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SOLVED PAPER-2014 (EL ECTRONI CS & COMMUNI CAT I ON ENGI NEERI NG - EC) 1

(Q. 1 – 5) Carry One Mark Each.1. Choose the most appropriate phrase from the

options given below to complete the followingsentence.

The aircraft _________ take off as soon as its flightplan was filed.

(a) is allowed to

(b) will be allowed to

(c) was allowed to

(d) has been allowed to

2. Read the statements:

All women are enterpreneurs.

Some women are doctors.

Which of the following conclusions can be logicallyinferred from the above statements?

(a) All women are doctors

(b) All doctors are enterpreneurs

(c) All enterpreneurs are women

(d) Some enterpreneurs are doctors

3. Choose the most appropriate word from theoptions given below to complete the followingsentence.

Many ancient cultures attributed disease tosupernatural causes. However, modern sciencehas largely helped _________ such notions.

(a) impel (b) dispel

(c) propel (d) repel

4. The statistics of runs scored in a series by fourbatsmen are provided in the follwing table. Whois the most consistent batsman of these four?

Batsman Average Standard deviationKLMN

31.246.054.417.9

5.216.356.225.90

(a) K (b) L

(c) M (d) N

5. What is the next number in the series?

12 35 81 173 357 ___

(Q. 6 – 10) Carry Two Marks Each.6. Find the odd one from the following

(a) W,E,K,O (b) I,Q,W,A

(c) F,N,T,X (d) N,V,B,D

7. For submitting tax returns, all resident maleswith annual income below ` 10 lakh should fillup Form P and all resident females with incomebelow ` 8 lakh should fill up Form Q. All peoplewith incomes above ̀ 10 lakh should fill up FormR, except non residents with income above ` 15lakhs, who should fill up Form S. All othersshould fill Form T, An example of a person whoshould fill From T is

(a) a resident male with annual income ` 9 lakh

(b) a resident female with annual income ̀ 9 lakh

(c) a non-resident male with annual income` 16 lakh

(d) a non-resident female with annual income` 16 lakh

8. A train that is 280 metres long, travelling at auniform speed, crosses a platform in 60 secondsand passes a man standing on the platform in20 seconds. What is the length of the platform inmetres?

9. The exports and imports (in crores of `) of acountry from 2000 to 2007 are given in thefollowing bar chart. If the trade deficit is definedas excess of imports over exports, in which yearis the trade deficit 1/5th of the exports?

120110100

9080706050403020100

2000 2001 2002 2003 2004 2005 2006 2007

Exports Imports

(a) 2005 (b) 2004

(c) 2007 (d) 2006

GATE - 2014EC : EL ECTRONI CS & COM M UNI CATI ON

ENGI NEERI NGNo. of Quest i ons : 65 Maximum Mar ks : 100

GENERAL APTITUDE

2 SOLVED PAPER-2014 (EL ECTRONI CS & COMMUNI CAT I ON ENGI NEERI NG - EC)

10. You are given three coins: on has heads on bothfaces, the second has tails on both faces, and thethird has a head on one face and a tail on theother. You choose a coin at random and toss it,and it comes up heads. The probability that theother face is tails is

(a) 1/4 (b) 1/3

(c) 1/2 (d) 2/3

ELECTRONICS AND COMMUNICATION

(Q. 1 – 25) Carry One Mark Each.

1. For matrices of same dimension M, N and scalarc, which one of these properties DOES NOTALWAYS hold?

(a) (MT)T = M (b) (cM)T = c(M)T

(c) (M + N)T = MT + NT (d) MN = NM

2. In a housing society, half of the families have asingle child per family, while the remaining halfhave two children per family. The probability thata child picked at random, has a sibling is _____

3. C is a closed path in the z- plane given by z = 3.

The value of the integral

2 4

2C

z z jz j dz is

(a) – 4 (1 + j2) (b) 4 (3 – j2)

(c) – 4 (3 + j2) (d) 4 (1 – j2)

4. A real (4 × 4) matrix A satisfies the equationA2 = I, where I is the (4 × 4) identity matrix. Thepositive eigen value of A is _______.

5. Let X1, X2 and X3 be independent and identicallydistributed random variables with the uniformdistribution on [0, 1]. The probability P {X1 is thelargest} is ______

6. For maximum power transfer between twocascaded sections of an electrical network, therelationship between the output impedance Z1 ofthe first section to the input impedance Z2 of thesecond section is

(a) Z2 = Z1 (b) Z2 = –Z1

(c) Z2 = Z *1 (d) Z2 = – Z *

1

7. Consider the configuration shown in the figurewhich is a protion of a larger electrical network

i5

i3

i2

i4

i1 i6

R

RR

For R = 1 and currents i1 = 2 A, i4 = –1 A,i5, = –4 A, which one of the following is TRUE?(a) i6 = 5 A

(b) i3 = –4 A

(c) Data is sufficient to conclude that the supposedcurrents are impossible

(d) Data is insufficient to identify the currentsi2, i3, and i6

8. When the optical power incident on a photodiodeis 10W and the responsivity is 0.8 A/W, thephotocurrent generated (in A) is ________.

