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GATE EC2004Brought to you by: Nodia and Company Visit us at: www.nodia.co.in

PUBLISHING FOR GATEQ.1 - 30 Carry One Mark EachMCQ 1.1 Consider the network graph shown in the figure. Which one of the following isNOTa ‘tree’ of this graph ?(A) a (B) b(C) c (D) dSOL 1.1 For a tree there must not be any loop. So a, c, and d don ’t have any loop. Only bhas loop.Hence (B) is correct option.MCQ 1.2 The equivalent inductance measured between the terminals 1 and 2 for the

circuitshown in the figure isPage 2 GATE EC 2004 www.gatehelp.com

Brought to you by: Nodia and Company Visit us at: www.nodia.co.inPUBLISHING FOR GATE

(A) L1+ L2+M (B) L1+ L2−M(C) L1+ L2+ 2M (D)L1+ L2− 2MSOL 1.2 The sign of M is as per sign of L If current enters or exit the dotted terminals ofboth coil. The sign of M is opposite of L If current enters in dotted terminal of acoil and exit from the dotted terminal of other coil.Thus Leq = L1+ L2− 2M

Hence (D) is correct option.MCQ 1.3 The circuit shown in the figure, with R , L H 3

14

= Ω = 1 and C = 3 F has inputvoltage v (t ) = sin2t . The resulting current i (t ) is(A) 5 sin(2t + 53.1c) (B) 5 sin(2t − 53.1c)(C) 25 sin(2t + 53.1c) (D) 25 sin(2t − 53.1c)SOL 1.3 Here ω = 2 and V = 1+0cYR

j C

j L1 ω 1ω

= + +

3 j 2 3 3 4 j j2



14

1 ##= + + = +5 tan .3 4= + -1 = 5+53 11cI = V * Y = (1+0c)(5+53.1c) = 5+53.1cThus i (t ) = 5 sin(2t + 53.1c)Hence (A) is correct option.MCQ 1.4 For the circuit shown in the figure, the time constant RC = 1 ms. The inputvoltageis v i (t ) 2 sin10 t= 3 . The output voltage v (t ) o is equal toPage 3 GATE EC 2004 www.gatehelp.com

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PUBLISHING FOR GATE

(A) sin(103t − 45c) (B) sin(103t + 45c)(C) sin(103t − 53c) (D) sin(103t + 53c)SOL 1.4 Hence (A) is correct option.v i (t ) = 2 sin103tHere ω = 103 rad and V i = 2 +0cNow V 0 .R

j C j C V j CR

V1111t i

ω

ω

ω

=+

=+

1 j 10 101 2 0 3 3 # #= + c+ -

= 1 −45cv 0(t ) = sin(103t − 45c)



MCQ 1.5 For the R − L circuit shown in the figure, the input voltage v i (t ) = u(t ). Thecurrent i (t ) isSOL 1.5 Hence (C) is correct option.Input voltage v i (t ) = u(t )Taking laplace transform V i (s)

s= 1Page 4 GATE EC 2004 www.gatehelp.com


Impedance Z (s) = s + 2I (s)( )s ( )V s2 s s 2= i 1+=

+

or I (s)2 s s1 12= − 1; + ETaking inverse laplace transformi (t ) 2 (1 e )u(t )= 1 − −2t

At t = 0, i (t ) = 0 At t 2

1 = , i (t ) = 0.31 At t = 3, i (t ) = 0.5Graph (C) satisfies all these conditions.MCQ 1.6 The impurity commonly used for realizing the base region of a silicon n − p − ntransistor is(A) Gallium (B) Indium(C) Boron (D) PhosphorusSOL 1.6 Trivalent impurities are used for making p type semiconductor. Boron is trivalent.

Hence option (C) is correctMCQ 1.7 If for a silicon npn transistor, the base-to-emitter voltage (V BE ) is 0.7 V and thecollector-to-base voltage (V CB) is 0.2 V, then the transistor is operating in the(A) normal active mode (B) saturation mode(C) inverse active mode (D) cutoff modeSOL 1.7 Here emitter base junction is forward biased and base collector junction isreversedbiased. Thus transistor is operating in normal active region.



Hence option (A) is correct.MCQ 1.8 Consider the following statements S1 and S2.S1 : The β of a bipolar transistor reduces if the base width is increased.S2 : The β of a bipolar transistor increases if the dopoing concentration in the baseis increased.

Which remarks of the following is correct ?(A) S1 is FALSE and S2 is TRUE(B) Both S1 and S2 are TRUE(C) Both S1 and S2 are FALSE(D) S1 is TRUE and S2 is FALSESOL 1.8 Hence option (D) is correct.We have β 1 α = α

− Page 5 GATE EC 2004 www.gatehelp.com


Thus α -" β -α ." β .If the base width increases, recombination of carrier in base region increases andα decreases & hence β decreases. If doping in base region increases, recombinationof carrier in base increases and α decreases thereby decreasing β . Thus S1 is trueand S2 is false.MCQ 1.9 An ideal op-amp is an ideal(A) voltage controlled current source (B) voltage controlled voltage source(C) current controlled current source (D) current controlled voltage sourceSOL 1.9 An ideal OPAMP is an ideal voltage controlled voltage source.Hence (B) is correct option.MCQ 1.10 Voltage series feedback (also called series-shunt feedback) results in(A) increase in both input and output impedances(B) decrease in both input and output impedances(C) increase in input impedance and decrease in output impedance(D) decrease in input impedance and increase in output impedanceSOL 1.10 In voltage series feed back amplifier, input impedance increases by factor (1 +

A β )and output impedance decreases by the factor (1 + A β ).R if = R i (1 + A β )R of

( A )R1o

β

=

+

Hence (C) is correct option.



MCQ 1.11 The circuit in the figure is a(A) low-pass filter (B) high-pass filter(C) band-pass filter (D) band-reject filterSOL 1.11 This is a Low pass filter, because

At ω = 3

VVin

0 = 0and at ω = 0VVin

0 = 1Hence (A) is correct option.Page 6 GATE EC 2004 www.gatehelp.com


PUBLISHING FOR GATEMCQ 1.12 Assuming V CEsat = 0.2 V and β = 50, the minimum base current (I B) required todrive the transistor in the figure to saturation is(A) 56 μ A (B) 140 mA(C) 60 mA (D) 3 mASOL 1.12 Applying KVL we getV CC− I C R C − V CE = 0or I CRV VC

= CC − CE . 2.81k= 3 − 0 2 = mANow I B I 2.8 5650C m A β

= = = μ

Hence option (A) is correct.MCQ 1.13 A master - slave flip flop has the characteristic that(A) change in the output immediately reflected in the output(B) change in the output occurs when the state of the master is affected

(C) change in the output occurs when the state of the slave is affected(D) both the master and the slave states are affected at the same timeSOL 1.13 A master slave D-flip flop is shown in the figure.In the circuit we can see that output of flip-flop call be triggered only by transitionof clock from 1 to 0 or when state of slave latch is affected.Hence (C) is correct answer.MCQ 1.14 The range of signed decimal numbers that can be represented by 6-bits 1’scomplement number is



(A) -31 to +31 (B) -63 to +63(C) -64 to +63 (D) -32 to +31Page 7 GATE EC 2004 www.gatehelp.com


SOL 1.14 The range of signed decimal numbers that can be represented by n − bits 1’scomplement number is −(2n- 1− 1) to +(2n- 1− 1).Thus for n = 6 we haveRange =− (26 - 1− 1) to (26 1 1) + - −

=− 31 to + 31Hence (A) is correct answer.MCQ 1.15 A digital system is required to amplify a binary-encoded audio signal. The usershould be able to control the gain of the amplifier from minimum to a maximumin 100 increments. The minimum number of bits required to encode, in straightbinary, is(A) 8 (B) 6(C) 5 (D) 7SOL 1.15 The minimum number of bit require to encode 100 increment is2n $ 100or n $ 7Hence (D) is correct answer.MCQ 1.16 Choose the correct one from among the alternatives A,B,C ,D after matching anitem from Group 1 most appropriate item in Group 2.Group 1 Group 2P. Shift register 1. Frequency divisionQ. Counter 2. Addressing in memory chipsR. Decoder 3. Serial to parallel data conversion(A) P − 3,Q − 2,R − 1 (B) P − 3,Q − 1,R − 2(C) P − 2,Q − 1,R − 3 (D) P − 1,Q − 2,R − 2SOL 1.16 Shift Register " Serial to parallel data conversionCounter " Frequency divisionDecoder " Addressing in memory chips.Hence (B) is correct answer.MCQ 1.17 The figure the internal schematic of a TTL AND-OR-OR-Invert (AOI) gate. Forthe inputs shown in the figure, the output Y is(A) 0 (B) 1(C) AB (D) ABSOL 1.17 For the TTL family if terminal is floating, then it is at logic 1.Thus Y = ( AB + 1) = AB.0 = 0Page 8 GATE EC 2004 www.gatehelp.com


Hence (A) is correct answer.MCQ 1.18 Given figure is the voltage transfer characteristic of(A) an NOMS inverter with enhancement mode transistor as load(B) an NMOS inverter with depletion mode transistor as load(C) a CMOS inverter



(D) a BJT inverterSOL 1.18 Hence option (C) is correctMCQ 1.19 The impulse response h[n] of a linear time-invariant system is given byh[n] = u[n + 3] + u[n − 2) − 2n[n − 7] where u[n] is the unit step sequence. Theabove system is

(A) stable but not causal (B) stable and causal(C) causal but unstable (D) unstable and not causalSOL 1.19 A system is stable if h(n) <n

333=−

/ . The plot of given h(n) is

Thus h(n)n 33=−

/ h(n)n 36

==−

/= 1 + 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2= 15 < 3Hence system is stable but h(n) ! 0 for n < 0. Thus it is not causal.Hence (A) is correct answer.MCQ 1.20 The distribution function F x ( x ) of a random variable x is shown in the figure.The

probability that X = 1 isPage 9 GATE EC 2004 www.gatehelp.com


(A) zero (B) 0.25(C) 0.55 (D) 0.30SOL 1.20 Hence (D) is correct option.F ( x 1 # X < x 2) = p( X = x 2) − P ( X = x 1)or P ( X = 1) = P ( X = 1+) − P ( X = 1-)= 0.55 − 0.25 = 0.30MCQ 1.21 The z -transform of a system is H (z ) z .

z

0 2 = − . If the ROC is z < 0.2, then theimpulse response of the system is(A) (0.2)nu[n] (B) (0.2)nu[− n − 1](C) −(0.2)nu[n] (D) −(0.2)nu[− n − 1]SOL 1.21 Hence (D) is correct answer.H (z )z .z



0 2=

− z < 0.2We know that−anu[− n − 1]

1 az11 *− − z < aThus h[n] =− (0.2)nu[− n − 1]MCQ 1.22 The Fourier transform of a conjugate symmetric function is always(A) imaginary (B) conjugate anti-symmetric(C) real (D) conjugate symmetricSOL 1.22 The Fourier transform of a conjugate symmetrical function is always real.Hence (C) is correct answer.MCQ 1.23 The gain margin for the system with open-loop transfer function

G(s)H (s)( )s2 1 s2 = +

, is(A) 3 (B) 0(C) 1 (D) −3SOL 1.23 The open loop transfer function isG(s)H (s)( )s2 1 s2 = +

Substituting s = j ω we haveG( j ω)H ( j ω)2(1 j )ω2

= ω

−

+

...(1)Page 10 GATE EC 2004 www.gatehelp.com


+G( j ω)H ( j ω) =− 180c + tan−1ω

The frequency at which phase becomes −180c, is called phase crossover frequency.Thus −180 =− 180c + tan 1ωφ

−

or tan 1ωφ

− = 0



or ωφ = 0The gain at ωφ = 0 isG( j ω)H ( j ω) 2 122

3

ω = + ω =

Thus gain margin is 1 03= = and in dB this is −3.Hence (D) is correct optionMCQ 1.24 Given ( ) ( )( )( )G s H ss s sK

1 3=

+ + .The point of intersection of the asymptotes ofthe root loci with the real axis is(A) −4 (B) 1.33(C) −1.33 (D) 4SOL 1.24 Centroid is the point where all asymptotes intersects.σ

No.of Open Loop Pole No.of Open Loop zeroReal of Open Loop Pole Real Part of Open Loop Pole= Σ Σ

Σ Σ −

−

3= −1 − 3 =− 1.33Hence (C) is correct option.MCQ 1.25 In a PCM system, if the code word length is increased from 6 to 8 bits, thesignalto quantization noise ratio improves by the factor(A)6

8 (B) 12(C) 16 (D) 8SOL 1.25 When word length is 6N S

N =6 ` j = 22#6 = 212

When word length is 8N S

N =8 ` j = 22#8 = 216



NowN SNN SN6=8

= ^^hh22 2 16 12

16

= = 4 =

Thus it improves by a factor of 16.Hence (C) is correct option.MCQ 1.26 An AM signal is detected using an envelop detector. The carrier frequency andmodulating signal frequency are 1 MHz and 2 kHz respectively. An appropriate

value for the time constant of the envelop detector is(A) 500 μsec (B) 20 μsecPage 11 GATE EC 2004 www.gatehelp.com


(C) 0.2 μsec (D) 1 μsecSOL 1.26 Hence (B) is correct option.Carrier frequency f c 1 106 = # HzModulating frequencyf m 2 103 = # HzFor an envelope detector

2π f cRc> 1 > 2π f m2 f1π c RC2 f< < 1π m

2 f1

π c RC2 f< < 1π m

2 101π 6

RC



2 10< < 1 3 #1.59 10 7 # - < RC < 7.96 10 5 # -

so, 20 μsec sec best lies in this interval.MCQ 1.27 An AM signal and a narrow-band FM signal with identical carriers, modulating

signals and modulation indices of 0.1 are added together. The resultant signal canbe closely approximated by(A) broadband FM (B) SSB with carrier(C) DSB-SC (D) SSB without carrierSOL 1.27 Hence (B) is correct option.S AM (t ) = Ac [1 + 0.1 cosωmt ]cosωmtsNBFM (t ) = Ac cos [ωc t + 0.1 sinωmt ]s(t ) = S AM (t ) + SNB f m(t )= Ac [1 + 0.1 cosωmt ]cosωc t + Ac cos (ωc t + 0.1 sinωmt )= Ac cosωc t + Ac 0.1 cosωmt cosωc t+ Ac cosωc t cos (0.1 sinωmt ) − Ac sinωc t . sin(0.1 sinωmt )

As 0.1 sinωmt ,+ 0.1 to −0.1so cos (0.1 sinωmt ) . 1 As when θ is small cos θ . 1 and sinθ , θ , thussin(0.1 sinωmt ) = 0.1 sincosωc t cosωmt + Ac cosωc t − Ac 0.1 sinωmt sinωc t2 A cos t 0.1 A cos ( ) tcosecc c c c mUSB

= ω + ω + ω

1442443 1444442444443Thus it is SSB with carrier.MCQ 1.28 In the output of a DM speech encoder, the consecutive pulses are of opposite

polarity during time interval t 1

# t # t 2

. This indicates that during this interval(A) the input to the modulator is essentially constant(B) the modulator is going through slope overloadPage 12 GATE EC 2004 www.gatehelp.com


(C) the accumulator is in saturation(D) the speech signal is being sampled at the Nyquist rateSOL 1.28 Consecutive pulses are of same polarity when modulator is in slope overload.Consecutive pulses are of opposite polarity when the input is constant.Hence (A) is correct option.MCQ 1.29 The phase velocity of an electromagnetic wave propagating in a hollow

metallicrectangular waveguide in the TE 10 mode is(A) equal to its group velocity(B) less than the velocity of light in free space(C) equal to the velocity of light in free space(D) greater than the velocity of light in free spaceSOL 1.29 We know that v p > c > v g .Hence (D) is correct option.



