DR.

BABASAHEB AMBEDKAR TECHNOLOGICAL UNIVERSITY, LONERE

Regular End Semester Examination – Winter 2023
Course: B. Tech. Branch : E & T C Semester : V
Subject Code & Name: BTETC503 - Analog Communication Engineering
Max Marks: 60 Date: Duration: 3.45 Hr.
Instructions to the Students:
1. All the questions are compulsory.
2. The level of question/expected answer as per OBE or the Course Outcome (CO) on
which the question is based is mentioned in ( ) in front of the question.
3. Use of non-programmable scientific calculators is allowed.
4. Assume suitable data wherever necessary and mention it clearly.
(Level/CO) Marks
Q. 1 Answer Any Two of the following.
A) Define Communication and Explain the complete Analog communication CO 01 6
system.
B) What is modulation? Give the classification Modulation with definition CO 01 6
C) Write the difference between i)FDM and TDM ii) Analog communication CO 01,07 6
and Digital communication

Q.2 Answer Any Two of the following.

A) Explain DSBFC Technique in detail with the help of (i) frequency spectrum CO 02 6
(ii) Time domain representation (iii) power relation with carrier wave
B) Calculate the percentage power saving when the carrier and one of the CO 02 6
sidebands are suppressed in an AM wave modulated to the depth of (i) 70%
and (ii) 65%
C) Explain generation of AM using Nonlinear resistance device CO 03 6

Q. 3 Answer/Solve Any Two of the following.

A) What is Angle modulation? Derive the mathematical expression for PM. CO 05 6
B) Explain Narrowband and Wideband FM Also compare the FM and AM. CO 05 6
C) In an FM system, when the audio frequency (AF) is 300 Hz, and the AF CO 05 6
voltage is 1.2 V, the deviation is 2.4 kHz. If the AF voltage is now increased
to 3.6 V, what is the new deviation? If the AF voltage is further raised to 10
V while the AF is dropped to 150 Hz, what is the deviation? Find the modu-
lation index in each case.

Q.4 Answer the following.

A) Define: CO 07 6
(i) Sensitivity, (ii) Selectivity, (iii) Fidelity, (iv) Image frequency and its
rejection.
B) Explain working of Superheterodyne receiver in detail. CO 04 6
Q. 5 Solve the following.
A) Two resistors 20 kΩ and 50 kΩ are at room temperature (290 K). Calculate CO 04 6
for bandwidth of 100 KHz, the thermal noise for the following conditions:
(i) For resistor 20 kΩ
(ii) For resistor 50 kΩ
(iii) For two resistors in series
(iv) For two resistors in parallel.
B) An amplifier operating over the frequency range from 3 to 10MHz has a CO 04, 06 6
20K input resistance. What is the rms noise voltage at the input to this
amplifier at room temperature?
*** End ***
 D
47 7D3 51C C36 6413 354 1E3 38D 94 210 076F
5 1 2
7D D35 1C3 364 4135 541 E38 8D9 942 107 76F CA
1 1 E
3 C 6 3 4 3 D 42 0 6F CA 10 1
35 51C 364 4135 541 1E38 8D9 942 107 76F CA 10D D47
3 1 C
51 1C3 641 1354 41E E38D D94 421 076 6FC A1 10D 47D D35
C3 64 35 1E 38 9 21 07 FC A1 0D 47 3 1C
C3 64 135 41E 38 D9 421 07 6FC A1 0D 47 D3 51C 36
1 4 D 4 6 D 5 4
36 6413 354 1E3 38D 94 210 076F FCA A10 0D4 47D 351 1C3 3641 135
4 1 2 7 7 3 6
64 135 541 E38 8D9 942 107 6F CA 10D D47 D3 51C C36 413 354 41E3
1 4 E D 4 10 6 C 10 4 D 5 3 4 5 1 8

