ADC

1. Analog and Digital
Communication
(ADC)
By Praveen Pawar
IIIT Bhopal
ADC - IIIT Bhopal 1

2. ADC - IIIT Bhopal 2

3. Vestigial Sideband Modulation (VSB)
• SSB-SC modulation is well suited for transmission of voice because of
frequency gap that exist in the spectrum of voice signal zero and a
few hundred Hz. When the baseband signal contains significant
components at extremely low frequencies as in case of T.V. picture
signal, the upper and lower side band need at carrier frequency. This
means that use of SSB modulation is inappropriate for transmission of
such baseband signals due to difficulty of isolating one side band.
ADC - IIIT Bhopal 3

4. VSB
300 Hz 3300 Hz
- 3300 Hz - 300 Hz
f
v(f)
fc+300 Hz fc+3300 Hz
fc-3300 Hz fc-300 Hz
f
V(f)
fc
p(f)
4.5 MHz
- 4.5 MHz
P(f)
fc+4.5 MHz
fc- 4.5 MHz fc
f
Modulation
Modulation
ADC - IIIT Bhopal 4

5. VSB
m(𝜔)
𝜔𝑚
−𝜔𝑚
𝜔
M(𝜔)
𝜔𝑐 − 𝜔𝑚
𝜔
𝜔𝑐 + 𝜔𝑚
𝜔𝑐
M(𝜔)
𝜔𝑐 − 𝜔𝑚
𝜔
𝜔𝑐 + 𝜔𝑚
𝜔𝑐
• In VSB Modulation, one sideband is passed
almost completely where as just a trace or
vestige of unwanted sideband retains.
• Specifically the vestige part of unwanted
sideband components for the part removed from
desired sideband.
Modulation
𝜔𝑣
−𝜔𝑣
𝜔𝑐 − 𝜔𝑣
𝜔𝑐 + 𝜔𝑣
ADC - IIIT Bhopal 5
BW = 𝜔𝑚 + 𝜔𝑣

6. VSB
VSB Filter
𝑚 𝑡
𝑐 𝑡 = cos 𝜔𝑐𝑡
𝑠 𝑡
ℎ 𝑡
Low Pass Filter
𝜔𝑐 = 𝜔𝑚
𝑠 𝑡
cos 𝜔𝑐𝑡
𝑦 𝑡
Transmitter Side
Receiver Side
𝑥 𝑡
𝑦1 𝑡
ADC - IIIT Bhopal 6

7. VSB
𝑥 𝑡 = 𝑚 𝑡 ⋅ cos 𝜔𝑐𝑡
𝑠 𝑡 = 𝑚 𝑡 ⋅ cos 𝜔𝑐𝑡 ∗ ℎ 𝑡
𝑆 𝜔 =
1
2
𝑀 𝜔 − 𝜔𝑐 + 𝑀 𝜔 + 𝜔𝑐 ⋅ 𝐻(𝜔)
∵ 𝑦1 𝑡 = 𝑠 𝑡 ⋅ cos 𝜔𝑐𝑡
𝑌1 𝜔 =
1
2
𝑆 𝜔 − 𝜔𝑐 + 𝑆(𝜔 + 𝜔𝑐)
𝑌1 𝜔
=
1
4
𝑀 𝜔 − 𝜔𝑐 − 𝜔𝑐 + 𝑀 𝜔 − 𝜔𝑐 + 𝜔𝑐 ⋅ 𝐻 𝜔 − 𝜔𝑐 + {𝑀 𝜔 + 𝜔𝑐 − 𝜔𝑐 + 𝑀 𝜔 + 𝜔𝑐 + 𝜔𝑐 } ⋅ 𝐻(𝜔 + 𝜔𝑐)
𝑌 𝜔 =
1
4
𝑀 𝜔 𝐻 𝜔 − 𝜔𝑐 + 𝐻 𝜔 + 𝜔𝑐 ; −𝜔𝑚 ≤ 𝜔 ≤ 𝜔𝑚
ADC - IIIT Bhopal 7

