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EC-Gate 2016 SET-3

ECE sEt-3 2016Q1 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 on-ion

(D) Apples are more costlier than onions

s1 Correct option is (C)Based on the given sentences option ‘C’ is the correct sentence which is in the comparative degree. Option A and B convey the wrong comparison and D has double comparative and so they are wrong.

Q2 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.”(A) Burning

(B) igniting

(C) clutching

(D) flinging

s1 Correct option is (C)The underlined word grasping means clutching or holding something tightly.

Q3 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

s1 Correct option is (B)Q and R are the son and daughter of M, E is the mother of P and daughter-in-law of M means Q and E are married couples in the family.

P is the grandchild of M.

Q4 The number that least fits this set: (324, 441, 97, and 64) is ______(A) 324

(B) 441

(C) 97

(D) 64

s1 Correct option is (C)

In the given set of number, all are perfect squares but 97 is not 324 is square of 18 & 18 2^ h 324= 441 is square of 21 & 21 2^ h 441= 64 is square of 8 & 8 2^ h 64=97 is not the square of any number

The number that least fits in gives set is 97.

Q5 It takes 10s and 15s 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

(B) 10.0

(C) 12.0

(D) 22.0

s1 Correct option is (A)Speed of the train (ST)

Time (T)

length of the train (LT) Distance (D)= +

(ST) TLT D= +

D = Distance (or) length of the platform 0= Speed of the first train ST1^ h

10120=

12= m/sSpeed of the second train ST2^ h

15150=

10= m/s The magnitude of the difference in the speed of the

two trains (m/s) 12 10= − 2=

Q.6-Q.10 carry two marks each

Q6 The velocity V of a vehicle along a straight line 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)?

EC-Gate 2016 SET-3

(A) 0

(B) 3

(C) 4

(D) 5

s1 Correct option is (D)The odometer reading increase from starting point to end pointArea of the velocity and time graph per second

1st sec & triangle 21 1 1# #=

21=

2nd sec & square 1 1 1#= =

3rd sec & square+triangle 1 1 21 1 1# # #= +

1 21=

4th sec & triangle 21 1 2# #=

1=5th sec & straight line 0=6th sec & triangle 2

1 1 1# #=

21=

7th sec & triangle 21 1 1# #=

21=

Total Odometer reading at 7 seconds

21 1 1 2

1 1 0 2121= + + + + + +

5=

Q7 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 source of rabies worldwide.

s1 Correct option is (A)Only option ‘A’ can be logically inferred from the information provided in the argument.

Q8 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

s1 Correct option is (C) Manu age = Sravan age + 2 months Manu age = Trideep age - 3 months Pavan age = Sravan’s age + 1 monthFrom this Trideep age>Manu>Pavan>Sravan

Trideep can occupy the extra spane in the flat.

Q9 Find the area bounded by the lines x y3 2 14+ = , x y2 3 5− = in the first quadrant.

(A) 14.95

(B) 15.25

(C) 15.70

(D) 20.35

s1 Correct option is (B)

A ,314 0= ; E

B ,0 7= 6 @ C ,2

5 0= : D D ,0 3

5= −: D E ,4 1= 6 @ F ,0 1= 6 @ Required area Area of Area ofleOAB leCEAT T= −

21

314 7 2

1314

25 1# # # #= − −b bl l; E

EC-Gate 2016 SET-3

.15 25= sq. unitsAnother method:Required area Area of Area ofle BFE FEOCT= + .2

1 4 6 21 4 2 5 1# # # #= + +^ h

.12 3 25= + .15 25= sq. units.

Q10 A straight line is fit to a data set ( ,lnx y ). This line intercepts the abscissa at .lnx 0 1= and has a slope of .0 02- . What is the value of y at x 5= from the fit?