Q.3.
13 3541 1E3 38D 942 210 76F FCA A10 D4 7D3 351 1C3 641 1354 41E E38D D94
9 7 D 7 6
35 541E E38 8D9 421 107 6FC CA1 10D 47 D35 51C C36 413 3541 1E3 38D 942 210
41 38 D9 42 07 6F A 0D 47 D3 1C 36 41 54 E 8D 94 10 76
41 E38 D9 42 107 6F CA 10D 47 D3 51C 36 41 354 1E 38D 94 210 76 FC
E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E 38 94 21 76 FC A1
8D 94 10 76 CA 10 D4 7D 351 C3 64 35 41E 38 D9 21 07 FC A1 0D
94 210 76 FC 10 D4 7D 35 C3 64 135 41 38 D9 421 07 6FC A 0D 47
21 76 FC A1 D 7D 35 1C 64 13 41 E3 D 42 07 6F A 10 47 D3
07 FC A 0D 47 3 1C 36 13 54 E3 8D 94 10 6F CA 10 D4 D 51
6F A 10 47 D3 51 36 41 54 1E 8D 94 21 76 C 1 D 7D 35 C
CA 10 D4 D 51 C3 41 35 1E 38 9 21 076 FC A1 0D 47D 35 1C 364

Date:- 16/05/2019
10 D4 7D 351 C3 641 35 41E 38 D94 421 076 FC A1 0D 47D 35 1C 364 13
D4 7D 35 C3 64 35 41 38 D9 21 07 FC A 0D 47 3 1C 36 13 54
7D 35 1C3 64 135 41 E38 D9 42 07 6F A 10D 47 D3 51C 36 413 54 1E3

frequency.
35 1C 64 135 41 E38 D9 42 107 6F CA 10D 47 D3 51C 36 41 54 1E 8D

Instructions to the Students

1C 36 13 41 E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E 38 94

transmitter.
36 413 54 E3 8D 94 10 76F CA 10 D4 7D 351 C3 641 35 41E 38 D9 21
41 54 1E 8D 94 21 76 C 1 D 7D 35 C 64 35 41 38 D9 42 07
35 1E 38 9 21 07 FC A1 0D 47 35 1C 36 13 41 E3 D 42 10 6F
41 38 D9 42 07 6F A 0D 47 D3 1C 36 41 54 E 8D 94 10 76 C
E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E 38 94 21 76 FC A1
8D 94 10 76 CA 10 D4 7D 351 C3 64 35 41E 38 D9 21 07 FC A1 0D
94 210 76 FC 10 D4 7D 35 C3 64 135 41 38 D9 421 07 6FC A 0D 47
21 76 FC A1 D 7D 35 1C 64 13 41 E3 D 42 07 6F A 10 47 D3
07 FC A 0D 47 3 1C 36 13 54 E3 8D 94 10 6F CA 10 D4 D 51
6F A 10 47 D3 51 36 41 54 1E 8D 94 21 76 C 1 D 7D 35 C 1. Each question carries 12 marks.
CA 10 D4 D 51 C3 41 35 1E 38 9 21 076 FC A1 0D 47D 35 1C 364
10 D4 7D 351 C3 641 35 41E 38 D94 421 076 FC A1 0D 47D 35 1C 364 13

other AM Techniques
D4 7D 35 C3 64 35 41 38 D9 21 07 FC A 0D 47 3 1C 36 13 54

assume it and should mention it clearly

7D 35 1C3 64 135 41 E38 D9 42 07 6F A 10D 47 D3 51C 36 413 54 1E3

basic reactance modulator.

35 1C 64 135 41 E38 D9 42 107 6F CA 10D 47 D3 51C 36 41 54 1E 8D
1C 36 13 41 E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E 38 94
36 413 54 E3 8D 94 10 76F CA 10 D4 7D 351 C3 641 35 41E 38 D9 21
2. Attempt any five questions of the following.

