communication systems, 5ebazuinb/ece4600/ch04_1.pdfphasor analysis am • given a tone message … s...
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![Page 1: Communication Systems, 5ebazuinb/ECE4600/Ch04_1.pdfPhasor Analysis AM • Given a tone message … s t A c 1 cos 2 f m t cos 2 f c t • A positive frequency phasor can be defined](https://reader033.vdocuments.us/reader033/viewer/2022043023/5f3e9e660b27b517f17b4bf4/html5/thumbnails/1.jpg)
Communication Systems, 5e
Chapter 4: Linear CW Modulation
A. Bruce CarlsonPaul B. Crilly
© 2010 The McGraw-Hill Companies
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Chapter 4: Linear CW Modulation
• Bandpass signals and systems• Double-sideband amplitude modulation• Modulation and transmitters• Suppressed-sideband amplitude modulation• Frequency conversion and demodulation
© 2010 The McGraw-Hill Companies
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Linear CW Modulation
• For CW types of modulation, we are concerned with the pre-multiplier of the cosine
• The message is carried by the signal envelope.
t
3f2p01 dm2tmtf2costm1Ats
tf2costAts 0
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MATLAB AM
• AM=(carrier+mu*message.*carrier);• AM=(1+μ*m(t))*cos(2π*fc*t)
4
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1Message
Time (sec)
Am
plitu
de
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8-1.5
-1
-0.5
0
0.5
1
1.5AM Waveform
Time (sec)
Am
plitu
de
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Signal Spectrum for AM
• Baseband Signal Spectrum• AM Signal Spectrum
5
0 0.5 1 1.5 2 2.5 3
x 104
-150
-100
-50
0Sequenctial FFTs of the message
Frequency (Hz)
Pow
er (d
B)
0 0.5 1 1.5 2 2.5 3
x 104
-150
-100
-50
0Sequenctial FFTs of the AM Waveform
Frequency (Hz)
Pow
er (d
B)
AM_Siggen.m
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Baseband vs. Bandpass Signals
• Baseband: A bandlimited waveform typically centered at f=0,
• Bandpass: A bandlimited waveform with a carrier
fW
WffAfV
,0
,
fWf,0
WffWf,ffA
Wff,0
fV
c
ccc
c
tAtv
tf2costAtv c
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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
(a) Spectrum; (b) Waveform
Bandpass signal
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CW Waveform Envelope and Phase
• The envelope is always positive• The phase includes the carrier phase and the
“sign” of the modulated envelope
tf2costAtv c
tAtenv
tAargtphase
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9
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
(a) Rotating phasor;
(b) Phasor diagram with rotation suppressed
Envelope and phase Quadrature Carriers
CW Phasor Diagrams
jtfjtAtv ccomplex 2exp
jtAtvcomplex exp
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Quadrature Decomposition
• The real signal may be decomposed into orthogonal (90 deg. separation) components whenever a phase offset exists.– Using a cos(a+b) expansion
tftvtftv
tftAtftA
tftAtv
cqci
cc
cbp
2sin2cos2sinsin2coscos
2cos
This defines the linear combination of two baseband orthogonal phasors (cos and sin) rotating in the same direction. (Figure b from the previous slide)
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Quadrature to Envelope-Phase
• Conversion is as expected
sintAtvq
costAtvi 2q
2i tvtvtA
tvtv
i
qarctan
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AMExample.m
• Tracking the “complex” AM waveform
12-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
Time (sec)
Am
plitu
de
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Lowpass Equivalent Signal
• Using the quadrature components, an equivalent low-pass version of the signal can be defined as
tvjtv21tv qilp
jexptA21tvlp
• Comparing this to the bandpass signal jtfjtvtv clpbp 2expRe2
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Complex Baseband Processing
• When the bandpass signal is represented as a low-pass equivalent, we can perform “equivalent” low-pass signal processing to validate performance.
• When built, the bandpass implementation will be performed with the system performance following the baseband signal analysis.
Mod FilterMessageIn
tf2jexp c
Mod FilterMessageOut
tf2jexp c
Filter
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Real and Complex FFTs
• Hilbert Transform or Complex AM FFT spectrum• Real AM FFT spectrum
150 1000 2000 3000 4000 5000 6000 7000 8000 9000-150
-100
-50
0
50
100
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A family of AM Waveforms
• Double-Sideband, Carrier (AM in text, AM radios)
• Double-Sideband, Suppressed Carrier (DSB in text)
tf2costAtv c
tmAtA c
tm1AtA c
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Conventional AM
• Baseband
• Bandpass
• Fourier Domain
tm1AtA c
tf2costmAtf2cosA
tf2costm1Atv
cccc
cc
ccc
ccc ffMffM
2Affff
2AfV
carrier message
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AM Sidebands
• Assume that the message is a cosine wave– messages are typically bounded by +/- 1.0
• The spectral response becomes tf2costm m
mcmcc
mcmcc
ccc
ffffff4
A
ffffff4
A
ffff2
AfV
mm ffff21fM
carrier
sidebands
sidebands
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AM Positive Frequencies
• For the positive frequency segment of the spectrum
mcc
cc
mcc fff
4Aff
2Afff
4AfV
Signal carrier with two sidebands
Lower sideband Upper sidebandCarrier
Mag
cfmc ff mc ff
2Ac
4Ac
4Ac Negative part
of spectrum not shown
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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Message
AM with aValid Mod Index
AM with anInvalid Mod Index
AM waveforms (a) Message;
(b) AM wave with < 1; (c) AM wave with > 1
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AM Modulation index
• For normalized message signals, the magnitude of the modulation index , , defines whether there is a “phase reversal” of the carrier wave.– =1, 100% modulation index, if the message ever
becomes -1, the baseband signal will become zero– <1, 0-99.9% modulation index, if the message is
bounded by +/-1, the baseband signal is always greater than zero
– >1, the baseband signal will have regions that are negative; thereby, causing a phase reversal of the carrier for the negative values. Envelope distortion will occur.