(A) .0 030-(B) .0 014-(C) .0 014

(D) .0 030

s1 Correct option is (A) Straight line equation y mx c= + m = slope .0 02=−set (log ,x y )If log x X= , then set ,x y^ h . ,x y0 1 0= =6 @ y mX C= + 0 . . C0 02 0 1#=− +

C .0 002= y mX C= + y . logx C0 02#=− + @x 5= y . .log0 02 5 0 002#=− + .0 030=−

Q.11-Q.35 Carry one mark each

Q11 Consider a 2 2# square matrix

A xσ

ω σ= > HWhere x is unknown. If the eigen values of the matrix A are jσ ω+^ h and jσ ω-^ h, then x is equal to(A) jω +(B) jω -(C) ω +(D) ω -

s1 Correct option is (D) det A^ h x2σ ω= − 2 2σ ω= + x2σ ω= − 2ω = xω =− x2σ ω+ 0=

x ω =−

Q12 For f z^ hsin

zz2= ^ h

, the residue of the pole at z 0= is ______

s1 Correct answer is 1

sin

zz2^ h

! ! ......zz z z13 52

3 2

= − + −' 1

! ! ......zz z13 5

3

= − + −

Res. f z^ h 1= z 0=

Q13 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 ______

s1 Correct answer is 0.07203 P . .0 7 0 34= ] ^g h .0 07203=

Q14 The integral x

dx10

1

-^ h# is equal to ______


x

dx10

1

-^ h# x2 1 0

1= − −" ,

2 0 1=− −^ h6 @ 2=

Q15 Consider the first order initial value problem yl y x x2 2= + − , y 0^ h 1= , x0 < <3^ hwith exact solution y x x ex2= +^ h . 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 step-size .h 0 1= is ______

s1 Correct answer is 0.6

dxdy y x x2 2= + −

y 0^ h 1= , x0 < <3^ hGiven ,f x y^ h y x x2 2= + − x0 0= y0 1= h .0 1= k1 ,hf x y0 0= ^ h

.0 1 1 2 0 0 2= + −^ ^_ h h i .0 1= k2 ,hg x h y k0 0 1= + +^ h . y k x h x h0 1 20 1 0 0

2= + + + − +^ ^ ^h h h

. . . .0 1 1 0 1 2 0 1 0 1 2= + + −^ ^ ^h h h . . . .0 1 1 1 0 2 0 01= + −^ h

.0 129= y1 y k k2

10 1 2= + +^ h

. .1 21 0 1 0 129= + +^ h

.1 0 1145= + .1 1145=Exact solution, y x^ h x ex2= + .y 0 1^ h . e0 1 .2 0 1= +^ h . .0 01 1 1052= +

EC-Gate 2016 SET-3

.1 1152= ERROR . .1 1152 1 1145= − .0 00062= Percentage Error .0 00062 100#= . %0 06

Q16 Consider the signal cos sinx t t t6 8p p= +^ ^ ^h h h, where t is in seconds. The Nyquist sampling rate (in samples/second) for the signal y t x t2 5= +^ ^h h is(A) 8

(B) 12

(C) 16

(D) 32

s1 Correct option is (C) x t^ h cos sint t6 8p p= +^ ^h h

y t^ h x t2 5= +^ h y t^ h cos sint t12 30 16 40p p p p= + + +^ ^h h

fm1 6= fm2 8= fm 8= Hz fs min^ h f2 m= 16= Hz

Q17 If the signal *x t sin sintt

tt= p p^

^ ^h

h h with * denoting the convolution operation, then x t^ h is equal to

(A) sin

tt

π ^ h

(B) sin

tt2

π ^ h

(C) sin

tt2

π ^ h

(D) sin

tt 2

π ^

chm

s1 Correct option is (A)

Convolution of two sinc pulses is sinc pulse.

x t1^ h sin

tt

π = ^ h

x t^ h *x t x t1 1^ ^h h X ω ^ h X X1 1$w w= ^ ^h h X1 ω = ^ h x t^ h x t1= ^ h

sin

tt

π = ^ h

Q18 A discrete-time signal x n6 @ n n3 2 5d d= − + −6 6@ @ has z -transform X z^ h. If Y z X z= −^ ^h h is the z -transform of another signal y n6 @, then

(A) y n x n=6 6@ @(B) y n x n= −6 6@ @(C) y n x n=−6 6@ @(D) y n x n=− −6 6@ @s1 Correct option is (C)

a x nn^ ^h h X a

z) c m

a 1=−but x n^ h n n3 2 5d d= − + −6 6@ @ y n^ h x n1 n= −^ ^h h n n1 3 2 5n d d= − − + −^ ^h h 68 @ y n^ h n n3 2 5d d=− − − −^ ^h h x n=− ^ h

