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GATE SOLVED PAPERElectronics & Communication 2014-2

Copyright © By NODIA & COMPANY

Information contained in this book has been obtained by authors, from sources believes to be reliable. However, neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor its authors shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that Nodia and its authors are supplying information but are not attempting to render engineering or other professional services.

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GATE SOLVED PAPER - EC2014-2

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General Aptitude

Q. 1 - Q. 5 carry one mark each.

Q. 1 Choose the most appropriate word from the options given below to complete the following sentence.Communication and interpersonal skills are ____important in their own ways.(A) each (B) both

(C) all (D) either

Sol. 1 Correct option is (B).

Q. 2 Which of the options given below best completes the following sentence ?She will feel much better if she ____.(A) will get some rest (B) gets some rest

(C) will be getting some rest (D) is getting some rest


Q. 3 Choose the most appropriate pair of words from the options given below to complete the following sentence. She could not _____ the thought of _________ the election to her bitter rival.(A) bear, loosing (B) bare, loosing

(C) bear, losing (D) bare, losing

Sol. 3 Correct option is (C).

Q. 4 A regular die has six sides with numbers 1 to 6 marked on its sides. If a very large number of throws show the following frequencies of occurrence : .1 0 167" ;

.2 0 167" ; .3 0 152" ; .4 0 166" ; .5 0 168" ; .6 0 180" . We call this die(A) irregular (B) biased

(C) Gaussian (D) insufficient


Q. 5 Fill in the missing number in the series.2 3 6 15 _____ 157.5 630

Sol. 5 Correct answer is 45.

Q. 6 - Q. 10 carry two mark each.

Q. 6 Find the odd one in the following groupQ, W, Z, B B, H, K, M W, C, G, J M, S, V, X(A) Q, W, Z, B (B) B, H, K, M

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GATE SOLVED PAPER - EC 2014-2

(C) W, C, G, J (D) M, S, V, X

Sol. 6 Correct option is (C).

Q. 7 Lights of four colors (red, blue, green, yellow) are hung on a ladder. On every step of the ladder there are two lights. If one of the lights is red, the other light on that step will always be blue. If one of the lights on a step is green, the other light on that step will always be yellow. Which of the following statements is not necessarily correct ?(A) The number of red lights is equal to the number of blue lights

(B) The number of green lights is equal to the number of yellow lights

(C) The sum of red and green lights is equal to sum of the yellow and blue lights

(D) The sum of red and blue lights is equal to the sum of green and yellow lights

Sol. 7 Correct option is (D).

Q. 8 The sum of eight consecutive odd numbers is 656. The average of four consecutive even numbers is 87. What is the sum of the smallest odd number and second largest even number ?


Q. 9 The total exports and revenues from the exports of a country are given in the two charts shown below. The pie chart for exports shows the quantity of each item exported as a percentage of the total quantity of exports. The pie chart for the revenues shows the percentage of the total revenue generated through export of each item. The total quantity of exports of all the items is 500 thousand tonnes and the total revenues are 250 crore rupees. Which item among the following has generated the maximum revenue per kg ?

(A) Item 2 (B) Item 3

(C) Item 6 (D) Item 5


Q. 10 It takes 30 minutes to empty a half-full tank by draining it at a constant rate. It is decided to simultaneously pump water into the half-full tank while draining it. What is the rate at which water has to be pumped in so that it gets fully filled in 10 minutes ?(A) 4 times the draining rate (B) 3 times the draining rate

(C) 2.5 times the draining rate (D) 2 times the draining rate

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Sol. 10 Correct option is (A).

END OF THE QUESTION PAPER

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Electronics and Communication

Q. 1 - Q. 25 carry one mark each.

Q. 1 The determinant of matrix A is 5 and the determinant of matrix B is 40. The determinant of matrix AB is ______.

Sol. 1 Correct answer is 200.Given the determinent of matrix A and B as | |A 5= , | |B 40= .From the property of matrix, we have

| |AB | || |A B= 5 40 200#= =

Q. 2 Let X be a random variable which is uniformly chosen from the set of positive odd numbers less than 100. The expectation, E X6 @, is _____.

Sol. 2 Correct answer is 50.Given that the random variable X is uniformly chosen from the set of positive odd number less than 100. So, we have

X , , , .....1 3 5 7 99=So, the total number of values in set of X is

n 50=Since, X is uniformly chosen from the set of numbers. So, the probability density function for each value of X is

( )f xX n1

501= =

Hence, we obtain the expected value of X as

[ ]E X ( )Xf xX= / ....50

1503

505

5099= + + + +

( ... )501 1 3 5 99= + + + +

( )501

250 2 1 50 1 2#= + -: D" ,

( )21 2 49 2 50#= + =

Q. 3 For t0 < 3# , the maximum value of the function f t e e2t t2= -- -^ h occurs at(A) logt 4e= (B) logt 2e=(C) t 0= (D) logt 8e=

Sol. 3 Correct option is (A).Given the function,

( )f f e e2t t2= -- -

For maximum value of the function we have

( )f tl 0=

& ( )dtd e e2t t2-- - 0=

& e e4t t2- +- - 0=& t log 4e=

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Now, we check this value for point of maxima. We have

( )f tll ( )dtd e e4t t2= - +- -

e e8t t2= -- -

Substituting logt 4e= in above expression,

( )logf t 4e=ll e e8log log4 2 4e e= -- -

41 8 4

141

212

= - = -b l

41 0<= -

Hence, logt 4e= is the point of maxima.