MCQ 1.30 Consider a lossless antenna with a directive gain of +6 dB. If 1 mW of power isfedto it the total power radiated by the antenna will be(A) 4 mW (B) 1 mW(C) 7 mW (D) 1/4 mW

SOL 1.30 Hence (A) is correct option.We have GD(θ ,φ)( , )P4 Urad

= π θ φ

For lossless antennaP rad = P inHere we have P rad = P in = 1 mWand 10 logGD(θ ,φ) = 6 dBor GD(θ ,φ) = 3.98

Thus the total power radiated by antenna is4π U (θ ,φ) = P rad GD(θ ,φ) = 1 m #3.98 = 3.98 mW

Q.31 - 90 Carry Two Marks EachMCQ 1.31 For the lattice shown in the figure, Z a = j 2 Ω and Z b = 2 Ω. The values of theopencircuit impedance parameters zzzzz11

211222

6 @ = = G arePage 13 GATE EC 2004 www.gatehelp.com


(A) j j j j

1111−

+

+

= + G (B)



j j j j1

111−

− ++

= − G(C)

j j j

j1111+−

+

= − G (D) j j

j j1111+

− +

− +

= + GSOL 1.31 We know that

V 1

= z 11

I 1

+ z 12

I 2

V 2 = z 11I 1+ z 22I 2where z 11 IV1 I12 0

==

z 21 I



V1 I21 0

==

Consider the given lattice network, when I 2 = 0. There is two similar path in the

circuit for the current I 1. So I I21= 1

For z 11 applying KVL at input port we getV 1 = I (Z a + Z b)Thus V 1 I (Z Z )21= 1 a + b

z 11 (Z Z )

21= a + b

For Z 21 applying KVL at output port we getV 2 Z I Z I2 2 a b

= 1 − 1

Thus V 2 I (Z Z )21= 1 a − b

z 21 (Z Z )21= a − b

For this circuit z 11 = z 22 and z 12 = z 21. Thuszzzz11211222

= GZ ZZ ZZ ZZ Z22



22a ba ba b

a b =

+− −

+

RTSSSSSVXWWWW

WHere Z a = 2 j and Z b = 2Ω

Thuszzzz11211222

= G j

j j j111= 1+

−

−

= + G

Hence (D) is correct option.Page 14 GATE EC 2004 www.gatehelp.com


MCQ 1.32 The circuit shown in the figure has initial current i L (0−) = 1 A through theinductorand an initial voltage v C (0−) =− 1 V across the capacitor. For input v (t ) = u(t ),the Laplace transform of the current i (t ) for t $ 0 is



(A)s ss2+ + 1(B)

s ss122+ ++

(C)s ss122

+ +− (D)s s 112+ +

SOL 1.32 Applying KVL,v (t ) ( )( )Ri t ( )dtLdi tC1 i t dt0

= + +3 #Taking L.T. on both sides,V (s) ( ) ( ) ( )( ) ( )RI s LsI s LisCI ssC

v0= + − + + + c 0+

v (t ) = u(t ) thus V (s)s= 1Hences



1 ( ) ( )( )I s sI ssI s

s= + − 1 + − 1s2 + 1( )sI s= 6s2+ s + 1@or I (s)s ss

122 =

+ ++

Hence (B) is correct option.MCQ 1.33 Consider the Bode magnitude plot shown in the fig. The transfer function H (s)is(A)( )( )( )s ss1 10010+ +

+

(B)( )( )( )s ss10 10010 1+ ++

(C)( )( )( )s s



s10 100102 1+ ++

(D)( )( )( )s ss1 10103 100+ +

+Page 15 GATE EC 2004 www.gatehelp.com


SOL 1.33 The given bode plot is shown below At ω = 1 change in slope is +20 dB " 1 zero at ω = 1 At ω = 10 change in slope is −20 dB " 1 poles at ω = 10 At ω = 100 change in slope is −20 dB " 1 poles at ω = 100Thus T (s)( )( )K (s )1 11s s10 100

=+ +

+

Now 20 log10K =− 20 " K = 0.1Thus T (s)( )( ). ( )( )( )s ( )s ss

1 10 1 110 100100 1s s10 100

=

+ +

+ =



+ +

+

Hence (C) is correct option.MCQ 1.34 The transfer function ( )( )

( )H sV sV si

= o of an RLC circuit is given byH (s)s 20s 10102 66

=

+ +The Quality factor (Q-factor) of this circuit is(A) 25 (B) 50(C) 100 (D) 5000SOL 1.34 Characteristics equation iss2+ 20s + 106 = 0Comparing with s 2 n s n 02+ ξω + ω2 = we haveωn = 106 = 103

2ξω = 20Thus 2ξ .1020 0 02 3 = =

Now Q2 .10 021 50 ξ = = =

Hence (B) is correct option.MCQ 1.35 For the circuit shown in the figure, the initial conditions are zero. Its transferfunction ( )( )( )H sV sV si

= c isPage 16 GATE EC 2004 www.gatehelp.com




PUBLISHING FOR GATE

(A)s 10 s 1012+ 6 + 6

(B)s 10 s 10102 3 66

+ +

(C)s 10 s 10102 3 63

+ +

(D)

s 10 s 10102 6 66

+ +

SOL 1.35 Hence (D) is correct option.H (s)( )( )V sV si

= 0

R sLsCsC1 s LC sCR1112 =

+ +=

+ +s (10 10 ) s(10 10 ) 112 2 4 4 4 # #

=− − + − +

10 s s 1 s s1



10 10106 2 2 6 66

=

+ +

=− + +

MCQ 1.36 A system described by the following differential equation ( )dtd ydt3dy 2y x t 2

2

+ + = isinitially at rest. For input x (t ) = 2u(t ), the output y (t ) is(A) (1 − 2e−t + e−2t )u(t ) (B) (1 + 2e−t− 2e−2t )u(t )(C) (0.5 + e−t + 1.5e−2t )u(t ) (D) (0.5 + 2e− __________ t + 2e−2t )u(t )SOL 1.36 Hence Correct Option is (A)Given, 3dtd ydtdy 2y 2

2

+ + = x ̂ t hTaking Laplace Transformation both sides, we have6s2+ 3s + 2@Y ̂ s h X s s

= ^ h = 2

or Y ̂ s hs s 1 s 2= 2^ + h^ + h s s s1122= − 1 + + +

Increasing Laplace transformation gives,y ̂ t h = 1 − 2e−t + e−2t u t ^ h ^ h

MCQ 1.37 Consider the following statements S1 and S2S1 : At the resonant frequency the impedance of a series RLC circuit is zero.S2 : In a parallel GLC circuit, increasing the conductance G results in increase inits Q factor.Page 17 GATE EC 2004 www.gatehelp.com


Which one of the following is correct?



(A) S1 is FALSE and S2 is TRUE(B) Both S1 and S2 are TRUE(C) S1 is TRUE and S2 is FALSE(D) Both S1 and S2 are FALSESOL 1.37 Impedance of series RLC circuit at resonant frequency is minimum, not zero.

Actually imaginary part is zero.Z R j LCω 1ω

= + ` − j At resonance LCω 1 0ω

− = and Z = R that is purely resistive. Thus S1 is false

Now quality factor Q RL= CSince GR 1= , QG L= 1 CIf G - then Q . provided C and L are constant. Thus S2 is also false.Hence (D) is correct option.MCQ 1.38 In an abrupt p − n junction, the doping concentrations on the p −side and n-

sideare N A 9 1016 = # /cm3 respectively. The p − n junction is reverse biased and thetotal depletion width is 3 μm. The depletion width on the p −side is(A) 2.7 μm (B) 0.3 μm(C) 2.25 μm (D) 0.75 μmSOL 1.38 We know thatW pN A = W nN Dor W pNW N

A

= n # D 0.39 103 101616

#= μ # = μmHence option (B) is correct.



MCQ 1.39 The resistivity of a uniformly doped n −type silicon sample is 0.5Ω - mc. If theelectron mobility ( μn) is 1250 cm2/V-sec and the charge of an electron is 1.6 10 19 # -

Coulomb, the donor impurity concentration (N D) in the sample is(A) 2 1016 # /cm3 (B) 1 1016 # /cm3

(C) 2.5 1015 # /cm3 (D) 5 1015 # /cm3

SOL 1.39 Hence option (B) is correct.Conductivity σ = nqun

or resistivity ρ nq1 1σ μn

= =

Thus nq1 ρμn

=1.6 10 0.5 12501 10 19

16

# # #= = - /cm3

For n type semiconductor n = N DPage 18 GATE EC 2004 www.gatehelp.com


MCQ 1.40 Consider an abrupt p − n junction. Let V bi be the built-in potential of this junction

and V R be the applied reverse bias. If the junction capacitance (C j ) is 1 pF forV bi + V R = 1 V, then for V bi + V R = 4 V, C j will be(A) 4 pF (B) 2 pF(C) 0.25 pF (D) 0.5 pFSOL 1.40 We know thatC j(V V )(N N )e N N2 bi R A D

S A D 2

1 = ε

; + + EThus C j(V V )1bi R

\ +

NowCC



j j12

( )( )V V

V V4121bi Rbi R2

= 1

++ = =

or C j 2 0.5

C2 2= j 1 = 1 = pFHence option (D) is correct.MCQ 1.41 Consider the following statements Sq and S2.S1 : The threshold voltage (V T ) of MOS capacitor decreases with increase in gateoxide thickness.S2 : The threshold voltage (V T ) of a MOS capacitor decreases with increase insubstrate doping concentration.Which Marks of the following is correct ?(A) S1 is FALSE and S2 is TRUE

(B) Both S1 and S2 are TRUE(C) Both S1 and S2 are FALSE(D) S1 is TRUE and S2 is FALSESOL 1.41 Increase in gate oxide thickness makes difficult to induce charges in channel.ThusV T increases if we increases gate oxide thickness. Hence S1 is false.Increase in substrate doping concentration require more gate voltage becauseinitially induce charges will get combine in substrate. Thus V T increases if weincrease substrate doping concentration. Hence S2 is false.Hence option (C) is correct.MCQ 1.42 The drain of an n-channel MOSFET is shorted to the gate so that V GS = V DS .

Thethreshold voltage (V T ) of the MOSFET is 1 V. If the drain current (I D) is 1 mA forV GS = 2 V, then for V GS = 3 V, I D is(A) 2 mA (B) 3 mA(C) 9 mA (D) 4 mAPage 19 GATE EC 2004 www.gatehelp.com




SOL 1.42 We know thatI D K (V GS V T )= − 2

ThusI

IDIDS

( )( )V VV VGS TGS T1222

=

− −

Substituting the values we haveIIDD12

( )( )2 13 14 2

2

=

−

− =or I D2 = 4I DI = 4 mAHence option (D) is correct.MCQ 1.43 The longest wavelength that can be absorbed by silicon, which has thebandgapof 1.12 eV, is 1.1 μm. If the longest wavelength that can be absorbed by anothermaterial is 0.87 μm, then bandgap of this material is

(A) 1.416 A/cm2 (B) 0.886 eV(C) 0.854 eV (D) 0.706 eVSOL 1.43 Hence option (A) is correct.E g \ 1 λ ThusEEg



g12

.

.0 87

1 121

λ

= λ =or E g 2

.

. . .0 87= 1 1 # 1 12 = 1 416 eVMCQ 1.44 The neutral base width of a bipolar transistor, biased in the active region, is 0.5 μ

m. The maximum electron concentration and the diffusion constant in the base are1014/cm3 and Dn = 25 cm2/sec respectively. Assuming negligible recombination inthe base, the collector current density is (the electron charge is 1.6 10 19 # - Coulomb)(A) 800 A/cm2 (B) 8 A/cm2

(C) 200 A/cm2 (D) 2 A/cm2

SOL 1.44 Concentration gradientdxdn0.5 1010 2 10 4

1418

#= = # -

q 1.6 10 19C = # -

Dn = 25dxdn0.5 1010414

#= -

J C qDdxdn= n

1.6 10 19 25 2 1018 = # - # # # = 8 A/cm2

Hence option (B) is correct.Page 20 GATE EC 2004 www.gatehelp.com




PUBLISHING FOR GATE

MCQ 1.45 Assume that the β of transistor is extremely large and V BE = 0.7V , I C and V CE inthe circuit shown in the figure(A) I C = 1mA,V CE = 4.7 V (B) I C = 0.5 mA, V CE = 3.75 V(C) I C = 1 mA, V CE = 2.5 V (D) I C = 0.5 mA, V CE = 3.9 V

SOL 1.45 The thevenin equivalent is shown belowV TR RR V C1 2

= 1

+ 54 1= 1 # 1+

= VSince β is large is large, I C . I E , I B . 0 and

I ERV VE

= T − BE .300= 1 − 0 7 = 3 mANow V CE = 5 − 2.2kI C− 300I E= 5 − 2.2k#1m− 300#1m= 2.5 VHence (C) is correct option

MCQ 1.46 A bipolar transistor is operating in the active region with a collector current of 1mA. Assuming that the β of the transistor is 100 and the thermal voltage (V T ) is25 mV, the transconductance (g m) and the input resistance (r π ) of the transistor inthe common emitter configuration, are(A) g m = 25 mA/V and r π = 15.625 kΩ (B) g m = 40 mA/V and r π = 4.0 kΩ (C) g m = 25 mA/V and r π = 2.5 k Ω

(D) g m = 40 mA/V and r π = 2.5 kΩ Page 21 GATE EC 2004 www.gatehelp.com


SOL 1.46 When I C >> I COg mVIT

C =

251mV



= mA = 0.04 = 40 mA/Vr π .g 40 10100 2 5m 3 #

= β = = - kΩ Hence (D) is correct option.

MCQ 1.47 The value of C required for sinusoidal oscillations of frequency 1 kHz in thecircuitof the figure is(A)21π μF (B) 2π μF(C)2 6

1π μF (D) 2π 6 μF]SOL 1.47 The given circuit is wein bridge oscillator. The frequency of oscillation is2π fRC= 1or C2 Rf1π =

2 10 1013 3 π # #=

21π

= μ

Hence (A) is correct option.MCQ 1.48 In the op-amp circuit given in the figure, the load current i L is(A)RV s2

− (B)RV s2

Page 22 GATE EC 2004 www.gatehelp.com




(C)RVL

− s (D)RV s1

SOL 1.48 The circuit is as shown belowWe know that for ideal OPAMPV - = V +

Applying KCL at inverting terminalRV VR

s V V1 1

-− + -− 0 = 0or 2V -− V o = V s ...(1)

Applying KCL at non-inverting terminalRV IRV VLo2 2

+ + + +− = 0or 2V +− V o + I LR 2 = 0 ...(2)Since V - = V +, from (1) and (2) we haveV s + I LR 2 = 0or I LRV s2

=−

Hence (A) is correct option.MCQ 1.49 In the voltage regulator shown in the figure, the load current can vary from 100mA to 500 mA. Assuming that the Zener diode is ideal (i.e., the Zener knee current

is negligibly small and Zener resistance is zero in the breakdown region), the valueof R is(A) 7 Ω (B) 70 Ω

(C)370 Ω (D) 14 Ω SOL 1.49 If I Z is negligible the load current isPage 23 GATE EC 2004 www.gatehelp.com




R12 − V z I = L

as per given condition

100 mAR# 12 V Z # 500 − mA

At I L = 100 mAR12 − 5 = 100 mA V 5 Z = Vor R = 70Ω

At I L = 500 mAR12 − 5 = 500 mA V 5 Z = Vor R = 14 Ω

Thus taking minimum we getR = 14 Ω

Hence (D) is correct option.MCQ 1.50 In a full-wave rectifier using two ideal diodes, V dc and V m are the dc and peakvaluesof the voltage respectively across a resistive load. If PIV is the peak inverse voltageof the diode, then the appropriate relationships for this rectifier are(A) V dc V , PIV 2Vm

π m

= = (B) I dc 2V , PIV 2Vm

π m= =

(C) V dc 2V , PIV Vm

π m

= = (D) V dc V , PIV Vm

π m

=

SOL 1.50 Hence (B) is correct option.MCQ 1.51 The minimum number of 2- to -1 multiplexers required to realize a 4- to -1multiplexers is

(A) 1 (B) 2(C) 3 (D) 4SOL 1.51 Number of MUX is3 4= 2 and2 2= 1. Thus the total number 3 multiplexers isrequired.