41 54 1E 8D 94 21 76 C 1 D 7D 35 C 64 35 41 38 D9 42 07
[BTEXC402]

35 1E 38 9 21 07 FC A1 0D 47 35 1C 36 13 41 E3 D 42 10 6F

Answer any two of the following

undefined

41 38 D9 42 07 6F A 0D 47 D3 1C 36 41 54 E 8D 94 10 76 C

OR
E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E 38 94 21 76 FC A1
8D 94 10 76 CA 10 D4 7D 351 C3 64 35 41E 38 D9 21 07 FC A1 0D
94 210 76 FC 10 D4 7D 35 C3 64 135 41 38 D9 421 07 6FC A 0D 47
21 76 FC A1 D 7D 35 1C 64 13 41 E3 D 42 07 6F A 10 47 D3
07 FC A 0D 47 3 1C 36 13 54 E3 8D 94 10 6F CA 10 D4 D 51
RAIGAD -402 103

6F A 10 47 D3 51 36 41 54 1E 8D 94 21 76 C 1 D 7D 35 C
CA 10 D4 D 51 C3 41 35 1E 38 9 21 076 FC A1 0D 47D 35 1C 364
10 D4 7D 351 C3 641 35 41E 38 D94 421 076 FC A1 0D 47D 35 1C 364 13

A10D47D351C36413541E38D9421076FC
D4 7D 35 C3 64 35 41 38 D9 21 07 FC A 0D 47 3 1C 36 13 54
7D 35 1C3 64 135 41 E38 D9 42 07 6F A 10D 47 D3 51C 36 413 54 1E3
35 1C 64 135 41 E38 D9 42 107 6F CA 10D 47 D3 51C 36 41 54 1E 8D
1C 36 13 41 E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E 38 94
36 413 54 E3 8D 94 10 76F CA 10 D4 7D 351 C3 641 35 41E 38 D9 21
Branch: Electronics and Telecommunication Engineering

41 54 1E 8D 94 21 76 C 1 D 7D 35 C 64 35 41 38 D9 42 07
Semester Examination – May - 2019

35 1E 38 9 21 07 FC A1 0D 47 35 1C 36 13 41 E3 D 42 10 6F
41 38 D9 42 07 6F A 0D 47 D3 1C 36 41 54 E 8D 94 10 76 C
E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E 38 94 21 76 FC A1
8D 94 10 76 CA 10 D4 7D 351 C3 64 35 41E 38 D9 21 07 FC A1 0D
4. If some part or parameter is noticed to be missing, you may appropriately

94 210 76 FC 10 D4 7D 35 C3 64 135 41 38 D9 421 07 6FC A 0D 47

21 76 FC A1 D 7D 35 1C 64 13 41 E3 D 42 07 6F A 10 47 D3
3. Illustrate your answers with neat sketches, diagram etc., wherever necessary.

07 FC A 0D 47 3 1C 36 13 54 E3 8D 94 10 6F CA 10 D4 D 51
6F A 10 47 D3 51 36 41 54 1E 8D 94 21 76 C 1 D 7D 35 C
CA 10 D4 D 51 C3 41 35 1E 38 9 21 076 FC A1 0D 47D 35 1C 36
10 D4 7D 351 C3 641 35 41E 38 D94 421 076 FC A1 0D 47D 35 1C 364
D4 7D 35 C3 64 35 41 38 D9 21 07 FC A 0D 47 3 1C 36

50 percent. Calculate the total power in modulated wave.

7D 35 1C3 64 135 41 E38 D9 42 07 6F A 10D 47 D3 51C 36 413
35 1C 64 135 41 E38 D9 42 107 6F CA 10D 47 D3 51C 36 41
1C 36 13 41 E3 D 42 10 6F CA 10 4 D 51 3 41 35
36 413 54 E3 8D 94 10 76F CA 10 D4 7D 351 C3 641 35 4

Assuming that a 750 W carrier is modulated to the depth of

now increased to 8V what is new deviation? If the AF voltage is

AF voltage is 2.6 V, the deviation is 5.2 kHz. If the AF voltage is
7500πt is to be sampled. Calculate the minimum sampling
Criteria? A continuous time signal x (t) = 7 sin 2500πt + 5 sin

Also explain generation of the same using Balanced Modulator.