Q. 4 The value of

lim x1 1x

x+

"3b l

is(A) ln2 (B) 1.0

(C) e (D) 3

Sol. 4 Correct option is (C).The exponential function arises whenever a quantity grows or decays at a rate proportional to its current value. Bernouli defined the exponential function as

e n1 1 n

n

= +"3

b l/

or e lim x1 1n

x

= +"3

b l

AlternativeThe function given is a standard exponential term, however we can also prove it by using Taylor’s series. Let

y x1 1 x

= +b l

Taking logarithm both the sides,

log ye log x1 1e

x

= +b l: D

& log ye logx x1 1e= +b l

& log ye ...x x x x1

21

31

2 3= - + -: D

& log ye ....x x1 2

131

2= - + -

& y e ...x x1 2

131

2= - + -c m

& limyx"3

lim e ....

xx x

1 21

31

2="3

- + -c m8 B

& lim x1 1x

x+

"3b l e e1= =

Q. 5 If the characteristic equation of the differential equation

dxd y

dxdy y22

2

a + + 0=

has two equal roots, then the values of a are(A) 1! (B) 0, 0

(C) j! (D) /1 2!

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Sol. 5 Correct option is (A).Given the differential equation,

dxd y

dxdy y22

2

a + + 0=

or D D2 12 a + + 0=The above differential equation has the equal roots if its discriminant is zero, i.e.

( )2 42a - 0=& a 1!=

Q. 6 Norton’s theorem states that a complex network connected to a load can be replaced with an equivalent impedance(A) in series with a current source (B) in parallel with a voltage source

(C) in series with a voltage source (D) in parallel with a current source

Sol. 6 Correct option is (D).Norton’s theorem states that a complex network connected to a load can be replaced with an equivalent impedance in parallel with a current source as shown in figure below.

Q. 7 In the figure shown, the ideal switch has been open for a long time. If it is closed at t 0= , then the magnitude of the current (in mA) through the k4 W resistance at t 0= + is _____.

Sol. 7 Correct answer is 1.25.Given the circuit

Before t 0= , the switch was open for a long time. So, the capacitor behaves as open circuit and inductor as short circuit. Hence, we get

( )V 0C-

( )V5 4 1

10 4 15#= + +

+ =

( )i 0L- mA5 4 1

10 1= + + =

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At t 0= , switch is closed. But, we know the properties:1. Capacitor opposes the change in voltage across it

2. Inductor opposes the, change in current through it.

So, we have the equivalent circuit at t 0= + as shown below

Applying KVL, we have

i5 4- 0=

i .45 1 25= =

Q. 8 A silicon bar is doped with donor impurities . /atoms cmN 2 25 10D15 3

#= . Given the intrinsic carrier concentration of silicon at KT 300= is . cmn 1 5 10i

10 3#= -

. Assuming complete impurity ionization, the equilibrium election and hole concentrations are(A) . cmn 1 5 100

6 3#= - , . cmp 1 5 100

5 3#= -

(B) . cmn 1 5 10010 3

#= - , . cmp 1 5 10015 3

#= -

(C) . cmn 2 25 10015 3

#= - , . cmp 1 5 10010 3

#= -

(D) . cmn 2 25 10015 3

#= - , cmp 1 1005 3

#= -

Sol. 8 Correct option is (D).Given the concentration of donor impurities,

ND . /atoms cm2 25 1015 3#=

Intrindic carrier concentration,

ni . cm1 5 1010 3#= -

Since, we have

ND n>> i

So, the equilibrium electron concentration is

n0 . cmN 2 25 10D15 3

#= = -

Hence, using mass action law, we obtain the hole concentration as

p0 nn

o

i2

= ..

cm2 25 101 5 10

1 1015

10 25 3

#

##= = -^ h

Q. 9 An increase in the base recombination of a BJT will increase(A) the common emitter dc current gain b (B) the breakdown voltage BVCEO

(C) the unity-gain cut-off frequency fT(D) the transconductance gm

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Q. 10 In CMOS technology, shallow P -well or N -well regions can be formed using(A) low pressure chemical vapour deposition

(B) low energy sputtering

(C) low temperature dry oxidation

(D) low energy ion-implantation


Q. 11 The feedback topology in the amplifier circuit ( the base bias circuit is not shown for simplicity) in the figure is

(A) Voltage shunt feedback (B) Current series feedback

(C) Current shunt feedback (D) Voltage series feedback

Sol. 11 Correct option is (B).Given the feedback topology in the amplifier circuit,

We define the current series feedback asCurrent is sampled from the outputVoltage is feedback to the source/input.In the given circuit, the feedback signal becomes zero by opening the output feedback. Hence, it is current series feedback.

Q. 12 In the differential amplifier shown in the figure, the magnitudes of the common-mode and differential-mode gains are Acm and Ad , respectively. If the resistance RE is increased, then

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(A) Acm increases

(B) common-mode rejection ratio increases

(C) Ad increases

(D) common-mode rejection ratio decreases


Q. 13 A cascade connection of two voltage amplifiers A1 and A2 is shown in the figure. The open-loop gain Av0, input resistance Rin , and output resistance Ro for A1 and A2 are as follows :

A1 : A 10v0 = , kR 10in W = , kR 1o W = A2 : A 5v0 = , kR 5in W = , R 200o W =The approximate overall voltage gain /v vout in is______.

Sol. 13 Correct answer is 34.722.Given the cascade connection

For the two stages of cascade, we have:A1 A 10v1 = , kR 10i1 W = , kR 1o1 W =:A2 A 5v2 = , kR 5i2 W = , R 200o1 W = .

So, we may redraw the equivalent 1st stage of cascade as

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So, we get

v vi oL2 = R RR A vin

o i

iv

1 2

21= +

v v1 55 10 6

50in in#= + =

Again, we obtain the equivalent 2nd stage circuit as

So, we obtain

vout Av v R RR

io L

L2 2

2= +

KK v200 1

1 5 650

in# #= +

Hence, vvin

o . .1 21 5 6

50 34 722# #= =

Q. 14 For an n -variable Boolean function, the maximum number of prime implicants is(A) n2 1-^ h (B) /n 2

(C) 2n (D) 2 n 1-^ h

Sol. 14 Correct option is (D).A prime implicant of a function is an implicant that can not be reduced to a more general form.For example, we consider the K-map of variables A, B .

