ELECTRONICS & COMMUNICATION ENGINEERING ESE TOPICWISE CONVENTIONAL SOLVED PAPER-II
1998-2019
Office: F-126, (Lower Basement), Katwaria Sarai, New Delhi-110 016 Phone: 011-2652 2064
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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
:
2018
Second Edition
:
2019
Typeset at : IES Master Publication, New Delhi-110016
PREFACE
Engineering Services Exam (ESE) is one of most coveted exams written by engineering students aspiring for reputed posts in the various departments of the Government of India. ESE is conducted by the Union Public Services Commission (UPSC), and therefore the standards to clear this exam too are very high. To clear the ESE, a candidate needs to clear three stages - ESE Prelims, ESE Mains and Personality Test. It is not mere hard work that helps a student succeed in an examination like ESE that witnesses lakhs of aspirants competing neck to neck to move one step closer to their dream job. It is hard work along with smart work that allows an ESE aspirant to fulfil his dream. After detailed interaction with students preparing for ESE, IES Master has come up with this book which is a one-stop solution for engineering students aspiring to crack this most prestigious engineering exam. The book includes previous years' solved conventional questions segregated subject-wise along with detailed explanation. This book will also help ESE aspirants get an idea about the pattern and weightage of questions asked in ESE. IES Master feels immense pride in bringing out this book with utmost care to build upon the exam preparedness of a student up to the UPSC standards. The credit for flawless preparation of this book goes to the entire team of IES Master Publication. Teachers, students, and professional engineers are welcome to share their suggestions to make this book more valuable.
IES Master Publication New Delhi
CONTENTS
1.
ANALOG & DIGITAL COMMUNICATION SYSTEM ................................. 01 – 71
2.
CONTROL SYSTEM .............................................................................. 72 – 185
3.
COMPUTER ORGANIZATION & ARCHITECTURE ............................... 186 – 214
4.
ELECTROMAGNETIC FIELD THEORY ................................................ 215 – 301
5.
ADVANCE ELECTRONICS ................................................................... 302 – 315
6.
ADVANCE COMMUNICATION SYSTEM .............................................. 316 – 366
7.
SIGNAL & SYSTEM ............................................................................ 367 – 425
8.
MICROPROCESSOR ........................................................................... 426 – 460
UNIT 1
ANALOG AND DIGITAL COMMUNICATION SYSTEM SYLLABUS
Random signals, noise, probability theory, information theory; Analog versus digital communication & applications; Systems-AM, FM, transmitters/receivers, theory/practice/standards, SNR comparison, Digital communication basics: Sampling, qunatizing, coding, PCM, DPCM, multiplexing-audio/video; Digital modulation: ASK, FSK, PSK: Multiple access: TDMA, FDMA, CDMA; Optical communication: fibre optics, theory, practice/standards.
CONTENTS
1.
Random Variables and Noise ..................................................................... 02-12
2.
Analog Communication System .................................................................. 13-29
3.
Digital Communication System ................................................................... 30-55
4.
Fundamentals of Information Theory .......................................................... 56-71
Chapter
1 Q–1:
Random Variables and Noise
A two stage amplifier has the following parameters : First stage
Second stage
Voltage gain
12
20
Input resistance
500 ohms
80 K ohms
Equivalent Noise Resistance
1500 ohms
10 K ohms
Output Resistance
25 K ohms
1 M ohms
Calculate :
Sol.
(i)
the equivalent noise resistance of the two stage amplifier;
(ii)
the noise figure of the amplifier if it is driven by a generator with output impedance 50 ohms. [15 Marks ESE–1998]
Given : The voltage gain of first stage = A1 = 12 The voltage gain of second stage = A2 = 20 Input resistance for first stage, Ri1 = 500 Input resistance for second stage, Ri2 = 10 K Now,
R1 = Ri1 + Rn1
[Noise resistance is in series with input resistance] R1 = 1500 + 500 R1 = 2000 Also,
R2 =
R01 Ri2 Rn2
[output resistance of first stage is parallel to input resistance of 2nd stage] + [equivalent noise resistance of 2nd stage] R2 =
25K 80K 10K
25 80 10 K R2 = 105 R2 = 29.04 K Again, R 3 R02 1M 1000K Now, equivalent input noise resistance is given as Req = R1
R2 R 23 2 2 A 1 A1 A 2
= 2000
29.04 103 10 6 2 2 12 12 20 2
Electronics & Communication Engineering
ESE Topicwise Conventional Solved Paper-II = 2000
3
29040 1000 144 144 4
Req = 2219.1 Noise figure given by, R1eq F = 1 Ra
...(1)
When R1eq R eq Ri1 = 2219.1 500 1719.1
Ra = output resistance of generator = 50 Hence from (1), F = 1
1719.1 35.38 50
F = 35.38 or 15.48 dB
[F in dB = 10log10 F]
Q–2:
Show that the input-to-output SNR gain of a matched filter depends on the product of the input signal duration and the noise bandwidth. [10 Marks ESE–2002]
Sol.
Impulse response of matched filter is h(t) = S * T t T = Time period of s(t) Noise power at output of the filter is given as Pn = where
= Noise power spectral density at output filter.. 2
Now
2 H f df 2
S0 T
2
=
2
H f Si f .e 2fT df
Signal to noise ratio at the output is given as 2
(SNR)o/p =
H f Si f e j2fT df 2 H f df 2
Si f 2 df E Energy of the signal SNR o/p 2E
SNR o/p
max
2E =
...(i)
Now the input signal to noise ratio is given as Si =
1 T 2 E S t dt T 0 i T
Ni = 2B .B 2
[B = Noise bandwidth] IES MASTER Publication
4
Analog and Digital Communication System Si E (SNR)i = N BT i
...(ii)
Now from (i) and (ii)
SNR o/p SNR i/p
=
2E BT 2BT E
SNR o/p
SNR i/p
= 2BT
Q–3:
An amplifier has a noise figure of 4 dB, a bandwidth of 500 kHz and an input resistance of 50 . Calculate the input signal voltage needed to yield an output SNR = 1 when the amplifier is connected to a signal source of 50 at 290 K. [8 Marks ESE–2006]
Sol.
Given that for amplifier Noise figure =
Fn dB
4dB
RS = 50 +
Bandwidth = Bn 5000 kHz
SNR 0
and,
VS
S0 = N 1 0
Vi –
Amp R in
Rin = 50 Let Si be the input signal power and Ni be input noise power then, Si =
Vi 2 Rin 2
1 Rin 2 Si = Vs Rin Rin R s
or
2
= Vs =
Vs2 1 50 2 .Vs = 50 50 50 200 Si 200
...(i)
Since
SNR in SNR i Si Noise figure = Fn = SNR i N 1 SNR out i Si = Fn Ni
= Fn KT0Bn Where K = Boltzman’s constant = 1.38 1023 Bn = Receiver bandwidth = 500 10 3 Hz To = Temperature = 290 K
or or
Fn dB
= 10log10 Fn
4 = 10log10 Fn Fn = 10 0.4 2.512
Hence from eq. (ii) Si = 2.512 1.38 1023 290 500 10 3
...(ii)