Q19 In the RLC circuit shown in the figure, the input voltage is given by cos sinv t t t2 200 4 500i = +^ ^ ^h h h. The output voltage v t0^ h is

(A) cos sint t200 2 500+^ ^h h

(B) cos sint t2 200 4 500+^ ^h h

(C) sin cost t200 2 500+(D) sin cost t2 200 4 500+^ ^h h

s1 Correct option is (B)Given V ti ^ h cos sint t2 200 4 500= +^ h

Let us apply SPT [Super Position Theorem] only consider cos t2 200 , then circuit becomes

So, V t0l h cos t2 200=Now only consider sin t4 500 , then circuit becomes

EC-Gate 2016 SET-3

So, again V t0ll h sin t4 500=Finally according to SPT V t0^ h V t V t0 0= +l ll^ ^h h V t0^ h cos sint t2 200 4 500= +^ ^h h

Q20 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(A) E E E> >gX gY gZ

(B) E E EgX gY gZ= =(C) E E E< <gX gY gZ

(D) no relationship among these band gaps exists

s1 Correct option is (C)Where Vr is cut in voltage Vr3 V V> >r r2 1

Vr Eg\

So, EgZ E E> >gY gX

Q21 The figure shows the band diagram of a Metal Oxide Semiconductor (MOS). The surface region of this MOS is in(A) inversion

(B) accumulation

(C) depletion

(D) flat band


Q22 The figure shows the I V- characteristic of a solar cell illuminated uniformly with solar light of power 100 mW/cm2. The solar cell has an area of cm3 2 and a fill factor of 0.7. The maximum efficiency (in%) of the device is ______

s1 Correct answer is 21 Fill factor .0 7=

PPmaxT

=

I VPmaxSC OC$

=

Pmax" 63 10 3#= − W (PT " Theoretical power)

maxη PPmaxin

=

cm

W100 10 3

63 10 100cmW3 2

3

2# #

##= −

−

%21=

Q23 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 ______

s1 Correct answer is 0V

EC-Gate 2016 SET-3

Diodes are ideal therefore during Positive cycle of input V0 10 10= − V0=During Negative cycle, the diodes are Reverse biased V V00 =

V0 V0= (always)

Q24 Consider the circuit shown in the figure. Assuming .V V 0 7BE EB1 2= = volt, value of the dc voltage VC2 (in volt)

is ______

s1 Correct answer is 0.5 V VE1 . .2 5 0 7= − .1 8= V VB2 V VE EB1 2= − . .1 8 0 7= − .1 1= V

IB2 kV101B2= −

.k10

1 1 1= −

.k100 1=

IC2 IB2β =

.k50 100 1= ; E

VC2 kI 1C2= ^ h

.k k10

50 0 11= ^^h

h

.0 5= V

Q25 In a stable multivibrator circuit shown in the figure, the frequency of oscillation (in kHz) at the output pin 3 is ______

s1 Correct answer is 5.65

T1 . R R C0 693 A B= +^ h . . . .k k0 693 2 2 4 7 0 022µ = +^ h .0 1052= msecFrequency of oscillations f^ h

T1=

.5 65= kHz

Q26 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(A) 4E and 0

(B) 4E and 1

(C) 4F and 0

(D) 4F and 1

s1 Correct option is (D)Given A A7H= 1 01 0 011

CY0=

After executing RLC A 1 0 0 0111 1= A F4 H= and cy 1=

Q27 The logic functionality realized by the circuit shown below is

EC-Gate 2016 SET-3

(A) OR

(B) XOR

(C) ANAD

(D) AND

s1 Correct option is (D)It is a AND gate

A B Y

0 0 0

0 1 0

1 0 0

1 1 1

Q28 The minimum number of 2-input NAND gates required to implement a 2-input XOR gate is (A) 4

(B) 5

(C) 6

(D) 7

s1 Correct option is (A)Minimum number of NAND gates required for 2-input EX-OR gate=4