In the K-map, pair or quadrants are not for opened for the two casesm m 11 4= = , m m 02 3= = ; Prime implication ,A B AB/ " ,

m m 01 4= = , m m 02 3= = ; Prime implication ,A B AB/ " ,

For all other cases of two or more implicants, the K-map has a pair or quadrant; i.e function is reduced. Hence, the maximum number of prime implicant is 2.Similarly, we may consider the cases for more number of variables. Thus, the maximum number of prime implicant is 2n 1- .

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Q. 15 The number of bytes required to represent the decimal number 1856357 in packed BCD (Binary Coded Decimal) form is_______.

Sol. 15 Correct answer is 4 bits.A digit of decimal number is represented into 4 bits of a BCD number. Since, the decimal number 1856357 have 7 digits. So, the total number of bits required to represent it in BCD form is bits7 4 28# = .Hence, the number of bytes byte bits1 8=^ h required is

number of bytes .828 3 5$ =

or number of bytes bits4=

Q. 16 In a half-subtractor circuit with X and Y as inputs, the Borrow (M ) and Difference N X Y= -^ h are given by(A) M X Y5= , N XY= (B) M XY= , N X Y5=(C) M XY= , N X Y5= (D) M XY= , N X Y5=

Sol. 16 Correct option is (C).We have the given difference function

N X Y= -and the borrow is M .

So, we may obtain the truth table as

X Y Diff erenceN

BorrowM

0 0 0 00 1 1 11 0 1 01 1 0 0

From the truth table, we obtain

N X Y X Y X Y5= + = M X Y=

Q. 17 An FIR system is described by the system function

H z^ h z z1 27

231 2= + +- -

The system is(A) maximum phase (B) minimum phase

(C) mixed phase (D) zero phase

Sol. 17 Correct option is (C).Given the FIR system function,

( )H z z z1 27

231 2= + +- -

The zeros of the function are obtained as

z z1 27

231 2+ +- - 0=

z z2 7 32 + + 3= z z z2 6 32 + + + 0= z z2 1 3+ +^ ^h h 0=

z ,21 3= - -

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Since, one zero /z 1 2=-^ h is inside the unit circle, and one zero z 3=-^ h is outside the unit circle. So, it is a mixed phase system.• For minimum phase system, all zeros are inside the unit circle.

• For maximum phase system all zeros are outside the unit circle.

Q. 18 Let x n x n= -6 6@ @. Let X z^ h be the z -transform of x n6 @. If . .j0 5 0 25+ is a zero of X z^ h, which one of the following must also be a zero of X z^ h.(A) . .j0 5 0 25- (B) / . .j1 0 5 0 25+^ h

(C) / . .j1 0 5 0 25-^ h (D) j2 4+

Sol. 18 Correct option is (B).From the time revorsal property of Z -transform, we know that

if [ ]x n [ ]x n= -Then ( )X Z ( )X Z 1= - ....(i)Since, ( . . )j0 5 0 25+ is a zero of ( )X Z , i.e.

( . . )X j0 5 0 25+ 0=So, using equation (i), we may also write

. .X j0 5 0 251

+b l 0=

i.e. . .j0 5 0 251

+b l is also a zero of ( )X Z .

Q. 19 Consider the periodic square wave in the figure shown.

The ratio of the power in the 7th harmonic to the power in the 5th harmonic for this waveform is closest in value to ______.

Sol. 19 Correct answer is 0.5.

Q. 20 The natural frequency of an undamped second-order system is 40 rad/s. If the system is damped with a damping ratio 0.3, the damped natural frequency in rad/s is ______.

Sol. 20 Correct answer is 4 rad/s.Given the natural frequency of an undamped second order system,

nw / secrad40=Damping ratio, d .0 3=So, the damped natural frequency is

dw 1n2w d= -

( . ) .40 1 0 3 40 1 0 092= - = - .40 0 1#= /rad s4=

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Q. 21 For the following system,

when X s 01 =^ h , the transfer function X sY s

2^

^

h

h is

(A) s

s 12

+ (B) s 11+

(C) s ss

12

++

^ h (D)

s ss

21

++

^ hSol. 21 Correct option is (D).

We have the block diagram

When ( )X s 01 = , the block diagram in its reduced form is shown below

Hence, we obtain the transfer function as

( )( )

X sY s

GHG

12= +

/( )

s ss

ss s

s1 1

1

121

#=

+ += +

+

Q. 22 The capacity of a band-limited additive white Gaussian noise (AWGN) channel

is given by logC WW

P12 2s = +b l bits per second (bps), where W is the channel

bandwidth, P is the average power received and 2s is the one sided power spectral density of the AWGN.For a fixed 1000P

2 =s , the channel capacity (in kbps) with infinite bandwidth W " 3^ h is approximately.(A) 1.44 (B) 1.08

(C) 0.72 (D) 0.36

Sol. 22 Correct option is (A).The channel capacity is defined as

C logW WP1

2 2s = +b l ...(i)Since, we know that

log ba log logc ba c=So, we may rewrite the equation (1) as

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C ( )log logW eW

P1e2 2s = +b l; E

( ) ...logW e WP

WP

WP

21

31

2 2 2

2

2

3

s s s= - + -b b bl l l; E

( ) ...log e PW

PW

P2 32 2 4

2

6 2

3

s s s= - + -; E ...(ii)

Again, We have

P2s 1000= , W " 3

Substituting these value in equation (ii), we get

C ( ) ....lim log eW

PW

P10002 3W

2 4

2

6 2

3

s s= - + -

"3 ; E

( )( )log e 1000 bits per sec2= . kbps1 44=

Q. 23 Consider sinusoidal modulation in an AM system. Assuming no overmodulation, the modulation index (m ) when the maximum and minimum values of the envelope, respectively, are 3 V and 1 V, is ______.