Hence (C) is correct answer.MCQ 1.52 The Boolean expression AC + BC is equivalent to(A) AC + BC + AC (B) BC + AC + BC + ACB(C) AC + BC + BC + ABC (D) ABC + ABC + ABC + ABCSOL 1.52 Hence (D) is correct answer.

AC + BC = AC 1 + BC 1= AC (B + B) + BC ( A + A)= ACB + ACB + BCA + BCAMCQ 1.53 11001, 1001, 111001 correspond to the 2’s complement representation ofwhich oneof the following sets of number(A) 25,9, and 57 respectively (B) -6, -6, and -6 respectivelyPage 24 GATE EC 2004 www.gatehelp.com


(C) -7, -7 and -7 respectively (D) -25, -9 and -57 respectivelySOL 1.53 Hence (C) is correct answer.11001 1001 11100100110 0110 000110+1 +1 +100111 0111 0001117 7 7Thus 2’s complement of 11001, 1001 and 111001 is 7. So the number given in thequestion are 2’s complement correspond to -7.MCQ 1.54 The 8255 Programmable Peripheral Interface is used as described below.(i) An A/D converter is interface to a microprocessor through an 8255.The conversion is initiated by a signal from the 8255 on Port C. A signal on PortC causes data to be stobed into Port A.(ii) Two computers exchange data using a pair of 8255s. Port A works as abidirectional data port supported by appropriate handshaking signals.The appropriate modes of operation of the 8255 for (i) and (ii) would be(A) Mode 0 for (i) and Mode 1 for (ii)(B) Mode 1 for (i) and Mode 2 for (ii)(C) Mode for (i) and Mode 0 for (ii)(D) Mode 2 for (i) and Mode 1 for (ii)SOL 1.54 For 8255, various modes are described as following.Mode 1 : Input or output with hand shakeIn this mode following actions are executed1. Two port (A & B) function as 8 - bit input output ports.

2. Each port uses three lines from C as a hand shake signal3. Input & output data are latched.Form (ii) the mode is 1.Mode 2 : Bi-directional data transferThis mode is used to transfer data between two computer. In this mode port A canbe configured as bidirectional port. Port A uses five signal from port C as handshake signal.For (1), mode is 2



Hence (D) is correct answer.MCQ 1.55 The number of memory cycles required to execute the following 8085instructions(i) LDA 3000 H(ii) LXI D, FOF1H

would bePage 25 GATE EC 2004 www.gatehelp.com


(A) 2 for (i) and 2 for (ii) (B) 4 for (i) and 3 for (ii)(C) 3 for (i) and 3 for (ii) (D) 3 for (i) and 4 for (ii)SOL 1.55 LDA 16 bit & Load accumulator directly this instruction copies data byte frommemory location (specified within the instruction) the accumulator.It takes 4 memory cycle-as following.1. in instruction fetch2. in reading 16 bit address1. in copying data from memory to accumulatorLXI D, (F0F1)4 & It copies 16 bit data into register pair D and E.It takes 3 memory cycles.Hence (B) is correct answer.MCQ 1.56 In the modulo-6 ripple counter shown in figure, the output of the 2- input gate isused to clear the J-K flip-flopThe 2-input gate is(A) a NAND gate (B) a NOR gate(C) an OR gate (D) a AND gareSOL 1.56 In the modulo - 6 ripple counter at the end of sixth pulse (i.e. after 101 or at110)all states must be cleared. Thus when CB is 11 the all states must be cleared. The

input to 2-input gate is C and B and the desired output should be low since theCLEAR is active lowThus when C and B are 0, 0, then output must be 0. In all other case the outputmust be 1. OR gate can implement this functions.Hence (C) is correct answer.MCQ 1.57 Consider the sequence of 8085 instructions given belowLXI H, 9258MOV A, MCMAMOV M, AWhich one of the following is performed by this sequence ?

(A) Contents of location 9258 are moved to the accumulatorPage 26 GATE EC 2004 www.gatehelp.com


(B) Contents of location 9258 are compared with the contents of the accumulator(C) Contents of location 8529 are complemented and stored in location 8529(D) Contents of location 5892 are complemented and stored in location 5892SOL 1.57 Hence (A) is correct answer.



LXI H, 9258H ; 9258H " HLMOV A, M ; (9258H) " ACMa ; A " AMOV M, A ; A " MThis program complement the data of memory location 9258H.

MCQ 1.58 A Boolean function f of two variables x and y is defined as follows :f (0,0) = f (0,1) = f (1,1) = 1; f (1,0) = 0 Assuming complements of x and y are not available, a minimum cost solution forrealizing f using only 2-input NOR gates and 2- input OR gates (each having unitcost) would have a total cost of(A) 1 unit (B) 4 unit(C) 3 unit (D) 2 unitSOL 1.58 Hence (D) is correct answer.We have f ( x ,y ) = xy + xy + xy = x (y + y ) + xy = x + xyor f ( x ,y ) = x + yHere compliments are not available, so to get x we use NOR gate. Thus desired

circuit require 1 unit OR and 1 unit NOR gate giving total cost 2 unit.MCQ 1.59 It is desired to multiply the numbers 0AH by 0BH and store the result in theaccumulator. The numbers are available in registers B and C respectively. A partof the 8085 program for this purpose is given below :MVI A, 00HLOOP -----------------HLTENDThe sequence of instructions to complete the program would be(A) JNX LOOP, ADD B, DCR C(B) ADD B, JNZ LOOP, DCR C(C) DCR C, JNZ LOOP, ADD B(D) ADD B, DCR C, JNZ LOOPSOL 1.59 Hence (D) is correct answer.MVI A, 00H ; Clear accumulatorLOOP ADD B ; Add the contents of B to APage 27 GATE EC 2004 www.gatehelp.com


DCR C ; Decrement CJNZ LOOP ; If C is not zero jump to loop

HLTENDThis instruction set add the contents of B to accumulator to contents of C times.Hence (D) is correct answer.MCQ 1.60 A 1 kHz sinusoidal signal is ideally sampled at 1500 samples/sec and thesampledsignal is passed through an ideal low-pass filter with cut-off frequency 800 Hz. Theoutput signal has the frequency.



(A) zero Hz (B) 0.75 kHz(C) 0.5 kHz (D) 0.25 kHzSOL 1.60 Hence Correct Option is (C)Here f s = 1500 samples/sec, f m = kHzThe sampled frequency are 2.5 kHz, 0.5 kHz, Since LPF has cut-off frequency

800 Hz, then only output signal of frequency 0.5 kHz would pass through itMCQ 1.61 A rectangular pulse train s(t ) as shown in the figure is convolved with the signalcos2(4 p 103t ) # . The convolved signal will be a(A) DC (B) 12 kHz sinusoid(C) 8 kHz sinusoid (D) 14 kHz sinusoidSOL 1.61 Hence Correct Option is (D)S^t h T 1 1 2 cos t 2 cos 2 t .....................s

= 6 + ωs + ωs + @cos24 103t π #cos t2

1 8 103 π #=

^ + hωs

0.1 102 2 10 10 3

3

#

# # = π = π −

S^t h* x ̂ t h = S τ # τ − t d τ 33

−

# ^ h ^ h10 10 1 2 cos s t 2 cos 2 s t ........ dt3 = # + ω + ω +33−

# 6 @

cos t21 8 103 π #

#

6 + @Page 28 GATE EC 2004 www.gatehelp.com


So, frequencies present will be f s!f m,2f s!3f s!f m; f s = 10 kHzf m 2 4 kHz8 103 #



π

= π =

Hence 14 kHz sinusoidal signal will be presentMCQ 1.62 Consider the sequence x [n] = [− 4 − j 51 + j 25]-

. The conjugate anti-symmetric partof the sequence is(A) [− 4 − j 2.5, j 2, 4 − j 2.5] (B) [− j 2.5, 1, j 2.5](C) [− j 2.5, j 2, 0] (D) [− 4, 1, 4]SOL 1.62 Hence (A) is correct answer.We have x (n) = [− 4 − j 5, 1 + 2 j , 4]-

x *( n) = [− 4 + j 5, 1 − 2 j , 4]-

x *( − n) = [4, 1 − 2 j , − 4 + j 5]-

x cas (n) x (n) x ( n)

2*

= − −

[ 4 j , 2 j 4 j ] 2

52

= − − − 5-

MCQ 1.63 A causal LTI system is described by the difference equation2y [n] = αy [n − 2] − 2 x [n] + β x [n − 1]The system is stable only if(A) α = 2, β < 2 (B) α > 2, β > 2

(C) α < 2, any value of β (D) β < 2, any value of α SOL 1.63 Hence (C) is correct answer.We have 2y (n) = αy (n − 2) − 2 x (n) + β x (n − 1)Taking z transform we get2Y (z ) = αY (z )z −2− 2 X (z ) + β X (z )z −1

or( )( )

X zY zz

z2221

α

= β

−

−



−

−

c m ...(i)or H (z )( )( )

zz z22

= 2

−

− α

β

It has poles at ! α/2 and zero at 0 and β /2. For a stable system poles must lieinside the unit circle of z plane. Thus2 α

< 1or α < 2But zero can lie anywhere in plane. Thus, β can be of any value.Page 29 GATE EC 2004 www.gatehelp.com


MCQ 1.64 A causal system having the transfer function H (s) = 1/(s + 2) is excited with10u(t ). The time at which the output reaches 99% of its steady state value is(A) 2.7 sec (B) 2.5 sec(C) 2.3 sec (D) 2.1 secSOL 1.64 Hence (C) is correct option.We have r (t ) = 10u(t )

or R (s)s= 10Now H (s)s 2= 1+C (s) ( ) ( )( )H s R ss 2 s s s

1 102= $ = $ 10+ +or C (s)s s52



= − 5+

c (t ) = 5[1 − e−2t ]The steady state value of c (t ) is 5. It will reach 99% of steady state value reachesat t , where

5[1 − e−2t ] = 0.99 # 5or 1 − e−2t = 0.99e−2t = 0.1or −2t = ln0.1or t = 2.3 secMCQ 1.65 The impulse response h[n] of a linear time invariant system is given ash[n],,nn

2 2 1 14 2 2 20 otherwise=− = −

* = −

If the input to the above system is the sequence e j π n/4, then the output is(A) 4 2 e j π n/4 (B) 4 2 e− j π n/4

(C) 4e j π n/4 (D) −4e j π n/4

SOL 1.65 Hence (D) is correct answer.We have x (n) = e j π n/4

and h(n) = 4 2 δ(n + 2) − 2 2 δ(n + 1) − 2 2 δ(n − 1)+4 2 δ(n − 2)Now y (n) = x (n)* h(n)

x (n k )h(k )k

= − 33=−

/ x (n k )h(k )k 22

= −

=−

/or y (n) = x (n + 2)h(− 2) + x (n + 1)h(− 1)+ x (n − 1)h(1) + x (n − 2)h(2)4 2 e j (n 2) 2 2 e j (n 1) 2 2 e j (n 1) 4 2 e j (n 2) = 4 +− 4 +− 4 − + 4− π π π π





4 2 e j (n 2) e j (n 2) 2 2 e j (n 1) e j (n 1) = 4 + + 4− − 4 + + 4− π π π π 6 @ 6 @

4 2 e j n e j e j 2 2 e j n e j e j = 4 2 + − 2− 2 4 + − 4 π π π π π π 6 @ 6 @4 2 e j n [0] 2 2 e j n [2 cos ]4

= 4− 4 π π π

or y (n) 4e j n =− 4r

MCQ 1.66 Let x (t ) and y (t ) with Fourier transforms F (f ) and Y (f ) respectively be related asshown in Fig. Then Y (f ) is(A) 2 X (f /2)e−1 − j π f (B) X (f / )e2−1 2 j 2π f

(C) − X (f /2)e j 2π f (D) − X (f /2)e− j 2π f

SOL 1.66 From given graph the relation in x (t ) and y (t ) isy (t ) =− x [2(t + 1)]

x (t ) X (f ) F

Using scaling we have x (at )a

XaF 1 f c mThus x (2t ) X f212F c mUsing shifting property we ge

( ) x t t 0− = e− j 2π ft 0 X (f )Thus x [2(t + 1)] e X f e X f212 2 2F j f ( ) j f2 12

π =π

− − b l b l

− x [2(t + 1)] e X f2 2F j 2 f

− π

c mHence (B) is correct answer.MCQ 1.67 A system has poles at 0.1 Hz, 1 Hz and 80 Hz; zeros at 5 Hz, 100 Hz and 200Hz.



The approximate phase of the system response at 20 Hz is(A) −90c (B) 0c(C) 90c (D) −180cSOL 1.67 Approximate (comparable to 90c) phase shift areDue to pole at 0.01 Hz "− 90c

Due to pole at 80 Hz "− 90cDue to pole at 80 Hz " 0Due to zero at 5 Hz " 90cPage 31 GATE EC 2004 www.gatehelp.com


Due to zero at 100 Hz " 0Due to zero at 200 Hz " 0Thus approximate total −90c phase shift is provided.Hence (A) is correct option.MCQ 1.68 Consider the signal flow graph shown in Fig. The gain

x x1

5 is(A)( )abcd1 − be + cf + dg(B)(be cf dg )bedg1− + +

(C)(be cf dg ) bedgabcd1− + + + (D)( )abcd1 − be + cf + dg + bedgSOL 1.68 Mason Gain FormulaT (s) pk k

3= Σ 3

In given SFG there is only one forward path and 3 possible loop. p1 = abcd31 = 13= 1 − (sum of indivudual loops) - (Sum of two non touching loops)= 1 − (L1+ L2+ L3) + (L1L3)Non touching loop are L1 and L3 whereL1L2 = bedgThus



( )( )R sC s(be cf dg ) bedg

p1= 131

− + + +

(be cf dg ) bedgabcd1=

− + + +

Hence (C) is correct optionMCQ 1.69 If A

212= 3−

= − G, then sin At is(A)( ) ( )( ) ( )( ) ( )( ) ( )

sin sinsin sinsin sinsin sint tt tt t3 t t1 4 242 4 22 4− + − − − + −

− − + −

= − + − G(B)( )( )



( )( )sinsinsin

sintttt2 23−

= − G(C)( ) ( )

( ) ()( ) ( )( ) ( )sin sinsin sinsin sinsin sint tt tt t3 t t

1 4 242 4 22 4+

− − +− − −

= + G(D)( ) ()( ) ( )( ) ( )( ) ()cos coscos coscos coscos cost tt t



t t3 t t1 242 4 2

2 4− +− − + −

− + −

= − − + GPage 32 GATE EC 2004 www.gatehelp.com


SOL 1.69 Hence (A) is correct optionWe have A212= 3−

= − GCharacteristic equation is[ λI − A] = 0or2123 λ λ

+

− −

+ = 0or ( λ + 2)( λ + 3) − 2 = 0or λ2+ 5 λ + 4 = 0Thus λ1 =− 4 and λ2 =− 1Eigen values are −4 and −1.Eigen vectors for λ1 =− 4( λ1I − A) X 1 = 0or

x x2123



111121

λ

λ + −

= + G= G = 0 x x21211121

−

−

− = − G= G = 0or −2 x 11− 2 x 21 = 0or x 11+ x 21 = 0We have only one independent equation x 11 =− x 21.Let x 21 = K , then x 11 =− K , the Eigen vector will be

x x1121

= GK

K K1= 1−

=

−

= G = GNow Eigen vector for λ2 =− 1( λ2I − A) X 2 = 0or

x

x2123221222



λ

λ

+

− −

= + G= G = 0or112− 2−

= G x x1222

= G = 0We have only one independent equation x 12 = 2 x 22

Let x 22 = K , then x 12 = 2K . Thus Eigen vector will be x x1222

= GKK2= = G K

2= =1GDigonalizing matrixM

x x x x112111211222

= =

−

= G = GNow M −1



31 112= 1 −

− −

` j= − GPage 33 GATE EC 2004 www.gatehelp.com


Now Diagonal matrix of sin At is D whereD( )( )sinsint0 t1 02

λ

= = λ G( )( )sinsint

t400 λ2 =

−

= GNow matrix B = sin At = MDM −1

( )( )sin

sint3 t1 11214



00 112=− 1

− − − − −

` j= G= G= − G( 4) 2 ( )( ) ()( ) ( )( ) ( )sin sinsin sinsin sin

sin sint tt tt t3 t t14 22 4 2=− 2 4− − − − − +

− − − ` j= − − − − G( ) ( )( ) ( )( ) ( )( ) ( )sin sinsin sinsin sinsin sint tt tt t3 t t1 4 242 4 2=− 2 4 2− − − −