Q29 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 G HG G1 1 1

1 2= +

(B) G G G G HG G

1 1 2 1 1

1 2= + +

(C) G G G HG G

1 1 2 1

1 2= +

(D) G G G G G HG G

1 1 2 1 2 1

1 2= + +

s1 Correct option is (B)From block diagram

X sY s^

^

h

h G s= ^ h

G H G GG G

1 1 1 1 2

1 2= + +

Q30 For the unity feedback control system shown in the

figure, the open-loop transfer function G s^ h is given as G s s s 1

2= +^^

hh. The steady state error ess due to a unit step

input is

(A) 0

(B) 0.5

(C) 1.0

(D) 3


Given G s^ h s s 12=+^ h

,

H s^ h 1=& Type 1- system, to the unit step input the e 0ss =

Q31 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 _____

s1 Correct answer is 3485 MHz fIf 15= MHz FLo 3500= MHz f fs Lo- fIf= fs f fLo If= + 3515= MHz

fsi = image frequency f f2s If= − 3515 2 15#= − 3485= MHz

Q32 An analog baseband signal, bandlimited to 100 Hz, is sampled at the Nyquist rate. The samples are quantized into four message symbols that occur independent with probabilities .p p 0 1251 4= = and p p2 3= . The information rate (bits/sec) of the message source is ______

s1 Correct answer is 362.255 P1 .0 125= P4 .0 125= P2 .0 375= P3 .0 375= H . . . .log log0 125 0 125

1 0 125 0 1251

2 2= +

. . . .log log0 375 0 3751 0 375 0 375

12 2+ +

.1 811=

EC-Gate 2016 SET-3

Information rate H 200#= .362 255=

Q33 A binary baseband digital communication system

employs the signal p t^ h otherwiset T

00 < <

T S1S= .

For transmission of bits. The graphical representation of the matched filter output y t^ h for this signal will be


Q34 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

(C) elliptically polarized with a tilt angle of 45c

(D) horizontally polarized


If the wave is incident on perfect conductor then reflection coefficient is given by

Γ EEi

r

0

0/

1=− Er0 E 180i0 c+=If incident wave is traveling along Z+ direction then the reflected wave will be traveling along Z- direction.

The reflected wave is left hand circularly polarized

(LHCP)

Q35 Faraday’s law of electromagnetic induction is mathematically described by which one of the following equations?(A) B 0:d =v

(B) D v:d ρ =v

(C) EtB

#d 22=−v v

(D) H EtD

#d 22σ = +v v v

s1 Correct option is (C)Differential form of Faraday’s law is given by

E#d v tB22=−v

or

E#d v tH22µ =−v

Q.36-Q.55 carry two marks each

Q36 The particular solution of the initial value problem given below is

dxd y

dxdy y12 362

2

+ + 0= with y 0 3=^ h and

dxdy

x 0= 36=−

(A) x e3 18 x6- -^ h

(B) x e3 25 x6+ −^ h

(C) x e3 20 x6+ −^ h

(D) x e3 12 x6- -^ h

s1 Correct option is (A) D D12 362+ + 0=& ,6 6=− −The solution is y C e C xex x

16

26= +− − ...(1)

y 0^ h 3= C3 1& =

(1) y& e C xex x62

6= +− −

dxdy e C xe e18 6x x x6

26 6=− + − +− − −" ,

& dxdy

x 0= C18 2=− +

& 36- C18 2=− +& C2 18=−

The solution is y e xe3 18x x6 6= −− −

Q37 If the vectors , ,e 1 0 21= ^ h, , ,e 0 1 02 = ^ h and , ,e 2 0 13 = −^ h form an orthogonal basis of the three

dimensional real space R3, then the vector , ,u R4 3 3 3ε = −^ h can be expressed as

(A) u e e e52 3 5

111 2 3=− − −

(B) u e e e52 3 5

111 2 3=− − +

(C) u e e e52 3 5

111 2 3=− + +

(D) u e e e52 3 5

111 2 3=− + −

EC-Gate 2016 SET-3

s1 Correct option is (D) u x e x e x e1 1 2 2 3 3= + + , ,4 3 3-^ h , , , , , ,x x x1 0 2 0 1 0 2 0 11 2 3= + + −^ ^ ^h h h x x21 3- 4= ...(3)On solving these equations, we get

x1 52=− ,

x2 3= ,

x3 511= −

u e e e52 3 5

1112 3=− + −

Q38 A triangle in the xy -plane is bounded by the straight lines x y2 3= , y 0= and x 3= . The volume above the triangle and under the plane x y z 6+ + = is ______


