These slides cover chapter 3 of "Fundamentals of Communication Systems" by John G. Proakis and Masoud Salehi

1. Amplitude Modulation
CLO 2, Chapter 3

2. Modulation

3. Why Modulate the Signal?
• (1) To translate the frequency of the lowpass signal to the passband of
the channel.
• (2) To simplify the structure of the transmitter by employing higher
frequencies. Smaller wavelength allows smaller antenna (𝜆 = 𝑐/𝑓).
• (3) To simultaneously transmit signals from several message sources,
by means of frequency-division multiplexing (FDM). (See Section
3.4.)
• (4) To expand the bandwidth of the transmitted signal in order to
increase its noise and interference immunity (Chapter 6).

4. Amplitude Modulation
• The message signal 𝑚(𝑡) is impressed on the amplitude of the carrier
signal 𝑐(𝑡) = 𝐴 𝑐 cos(2𝜋𝑓𝑐 𝑡).
• This results in a sinusoidal signal whose amplitude is a function of the
message signal 𝑚(𝑡).
• Various flavors of AM:
• (a) Double Sideband Suppressed-Carrier (DSB-SC) AM,
• (b) Conventional Double-Sideband AM
• (c) Single-Sideband (SSB) AM
• (d) Vestigial-Sideband (VSB) AM

5. Double-Sideband
Suppressed-Carrier AM
• A DSB-SC AM signal is obtained by
multiplying the message signal m(t)
with the carrier signal
• c t = 𝐴 𝑐 cos(2𝜋𝑓𝑐 𝑡).
• The DSB-SC modulated signal is
• 𝑢 𝑡 = 𝑚 𝑡 𝑐 𝑡
• 𝑢(𝑡) = 𝐴 𝑐 𝑚(𝑡) cos(2𝜋𝑓𝑐 𝑡)

6. DSB-SC Spectrum
• The spectrum of the modulated signal
using the result of Example 2.3.14. Thus,
we obtain
• 𝑈(𝑓) =
𝐴 𝑐
2
𝑀 𝑓 − 𝑓𝑐 + 𝑀 𝑓 + 𝑓𝑐
• The magnitude of the spectrum of the
message signal 𝑚(𝑡) has been translated
or shifted in frequency by an amount 𝑓𝑐 ·
• The bandwidth occupancy of the
amplitude-modulated signal is 2𝑊,
whereas the bandwidth of the message
signal 𝑚(𝑡) is 𝑊.
• Channel bandwidth required to transmit
the modulated signal 𝑢(𝑡) is 𝐵𝑐 = 2𝑊.

7. DSB-SC Spectrom
• If 𝑚 𝑡
𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚
𝑀 𝑓
• By frequency shifting property of Fourier transform:
• 𝑚 𝑡 𝑒 𝑗2𝜋𝑓𝑐 𝑡
𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚
𝑀 𝑓 − 𝑓𝑐
• ∵ cos(𝜃) =
1
2
(𝑒 𝑗𝜃 + 𝑒−𝑗𝜃)
• 𝑢(𝑡) = 𝐴 𝑐 𝑚(𝑡) cos(2𝜋𝑓𝑐 𝑡)
•
𝐴 𝑐
2
𝑒 𝑗2𝜋𝑓𝑐 𝑡
+ 𝑒−𝑗2𝜋𝑓𝑐 𝑡
𝑚 𝑡
𝐹 1
2
𝐴 𝑐 𝑀 𝑓 − 𝑓𝑐 + 𝑀 𝑓 + 𝑓𝑐 = 𝑈(𝑓)

10. Assignment
• How can we demodulate this signal?

11. • Let the message signal be 𝑚 𝑡 = sinc(104 𝑡). Determine the DSB-SC-
modulated signal and its bandwidth when the carrier is a sinusoid
with a frequency of 1 MHz
• Carrier: 𝑐 𝑡 = cos 2𝜋106 𝑡
• 𝑢 𝐷𝑆𝐵−𝑆𝐶 𝑡 = sinc 104 𝑡 cos 2𝜋106 𝑡
• 𝑀 𝑓 = 10−4 Π 10−4 𝑓
• The Fourier transform is constant from -5kHz
to 5kHz, and zero at other frequencies.
• Message bandwidth 𝑊 = 5𝑘𝐻𝑧
• Modulated bandwidth = 2𝑊 = 10𝑘𝐻𝑧

12. Power Content of DSB-SC Signals

13. • Determine the DSB-SC AM signal and its upper and lower
sidebands. 𝑚 𝑡 = 𝑎 cos 2𝜋𝑓𝑚 𝑡
• Determine the power in the modulated signal and the power in
each of the sidebands.
• 𝑢 𝑡 = 𝑚 𝑡 𝑐 𝑡 = 𝐴 𝑐 𝑎 cos 2𝜋𝑓𝑚 𝑡 cos 2𝜋𝑓𝑐 𝑡
• 𝑃𝑚 =
𝑎2
2
• 𝑃𝑢 =
𝐴 𝑐
2
2
𝑃𝑚 =
𝐴 𝑐
2 𝑎
4
• 𝑃𝑢𝑠 = 𝑃𝑙𝑠 =
𝐴 𝑐
2 𝑎
8

