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1991-2019

ELECTRONICS AND

COMMUNICATION ENGINEERING

ESE TOPICWISE OBJECTIVESOLVED PAPER-II

Typeset at : IES Master Publication, New Delhi-110016

All rights reserved.Copyright © 2019, by IES MASTER Publication. No part of this booklet may be reproduced, ordistributed in any form or by any means, electronic, mechanical, photocopying, recording,or otherwise or stored in a database or retrieval system without the prior permission ofIES MASTER Publication, New Delhi. Violates are liable to be legally prosecuted.

First Edition : 2017

Second Edition : 2018

Third Edition : 2019

IES MASTER PUBLICATIONF-126, (Lower Basement), Katwaria Sarai, New Delhi-110016

Phone : 011-26522064, Mobile : 8130909220, 9711853908

E-mail : info.publications@iesmaster.org

Web : iesmasterpublications.com

Engineering Services Examination is the gateway to an immensely satisfying job in the engineering sectorof India that offers multi-faceted exposure. The exposure to challenges and opportunities of leading thediverse field of engineering has been the main reason behind engineering students opting for EngineeringServices as compared to other career options. To facilitate selection into these services, availability ofarithmetic solution to previous years’ paper is the need of the day.

It is an immense pleasure to present previous years’ topic-wise objective solved papers of EngineeringServices Examination (ESE). This book is an outcome of regular and detailed interaction with the studentspreparing for ESE every year. It includes solutions along with detailed explanation to all questions. Theprime objective of bringing out this book is to provide explanation to each question in such a mannerthat just by going through the solutions, students will be able to understand the basic concepts and havethe capability to apply these concepts in solving other questions that might be asked in future exams.Towards the end, this book becomes indispensable for every ESE aspiring candidate.

IES Master PublicationNew Delhi

PREFACE

Note: Direction of all Assertion Reasoning (A–R) type of questions coveredin this booklet is as follows:

DIRECTIONS:

The following four items consist of two statements, one labelled as ‘AssertionA’ and the other labelled as ‘Reason R’. You are to examine these twostatements carefully and select the answer to these two statements carefullyand select the answer to these items using the codes given below:

(a) Both A and R are individually true and R is the correct explanation ofA

(b) Both A and R are individually true but R is not the correct explanationof A

(c) A is true but R is false

(d) A is false but R is true.

Note: Direction of all Statement-I and Statement-II type of questions coveredin this booklet is as follows:

DIRECTION:

Following items consists of two statements, one labelled as ‘Statement (I)’and the other as ‘Statement (II)’. You are to examine these two statementscarefully and select the answers to these items using the code given below:

(a) Both Statement : (I) and Statement (II) are individually true andStatement (II) is the correct explanation of Statement (I).

(b) Both Statement (I) and Statement (II) are individually true butStatement (II) is not the correct explanation of Statement (I).

(c) Statement (I) is true but Statement (II) is false

(d) Statement (I) is false but Statement (II) is true.

https://iesmaster.org/genius-batch

1. Control System ........................................................................................... 01–157

2. Electromagnetic Field Theory ................................................................... 158–332

3. Communication Systems .......................................................................... 333–415

4. Computer Organization and Architecture................................................... 416–460

5. Advanced Electronics ............................................................................... 461–466

6. Microprocessor ........................................................................................ 647–510

7. Advance Communication Systems ........................................................... 511–553

8. Signal and Systems .................................................................................. 554–670

CONTENT

SYLLABUS

Signal flow graphs, Routh-Hurwiz criteria, root loci, Nyquist/Bode plots Feedback systems-open & close looptypes, stability analysis, steady state, transient and frequency response analysis; Design of control systems,Compensators, elements of lead/lag compensation, PID and industrial controllers.

CONTENTS

1. Basics of Control System, Block Diagram and Signal Flow Graph ................ 1-16

2. Modelling and Response of Physical Systems ............................................. 17-28

3. Time Response Analysis .............................................................................. 29-62

4. Stability and Routh-Hurwitz Criterion ............................................................. 63-78

5. Root Locus Technique .................................................................................. 79-91

6. Frequency Response Analysis .................................................................... 92-125

7. Controllers and Compensators .................................................................. 126-150

8. State Variable Analysis ............................................................................. 151-157

UNIT-1 Control System

1 BASICS OF CONTROLSYSTEM, BLOCK DIAGRAMAND SIGNAL FLOW GRAPH

IES – 20191. A transfer function having all its poles and zeros

only in the left-half of the s-plane is called(a) a minimum-phase transfer function(b) a complex transfer function(c) an all-pass transfer function(d) a maximum-phase transfer function

