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Chapter 4
Amplitude Modulation
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Baseband vs Passband Transmission
Baseband signals:
Voice (0-4kHz)
TV (0-6 MHz)
A signal may be sent inits baseband format
when a dedicated wired
channel is available.
Otherwise, it must be
converted to passband.
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Modulation: What and Why?
The process of shifting the baseband signal topassband range is calledModulation.
The process of shifting the passband signal to
baseband frequency range is calledDemodulation.
Reasons for modulation:
Simultaneous transmission of several signals
Practical Design of Antennas
Exchange of power and bandwidth
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Types of (Carrier) Modulation
In modulation, one characteristic of a signal(generally a sinusoidal wave) known as thecarrieris changed based on the information
signal that we wish to transmit (modulatingsignal).
That could be the amplitude, phase, or frequency,which result in Amplitude modulation (AM),
Phase modulation (PM), or Frequencymodulation (FM). The last two are combined asAngle Modulation
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Types of Amplitude Modulation (AM)
Double Sideband with carrier (we will call it AM):This is the most widely used type of AM modulation.In fact, all radio channels in the AM band use this typeof modulation.
Double Sideband Suppressed Carrier (DSBSC):This is the same as the AM modulation above butwithout the carrier.
Single Sideband (SSB): In this modulation, only halfof the signal of the DSBSC is used.
Vestigial Sideband (VSB): This is a modification ofthe SSB to ease the generation and reception of thesignal.
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Double Sideband Suppressed Carrier (DSBSC)
Assume that we have a message signal m(t) withbandwidth 2B rad/s (orB Hz). m(t) M().
Let c(t) be a carrier signal, c(t) = cos(c
t), c
>> 2B
gDSBSC(t) = m(t)cos(ct)
(1/2) [M(c) +M(+ c)].
Xm(t)
c(t)
gDSBSC
(t)
DSBSC Modulator (transmitter)
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Time and Frequency Representation of DSBSC
Modulation Process
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DSBSC Demodulation
e (t)=gDSBSC(t)cos(ct)= m(t)cos2(ct)
= (1/2) m(t) [1 + cos(2ct)]
= (1/2) m(t) + (1/2) m(t) cos(2 ct)
E() (1/2)M() + (1/4) [M(
2 c) +M(+ 2 c)].
The output signalf(t) of the LPF will be
f(t) = (1/2) m(t) (1/2)M().
X
c(t)
gDSBSC
(t)e(t)
HLPF
()
BW = 2Bf(t)
DSBSC Demodulator (receiver)
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Time and Frequency Representation of DSBSC
Demodulation Process
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Modulator Circuits
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Switching Modulators
Any periodic function can be expressed as a
series of cosines (Fourier Series).
The information signal, m(t), can therefore be,equivalently, multiplied by any periodic
function, and followed by BPF.
Let this periodic function be a train of pulses.
Multiplication by a train of pulses can be
realized by simple switching.
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Switching Modulator Illustration
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Switching Modulator: Diode Bridge
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Switching Modulator: Ring
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Demodulation of DSBSC
The modulator circuits can be used for demodulation, but
replacing the BPF by a LPF of bandwidthB Hz.
The receiver must generate a carrier frequency in phase
and frequency synchronization with the incoming carrier.
This type of demodulation is therefore called coherent
demodulation (or detection).
X
c(t)
gDSBSC
(t)e(t)
HLPF
()
BW = 2B
f(t)
DSBSC Demodulator (receiver)
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From DSBSC to DSBWC (AM)
Carrier recovery circuits, which are required forthe operation of coherent demodulation, aresophisticated and could be quite costly.
If we can let m(t) be the envelope of themodulated signal, then a much simpler circuit,the envelope detector, can be used fordemodulation (non-coherent demodulation).
How can we make m(t) be the envelope of themodulated signal?
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Definition of AM
Shift m(t) by some DC value A
such thatA+m(t) 0. OrA mpeak
Called DSBWC. Here will refer to
it as Full AM, or simply AM
Modulation index m= mp/A.
0 m 1
)cos()()cos(
)cos()]([)(
ttmtA
ttmAtg
CC
CAM
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Spectrum of AM
)()(2
1)()()( CCCCAM MMAtg
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The Buy and Price of AM
Buy: Simplicity in demodulation.
Price: Waste in Power
gAM(t) =Acosct + m(t) cosct
Carrier Power Pc =A2/2 (carries no information)
Sideband Power Ps
= m2(t) /2=Pm
/2 (useful)
Power efficiency = h= Ps/(Pc + Ps)= Pm/(A2 +Pm)
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Tone Modulation
m(t) = mcos(mt) and m2(t) = (mA)2/2
g(t)=[A+Bcos(mt)] cosct=A[1+mcos(mt)] cosct
h= m2/(2+m2) 100%
Under best conditions, m=1hmax =1/3 =33%
For m= 0.5, h= 11.11%
For practical signals, h< 25%
? Would you use AM or DSBSC?