Sol. 23 Correct answer is 0.5.An amplitude modulated system is defined as

( )S t [ ( )]cosA m t t1c a n cm w= +where Ac is amplitude of carrier signal, am is modulation index, ( )m tn is normalized message signal; [ ( )]m t1 1n# #- .So, the maximum and minimum amplitude of AM signal are

| ( ) |S t max ( ) VA 1 3c am = + = | ( ) |S t min ( ) VA 1 1c am = - =Hence, We get

r11

a

a

m -+ 1

3=

22 am 4

2= am .0 5=

Q. 24 To maximize power transfer, a lossless transmission line is to be matched to a resistive load impedance via a /4l transformer as shown.

The characteristic impedance (in W ) of the /4l transformer is ______.

Sol. 24 Correct answer is .70 7 W .Given the input impedance,

Zin 50 W =

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Load impedance,

ZL 100 W =For a /4l transformer, the input impedance is given as

Zin ZZ

L

o2

=

So, we obtain the characteristic impedance as

Zo Z Z 50 100 5000in L #= = = .70 7 W =

Q. 25 Which one of the following field patterns represents a TEM wave travelling in the positive x direction ?(A) E y8=+ t, H z4=- t (B) E y2=- t, H z3=- t

(C) E z2=+ t, H y2=+ t (D) E y3=- t, H z4=+ t

Sol. 25 Correct option is (B).The direction of propagation of a TEM wave is given as

aTEMv a aE H#= v v

where aEv is direction of electric field, aHv is direction of magnetic field.From the given options, we have the option (B) with

Ev y2=- t ; a yE =-v t

Hv z3=- t ; a zH =-v t

So, the direction of TEM wave is

aTEMv ( ) ( )y z x#= - - =t t t

i.e. the TEM wave travelling in positive x -direction.

Q. 26 - Q. 55 carry two mark each.

Q. 26 The system of linear equations

abc

231

102

315

J

L

KKK

J

L

KKK

N

P

OOO

N

P

OOO

54

14= -

J

L

KKK

N

P

OOO has

(A) a unique solution (B) infinitely many solutions

(C) no solution (D) exactly two solutions

Sol. 26 Correct option is (B).Given the system of linear equation.

abc

231

102

315

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

54

14= -

R

T

SSSS

V

X

WWWW

Comparing it to general form of equation

[ ] [ ]A X [ ]B=We have

[ / ]A B RRR

231

102

315

54

14

1

2

3

"

"

"

= -

R

T

SSSS

V

X

WWWW

To make zero the first term of R2 and R3 using R1, we perform

R R R3 22 1 2" -

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R R R23 1 3" -So, we get

A B8 B 200

133

371

52323

=- -

R

T

SSSS

V

X

WWWW

Again, to make zero, the second term of R3, we perform R R R3 3 2" +Hence, we get

A B8 B 200

130

370

5230

=

R

T

SSSS

V

X

WWWW

So, we have the rank of matrices,

Rank A6 @" , Rank A B number of variables<= 8 B$ .

Thus, the system has infinite number of solution.

Q. 27 The real part of an analytic function f z^ h where z x jy= + is given by cose xy- ^ h. The imaginary part of f z^ h is(A) cose xy ^ h (B) sine xy- ^ h

(C) sine xy- ^ h (D) sine xy- - ^ h

Sol. 27 Correct option is (B).The analytic function is defined as

( )f z ( )f x jy u iv= + = +where v is the real part and v is imaginary part of ( )f z that satisfies C-R equations, i.e.

xu

22

yv

22= ...(i)

yu

22

xv

22=- ...(ii)

Given that real part of ( )f z is

u ( )cose xy= -

So, we get

xu

22 ( )sin sine x e xy y= - =-- -

Substituting is in equation (1), we have

yv

22 sine xy=- -

& v ( ) ( )sin sine xdy x e x xy yf f=- + = +- -#where ( )xf is a function of x .Again, we obtain

xv

22 ( )cose x xy f = +- l

and yu

22 cose xy=- -

Substituting it in equation (2), we get

cose xy- - ( )cose x xy f =- +- l6 @

( )xf l 0=Hence, the imaginary part of ( )f z is

v sine xy= -

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Q. 28 The maximum value of the determinant among all 2 2# real symmetric matrices with trace 14 is _____.

Sol. 28 Correct answer is 49.Let the 2 2# symmetric matrix be

A ab

bc= > H

Given that the trace of matrix is 14, i.e.

a c+ 14=The determinant of matrix is given as

| |A ac b2= -For maximum value of | |A , b2 must be minimum. Since, b2 is squared term (non-negative number) so, the minimum value is

b2 0=Therefore, we have (for maximum determinant)

| |A ( )ac a a a a14 14 2= = - = -For maximum value of | |A , we may write

| |da

d A 0=

& a14 2- 0=& a 7= , c 7=Hence, the maximum determinant is

| |A ac 7 7 49#= = =

Q. 29 If r xa ya zax y z= + +v t t t and r r=v , then lndiv r r2d =^^ hh ____.

Sol. 29 Correct answer is 3.Given the vector field,

rv xa ya zax y z= + +t t t

So, its magnitude is given as

| |rv r x y z2 2 2= = + +Therefore, we obtain

( )ln r4 ( )r r1 4=

r xr a

yr a

zr a1

x y z22

22

22

= + +t t t; E

r x y zx a

x y zy a

x y zz a1

22

22

22

x y z2 2 2 2 2 2 2 2 2= + ++

+ ++

+ +t t t= G

r rxa ya za

rr1 x y z2=

+ +=

t t t v: D

So, we have

( )lnr r24 rrr r22= =v v

Hence, we get

( )lndiv r r24^ h ( )div r= v

x

xy

yz

z 1 1 122

22

22= + + = + +

3=

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Q. 30 A series LCR circuit is operated at a frequency different from its resonant frequency. The operating frequency is such that the current leads the supply voltage. The magnitude of current is half the value at resonance. If the values of L , C and R are H1 , F1 and 1 W , respectively, the operating angular frequency (in rad/s) is ______.

