EKT 343 Principle of Communication Engineering 2015/2016
Question 1
Frequency Modulation and Phase Modulation
FM
As the amplitude of the information signal varies, the carrier frequency shifts proportionately.
As the modulating signal amplitude increases, the carrier frequency increases.
PM
When the amount of phase shift of a constant-frequency carrier is varied in accordance with a modulating signal, the resulting output is a phase-modulation (PM) signal.
Phase modulators produce a phase shift which is a time separation between two sine waves of the same frequency.
The greater the amplitude of the modulating signal, the greater the phase shift.
(i) Modulation index, m = 8 k / 2 k = 4;
(ii) From the Bessel function Table for m = 4;
(iii) J
0
= -0.4, J
1
= - 0.07, J
2
= 0.36, J
3
= 0.43, J
4
= 0.28, J
5
= 0.13, J
6
= 0.05, J
7
=
0.02
Let Pc = 100 W and
P
0
= Pc (J
0
)
2
= 100 (-0.4)
2
= 16 Watts.
P
1
= Pc (J
1
)
2
= 100 (-0.07)
2
= 0.490 W.
P
2
= Pc (J
2
) 2 = 100 (0.36) 2 = 12.960 W.
P
3
= Pc (J
3
)
2
= 100 (0.43)
2
= 18.490 W.
P
4
= Pc (J
4
)
2
= 100 (0.28)
2
= 7.840 W.
P
5
= Pc (J
5
)
2
= 100 (0.13)
2
= 1.690 W.
P
6
= Pc (J
6
)
2
= 100 (0.05)
2
= 0.250 W.
Tutorial 3 Chapter 3: Angle Modulation

EKT 343 Principle of Communication Engineering 2015/2016
(iv) Filter Bandwidth = 10.5 kHz
The output of band pass filter will contain the following frequency components: f c
, (f c
+ f m
), and (f c
+ 2f m
)
Power delivered to the load, P d
= P
0
+ 2P
1
+ 2P
2
= 42.9 Watts.
Filter Bandwidth = 21 kHz
The output of band pass filter will contain the following frequency components: f c
, (f c
+ f m
), (f c
+ 2f m
), (f c
+ 3f m
), (f c
+ 4f m
), and (f c
+ 5f m
),
Power delivered to the load, P d
= P
0
+ 2P
1
+ 2P
2
+ 2P
3
+ 2P
4
+ 2P
5
= 98.94 Watts.
Question 2
Solution:
Advantages of FM over AM are:
1. Less radiated power.
2. Low distortion due to improved signal to noise ratio (about 25dB) w.r.t. to man made interference.
3. Smaller geographical interference between neighbouring stations.
4. Well defined service areas for given transmitter power.
Disadvantages of FM:
1. Much more Bandwidth (as much as 20 times as much).
2. More complicated receiver and transmitter. b.
(i)
[1 Mark]
∆ f = 9 kHz,
∆𝒇 m f
= 𝒇 𝒎
, fm =
𝟗 𝒌
𝟑
= 𝟑 𝒌𝑯𝒛
(ii) [2 Marks]
V
FM
(t) = V c
cos (2
π
f c
t + m f sin 2
π
f m
t)
= 30 cos (1140k
π t + 3 sin 6k
π
t)
Tutorial 3 Chapter 3: Angle Modulation

EKT 343 Principle of Communication Engineering 2015/2016
(iii)
[1 mark]
(iv) n= 6
[2 Marks]
V o
= V c
J o
= 30x -0.26 = -7.8 V
V
1
= V c
J
1
= 30 x 0.34 = 10.2 V
V2 = V c
J
2
= 30 x 0.49 = 14.7 V
V3 = V c
J
3
= 30 x 0.31 = 9.3 V
V4 = V c
J
4
= 30 x 0.13 = 3.9 V
V5 = V c
J
5
= 30 x 0.04 = 1.2 V
(v)
V6 = V c
J
6
= 30 x 0.01 = 0.3 V
[3 Marks]
0.3 V
552
3.9 V
1.2 V
555 558
9.3 V
561
14.7 V
564
10.2 V
567
10.2 V
570 573
14.7 V
576
9.3 V
3.9 V
1.2 V
0.3 V
579 582 585 588 𝒎 𝒇
= 𝒌
𝟏 𝒗 𝒎 𝒇 𝒎
=
𝟖 𝒌 𝒙𝟐.𝟓
𝟏𝟎 𝒌
= 2
So, n = 4,
V o
= V c
J o
= 0.22 V c
V
1
= V c
J
1
= 0.58 V c
V
2
= V c
J
2
= 0.35 V c
V
3
= V c
J
3
= 0.13 V c
V
4
= V c
J
4
= 0.03 V c
-7.8 V
Tutorial 3 Chapter 3: Angle Modulation