2. Consider the following open-loop transfer function :

G =

K s 2

s 1 s 4The characteristic equation of the unity negativefeedback will be(a) (s + 1) (s + 4) + K (s + 2) = 0(b) (s + 2) (s + 1) + K (s + 4) = 0(c) (s + 1) (s – 2) + K (s + 4) = 0(d) (s + 2) (s + 4) + K (s + 1) = 0

3. Consider the following statements regarding aparabolic function :1. A parabolic function is one degree faster

than the ramp function.2. A unit parabolic function is defined as

2t , for t 0f t 20, otherwise

3. Laplace transform of unit parabolic function

is 31 .s

Which of the above statements are correct?(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3

4. The price for improvement in sensitivity by theuse of feedback is paid in terms of(a) Loss of system gain(b) Rise of system gain(c) Improvement in transient response, delayed

response(d) Poor transient response

IES – 20185. Consider the following statements for signal flow

graph:1. It represents linear as well as non linear

systems.2. It is not unique for a given system.Which of the above statements is/are correct?(a) 1 only(b) 2 only(c) Both 1 and 2(d) Neither 1 nor 2

6. The closed-loop transfer function C(s) R(s) ofthe system represented by the block diagram inthe figure is

1

1

s+1

s+1

R(s) +

–

+

–

C(s)

(a) 21

(s +1) (b)1

(s +1)(c) s+1 (d) 1

IES – 20157. The Laplace transform of e–2t sin 2 t is

(a) 2 22s

(s 2) 2 (b) 2 22

(s 2) 4

(c) 2 22

(s 2) 4

(d) 2 2

2s(s 2) 2

IES – 20138. The open-loop transfer function of a unity

feedback control system is

2

1G s =s + 2

The closed loop transfer function poles are locatedat:

IES MASTER Publication

CONTROL SYSTEM 3E&T Engineering

(a) –2, –2 (b) –2, –1(c) –2, +2 (d) 2 j1

9. In control systems, excessive bandwidth is NOTemployed because:(a) noise is proportional to bandwidth(b) it leads to low relative stability(c) it leads to slower response(d) noise is proportional to the square of the

bandwidth

IES – 2012

10. A system is described by the transfer function

2s + 5G =ss + 5 s + 4 . The dc gain of the system

is(a) 0.25 (b) 0.5(c) 1 (d)

11. The sensitivity TKS of transfer function

1 + 2KT =3 + 4K with respect to the parameter K

is given by

(a) 2K

3 + K (b) 23K

2 + 4K + K

(c) 22K

3 +10K + 8K (d) 24K

2 + 5K + 7K

IES – 2011

12. What is the unit impulse response of the systemshown in figure for t 0 ?

R(s) C(s)1s

1s 1

(a) –t1+ e (b) –t1 – e(c) –te (d) –t–e

13. Given the differential equation smodel of aphysical system, determine the time constantof the system

dx40 + 2x = f tdt(a) 10 (b) 20(c) 1/10 (d) 4

IES – 201014. A linear time-invariant system is initially at

rest, when subjected to a unit-step input, givesa response –ty t = te ,t > 0 . The transfer

function of the system is:

(a) 21

s +1 (b) 21

s s +1

(c) 2s

s +1 (d)1

s +1

15.

R(s)3

y(s)

d(s)

23s 1

The transfer function from d(s) to y(s) is:

(a)2

3s 7 (b)2

3s 1

(c)6

3s 7 (d)2

3s 6

IES – 2009

16. Consider the function F(s) = 2 2s + where F(s)

is the Laplace transform of f(t). What is thesteady state value of f(t)?(a) Zero(b) One(c) Two(d) A value between –1and +1

17. What is the characteristic of a good controlsystem ?(a) Sensitive to parameter variation(b) Insensitive to input command(c) Neither sensitive to parameter variation nor

sensitive to input commands.(d) Insensitive to parameter variation but

sensitive to input commands.