9. In the figure, assume that the forward voltagedrops of the PN diode D1 and Schottky diode D2

are 0.7 V and 0.3 V, respectively. If ON denotesconducting state of the diode and OFF denotesnon-conducting state of the diode, then in thecircuit,

1 k 20

D2D110 V

(a) both D1 and D2 are ON

(b) D1 is ON and D2 is OFF

(c) both D1 and D2 are OFF

(d) D1 is OFF and D2 is ON

10. If fixed positive charges are present in the gateoxide of an n-channel enhancement type MOSFET,it will lead to

(a) a decrease in the threshold voltage

(b) channel length modulation

(c) an increase in substrate leakage current

(d) an increase in accumulation capacitance

11. A good current buffer has

(a) low input impedance and low outputimpedance

(b) low input impedance and high outputimpedance

(c) high input impedance and low outputimpedance

(d) high input impedance and high outputimpedance


12. In the ac equivalent circuit shown in the figure,if iin is the input current and RF is very large, thetype of feedback is

RD

RD

Vout

M1

M2

RFi insmall signalinput

(a) voltage-voltage feedback

(b) voltage-current feedback

(c) current-voltage feedback

(d) current-current feedback

13. In the low-pass filter shown in the figure, for acut-off frequency of 5 kHz, the value of R2 (in k)is __________.

R2

1 k

C10 nF

VoR1

V i

+

–

14. In the following circuit employing pass transistorlogic, all NMOS transistors are identical with athreshold voltage of 1 V. Ignoring the body-effect,the output voltages at P, Q and R are,

5 V5 V5 V

5 V

P Q R

(a) 4V, 3V, 2V (b) 5V, 5V, 5V

(c) 4V, 4V, 4V (d) 5V, 4V, 3V

15. The Boolean expression (X + Y) (X + Y ) +

(XY)+X simplifies to

(a) X (b) Y

(c) XY (d) X+Y

16. Five JK flip-flops are cascaded to form the circuitshown in figure. Clock pulses at a frequency of 1MHz are applied as shown. The frequency (in kHz)of the waveform at Q3 is ________.

1

1 1

1 1

1 1

11

1J4 Q4clk

K4

1 J3

K3clk clk

Q3 J2 Q2

K2

J1clk clk

K1

Q1 J0 Q0

K0

clock

17. A discrete-time signal x[n] = sin(2n),n being aninterger, is

(a) periodic with period .

(b) periodic with period 2.

(c) periodic with period /2.

(d) not periodic.

18. Consider two real valued signals, x (t) band-limitedto [– 500 Hz, 500 Hz] and y(t) band-limited to [–1kHz, 1 kHz]. For z(t) = x(t) . y(t), the Nyquistsampling frequency (in kHz) is _____.

19. A continuous, linear time-invariant filter has animpulse reponse h(t) described by

3for0 3( ) =

0 otherwise

th t

When a constant input of value 5 is applied tothis filter, the steady state outout is _____.

20. The forward path transfer function of a unitynegative feedback system is given by

KG( )

( 2)( 1)s

s s

The value of K which will place both the poles ofthe closed-loop system at the same location,is ______.

21. Consider the feedback system shown in the figure.The Nyquist plot of G (s) is also shown. Whichone of the following conclusions is correct?

+ k G(s)–

Im G(j )

Re G(j )+1–1

(a) G(s) is an all-pass filter

(b) G(s) is a strictly proper transfer function

(c) G(s) is a stable and minimum-phase transferfunction

(d) The closed-loop system is unstable forsufficiently large and positive k


22. In a code-division multiple access (CDMA) systemwith N = 8 chips, the maximum number of userswho can be assigned mutually orthogonalsignature sequences is ______

23. The capacity of a Binary Symmetric Channel(BSC) with cross-over probability 0.5 is _______

24. A two-port network has scattering parameters

given by [S]=

11 12

21 22

s ss s . If the port-2 of the two-

port is short circuited, the s11 parameter for theresultant one-port network is

(a)11 11 22 12 21

221s s s s s

s

(b)11 11 22 12 21

221s s s s s

s

(c)11 11 22 12 21

221s s s s s

s

(d)11 11 22 12 21

221s s s s s

s

25. The force on a point charge +q kept at a distanced from the suface of an infinite grounded metalplate in a medium of permittivity is

(a) 0

(b) 2

216q

d away from the plate

(c)2

216q

d towards the plate

(d) 2

24q

dtowards the plate

(Q. 26 – 55) Carry Two Marks Eack.26. The Taylor series expansion of 3 sin x + 2 cos x is

(a) 2 + 3x – x2 – 3

......2x

(b) 2 – 3x + x2 – 3

......2x

(c) 2 + 3x + x2 + 3

......2x

(d) 2 – 3x – x2 +3

......2x

27. For a function g(t), it is given that

g(t) e–jt dt = e–22 for any real value . If

y(t) =

g( ) d , then

y(t) dt is

(a) 0 (b) – j

(c) –2j

(d)2j

28. The volume under the surface z(x, y) = x + y andabove the triangle in the x-y plane defined by{0 y x and 0 x 12} is _______.

