Ece Practice Book PDF

ECE

GATE Practice Book

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GATE ELECTRONICS COMMUNICATION ENGINEERING (ECE)

PRACTICE BOOK 

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Register at www.engineersinstitute.com & get 3 Full length GATE-Mock Test

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Er. KUNAL SRIVASTAVA AIR-1 ESE 2012, AIR-44 GATE 2013

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GATE Practice Book

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� 2013 By Engineers Institute of India ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored or used in any form or by any means graphic, electronic, or mechanical & chemical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems.

Engineers Institute of India

28-B/7, Jia Sarai, Near IIT Hauz Khas New Delhi-110016 Tel: 011-26514888 For publication information, visit www.engineersinstitute.com/publication ISBN: 978-93-5137-760-3 Price: Rs. 299

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ECE

GATE Practice Book

This book is dedicated to all Electronics C Communication ommunication Engineers preparing for GATE & Public sector e examination xaminations xaminations.

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ECE

GATE Practice Book

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GATE exa ination is one of the most prestigious competitive examination conducted or graduate engineers. Over the past few y ars, it has become more com etitive as a number of aspirants are incr asingly becoming interested in M.Tech & government jobs due to decl ne in other career options. In my opi ion, GATE exam test candidates’ basics understanding of concepts a d ability to apply the same in a numerical approach. A candidate is supposed to s artly deal with the syllabus not just mu ging up concepts. Thorough understanding wit critical analysis of topics and ability to xpress clearly are some of the pre-requisites to crack this exam. The questioning & exam ination pattern has changed over the past few y ars, as numerical answer type questions p ay a major role to score a good rank. Keeping in mind, the difficulties of an average student, we have composed this booklet. We are thankf l to Er. KUNAL SRIVASTAVA (AIR 1 SE 2012, AIR 44 GATE 2013 in ECE) for preparing this booklet. In Electronics Communicati n Engineering, over 2,50,000 candidates ppeared in GATE 2013 with 36,394 finally qualified. The cut-off marks for general categ ry is 25. For more details log onto our institute’s website http://www.engineersinstitute.com/ lectronics Established in 2006 by a tea of IES and GATE toppers, we at Engineer Institute of India have consistently provided ri orous classes and proper guidance to engine ering students over the nation in successfully a ccomplishing their dreams. We believe i providing examoriented teaching methodolo gy with updated study material and test series so that our students stay ahead in the competition. The faculty at EII are a te m of experienced professionals who have gui ed thousands to aspirants over the years. They are readily available before and after cla ses to assist students and we maintain a hea thy student-faculty ratio. Many current and past ear toppers associate with us for contributing towards our goal of providing quality education and share their success with the future as irants. Our results speak for themselves. Past tudents of EII are currently working in various government departments and PSU’s and ursuing higher specializations at IISc, IITs , NITs & reputed institutions. Best wishes for future career R.K. Rajesh Director Engineers Institute of India [email protected]

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ECE

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GATE Practice Book

CONTENTS 1.

2.

3.

4.

5.

6.

7.

8.

ANALOG ELECTRONICS........................................................... 01-70 Questions ………………………………………………………..

01-38

Solutions ……………………………………………………......

39-70

COMMUNICATION ................................................................... 71-106 Questions ………………………………………………………..

71-87

Solutions ……………………………………………………......

88-106

CONTROL SYSTEM .................................................................. 107-152 Questions ………………………………………………………..

107-132

Solutions ……………………………………………………......

133-152

DIGITAL ELECTRONICS ........................................................... 153-188 Questions ………………………………………………………..

153-173

Solutions ……………………………………………………......

174-188

ELECTRONICS DEVICES AND CIRCUITS (EDC) ..................... Questions ………………………………………………………..

189-206

Solutions ……………………………………………………......

207-220

189-220

ELECTROMAGNETIC THEORY-EMT......................................... 221-248 Questions ………………………………………………………..

221-236

Solutions ……………………………………………………......