14. Demodulation of DSB-SC AM Signals
• Suppose that the DSB-SC AM signal 𝑢(𝑡) is transmitted
through an ideal channel (with no channel distortion and
no noise)
• We demodulate by multiplying 𝑟(𝑡) by a locally
generated sinusoid cos 2𝜋𝑓𝑐 𝑡 + 𝜙
• Since 𝑚(𝑡) is limited to W Hz, low pass filtering gives:

15. • Power of received signal depends upon phase of locally generated
carrier, 𝜙. 𝑦𝑙 𝑡 =
1
2
𝐴 𝑐 𝑚 𝑡 cos 𝜙
• If 𝜙 = 45°, the amplitude of the desired signal is reduced by 2 and
the signal power is reduced by a factor of 2
• If 𝜙 = 90°, the desired signal component vanishes
• DSB-SC requires phase-coherent or synchronous demodulator
• Solutions ?

16. • Solutions
• 1) Add pilot tone
• 2) Use Phase Locked Loop

17. 3.2.2 Conventional Amplitude Modulation
• A conventional AM signal consists of a large carrier component, in
addition to the DSB AM-modulated signal. 𝑚 𝑡 < 1

21. Power for the Conventional AM Signal

23. Demodulation of Conventional DSB·AM
Signals
• There is no need for a synchronous demodulator. Since the message signal m(t)
satisfies the condition l m(t) I < 1, the envelope (amplitude) 1 + m(t) > 0.
• Envelope Detection:
• 1) Rectify the received signal
• 2) Pass the signal through a low pass filter of bandwidth matched to message
signal
• d(t)= g1+g2 m ( t )
• 3) Eliminate g1 by passing d(t) through a transformer

25. Single-Sideband AM
• 𝑢 𝑡 = 𝐴 𝑐 𝑚 𝑡 cos 2𝜋𝑓𝑐 𝑡 ± 𝐴 𝑐 𝑚 𝑡 sin2𝜋𝑓𝑐 𝑡 ,

28. SSB Demodulation

29. Vestigial-Sideband AM

32. IMPLEMENTATION OF AMPLITUDE
MODULATORS AND DEMODULATORS
• Power-Law Modulation

34. Switching Modulator

36. Ring Modulator

37. Envelope Detector

40. Demodulation of DSB-SC AM Signals.

41. Frequency-Division Multiplexing
• Multiplexing: Combining separate message signals into a composite
signal for transmission over a common channel is called
• Types of Multiplexing:
• (1) Time-division multiplexing (TDM). Used for digital signals.
• (2) Frequency-division multiplexing (FDM). Used for analog/digital

42. • K-message signals at the transmitter
• Transmitter lowpass filters (LPFs) ensure message bandwidth is W-Hz

43. Quadrature-Carrier Multiplexing
• Use single carrier to multiplex two signals.
• 𝑢 𝑡 = 𝐴 𝑐 𝑚1 𝑡 cos 2𝜋𝑓𝑐 𝑡 + 𝐴 𝑐 𝑚2 𝑡 sin 2𝜋𝑓𝑐 𝑡
• Requires synchronous demodulator
• Each message transmitted by DSB-SC AM
• 𝑢 𝑡 cos 2𝜋𝑓𝑐 𝑡 = 𝐴 𝑐 𝑚1 𝑡 cos2
(2𝜋𝑓𝑐 𝑡) +
𝐴 𝑐 𝑚2(𝑡) cos 2𝜋𝑓𝑐 𝑡 sin 2𝜋𝑓𝑐 𝑡
• =
𝐴 𝑐
2
𝑚1 𝑡 +
𝐴 𝑐
2
𝑚1 𝑡 cos 4𝜋𝑓𝑐 𝑡 +
𝐴 𝑐
2
𝑚2 𝑡 sin 4𝜋𝑓𝑐 𝑡
• Which is the low pass component in above expression?

45. Superheterodyne Receiver
• Prior to the superhet receiver, radio listeners had to constantly play
with a set of knobs on a radio to keep locked on to radio programs of
the day
• The receiver used in AM radio broadcast is super heterodyne receiver
• 540-1600 kHz with 10 kHz spacing, 𝑓𝐼𝑓 = 455 𝑘𝐻𝑧,
• Economy: lower frequency components can be used,
• many components can be designed for a fixed frequency
• Components can even be shared between different receiver designs
• Sensitivity: Filtering out unwanted signals at IF is a much easier job
than filtering them out at RF

46. • 𝑓𝐿𝑂 = 𝑓𝑐 + 𝑓𝐼𝐹
• The tuning range of the local
oscillator is 995-2055 kHz
• image frequency 𝑓𝑐
′
= 𝑓𝐿𝑂 +
𝑓_𝐼𝐹