29. Consider the matrix

j6=

0 0 0 0 0 10 0 0 0 1 00 0 0 1 0 0

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

Which is obtained by reversing the order of thecolumns of the identity matrix I6. LetP = I6 + J6, where is a non-negative realnumber. The value of for which det(P) = 0is ______.

30. A Y-network has resistances of 10 each in twoof its arms, while the third arm has a resistanceof 11. In the equivalent -network, the lowestvalue (in ) among the three resistancesis ________.

31. A 230 V rms source supplies power to two loadsconnected in parallel. The first load draws 10 kWat 0.8 leading power factor and the second onedraws 10 kVA at 0.8 lagging power factor. Thecomplex power delivered by the source is

(a) (18 + j 1.5) kVA (b) 18 – j 1.5) kVA

(c) (20 + j 1.5) kVA (d) (20 – j 1.5) kVA

32. A periodic variable x is shown in the figure as afunction of time. The root-mean-square (rms)value of x is _______.

t0

1x

T/2 T/2

33. In the circuit shown in the figure, the value ofcapacitor C (in mF) needed to have criticallydamped response i(t) is _______.

C4 H40

i(t)+

Vo

–

34. A BJT is biased in forward active mode. AssumeVBE = 0.7 V, kT/q = 25 mV and reverse saturationcurrent Is = 10-13 A. The transconductance of theBJT (in mA/V) is ________.


35. The doping concentrations on the p-side andn-side of a silicon diode are 1 × 1016 cm–3 and1 × 1017 cm–3, respectively. A forward bias of0.3 V is applied to the diode. At T = 300 K,the intrinsic carrier concentration of silicon

ni = 1.5 ×1010 cm–3 and kTq

= 26 mV. The electron

concentration at the edge of the depletion on thep-side is

(a) 2.3 × 109 cm–3

(b) 1 × 1016 cm–3

(c) 1 × 1017 cm–3

(d) 2.25 × 106 cm–3

36. A depletion type N-channel MOSFET is biased inits linear region for use as a voltage controlledresistor. Assume threshold voltage VTH = –0.5 V,VGS = 2.0V, VDS = 5V, W/L = 100, COX = 10–8 F/cm2

and n 800cm2/V–s. The value of the resistance ofthe voltage controlled resistor (in) is _______.

37. In the voltage regulator circuit shown in the figure,the op-amp is ideal. The BJT has VBE = 0.7 V and = 100, and the zener voltage is 4.7 V. For aregulated output of 9 V, the value of R (in ) is ____.

V = 12 VI V = 9 VO

1 k

RV = 4.7 VZ

1 k+

–

38. In the circuit shown, the op-amp has finite inputimpedance, infinite voltage gain and zero inputoffset voltage. The output voltage Vout is

R2

R1

Vout

I2

I1 –+

(a) –I2 (R1 + R2) (b) I2R2

(c) I1R2 (d) –I1(R1 + R2)

39. For the amplifier shown in the figure, the BJTparameters are VBE = 0.7 V, = 200, and thermalvoltage VT = 25 mV. The voltage gain (vo/vi) of theamplifier is _______.

V = + 12 VCC

RC

R1

33 k

5 k

1 F1 F

v0

vi

R2

11 k RS

10

RE

1 kCE

1 mF

40. The output F in the digital logic circuit shown inthe figure is

F

AND

XORXY

Z

XNOR

(a) F = X YZ + X Y Z

(b) F = X Y Z + X Y Z

(c) F = X Y Z + XYZ

(d) F = X Y Z + XYZ

41. Cosider the Boolean function, F(w,x,y,z) = wy +xy + w xyz + w x y + xy + x y z . Which one of

the following is the complete set of essential primeimplicants?

(a) w, y, xz, x z (b) w, y, xz

(c) y, x y z (d) y, xz, x z

42. The digital logic shown in the figure satisfies thegiven state diagram when Q1 is connected to inputA of the XOR gate.

D1 Q1

Q1CLKS

AD2 Q2

Q2

S=1 S=0S=0

S=1

S=1

S=1

S=0

S=0

00 01

1110


Suppose the XOR gate is replaced by an XNORgate. Which one of the following options preservesthe state diagram?

