ELECTRONICS AND COMMUNICATION ENGINEERING
ESE TOPICWISE OBJECTIVE SOLVED PAPER-II
1991-2019
Office: F-126, (Lower Basement), Katwaria Sarai, New Delhi-110 016 Phone: 011-26522064
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IES MASTER PUBLICATION F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 Phone : 011-26522064, Mobile : 8130909220, 9711853908 E-mail : info.publications@iesmaster.org Web : iesmasterpublications.com
All rights reserved. Copyright Š 2019, by IES MASTER Publication. No part of this booklet may be reproduced, or distributed 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 of IES MASTER Publication, New Delhi. Violates are liable to be legally prosecuted.
First Edition
: 2017
Second Edition : 2018 Third Edition
: 2019
Typeset at : IES Master Publication, New Delhi-110016
PREFACE
Engineering Services Examination is the gateway to an immensely satisfying job in the engineering sector of India that offers multi-faceted exposure. The exposure to challenges and opportunities of leading the diverse field of engineering has been the main reason behind engineering students opting for Engineering Services as compared to other career options. To facilitate selection into these services, availability of arithmetic 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 Engineering Services Examination (ESE). This book is an outcome of regular and detailed interaction with the students preparing for ESE every year. It includes solutions along with detailed explanation to all questions. The prime objective of bringing out this book is to provide explanation to each question in such a manner that just by going through the solutions, students will be able to understand the basic concepts and have the 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 Publication New Delhi
Note: Direction of all Assertion Reasoning (A–R) type of questions covered in this booklet is as follows: DIRECTIONS: The following four items consist of two statements, one labelled as ‘Assertion A’ and the other labelled as ‘Reason R’. You are to examine these two statements carefully and select the answer to these two statements carefully and 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 of A (b) Both A and R are individually true but R is not the correct explanation of 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 covered in 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 statements carefully and select the answers to these items using the code given below: (a) Both Statement : (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I). (b) Both Statement (I) and Statement (II) are individually true but Statement (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.
CONTENT
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
UNIT-1
Control System
SYLLABUS Signal flow graphs, Routh-Hurwiz criteria, root loci, Nyquist/Bode plots Feedback systems-open & close loop types, 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
1
BASICS OF CONTROL SYSTEM, BLOCK DIAGRAM AND SIGNAL FLOW GRAPH IES – 2019
1.
2.
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
IES – 2018 5.
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
Consider the following open-loop transfer function : G =
Consider the following statements for signal flow graph:
(b) 2 only
K s 2
(c) Both 1 and 2
s 1s 4
The characteristic equation of the unity negative feedback will be
(d) Neither 1 nor 2 6.
(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
The closed-loop transfer function C(s) R(s) of the system represented by the block diagram in the figure is R(s)
(d) (s + 2) (s + 4) + K (s + 1) = 0 3.
Consider the following statements regarding a parabolic function :
(a)
(b) 1 and 3 only
(c) 2 and 3 only
(d) 1, 2 and 3
The price for improvement in sensitivity by the use 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
1 (s +1)2
–
(b)
(c) s+1
t2 , for t 0 f t 2 0, otherwise
(a) 1 and 2 only
+
1 s+1
2. A unit parabolic function is defined as
3. Laplace transform of unit parabolic function 1 is 3 . s Which of the above statements are correct?
C(s)
1 s+1
–
1. A parabolic function is one degree faster than the ramp function.
4.
+
1 (s +1)
(d) 1
IES – 2015 7.
The Laplace transform of e–2t sin 2 t is (a) (c)
2s (s 2)2 22 2 (s 2)2 42
(b)
2 (s 2)2 42
(d)
2s (s 2)2 22
IES – 2013 8.
The open-loop transfer function of a unity feedback control system is G s =
1
s + 2 2
The closed loop transfer function poles are located at:
E&T Engineering
CONTROL SYSTEM
(a) –2, –2
(b) –2, –1
(c) –2, +2
(d) 2 j1
function of the system is: 1
(a) 9.
3
In control systems, excessive bandwidth is NOT employed because:
(b)
s s +1 2
s +12
(d)
1 s +1
s +1 s
(c)
(a) noise is proportional to bandwidth (b) it leads to low relative stability
1 2
15.
d(s)
(c) it leads to slower response
R(s)
(d) noise is proportional to the square of the bandwidth
2 3s 1
3
y(s)
IES – 2012 10.
11.
A system is described by the transfer function 2s + 5 G s = s + 5 s + 4 . The dc gain of the system is
The transfer function from d(s) to y(s) is: (a)
2 3s 7
(b)
2 3s 1
(a) 0.25
(b) 0.5
(c)
6 3s 7
(d)
2 3s 6
(c) 1
(d)
1 + 2K T= 3 + 4K with respect to the parameter K is given by (a)
K 3 + K2
1
(c) Two (d) A value between –1and +1 17.
1 s
s 1
(a) 1 + e –t
(b) 1 – e –t
(c) e –t
(d) –e–t
(b) 20
(c) 1/10
(d) 4
(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.
40
19.
IES – 2010 14.
What is the characteristic of a good control system ? (a) Sensitive to parameter variation
C(s)
Given the differential equation smodel of a physical system, determine the time constant of the system dx + 2x = f t dt (a) 10
s2
(b) One
What is the unit impulse response of the system shown in figure for t 0 ? R(s)
where F(s) + 2 is the Laplace transform of f(t). What is the steady state value of f(t)?