Volume zdxdy x y dxdy6y

x

x 0

32

0

3= − −

==^ h####

10=

Q39 The values of the integral dzj ze

c

21

2

z

π -# along a closed contour c in anti-clockwise direction for(i) the point z 20 = inside the contour c , and(ii) the point z 20 = outside the contour c , respectively, are(A) (i) 2.72 (ii) 0

(B) (i) 7.39 (ii) 0

(C) (i) 0 (ii) 2.72

(D) (i) 0 (ii) 7.39


(i) j ze dz2

12

z

cπ -^ h# j jf2

1 2 2p p= ^ h

& e2 .7 39=

(ii) j ze dz2

12

z

cπ -^ h# 0= ( z 2a = lies outside c)

Q40 A signal cos cost t2 32 π -π ^ ^h h is the input to an LTI

system with the transfer function H s e es s= + −^ h . 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

s1 Correct option is (B) H ejω

^ h e ej j= +w w−

cos2 ω =

It x t^ h cos t2 32π = b l

0ω 32π =

H j 0ω ^ h cos2 32π = b l

2 21 1= − =−b l

y t^ h cos t2 32 180cπ = +b l

x t^ h cos tπ = 0ω π = H j 0ω ^ h cos2 π = ^ h 2=− y t^ h cos t2 180cπ = +^ h

y t^ h cos cost t2 32 2p p p p= + − +b ^l h

1ω 32π =

2ω π = T1 3= T2 2= T0 6= 0ω T

20

π =

3π =

y t^ h cos cost t2 2 2 20 0ω π ω π= + − +^ ^h h

y t^ h e e e ej t j t j t j t2 2 3 30 0 0 0= + − −ω π ω π ω π ω π+ − + + − +^ ^ ^ ^h h h h

y t^ h e e e ej t j t j t j t2 2 3 30 0 0 0=− − + +w w w w− −^ ^ ^ ^h h h h

C3 1=

Q41 The ROC (region of convergence) of the z -transform of a discrete-time signal is represented by the shaded region is the z -plane. If the signal .x n 2 0 n= ^ h6 @ , n< <3 3− + then the ROC of its z -transform is represented by

EC-Gate 2016 SET-3

s1 Correct option is (D)

x y^ h u n u n2 21 1nn

= + − −^ ^ b ^h h l h

roc z z2> < 21k= ^ _h i

φ = No ROC

Q42 Assume that the circuit in the figure has reached the steady state before time t 0= when the 3Ω resistor suddenly burns out, resulting in an open circuit. The current i t^ h (in ampere) at t 0= + is ______

s1 Correct answer is 0.98 to 1.02Here direction of current & correct component was not mentionedCircuit at t 0= −

Now t 0= +

So, i 0 +] g 44= −

A1=− The magnitude of the current is 1 Amp

Q43 In the figure shown, the current i (in ampere is

______

s1 Correct answer is i A1=−

Nodal

V V V V18

1 18

1−

+ +−

+^ ^h h 0=

V4 16= V 4= VoltsNow KCL i 4 4 1− + + 0= i A1=−

Q44 The z -parameter matrix zz

zz

11

21

12

22> H for the two-port

network shown is

(A) 22

22--> H

(B) 22

22> H

(C) 9639-> H

(D) 9639> H

s1 Correct option is (A)This is in Lattice formwhere Za 3Ω = Zb 0Ω = Zc 0Ω =

EC-Gate 2016 SET-3

Zd 6Ω =But it is not symmetrical & balanced

So, Z11 IV

I1

1

02

==

//3 6 2Ω = =

Z21 IV

I1

2

02

==

KVL V I2 02 1− − + 0= V2- I2 1=−& Z21 2Ω =−Also

So, Z22 IV

I2

2

01

==

//3 6 2Ω = =

Z21 IV

I2

1

01

==KVL

V I21 2- - 0= V1 I2 1=−& Z21 2Ω =−& Z12 2Ω =−So, final answer

Z 22

22 Ω = −

−> HQ45 A continuous-time speech signal x ta ^ h 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 radix-2 decimation-in-frequency FFT algorithm. If the processor performs all operations sequentially, and takes s20 µ 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 ______

s1 Correct answer is 8The number of complex multiplications required for DIF-FFT

logN N2 2= b l

log secN N2 202 µ b ^l h sec125 µ =

Q46 The direct form structure of an FIR (finite impulse response) filter is shown in the figure.