8. AM Receiver
Functions of Receivers
1. To collect the electromagnetic waves transmitted by the
transmitter.
2. To select the desired signal and reject all others. This is known as
selectivity of the receiver.
3. To amplify the selected modulated signal. This is known as
sensitivity of the receiver.
4. To detect the baseband modulating signal from the modulated
Radio Frequency (RF) signal.
5. To amplify the baseband signal so as to operate loudspeaker.
ADC - IIIT Bhopal 8
Date: 28-01-2020

9. Definitions
• Selectivity :
The ability of receiver to distinguish between two adjacent carrier frequencies.
• Sensitivity:
The ability of receiver to detect the weakest possible signal.
• Fidelity:
The ability of receiver to reproduce faithfully all frequency components of the
baseband signal.
ADC - IIIT Bhopal 9

10. Tuned Radio Frequency (TRF) Receiver
ADC - IIIT Bhopal 10
RF
Amplifier
Demodul
ator
AF
Amplifier
Antenna
LS
RF: Radio Frequency
AF: Audio Frequency
LS: Loud-Speaker
𝑓𝑆 : Signal Frequency
𝑓𝑆
Quality Factor,
𝑄 =
𝐶𝑒𝑛𝑡𝑒𝑟 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝐵𝑎𝑛𝑑𝑤𝑖𝑑𝑡ℎ
=
𝑓𝑆
𝐵𝑊
AM: 535 – 1605 KHz
𝑓𝑠 = 1000 KHz (consider)
𝐵𝑊 = 10 KHz (for 5 KHz message signal)
𝑄 =
1000
10
= 100
If 𝑓𝑆 = 1600 KHz, to maintain the 𝑄 value, required 𝐵𝑊 is 16 KHz

11. Superheterodyne Receiver
ADC - IIIT Bhopal 11
RF
Amplifier
Mixer
IF
Amplifier
Detector/
Demodul
ator
AF
Amplifier
Local
Oscillator
Ganged Tuning
𝑓𝑆
𝑓𝐿𝑂 = 𝑓𝑆 + 𝑓𝐼𝐹
LS
RF: Radio Frequency
IF: Intermediate Frequency
AF: Audio Frequency
LS: Loud-Speaker
𝑓𝑆 : Signal Frequency
𝑓𝐿𝑂 : Local-Oscillator Frequency
Antenna 𝑓𝐼𝐹
𝑓𝑆
2𝑓𝑆 + 𝑓𝐼𝐹
OR
𝑓𝐼𝐹
𝑓𝐿𝑂 = 𝑓𝑆 + 𝑓𝐼𝐹
Mixer

12. Superheterodyne Receiver
• 𝑓𝐿𝑂 =
1
2𝜋 𝐿𝐶
𝐶𝑚𝑎𝑥
𝐶𝑚𝑖𝑛
=
𝑓𝑚𝑎𝑥
𝑓𝑚𝑖𝑛
2
• 𝑓𝐿𝑂 > 𝑓𝑆
• Choice of IF:
• High value of IF results in poor adjacent channel selectivity.
• Low value of IF, Image frequency rejection is poor.
• IF is fixed to 455 KHz.
ADC - IIIT Bhopal 12
𝑓 ∝
1
𝐶
𝑓𝑚𝑎𝑥 =
1
𝐶𝑚𝑖𝑛
𝑓𝑚𝑖𝑛 =
1
𝐶𝑚𝑎𝑥
𝑓𝑚𝑎𝑥
𝑓𝑚𝑖𝑛
=
𝐶𝑚𝑎𝑥
𝐶𝑚𝑖𝑛
𝐶𝑚𝑎𝑥
𝐶𝑚𝑖𝑛
=
𝑓𝑚𝑎𝑥
𝑓𝑚𝑖𝑛
2

