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0% found this document useful (0 votes)
56 views2 pages

May 24

Hhhhhhhhhvvvvhbhhhh

Uploaded by

Akanksha
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 2

Y1

X5
A0

5
0

8X
3A
Paper / Subject Code: 51424 / Principle of Communication

25
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A0
5Y
June 12, 2024 02:30 pm - 05:30 pm 1T01233 - S.E.(Information Technology Engineering)(SEM-III)

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(Choice Base Credit Grading System ) (R- 19) (C Scheme) / 51424 - Principle of Communication

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QP CODE: 10056409

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Time: 3 Hours Marks: 80

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N.B. (1). Question No.1 is compulsory.

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(2). Out of remaining attempt any three.

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3A
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2

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0
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(3). Assume & mention suitable data wherever required.

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5Y

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(4). Figures to right indicates full marks.

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5Y
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00
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1
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Q.1. Solve any four [20]

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a). The signal power & noise power measured at the input of an amplifier are 150 µw & 1.5 µw

08
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X5

1
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respectively. If the signal power at the o/p is 1.5w and noise power is 40mw, calculate amplifier
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13

25
3

3
2
0
1

0
5

1
0

5
5Y

X5
5Y

A0
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5Y
A

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A
noise factor & noise figure.
08
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1
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5Y

0
8X

A0
8X

5Y
A

8X
3A
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52

13
00

b). Calculate the percentage power saving for DSB-SC signal for percentage modulation

2
00
5

Y1

5
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3A

8X

5Y
3A

X
3A
5

25

8
2
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1

0
X5

Y1
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5
A0
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A0
3A

of a) 100 % b) 50 %
X

8X
25

25
8

25
8
3

13
0
Y1

00
X5

00
X5
0

X5
5Y

5Y
3A

A
25

3A
08

c). Compare PAM, PWM & PPM


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0

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0
5Y
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8X
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5

Y1
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5
5Y

d). State advantages of digital transmission.


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3A

e). Explain in brief different types of communication channels.


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5Y
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3A

5Y
3A

3A
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Y1

f). Explain the principle of reflection and refraction.


52
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A0
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3A

8X
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25
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Y1
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X5

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5
A0
Y
3A

X
25

3A
08

25
08

5
8
3

52
Y1

0
X5

1
A0

Y1
A0

0
5Y

Q.2 a) Explain FDM with neat block diagram . [10]


X

8X
A
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3
2

00
X5

Y1
A0

5
A0

X5
5Y

8X

3A
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5
13

b). State and prove the following properties of Fourier transform with example
13

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52

52
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A0
5Y

5Y

A0
8X

8X
A

25
3
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i) Convolution in time domain ii) Time scaling [10]


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1
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A0
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Q.3. a) In an AM radio receiver, loaded Q of an antenna circuit at the input to the mixer
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5Y

5Y
X

3A
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13

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52

Y1
A0
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0
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Is 100.if the intermediate frequency is 455 KHz. calculate the image frequency &
8X
A

25
13

13
52
0

00
A0

X5
5Y

5Y
8X

3A

Its rejection at 1 MHz . [10]


13

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5Y

A0
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8X
3A

25
52

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Y1

X5

b).With the help of neat circuit diagram explain varactor diode method of FM Generation
8X

5Y
3A

3A
25

08
00

52
Y1
X5

[10]
A0
3A

8X
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3
Y1

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3A
25

56409
25

Page 1 of 2
08
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X5

Y1
X5
A0
5Y
08

25
08
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52

X5
A0
5Y
8X

X525Y13A008X525Y13A008X525Y13A008X525Y13A008
08
13
52
00
08 3A 5Y 8X A0
X5 00 13 52 08
25 8X A0 5Y X5
00 Y1 52 08 13 25
8X 3A 5Y X5 A0 Y1
52 00 13 25 08 3A
5Y 8X A0 Y1 X5 00
13 52 08 3A 25 8X
5Y Y1

Q6.
Q.5
Q.4
A0 X5 00 52
52 08 13 25 8X 3A 5Y
5Y A0 00

56409
X5 Y1 52 13
13 25 08 3A 5Y 8X A0
A0 Y1 X5 00 13 52 08
08 3A 25 8X A0 5Y X5
X5 00 Y1 52 08 13 25
13 25 8X 3A 5Y X5 A0 Y1