Sol. 30 Correct answer is 0.45 r/sec.

Q. 31 In the h -parameter model of the 2-port network given in the figure shown, the value of h22 (in S ) in _____.


Q. 32 In the figure shown, the capacitor is initially uncharged. Which one of the following expressions describes the current I t^ h (in mA) for t 0> ?

(A) I t e1 /t35= - t -^ ^h h, secm3

2t =(B) I t e1 /t

25= - t -^ ^h h, secm3

2t =(C) I t e1 /t

35= - t -^ ^h h, secm3t =

(D) I t e1 /t25= - t -^ ^h h, secm3t =

Sol. 32 Correct option is (A).Given the circuit,

Initially, the capacitor is uncharged, i.e.

( )V 0c 0=For t 0> the capacitor is being charged by the voltage source of 5 V. The maximum voltage across capacitor ( )at t 3= is

( )Vc 3 R RR5 5 2 1

2310

1 2

2# #= + = + =

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Therefore, the voltage across capacitor at any particular time t is given as

( )V tc ( ) ( ) ( )V V V e0 /c c c

t3 3= + - t -6 @

e310 0 3

10 /t= + - t -: D

( )e310 1 /t= - t -

Hence, the current ( )I t through resistance R2 is

( )I t ( )

( )RV t

e35 1 /c t

2= = - t -

where t is the charging time given as R Ceqt =where Req is the equivalent resistance across capacitor, obtained as

Req kR RR R

1 21 2

32

1 2

1 2 # W = + = + =

Hence, t ( ) secmk32

1 32

# m = =b l

Q. 33 In the magnetically coupled circuit shown in the figure, %56 of the total flux emanating from one coil links the other coil. The value of the mutual inductance in H^ h is ______.

Sol. 33 Correct answer is 2.50 H.We have the magnetically coupled circuit as shown below

Given that 56% of the total flux emanating from one coil links the other coil. So, we have the coupling factor

k %56=or k .0 56=So, the value of mutual inductance is

M k L L1 2=

. . H0 56 4 5 2 50#= =

Q. 34 Assume electronic charge . Cq 1 6 10 19#= - , /kT q mV25= and electron mobility

/cm V s1000n2m -= . If the concentration gradient of electrons injected into a P

-type silicon sample is /cm1 1021 4# , the magnitude of electron diffusion current

density /in A cm2^ h is _____.

Sol. 34 Correct answer is 4000 /A cm2.Given the electronic charge, . Cq 1 6 10 19

#= -

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electron mobility, S-/cm V1000n2m =

electron concentration gradient, /dxdn 1 10 cm21 4

#=

Thermal voltage, mVqkT 25=

Now, from einstein equation, we have

Dn

n

m qkT=

& S-/cm

D1000 V

n2 Volt25 10 3

#= -

& Dn /cm S25 2=Hence, the diffusion current density is

Jn qD dxdn

n=

.1 6 10 25 1019 21# # #= -

/A cm4000 2=

Q. 35 Consider an abrupt PN junction (at KT 300= ) shown in the figure. The depletion region width Xn on the N -side of the junction is . m0 2 m and the permittivity of silicon sie ^ h is . /F cm1 044 10 12

#- . At the junction, the approximate value of the

peak electric field (in kV/cm) is_____.

Sol. 35 Correct answer is 30.7.Given the depletion region width of N-side of the function,

xn . .m cm0 2 0 2 10 4#m = = -

permittivity of silicon,

e . /F cm1 044 10 12#= -

Donor concentration,

ND /cm1016 3=Since, we have N N>>A D and we know that

x Nn D x Np A=So, the deption region width in p-side is much less than that on N -side, i.e.x x<<p n

Therefore, the electric field is given as

E eN xD n

e =

.

( . )( )( . )1 044 10

1 6 10 10 0 2 1012

19 16 4

#

# #= -

- -

. /V cm0 3065 105#=

. /kV cm30 7=

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Q. 36 When a silicon diode having a doping concentration of cmN 9 10A16 3

#= - on p-side and cmN 1 10D

16 3#= - on n -side is reverse biased, the total depletion width

is found to be m3 m . Given that the permittivity of silicon is . /F cm1 04 10 12#

- , the depletion width on the p-side and the maximum electric field in the depletion region, respectively, are(A) . m2 7 m and . /V cm2 3 105

# (B) . m0 3 m and . /V cm4 15 105#

(C) . m0 3 m and . /V cm0 42 105# (D) . m2 1 m and . /V cm0 42 105

#

Sol. 36 Correct option is (B).Given the doping concentrations, cmN 9 10A

16 3#= - , cmN 1 10D

16 3#= -

The permittivity of silicon is . /F cm1 04 10 12#e = -

Total depletion width, m cmW 3 3 10 4#m = = -

Let the depletion width on n -side and p-side be xn and xp respectively. So, we have

x xn p+ cmW 3 10 4#= = - ...(1)

Again, we have the relation

x Nn D x Np A=& ( )x 1 10n

16# ( )x 9 10p

16#=

& xn x9 p= ...(2)Substituting it in equation (i), we get

x x9 p p+ 3 10 4#= -

xp 103 10 4#=

-

. .cm m0 3 10 0 34# m = =-

Hence, the maximum electric field in depletion region is given as

Emax eN xA p

e =

.

( . ) ( ) ( . )1 04 10

1 6 10 9 10 0 3 102

19 16 4

#

# # # # #= -

- -

. /V cm4 15 105#=

Q. 37 The diode in the circuit shown has .V 0 7on = Volts but is ideal otherwise. If sinV t5i w = ^ h Volts, the minimum and maximum values of Vo (in Volts) are,

respectively,

(A) 5- and 2.7 (B) 2.7 and 5

(C) 5- and 3.85 (D) 1.3 and 5

Sol. 37 Correct option is (C).Given the diode circuit,

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and the input voltage is ( )sinV t5i w =So, we may sketch the waveform of input voltage as

Therefore, the maximum and minimum input voltage is

V , maxi V5= V , mini V5=-For the maximum voltage, VV 5, maxi =The diode is forward biased ( . )VV 0 7on = . Hence, The maximum output voltage corresponding to maximum input voltage is given as

Rv v, ,max maxi o

1

-

.R R

v 0 7 2, maxi

1 2= +

- -

& v ,maxo .

