ESE 2020 - Electronic and Communication Engineering ESE Topicwise Conventional Solved Paper 2 https://iesmasterpublications.com/ese-electronic-and-communication-engineering-ese-topicwise-conventional-solved-paper-2

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

slide 5:

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

slide 6:

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
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

slide 7:

Q–1: 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 :
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
Sol. Given :
The voltage gain of first stage A
1
12
The voltage gain of second stage A
2
20
Input resistance for first stage R
i1
500
Input resistance for second stage R
i2
10 K
Now R
1
R
i1
+ R
n1
Noise resistance is in series with input resistance
R
1
1500 + 500
R
1
2000
Also R
2
01 i2 n2
R R R
output resistance of first stage is parallel to input resistance of 2
nd
stage + equivalent noise resistance
of 2
nd
stage
R
2
10K 25K 80K
R
2
25 80
K 10
105
R
2
29.04 K
Again
3 02
R R 1M 1000K
Now equivalent input noise resistance is given as
R
eq
3 2
1
2 2 2
1 1 2
R R
R
A A A
3 6
2 2 2
29.04 10 10
2000
20 12 12
1
Random Variables and
Noise
Chapter

slide 8:

ESE Topicwise Conventional Solved Paper-II 3
IES MASTER Publication
Electronics Communication Engineering
29040 1000
2000
144 144 4
R
eq
2219.1
Noise figure given by
F
1
eq
a
R
1
R
...1
When
1
eq eq i1
R R R
1719.1
2219.1 500
R
a
output resistance of generator 50
Hence from 1
F
1719.1
1 35.38
50
F 35.38 or 15.48 dB F in dB 10log
10
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
ht S
T t
T Time period of st
Noise power at output of the filter is given as
P
n
2
df H
f
2
where
2
Noise power spectral density at output filter. .
Now
2
0
S
T
2
2 fT
i
H S .e df
f f
Signal to noise ratio at the output is given as
SNR
o/p
2
j2 fT
i
2
H S e df
f f
df H
f
2
2
i
df E Energy of the signal
S
f
o/p
SNR
2E
o/p
max
SNR
2E
...i
Now the input signal to noise ratio is given as
S
i
T
2
i
0
1 E
S dt
t
T T
N
i
.B 2B
2
B Noise bandwidth

slide 9:

Analog and Digital Communication System 4
SNR
i
i
i
E S
BT N
...ii
Now from i and ii
o/p
i/p
SNR
SNR
2E BT
2BT
E
o/p
i/p
SNR
SNR
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
n
dB
4dB F
Bandwidth
n
B 5000 kHz
and
0
SNR
0
0
S
1
N
Amp
R 50
S
+
–
V
S
V
i
R
in
R
in
50
Let S
i
be the input signal power and N
i
be input noise power then
S
i
2
i
in
V
R
or S
i
2
in
2
s
in s in
R 1
V
R R R
2
2
s
1 50
.V
50 50 50
2
s
V
200
V
s
i
S 200 ...i
Since
Noise figure
in i
n
out
SNR SNR
F
1
SNR
i
i
i
S
SNR
N
S
i
n i
F N
n 0 n
F KT B ...ii
Where K Boltzman’s constant
23
1.38 10
B
n
Receiver bandwidth
3
500 10 Hz
T
o
Temperature 290 K
n
dB
F
10 n
10log F
or 4
10 n
10log F
or F
n
0.4
2.512
10
Hence from eq. ii
S
i
23 3
2.512 1.38 10 290 500 10

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