(4) TDM
A0 Y1 52 00 13 25 08 3A
08 3A 5Y 8X A0 Y1 X5 00
X5 00 13 52 08 3A 25 8

waveforms
25 8X A0 5Y X5 00 Y1 X5
Y1 52 08 13 25 8X 3A 25
08 3A 5Y X5 A0 Y1 00 Y1
X5 00 13 08 3A 52 8 3A
25 5Y
(v) Skip Zone
25 8X A0 Y1 X5 00 X5 00
Y1 52 8X 13 25
(i) virtual height

3A 08 3A 25 8X A0 Y1
5Y

3. Delta modulation
X5 52 00 Y1 52
00 1 3 2 8 3 A 5 X 3A 5Y 08

1. Need of modulation
8X A0 5 Y X5 0 Y1 5 00 13
52 0 13 2 0 8 3A 2 5 8X

(1) Aliasing or fold over error


5

4. Friss Formula of noise.


5Y 8 A0 X 5 0 Y 1 52
A0
13
X5
2 08
Y1
3 25 0 8 3 A 5Y 08 a) define/Explain the following
A X

2. Role of balance modulator


5 X5 0 Y1 5 00 1
X5
A0
08
Y1
3A 25 0 8X 3A 2 5Y 8X 3 A0 25
Y1
0 Y 5 0 1 5 2 0
.

X5
08 1 25 0 3 5Y 8 3A

.
25 3A 8X A0 X5 00
Y1 X5 0 0
Y1 5 2 08 1 3 2 5
2 8 3 5 A Y 8X

1. Unipolar RZ 2. Polar RZ 3. AMI

b) . Write short note on following (any two)


3A 5Y X A0 Y 1
X5 00 1 52
00 1 52 0 8 3 25 8 3A 5Y

Page 2 of 2
3 A X
(iii) maximum usable frequency (MUF)

8X A0 5Y X5 0 Y1 5 00 13
52 0 13 2 08 3 A 2 5 8X
8 5 0 Y1 5
A0
a). With reference to sky wave propagation explain

5Y X5 A0 Y1 X5
0 2 08
13 2 5 08 3 A 25 8 X 3A 5Y X5
A0 Y1 X5 0 08
Y1 5 2 00
8X 13 25
08 25 3A
(5) Inter symbol interference (ISI)

3A X 5Y A0 Y1
X5 0 Y 1 52 0 0 13 5 2 08 3A
25 0 3 5Y 8 5 X 0
8X
5
A0 X5 A0 Y1
3 5
(2) Slope overload error

Y1
3A 2 0 8 13 2 5 08 A 25
Y1
5Y A0 Y1 X5 00
b). Derive the mathematical expression for FM with neat sketch.

X5
00 1 3 2 5Y 0 8 3 A 25 8X 3A
8X A0 X5 0 Y1
3 5 00
0 13 2 08 25 8X

X525Y13A008X525Y13A008X525Y13A008X525Y13A008
Draw the resulting waveform if the sequence is transmitted using
52 8 5 A Y

******************************************
(iv) skip distance

5Y X5 0 1 5

4. Split phase Manchester 5. M-ary where M=4 (Polar quaternary)


X5 A0 Y1 0
13 2 5Y 0 8 3 A 25 8 X 3A
X5 Y1 00
(ii) critical Frequency

A0 52
Paper / Subject Code: 51424 / Principle of Communication

2 00 3
08 13 5 8 A 5 Y 8X
X5 A0 Y1 X5 0 0 1 52
25 08 3 A 25 8 X 3A 5Y
Y1 X5 0 Y1 5 0 0 1
3A 25 08
X5 3A 25
Y 8X
00 Y1 00 13 52
8 A
[10]

25 X 5Y
[10]

3A
[10]
[10]

8X 5 00
a). Consider that bit sequence given below is to be transmitted Bit sequence =10110011.

Y1 13
52 00 2 [10]
b). Draw the block diagram of BSK generation & detection explain the working giving

5Y 8X 3A 5Y 8X A0
52 00 1 52
(3) quantization process

13 5Y 8X 3A 5Y
A0
08 1 3A 5 2 00
8X 13
5Y A0
X5 0 0 1 52 08
25 8X 3A 5Y X5
Y1 5 2 00
8X 1 3
3A 5Y A0
00 1 5 2 08
8X 3A 5Y X

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