− − −

− − −

` j= − − + − G( ) ( )( ( )

( ) ( )( ) ( )sin sinsin sinsin sinsin sint tt tt tt t s3

1 4 242 4 2= 2 4− + −

− − + − − − + −

` j= − + − GMCQ 1.70 The open-loop transfer function of a unity feedback system isG(s)s(s s )(s )

K2 2 3 =+ + +

The range of K for which the system is stable is(A) K421 > > 0 (B) 13 > K > 0(C) K421 < < 3 (D) −6 < K < 3SOL 1.70 For ufb system the characteristic equation is1

+G(s)

=0

s(s s )(s )1 K2 2 3G(s)21

+

+ + ++



= 0s4+ 4s3+ 5s2+ 6s + K = 0The routh table is shown below. For system to be stable,0 < K and/

( ) K02 7< 21 − 4This gives 0 K4< < 21s4 1 5 Ks3 4 6 0s2

2

7 K

s1/K7 2

21−4 0s0 KPage 34 GATE EC 2004 www.gatehelp.com


Hence (A) is correct optionMCQ 1.71 For the polynomial P (s) = s2+ s4+ 2s3+ 2s2+ 3s + 15 the number of roots whichlie in the right half of the s −plane is

(A) 4 (B) 2(C) 3 (D) 1SOL 1.71 Hence (B) is correct option.We have P (s) = s5+ s4+ 2s3+ 3s + 15The routh table is shown below.If ε " 0+ then 2 12

ε

ε+ is positive and 2 12

15 2 24 144ε

ε ε

+

− − − is negative. Thus there are two sign

change in first column. Hence system has 2 root on RHS of plane.s5 1 2 3s4 1 2 15s3 ε −12 0s2 2 12

ε

ε+ 15 0s12 12



15 2 24 144ε

ε ε

+

− − −

s0 0MCQ 1.72 The state variable equations of a system are : x o1 =− 3 x 1− x 2 = u, x o2 = 2 x 1 and

y = x 1

+ u. The system is(A) controllable but not observable(B) observable but not controllable(C) neither controllable nor observable(D) controllable and observableSOL 1.72 Hence (D) is correct option.We have

x x12

= G

x x u32101012

= +− −

= G= G = Gand Y = [1 0]

x x u1212

= G+= GHere A321= 0− −

= G, B1= =0G and C = [1 0]The controllability matrix is



QC = [B AB]103= 2

− = GdetQC ! 0 Thus controllableThe observability matrix isPage 35 GATE EC 2004 www.gatehelp.com


Q0 = [C T AT C T ]103= 1 ! 0−

= − GdetQ0 ! 0 Thus observableMCQ 1.73 Given A100= = 1G, the state transition matrix e At is given by(A)e0 e

t 0t−

−

> H (B)e0 et 0= t G(C)e0 et 0t−

> − H (D)e0 et 0t

= G



SOL 1.73 Hence (B) is correct option.(sI − A)s0 s0 1

00= = G−= 1Gss100= 1−

= − G

(sI − A)−1( )( )s ( )s1 s1 1002 1 =−

− = − G 00 s

s111

= 1 −

−

> He At = L−1[(sI − A)]−1

e0 et

0= = t GMCQ 1.74 Consider the signal x (t ) shown in Fig. Let h(t ) denote the impulse response ofthefilter matched to x (t ), with h(t ) being non-zero only in the interval 0 to 4 sec. Theslope of h(t ) in the interval 3 < t < 4 sec is(A) sec2



1 -1 (B) −1 sec-1

(C) sec2−1 -1 (D) 1 sec-1

SOL 1.74 The impulse response of matched filter is

h(t ) = x (T − t )Since here T = 4, thush(t ) = x (4 − t )The graph of h(t ) is as shown below.Page 36 GATE EC 2004 www.gatehelp.com


From graph it may be easily seen that slope between 3 < t < 4 is −1.Hence (B) is correct option.MCQ 1.75 A 1 mW video signal having a bandwidth of 100 MHz is transmitted to areceiverthrough cable that has 40 dB loss. If the effective one-side noise spectral density atthe receiver is 10-20 Watt/Hz, then the signal-to-noise ratio at the receiver is(A) 50 dB (B) 30 dB(C) 40 dB (D) 60 dBSOL 1.75 The SNR at transmitter isSNR trBP N= tr

10 100 101020 63

# # -

-

109 =In dB SNR tr = 10 log 109 = 90 dBCable Loss = 40 db

At receiver after cable loss we haveSNR Rc = 90 − 40 = 50 dBHence (A) is correct option.MCQ 1.76 A 100 MHz carrier of 1 V amplitude and a 1 MHz modulating signal of 1 Vamplitude are fed to a balanced modulator. The ourput of the modulator is passed

through an ideal high-pass filter with cut-off frequency of 100 MHz. The outputof the filter is added with 100 MHz signal of 1 V amplitude and 90c phase shift asshown in the figure. The envelope of the resultant signal is(A) constant (B) 1 sin(2 106t ) + π #(C) sin( t )45 − 2π − 106 (D) cos ( t )4



5 2 106 + π #SOL 1.76 Hence (C) is correct option.We have f c = 100 MHz 100 106 = # and f m = 1 MHz1 106 = #Page 37 GATE EC 2004 www.gatehelp.com


The output of balanced modulator isV BM (t ) = [cosωc t ][cosωc t ][cos ( ) t cos ( ) t ]21= ωc + ωm + ωc − ωm

If V BM (t ) is passed through HPF of cut off frequency f H 100 106 = # , then only(ωc + ωm) passes and output of HPF isV HP (t ) cos ( ) t21= ωc + ωm

Now V 0(t ) V HP (t ) sin(2 100 10 ) t6 = + π # #cos [2 100 10 2 1 10 t ] sin(2 100 10 ) t2= 1 π # 6+ π # # 6 + π # # 6

cos [ t ] sin( ) t2= 1 2π 108+ 2π 106 + 2π 108

[cos (2 10 t ) t cos (2 10 t )] sin[2 10 t sin(2 10 t ) sin2 10 t ]2= 1 π 8 π 6 − π 8 π 6 + π 8

cos ( t )cos t sin t sin t21 2 10 2 10 12= π 6 π 8 +` − 1 2π 106 j 2π 108

This signal is in form= Acos 2π 108t + Bsin2π 108t

The envelope of this signal is= A2+ B2

cos ( t ) sin( t21 2 10 126 2 1 2 106 2 = ` π j +` − π jcos ( t ) sin ( t ) sin( t )4



1 2 10 14= 2 π 6 + + 1 2 2π 106 − 2π 106

sin( t )4

= 1 + 1 − 2π 106sin( t )4= 5 − 2π 106

MCQ 1.77 Two sinusoidal signals of same amplitude and frequencies 10 kHz and 10.1kHz areadded together. The combined signal is given to an ideal frequency detector. Theoutput of the detector is(A) 0.1 kHz sinusoid (B) 20.1 kHz sinusoid(C) a linear function of time (D) a constantSOL 1.77 Hence (A) is correct option.

s(t ) Acos [2 10 103

t ] Acos [2 10.1 103

t ] = π

# + π

#Here T 1 sec10 101 100 3 #= = μ

and T 2.sec10 1 101 99 3 #= = μ Period of added signal will be LCM [T 1,T 2]Page 38 GATE EC 2004 www.gatehelp.com


Thus T = LCM [100,99] = 9900 μsecThus frequency f .99001 0 1 μ

= = kHzMCQ 1.78 Consider a binary digital communication system with equally likely 0’s and 1’s.When binary 0 is transmitted the detector input can lie between the levels −0.25V and +0.25 V with equl probability : when binary 1 is transmitted, the voltage

at the detector can have any value between 0 and 1 V with equal probability. If thedetector has a threshold of 0.2 V (i.e., if the received signal is greater than 0.2 V,the bit is taken as 1), the average bit error probability is(A) 0.15 (B) 0.2(C) 0.05 (D) 0.5SOL 1.78 The pdf of transmission of 0 and 1 will be as shown below :Probability of error of 1P (0 # X # 0.2) = 0.2



Probability of error of 0 :P (0.2 # X # 0.25) = 0.05 # 2 = 0.1

Average errorP ( X . ) P ( . X . )2

0 # # 0 2 0 2 # # 0 25 = +. . .0= 0 2 + 0 1 = 0 15Hence (A) is correct option.MCQ 1.79 A random variable X with uniform density in the interval 0 to 1 is quantized asfollows :If 0 # X # 0.3, x q = 0If 0.3 < X # 1, x q = 0.7where x q is the quantized value of X .The root-mean square value of the quantization noise is

(A) 0.573 (B) 0.198(C) 2.205 (D) 0.266SOL 1.79 Hence (B) is correct option.The square mean value isσ 2 ( x x q) f ( x )dx= − 233

- #Page 39 GATE EC 2004 www.gatehelp.com


( x x q) f ( x )dx201

= # −

( x 0) f ( x )dx ( x 0.7) f ( x )dx.. 2 . 20 30 100 3

= # − + # −

x x . x x3 30 49 142..300 3 3 20 3



1

= ; E +; + − Eor σ 2 = 0.039RMS = σ 2 = 0.039 = 0.198MCQ 1.80 Choose the current one from among the alternative A,B,C ,D after matching anitem from Group 1 with the most appropriate item in Group 2.Group 1 Group 21. FM P. Slope overload2. DM Q. μ-law3. PSK R. Envelope detector4. PCM S. Hilbert transformT. Hilbert transformU. Matched filter(A) 1 - T, 2 - P, 3 - U, 4 - S (B) 1 - S, 2 - U, 3 - P, 4 - T(C) 1 - S, 2 - P, 3 - U, 4 - Q (D) 1 - U, 2 - R, 3 - S, 4 - QSOL 1.80 Hence (C) is correct option.FM $ Capture effect

DM $ Slope over loadPSK $ Matched filterPCM $ μ − lawMCQ 1.81 Three analog signals, having bandwidths 1200 Hz, 600 Hz and 600 Hz, aresampledat their respective Nyquist rates, encoded with 12 bit words, and time divisionmultiplexed. The bit rate for the multiplexed. The bit rate for the multiplexedsignal is(A) 115.2 kbps (B) 28.8 kbps(C) 57.6 kbps (D) 38.4 kbpsSOL 1.81 Since f s = 2f m, the signal frequency and sampling frequency are as follows

f m1 = 1200 Hz $2400 samples per secf m2 = 600 Hz $1200 samples per secf m3 = 600 Hz $1200 samples per secThus by time division multiplexing total 4800 samples per second will be sent.Since each sample require 12 bit, total 4800 # 12 bits per second will be sentThus bit rate R b = 4800 # 12 = 57.6 kbpsHence (C) is correct option.MCQ 1.82 Consider a system shown in the figure. Let X (f ) and Y (f ) and denote theFourierPage 40 GATE EC 2004 www.gatehelp.com


PUBLISHING FOR GATEtransforms of x (t ) and y (t ) respectively. The ideal HPF has the cutoff frequency10 kHz.The positive frequencies where Y (f ) has spectral peaks are(A) 1 kHz and 24 kHz (B) 2 kHz and 244 kHz(C) 1 kHz and 14 kHz (D) 2 kHz and 14 kHzSOL 1.82 The input signal X (f ) has the peak at 1 kHz and −1 kHz. After balancedmodulator



the output will have peak at f c ! 1 kHz i.e. :10 ! 1 $11 and 9 kHz10 ! (− 1) $9 and 11 kHz9 kHz will be filtered out by HPF of 10 kHz. Thus 11 kHz will remain. Afterpassing through 13 kHz balanced modulator signal will have 13 ! 11 kHz signal i.e.

2 and 24 kHz.Thus peak of Y (f ) are at 2 kHz and 24 kHz.Hence (B) is correct option.MCQ 1.83 A parallel plate air-filled capacitor has plate area of 10-4 m2 and plateseparationof 10-3 m. It is connect - ed to a 0.5 V, 3.6 GHz source. The magnitude of thedisplacement current is ( 10 36

ε = 1 9 π

- F/m)(A) 10 mA (B) 100 mA(C) 10 A (D) 1.59 mA

SOL 1.83 The capacitance isC .d

A10o 8 85 10 103

12 4 = ε = # # -

- -

8.85 10 13 = # -

The charge on capacitor isQ = CV 8.85 10 13 4.427 10 13 = # - = # -

Displacement current in one cycleIT= Q = fQ 4.427 10 13 3.6 109 1.59 = # - # # = mAHence (D) is correct option.Page 41 GATE EC 2004 www.gatehelp.com


MCQ 1.84 A source produces binary data at the rate of 10 kbps. The binary symbols arerepresented as shown in the figure given below.The source output is transmitted using two modulation schemes, namely Binary

PSK (BPSK) and Quadrature PSK (QPSK). Let B1

and B2

be the bandwidthrequirements of BPSK and QPSK respectively. Assume that the bandwidth of heabove rectangular pulses is 10 kHz, B1 and B2 are(A) B1 = 20 kHz,B2 = 20 kHz (B) B1 = 10 kHz,B2 = 20 kHz(C) B1 = 20 kHz,B2 = 10 kHz (D) B1 = 20 kHz,B2 = 10 kHzSOL 1.84 The required bandwidth of M array PSK isBWn



= 2R bwhere 2n = M and R b is bit rateFor BPSK, M = 2 = 2n$n = 1Thus B1 R 1= 2 b = 2 # 10 = 20 kHz

For QPSK, M = 4 = 2n$n = 2Thus B2 R 2= 2 b = 10 kHzHence (C) is correct option.MCQ 1.85 Consider a 300 Ω, quarter - wave long (at 1 GHz) transmission line as showninFig. It is connected to a 10 V, 50 Ω source at one end and is left open circuited atthe other end. The magnitude of the voltage at the open circuit end of the line is(A) 10 V (B) 5 V(C) 60 V (D) 60/7 VSOL 1.85 Hence (C) is correct option.