(a) Input A is connected to 2Q

(b) Input A is connected to Q2

(c) Input A is connected to 1Q and S iscomplemented

(d) Input A is connected to 1Q

43. Let x[n] =

1 1( ) ( 1)

9 3

n n

u n u n . The

Region of Convergence (ROC) of the z-transformof x[n]

(a) is 19

z (b) is 13

z

(c) is 1 13 9

z (d) does not exist

44. Consider a discrete time periodic signal

x[n] = sin

5

n. Let ak be the complex Fourier

series coefficients of x[n]. The cofficients {ak} arenon-zero where k = Bm ± 1, when m is any integer.The value of B is _____.

45. A system is described by the following differentialequation, where u(t) is the input to the systemand y(t) is the output of the system.

( ) 5 ( ) ( )y t y t u t When y (0) = 1 and u(t) is a unit step function ,y(t) is(a) 0.2 + 0.8e–5t (b) 0.2 – 0.2e–5t

(c) 0.8 + 0.2e–5t (d) 0.8 – 0.8e–5t

46. Consider the state space model of a system, asgiven below

1

2

3

xxx

=

1 1

2 2

3 3

1 1 0 0

0 1 0 4 ; 1 1 1 .

0 0 2 0

x xx u y xx x

The system is(a) controllable and observable(b) uncontrollable and observable(c) uncontrollable and unobservable(d) controllable and unobservable

47. The phase margin in degrees of

G(s) = 10

0.1 1 10s s s calculated using the

asymptotic Bode plot is ______.

48. For the following feedback system

G(s) =

11 2s s

. The 2%-settling time of the

step response is required to be less than 2 seconds.

r + yC(s) G(s)

–

Which one of the following compensators C(s)achieves this?

(a) 31

5s

(b) 5

0.031

s

(c) 2(s + 4) (d)

84

3ss

49. Let X be a real-valued random variable with E[X]and E[X2] denoting the mean value of X and X2,respectively. The relation which always holds trueis

(a) (E[X])2 > E[X2] (b) E[X2] > (E[X])2

(c) E[X2] = (E[X])2 (d) E[X2] > (E[X])2

50. Consider a random process X(t) = 2 sin(2 t ),where the random phase is uniformlydistributed in the interval [0, 2]. The auto-correlation E [X(t1)X(t2)] is

(a) cos (2(t1 + t2)) (b) sin (2(t1 – t2))

(c) sin (2(t1 + t2)) (d) cos (2(t1 – t2))

51. let Q( ) be the BER of a BPSK system over an AWGN channel with two-sided noise power spectral

density N0 /2. The parameter is a function of bit energy and noise power spectral density.A system with two independent and identical AWGN channels with noise power spectral density N0 /2 isshown in the figure. The BPSK demodulator receives the sum of outputs of both the channels.

BPSKDemodulator

AWGNChannel 1

AWGNChannel 2

BPSKModulator

0/1 0/1

If the BER of this system is Q(b ), then the value of b is _____.


52. A fair coin is tossed repeatedly until a ‘Head’appears for the first time. Let L be the number oftosses to get this first ‘Head’. The entropy H(L)in bits is _____.

53. In spherical coordinates, let ˆ ˆ,a a denote unit

vectors the , directions.

100 ˆE sin cos( ) /t r a V m

r

and 0.265 ˆH sin cos( ) /t r a A m

rrepresent the electric and magnetic fieldcomponents of the EM wave at large distances rfrom a dipole antenna, in free space. The averagepower (W) crossing hemispherical shell located

at r = 1km, 0 /2 is _____

54. For a parallel plate transmission line, let v be thespeed of propagation and Z be the characteristicimpedance. Neglecting fringe effects, a reductionof the spacing between the plates by a factor oftwo results in

(a) halving of v and no change in Z

(b) no changes in v and halving of Z

(c) no change in both v and Z

(d) halving of both v and Z

55. The input impedance of a 8

section of a lossless

transmission line of characteristic impedance 50is found to be real when the other end isterminated by a load ZL (= R + jX). If X is 30,the value of R (in ) is ____

ANSWERS

GENERAL APTITUDE

1. (c) 2. (d) 3. (b) 4. (a) 5. (725 to 725) 6. (d) 7. (b) 8. (560 to 560)

9. (d) 10. (b)

ELECTRONICS AND COMMUNICATION

1. (d) 2. (0.65 to 0.68) 3. (c) 4. (0.99 to 1.01) 5. (0.32 to 0.34)

6. (c) 7. (a) 8. (7.99 to 8.01) 9. (d) 10. (a)

11. (b) 12. (b) 13. (3.1 to 3.26) 14. (c) 15. (a)

16. (62.4 to 62.6) 17. (d) 18. (2.99 to 3.01) 19. (44 to 46) 20. (2.24 to 2.26)

21. (d) 22. (7.99 to 8.01) 23. (– 0.01 to 0.01) 24. (b) 25. (c)

26. (a) 27. (b) 28. (862 to 866) 29. (0.99 to 1.01) 30. (29.08 to 29.10)