R Rv

R Rv 2 7, ,max maxi i

11 1 2

= - +-

; E

. .1 1

51 1

5 2 75 3

2 3= - +-

= -: D

. V3 85=For minimum input voltage, ( )Vv 5, mini =- , the diode is reverse biased; i.e. OFF. So, no current flows through the resistance. Hence, the minimum output voltage corresponding to minimum input is Vv 5o =- .

Q. 38 For the n -channel MOS transistor shown in the figure, the threshold voltage VTh is . V0 8 . Neglect channel length modulation effects. When the drain voltage

. VV 1 6D = , the drain current ID was found to be 0.5 mA. If VD is adjusted to be V2 by changing the values of R and VDD , the new value of ID (in mA) is

(A) 0.625 (B) 0.75

(C) 1.125 (D) 1.5

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Sol. 38 Correct option is (C).We have the MOS transistor circuit as shown below

The gate and drain terminals are common, So we have

VGS VDS=Also, we have

VD . V1 6= ; Vin . V0 8= ; ID . mA0 5=So, we may deduce that . VV V 1 6DS GS= = or V V>GS Th , V V V>DS GS Th-Therefore, the MOSFET is in saturation, and hence we write the drain current equation as

Io ( )C LW V V2

1n OX GS Th

2m = -

& . m0 5 ( . . )C LW

21 1 6 0 8n OX

2m = - ....(i)Now, the drain voltage changes to

VD V2=So, VDS VV 2GS= =Therefore, the current equation becomes

ID ( )C LW V V2

1n OX GS Th

2m = -

& ID ( . )C LW

21 2 0 8n OX

2m = - ...(ii)Dividing equation (2) by equation (1), we get

. mI

0 5D

( . . )( . )1 6 0 82 0 8

2

2

=-

-

& ID .( . )( . )

m0 50 81 2

2

2

=

. mA1 125=

Q. 39 For the MOSFETs shown in the figure, the threshold voltage VV 2t = and . /mA VK C 0 1ox L

W21 2m = =^ h . The value of ID (in mA) is ____.

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Sol. 39 Correct answer is 0.9.We have the MOSFET circuit as shown below.

From the circuit, we have

VG2 0=

and VG1 V R RR

DD1 2

2= +

12 10 1010

#= + V6=So, both the p-MOS and n -MOS are in saturation region.Also, we have the drain current, I I ID D D1 2= =So, we obtain the drain current as

Io2 ( )C LW V V2

1n OX G T2

2m = -

. [ ( ) ]0 1 0 5 2 2#= - - -

.0 1 9#= . mA0 9=or ID . mA0 9=

Q. 40 In the circuit shown, choose the correct timing diagram of the output y^ h from the given waveforms W1, W2, W3 and W4.

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(A) W1 (B) W2

(C) W3 (D) W4

Sol. 40 Correct option is (C).The two flip-flops in the circuit are negative edge triggered D -flip-flop. So, the output of the flip-flop changes when.Clock pulse changes from 1 to 0; i.e.

Q1 X1= , when CLK switches from 1 to 0

Q2 X2= , when CLK switches from 1 to 0

The output Y is Y Q Q1 2=Hence, we obtain the output waveform as

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Q. 41 The outputs of the two flip-flops Q1, Q2 in the figure shown are initialized to 0, 0. The sequence generated at Q1 upon application of clock signal is

(A) 01110.... (B) 01010....

(C) 00110.... (D) 01100....

Sol. 41 Correct option is (D).From the given circuit, we may deduce that

J1 Q2= , K Q1 2= J2 Q1= , K Q2 1=Initially, the output of flip-flops are

Q1 0= Q2 0=So, we obtain the truth table for the sequential circuit as

J1 K1 Q1 J2 K2 Q2

- - 0 - - 01 0 1 0 1 01 0 1 1 0 10 1 0 1 0 10 1 0 1 0 10 1 0 1 0 1

Thus, the sequence generated at Q1 is 01100...

Q. 42 For the 8085 microprocessor, the interfacing circuit to input 8-bit digital data DI DI0 7-^ h from an external device is shown in the figure. The instruction for

correct data transfer is

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(A) MVI A, F8H (B) IN F8H

(C) OUT F8H (D) LDA F8F8H


Q. 43 Consider a discrete-time signal

x n6 @ for

otherwise

n n0 10

0

# #= )

If y n6 @ is the convolution of x n6 @ with itself, the value of y 46 @ is _____.

Sol. 43 Correct answer is 10.Given the discrete time signal,

[ ]x n ,

,

for

otherwise

n n0 10

0

# #= *

and the signal [ ]y n is convolution of [ ]x n with itself i.e.

[ ]y n [ ] [ ]x n x n= *

[ ] [ ]x k x n kk

n

0

= -=/

Therefore, we obtain

[ ]y 4 [ ] [ ]x x x k4k 0

4

= -=/

[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]x x x x x x x x x0 4 1 3 3 2 2 3 1 4 0= + + + + 0 1 3 2 2 3 1 0# # #= + + + + 10=

Q. 44 The input-output relationship of a causal stable LTI system is given as

y n6 @ y n x n1a b= - +6 6@ @

If the impulse response h n6 @ of this system satisfies the condition h n 2n 0S =3= 6 @ ,

the relationship between a and b is(A) /1 2a b= - (B) /1 2a b= +(C) 2a b= (D) 2a b=-

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Sol. 44 Correct option is (A).Given the input-output relationship of the system,

[ ]y n [ ] [ ]y y x n1a b= - +Taking z -transform both the sides,

( )Y z ( ) ( )z Y z X z1a b= +-

& ( )( )

X zY z

z1 1a

b=- -

& ( )H z z1 1a

b=- - ...(i)

This is the z -transform of impulse response [ ]h n ; which can also be expressed as

( )H z [ ]h n z n

0

=3

-/Substituting z 1= in above expression, we have

( )H 1 [ ]h n 20

= =3

/ (given)

Again, Substituting z 1= in equation (1), we get

( )H 1 1 ab= -

& 2 1 ab= -

& 1 a - 2b =

& a 1 2b = -

Q. 45 The value of the integral sinc t dt52

3

3

-^ h# is ____.