41 54 1E 8D 94 21 76 C 1 D 7D 35 C 64 35 41
35 1E 38 9 21 07 FC A1 0D 47 35 1C 36 13 41
41 38 D9 42 07 6F A 0D 47 D3 1C 36 41 54 E
Sem.:- IV

E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E
DR. BABASAHEB AMBEDKAR TECHNOLOGICAL UNIVERSITY, LONERE –

Subject with Subject Code:- Analog Communication Engineering Marks: 60

8D 94 10 76 CA 10 D4 7D 351 C3 64 35 41E 38

B) In an FM system, when the audio frequency (AF) is 750 Hz, and (06)
A) Explain generation of Frequency modulation with the help of (06)
2. Explain VSB technique of AM, What are its advantages over (06)
B) 1. Derive an expression for power relation in an AM wave. (06)
Q.2. A) Explain double sideband suppressed carrier technique of AM. (12)
B) With the help of neat diagram explain a typical radio (06)
Q.1. A) State and explain Sampling Theorem. What is Nyquist (06)

94 210 76 FC 10 D4 7D 35 C3 64 135 41 38
============================================================

21 76 FC A1 D 7D 35 1C 64 13 41 E3 D
-------------------------------------------------------------------------------------------------------

_____________________________________________________________________
Time:- 3 Hr.

(Marks)

07 FC A 0D 47 3 1C 36 13 54 E3 8D
6F A 10 47 D3 51 36 41 54 1E 8D 94
CA 10 D4 D 51 C3 41 35 1E 38 9
10 D4 7D 351 C3 641 35 41E 38 D94 42
D4 7D 35 C3 64 35 41 38 D9 21
7D 35 1C 64 13 41 E3 D9 42 0
 D
47 7D3 51C C36 6413 354 1E3 38D 94 210 076F
5 1 2
7D D35 1C3 364 4135 541 E38 8D9 942 107 76F CA
1 1 E
3 C 6 3 4 3 D 42 0 6F CA 10 1
35 51C 364 4135 541 1E38 8D9 942 107 76F CA 10D D47
3 1 C
51 1C3 641 1354 41E E38D D94 421 076 6FC A1 10D 47D D35
C3 64 35 1E 38 9 21 07 FC A1 0D 47 3 1C
C3 64 135 41E 38 D9 421 07 6FC A1 0D 47 D3 51C 36
1 4 D 4 6 D 5 4
36 6413 354 1E3 38D 94 210 076F FCA A10 0D4 47D 351 1C3 3641 135
4 1 2 7 7 3 6
64 135 541 E38 8D9 942 107 6F CA 10D D47 D3 51C C36 413 354 41E3
1 4 E D 4 10 6 C 10 4 D 5 3 4 5 1 8

Q.6.
13 3541 1E3 38D 942 210 76F FCA A10 D4 7D3 351 1C3 641 1354 41E E38D D94
9 7 D 7 6
35 541E E38 8D9 421 107 6FC CA1 10D 47 D35 51C C36 413 3541 1E3 38D 942 210
41 38 D9 42 07 6F A 0D 47 D3 1C 36 41 54 E 8D 94 10 76
41 E38 D9 42 107 6F CA 10D 47 D3 51C 36 41 354 1E 38D 94 210 76 FC
E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E 38 94 21 76 FC A1
8D 94 10 76 CA 10 D4 7D 351 C3 64 35 41E 38 D9 21 07 FC A1 0D
94 210 76 FC 10 D4 7D 35 C3 64 135 41 38 D9 421 07 6FC A 0D 47
21 76 FC A1 D 7D 35 1C 64 13 41 E3 D 42 07 6F A 10 47 D3
07 FC A 0D 47 3 1C 36 13 54 E3 8D 94 10 6F CA 10 D4 D 51
6F A 10 47 D3 51 36 41 54 1E 8D 94 21 76 C 1 D 7D 35 C