18. A negative-feedback closed-loop system issupplied with an input of 5V. The system hasa forward gain of 1 and a feedback gain of 1.What is the output voltage ?(a) 1.0 V (b) 1.5 V(c) 2.0 V (d) 2.5 V

19. In closed loop control system, what is thesensitivity of the gain of the overall system, Mto the variation in G ?

(a) 1

1 + G Hs s (b) 1

1 + G s

IES MASTER Publication

CONTROL SYSTEM 9E&T Engineering

ANSWER KEY

1 (a)

2 (a)

3 (c)

4 (a)

5 (b)

6 (b)

7 (c)

8 (d)

9 (a)

10 (a)

11 (c)

12 (b)

13 (b)

14 (c)

15 (a)

16 (d)

17 (d)

18 (d)

19 (a)

20 (c)

21 (c)

22 (a)

23 (d)

24 (c)

25 (b)

26 (c)

27 (b)

28 (b)

29 (d)

30 (b)

31 (b)

32 (b)

33 (c)

34 (a)

35 (a)

36 (b)

37 (a)

38 (a)

39 (b)

40 (c)

41 (b)

42 (a)

43 (d)

44 (d)

45 (d)

46 (c)

47 (d)

48 (a)

49 (b)

50 (b)

51 (a)

52 (a)

53 (a)

EXPLANATIONS

Sol–1: (a)A minimum phase transfer function haspoles & zeros on left half of s-plane.A maximum phase transfer function haspoles & zeros on right half of s-plane.An all pass transfer function has poles &zeros symmetrically located with respect toj axis.A complex transfer function has complex poles& zeros, not necessarily on a particular half.

Sol–2: (a)Given open loop transfer function

G =

K(s 2)

(s 1)(s 4)with unity feedback, H(s) = 1Characteristic equation : 1 + GH = 0

K(s 2)1

(s 1)(s 4)

= 0

(s+1) (s+4) + K (s + 2) = 0Sol–3: (c)

(1) Unit Ramp function, r(t) = tu(t)

Unit parabolic function, f(t) = 2t u(t)

2Parabolic function grows faster thanramp function by 1 degree

(1) is true.(2) is true

(3) Laplace transform of 2

3t 1u(t)2 s

istrue.

Hence all statements are true.Sol–4: (a)

With the use of negative feedback, sensitiv-ity of system is improved but gain is decreased.

Sol–5: (b)Signal flow graph is applicable to linearsystems and not applicable to non linearsystems. Therefore, statement 1 is falseOne system can have different signal flowgraph according to the order in which theequations are used to define the variablewritten on the left hand side. So it is notunique for a given system. Statement-2 iscorrect.

10 ESE Topicwise Objective Solved Paper-II 1991-2019 E&T Engineering

Sol–6: (b)1

s +1

1s +1

R(s) C(s)

+–

–

1s +1

1s +1

R(s) C(s)+–

1s +1

s +1s

R(s) C(s)–

1s

R(s) C(s)–

C sR s =

11s =1 s +11 +

s

Sol–7. (c)–2tL e sin 2 t = ?

L sin2 t = 2 2 22

2 2s 4s 2

–2tL e sin 2 t = 2 22

s 2 4

Sol–8. (d)

Closed loop poles are roots of characteristicequation 1 + G(s)H(s) = 0or, 1 + G(s) = 0 {as H(s) = 1}

or, 211 +

(s + 2) = 0

or, (s + 2)2 + 1 = 0or, s2 + 4s + 5 = 0

or, s = –4 ± 16 – 4 ×1×52

= –4 ± –4 = (–2 ± j)2

Sol–9. (a)Noise is proportional to bandwidth as

P = K.T(B)Where B = BandwidthIt is clear that as Bandwidth increases, noisealso increases.

Sol–10. (a)

G(s) =2s + 5

(s +5)(s + 4)

=

25 1 + s5

s s5 × 4 1 + 1 +5 4

To find DC gain – first arrange system intime constant form and then puts = 0 as for DC, f = 0 = 0 s = j =0

|G(s)|s = 0=5(1 + 0) = 0.25

5 ×4(1 + 0)(1 + 0)Sol–11. (c)

TKs =

T / T K T= .K / K T K

= 2K(3 + 4K) (0 + 2)(3 + 4K) – (1 + 2K)(0 + 4)×

1 + 2K (3 + 4K)

= 2K(3 + 4K) 6 + 8K – 4 – 8K×(1 + 2K) (3 + 4K)

= 2K

(1 + 2K) 3 + 4K)

= 22K

8K +10K + 3Sol–12. (b)

Transfer Function =1 1 1

(s 1) s s(s 1)

Impulse response = –1 1Ls(s 1)

= –1 1 1L –s (s 1)

= –t(1 – e )u(t)Sol–13. (b)

dx40. + 2xdt = f(t)

Applying Laplace Transform on both sides40sX(s) + 2 X(s) = F(s)