31. (b) 32. (0.39 to 0.42) 33. (9.99 to 10.01) 34. (5.7 to 5.9) 35. (a)

36. (499 to 501) 37. (1092 to 1094) 38. (c) 39. (– 240 to – 230) 40. (a)

41. (d) 42. (d) 43. (c) 44. (9.99 to 10.01) 45. (a)

46. (b) 47. (42 to 48) 48. (c) 49. (b) 50. (d)

51. (1.4 to 1.42) 52. (1.99 to 2.01) 53. (55.4 to 55.6) 54. (b) 55. (39 to 41)


EXPLANATIONS

GENERAL APTI TUDE

4. I f the standard deviat ion is less, there wil l beless deviat ion or batsman is more consistent .

5. The given ser ies is

12, 35, 81, 173, 357, .......

The given ser ies fol lows the fol lowing pat tern

12 2 + 11 = 35

35 2 + 11 = 81

81 2 + 11 = 173

173 2 + 11 = 357

357 2 + 11 = 725

6. W, E, K, O,

+8 +6 +4

– 2– 2

I , Q, W, A,

+8 +6 +4

– 2– 2

F, N, T, X,

+8 +6 +4

– 2– 2

N, V, B, D,

+8 +6 +2

– 4– 2Hence the odd one from the fol lowing group is N,V, B, D.

7. By reading the instructions, it is clear that personwho fi l ls form T is a resident female with annualincome of ` 9 Lakh.

8. Let the length of plat form = x.m

and t rain length = 280 m (given)

According to Quest ion

28060

x =28020

x = 560 m.

9. For 2005,t rade deficit = (90 – 70) crores

= 20 crores

Now1

th5

of expor t = 15

(70) crores

= 14 crores

20 crores.

Hence opt ion (a) is wrong.

For 2004,Trade deficit = (80 – 70) crores

= 10 crores

Now 1

th5

of expor t = 15

(70) crores

= 14 crores

20 crores.

Hence opt ion (b) is wrong.

For 2007,Trade deficit = (120 – 110) crores

= 10 crores

Now 1th

5 of expor t =

15

(110) crores

= 22 crores

10 crores.

Hence opt ion (c) is wrong.For 2006

Trade deficit = (120 – 100) crores

= 20 crores

Now 1th

5 of expor t =

15

(100) crores

= 20 crores

= Trade deficits

Hence opt ion (d) is cor rect .

10. The probail i ty that the other face is tai l , is 13

.

EL ECTRONI CS AND COMM UNI CATI ON

1. M at r ix mult ipl icat ion is not commutat ive ingeneral.

2. Let the number of famil ies in housing societybe x.

2x

family have single chi ld.

Total 2x

chi ldren

Now remaining 2x

famil ies have 2 chi ldren.

Total 2x

2 = x, sibl ing chi ldren

Probabil i ty that a chi ld picked at random has a

sibl ing =

2

xx

x

= 2

3x

x

= 23

= 0.667

3. Z = – 2j is a singular ity l ies inside C : | Z| = 3

By Cauchy’s integral formula,2

C

Z Z 4Z 2 j

dzj

= 2Z 22 .[Z Z 4 ] jj j

= 2j [– 4 + 2j + 4j ]

= – 4[3 + j2]


4. A real (4 4) matr ix A sat isfies the equat ionA2 = I where I is the (4 4) ident ity matr ix. Theposit ive eigen value of A is 1.

5. X1, X 2, X 3 ar e i ndependent and i dent i cal l ydist r ibuted random var iables.

So P{X1 is the largest} = 1/3 = 0.33

6. Two cascaded sect ions

Z1 = Output impedance of fir st sect ion

Z2 = Input impedance of second sect ion

For maximum power t ransfer, upto 1st sect ion is

ZL = Z1*

ZL = Z2 Z1*

7. Using KVL at al l the three nodes.

i 4 + i 1 – i 2 = 0

i 6 + i 3 – i 1 = 0

i 5 + i 2 – i 3 = 0

By solving the equat ion, i 6 = 5A

8. Responsivity (R) = P

0

IP

where IP = photo cur rent

P0 = Incident power

IP = R P0

= 8µA

9. Assume both the diode ON.

Then circuit wil l be as per figure (2)

Figur e (2)

I = 10 0.7

1k

= 9.3 mA

I D2 = 0.7 0.3

20

= 20 mA

Now, I D1 = I – I D2

= – 10.7 mA (Not Possible)

D1 is OFF and hence D2 – ON

10. I f fixed posit ive charges are present in the gateox ide of an n -channel enhancemen t t ypeM OSFETS i t wi l l l ead to a decr ease in t hethreshold voltage.