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Generation of AM
AM signals can be generated by any DSBSC
modulator, by usingA+m(t) as input instead of
m(t).
In fact, the presence of the carrier term can
make it even simpler. We can use it for
switching instead of generating a local carrier.
The switching action can be made by a single
diode instead of a diode bridge.
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AM Generator
A >> m(t)(to ensure switchingat every period).
vR=[cosct+m(t)][1/2 + 2/(cosct-1/3cos3ct+ )]
=(1/2)cosct+(2/)m(t) cosct+ other terms (suppressed by BPF)
vo(t) = (1/2)cosct+(2/)m(t) cosct
cos(ct)
m(t)
R BPF vo(t)
A
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AM Modulation Process (Frequency)
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AM Demodulation: Rectifier Detector
Because of the presence of a carrier term in the
received signal, switching can be performed in
the same way we did in the modulator.
[A+m(t)]cos(ct)
LPF m(t)
C
R
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Rectifier Detector: Time Domain
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Rectifier Detector (Frequency Domain)
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Envelope Detector
When D is forward-biased, the capacitor charges andfollows input.
When D is reverse-biased, the capacitor dischargesthroughR.
[A+m(t)]cos(ct)
vo(t)RC
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Envelope Detection
The operations of the circuit requires
careful selection of t=RC
IfRCis too large, discharging will be
slow and the circuit cannot follow a
decreasing envelope.
WhenRCis too small the ripples will
be high.
1/(2B)
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Quadrature Amplitude Modulation (QAM)
In DSBSC or AM the modulated signal
occupies double the bandwidth of the baseband
signal.
It is possible to send two signals over the same
band, one modulated with a cosine and one
with sine.
Interesting enough, the two signals can bereceived separately after demodulation.
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m1(t)cos(ct) HLPF
()
BW = 2BXm1(t)
cos(ct)
QAM Modulator/Demodulator
m2(t)sin(
ct)
Xm2(t)
sin(ct) Phase Shifter/2
m1(t)cos(ct) + m2(t)sin(ct)
X m1(t)/2
cos(ct)
X m2(t)/2
sin(ct) Phase Shifter/2
m1(t)cos2(
ct) + m
2(t)sin(
ct)cos(
ct)
=m1(t)/2+m
1(t) cos(2
ct)/2 + m
2(t)sin(2
ct)/2
HLPF
()
BW = 2B
Baseband Around c Around c
m1
(t)sin(c
t)cos(c
t) + m2
(t)sin2(c
t)
=m1(t)sin(2
ct)/2 + m
2(t)/2 m
2(t)cos(2
ct)/2
BasebandAround c Around c
QUADRATURE
modulator branch
IN-PHASE
modulator branch
QUADRATUREdemodulator branch
IN-PHASE
demodulator branch
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m1(t)cos(
ct)
HLPF
()
BW = 2BXm1(t)
cos(ct)
QAM Modulator/Demodulator with Demodulator Carrier Phase and/or Frequency Error
m2(t)sin(
ct)
Xm2(t)
sin(ct)
Phase Shifter
/2
m1(t)cos(ct) + m2(t)sin(ct)
X (1/2)[m1(t)cos(t+)m2(t)sin(t+)]
cos[(c+)t+
X (1/2)[m1(t)sin(t+) + m2(t)cos(t+)]
sin[(c+)t+
Phase Shifter
/2
m1(t)cos(
ct)cos[(
c+)t+ + m
2(t)sin(
ct)cos[(
c+)t+
=(1/2)[m1(t)cos(t+) + m
1(t) cos(2
ct+t+) m
2(t)sin(t+) + m
2(t)sin(2
ct+t+)]
HLPF
()
BW = 2B
Baseband Around c Around c
BasebandAround c Around c
Baseband
m1(t)cos(ct)sin[(c+)t+ + m2(t)sin(ct)sin[(c+)t+=(1/2)[m
1(t)sin(t+) + m
1(t) sin(2
ct+t+) + m
2(t)cos(t+) m
2(t)cos(2
ct+t+)]
Baseband
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Single-Side Band (SSB) Modulation
DSBSC (as well as AM) occupies double the
bandwidth of the baseband signal, although the two
sides carry the same information.
Why not send only one side, the upper or the lower? Modulation: similar to DSBSC. Only change the
settings of the BPF (center frequency, bandwidth).
Demodulation: similar to DSBSC (coherent)
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SSB Representation
How would we
represent the SSB signal
in the time domain?
gUSB(t) = ?gLSB(t) = ?