The filter can be used to approximate a(A) low-pass filter

(B) high-pass filter

(C) band-pass filter

(D) band-stop filter

s1 Correct option is (C) y n^ h x n x n5 2= − −^ ^h h6 @ Y ejω ^ h e X e5 1 j j2= − w w−

^ h8 B H ejω ^ h e5 1 j2= − ω −6

ω H ejω ^ h

0 0

2π 10

π 0

So it is Band pass filter

Q47 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, cm800n

2µ = /(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 V 26T = mV at room temperature, electronic charge . Cq 1 6 10 19#= − )

EC-Gate 2016 SET-3

s1 Correct answer is 6.656 MA

IC AeD dxdnn=

Ae dxdnnµ =

IC . . .0 001 1 6 10 800 0 02619# # # #= −

.0 5 1010 0

4

4

##

--d n

IC .6 656= mA

Q48 Figures I and II show two MOS capacitors of unit area. The capacitor in Figure I has insulator materials X (of thickness t 11 = nm and dielectric constant 41ε =) and Y (of thickness t 32 = nm and dielectric constant

202ε = ). The capacitor in Figure II has only insulator material X of thickness t .Eq . If the capacitors are of equal capacitors, then the value of tEq (in nm) is ______

s1 Correct answer is 1.55 to 1.65

C dAε =

C C CC C1 2

1 2= +

.8 8521 101 104

3 1020

1 104

3 1020

12

9 9

9 9##=

+# #

# # −

− −

− −> H .2 5 109 0# ε =

C teqr 0e e=

teq .2 5 10r

90

0

# ee e=

.2 5 104

90

0

#

#

ee=

.1 6 10 9#= − m .1 6= m

Q49 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

(B) D2 only

(C) Both D1 & D2

(D) None of D1 & D2


Here both zener diodes are in RB VBZ1 80= V VBZ2 70= VD1 have list saturation currentWhen we will vary the voltage above 80 VD1 get breaks down and will replaced by 80 V and through it ‘3’ current can flow through it. But because of D2 we will take minimum current i.e. net current equals to reverse saturation current of D2 as we know

The diode have least saturation will break down first and it will replaced by its break down voltage and the net current equal upto other diode reverse saturation current.

Q50 For the circuit shown in the figure, R R R 11 2 3 Ω = = =, HL 1µ = and FC 1µ = . If the input cosV t10in

6= ^ h, then the overall voltage gain V

Vin

out_ i of the circuit is ______

EC-Gate 2016 SET-3

s1 Correct answer is -1

VVin

x s

11016= + −

s ss1 10 106 6

= + + +

V0 V11s

x106 $=+−< F

VVx

0=

s

s106

=+−

VVin

0 VVVV

x in

x0 $=

ss

ss

1010

6

6

= +− +: :D D

1=−

VVin

0 1=−

Q51 In the circuit shown in figure, the channel length modulation of all transistor is non-zero 0!λ ^ h. Also all transistors operate in saturation and have negligible body effect. The ac small signal voltage gain V

Vin

0_ i of the circuit

is

(A) // //g r r rm1 01 02 03- ^ h

(B) // //g r g r1m

m1 0

303- b l

(C) // // //g r g r r1m

m1 01

202 03- b l; E

(D) // // //g r g r r1m

m1 01

303 02- b l; E


VVin

0 | | | | | |g r g r r1m

m1 01

202 03= b l; E

Q52 In the circuit shown in the figure,transistor M1 is in saturation and has transconductance

.g 0 01m = seimens. Ignoring internal parasitic capacitances and assuming the channel length modulation λ to be zero, the small signal input pole frequency (in kHz)is ______

s1 Correct answer is 57.8745 kHz

CM1 PF A50 1 V= −6 @ AV g Rm D=− .0 01 1#=− AV 10=− CM1 PF50 1 10= +6 @ . F0 55 10 9#= −

.0 55= nF

fp R C21i M1π =

fp .K2 5 0 55 101

9# # #π

= −

.57 8745= kHz

Q53 Following is the K-map of a Boolean function of five variables, , , ,P Q R S and X . The minimum sum-of-produce (SOP) expression for the function is