13. ADC - IIIT Bhopal 13

14. Superheterodyne Receiver
Problem:
For a superheterodyne receiver, the intermediate frequency is 15 MHz
and the local oscillator frequency is 3.5 GHz. If the frequency of the
received signal is less than the local oscillator frequency, then the
image frequency (in MHz) is______________.
ADC - IIIT Bhopal 14
Given,
𝑓𝐼𝐹 = 15 MHz and 𝑓𝐿𝑂 = 3.5 GHz = 3500 MHz
∵ 𝑓𝐿𝑂 > 𝑓𝑆
∴ 𝑓𝑖𝑚𝑎𝑔𝑒 = 𝑓𝑆 + 2𝑓𝐼𝐹 𝑓𝐿𝑂 = 𝑓𝑆 + 𝑓𝐼𝐹 ⇒ 𝑓𝑆 = 3500 − 15 = 3485 MHz
𝑓𝑖𝑚𝑎𝑔𝑒 = 3485 + 2 × 15 = 3515 MHz

15. Angle Modulation
FM & PM
ADC - IIIT Bhopal 15

16. ADC - IIIT Bhopal 16

17. Angle Modulation
• Frequency Modulation (FM)
In FM, the frequency of high frequency carrier signal is varied in accordance
with the instantaneous value of modulating signal keeping amplitude and phase
constant.
• Phase Modulation (PM)
In PM, the phase angle is varied directly in accordance with instantaneous value
of modulating signal keeping amplitude and frequency constant.
ADC - IIIT Bhopal 17

18. ADC - IIIT Bhopal 18
FM
Let m(t) -> any arbitrary modulating signal with
maximum frequency 𝜔𝑚
𝑐 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡
𝜃𝑐 𝑡 = 𝜔𝑐𝑡
𝑋𝐹𝑀 𝑡 = 𝐴𝑐 cos 𝜃𝑖 𝑡
where,
𝜃𝑖 𝑡 = 0
𝑡
𝜔𝑖 𝑡 𝑑𝑡
𝜃𝑖 𝑡 = 0
𝑡
𝜔𝑐 + 𝑘𝑓𝑚 𝑡 𝑑𝑡
𝜃𝑖 𝑡 = 𝜔𝑐𝑡 + 𝑘𝑓 0
𝑡
𝑚 𝑡 𝑑𝑡
PM
Let m(t) -> any arbitrary modulating signal
𝑐 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡
𝜃𝑐 𝑡 = 𝜔𝑐𝑡
𝑋𝑃𝑀 𝑡 = 𝐴𝑐 cos 𝜃𝑖 𝑡
where,
𝜃𝑖 𝑡 = 𝜔𝑐𝑡 + 𝑘𝑝𝑚 𝑡
𝑘𝑓 -> frequency sensitivity in Hz/volt
𝑘𝑝 -> phase sensitivity in radians/volts

19. ADC - IIIT Bhopal 19
𝑋𝐹𝑀 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡 + 𝑘𝑓
0
𝑡
𝑚 𝑡 𝑑𝑡
𝑋𝑃𝑀 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡 + 𝑘𝑝𝑚 𝑡
𝑚 𝑡
• Relation between FM & PM
𝑑
𝑑𝑡
Frequency
Modulator
PM Signal
𝑚 𝑡
0
𝑡
𝑑𝑡
Phase
Modulator
FM Signal
• In PM, phase angle varies linearly with 𝑓 𝑡 whereas in FM phase angle varies linearly with the integral of 𝑓 𝑡 .
𝑐 𝑡
𝑐 𝑡