11. I deal cur rent Buffer has Zi = 0

Z0 =

12. Output sample is voltage and is added at the inputor cur rent

I t is voltage – shunt negat ive feedback i .e.,voltage-cur rent negative feedback

13. f = 5KHz

Cut off frequency (LPF) = 2

12 R C

= 5 KHz

R2 = 3 91

2 5 10 10 10 = 3.18 k

14. Assume al NMOS are in saturat ion

VDS (VGS – VT)

For m1,

(5 – Vp) (5 – Vp – 1)

(5 – Vp) > (4 – Vp) Sat

I D1 = k(VGS – VT)2

I D1 = K(4 – VP)2 ...(1)

For m2,

I D1 = K(5 – VQ – 1)2

I D2 = K(4 – VQ)2 ...(2)

I D1 = I D2

(4 – VP)2 = (4 – VQ)2

VP = VQ & VP + VQ = 8

VP = VQ = 4V

For m3,

I D3 = K(5 – VR – 1)2

I D2 = I D3

(4 – VQ)2 = (4 – VR)2

VR = VQ = 4V

VP = VQ = VR = 4V


15. f = ( )( ) ( )x y x y xy x

= ( )x xy xy xy x

= ( )x xy xy x y x

= (1 )x y y

= x

16. 3Qf = input frequencymodulus of counter

= 1MH

16z

3Qf = 62.5 kHz.

17. Time per iod of a discrete signal,

0

2w

= KN

N =

0

2 Kw

=

2

2 K 2K

N is a ir rat ional number so signal is not per iodic.

18. Multiplication in the t ime domain

= Convolution in frequency domain

x1(t) x2(t) = 1 2X ( )X ( )j j

Fundamental frequencies = f1, f1 ± f2, f1 ±2 f2 ...

= 500, 1500....

Nyquist rate = 2 × 1500

= 3000 Hz

= 3 kHz19.

y(t ) = x (t ) * h(t )

x(t ) =

h (t ) =

y(t ) = 3

0

3.5 d = 45(steady state output)

20. Using root locus,

XX–2 1

Break away point ,

K1

( 2)( 1)s s = 0

K = ( 2)( 1)s s

Kdds

= – 2s – 1 = 0

s = – 0.5

G(s)s = – 0.5 = 1

s 0.5

K( 2)( 1)s s = 1

21. For al l pass system the pole zero pair must besymmetr ical about imaginary axis with zero onthe RHS and pole on the LHS of s-plane.

This is not minimum phase system.

Encirclement to the cr i t ical point (– 1, 0) = N = 0

Open loop pole at RHS = P = 1

N = P – z

z = 1 (Close loop pole at RHS)

Thus system is unstable.

22. Spreading factor (SF) = chip rate

symbol rate

This i f a single symbol is represented by a codeof 8 chips

Chip rate = 80 symbol rate

S.F (Spreading Factor )= 8 symbol rate

symbol rate

= 8

Spread factor (or) process gain and determine toa cer tain extent the upper l imit of the totalnumber of uses suppor ted simultaneously by astat ion.

23. Channel capacity of BSC is

C = 2 2P log P (1 P) log (1 P) 1

= 0.5 log2 0.5 + 0.5 log2 0.5 + 1

= 0 [ log20.5 = 1]

I t is the case of channel with independent inputand output , hence C = 0

24.

b1 = s11a1 + s12a2

b2 = s21a1 + s22a2

1

2

bb

= 11 12 1

21 22 2

s s a;

s s a

s1 = 2

1

1 a 0

ba

By ver ificat ion Answer (b) sat isfies.


25. Q

– Q

d

d

F =

1 22

Q Q4 R

=

2

24 (2 )q

d =

2

216q

d26. Taylor series expansion of 3 sin x + 2 cos x

Taylor ser ies is given by

f(x) = f (x0) + f (x0) (x – x0) + 20

0( )

( )2!

f xx x

+ 300

( )( ) ...

3!f x

x x

Here x0 = 0

f(x) = f(0) + f (0)(x) + 2(0)2

fx

3(0)( ) ...

6f

x

...(i )

Now f(x) = 3 sin x + 2 cos x

f(0) = 2

Now f (x) = 3 cos x + 2 (– sin x)

= 3 cos x – 2 sin x

f (0) = 3

f (x) = – 3 sin x – 2 cos x

f (0) = – 2

f (x) = – 3 cos x + 2 sin x

f (0) = – 3

From equat ion (i ),

f(x) = f(0) + f (0) x + 2 3(0) (0)( ) ...

2 6f f

x x

= 2 3( 2) 32 3 ...

2 6x x x

= 2 + 3x – x2 – 31...

2x

27. Given,

jwtg(t ).e dt

= 22w.e (let G(j ))

g(t )dt

= 0

y(t ) = t

g(z).dz

y(t ) = g(t ) * u(t ) [u(t ) in unit step funct ion] Y(j) = G(j). U(j)

Y(j) =j ty(t ).e dt

Y(j0) = y(t ) dt

=22w 1

.e ( ) 0j

=1

jj

28. The volume under the sur face z(x, y) = x + y andabove the t r iangle in the x – y plane is given

Volume = 12

0 0

( )x

x y dy dx

= 12 2

0 02

xy

xy dx

= 12 2

2

0 2x

x dx

Volume = 123 3

03 6x x

= 123 3

0

26

x x

= 123

02x

= 31(12)

2= 864.