Given the integral, ( )sinc t dt52

3

3

-

#Let the function

( )x t ( )sin sinC t tt5 5

5p

p= =

Now, we have the fourier transform pair

( ), | |

, | |sin

tt X j

W

W

1

0

<>

*pw w

ww

= *

So, for the signal ( ) ( )sinx t t5= , we get

( )X jw , | |

, | |51 5

0 5

<

>

w p

w p= *

or ( )X f / , | | .

, | | .

f

f

1 5 2 5

0 2 5

<>

= *

The signal in the frequency domain can be represented as

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From Parsevall’s Theorem, we have

( )x t dt2

3

3

-/ ( )X f df2=

3

3

-/

Hence, we get

( )sin t dt52

3

3

-# df5

1.

. 2

2 5

2 5=

-b l#

( . . )251 2 5 2 5 5

1#= + =

.0 2=

Q. 46 An unforced linear time invariant (LTI) system is represented by

xx

1

2

o

o> H xx

10

02

1

2=

--> >H H

If the initial conditions are x 0 11 =^ h and x 0 12 =-^ h , the solution of the state equation is(A) x t 11 =-^ h , x t 22 =^ h (B) x t e t

1 =- -^ h , x t e2 t2 = -^ h

(C) x t e t1 = -^ h , x t e t

22=- -^ h (D) x t e t

1 =- -^ h , x t e2 t2 =- -^ h

Sol. 46 Correct option is (C).We have the equation of LTI system as

xx

1

2

o

o> H xx

10

02

1

2=

--> >H H

and the initial conditions are

( )x 01 1= , ( )x 0 12 =-So, we have the matrix

A 1

002=

--> H

Now, we may define the Laplace transform of signal ( )x t as

( )X s [ ] ( )SI A x 01= - - ...(i)So, we obtain

[ ]SI A- s

s1

00

2=+

+> H

Therefore,

[ ]SI A 1- - ( )( )s s

ss1 2

1 20

01= + +

++> H

s

s

11

0

0

21= +

+

R

T

SSSS

V

X

WWWW

Also, we have the initial condition

[ ]x 0 ( )( )

xx

00

11

1

2= = -> >H H

Substituting these values in equation (1), we obtain

( )X s s

s

s

s

11

0

0

21

11

11

21= +

+- = +

- +

R

T

SSSS

R

T

SSSS

>

V

X

WWWW

V

X

WWWW

H

Hence, the signal in time domain is

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( )x t ( )//L X s Lss

ee

1 11 2

t

t1 1

2= =+

- + = -- -

-

-> H" *, 4

or ( )x t1 e t= - , ( )x t e t2

2=- -

Q. 47 The Bode asymptotic magnitude plot of a minimum phase system is shown in the figure

If the system is connected in a unity negative feedback configuration, the steady state error of the closed loop system, to a unit ramp input, is_____.


Q. 48 Consider the state space system expressed by the signal flow diagram shown in the figure.

The corresponding system is(A) always controllable (B) always observable

(C) always stable (D) always unstable

Sol. 48 Correct option is (A).We have the signal flow diagram of state space system as

From the signal flow graph, we obtain

x1o x2= x2o x3= x3o a x a x a x u1 1 2 2 3 3= + + + y c x c x c x1 1 2 2 3 3= + +In matrix form, we can write

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xxx

1

2

3

o

o

o

R

T

SSSS

V

X

WWWW

a a a

xxx

u00

10

01

0011 2 3

1

2

3

= +

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

yyy

1

2

3

R

T

SSSS

V

X

WWWW c c c

xxx

1 2 3

1

2

3

=

R

T

SSSS

8

V

X

WWWW

B

So, we have the matrices

[ ]A a a a

00

10

01

1 2 3

=

R

T

SSSS

V

X

WWWW

[ ]B 001

=

R

T

SSSS

V

X

WWWW

[ ]C C C C1 2 3= 8 B

Therefore, we obtain the contrvability matrix

U B AB A B2= 9 C

a

aa a

001

01

1

3

3

2 32

=+

R

T

SSSS

V

X

WWWW

So, | |U 1 0!=-i.e. the rank of controllability matrix is 3; which is equal to number of variable.

Hence, the system is always controlable matrix

[ ]V C

CACA2

=

R

T

SSSS

V

X

WWWW

( ) ( ) ( )

CC a

C C a C

CC C a

C a C C a a

CC C a

C C a C C a a

1

3 1

2 3 3 1

2

1 2 2

3 1 2 3 3 2

3

2 3 3

1 2 2 2 3 3 3

=+

++ +

++ + +

R

T

SSSS

V

X

WWWW

Therefore | |V depends on the value of the unknown.Hence, the system is not observable always.Note:By Checking controlability only. We can deduce the answer. Although the observability matrix is solved to verify the answer.

Q. 49 The input to a 1-bit quantizer is a random variable X with pdf f x e2Xx2= -^ h

for x 0$ and f x 0X =^ h for x 0< . For outputs to be of equal probability, the quantizer threshold should be ____.

Sol. 49 Correct answer is 0.35.Given the pdf of random variable.