237-248

NETWORK THEORY.................................................................. 249-288 Questions ………………………………………………………..

249-273

Solutions ……………………………………………………......

274-288

SIGNAL & SYSTEM .................................................................. 289-330 Questions ………………………………………………………..

289-311

Solutions ……………………………………………………......

312-330

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ECE

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GATE Practice Book

1. ANALOG ELECTRONICS (1.)

16

3

15

3

Consider a Si p.n diode with doping concentrations, N a = 10 / cm and N d = 10 / cm . If 10

3

and zero bias junction capacitance = 0.5pF. The junction capacitance at ni = 1.5 ×10 / cm an applied reverse bias of 2V is: (a.) 0.5pF

(b.) 0.245pF 17

(c.) 0.168pF 3

16

(d.) 2.0698pF

3

(2.)

If at the doping levels of N D = 10 /cm . And NA = 10 /cm and at a reverse bias of 1V, a Si diode shows a junction capacitance of 0.5pF, find the zero bias junction. (a.) 0.76pF (b.) 1.16pF (c.) 0.86pF (d.) 1.31pF

(3.)

Consider a silicon p – n diode with reverse saturation current, I o = 10 A. Find the power

-13

dissipated in the 2k Ω resistor.

(a.) 20mW

(b.) 19.2mW

(c.) 8mW

(d.) 9.6mW

2 Linked Questions: (4.)

Find the power in the 2k Ω resistor if the diode has a cut in voltage = 0.7V and d.c. forward resistance = 10Ω

(5.) (6.)

(a.) 8mW (b.) 16.7mW (c.) 9.15mW (d.) 19.1mW Find the power consumed in the diode in the previous question (a.) 1.5mW (b.) 9.15mW (c.) 1.4mW (d.) 0.046mW If in the following circuit, diode has a power rating of 1mW, with a cut in voltage = 0.7V and zero forward resistance, find which of the following values of “R” will meet the requirement.

(a.) 3k Ω

(b.) 7k Ω

(c.) 3.97k Ω

(d.) 6.5k Ω

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ECE

GATE Practice Book

SOLUTION (1.)

ANS: (b) EXP:

C jo

C j =

1+

V bi = (2.)

kT q

V bi

ln

N a .N d ni

2

ANS: (a) EXP:

c j =

C jo

1+ (3.)

V R

V R V bi

ANS: (d) EXP: Using diode equation and K.V.L

⇒ 5 = 2 ×10 ×10 3

V  0.026  − 1 e   + V D   D

−13

⇒ V D = 0.62V (by hit and trial) ∴ I D = 2.2mA ∴ Power = (2.2mA) 2 × 2 kΩ = 9.68mW (4.)

ANS: (c) EXP: Replace the diode by its piecewise linear model

∴ I D =

5V − 0.7V

= 2.14mA 2.01k Ω ∴ Power = 9.15mW

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ECE

(5.)

GATE Practice Book

ANS: (a) EXP: Power consumed in diode = sum of powers consumed in the V γ and rf .

= Vγ .I D + ( I D )2 .rf = 1.54mW

(6.)

Mind that 1.54mW + 9.15mW = 10.69mW = power delivered by the 5V source ANS: (b) EXP: The power consumed in the diode must not exceed 1mW.

∴ I D ≤

∴ R ≥

1mW 0.7V

= 1.43mA

10V − 0.7V

1.43mA thus, R ≥ 6.51k Ω

= 6.51k Ω

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ECE

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GATE Practice Book

2. COMMUNICATION (1.)

For an A.M. signal x(t ) = Ac 1 + (a.) 25%

1 2

6 sin10t .cos10 t find the power efficiency?

(b.) 20%

(c.) 70.7% 3

(d.) 56.56%

6

(2.)