20. ADC - IIIT Bhopal 20
For sinusoidal FM,
𝑚 𝑡 = 𝐴𝑚 cos 𝜔𝑚𝑡
𝑐 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡
∴ 𝑋𝐹𝑀 𝑡 = 𝐴𝑐 cos 𝜃𝑖 𝑡
𝜔𝑖 𝑡 = 𝜔𝑐 + 𝑘𝑓𝐴𝑚 cos 𝜔𝑚𝑡
Peak frequency deviation Δ𝜔 = 𝑘𝑓𝐴𝑚
𝜔𝑖 𝑡 = 𝜔𝑐 + Δ𝜔 cos 𝜔𝑚𝑡
Hence,
𝜃𝑖 𝑡 =
0
𝑡
𝜔𝑐 + Δ𝜔 cos 𝜔𝑚𝑡 𝑑𝑡
𝜃𝑖 𝑡 = 𝜔𝑐𝑡 +
𝚫𝝎
𝝎𝒎
sin 𝜔𝑚𝑡
Modulation Index
𝜷𝑭𝑴 =
𝚫𝝎
𝝎𝒎
=
𝚫𝒇
𝒇𝒎
𝜃𝑖 𝑡 = 𝜔𝑐𝑡 + 𝛽𝐹𝑀 sin 𝜔𝑚𝑡

21. ADC - IIIT Bhopal 21
For sinusoidal PM,
𝑚 𝑡 = 𝐴𝑚 cos 𝜔𝑚𝑡
𝑐 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡
∴ 𝑋𝑃𝑀 𝑡 = 𝐴𝑐 cos 𝜃𝑖 𝑡
𝜃𝑖 𝑡 = 𝜔𝑐𝑡 + 𝑘𝑝𝐴𝑚 cos 𝜔𝑚𝑡
Peak phase deviation,
𝛽𝑃𝑀 = 𝑘𝑝𝐴𝑚
𝜃𝑖 𝑡 = 𝜔𝑐𝑡 + 𝛽𝑃𝑀 cos 𝜔𝑚𝑡

22. ADC - IIIT Bhopal 22
• Difference between FM and PM Signal
𝑋𝐹𝑀 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡 + 𝛽𝐹𝑀 sin 𝜔𝑚𝑡
𝜃𝑖 𝑡 = 𝜔𝑐𝑡 +
𝚫𝝎
𝝎𝒎
sin 𝜔𝑚𝑡
𝜔𝑖 𝑡 = 𝜔𝑐 + Δ𝜔 cos 𝜔𝑚𝑡
Δ𝜔 = 𝑘𝑓 ⋅ 𝐴𝑚
𝑋𝑃𝑀 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡 + 𝛽𝑃𝑀 cos 𝜔𝑚𝑡
𝜃𝑖 𝑡 = 𝜔𝑐𝑡 + 𝑘𝑝𝐴𝑚 cos 𝜔𝑚𝑡
𝜔𝑖 𝑡 = 𝜔𝑐 − 𝑘𝑝𝐴𝑚𝜔𝑚 sin 𝜔𝑚𝑡
Δ𝜔 = 𝑘𝑝𝐴𝑚𝜔𝑚
𝛽𝑃𝑀 =
Δ𝜔
𝜔𝑚
= 𝑘𝑝𝐴𝑚
𝛽𝐹𝑀 =
Δ𝜔
𝜔𝑚
=
𝑘𝑓𝐴𝑚
𝜔𝑚

23. ADC - IIIT Bhopal 23
FM Modulation

24. Problems:
ADC - IIIT Bhopal 24
1. A sinusoidal modulating waveform of amplitude 5 V and a frequency of 2 KHz is
applied to FM generator, which has a frequency sensitivity of 40 Hz/volt. Calculate the
frequency deviation and modulation index.
2. An FM wave is given by 𝑠 𝑡 = 20 cos 8𝜋 × 106
𝑡 + 9 sin 2𝜋 × 103
𝑡 . Calculate the
frequency deviation, modulation index, and power of FM wave.
Frequency deviation, Δ𝑓 = 𝑘𝑓𝐴𝑚 = 40 × 5 = 200 Hz
Modulation Index, 𝛽𝐹𝑀 =
Δ𝑓
𝑓𝑚
=
200
2∗1000
= 0.1
Given,
𝐴𝑚 = 5 V,
𝑓𝑚 = 2 KHz = 2000 Hz,
𝑘𝑓 = 40 Hz/V
Standard Equation of FM is given as,
𝑋𝐹𝑀 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡 + 𝛽𝐹𝑀 sin 𝜔𝑚𝑡
𝑠 𝑡 = 20 cos 8𝜋 × 106𝑡 + 9 sin 2𝜋 × 103𝑡
𝐴𝑐 𝜔𝑐 = 2𝜋𝑓𝑐
𝜔𝑚 = 2𝜋𝑓𝑚
𝛽𝐹𝑀
Frequency deviation, Δ𝑓 = 𝛽𝐹𝑀𝑓𝑚 = 9 × 103
= 9 KHz
∵ 𝛽𝐹𝑀=
Δ𝑓
𝑓𝑚
𝑓𝑐 = 4 × 106
Hz 𝑓𝑚 = 103Hz
Modulation Index, 𝛽𝐹𝑀 = 9
Power of FM wave, 𝑝𝐹𝑀 =
𝐴𝑐
2
2
=
202
2
= 200 Watts