Consider the function F(s) =
(a) Zero
4K (d) 2 + 5K + 7K 2
IES – 2011
13.
16.
3K 2 + 4K + K 2
(b)
2K (c) 3 +10K + 8K 2
12.
IES – 2009
The sensitivity STK of transfer function
A linear time-invariant system is initially at rest, when subjected to a unit-step input, gives a response y t = te –t ,t > 0 . The transfer
A negative-feedback closed-loop system is supplied with an input of 5V. The system has a 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
In closed loop control system, what is the sensitivity of the gain of the overall system, M to the variation in G ? 1 (a) 1 + G H s s
1 (b) 1 + G s
IES MASTER Publication
E&T Engineering
9
CONTROL SYSTEM
ANSWER KEY 1
(a)
12 (b)
23 (d)
34 (a)
45 (d)
2
(a)
13 (b)
24 (c)
35 (a)
46 (c)
3
(c)
14 (c)
25 (b)
36 (b)
47 (d)
4
(a)
15 (a)
26 (c)
37 (a)
48 (a)
5
(b)
16 (d)
27 (b)
38 (a)
49 (b)
6
(b)
17 (d)
28 (b)
39 (b)
50 (b)
7
(c)
18 (d)
29 (d)
40 (c)
51 (a)
8
(d)
19 (a)
30 (b)
41 (b)
52 (a)
9
(a)
20 (c)
31 (b)
42 (a)
53 (a)
10 (a)
21 (c)
32 (b)
43 (d)
11
22 (a)
33 (c)
44 (d)
(c)
EXPLANATIONS Sol–1:
(a)
2
t u(t) 2 Parabolic function grows faster than ramp function by 1 degree
Unit parabolic function, f(t) =
A minimum phase transfer function has poles & zeros on left half of s-plane. A maximum phase transfer function has poles & zeros on right half of s-plane.
Sol–2:
(1) is true.
An all pass transfer function has poles & zeros symmetrically located with respect to j axis.
(2) is true
A complex transfer function has complex poles & zeros, not necessarily on a particular half.
true. Hence all statements are true.
(3) Laplace transform of
Sol–4:
(a)
K(s 2) (s 1)(s 4)
with unity feedback, H(s) = 1 Characteristic equation : 1 + GH = 0
Sol–3:
K(s 2) 1 (s 1)(s 4) = 0 (s+1) (s+4) + K (s + 2) = 0 (c) (1) Unit Ramp function, r(t) = tu(t)
(a) With the use of negative feedback, sensitivity of system is improved but gain is decreased.
Given open loop transfer function G =
t2 1 u(t) 3 is 2 s
Sol–5:
(b) Signal flow graph is applicable to linear systems and not applicable to non linear systems. Therefore, statement 1 is false One system can have different signal flow graph according to the order in which the equations are used to define the variable written on the left hand side. So it is not unique for a given system. Statement-2 is correct.
IES MASTER Publication
10 Sol–6:
E&T Engineering
ESE Topicwise Objective Solved Paper-II 1991-2019 (b)
Sol–9.
1 s +1
R(s) –
1 s +1
C(s) –
+
(a) Noise is proportional to bandwidth as P = K.T(B) Where
B = Bandwidth
It is clear that as Bandwidth increases, noise also increases. Sol–10. (a)
R(s) –
+
1 s +1
C(s)
1 s +1
C(s)
1 s +1
R(s) –
s +1 s
G(s) =
2 5 1 + s 5 = s s 5 × 4 1 + 1 + 5 4 To find DC gain – first arrange system in time constant form and then put
s = 0 as for DC, f = 0 = 0 s = j = 0 |G(s)|s = 0=
1 s
R(s) –
C(s)
5(1 + 0) = 0.25 5 × 4(1 + 0)(1 + 0)
Sol–11. (c)
s TK =
1 C s = = 1 s +1 R s 1+ s
K(3 + 4K) 6 + 8K – 4 – 8K = (1 + 2K) × (3 + 4K)2 2K = (1 + 2K) 3 + 4K)
(c) L e
–2t
sin 2t = ?
2 2 L sin 2t = 2 2 s 42 s 2 2 2 L e –2t sin 2t = s 2 4 2
=
2
Sol–8.
(d) Closed loop poles are roots of characteristic equation 1 + G(s)H(s) = 0 or, 1 + G(s) = 0 {as H(s) = 1} or, 1 +
s=
–4 ± 16 – 4 ×1 × 5 2
=
–4 ± –4 = (–2 ± j) 2
2K 8K 2 +10K + 3
Sol–12. (b) Transfer Function =
1 1 1 (s 1) s s(s 1)
1 –1 Impulse response = L s(s 1) 1 –1 1 = L – s (s 1)
1
= 0 (s + 2)2 or, (s + 2)2 + 1 = 0 or, s2 + 4s + 5 = 0 or,
T / T K T = . K / K T K
K(3 + 4K) (0 + 2)(3 + 4K) – (1 + 2K)(0 + 4) = 1 + 2K × (3 + 4K)2
1 s
Sol–7.
2s + 5 (s + 5)(s + 4)
= (1 – e–t )u(t) Sol–13. (b) dx + 2x = f(t) dt Applying Laplace Transform on both sides 40.
40sX(s) + 2 X(s) = F(s)