EC-Gate 2016 SET-3

(A) PQSX PQSX QRSX QRSX+ + +(B) QSX QSX+(C) QSX QS X+(D) QS QS+

s1 Correct option is (B)It is a 5-variable K-map QS X QSX= +

Q54 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 ______

s1 Correct answer is 6 T NOR MUXI MUX0 2

. .ns ns ns2 1 5 1 5" " "=

Delay . .2 1 5 1 5= + + ns5= T NOT MUXI NOR MUX1 2

. .ns ns ns ns1 1 5 2 1 5" " " "=

Delay . .1 1 5 2 1 5= + + + ns6=

Maximum delay secn6=

Q55 For the circuit shown in the figure, the delay of the bubbled NAND gate is 2ns and that of the counter is assumed to be zero

If the clock (C1k) frequency is 1Ghz, the counter behaves as a (A) mod-5 counter

(B) mod-6 counter

(C) mod-7 counter

(D) mod-8 counter

s1 Correct option is (D)At 6th C1k pulse (i.e. at 6ns) Q Q Q 1102 1 0& &= It makes NAND gate output ‘0’ at 8ns due to its delay. By that time counter receives ,7 8th th C1k pulse and counts 111, 000. Thus it is a mod 8- counter.

Q56 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) . .K2 0 0 5< <− −(B) .K0 0 5< <(C) K0 < <3(D) . K0 5 < <3

s1 Correct option is (D)

for stability

s

s

s

s

kk1

2

4

2 34

3

2

1

0

+

k k4 6 42+ − 0> ,k 2> >− .k 0 5> . k0 5 < <3

Q57 A second-order linear time-invariant system is described by the following state equations

dtd x t x t21 1+^ ^h h u t3= ^ h

dtd x t x t2 2+^ ^h h u t= ^ h

where x t1^ h and x t2^ h are the two state variables and u t^ h denotes the input. If the output c t x t1=^ ^h h, then the system is (A) controlled but not observable

(B) observable but not controllable

(C) both controllable and observable

(D) neither controllable nor observable

s1 Correct option is (A) x1o x U2 31=− + x2o x U2=− +

EC-Gate 2016 SET-3

c x1=

xx1

2

o

o> H xx U

2001

11

1

2=

−− +> > >H H H

c6 @ xx

1 0 1

2= 8 >B H

By applying Gilbert’s test, the system is controlled but not observable.

Q58 The forward-path transfer function and the feedback-path transfer function of a single loop negative feedback control system are given as G s s s

K s

2 2

22= + +

+^

^h

h and H s 1=^ h 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 ______

s1 Correct answer is .3 41-

Given G s^ h s sK s

2 22

2=+ +

+^ h,

H s^ h 1=

Break away point& dsdk 0=

dsds ss

2 22

2 + ++

d n 0=

& s s

s s s s2 2

1 2 2 2 2 22 2

2

+ ++ + − + +^

^ ^ ^

h

h h h> H 0=

& s s4 22- - - 0=& .0 58- , .3 41-Valid BAP is .3 41-

Q59 A wide sense stationary random process X t^ h passes through the LTI system shown in the figure. If the autocorrelation function of X t^ h is RX τ ^ h, then the autocorrelation function RY τ ^ h of the output y t^ h is equal to

(A) R R T R T2 X X X0 0t t t+ − + +^ ^ ^h h h

(B) R R T R T2 X X X0 0t t t− − − +^ ^ ^h h h

(C) R R T2 2X X 0t t+ −^ ^h h

(D) R R T2 2 2X X 0t t- -^ ^h h

s1 Correct option is (B) Y t^ h X t X t T0= − −^ ^h h

ACf of o/p

Ry τ = ^ h

E y t Y t τ = +^ ^h h7 A Ry τ ^ h E X t X t T X t X t T0 0t t= − − + − + −^ ^ ^ ^h h h h68 B Ry τ ^ h E X t X t X t X t T X t T0 0t t= + − + − − −^ ^ ^ ^ ^h h h h h6 X t X t T X t T0 0t t+ + − + −^ ^ ^h h hA Ry τ ^ h R R T R T RX X X X0 0t t t t= − − − + +^ ^ ^ ^h h h h8 B Ry τ ^ h R R T R T2 X X X0 0t t t= − − − +^ ^ ^h h h