For an A.M signal x (t ) = 10 1 + 0.5 cos(2π .10 .t ) cos10 t find the upper side band power

(3.)

and modulated signal bandwidth. (a.) 3.125W, 2kHz (b.) 6.25W, 2kHz (c.) 3.125W, 1kHz (d.) 6.25W, 1kHz In commercial T.V, picture signals are transmitted by …………….. and speech signals by ……………… modulation. (a.) S.S.B., F.M (b.) F.M, V.S.B (c.) F.M., S.S.B (d.) V.S.B., F.M.

2 Linked Questions (4.)

(5.) (6.)

A signal by f (t ) = 2cos(1002πt ) + 20cos(1000πt ) + 2 cos(998π t ) has been radiated with an antenna of resistance 10Ω. Find modulation index and total power transmitted. (a.) µ = 0.1 (b.) µ = 0.1 (c.) µ = 0.1 (d.) µ = 0.2 Pt = 201W Pt = 201W Pt = 20.1W Pt = 20.4W The bandwidth and the power of the upper sideband in the previous question are (a.) 2Hz, 0.2W (b.) 4Hz, 0.4W (c.) 2Hz, 0.4W (d.) 4Hz, 0.2W An A.M. transmitter current is given by 10A with 40% single tone modulation, Find A.M. transmitter current with 80% single tone modulation. (a.) 11.05A (b.) 8.18A (c.) 9.045A (d.) 12.22A

2 Linked Questions 6

A carrier signal c(t) = 10.cos(2π × 10 t ) is simultaneously amplitude modulated with two message signals of frequencies 1kHz and 2kHz with modulation indices of 0.4 and 0.3 respectively (7.) (8.)

(9.)

Find overall modulation index and total power transmitted if antenna resistance = 1Ω (a.) 0.5, 56.25W (b.) 0.5, 62.25W (c.) 0.7, 56.25W (d.) 0.7, 62.25W Find the bandwidth and power in the two sidebands corresponding to 2kHz modulating frequency. (a.) 4kHz, 2.25W (b.) 3kHz, 4W (c.) 3kHz, 2.25W (d.) 4kHz, 4W An amplitude modulated signal is plotted.

Find the modulation index and amplitude of the carrier wave (a.) 0.6, 60V (b.) 0.4, 50V (c.) 0.4, 60V

(d.) 0.6, 50V

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ECE

GATE Practice Book

SOLUTION

(1.)

ANS: b EXP: 2

η =

µ

2 + µ 2

(2.) (3.)

ANS: a ANS: d

(4.)

ANS: c EXP:

P=

Ac

2

2 R

(5.) (6.)

ANS: a ANS: a

(7.) (8.) (9.)

ANS: a ANS: a ANS: d

2

1+

µ

2

watt

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ECE

GATE Practice Book

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3. CONTROL SYSTEM (1.) (2.)

The transfer function of a system in Laplace transform of (a.) Impulse response (b.) Step response (c.) Ramp response (d.) None of these The open loop D.C. gain of a unity negative feedback system with closed loop transfer

s+4

function (a.) (3.)

s + 7 s + 13

is:

4

(b.)

13

4

(c.) 4

9

If unit step response of a system is C (t ) = 1 − e (a.)

(4.)

2

3

(b.)

s (s + 3)

3 ( s + 3)

(a.)

3 s

(d.) ( s + 3)

s+6 s( s + 4s + 3)

5 −t 1 −3t    2 + .e + .e  .u(t ) 2 2  

5 − t 1 −3t   (b.)  2 − .e + .e  .u(−t ) 2 2  

 

(d.)  2 +

1 −3t  −t .e − .e  .u (t ) 2 2  5

Find the inverse Laplace transform

F ( s) =

( (c.) ( −e (a.) e

(7.)

(c.) 3

2

5 − t 1 −3t   (c.)  2 − .e + .e  .u (t ) 2 2  

(6.)

for t ≥ 0 , its transfer function is,

The inverse Laplace transform of

F ( s) =

(5.)