29. J6 =

0 0 0 0 0 10 0 0 0 1 00 0 0 1 0 00 0 1 0 0 00 1 0 0 0 01 0 0 0 0 0

I 6 =

1 0 0 0 0 00 1 0 0 0 0

0 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1


Now P = I 6 + J6

=

1 0 0 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0 00 0 0 1 0 0 0 0 0 0 00 0 0 0 1 0 0 0 0 0 00 0 0 0 0 1 0 0 0 0 0

P =

1 0 0 0 00 1 0 0 0

0 0 1 0 00 0 1 0 00 0 0 1 0

0 0 0 0 1

P = 1

1 0 0 0 0 1 0 0

0 1 0 0 0 0 1 00 1 0 0 0 0 1 0

0 0 1 0 0 0 0 10 0 0 0 1 0 0 0 0

=

1 0 0 0 1 01 0 0 0 1 0

0 0 1 0 0 0 0 1

0 0 0 1 0 0 0

= 1 0 0 0 0 0 1 0

0 1 0 0 1 0 0 0 1

0 0 1 0 0 1 0 0

2

0 0

0 0 1

0 0

P = 1(1) – () – ((– 1)(– )) + 2[(– ) (– )] = 0

1 – 2 – 2 + 4 = 0

4 – 22 + 1 = 0

(2 – 1)2 = 0

2 = 1

= 1 = 1 [ is non-negat ive real number ]

30.

X = 29.09y = 32z = 32

X = (10)(10) (10)(11) (10)(11)

11

y = (10)(10) (10)(11) (10)(11)

10

z = (10)(10) (10)(11) (10)(11)

10

i .e, lowest value among three r esistances is29.09�

31.

Load I :

P = 10kw

cos = 0.8

Q = P tan = 7.5KVAR 1S P jQ 10 j7.5 KVA

Load I I : S =10 KVA

cos = 0.8 sin = QS

cos = PS

0.8 = P

P 8kw10

Q = 6 KVAR

S1 = P + jQ = 8 + j6

Complex power del i ver ed by t he sour ce i sSI + SI I =18 – j1.5 KVA

32. RMS value =T

2

0

1( )

Tf t dt

where T is t ime per iod

=

2T 2 T2

0 T / 2

1 2(0)

T Tt dt dt

=

T / 22

20

1 4T T

t dt

= 0.408


33. For cir i t ical damping,

= 1

12Q

where Q = Quality factor

For ser ies circuit ,

Q = 1 LR C

1

2 LR C

= 1

C = 2

2L

R

= 22

440

= 10 mF..

34. gm = C

T

IV

I C =

BE

T

VV

SI 1e

I C =

0.713 0.02510 1e

I C = 144.6257 mA

gm = 0.1440.025

= 5.78 mAV

35. Electron concentrat ion,

n i in T

2V / V

A

ne

N

=10 2

0.3 / 26mV16

(1.5 10 )e

1 10

= 2.3 109/ cm3

36. r DS =

n on GS T

1W( C ) (V V )L

r DS = 4 8

1800 10 10 100 (20 0.5)

r DS = 500 37. Given circuit is an op-amp ser ies regulator, V0 is

given by

V0 = 1

2

R1 V

R z

9V = 2

11 4.7

Rk

R2 = 1093.02

38. Given, Zi =

L0A =

i0V = 0

V2 = (R1//R2) I 1

= 1 2

11 2

R RI

R R ...(1)

KCL at inver t ing node

2 02

1 2

V – VV+ 0

R R ( Z )i

0

2

VR

=

21 2

1 1V

R R

0

2

VR

=

1 2 2 11

1 2 1 2

R R R RI

R R R R

V0 = I 1R2

39. VBE = 0.7V, = 200, VT = 25mVDC Analysis:

VB =11

12 3V11 33K

kk

VE = 3 – 0.7 = 2.3V

I E =2.3

2.277mA10 1K

I B = 11.34 µA

I C = 2.26 mA

r e = 25 mV

10.982.277 mA

AV =

0 CV – R

V (1 )( )i e sr R

=– 200 5K

200 10.98 (201)10

AV = – 237.76


40.