detail.
CA 10 D4 D 51 C3 41 35 1E 38 9 21 076 FC A1 0D 47D 35 1C 364
10 D4 7D 351 C3 641 35 41E 38 D94 421 076 FC A1 0D 47D 35 1C 364 13
D4 7D 35 C3 64 35 41 38 D9 21 07 FC A 0D 47 3 1C 36 13 54
7D 35 1C3 64 135 41 E38 D9 42 07 6F A 10D 47 D3 51C 36 413 54 1E3
35 1C 64 135 41 E38 D9 42 107 6F CA 10D 47 D3 51C 36 41 54 1E 8D
1C 36 13 41 E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E 38 94
36 413 54 E3 8D 94 10 76F CA 10 D4 7D 351 C3 641 35 41E 38 D9 21
41 54 1E 8D 94 21 76 C 1 D 7D 35 C 64 35 41 38 D9 42 07
35 1E 38 9 21 07 FC A1 0D 47 35 1C 36 13 41 E3 D 42 10 6F
41 38 D9 42 07 6F A 0D 47 D3 1C 36 41 54 E 8D 94 10 76 C
E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E 38 94 21 76 FC A1
8D 94 10 76 CA 10 D4 7D 351 C3 64 35 41E 38 D9 21 07 FC A1 0D
94 210 76 FC 10 D4 7D 35 C3 64 135 41 38 D9 421 07 6FC A 0D 47
21 76 FC A1 D 7D 35 1C 64 13 41 E3 D 42 07 6F A 10 47 D3

parallel connections.
07 FC A 0D 47 3 1C 36 13 54 E3 8D 94 10 6F CA 10 D4 D 51
6F A 10 47 D3 51 36 41 54 1E 8D 94 21 76 C 1 D 7D 35 C
CA 10 D4 D 51 C3 41 35 1E 38 9 21 076 FC A1 0D 47D 35 1C 364
10 D4 7D 351 C3 641 35 41E 38 D94 421 076 FC A1 0D 47D 35 1C 364 13
D4 7D 35 C3 64 35 41 38 D9 21 07 FC A 0D 47 3 1C 36 13 54
7D 35 1C3 64 135 41 E38 D9 42 07 6F A 10D 47 D3 51C 36 413 54 1E3
35 1C 64 135 41 E38 D9 42 107 6F CA 10D 47 D3 51C 36 41 54 1E 8D
1C 36 13 41 E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E 38 94
36 413 54 E3 8D 94 10 76F CA 10 D4 7D 351 C3 641 35 41E 38 D9 21
41 54 1E 8D 94 21 76 C 1 D 7D 35 C 64 35 41 38 D9 42 07
35 1E 38 9 21 07 FC A1 0D 47 35 1C 36 13 41 E3 D 42 10 6F
undefined

Answer any two of the following

41 38 D9 42 07 6F A 0D 47 D3 1C 36 41 54 E 8D 94 10 76 C
E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E 38 94 21 76 FC A1

C) Define: Noise Figure, Noise Factor

8D 94 10 76 CA 10 D4 7D 351 C3 64 35 41E 38 D9 21 07 FC A1 0D
94 210 76 FC 10 D4 7D 35 C3 64 135 41 38 D9 421 07 6FC A 0D 47
21 76 FC A1 D 7D 35 1C 64 13 41 E3 D 42 07 6F A 10 47 D3
07 FC A 0D 47 3 1C 36 13 54 E3 8D 94 10 6F CA 10 D4 D 51

amplifier noise factor and noise figure

6F A 10 47 D3 51 36 41 54 1E 8D 94 21 76 C 1 D 7D 35 C
CA 10 D4 D 51 C3 41 35 1E 38 9 21 076 FC A1 0D 47D 35 1C 364
enlist advantages of superheterodyning.

10 D4 7D 351 C3 641 35 41E 38 D94 421 076 FC A1 0D 47D 35 1C 364 13

A10D47D351C36413541E38D9421076FC
D4 7D 35 C3 64 35 41 38 D9 21 07 FC A 0D 47 3 1C 36 13 54
7D 35 1C3 64 135 41 E38 D9 42 07 6F A 10D 47 D3 51C 36 413 54 1E3
35 1C 64 135 41 E38 D9 42 107 6F CA 10D 47 D3 51C 36 41 54 1E 8D
1C 36 13 41 E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E 38 94
36 413 54 E3 8D 94 10 76F CA 10 D4 7D 351 C3 641 35 41E 38 D9 21
deviation? Find Modulation index in each case.