Q60 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 5 102

5#=η −

Watt per Hz. If information at the rate 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 ______

s1 Correct answer is 31.503 C 52= kbps B 4= kHz

No2 .2 5 10 5#= −

N .4 10 2 5 10 23 5# # # #= −

C logB NS12= +: D

S .1638 2= Eb R

Sb

=

//secsec

bitsJ=

.31 503= B

C log NS11= +b l

& log NS12 +b l B

C=

& NS1 +b l 2 /C B=

213= 8192=& N

S=

8191=& S .8191 4 10 2 5 10 23 5

# # # # #= −

.819 1 2#= Eb

.R

819 1 2b

#=

.31 503=

Q61 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

EC-Gate 2016 SET-3

will be in error is ______

s1 Correct answer is 0.735 P N10 5= −

105=Method 1:Binomial Method: n p qCx

x n x-

P x P x0 1= + =^ h 6 @ c10 10 1 1050

5 0 5 105= −− −^ ^h h

c10 10 1 1051

5 1 5 10 15

+ −− − −^ ^h h

. .1 1 0 367 0 367#= +^ ^h h .0 735=Method 2:

Poisson method !xe xλ =

λ −

λ np 10 105 5#= = −

1=

!e e111 1

1= +−

−] g

& e2 1#

- .0 735=

Q62 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 ______

s1 Correct answer is 158.07GivenAir filled RWG, a .2 286= cm b .1 016= cm f 10= GHzAssume dominant mode TE10^ h is propagating in the waveguide, cut-off frequency of TE10 mode is given by

TEfc 10^ h ac2=

.2 2 2863 1010

##=

fc .6 56= GHzPropagation constant γ is given by γ jβ = v

i ff1 c

0 0

2

ω µ ε= − c m

.j2 10 103 101 1 10

6 5698

2

# # ##

π = − b l

γ . mj158 07 1= −

Therefore the value of propagation constant is given by γ . m158 07 1= −

Q63 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(A) TE TE TE TE< < <01 10 11 20

(B) TE TE TE TE< < <20 11 10 01

(C) TE TE TE TE< < <10 20 01 11

(D) TE TE TE TE< < <10 11 20 01

s1 Correct option is (C) a .2 286= cm b .1 016= cmair filled RWG

f TEc 11^ h ca b21 12 2= + (a m 1= , n 1= )

. .2

3 102 2161

1 016110

2 2#= +

^ ^c h h m

f TEc 11^ h .16 15= GHz

f TEc 11^ h bc2=

.201 0163 1010#=

.14 76= GHz

f TEc 20^ h ac=

.2 2863 1010#=

.13 12= GHz

f TEc 20^ h ac2=

.2 2 2863 1010

##=

.6 56= GHz Increasing order of the cut-off frequency is given by

TE TE TE TE< < <10 20 01 11

Q64 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 regret 1 km away having radar cross-section of m3 2. If it transmit 100 kW, then the received power (in Wµ ) is

s1 Correct answer is 0.0122Givenfrequency, f 5= GHz 5 109#= Hz

wave length, λ fc=

5 103 10

9

8

#

#=

.0 06= mgain of antenna, G 150=Range of target, Rmax 1= km 103= mradar cross-section, σ m3 2=transmitted power, Pt 100= kWThe RADAR range equation is given by

Rmax PP G G

4 4 R

t24

2

#

# # # #

πσ

= πλ

^ h> H

A G4e

2

a πλ=c m

The received power, PR is given by

PR .

4 4 10100 10 150 150 0 06 3

3 3 4

3 2

#

# # # # #

π =

^

^

h

h

.1 22 10 8#= −

. W0 0122 µ =

EC-Gate 2016 SET-3

Q65 Consider the charge profile shown in figure. The resultant potential distribution is best described by

s1 Correct answer is (D)Let us consider b 1=− and a 1=For line (1):Here (-1,0) to (0,-1) the line equation is

y 0- t11 1= − +^ h

y t 1=− −

t dt1t

1− +

−^ h#

t211 1 0< <2

=− +^ h

For line (2)Here ,0 1-^ h to (1, 0) the line equation is y t 1= −

t dt1t

0-^ h#

tt2

10 1< <2

=−^ h

At t 0= −:

y t21 2

=− +^ h

y 21= −

At t 0= +:

y t21 2

=−^ h

y 21=

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