−3t

(d.) 13

s +6

( s + 2)( s 2 + 2 s + 1)

−2 t

−t

) .u(t ) − t.e ) .u (t ) − t

− e + t.e

−2 t

+e

−t

− t

Find the inverse Laplace transform of F ( s) =

−2 t −t − t (b.) ( e + e + t.e ) .u (t ) − t −t − t (d.) ( e 2 − e + t 2 .e ) .u (t )

s+9 2 s + 6s + 25

(a.)

3 −3t  −3t  e t e t .cos 4 + . .sin 4   .u( t ) 2  

2 −3t  −3 t  (b.)  e .cos 4t + .e .sin 4t  .u( t) 3  

(c.)

 2 −3 t  −3t  e .cos 4t + .e .sin 4t  .u( t ) 3 

 3 −3t  −3t (d.)  e .cos 4t + .e .sin 4t  .u( t) 2 

Find the inverse Laplace transform of f1 ( s) = (a.) cos h3t.u (t ); (c.) cos 3t.u (t );

2 3 2 3

s 2 s −9

; and f 2 ( s) =

sin 3t.u (t )

(b.) cos h3t .u (t );

sin h3t.u(t )

(d.) cos 3t.u (t );

2 2 s −9

2 3 2 3

sin h3t.u(t ) sin 3t.u(t )

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ECE

GATE Practice Book

SOLUTION (1.) (2.)

ANS: (a) ANS: (b) EXP:

G(s)

1 + G ( s).H ( s) (3.)

=

s+4 2

s + 7 s + 13

Put H(s) = 1, find G(s) and put s = 0 (for d.c.) ANS: (b) EXP: Impulse response, h(t) =

d dt

c(t )

H (s) = L [ h(t ) ] (4.)

ANS: (c) EXP:

F ( s) = A = B =

6 3

s+6 s ( s + 1)( s + 3)

=

A s

+

B s +1

+

C

s+3

=2 5

−5

2 1 C = = −3 × −2 2

(5.)

−1× 2 3

=

ANS: (a) EXP:

F ( s) = A = B =

C =

1 ( s + 2)( s + 1)

1

=

A

( s + 2)

+

B

(s + 1)

+

C

(s + 1)2

=1

( −1) 2 −1

( −1 + 2) 1 ( −1 + 2)

∴ F (s) =

2

= −1

=1

1 ( s + 2)

−

1 (s + 1)

+

1 ( s + 1)2

∴ f (t ) = ( e−2 t − e− t + t.e− t ) .u(t )

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ECE

(6.)

GATE Practice Book

ANS: (a) EXP:

F ( s) =

=

s+9 2

s + 6 s + 25

=

s+3 2

( s + 3) + (4)

2

s + 3+ 6 2

( s + 3) + (4) +

3 2

×

2

4 2

( s + 3) + (4)

2

3 −3t −3 t = e .cos 4t + .e .sin 4t .u( t ) 2 (7.)

ANS: (b)

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ECE

GATE Practice Book

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4. Digital Electronics (1.)

Assuming that initially Qn = 0, now the clock pulses are given, fine the resulting sequence at

Qn .

(2.)

(a.) 0, 0, 0, 0… (b.) 1, 1, 1, 1… (c.) 0, 1, 0, 1, 0, 1… (d.) 1, 0, 1, 0… Fine the output frequency of the circuit.

(a.) 49.76 MHz (b.) 995.22 Hz (c.) 6.25 KHz (d.) 7.961 KHz

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ECE

(3.)

GATE Practice Book

Find the expression for the output logic function.

(a.) AB + C(D + E) (b.) (A+B) .C + (D + E) (c.) AB + C. (D+E) (d.) AB + C. ( D + E )

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ECE

GATE Practice Book

SOLUTION (1.)

ANS: c EXP: The input changes as 11 01 11 01, use JKFF truth table. ANS: b EXP: The frequency of Hartley’s oscillator = 7961.8Hz. The output of mod – 8 counter →

(2.)

= (3.)

→

→

7961.8 Hz 8

ANS: d EXP: NMOS Parallel OR NMOS series AND Reverse for PMOS Put complement on the final expression →

→

→

→

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Ece Practice Book PDF

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