Modulation for Analog
Communication
Yao Wang
Polytechnic University, Brooklyn, NY11201
http://eeweb.poly.edu/~yao
Outline
• Baseband communication: bandwidth requirement
• Modulation of continuous signals
– Amplitude modulation
– Quadrature amplitude modulation
– Other modulation techniques: frequency/phase modulation
• Frequency division multiplexing
• Application of modulation
• Demo of AM and QAM
©Yao Wang, 2006
EE3414: Analog Communications
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Baseband Communications
• Signal strength attenuates with distance. Needs
repeaters to amplify the signals in stages
• Received signal is corrupted by noise
– R(t)=A S(t)+ n(t)
• Received signal quality depends on channel noise and
noise between repeaters accumulate
• To transmit a signal with bandwidth B, we need >=B Hz
in channel bandwidth
• If the signal is low-pass (0-B), must the channel operate
at 0-B range of frequency?
• How do we send multiple signals over the channel?
©Yao Wang, 2006
EE3414: Analog Communications
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A Typical Communication System
Modulator
Transmitter
Demodulator
Receiver
Signal to be
transmitted
(analog or
digital)
Received
signal
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Modulation = Frequency Shifting
©Yao Wang, 2006
Baseband
signal
Modulated
signal
0
fc
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Frequency
5
Why do we need “modulation”?
– A communication channel only operates at a certain frequency
range
• telephone cables, terrestrial (over the air broadcast), ethernet,
optical fiber, etc.
– Modulation translates a signal from its baseband to the operating
range of the channel
– By modulating different signals to different frequency bands,
they can be transmitted simultaneously over the same channel
frequency division multiplexing
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Frequency Division Multiplexing
•
To transmit the
three signals over
the same channel,
each signal is
shifted to a
different carrier
frequency and
then summed
together.
•
From Figure 7.22
in Signals and
Systems
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How do we shift the frequency of a
signal?
• By multiplying with a sinusoid signal !
y (t ) = x(t ) cos(ω c t )
x(t )
cos(ω c t )
carrier signal
ω c : carrier frequency
©Yao Wang, 2006
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Basic Equalities
• Basic equality
x(t )e j 2πf ct ↔ X ( f − f c )
x(t )e − j 2πf ct ↔ X ( f + f c )
x(t )cos(2πf c t ) ↔
1
( X ( f − f c ) + X ( f + f c ))
2
• Proof on the board
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Frequency Domain Interpretation of
Modulation
From Figure
7.5 in
Signals/Syste
ms
x(t )
cos(ω c t )
y (t ) = x(t ) cos(ω c t )
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How to get back to the baseband?
(Demodulation)
• By multiplying with the same sinusoid + low pass
filtering!
H (ω )
w(t )
2
y (t )
x (t )
− ωm
cos(ω c t )
©Yao Wang, 2006
− ωm
LPF
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Frequency Domain Interpretation of
Demodulation
Figure 7.7 in
Signals and
Systems
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Temporal Domain Interpretation
Modulation :
y (t ) = x(t )cos(2πf c t )
Demodulation :
w(t ) = y (t ) cos(2πf c t ) = x(t )cos 2 (2πf c t )
1
(1 + cos(2θ ) )
2
1
1
1
w(t ) = (1 + cos(4πf c t ) )x(t ) = x(t ) + x(t ) cos(4πf c t )
2
2
2
The LPF will retain the first term and remove the second term.
Using the equality cos 2 (θ ) =
©Yao Wang, 2006
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Example
•
How to transmit a signal with frequency ranging in (-5KHz,5KHz) using a channel
operating in (100KHz,110KHz)? What should be the carrier frequency ? Draw the
block diagrams for the modulator and demodulator, and sketch the spectrum of the
modulated and demodulated signals.
©Yao Wang, 2006
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Frequency Division Multiplexing:
Frequency domain interpretation
Figure 7.22 in
Signals and
Systems
ya (t ) = xa (t ) cos(ω a t )
ya (t ) = xa (t ) cos(ω a t )
ya (t ) = xa (t ) cos(ω a t )
w(t ) = ya (t ) + yb (t ) + yc (t )
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FDM Transmitter
Figure 7.21 in Signals and Systems
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FDM Receiver
Demultiplexing
Demodulation
cos(ω a t )
Figure 7.23 in Signals and Systems
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Example
•
How to transmit two signals each with frequency ranging in (-10KHz,10KHz) over a
channel operating in the frequency range (300KHz,340KHz)? Draw the block
diagrams for the modulator and demodulator, and sketch the spectrum of the
modulated and demodulated signals.