( )f xX ,

,

e x

x

2 0

0 0<

x2 $=-

*

Also, we have the 1 bit quantizer, which can be defined as

xq ,

,

x x

x x

1

0 <T

T

$= *

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Where xT is threshold value. Since, the output should be of equal probability, i.e.

( )P x 1q = ( )P x 0q= =So, ( )P X xT$ ( )P x x< T=Therefore, we obtain the required value of threshold as

( )f x dxX

xT

3

# ( )f x dxX

x

0

T

= #

& e dx2 x

x

2

T

3 -# e dx2 xx

2

0

T

= -#

e2

2 x

x

2

T-

3-

: D e2

2 x x2

0

T

= --

: D

& e0 x2 T+ - e 1x2 T=- +-

& e2 x2 T- 1=

& xT ln21 2=

.0 35=

Q. 50 Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol waveforms coss t f t21 1a p=^ h and coss t f t22 2a p=^ h , where

mV4a = . Assume an AWGN channel with two sided noise power spectral density . /W Hz0 5 10N

2120 #= - . Using an optimal receiver and the relation

Q v e du/u

v21 22

=3

p -^ h # , the bit error probability for a data rate of 500 kbps is

(A) Q 2^ h (B) Q 2 2^ h

(C) Q 4^ h (D) Q 4 2^ h

Sol. 50 Correct option is (C).Given the two equiprobable symbol waveforms,

( )t1d ( )cos f t2 1a p= ( )t2d ( )cos f t2 2a p= ; mV V4 4 10 3

#a = = -

Data rate,

Rb / seckbps bits500 5 105#= =

Noise power spectral density,

N2

o . /W Hz0 5 10 12#= -

or No . /W Hz2 0 5 10 1012 12# #= =- -

So, we obtain energy per bit,

Eb T

R2 2 b

2 2a a= =

( )

( )2 5 10

4 1016 105

3 212

# #

##= =

--

Hence, The bit error probability is given by

Pe Q NE

Q 1016 10

o

b12

12#= = -

-

c cm m

( )Q 4=

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Q. 51 The power spectral density of a real stationary random process X t^ h is given by

S fX ^ h ,

,

f W

f W0 >W1 #

= *

The value of the expectation E X t X t W41p -^ bh l; E is _____.

Sol. 51 Correct answer is 4.Given the power spectral density

( )S fX , | |

, | |W f W

f W

1

0 >

#= *

So, we obtain the autocorrelation function as

( )RX t ( )S f e dtXj ft2=

3

3 p

-#

W e df W je1 1j f

W

W j f

W

W2

2

2

= =p tpt

p t

- -= G#

( )sin

j We e

WW

21 2j W j W2 2

pt ptpt= - =pt pt-^ h

Now, we have to obtain

( )E X t X t N41

p -b l: D ( )E X t X t W41p = -b l; E

Since, we define the autocorrelation function as

( )RX t [ ( ) ( )]E X t X t t = -

So, R W41

X b l ( )E X t X t W41

= -b l; E

Hence, We get

( )E X t X t W41p -b l; E R W4

1Xp = b l

sin

W W

W W

41

2 41

pp

p=

b

b

l

l; E

( / )sin4 2p = 4=

Q. 52 If the figure, M f^ h is the Fourier transform of the message signal m t^ h where HzA 100= and HzB 40= . Given cosv t f t2 cf =^ ^h h and cosw t f A t2 cp = +^ ^^h h h

, where f A>c . The cutoff frequencies of both the filters are fc .

The bandwidth of the signal at the output of the modulator (in Hz) is____.

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Sol. 52 Correct answer is 60.We have the waveform of fourier transform of message signal as

Now, we have the signal

( )v t ( )cos f t2 cp =So, the frequency spectrum for the signal ( ) ( )v t m t is

Now, the signal passes through high pass filter with cut off frequency fc . So, it stops the signal component with frequency higher than fc . Therefore, the frequency spectrum for filtered signal is obtained as

Again, the signal is multiplied with ( )w t defined as

( )w t [ ( ) ]cos f A t2 cp = +So, we get the frequency spectrum as

Again, the signal is passed through low pass filter with cut off frequency fc .Hence, the waveform of output signal is

Therefore, the bandwidth of output signal is

A B- 100 40= - Hz60=

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Q. 53 If the electric field of a plane wave is

,E z tv^ h cos sinx t kz y t kz3 30 4 45c cw w= - + - - +t t^ ^h h

(mV/m),the polarization state of the plane wave is(A) left elliptical (B) left circular

(C) right elliptical (D) right circular

Sol. 53 Correct option is (A).

Q. 54 In the transmission line shown, the impedance Zin (in ohms) between node A and the ground is _____.

Sol. 54 Correct answer is 33.33 ohm.

Q. 55 For a rectangular waveguide of internal dimensions a b a b># ^ h, the cur-off frequency for the TE11 mode is the arithmetic mean of the cut-off frequencies for TE10 mode and TE10 mode and TE20 mode. If a 5= cm, the value of b (in cm) is______.


END OF THE QUESTION PAPER

************

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ANSWER KEY

General Aptitude

1 2 3 4 5 6 7 8 9 10

(B) (B) (C) (B) (45) (C) (D) (163) (D) (A)

Electronics & Communication

1 2 3 4 5 6 7 8 9 10

(200) (50) (A) (C) (A) (D) (1.25) (D) (B) (D)

11 12 13 14 15 16 17 18 19 20

(B) (B) (34.722) (D) (4) (C) (C) (B) (0.5) (4)

21 22 23 24 25 26 27 28 29 30

(D) (A) (0.5) (70.7) (B) (B) (B) (49) (3) (0.45)

31 32 33 34 35 36 37 38 39 40

(1.24) (A) (2.50) (4000) (30.7) (B) (C) (C) (0.9) (C)

41 42 43 44 45 46 47 48 49 50

(D) (D) (10) (A) (0.2) (C) (0.50) (A) (0.35) (C)

51 52 53 54 55

(4) (60) (A) (33.33) (2)

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