25. Types of frequency Modulation
ADC - IIIT Bhopal 25
𝑋𝐹𝑀 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡 + 𝛽 sin 𝜔𝑚𝑡
𝑋𝐹𝑀 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡 cos 𝛽 sin 𝜔𝑚𝑡 − sin 𝜔𝑐𝑡 sin 𝛽 sin 𝜔𝑚𝑡
Depending upon value of 𝛽, there are two types of frequency modulation: Narrowband FM
and Wideband FM
1. Narrowband FM
When 𝛽 is very-very small (𝛽 → 0)
cos
𝛽→0
𝛽 sin 𝜔𝑚𝑡 → 1
sin
𝛽→0
𝛽 sin 𝜔𝑚𝑡 → 𝛽 sin 𝜔𝑚𝑡
and
(1)
(2)
On putting the above values in (2),

26. ADC - IIIT Bhopal 26
1. Narrowband FM (continue…)
𝑋𝑁𝐵𝐹𝑀 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡 − sin 𝜔𝑐𝑡 ⋅ 𝛽 sin 𝜔𝑚𝑡
𝑋𝑁𝐵𝐹𝑀 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡 − 𝐴𝑐𝛽 sin 𝜔𝑐𝑡 ⋅ sin 𝜔𝑚𝑡
𝑋𝑁𝐵𝐹𝑀 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡 +
𝐴𝑐𝛽
2
cos 𝜔𝑐 + 𝜔𝑚 𝑡 −
𝐴𝑐𝛽
2
cos 𝜔𝑐 − 𝜔𝑚 𝑡
The narrowband FM signal is similar to AM except that in narrowband FM,
the lower-side-band is 1800
out of phase with respect to carrier as well as upper-side-band.
Frequency domain
representation:
𝐴𝑐𝜋
𝐴𝑐𝛽
2
𝜋
−𝐴𝑐𝛽
2
𝜋
𝜔
𝑋𝑁𝐵𝐹𝑀 𝜔
𝜔𝑐 − 𝜔𝑚
− 𝜔𝑐 − 𝜔𝑚
− 𝜔𝑐 + 𝜔𝑚 𝜔𝑐 + 𝜔𝑚
−𝜔𝑐 𝜔𝑐
𝐵𝑊 = 𝜔𝑐 + 𝜔𝑚 − 𝜔𝑐 − 𝜔𝑚
𝐵𝑊 = 2𝜔𝑚
𝐵𝑊 = 2𝑓𝑚

27. ADC - IIIT Bhopal 27
2. Wideband FM
When 𝛽 is large (𝛽 → ∞)
When 𝛽 is large, FM produces a large number of sidebands and the bandwidth of FM is quite large.
Such systems are called wideband FM.
𝑋𝐹𝑀 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡 + 𝛽 sin 𝜔𝑚𝑡
𝑋𝐹𝑀 𝑡 = 𝐴𝑐 cos 𝜔𝑐𝑡 cos 𝛽 sin 𝜔𝑚𝑡 − sin 𝜔𝑐𝑡 sin 𝛽 sin 𝜔𝑚𝑡
Here, cos 𝛽 sin 𝜔𝑚𝑡 and sin 𝛽 sin 𝜔𝑚𝑡 are periodic signals with fundamental frequency 𝜔𝑚.
Thus the Fourier transform for each of these signals is an impulse train with impulses at integer
multiple of 𝜔𝑚 and amplitude proportional to Bessel’s Function of first kind,
(1)
(2)