41 54 1E 8D 94 21 76 C 1 D 7D 35 C 64 35 41 38 D9 42 07
detector. How AGC obtained from the detector

35 1E 38 9 21 07 FC A1 0D 47 35 1C 36 13 41 E3 D 42 10 6F
41 38 D9 42 07 6F A 0D 47 D3 1C 36 41 54 E 8D 94 10 76 C
E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E 38 94 21 76 FC A1
C) Explain concept of Pre-emphasis and De-emphasis.

8D 94 10 76 CA 10 D4 7D 351 C3 64 35 41E 38 D9 21 07 FC A1 0D
are advantages and disadvantages of ratio detector?

94 210 76 FC 10 D4 7D 35 C3 64 135 41 38 D9 421 07 6FC A 0D 47

21 76 FC A1 D 7D 35 1C 64 13 41 E3 D 42 07 6F A 10 47 D3
07 FC A 0D 47 3 1C 36 13 54 E3 8D 94 10 6F CA 10 D4 D 51
6F A 10 47 D3 51 36 41 54 1E 8D 94 21 76 C 1 D 7D 35 C
CA 10 D4 D 51 C3 41 35 1E 38 9 21 076 FC A1 0D 47D 35 1C 36
B) Define: Sensitivity, Selectivity, Fidelity, Image frequency

10 D4 7D 351 C3 641 35 41E 38 D94 421 076 FC A1 0D 47D 35 1C 364

D4 7D 35 C3 64 35 41 38 D9 21 07 FC A 0D 47 3 1C 36
7D 35 1C3 64 135 41 E38 D9 42 07 6F A 10D 47 D3 51C 36 413
35 1C 64 135 41 E38 D9 42 107 6F CA 10D 47 D3 51C 36 41
1C 36 13 41 E3 D 42 10 6F CA 10 4 D 51 3 41 35
36 413 54 E3 8D 94 10 76F CA 10 D4 7D 351 C3 641 35 4
The signal power and noise power measured at the input of an
amplitude limiting can be achieved using ratio detector? What
superheterodyne receiver with the help of block diagram. Also

amplifier are 200 μW and 2μW respectively. If the signal power

the bandwidth of 100kHz, calculate thermal noise for following
further raised to 10V while AF is dropped to 200 Hz, what is the

at the output is 1.5W and noise power is 40mW, calculate the

conditions. (a) for each resistor (b) for series connection (c) for

41 54 1E 8D 94 21 76 C 1 D 7D 35 C 64 35 41
35 1E 38 9 21 07 FC A1 0D 47 35 1C 36 13 41
41 38 D9 42 07 6F A 0D 47 D3 1C 36 41 54 E
E3 D 42 10 6F CA 10 4 D 51 3 41 35 1E
8D 94 10 76 CA 10 D4 7D 351 C3 64 35 41E 38
(06)
B) Two Resistors 15KΩ and 60KΩ are at room temperature, for (06)
A) Define Noise. What are different sources of noise? Explain in (06)
B) Draw and explain the circuit diagram of ratio detector. How (06)
Q.5. A) Draw and explain the circuit diagram of practical diode (06)
(04)
Q.4. A) What is superheterodyne principle? Explain operation of (08)
(06)

94 210 76 FC 10 D4 7D 35 C3 64 135 41 38
21 76 FC A1 D 7D 35 1C 64 13 41 E3 D
07 FC A 0D 47 3 1C 36 13 54 E3 8D
6F A 10 47 D3 51 36 41 54 1E 8D 94
CA 10 D4 D 51 C3 41 35 1E 38 9
10 D4 7D 351 C3 641 35 41E 38 D94 42
D4 7D 35 C3 64 35 41 38 D9 21
7D 35 1C 64 13 41 E3 D9 42 0