VVinL

ZZin

= O



or V LZZ VinO

= in 650= 10 # 300 = 0 VMCQ 1.86 In a microwave test bench, why is the microwave signal amplitude modulatedat 1kHz(A) To increase the sensitivity of measurement(B) To transmit the signal to a far-off place

(C) To study amplitude modulations(D) Because crystal detector fails at microwave frequenciesSOL 1.86 Hence (D) is correct option.MCQ 1.87 If E (a x jay )e= t + t jkz - k ωt and H (k / ) (a ka )e y x

= ωμ t + t jkz - j ωt , the time-averaged Poyntingvector is(A) null vector (B) (k /ωμ)atz



(C) (2k /ωμ)atz (D) (k /2ωμ)atzSOL 1.87 Hence (A) is correct option.R avg Re[ ]21 E H * = #

E H * # (a x jay )e k ( ja a )e jkz j t x y jkz j t

ωμ = t + t − ω # − t + t − + ω

a k ( j ) ( j ) k 0z ωμ ωμ

= t ; − − E =

Thus R avg Re[ ]21 E H * = # = 0MCQ 1.88 Consider an impedance Z = R + jX marked with point P in an impedance Smithchart as shown in Fig. The movement from point P along a constant resistancecircle in the clockwise direction by an angle 45c is equivalent to(A) adding an inductance in series with Z(B) adding a capacitance in series with Z(C) adding an inductance in shunt across Z(D) adding a capacitance in shunt across ZPage 43 GATE EC 2004 www.gatehelp.com


SOL 1.88 Suppose at point P impedance isZ = r + j (− 1)If we move in constant resistance circle from point P in clockwise direction by an

angle 45c, the reactance magnitude increase. Let us consider a point Q at 45c frompoint P in clockwise direction. It’s impedance isZ 1 = r − 0.5 jor Z 1 = Z + 0.5 jThus movement on constant r - circle by an +45c in CW direction is the additionof inductance in series with Z .Hence (A) is correct option.MCQ 1.89 A plane electromagnetic wave propagating in free space is incident normallyon alarge slab of loss-less, non-magnetic, dielectric material with ε > ε0. Maxima andminima are observed when the electric field is measured in front of the slab. The

maximum electric field is found to be 5 times the minimum field. The intrinsicimpedance of the medium should be(A) 120π Ω (B) 60π Ω (C) 600π Ω (D) 24π Ω

SOL 1.89 Hence (D) is correct option.We have VSWREE 5



11minmax

Γ

Γ

= = =+−

or Γ

3= 2Thus Γ

3=−2Now Γ 2 12 1

η η

= η η + − or3−212012022

η π

= η π

+ − or

η2

=24

π

MCQ 1.90 A lossless transmission line is terminated in a load which reflects a part of theincident power. The measured VSWR is 2. The percentage of the power that isreflected back is(A) 57.73 (B) 33.33(C) 0.11 (D) 11.11SOL 1.90 Hence (D) is correct option.The VSWR 211Γ

Γ

=

+−

or Γ

3= 1Page 44 GATE EC 2004 www.gatehelp.com




PUBLISHING FOR GATE

ThusPPincref

9= Γ 2 = 1or P ref P9= inc

i.e. 11.11% of incident power is reflected.Answer Sheet

1. (B) 19. (A) 37. (D) 55. (B) 73. (B)2. (D) 20. (D) 38. (B) 56. (C) 74. (B)3. (A) 21. (D) 39. (B) 57. (A) 75. (A)4. (A) 22. (C) 40. (D) 58. (D) 76. (C)

5. (C) 23. (D) 41. (C) 59. (D) 77. (A)6. (C) 24. (C) 42. (D) 60. (C) 78. (A)7. (A) 25. (C) 43. (A) 61. (D) 79. (B)8. (D) 26. (B) 44. (B) 62. (A) 80. (C)9. (B) 27. (B) 45. (C) 63. (C) 81. (C)10. (C) 28. (A) 46. (D) 64. (C) 82. (B)11. (A) 29. (D) 47. (A) 65. (D) 83. (D)12. (A) 30. (A) 48. (A) 66. (B) 84. (C)13. (C) 31. (D) 49. (D) 67. (A) 85. (C)14. (A) 32. (B) 50. (B) 68. (C) 86. (D)15. (D) 33. (C) 51. (C) 69. (A) 87. (A)

16. (B) 34. (B) 52. (D) 70 (A) 88. (A)17. (A) 35. (D) 53. (C) 71 (B) 89. (D)18. (C) 36. (A) 54. (D) 72 (D) 90. (D)

GATE EC2003Brought to you by: Nodia and Company Visit us at: www.nodia.co.inPUBLISHING FOR GATE

Q. 1-30 Carry One Mark Each

MCQ 1.1 The minimum number of equations required to analyze the circuit shown in thefigure is(A) 3 (B) 4(C) 6 (D) 7SOL 1.1 Hence (B) is correct option.Number of loops = b − n + 1= minimum number of equationNumber of branches = b = 8



Number of nodes = n = 5Minimum number of equation= 8 − 5 + 1 = 4MCQ 1.2 A source of angular frequency 1 rad/sec has a source impedance consisting of1 Ω

resistance in series with 1 H inductance. The load that will obtain the maximumpower transfer is(A) 1 Ω resistance(B) 1 Ω resistance in parallel with 1 H inductance(C) 1 Ω resistance in series with 1 F capacitor(D) 1 Ω resistance in parallel with 1 F capacitorSOL 1.2 For maximum power transferZ L Z *= S = R s − jX sPage 2 GATE EC 2003 www.gatehelp.com


Thus Z L = 1 − 1 jHence (C) is correct option.MCQ 1.3 A series RLC circuit has a resonance frequency of 1 kHz and a quality factorQ = 100. If each of R ,L and C is doubled from its original value, the new Q of thecircuit is(A) 25 (B) 50(C) 100 (D) 200SOL 1.3 Hence (B) is correct option.QR C= 1 L

When R ,L and C are doubled,Q'R CLR CL Q21222

12= = =Thus Q'2= 100 = 50MCQ 1.4 The Laplace transform of i (t ) is given byI (s)



s(1 s)= 2+

At t " 3, The value of i (t ) tends to(A) 0 (B) 1

(C) 2 (D) 3SOL 1.4 From the Final value theorem we havelimi (t )t "3

limsI (s)s 0

="

( ) ( )lims lims 1 s s2

12 2s 0 s 0

=

+

=+

=" "

Hence (C) is correct answerMCQ 1.5 The differential equation for the current i (t ) in the circuit of the figure is(A) ( ) sin

dtd idt2 2 di i t t 2

2

+ + = (B) ( ) cosdtd idt2 di 2i t t 2

2

+ + =

(C) ( ) cosdtd idt2 2 di i t t 2

2

+ + = (D) ( ) sindt



d idt2 di 2i t t 2

2

+ + =

SOL 1.5 Applying KVL we get,sin t ( )( )Ri t L ( )dtdi tC

= + + 1 #i t dtPage 3 GATE EC 2003 www.gatehelp.com


or sin t ( )( )i t ( )dtdi t

= 2 + 2 + #i t dt

Differentiating with respect to t , we getcos t( ) ( )( )dt

di tdtd i ti t2 222

= + +

Hence (C) is correct option.MCQ 1.6 n-type silicon is obtained by doping silicon with(A) Germanium (B) Aluminium(C) Boron (D) PhosphorusSOL 1.6 Pentavalent make n −type semiconductor and phosphorous is pentavalent.Hence option (D) is correct.MCQ 1.7 The Bandgap of silicon at 300 K is(A) 1.36 eV (B) 1.10 eV(C) 0.80 eV (D) 0.67 eVSOL 1.7 Hence option (B) is correct.For silicon at 0 K E g 0 = 1.21 eV



At any temperatureE gT E g 0 3.6 10 T4 = − # -

At T = 300 K,E g 300 1.21 3.6 10 4 300 = − # - # = 1.1 eV

This is standard value, that must be remembered.MCQ 1.8 The intrinsic carrier concentration of silicon sample at 300 K is 1.5 1016 # /m3. Ifafter doping, the number of majority carriers is 5 1020 # /m3, the minority carrierdensity is(A) 4.50 1011 # /m3 (B) 3.333 104 # /m3

(C) 5.00 1020 # /m3 (D) 3.00 10 5 # - /m3

SOL 1.8 By Mass action lawnp ni

= 2

pn

ni

2

= . .5 101 5 10 1 5 102016 16

#= # # # 4.5 1011 = #Hence option (A) is correct.MCQ 1.9 Choose proper substitutes for X and Y to make the following statement correctTunnel diode and Avalanche photo diode are operated in X bias ad Y biasrespectively(A) X : reverse, Y : reverse (B) X : reverse, Y : forward(C) X : forward, Y : reverse (D) X : forward, Y : forwardSOL 1.9 Tunnel diode shows the negative characteristics in forward bias. It is used inforwardPage 4 GATE EC 2003 www.gatehelp.com


bias. Avalanche photo diode is used in reverse bias.Hence option (C) is correct.MCQ 1.10 For an n − channel enhancement type MOSFET, if the source is connected at

ahigher potential than that of the bulk (i.e. V SB > 0), the threshold voltage V T of theMOSFET will(A) remain unchanged (B) decrease(C) change polarity (D) increaseSOL 1.10 Hence option (D) is correct.MCQ 1.11 Choose the correct match for input resistance of various amplifierconfigurations



shown below :Configuration Input resistanceCB : Common Base LO : LowCC : Common Collector MO : ModerateCE : Common Emitter HI : High

(A) CB − LO, CC −MO, CE − HI(B) CB − LO, CC − HI, CE −MO(C) CB −MO, CC − HI, CE − LO(D) CB − HI, CC − LO, CE −MOSOL 1.11 For the different combinations the table is as followsCE CE CC CB

Ai High High Unity Av High Unity HighR i Medium High LowR o Medium Low HighHence (B) is correct option.

MCQ 1.12 The circuit shown in the figure is best described as a(A) bridge rectifier (B) ring modulator(C) frequency discriminator (D) voltage doubleSOL 1.12 This circuit having two diode and capacitor pair in parallel, works as voltagePage 5 GATE EC 2003 www.gatehelp.com


doubler.Hence (D) is correct option.MCQ 1.13 If the input to the ideal comparators shown in the figure is a sinusoidal signal of8V (peak to peak) without any DC component, then the output of the comparatorshas a duty cycle of(A) 1/2 (B) 1/3(C) 1/6 (D) 1/2SOL 1.13 If the input is sinusoidal signal of 8 V (peak to peak) thenV i = 4 sinωtThe output of comparator will be high when input is higher than V ref = 2 V andwill be low when input is lower than V ref = 2 V. Thus the waveform for input isshown belowFrom fig, first crossover is at ωt 1 and second crossover is at ωt 2 where4 sinωt 1 = 2VThus ωt 1 sin

216= -1 = π

ωt 2

6 6= π − π = 5π Duty Cycle



2 36 156

π

=

− =π π

Thus the output of comparators has a duty cycle of31 .Hence (B) is correct option.MCQ 1.14 If the differential voltage gain and the common mode voltage gain of adifferentialamplifier are 48 dB and 2 dB respectively, then common mode rejection ratio is(A) 23 dB (B) 25 dBPage 6 GATE EC 2003 www.gatehelp.com


(C) 46 dB (D) 50 dBSOL 1.14 Hence (C) is correct option.CMMR

A Ac

= d

or 20 logCMMR = 20 log Ad − 20 log Ac

= 48 − 2 = 46 dB

Where Ad "Differential Voltage Gainand AC " Common Mode Voltage GainMCQ 1.15 Generally, the gain of a transistor amplifier falls at high frequencies due to the(A) internal capacitances of the device(B) coupling capacitor at the input(C) skin effect(D) coupling capacitor at the outputSOL 1.15 The gain of amplifier is

Ai

g j Cg

bm

ω

=

+−

Thus the gain of a transistor amplifier falls at high frequencies due to the internalcapacitance that are diffusion capacitance and transition capacitance.Hence (B) is correct option.



MCQ 1.16 The number of distinct Boolean expressions of 4 variables is(A) 16 (B) 256(C) 1023 (D) 65536SOL 1.16 The number of distinct boolean expression of n variable is 22n . Thus224= 216 = 65536

Hence (D) is correct answer.MCQ 1.17 The minimum number of comparators required to build an 8-bits flash ADC is(A) 8 (B) 63(C) 255 (D) 256SOL 1.17 In the flash analog to digital converter, the no. of comparators is equal to 2n- 1,where n is no. of bit.sSo, 2n- 1 = 28− 1 = 255Hence (C) is correct answer.MCQ 1.18 The output of the 74 series of GATE of TTL gates is taken from a BJT in(A) totem pole and common collector configuration(B) either totem pole or open collector configuration

(C) common base configurationPage 7 GATE EC 2003 www.gatehelp.com


(D) common collector configurationSOL 1.18 When output of the 74 series gate of TTL gates is taken from BJT then theconfiguration is either totem pole or open collector configuration .Hence (B) is correct answer.MCQ 1.19 Without any additional circuitry, an 8:1 MUX can be used to obtain(A) some but not all Boolean functions of 3 variables(B) all functions of 3 variables but non of 4 variables(C) all functions of 3 variables and some but not all of 4 variables(D) all functions of 4 variablesSOL 1.19 A 2n:1 MUX can implement all logic functions of (n + 1) variable without andyadditional circuitry. Here n = 3. Thus a 8 : 1 MUX can implement all logic functionsof 4 variable.Here (D) is correct answer.MCQ 1.20 A 0 to 6 counter consists of 3 flip flops and a combination circuit of 2 input gate(s). The common circuit consists of(A) one AND gate (B) one OR gate(C) one AND gate and one OR gate (D) two AND gatesSOL 1.20 Counter must be reset when it count 111. This can be implemented by followingcircuitry

Hence (D) is correct answer.MCQ 1.21 The Fourier series expansion of a real periodic signal with fundamentalfrequencyf 0 is given by g p (t ) c enn

j 2 f t = 0

3π



=- / . It is given that c 3 = 3 + j 5. Then c −3 is

(A) 5 + j 3 (B) −3 − j 5(C) −5 + j 3 (D) 3 − j 5SOL 1.21 Hence (D) is correct answer.HereC 3 = 3 + j 5

For real periodic signalC −k C *= k

Thus C −3 = C k = 3 − j 5MCQ 1.22 Let x (t ) be the input to a linear, time-invariant system. The required output is4π (t − 2). The transfer function of the system should be(A) 4e j 4π f (B) 2e− j 8π f

(C) 4e− j 4π f (D) 2e j 8π f



SOL 1.22 Hence (C) is correct answer.y (t ) = 4 x (t − 2)Taking Fourier transform we getY (e j 2π f ) = 4e− j 2π f 2 X (e j 2π f ) Time Shifting propertyor( )( )

X eY e

j f j f2

2π π

= 4e−4 j π f

Thus H (e j 2π f ) = 4e−4 j π f

MCQ 1.23 A sequence x (n) with the z −transform X (z ) = z 4+ z 2− 2z + 2 − 3z −4 is applied asaninput to a linear, time-invariant system with the impulse response h(n) = 2δ(n − 3)whereδ(n),,

1 n 00 otherwise=

= )The output at n = 4 is(A) −6 (B) zero(C) 2 (D) −4SOL 1.23 Hence (B) is correct answer.



We have h(n) = 3δ(n − 3)or H (z ) = 2z −3 Taking z transform

X (z ) = z 4+ z 2− 2z + 2 − 3z −4

Now Y (z ) = H (z ) X (z )= 2z −3(z 4+ z 2− 2z + 2 − 3z −4)

= 2(z + z −1− 2z −2+ 2z −3− 3z −7)Taking inverse z transform we havey (n) = 2[δ(n + 1) + δ(n − 1) − 2δ(n − 2)+2δ(n − 3) − 3δ(n − 7)]

At n = 4, y (4) = 0MCQ 1.24 Fig. shows the Nyquist plot of the open-loop transfer function G(s)H (s) of asystem.If G(s)H (s) has one right-hand pole, the closed-loop system is(A) always stable(B) unstable with one closed-loop right hand pole(C) unstable with two closed-loop right hand poles(D) unstable with three closed-loop right hand polesPage 9 GATE EC 2003 www.gatehelp.com


SOL 1.24 Hence (A) is correct option.Z = P − NN " Net encirclement of (− 1 + j 0) by Nyquist plot,P " Number of open loop poles in right hand side of s − planeZ " Number of closed loop poles in right hand side of s − planeHere N = 1 and P = 1Thus Z = 0Hence there are no roots on RH of s −plane and system is always stable.MCQ 1.25 A PD controller is used to compensate a system. Compared to theuncompensatedsystem, the compensated system has(A) a higher type number (B) reduced damping(C) higher noise amplification (D) larger transient overshootSOL 1.25 PD Controller may accentuate noise at higher frequency. It does not effect thetypeof system and it increases the damping. It also reduce the maximum overshoot.Hence (C) is correct option.MCQ 1.26 The input to a coherent detector is DSB-SC signal plus noise. The noise at thedetector output is(A) the in-phase component (B) the quadrature - component

(C) zero (D) the envelopeSOL 1.26 The input is a coherent detector is DSB - SC signal plus noise. The noise at thedetector output is the in-phase component as the quadrature component nq (t ) ofthe noise n(t ) is completely rejected by the detector.Hence (A) is correct option.MCQ 1.27 The noise at the input to an ideal frequency detector is white. The detector isoperating above threshold. The power spectral density of the noise at the output is(A) raised - cosine (B) flat



(C) parabolic (D) GaussianSOL 1.27 The noise at the input to an ideal frequency detector is white. The PSD of noiseatthe output is parabolicHence (C) is correct option.