Assume dummy var iable K as a output of XOR

gate K X Y XY XY

F = K.(K Z)

= (KZ + K.Z)

= K.KZ + K.K.Z)

= 0 + K.Z ( K. K 0 and K.K K)

Put the value of K in above expression

F = (XY + XY) Z

= XYZ + XYZ

41.EPJ = xz

EPI = y

WX4z

00

01

11

10

EPI = x z

00 01 11 10

1

1 1

1

1

1

1

1

1

1 1

1

42. D2 = A S

D2 = Q1 S

But D2 = A S

if A = 1Q

then D2 = 1Q S

D2 = 1Q S

So input A is connected to 1Q

43. x(n) = Right side Signal Left side signal

1 14( ) 4( 1)

9 3

n

x n

ROC is | z| > 19

ROC is | z| < 13

So ROC is 1 1

| |3 9

z

45. 5 ( )dy

y tdt

= u(t)

y(0) = 1

( ) (0) 5 ( )sy s y y s = 1s

5 ( ) 1 ( )y s sy s = 1s

( )[ 5]y s s = 1

1s

( )y s = 1

( 5)s

s s

y(s) =

1 45 5( 5)s s

Applying inverse laplace t ransform

y(t) = 51 4( ) ( )

5 5tu t e u t

y(t) = (0.2 + 0.8 e– 5t)46. From the given state model.

A =

– 1 1 0 0

0 – 1 0 B 4 c = [1 11]

0 0 – 2 0

Control lable : Qc = c = [B AB A2B]

if | Q | 0c control lable

Qc =

0 4 – 8

4 – 4 4 | Q | 0

0 0 0c

Uncontrol lable

Observable : Q0 =

2

C

CA

CA

i f 0| Q | 0 observable

Q0 =

0

1 1 1

– 1 0 – 2 | Q | 1

1 1 4

ObservableThe system is uncontrol lable and observable.

47. G(S) = 10

( 0.1)( 1)( 10)s s s

G(S) =

10

0.1 1 [1 ] 1 .100.1 10s s

s

G(S) =

10[1 10 ][1 ][1 0.1 ]s s s

By Approximation, G(S) =

10[10 1]s


48. G(s) = 2( 4)

( 1)( 2)s

s s

T(s) = G(S)

1+G(S)+H(S)

= 2(S 4)

(S 1)(S 2) 2(S 4)

Compare with S2 + 2Wn S + wn2 = 0

Wn = 2.5

set t =

20.8 sec.

nw

49. x2 = E(x2) – [E(x)]2

2x can never be negat ive

E[x2] [E(x)]2

50. Given X( ) 2sin (2 )t t

0

12

f ( )

2

in uniformly dist r ibuted in the interval [0, 2]E [X(t1) X (t2)]

=

2

1 202sin (2 ) 2 sin(2 ) ( )t t f d

=

2

1 20

12 sin(2 )sin(2 ). .

2t t d

=

2

1 20

1sin(2 ( ) 2 )

2t t d

2

1 20

1cos(2 ( – ))

2t t d

First integral wil l result into zero as we areintegrat ing from 0 to 2.Second integral result into cos[2(t1 – t2)]

E [X(t1) X(t2)] = cos (2(t1 – t2))

51. Bit error rate for BPSK = O

2E EQ Q

NO N 2

Y = O

2EN

Funct ion of bit energy and noise OSD

NP

2Counter l lation diagram of BPSKChannel is AWGN which implies noise sample asindependentLet 2x + n1 + n2 = x1 + n1

where x1 = 2xn1 = n1 + n2

Now Bit er ror rate = Q

1

1O

2EN

E1 is energy in x1

1ON is PSD of h1

E1 = 4E [as amplitudes are get t ing doubled]1ON = NO [independent and ident ical channel]

Bit er ror rate = Q

OO

4E 2EQ 2

NN

2 or 1.414b

52. I n this problem random var iable is LL can be 1, 2, .................

1

P{L 1}2

1

P{L 2}4

1

P{L 3}8

2 2 2

1 1 1 1 1 1H{L } log log log ....

2 1 / 2 4 1 / 4 8 1 / 8

= 1 1 1

0 1. 2. 3. ....2 4 8

[Ar ithmat ic geometr ic ser ies summation]

=

2

1 12 2 21 11 – 1 –2 2


53. E = j r100sin e

r

H Q = j r0.265sin e

r

Pavg = *Qs

1E H . ds

2

= 2 2

2s

1 100(0.265)sin r sin d d

2 r

Paet = 2s

1(26.5)sin d d

2

= /2 2

3

0 Q 0

13.25 sin d d

= 213.25 (2 )

3

P = 55.5 w

54. Zo = r

276 dlog

r

d distance between the two plates

so, zo – changes, if the spacing between the plateschanges.

V = 1

LC independent of spacing between the

plates

55. Given, = s

Zo = 50

inZ 8 =

L oo

o L

Z JZZ

Z KZ

Zin = L

L

Z J5050

50 JZ

=L L

L L

Z J50 50 JZ50

50 JZ 50 JZ

Zin = 2 2

L L L2 2

L

50Z 50Z J(50 Z )50

50 Z

Given, Zin Real

So, I mg (Zin) = 0

2 2L50 Z = 0

2LZ = 502

R2 + X2 = 502

R2 = 502 – X2 = 502 – 302

R = 40

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