EKT 343 Principle of Communication Engineering 2015/2016
𝟏𝟓
Since, V>
𝟏𝟎𝟎
𝑽𝒄 ; V> 0.15 V c
, so, n = 2
Bandwidth, B = 2n f m
= 2(2) 10k = 40 kHz a.
Table 1
FM Modulator: PM Modulator:
K
1
= 28 kHz/V K = 0.1 rad/V
[5 Marks]
For FM,
∆𝒇 = 𝑲
𝟏
𝑽 𝒎
= 𝟐𝟖𝒌 𝒙 𝟓 = 𝟏𝟒𝟎 𝒌𝑯𝒛
Bc = 2 (
∆ f + f m
) = 2 (140 k + 20 k) = 320 kHz
For PM, m p
= ∆𝜽 = 𝒌𝑽 𝒎 m p
= m f
= 0.5;
= 𝟎. 𝟏 𝒙 𝟓 = 𝟎. 𝟓 𝒓𝒂𝒅
∆𝒇 m = 𝒇 𝒎
, ∆𝒇 = 𝒎𝒇 𝒎
= 𝟎. 𝟓 𝒙 𝟐𝟎𝒌 = 𝟏𝟎 𝒌𝑯𝒛
Bc = 2 (
∆ f + f m
) = 2 (10 k + 20k) = 60 kHz
Tutorial 3 Chapter 3: Angle Modulation

EKT 343 Principle of Communication Engineering 2015/2016
Question 3 a.
(i) modulation index
[4 marks] f c

12 x 10
5
2



600 kHz , f m

12 x 10
3
2



6 kHz
 f

5 % xf c

0 .
05 x 600 k

30 kHz m

 f f m

30 k
6 k

5
(ii) frequency sensitivity
[2 marks]
 f

K
1
V m
K
1

 f
V m

30 k
2

15 kHz / V
(iii) Carson’s bandwidth and actual transmission bandwidth
BW c

2 (
 f
 f m
)

2 ( 30 k

6 k )

72 kHz
BW
B

2 nf m

2 ( 8 )( 6 k )

96 kHz b.
deviation sensitivity.
 
K


KV m ,
 
V m

2 rad
10 V

0 .
2 rad / V
[2 marks]
[2 marks]
Tutorial 3 Chapter 3: Angle Modulation

EKT 343 Principle of Communication Engineering c.
Armstrong indirect FM [10 Marks]
Solution:
Information given: fc
1
= 20 kHz f
LO
= 440 kHz
∆ f
1
=5 Hz
FM output: fc
4
= 97.3 MHz
∆ f
4
=10.24 kHz n
1
x n
2
=
∆𝒇
𝟒
∆𝒇𝟏
=
𝟏𝟎.𝟐𝟒 𝒌
= 2048
𝟓
n
1
=
𝟐𝟎𝟒𝟖 𝒏𝟐
…… (1) fc =[( 𝒇 𝒄𝟏
= 𝒏
𝟏 𝒏
𝟐 𝒙 𝒏 𝒇 𝒄𝟏
𝟏
) − 𝒇
𝑳𝑶
− 𝒏
𝟐
𝒇
𝑳𝑶
] x 𝒏
𝟐
97.3 M = 𝒏
𝟏 𝒏
𝟐
(𝟐𝟎𝒌) − 𝒏
𝟐
(𝟒𝟒𝟎 𝒌) ….. (2)
Subs (1) into (2);
97.3 M =
𝟐𝟎𝟒𝟖 𝒏𝟐 𝒏
𝟐
(𝟐𝟎𝒌) − 𝒏
𝟐
(𝟒𝟒𝟎 𝒌)
97.3 M = 𝟐𝟎𝟒𝟖(𝟐𝟎𝒌) − 𝒏
𝟐
(𝟒𝟒𝟎 𝒌) 𝒏
𝟐
= 128.05
Solving 𝒏
𝟏
using (1) n
1
=
𝟐𝟎𝟒𝟖 𝒏𝟐
𝟐𝟎𝟒𝟖
=
𝟏𝟐𝟖.𝟎𝟓
=15.99
Due to frequency doublers were used , n
1
= 128 and n
2
= 16
Tutorial 3
2015/2016
Chapter 3: Angle Modulation