MCQ 1.28 At a given probability of error, binary coherent FSK is inferior to binarycoherentPSK by.(A) 6 dB (B) 3 dB(C) 2 dB (D) 0 dBSOL 1.28 Hence (B) is correct option.We have P e E212= erfc d c η mPage 10 GATE EC 2003 www.gatehelp.com


Since P e of Binary FSK is 3 dB inferior to binary PSKMCQ 1.29 The unit of 4# H is(A) Ampere (B) Ampere/meter(C) Ampere/meter 2 (D) Ampere-meterSOL 1.29 By Maxwells equations4# HtD J2= 2 +Thus 4# H has unit of current density J that is A/m2

Hence (C) is correct optionMCQ 1.30 The depth of penetration of electromagnetic wave in a medium havingconductivityσ at a frequency of 1 MHz is 25 cm. The depth of penetration at a frequency of 4MHz will be(A) 6.25 dm (B) 12.50 cm(C) 50.00 cm (D) 100.00 cmSOL 1.30 Hence (B) is correct option.We know that δ

f\ 1Thus12

δ

δ f



f2

= 1

25δ2

4= 1or δ2 .4= 1 # 25 = 12 5 cm

Q.31-90 Carry Two Marks EachMCQ 1.31 Twelve 1 Ω resistance are used as edges to form a cube. The resistancebetween twodiagonally opposite corners of the cube is(A) 65 Ω (B) 1Ω

(C) 56 Ω (D) 23 Ω SOL 1.31 For current i there is 3 similar path. So current will be divide in three pathPage 11 GATE EC 2003 www.gatehelp.com


so, we getV i i3 1 6 1 31 ab−b # l−b # l−b #1l = 0i

V ab R316131= eq = + +

6 5= Ω

Hence (A) is correct option.

MCQ 1.32 The current flowing through the resistance R in the circuit in the figure has theform P cos 4t where P is(A) (0.18 + j 0.72) (B) (0.46 + j 1.90)(C) −(0.18 + j 1.90) (D) −(0.192 + j 0.144)SOL 1.32 Data are missing in question as L1&L2 are not given

The circuit for Q. 33 & 34 is given below. Assume that the switch S is in position 1 for a long time and thrown to position2 at t = 0.



MCQ 1.33 At t = 0+, the current i 1 is(A)RV2 −

(B)R−VPage 12 GATE EC 2003 www.gatehelp.com


(C)RV4− (D) zeroSOL 1.33 Data are missing in question as L1&L2 are not givenMCQ 1.34 I 1(s) and I 2(s) are the Laplace transforms of i 1(t ) and i 2(t ) respectively. Theequations for the loop currents I 1(s) and I 2(s) for the circuit shown in the figure,after the switch is brought from position 1 to position 2 at t = 0, are(A)( )( )R LsLsLsRI sI s 0CsCss V1112

+ +

−

−

> + H= G = = G(B)

( )( )R LsLsLsRI sI s 0



CsCss V1112

+ +

− − + =

−

> H= G = G(C)( )( )R LsLsLsR LsI sI s 0CsCss V1112

+ +

− −

+ + =−

> H= G = G(D)( )( )R LsLsCsR LsI s

I s 0CsCss V1112

+ +

−



−

> + + H= G = = GSOL 1.34 At t = 0- circuit is in steady state. So inductor act as short circuit and capacitoract as open circuit.

At t = 0-, i 1(0 )

- i (0 ) 0 = 2 = -v c (0 )- = V

At t = 0+ the circuit is as shown in fig. The voltage across capacitor and current ininductor can’t be changed instantaneously. Thus

At t = 0+, i 1 iRV2 = 2 =−

Hence (A) is correct option.Page 13 GATE EC 2003 www.gatehelp.com


MCQ 1.35 An input voltage v (t ) = 10 2 cos (t + 10c) + 10 5 cos (2t + 10c) V is applied to aseries combination of resistance R = 1Ω and an inductance L = 1 H. The resultingsteady-state current i (t ) in ampere is(A) 10 cos (t + 55c) + 10 cos (2t + 10c + tan−12)(B) 10 cos (t 55 ) 10 cos (2t 55 ) 2

+ c + 3 + c(C) 10 cos (t − 35c) + 10 cos (2t + 10c − tan−12)(D) 10 cos (t 35 ) cos (2t 35 ) 2

− c + 3 − cSOL 1.35

Hence (C) is correct optionv (t ) 10 2 cos (t 10 ) 10 5 cos (2t 10 )v v 1 2

= + c + + c14444244443 14444244443Thus we get ω1 = 1 and ω2 = 2Now Z 1 = R + j ω1L = 1 + j 1Z 2 = R + j ω2L = 1 + j 2i (t )( ) ( )Zv t

Zv t112

= + 2

cos ( ) cos ( ) jt



jt110 2 101 2

c 10 5 2 10c = ++ + ++

( ) ( )tancostant cos t1 2 110 2 101 2 2

10 5 2 102 2 1 2 2 1

c c+

=

+

+ ++

+− −

( ) ( )tan

costant cos t2 4510 2 105 210 5 2 101 c 1

c c+

= + + +− −

i (t ) = 10 cos (t − 35c) + 10 cos (2t + 10c − tan−12)MCQ 1.36 The driving point impedance Z (s) of a network has the pole-zero locations asshown in the figure. If Z (0) = 3, then Z (s) is(A)( )s ss2 3



3 32+ +

+

(B)( )

s ss2 22 32+ ++

(C)( )s ss2 2

3 32+ +

+

(D)( )s ss2 32 32− −

−

SOL 1.36 Hence (B) is correct option.Zeros =− 3Pole1 =− 1 + jPole2 =− 1 − jPage 14 GATE EC 2003 www.gatehelp.com


Z (s)( )( )( )s j s jK s

1 1= 3+ + + −

+

( )( )( )( )



s jK ssK s1

31 132 2 2 =

+ − + =

+ +

+

From problem statement Z (0) 3 0 = ω=

Thus K2

3 = 3 and we get K = 2Z (s)( )s ss2 22 32 =+ +

+

MCQ 1.37 The impedance parameters z 11 and z 12 of the two-port network in the figure are(A) z 11 = 2.75 Ω and z 12 = 0.25 Ω (B) z 11 = 3 Ω and z 12 = 0.5 Ω (C) z 11 = 3 Ω and z 12 = 0.25 Ω (D) z 11 = 2.25 Ω and z 12 = 0.5 Ω SOL 1.37 Using 3− Y conversionR 1 .2 1 12 14= # 2 0 5+ +

= =R 2 .2 1 11 14= # 1 0 25+ +

= =R 3 .2 1 1



= 2 # 1 0 5+ +

=

Now the circuit is as shown in figure below.Now z 11 . . .

I V2 05 025 2751 I10 2

= =+ + ==

z 12 = R 3 = 0.25Page 15 GATE EC 2003 www.gatehelp.com


Hence (A) is correct option.MCQ 1.38 An n −type silicon bar 0.1 cm long and 100 μm2 i cross-sectional area has a

majoritycarrier concentration of 5 1020 # /m2 and the carrier mobility is 0.13 m2/V-s at 300K. If the charge of an electron is 1.5 10 19 # - coulomb, then the resistance of the baris(A) 106 Ohm (B) 104 Ohm(C) 10-1 Ohm (D) 10-4 OhmSOL 1.38 Hence option (A) is correct.We that R ,

A ρl ρ 1σ

= = and α = nqun

From above relation we haveRnq A1 μn

=

. .

.5 10 1 6 10 0 13 100 100 1 10

20 19 122

# # # # # #= #- -

-

= 106Ω MCQ 1.39 The electron concentration in a sample of uniformly doped n-type silicon at 300K varies linearly from 1017/cm3 at x = 0 to 6 1016 # /cm3 at x = 2 μm. Assume asituation that electrons are supplied to keep this concentration gradient constant



with time. If electronic charge is 1.6 10 19 # - coulomb and the diffusion constantDn = 35 cm2/s, the current density in the silicon, if no electric field is present, is(A) zero (B) -112 A/cm2

(C) +1120 A/cm2 (D) -1120 A/cm2

SOL 1.39 Hence option (D) is correct.

dxdn2 10 06 10 10416 17

#= #−

− -

2 1020 =− #

Now J n

nq E D qdxdn= μe + n

Since no electric field is present, E = 0 and we getSo, J n qDdxdn= n

1.6 10 19 35 ( 2 1020) = # - # # − # =− 1120 A/cm2

MCQ 1.40 Match items in Group 1 with items in Group 2, most suitably.Group 1 Group 2P. LED 1. Heavy dopingQ. Avalanche photo diode 2. Coherent radiationR. Tunnel diode 3. Spontaneous emissionS. LASER 4. Current gainPage 16 GATE EC 2003 www.gatehelp.com


(A) P - 1, Q - 2, R - 4, S - 3(B) P - 2, Q - 3, R - 1, S - 4(C) P - 3 Q - 4, R - 1, S - 2(D) P - 2, Q - 1, R - 4, S - 3SOL 1.40 LED works on the principal of spontaneous emission.In the avalanche photo diode due to the avalanche effect there is large current gain.Tunnel diode has very large doping.LASER diode are used for coherent radiation.Hence option (C) is correct.MCQ 1.41 At 300 K, for a diode current of 1 mA, a certain germanium diode requires aforward bias of 0.1435 V, whereas a certain silicon diode requires a forward bias of0.718 V. Under the conditions state above, the closest approximation of the ratio



of reverse saturation current in germanium diode to that in silicon diode is(A) 1 (B) 5(C) 4 103 # (D) 8 103 #SOL 1.41 Hence option (C) is correct.We know that I I o e V 1Vsi T

D1 = ` η − jwhere η = 1 for germanium and η = 2 silicon. As per questionI o ee 1VnVDsi

` h T− j I o e V 1VGe T

DGe = ` η − jorIIoosisi

ee11VVVVTDGeTDsi

=

− − η

η

ee11 4 10 .

.26 100 14352 26 100 718

333

= #−

− =## #--

MCQ 1.42 A particular green LED emits light of wavelength 5490 Ac. The energybandgap of



the semiconductor material used there is(Plank’s constant 6.626 10 34J s = # - − )(A) 2.26 eV (B) 1.98 eV(C) 1.17 eV (D) 0.74 eVSOL 1.42 Hence option (A) is correct

E g hc λ

= .54900 106 626 10 3 101034 8

#= # # # −

−

= 3.62 JIn eV E g (eV )( )e= E g J

.

. 2.261 6 103 62 101919

#

= # = −

−

eV

AlternativelyE g( )1.24 λ μm= eV . 2.265490 101 244 m # μ

= = − eVMCQ 1.43 When the gate-to-source voltage (V Gs) of a MOSFET with threshold voltage of

400mV, working in saturation is 900 mV, the drain current is observed to be 1 mA.Page 17 GATE EC 2003 www.gatehelp.com


Neglecting the channel width modulation effect and assuming that the MOSFET isoperating at saturation, the drain current for an applied V GS of 1400 mV is(A) 0.5 mA (B) 2.0 mA



(C) 3.5 mA (D) 4.0 mASOL 1.43 We know thatI D K (V GS V T )= − 2

Thus

IIDD12

( )( )V VV VGS TGS T12

22

=−

−

Substituting the values we haveIIDD12

( . . )

( . . )0 9 0 41 4 0 44 2

2

=

−

− =

or I D2 = 4I DI = 4 mAHence option (D) is correct.MCQ 1.44 If P is Passivation, Q is n −well implant, R is metallization and S is source/draindiffusion, then the order in which they are carried out in a standard n −well CMOSfabrication process, is(A) P − Q − R − S (B) Q − S − R − P(C) R − P − S − Q (D) S − R − Q − PSOL 1.44 In n −well CMOS fabrication following are the steps :(i) n − well implant (Q)(ii) Source drain diffusion (S)(iii) Metalization (R)



(iv) Passivation (P)Hence option (B) is correct.MCQ 1.45 An amplifier without feedback has a voltage gain of 50, input resistance of 1 kΩ and output resistance of 2.5 kΩ. The input resistance of the current-shunt negative

feedback amplifier using the above amplifier with a feedback factor of 0.2, is(A) k111 Ω (B) k51 Ω (C) 5 kΩ (D) 11 kΩ

SOL 1.45 Hence (A) is correct option.We have R i = 1kΩ, β = 0.2, A = 50Thus, R if( A )

R1i

β

=+

11= 1 kΩ MCQ 1.46 In the amplifier circuit shown in the figure, the values of R 1 and R 2 are such thatthe transistor is operating at V CE = 3 V and I C = 1.5 mA when its β is 150. For atransistor with β of 200, the operating point (V CE , I C ) isPage 18 GATE EC 2003 www.gatehelp.com


(A) (2 V, 2 mA) (B) (3 V, 2 mA)(C) (4 V, 2 mA) (D) (4 V, 1 mA)SOL 1.46 The DC equivalent circuit is shown as below. This is fixed bias circuit operatinginactive region.In first caseV CC− I C 1R 2− V CE 1 = 0or 6 − 1.5mR 2− 3 = 0or R 2 = 2k Ω

I B1 I 1.5150C m11

β

= = = 0.01 mAIn second case I B2 will we equal to I B1 as there is no in R 1.Thus I C 2 = β 2I B2 = 200 # 0.01 = 2 mA



V CE 2 = V CC − I C 2R 2 = 6 − 2m# 2 kΩ = 2 VHence (A) is correct option.MCQ 1.47 The oscillator circuit shown in the figure has an ideal inverting amplifier. Itsfrequency of oscillation (in Hz) is(A)

(2 6 RC )1π

(B)(2 RC )1π

(C)( 6 RC )1 (D)(2 RC )

6π Page 19 GATE EC 2003 www.gatehelp.com


SOL 1.47 The given circuit is a R − C phase shift oscillator and frequency of its oscillationisf2 6 RC1π

=

Hence (A) is correct option.MCQ 1.48 The output voltage of the regulated power supply shown in the figure is(A) 3 V (B) 6 V(C) 9 V (D) 12 VSOL 1.48 If we see th figure we find that the voltage at non-inverting terminal is 3 V by thezener diode and voltage at inverting terminal will be 3 V. Thus V o can be get byapplying voltage division rule, i.e.V20 4020+ o = 3

or V 0 = 9 VHence (C) is correct option.MCQ 1.49 The action of JFET in its equivalent circuit can best be represented as a(A) Current controlled current source(B) Current controlled voltage source(C) Voltage controlled voltage source(D) Voltage controlled current sourceSOL 1.49 For a JFET in active region we have



I DS IV1 V DSS

P

GS 2 = c − mFrom above equation it is clear that the action of a JFET is voltage controlledcurrent source.Hence option (D) is correct.MCQ 1.50 If the op-amp in the figure is ideal, the output voltage V out will be equal toPage 20 GATE EC 2003 www.gatehelp.com


(A) 1 V (B) 6 V(C) 14 V (D) 17 VSOL 1.50 The circuit is as shown belowV + (3)1 8

83= = 8 kΩ +V + = V - V

3= 8Now applying KCL at inverting terminal we getV V V12

5-− + -− o = 0or V o = 6V -− 1063= # 8 − 10 = 6 VHence (B) is correct option.MCQ 1.51 Three identical amplifiers with each one having a voltage gain of 50, inputresistanceof 1 kΩ and output resistance of 250 Ω are cascaded. The opened circuit voltagesgain of the combined amplifier is

(A) 49 dB (B) 51 dB(C) 98 dB (D) 102 dBSOL 1.51 The equivalent circuit of 3 cascade stage is as shown in fig.Page 21 GATE EC 2003 www.gatehelp.com


V 2k . k



k V1 025= 1 50 1

+ = 40V 1Similarly V 3

k . kk V1 025= 1 50 2

+ = 40V 2or V 3 = 40 # 40V 1V o = 50V 3 = 50 # 40 # 40V 1or AV

VV o 50 40 40 80001

= = # # =

or 20 log AV = 20 log 8000 = 98 dBHence (C) is correct option.MCQ 1.52 An ideal sawtooth voltages waveform of frequency of 500 Hz and amplitude 3Vis generated by charging a capacitor of 2 μF in every cycle. The charging requires(A) Constant voltage source of 3 V for 1 ms(B) Constant voltage source of 3 V for 2 ms(C) Constant voltage source of 1 mA for 1 ms(D) Constant voltage source of 3 mA for 2 msSOL 1.52 If a constant current is made to flow in a capacitor, the output voltage isintegration

of input current and that is sawtooth waveform as below :V CC1 idt t

0

= #The time period of wave form isTf 1500= = 1 = 2 m secThus 3 idt2 1016020 10 3

#

= # -#



or i (2 10 3 0) # - − 6 10 6 = # -

or i = 3 mAThus the charging require 3 mA current source for 2 msec.Hence (D) is correct optionMCQ 1.53 The circuit in the figure has 4 boxes each described by inputs P ,Q,R and

outputsY ,Z with Y = P 5 Q 5 R and Z = RQ + PR + QPThe circuit acts as aPage 22 GATE EC 2003 www.gatehelp.com