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Time-Domain Representation of SSB (1/2)
M() =M+() +M-()
Let m+(t)M+() and m-(t)M-()
Then: m(t) = m+(t) + m-(t) [linearity]
Because M+
(),M-
() are not even
m+(t), m-(t) are complex.
Since their sum is real they must be
conjugates.
m+(t) = [m(t) +jmh(t)]
m-(t) = [m(t) -jmh(t)]
What is mh(t) ?
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Time-Domain Representation of SSB (2/2)
M() =M+() +M-()
M+() =M()u();M-() =M()u(-)
sgn()=2u() -1u()= + sgn(); u(-) = - sgn()
M+() = [M() +M()sgn()]M-() = [M() - M()sgn()]
Comparing to:
m+(t) = [m(t) +jmh(t)] [M() +j Mh()]
m-(t) = [m(t) -jmh(t)] [M() - jMh()]We find
Mh() = - j M()sgn() where mh(t)Mh()
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Hilbert Transform
mh(t) is known as the Hilbert Transform (HT) ofm(t).
The transfer function of this transform is given by:H() = -j sgn()
It is basically a /2 phase shifter
H() = jsgn()
j
j
|H()| = 1
1
/2
/2
) ) sgn)
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Hilbert Transform of cos(ct)
cos(ct) (c) + (+ c)]
HT[cos(ct)] -j sgn() (c) + (+ c)]= j sgn() (c) (+ c)]
= j (c) + (+ c)]
= j (+ c) - (- c)] sin(ct)
Which is expected since:
cos(ct-/2) = sin(ct)
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Time-Domain Operation for Hilbert
Transformation
For Hilbert Transformation H() = -j sgn().
What is h(t)?
sgn(t) 2/(j) [From FT table]
2/(jt) 2sgn(-) [symmetry]
1/(t) -j sgn()
SinceMh() = - j M()sgn() =H() M()
Then
dt
m
tmt
tmh
)(1
)(*1
)(
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)sin()()cos()(
)(2
1)(
2
1
)(2
1)(
2
1)(
)sin()()cos()(
)(2
1)(
2
1
)(2
1)(
2
1)(
ttmttm
etjmetm
etjmetmtg
ttmttm
etjmetm
etjmetmtg
ChC
tj
h
tj
tj
h
tj
LSB
ChC
tj
h
tj
tj
h
tj
USB
CC
CC
CC
CC
)()()(
)()()(
CCLSB
CCUSB
MMG
MMG
tjtj
LSB
tjtj
USB
CC
CC
etmetmtg
etmetmtg
)()()(
)()()(
Finally
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Generation of SSB
Selective Filtering Method
Realization based on spectrum analysis
Phase-Shift Method
Realization based on time-domain expression
of the modulated signal
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Selective Filtering
GDSBSC
()
C+2B
C2B
C
C
C+2B
C2B
USBLSBLSBUSB
M()
+2B
2B CC
GUSB
()
C+2B
C
C
C2B
USBUSB
GLSB
()
C2B CC C+2B
LSBLSB
HUSB
()
C+2B
C2B
C
C
C+2B
C2B
HLSB
()
C+2B
C2B
C
C
C+2B
C2B
BW = 2B (B Hz)
Center Freq = c+B
BW = 2B (B Hz)
Center Freq = cB
M() (an example of an
audio signal)
5000 Hz 300 Hz 300 Hz 5000 Hz
Guard Band
of 600 Hz
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Phase Shifting
X
cos(ct)
SSB Modulator
X
sin(ct)
Phase Shifter
/2
Phase Shifter
/2
m(t)
mh(t)
mh(t)sin(
ct)
m(t)cos(ct)
gSSB
(t)
gUSB
(t) if
gLSB
(t) if ++ or
(a)
(b) (c)
(d)
)sin()()cos()()(
)sin()()cos()()(
ttmttmtg
ttmttmtg
ChCLSB
ChCUSB
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Phase-shifting Method:
Frequency-Domain Illustration
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SSB Demodulation (Coherent)
X
cos(ct)
gSSB
(t)
(Upper or Lower
Side bands)
HLPF
()
BW = 2Bm(t)
SSB Demodulator (receiver)
)(21OutputLPF
)2sin()(2
1)]2cos(1)[(
2
1)cos()(
)sin()()cos()()(
tm
ttmttmttg
ttmttmtg
ChCCSSB
ChCSSB
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FDM in Telephony
FDM is done in stages
Reduce number of carrier frequencies
More practical realization of filters
Group: 12 voice channels 4 kHz = 48 kHzoccupy the band 60-108 kHz
Supergroup: 5 groups 48 kHz = 240 kHzoccupy the band 312-552
Mastergroup: 10 S-G 240 kHz = 2400 kHzoccupy the band 564-3084 kHz
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FDM Hierarchy
4
0
54
321
109
876
5432
1
1112
60 k
108 k
312 k
552 k
Group
Supergroup
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Vestigial Side Band Modulation (VSB)
What if we want to generate SSB using
selective filtering but there is no guard band
between the two sides?