28. ADC - IIIT Bhopal 28
cos 𝛽 sin 𝜔𝑚𝑡 = 𝐽0 𝛽 + 2𝐽2 𝛽 cos 2𝜔𝑚𝑡 + 2𝐽4 𝛽 cos 4𝜔𝑚𝑡 + 2𝐽6 𝛽 cos 6𝜔𝑚𝑡 + ⋯
sin 𝛽 sin 𝜔𝑚𝑡 = 2𝐽1 𝛽 cos 𝜔𝑚𝑡 + 2𝐽3 𝛽 cos 3𝜔𝑚𝑡 + 2𝐽5 𝛽 cos 5𝜔𝑚𝑡 + ⋯
cos 𝜔𝑐𝑡 cos 𝛽 sin 𝜔𝑚𝑡 = 𝐽0 𝛽 cos 𝜔𝑐𝑡 + 2𝐽2 𝛽 cos 𝜔𝑐𝑡 cos 2𝜔𝑚𝑡 + 2𝐽4 𝛽 cos 𝜔𝑐𝑡 cos 4𝜔𝑚𝑡 +
2𝐽6 𝛽 cos 𝜔𝑐𝑡 cos 6𝜔𝑚𝑡 + ⋯
sin 𝜔𝑐𝑡 sin 𝛽 sin 𝜔𝑚𝑡 = 2𝐽1 𝛽 sin 𝜔𝑐𝑡 cos 𝜔𝑚𝑡 + 2𝐽3 𝛽 sin 𝜔𝑐𝑡 cos 3𝜔𝑚𝑡 +
2𝐽5 𝛽 sin 𝜔𝑐𝑡 cos 5𝜔𝑚𝑡 + ⋯
(3)
(4)
from (2), (5), and (6),
𝑋𝑊𝐵𝐹𝑀 𝑡 = 𝐴𝑐[𝐽0 𝛽 cos 𝜔𝑐𝑡 + 𝐽1 𝛽 cos 𝜔𝑐 + 𝜔𝑚 𝑡 − cos 𝜔𝑐 − 𝜔𝑚 𝑡 +
𝐽2 𝛽 cos 𝜔𝑐 + 2𝜔𝑚 𝑡 + cos 𝜔𝑐 − 2𝜔𝑚 𝑡 +
𝐽3 𝛽 cos 𝜔𝑐 + 3𝜔𝑚 𝑡 − cos 𝜔𝑐 − 3𝜔𝑚 𝑡 + ⋯ ]
(5)
(6)
(7)
cos 𝜔𝑐𝑡 × (3) and sin 𝜔𝑐𝑡 × 4
2. Wideband FM (continue...)

29. ADC - IIIT Bhopal 29
2. Wideband FM (continue...)
Frequency domain representation:
|𝑋𝑊𝐵𝐹𝑀 𝜔 |
𝜔𝑚
𝜔𝑚
𝜔𝑐 𝜔𝑐 + 𝜔𝑚 𝜔𝑐 + 2𝜔𝑚 𝜔𝑐 + 2𝜔𝑚
𝜔
. . .
. . .
𝜔𝑐 − 𝜔𝑚
𝜔𝑐 − 2𝜔𝑚
𝜔𝑐 − 3𝜔𝑚

30. ADC - IIIT Bhopal 30
Bandwidth for Wideband FM
𝐵𝑊𝐹𝑀 = 2 Δ𝜔 + 𝜔𝑚 rad/sec
= 2 Δ𝑓 + 𝑓𝑚 Hz
= 2 𝛽 + 1 𝑓𝑚 Hz ∵ 𝛽 =
Δ𝑓
𝑓𝑚
Carson’s Rule