(A) 4 bit adder giving P + Q(B) 4 bit subtractor giving P − Q(C) 4 bit subtractor giving Q-P(D) 4 bit adder giving P + Q + RSOL 1.53 Hence (B) is correct answer.We have Y = P 5 Q 5 RZ = RQ + PR + QPHere every block is a full subtractor giving P − Q − R where R is borrow. Thuscircuit acts as a 4 bit subtractor giving P − Q.MCQ 1.54 If the function W , X ,Y and Z are as followsW = R + PQ + RS X = PQRS + PQRS + PQRSY = RS + PR + PQ + P .Q Z = R + S + PQ + P .Q.R + PQ.SThen,(A) W = Z , X = Z (B) W = Z , X = Y(C) W = Y (D) W = Y = Z

SOL 1.54 Hence (A) is correct answer.W = R + PQ + RS

X = PQRS + PQRS + PQRSY = RS + PR + PQ + PQ= RS + PR $ PQ $ PQ

= RS + (P + R )(P + Q)(P + Q)= RS + (P + PQ + PR + QR )(P + Q)= RS + PQ + QR (P + P ) + QR

= RS + PQ + QRZ = R + S + PQ + PQR + PQS= R + S + PQ $ PQR $ PQS= R + S + (P + Q)(P + Q + R )(P + Q + S)= R + S + PQ + PQ + PQS + PR + PQR+PRS + PQ + PQS + PQR + QRS= R + S + PQ + PQS + PR + PQR + PRSPage 23 GATE EC 2003 www.gatehelp.com


+PQS + PQR + QRS= R + S + PQ(1 + S) + PR (1 + P ) + PRS+PQS + PQR + QRS



= R + S + PQ + PR + PRS + PQS+PQR + QRS= R + S + PQ + PR (1 + Q) + PQS + QRS= R + S + PQ + PR + PQS + QRSThus W = Z and X = Z

MCQ 1.55 A 4 bit ripple counter and a bit synchronous counter are made using flip flopshaving a propagation delay of 10 ns each. If the worst case delay in the ripplecounter and the synchronous counter be R and S respectively, then(A) R = 10 ns, S = 40 ns (B) R = 40 ns, S = 10 ns(C) R = 10 ns S = 30 ns (D) R = 30 ns, S = 10 nsSOL 1.55 Propagation delay of flip flop ist pd = 10 nsecPropagation delay of 4 bit ripple counterR = 4t pd = 40 nsand in synchronous counter all flip-flop are given clock simultaneously, soS = t pd = 10 ns

Hence (B) is correct answer.MCQ 1.56 The DTL, TTL, ECL and CMOS famil GATE of digital ICs are compared in thefollowing 4 columns(P) (Q) (R) (S)Fanout is minimum DTL DTL TTL CMOSPower consumption isminimumTTL CMOS ECL DTLPropagation delay isminimumCMOS ECL TTL TTL

The correct column is(A) P (B) Q(C) R (D) SSOL 1.56 The DTL has minimum fan out and CMOS has minimum power consumption.Propagation delay is minimum in ECL.Hence (B) is correct answer.MCQ 1.57 The circuit shown in the figure is a 4 bit DACPage 24 GATE EC 2003 www.gatehelp.com


The input bits 0 and 1 are represented by 0 and 5 V respectively. The OP AMPis ideal, but all the resistance and the 5 v inputs have a tolerance of !10%. The

specification (rounded to nearest multiple of 5%) for the tolerance of the DAC is(A) !35% (B) !20%(C) !10% (D) !5%SOL 1.57 Hence (A) is correct answer.V o V RR b RR b RR b R



R b =− 1: o + 2 1+ 4 2+ 4 3DExact value when V 1 = 5, for maximum outputV oExact 5 1 2 9.37514

18=− : + + + 1D =− Maximum V out due to toleranceV omax 5.5 901102 901104 901108 90

110# # #

=− : + + + D

=− 12.604Tolerance = 34.44% = 35%MCQ 1.58 The circuit shown in figure converts(A) BCD to binary code (B) Binary to excess - 3 code(C) Excess -3 to gray code (D) Gray to Binary codeSOL 1.58 Hence (D) is correct answer.Let input be 1010; output will be 1101Let input be 0110; output will be 0100Page 25 GATE EC 2003 www.gatehelp.com


Thus it convert gray to Binary code.MCQ 1.59 In the circuit shown in the figure, A is parallel-in, parallel-out 4 bit register,whichloads at the rising edge of the clock C . The input lines are connected to a 4 bit bus,W . Its output acts at input to a 16 # 4 ROM whose output is floating when theinput to a partial table of the contents of the ROM is as followsData 0011 1111 0100 1010 1011 1000 0010 1000

Address 0 2 4 6 8 10 11 14The clock to the register is shown, and the data on the W bus at time t 1 is 0110.

The data on the bus at time t 2 is(A) 1111 (B) 1011(C) 1000 (D) 0010SOL 1.59 After t = t 1, at first rising edge of clock, the output of shift register is 0110, whichin input to address line of ROM. At 0110 is applied to register. So at this time datastroed in ROM at 1010 (10), 1000 will be on bus.When W has the data 0110 and it is 6 in decimal, and it ’s data value at that addis 1010





then 1010 i.e. 10 is acting as odd, at time t 2 and data at that movement is 1000.Hence (C) is correct answer.MCQ 1.60 In an 8085 microprocessor, the instruction CMP B has been executed whilethecontent of the accumulator is less than that of register B. As a result(A) Carry flag will be set but Zero flag will be reset(B) Carry flag will be rest but Zero flag will be set(C) Both Carry flag and Zero flag will be rest(D) Both Carry flag and Zero flag will be setSOL 1.60 CMP B & Compare the accumulator content with context of Register BIf A < R CY is set and zero flag will be reset.Hence (A) is correct answer.MCQ 1.61 Let X and Y be two statistically independent random variables uniformlydistributed in the ranges (− 1,1) and (− 2,1) respectively. Let Z = X + Y . Thenthe probability that (z #− 1) is(A) zero (B)61(C)31 (D)121SOL 1.61 The pdf of Z will be convolution of pdf of X and pdf of Y as shown below.Now p[Z # z ] f Z (z )dzz =

3 - # p[Z #− 2] f Z (z )dz2

=-3

- #= Area [z #− 2]21

61 1121 = # # =Page 27 GATE EC 2003 www.gatehelp.com


Hence (D) is correct option.



MCQ 1.62 Let P be linearity, Q be time-invariance, R be causality and S be stability. Adiscrete time system has the input-output relationship,y (n)( ) 10, 0

( ) x n nn

x n 1 n 1$#= =+ −

*where x (n) is the input and y (n) is the output. The above system has the properties(A) P, S but not Q, R (B) P, Q, S but not R

(C) P, Q, R, S (D) Q, R, S but not PSOL 1.62 System is non causal because output depends on future valueFor n # 1 y (− 1) = x (− 1 + 1) = x (0)( ) y n n0 − = x (n − n0+ 1) Time varyingy (n) = x (n + 1) Depends on Futurei.e. y (1) = x (2) None causalFor bounded input, system has bounded output. So it is stable.y (n) = x (n) for n $ 1= 0 for n = 0= x ( x + 1) for n #− 1So system is linear.

Hence (A) is correct answer.Common data for Q 63 & 64 :Page 28 GATE EC 2003 www.gatehelp.com


The system under consideration is an RC low-pass filter (RC-LPF) withR = 1 kΩ and C = 1.0 μF.MCQ 1.63 Let H (f ) denote the frequency response of the RC-LPF. Let f 1 be the highestfrequency such that( )( )

f f .HH f001 0 951

# # $ . Then f 1 (in Hz) is



(A) 324.8 (B) 163.9(C) 52.2 (D) 104.4SOL 1.63 The frequency response of RC-LPF isH (f )1 j 2 fRC

1π =

+

Now H (0) = 1( )( )HH f01 .

1 4 f R C1 0 95 2

12 2 2

$π

=+

or 1 4 2f R C1

+ π 2 2 2 # 1.108or 4 2f R C1

π 2 2 2 # 0.108or 2π f 1RC # 0.329or f 1 .2 RC# 0 329 π

or f 1 .2 RC# 0 329 π or f 1 0.2 1k 1329

## π μ or f 1 # 52.2 HzThus f 1max = 52.2 HzHence (C) is correct answer.MCQ 1.64 Let t g (f ) be the group delay function of the given RC-LPF and f 2 = 100 Hz.Thent g (f 2) in ms, is



(A) 0.717 (B) 7.17(C) 71.7 (D) 4.505SOL 1.64 Hence (A) is correct answerH (ω)1 j RC

1ω

=

+

θ (ω) =− tan−1ωRCt g ( )ddR CRCω 1 2 2 2

θ ω

ω

=− =

+

0.7171 4 10 10102 4 63

π # #

=

+

= − −

ms

Common Data for Questions 65 & 66 :Page 29 GATE EC 2003 www.gatehelp.com


X (t ) is a random process with a constant mean value of 2 and the autocorrelation function R ( ) 4(e . 1)

xx

τ = -0 2 τ + .MCQ 1.65 Let X be the Gaussian random variable obtained by sampling the process at t

= t iand letQ(α) e dy21 x

2

2

π

= −3



α #The probability that 6 x # 1@ is(A) 1 − Q(0.5) (B) Q(0.5)(C) Q2 2

1c m (D) 1 Q2 2− 1 c mSOL 1.65 Hence (D) is correct option.We have R XX (τ ) = 4(e-0.2 τ + 1)R XX (0) = 4(e-0.2 0 + 1) = 8 = σ 2

or σ = 2 2 Givenmean μ = 0Now P ( x # 1) = F x (1)1 Q X

σ = − − μ

c m at x = 11 Q2 2= − 1 − 0c m 1 Q2 2= − 1 c mMCQ 1.66 Let Y and Z be the random variable obtained by sampling X (t ) at t = 2 and t = 4respectively. Let W = Y − Z . The variance of W is

(A) 13.36 (B) 9.36(C) 2.64 (D) 8.00SOL 1.66 Hence (C) is correct option.W = Y − ZE [W 2] = E [Y − Z ]2= E [Y 2] + E [Z 2] − 2E [YZ ]w 2

= σ

We have E [ X 2(t )] = R x (10)= 4[e-0.2 0 + 1] = 4[1 + 1] = 8E [Y 2] = E [ X 2(2)] = 8

E [Z 2] = E [ X 2(4)] = 8E [YZ ] = R XX (2) = 4[e−0.2(4−2)+ 1] = 6.68[ ] E W 2 . 8 8 2 6 68 w

2 = σ = + − # = 2.64MCQ 1.67 Let x (t ) = 2 cos (800π ) + cos (1400π t ). x (t ) is sampled with the rectangular pulsetrain shown in the figure. The only spectral components (in kHz) present in thesampled signal in the frequency range 2.5 kHz to 3.5 kHz arePage 30 GATE EC 2003 www.gatehelp.com




(A) 2.7, 3.4 (B) 3.3, 3.6(C) 2.6, 2.7, 3.3, 3.4, 3.6 (D) 2.7, 3.3SOL 1.67 Hence (D) is correct option.

The frequency of pulse train isf101-3 = 1 k HzThe Fourier Series coefficient of given pulse train isC n T 1 Ae dt//o jn tTT22ooo

= − ω

−

− #T 1 Ae dt//o j tTT66ooo

= − ηω

−

− #( )[ ]T j

A e /

/o o j tT T6 6ooo

ηω = − − ω

− −

( )( )

j n A e e2 j o t j oT o/6



π = − − ω − ηω

j n ( ) A e e 2 j /3 j /3

= π − ηπ − ηπ

or C n sin

n A nπ 3= ` π jFrom C n it may be easily seen that 1,2,4,5,7, harmonics are present and 0,3,6,9,..are absent. Thus p(t ) has 1 kHz, 2 kHz, 4 kHz, 5 kHz, 7 kHz,... frequency componentand 3 kHz, 6 kHz.. are absent.The signal x (t ) has the frequency components 0.4 kHz and 0.7 kHz. The sampledsignal of x (t ) i.e. x (t )* p(t ) will have1 ! 0.4 and 1 ! 0.7 kHz2 ! 0.4 and 2 ! 0.7 kHz

4 ! 0.4 and 4 ! 0.7 kHzThus in range of 2.5 kHz to 3.5 kHz the frequency present is2 + 0.7 = 2.7 kHz4 − 0.7 = 3.3 kHzMCQ 1.68 The signal flow graph of a system is shown in Fig. below. The transfer functionC (s)/R (s) of the system isPage 31 GATE EC 2003 www.gatehelp.com


(A)s 29s 6

62+ +

(B)s ss29 662+ +

(C)( )s s

s s29 622+ ++

(D)( )s s



s s29 6272+ ++

SOL 1.68 Mason Gain FormulaT (s) pk k

3= Σ 3In given SFG there is only forward path and 3 possible loop.

p1 = 131

s s s= 1 + 3 + 24 = s + 27L1 ,s

Ls2 24= 2 − = − and Ls33 = −

where L1 and L3 are non-touchingThis( )( )R sC s1 (

ploop gain) pair of non touching loops131

+=

− −

1 s s s s . s

ss3 24 2 2 3

271 s s

ss29 6272

=

− − − +

= − − −

+

+ ++



^^ ^

hh h

( )

s ss s29 6272 =

+ +

+

Hence (D) is correct option.MCQ 1.69 The root locus of system G(s)H (s)s(s )(s )K

2 3=

+ + has the break-away pointlocated at(A) (− 0.5,0) (B) (− 2.548,0)(C) (− 4,0) (D) (− 0.784,0)SOL 1.69 We have1 + G(s)H (s) = 0ors(s )(s )1 K

2 3+

+ + = 0or K =− s(s2+ 5s2+ 6s)dsdK =− (3s2+ 10s + 6) = 0which gives s 0.784, 2.5486= −10 ! 100 − 72 =− − The location of poles on s − plane isPage 32 GATE EC 2003 www.gatehelp.com


Since breakpoint must lie on root locus so s =− 0.748 is possible.Hence (D) is correct option.MCQ 1.70 The approximate Bode magnitude plot of a minimum phase system is shown inFig.below. The transfer function of the system is(A)



( )( )( .)s ss10

10 1008 0 123

+ ++

(B)( )( )( .)s ss10

10 1007 0 13

+ +

+

(C)( )( )( .)s ss10 1000 122

+ ++

(D)( )( )( .)s ss10 1000 123

+ +

+

SOL 1.70 The given bode plot is shown below At ω = 0.1 change in slope is +60 dB " 3 zeroes at ω = 0.1 At ω = 10 change in slope is −40 dB " 2 poles at ω = 10 At ω = 100 change in slope is −20 dB " 1 poles at ω = 100Thus T (s)



( )( )K ( )1 11 .

s ss

1021000 13

=

+ ++

Now 20 log10K = 20or K = 10Thus T (s)( )( )( )

1 110 1 .

s ss1021000 13

=

+ +

+

( )( )

( .)s ss10 10010 0 128 3

=

+ ++

Hence (A) is correct option.Page 33 GATE EC 2003 www.gatehelp.com


MCQ 1.71 A second-order system has the transfer function( )( )R sC ss s4 4



42 =

+ +

With r (t ) as the unit-step function, the response c (t ) of the system is representedby