©Yao Wang, 2006
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Demo: modulating a sound signal
(amplitude_modulation.m)
X1 Spectrum
X1 Waveform
0
10
0.2
0
-10
fs=22k
10
-0.2
5
5.01 5.02 5.03 5.04 5.05
0
4
x 10
Modulated X1: Waveform
5
10
4
x 10
Modulated X1: Spectrum
0
10
0.2
0
-10
fc=50k
10
-0.2
5
©Yao Wang, 2006
5.01 5.02 5.03 5.04 5.05
4
0
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x 10Communications
5
10
4
x 10
19
DeModulated X1: Waveform
DeModulated X1: Spectrum
0
10
0.5
0
-10
-0.5
10
5
5.01 5.02 5.03 5.04 5.05
0
4
5
10
4
x 10
Reconstructed X1: Waveform
x 10
Reconstructed X1: Spectrum
0
10
0.2
0
-10
10
-0.2
5
©Yao Wang, 2006
5.01 5.02 5.03 5.04 5.05
0
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Communications
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10
5
10
4
10
20
Lowpass Filter
0.12
Magnitude (dB)
0
0.1
0.08
-50
-100
-150
0
2
4
0.06
6
Frequency (Hz)
8
6
Frequency (Hz)
8
10
4
x 10
0
Phase (degrees)
0.04
0.02
0
-200
-400
-600
-800
0
5
10
15
20
25
0
2
4
10
4
x 10
Length=20, Cut-off freq=11k
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Original and Reconstructed Waveform
0.3
original
reconstructed
0.2
original
0.1
0
-0.1
-0.2
-0.3
reconstructed
©Yao Wang, 2006
5000
5005
5010
5015
5020
5025
5030
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5035
5040
5045
5050
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Quadrature Amplitude Modulation
• With amplitude modulation: a signal with bandwidth B
needs 2B channel bandwidth
– This is called double sideband (DSB) AM
– Other techniques can reduce the bandwidth requirement
• Single sideband (SSB)
• Vestigial sideband (VSB)
• By using QAM, we can send 2 signals each with
bandwidth B over a channel bandwidth of 2B
– Equivalent to each signal with bandwidth B
©Yao Wang, 2006
EE3414: Analog Communications
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Quadrature Amplitude Modulation
(QAM)
• A method to modulate two signals onto the same carrier
frequency, but with 90o phase shift
cos( 2π f1t )
cos( 2π f1t )
s1 ( t )
s1 ( t )
m (t )
LPF
m (t )
LPF
s2 (t )
s2 (t )
sin( 2π f1t )
sin( 2π f1t )
QAM modulator
©Yao Wang, 2006
EE3414: Analog Communications
QAM demodulator
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QAM in more detail
Proof (in time domain) the demodulator can separate the signal on board!
Discuss the sensitivity of the system to synchronization of the carrier signal.
©Yao Wang, 2006
EE3414: Analog Communications
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Other Modulation Methods
y (t ) = x(t ) cos(2πf c t +θ 0)
• Amplitude modulation
– The amplitude of the carrier signal is controlled by the modulating signal
– Pitfall of AM: channel noise can corrupt the amplitude easily.
• Frequency modulation
y (t ) = cos(θ (t )),
dθ (t )
= 2πf c t + k f x(t )
dt
– The frequency of the carrier signal is proportional to the modulating
signal
• Phase modulation
y (t ) = cos(2πf c t + θ 0 + k p x(t ))
– The phase of the carrier signal is proportional to the modulating signal
©Yao Wang, 2006
EE3414: Analog Communications
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Application of Modulation and FDM
• AM Radio (535KHz--1715KHz):
– Each radio station is assigned 10 KHz, to transmit a mono-channel
audio (bandlimited to 5KHz)
– Using Amplitude modulation to shift the baseband signal
• FM Radio (88MHz--108 MHz):
– Each radio station is assigned 200 KHz, to transmit a stereo audio.
– The left and right channels (each limited to 15KHz) are multiplexed into
a single baseband signal using amplitude modulation
– Using frequency modulation to shift the baseband signals
• TV broadcast (VHF: 54-88,174-216MHz, UHF:470-890MHz)
– Each station is assigned 6 MHz
– The three color components and the audio signal are multiplexed into a
single baseband signal
– Using vestigial sideband AM to shift the baseband signals.
©Yao Wang, 2006
EE3414: Analog Communications
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What Should You Know
• Understand the bandwidth requirement
– Channel bandwidth > signal bandwidth
• Understand the principle of amplitude modulation
– Know how to modulate a signal to a certain frequency
– Know how to demodulate a signal back to the baseband
– Can write the equation and draw block diagram for both modulation and
demodulation
– Can plot the signal spectrum after modulation and demodulation
• Understand the principle of frequency division multiplexing
– Can write the equation and draw block diagram for both modulation and
demodulation, for multiplexing of two to three signals.
• Understand how do AM and FM radio and analog TV work in terms
of modulation and multiplexing.
©Yao Wang, 2006
EE3414: Analog Communications
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References
• A. M. Noll, Chapter 10.
• A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd
edition, Chapter 8, Sec. 8.1-8.3 (copies provided)
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