/2019 EC Forenoon New1

15/17

8/9/2019 EC Forenoon New1

16/17

8/9/2019 EC Forenoon New1

17/17

EC-GATE-2015

17

28. The max area (in sq units) of a rectangle whose vertices lie on the ellipse 2 2x 4y 1+ = is_________.

Key: 1Exp:

Let 2x, 2y be the length, breadth respectively of the rectangle inscribed in the ellipse2 2x 4y 1,+ = then

Area of the rectangle (2x) (2y) i.e., 4xy

Consider, ( )2f Area= 2 216x y=

( )2

2 2 2 1 x4x 1 x y4

−= − =

( ) ( )2 1f ' x 0 x 1 2x 0 x2

= − = =

2 1 1y y8 8

= =

( ) 2 1f '' x 8 48x 0 when x2

= − < =

1f is maximum at x =

2

Area is maximum and the maximum area is1 1

42 8

i.e., 1

29. A 16kb (=16,384 bit) memory array is designed as a square with as aspect ratio of one(no. of rows= no. of columns). The minimum no. of address lines needed for the rowdecoder is _______.

Key: 7 Exp: Generally the structure of a memory chip = Number of Row × Number of column

= M×N

→ The number of address line required for row decoder is n where M = 2 n orn = log 2M

→

As per information given in question : M = NSo M×N = M×M = M 2 = 16k = 2 4×2 10

M2 = 2 14

M = 128

→ n = log 2128 = 7

o

rectangle

ellipse