SOL 1.71 The characteristics equation iss2+ 4s + 4 = 0Comparing withs 2 n n

2+ ξω + ω2 = 0we get 2 n ξω 4 = and 4 n

2 ω =

Thus ξ = 1 Critically dampedt s 41 24 2

ξωn #= = =

Hence (B) is correct option.MCQ 1.72 The gain margin and the phase margin of feedback system withG(s)H (s)(s 100)83 =

+

are(A) dB,0c (B) 3,3(C) 3,0c (D) 88.5 dB, 3SOL 1.72 Hence (B) is correct option.MCQ 1.73 The zero-input response of a system given by the state-space equation

x x x x11011

212

=

oo= G = G= G and( )( )

x



x00101

2

= G = = G isPage 34 GATE EC 2003 www.gatehelp.com


(A)tett

= G (B)ett

= G(C)etet

= t G (D)t

=tet GSOL 1.73 We have

x

x12 oo= G

x x11011

2== =G G and( )( )

x x00



1012

= G = = G A

110= = 1G(sI − A)sss0 s0 11

011101 = − =−

= G = G = − − G(sI − A)−1

( )( )

s ( )s1 s1 11010( )ss s21111112

=

−

−

+ − = −

−

+



−

> H > HL−1[(sI − A)−1] eete e0 At

t

= = = t t G x (t ) e [ x (t )]ete eete0 10

Attt t

t= # 0 = = G= G = = t GHence (C) is correct option.MCQ 1.74 A DSB-SC signal is to be generated with a carrier frequency f c = 1 MHz using anon-linear device with the input-output characteristic V a v a v i i 0 0 1

3 = + where a0

and a1 are constants. The output of the non-linear device can be filtered by anappropriate band-pass filter.Let V i Ac cos (2 f c ) m(t )= i π i t + is the message signal. Then the value of f ci (in MHz)

is(A) 1.0 (B) 0.333(B) 0.5 (D) 3.0SOL 1.74 Hence (C) is correct option.v i Ac cos (2 f c t ) m(t )= 1 π +v 0 a v av o i i

3 = +

v 0 a [ A' cos (2 f ' t ) m(t )] a [ A' cos (2 f ' t ) m(t )]0 c c 1 c c

= π + + π + 3

a A' cos (2 f ' t ) a m(t ) a [( A' cos 2 f ' t )0 c c 0 1 c c

3 = π + + π

( A' cos (2 f ') t ) m(t ) 3 A' cos (2 f ' t )m (t ) m (t )]c c c c

2 2 3 + π + π +a A' cos (2 f ' t ) a m(t ) a ( A' cos 2f ' t )0 c c 0 1 c c

3 = π + + 3( )



( )cosa Af tm t

2' 1 4'cc1

2 ++ π

; E3a A' cos (2 f ' t )m (t ) m (t )1 c c

= π 2 + 3

The term 3a A' ( cos4 )m(t )

cf t1 2'

cπ is a DSB-SC signal having carrier frequency 1. MHz.Thus f 2 1 '

c = MHz or . f 0 5 '

c = MHzPage 35 GATE EC 2003 www.gatehelp.com


Common Data for Question 75 & 76 :Let m(t ) cos [(4 103) t ] = π # be the message signal &

c (t ) 5 cos [(2 106t )] = π # be the carrier.MCQ 1.75 c (t ) and m(t ) are used to generate an AM signal. The modulation index of thegenerated AM signal is 0.5. Then the quantityCarrier powerTotal sideband power is(A)21 (B)41(C)

31 (D)81SOL 1.75 Hence (D) is correct option.P T P 12 c

α2 = c + m



P sb

P P ( . )2 2c c 0 52 2 = α =

orPPcsb

8= 1MCQ 1.76 c (t ) and m(t ) are used to generated an FM signal. If the peak frequencydeviationof the generated FM signal is three times the transmission bandwidth of the AMsignal, then the coefficient of the term cos [2 (1008 103t )] π # in the FM signal (interms of the Bessel coefficients) is

(A) 5J 4(3) (B) J ( )2 58 3(C) J ( )2 58 4 (D) 5J 4(6)SOL 1.76 Hence (D) is correct option.

AM Band width = 2f mPeak frequency deviation = 3(2f m) = 6f mModulation index β f

6f 6m

= m =

The FM signal is represented in terms of Bessel function as x FM (t ) Ac J n ( )cos ( c n n) tn

= β ω − ω 33

=- /ωc + nωm 2 (1008 103) = π #2 106 n4 103 π + π # 2 (1008 103),n 4 = π # =

Thus coefficient = 5J 4(6)MCQ 1.77 Choose the correct one from among the alternative A,B,C ,D after matching anitem in Group 1 with most appropriate item in Group 2.Group 1 Group 2P. Ring modulator 1. Clock recoveryPage 36 GATE EC 2003 www.gatehelp.com




Q. VCO 2. Demodulation of FMR. Foster-Seely discriminator 3. Frequency conversionS. Mixer 4. Summing the two inputs5. Generation of FM6. Generation of DSB-Sc

(A) P − 1;Q − 3;R − 2;S − 4 (B) P − 6;Q = 5;R − 2;S − 3(C) P − 6;Q − 1;R − 3;S − 2 (D) P − 5;Q − 6;R − 1;S − 3SOL 1.77 Hence (B) is correct option.Ring modulation $ Generation of DSB - SCVCO $ Generation of FMFoster seely discriminator $ Demodulation of fmmixer $ frequency conversionMCQ 1.78 A superheterodyne receiver is to operate in the frequency range 550 kHz -1650kHz, with the intermediate frequency of 450 kHz. Let R = C max/C min denote therequired capacitance ratio of the local oscillator and I denote the image frequency

(in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, then(A) R = 4.41, I = 1600 (B) R = 2.10, I − 1150(C) R = 3.0, I = 600 (D) R = 9.0, I = 1150SOL 1.78 Hence (A) is correct option.f max = 1650 + 450 = 2100 kHzf min = 550 + 450 = 1000 kHzor f2 LC1π

=

frequency is minimum, capacitance will be maximumR ( . )CCff 2 1minmaxminmax22

= = = 2

or R = 4.41f i = f c + 2f IF = 700 + 2(455) = 1600 kHzMCQ 1.79 A sinusoidal signal with peak-to-peak amplitude of 1.536 V is quantized into128levels using a mid-rise uniform quantizer. The quantization-noise power is(A) 0.768 V (B) 48 10 6V 2 # -

(B) 12 10 6V 2 # - (D) 3.072 VSOL 1.79 Hence (C) is correct option.



Step size δ . .L2m128= p = 1 536 = 0 012 V

Quantization Noise power( . )12 122 0 012 2 = δ =12 10 6 = # − V 2Page 37 GATE EC 2003 www.gatehelp.com


MCQ 1.80 If E b , the energy per bit of a binary digital signal, is 10-5 watt-sec and the one-sidedpower spectral density of the white noise, N 0 10= -6 W/Hz, then the output SNRof the matched filter is(A) 26 dB (B) 10 dB(C) 20 dB (D) 13 dBSOL 1.80 Hence (D) is correct option.E b = 10-6 watt-secN o = 10-5 W/Hz(SNR) matched filler E .2 1010 05 N

o25

6

o#= = = -

(SNR)dB = 10 log 10(0.05) = 13 dBMCQ 1.81 The input to a linear delta modulator having a step-size 3= 0.628 is a sinewavewith frequency f m and peak amplitude E m. If the sampling frequency f x = 40 kHz,the combination of the sine-wave frequency and the peak amplitude, where slopeoverload will take place isE m f m(A) 0.3 V 8 kHz

(B) 1.5 V 4 kHz(C) 1.5 V 2 kHz(D) 3.0 V 1 kHzSOL 1.81 Hence (B) is correct option.For slopeoverload to take place Eff2 m



m

3 s

$ π This is satisfied with E m = 1.5 V and f m = 4 kHzMCQ 1.82 If S represents the carrier synchronization at the receiver and ρ represents thebandwidth efficiency, then the correct statement for the coherent binary PSK is(A) ρ = 0.5,S is required (B) ρ = 1.0,S is required(C) ρ = 0.5,S is not required (D) ρ = 1.0,S is not requiredSOL 1.82 Hence (A) is correct option.If s " carrier synchronization at receiver ρ " represents bandwidth efficiencythen for coherent binary PSK ρ = 0.5 and s is required.MCQ 1.83 A signal is sampled at 8 kHz and is quantized using 8 - bit uniform quantizer.

Assuming SNRq for a sinusoidal signal, the correct statement for PCM signal witha bit rate of R is(A) R = 32 kbps, SNR q = 25.8 dB (B) R = 64 kbps, SNR q = 49.8 dB(C) R = 64 kbps, SNR q = 55.8 dB (D) R = 32 kbps, SNR q = 49.8 dBSOL 1.83 Hence (B) is correct option.Page 38 GATE EC 2003 www.gatehelp.com


Bit Rate = 8k # 8 = 64 kbps(SNR)q = 1.76 + 6.02n dB= 1.76 + 6.02 # 8 = 49.8 dBMCQ 1.84 Medium 1 has the electrical permittivity ε1 = 1.5ε0 farad/m and occupies theregion to the left of x = 0 plane. Medium 2 has the electrical permittivity ε2 = 2.5ε0

farad/m and occupies the region to the right of x = 0 plane. If E 1 in medium 1 isE 1 = (2u x − 3uy + 1uz ) volt/m, then E 2 in medium 2 is

(A) (2.0u x − 7.5uy + 2.5uz ) volt/m (B) (2.0u x − 2.0uy + 0.6uz ) volt/m(C) (2.0u x − 3.0uy + 1.0uz ) volt/m (D) (2.0u x − 2.0uy + 0.6uz ) volt/mSOL 1.84 Hence (C) is correct option.We have E 1 = 2u x − 3uy + 1uz

E 1t =− 3uy + uy and E 1n = 2u x

Since for dielectric material at the boundary, tangential component of electric fieldare equalE 1t =− 3uy + uy = E 2t ( x = 0 plane)E 1n = 2u x

At the boundary the for normal component of electric field areD1n = D2n

or ε1E 1n = ε2E 2n

or 1.5εo2u x = 2.5εoE 2n

or E 2n

.u . u2 53 1 2 = x = x

Thus E 2 = E 2t + E 2n =− 3uy + uz + 1.2u x



MCQ 1.85 If the electric field intensity is given by E = ( xu x + yuy + zuz ) volt/m, the potentialdifference between X (2,0,0) and Y (1,2,3) is(A) +1 volt (B) −1 volt(C) +5 volt (D) +6 voltSOL 1.85 Hence (C) is correct option.

We have E = xu x + yuy + zuzdl = ut x dx + uty dy + utz dzV XY E .dl

XY

=− # xdxu x ydyuz zdzuz201230

= # t + # t + # t x y z2 2 2212 220 230

=−= + + G[ ]2

=−1 22− 12+ 02− 22+ 02− 32 = 5MCQ 1.86 A uniform plane wave traveling in air is incident on the plane boundarybetweenPage 39 GATE EC 2003 www.gatehelp.com


air and another dielectric medium with εr = 4. The reflection coefficient for thenormal incidence, is(A) zero (B) 0.5+180c(B) 0.333+0c (D) 0.333+180cSOL 1.86 Hence (D) is correct option.η ε

μ =

Reflection coefficientτ 2 12 1

η η

= η η + −



Substituting values for η1 and η2 we haveτ o roooo roo

0

=

+− ε ε

μ ε μ

ε ε

μ ε μ

111 41 4r

rε

= ε

+

− =+

− since 4 εr =

.3= −1 = 0 333+180cMCQ 1.87 If the electric field intensity associated with a uniform plane electromagneticwavetraveling in a perfect dielectric medium is given by E (z , t ) = 10 cos (2π 107 t − 0.1π z )V/m, then the velocity of the traveling wave is(A) 3.00 108 # m/sec (B) 2.00 108 # m/sec(C) 6.28 107 # m/sec (D) 2.00 107 # m/secSOL 1.87 Hence (B) is correct option.We have E (z , t ) 10 cos (2 107 t 0.1 z ) = π # − π where ω 2 107 t = π # β = 0.1π Phase Velocity u0.12 10 2 107

# # 8 β

ω

π

= = π = m/sMCQ 1.88 A short - circuited stub is shunt connected to a transmission line as shown infig. If Z 0 = 50 ohm, the admittance Y seen at the junction of the stub and thetransmission line is





(A) (0.01 − j 0.02) mho (B) (0.02 − j 0.01) mho(C) (0.04 − j 0.02) mho (D) (0.02 + j 0) mhoSOL 1.88 The fig of transmission line is as shown below .We know that Z in[ ][ ]tantanZZ jZ lZ jZ loo LL o

β β

=

++

For line 1, l2 λ = and β 2 ,Z L1 100 λ

= π = Ω

Thus Z in1

[ ][ ]tantanZZ jZZ jZo Z 100o LL o

L π

π = Ω

+

+ = =For line 2, l8 λ = and β 2 ,Z L2 0 λ

= π = (short circuit)Thus Z in2

[ 0]



[ tan ]ZZ

jZ jZ j

0o 50ooo4

+= Ω

+

= =π

YZ Z

1 1in1 in2

= + . . j j100150= + 1 = 0 01 − 0 02Hence (A) is correct option.Page 41 GATE EC 2003 www.gatehelp.com


MCQ 1.89 A rectangular metal wave guide filled with a dielectric material of relativepermittivityεr = 4 has the inside dimensions 3.0 cm # 1.2 cm. The cut-off frequency for thedominant mode is(A) 2.5 GHz (B) 5.0 GHz(C) 10.0 GHz (D) 12.5 GHzSOL 1.89 Hence (A) is correct option.u c .23 10 1 5 1008

# # 8

ε

= = =

In rectangular waveguide the dominant mode is TE 10 andf C va m



b n2= ` j2+` j2.2 . b

1 5 100 031 0 8

= # ` j2+` j2.. .0 061 5 10 2 58 = # = GHzMCQ 1.90 Two identical antennas are placed in the θ = π /2 plane as shown in Fig. Theelements have equal amplitude excitation with 180c polarity difference, operating

at wavelength λ. The correct value of the magnitude of the far-zone resultantelectric field strength normalized with that of a single element, both computed forφ = 0, is(A) 2 cos 2 s λ

b π l (B) 2 sin 2 s λ b π l(C) 2cos s λ

aπ k (D) 2 sin s

λ

aπ kSOL 1.90 Hence (D) is correct option.Normalized array factor 2 cos2 ψ

=

ψ = β d sinθ cosφ + δ

θ = 90c,d = 2 s ,φ = 45c,δ = 180c

Now 2 cos2 ψ 2 cos d sin cos2= β θ φ + δ

; E.2 cos s cos



22 2 452c 180 λ

= 8 π + B

Page 42 GATE EC 2003 www.gatehelp.comBrought to you by: Nodia and Company Visit us at: www.nodia.co.inPUBLISHING FOR GATE

2 cos s 90c λ

= 8 π + B 2 sin s λ

= π ` jAnswer Sheet

1. (B) 19. (D) 37. (A) 55. (B) 73. (C)2. (C) 20. (D) 38. (A) 56. (B) 74. (C)3. (B) 21. (D) 39. (D) 57. (A) 75. (D)4. (C) 22. (C) 40. (C) 58. (D) 76. (D)5. (C) 23. (B) 41. (C) 59. (C) 77. (B)6. (D) 24. (A) 42. (A) 60. (A) 78. (A)7. (B) 25. (C) 43. (D) 61. (D) 79. (C)8. (A) 26. (A) 44. (B) 62. (A) 80. (D)9. (C) 27. (C) 45. (A) 63. (C) 81. (B)10. (D) 28. (B) 46. (A) 64. (A) 82. (A)11. (B) 29. (C) 47. (A) 65. (D) 83. (B)12. (D) 30. (B) 48. (C) 66. (C) 84. (C)13. (B) 31. (A) 49. (B) 67. (D) 85. (C)14. (C) 32. (*) 50. (B) 68. (D) 86. (D)15. (B) 33. (*) 51. (C) 69. (D) 87. (B)16. (D) 34. (A) 52. (D) 70 (A) 88. (A)17. (C) 35. (C) 53. (B) 71 (B) 89. (A)18. (B) 36. (B) 54. (A) 72 (B) 90. (D)

gatye temp

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