We will filter-in a vestige of the other band.
Can we still recover our message, without
distortion, after demodulation?
Yes. If we use a proper LPF.
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Filtering Condition of VSB
X
2cos(ct)
m(t)H
VSB()
(BPF)g
VSB(t)
VSB Modulator (transmitter)
gDSBSC
(t)
X
2cos(ct)
gVSB
(t)H
LPF()
BW = 2Bm(t)
VSB Demodulator (receiver)
x(t)
)cos()(2)( ttmtg CDSBSC
)()()( CCDSBSC MMG
)()()()( CCVSBVSB MMHG
C
C
at
C
baseband
CVSB
Basebandat
CCVSB
MMH
MMHX
2
2
)2()()(
)()2()()(
)()()()()( MHHHZ CVSBCVSBLPF
)()(
1)(
CVSBCVSB
LPFHH
H
; || 2 B
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VSB FilteringH
VSB()
C
C
HVSB
(c)+H
VSB(
c)
Shifted filter
components
Band of Signal
HVSB
(c) = 1/[H
VSB(
c)+H
VSB(
c)]
over the band of the signal only
Band of Signal
What happens outside the
band of the demodulated
signal is not important . So,
the LPF does not have to
inverse this part.
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VSB Filter: Special Case
Condition For distortionless demodulation:
If we impose the condition on the filter at the modulator:
HVSB(c) +HVSB(c) = 1 ; || 2 B
Then HLPF
= 1 for|| 2 B (Ideal LPF)
HVSB() will then have odd symmetry around c over thetransition period.
)()(
1)(
CVSBCVSB
LPFHH
H
; || 2 B
M()
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GDSBSC
()
C
C
+2B
2B C
C
HVSB
()
2B (B Hz) < BW < 4B (2B Hz)
C
C
GVSB
()
C
C
X()
C
C
C
M()
C
C
HLPF
()
C
C
C
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AM Broadcasting
Allocated the band 530 kHz1600 kHz (with
minor variations)
10 kHz per channel. (9 kHz in some countries)
More that 100 stations can be licensed in the
same geographical area.
Uses AM modulation (DSB + C)
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AM station Reception
In theory, any station can be extracted from the stream of spectra by
tuning the receiver BPF to its center frequency. Then demodulated.
Impracticalities:
Requires a BPF with very high Q-factor (Q =fc/B).
Particularly difficult if the filter is to be tunable.
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Solution: Superheterodyne receiver
Step 1: Frequency Translation from RF to IF
Shift the desired station to another fixed pass band
(called Intermediate Frequency IF = 455 kHz)
Step 2: Bandpass Filtering at IFBuild a good BPF around IF to extract the desired
station. It is more practical now, because IF is
relatively low (reasonable Q) and the filter is not
tunable. Step 3: Demodulation
Use Envelope Detector
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The Local Oscillator
What should be the frequency of the local
oscillator used for translation from RF to IF?
fLO =fc +fIF (up-conversion)
or fLO =fcfIF (down-conversion)
Tuning ratio =fLO, max /fLO, min
Up-Conversion: (1600 + 455) / (530+455) 2
Down-Conversion: (1600455) / (530455) 12
Easier to design oscillator with small tuning ratio.
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Image Station Problem
While up-converting the desired station to IF, we are,at the same time, down-converting another station to IFas well.
These two stations are called image stations, and theyare spaced by 2x455=910kHz.
Solution:Before conversion, use a BPF (at RF) centered atfc ofthe desired station.
The purpose of the filter is NOT to extract the desiredstation, but to suppress its image. Hence, it does nothave to be very sharp.
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Superheterodyne Receiver Block Diagram
Antenna
IF Stage(intermediate frequency)
IF Amplifier
& IF BPF
X
Converter
(Multiplier)
a(t) b(t) d(t)
c(t)
Envelope Detector
Diode, Capacitor,
Resistor, & DC blocker
Audio Stage
Power amplifier
d(t) e(t) f(t) g(t)
Ganged RFBPF and
Oscillator
RF Stage(radio frequency)
RF Amplifier
& RF BPF
Local
Oscillator
cos[(c+
IF)t]
Notes:
With one knob, we are tuning the RF Filter
and the local oscillator.
The filter are designed with high gain
to provide amplification as well.