tm1AtA c
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DSB (with Suppressed Carrier)
• Baseband
• Bandpass
• Fourier Domain
tmAtA c
tf2costmAtv cc
ccc ffMffM
2AfV
Message only no carrier !
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DSB Sidebands
• Assume that the message is a cosine wave– messages are typically bounded by +/- 1.0
• The spectral response becomes tf2costm m
mcmcc
mcmcc
ffffff4
A
ffffff4
AfV
mm ffff21fM
sidebands
sidebands
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DSB Positive Frequencies
• For the positive frequency segment of the spectrum
mcc
mcc fff
4Afff
4AfV
No signal carrier, two sidebands
Lower sideband Upper sideband
Mag
cfmc ff mc ff
4Ac
4Ac
Negative part of spectrum not shown
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25
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Message
DSB
DSB-SC waveforms
DSB-SC has lots of phase reversals !
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Phasor Analysis AM
• Given a tone message …
tf2costf2cos1Ats cmc
• A positive frequency phasor can be defined and drawn
tf2cos1tA mm
tff2jexp2
Atff2jexp2
Atf2jexp2
Ats mcc
mcc
ccC
fpos
tff2cos2
Atff2cos2
Atf2cosAts mcc
mcc
cc
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Phasor Analysis AM (2)
• A positive frequency phasor can be defined and drawn
2A c
4
A c
4
A c
cf
mf
mf
tff2jexp2
Atff2jexp2
Atf2jexp2
Ats mcc
mcc
ccC
fpos
The message phasors add so only the magnitude of the carrier phasor appears to change.
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Phasor Analysis DSB
• Given a tone message … tf2cosAtm mm
tf2costf2cosAAts cmmc
tff2costff2cos2AAts mcmc
mc
• A positive frequency phasor can be defined and drawn
tff2jexptff2jexp
4AAts mcmc
mcCfpos
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Phasor Analysis DSB (2)
tff2jexptff2jexp
4AAts mcmc
mcCfpos
• A positive frequency phasor can be defined and drawn
4AA
m
c
4AA
m
c
cfmf
mf
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Power in AM Waveform tf2costm1Ats cc
2cc
2sT tf2costm1AtsPS
2c222
cs tf2costmtm21AP
2c222
c
2c
2c
2c
2cs
tf2costmA
tf2costm2A
tf2cosAP
zero mean
message
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Power in AM• Assume that the message is an independent, zero mean, random process
2T
2T
2cT
222c
2T
2T
2cT
2cs
dttf2coslimtmA
dttf2coslimAP
222
c2
cs tm
2A
2AP
sbcm2
2c
2c
s P2PP2
A2
AP
Carrier and symmetric subbands
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Power Levels
• Carrier to one sideband power
• Maximum magnitude of mod index and messagem
22
c
2c
sb
c
P2
A21
2A
PP
m2
sb
c
P2
PP
2112
PP
sb
c or csb P21P
Pow
er
cfmc ff mc ff
4A 2
c
8A 22
c 8
A 22c
Carrier and symmetric subbands
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Power Levels
• Total Signal power to one sideband power ratio
• Maximum magnitude of mod index and message
42112
PP
sb
s or ssb P41P
m2
2c
m2
2c
2c
sb
s
P2
A21
P2
A2
A
PP
2
P2
PP
m2
sb
s
Pow
er
cfmc ff mc ff cfmc ff mc ff
4A 2
c
8A 22
c 8
A 22c
4A 2
c
8A 22
c 8
A 22c
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AM Signal Power Summary
• Most of the AM signal power is in the carrier.– Less than 25% of the total signal power is in the
message subbands !
• This seams like a waste of power, but it makes it easier to design an AM radio receiver !– Lock on to the carrier– Then demodulate the AM signal
34
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DSB Power Levels tf2costmAts cc
2cc
2DSBT tf2costmAtsPS
2c22
c2
c22
cDSB tf2costmAtf2costmAP
21PAP m
2cDSB
• All the power is in the message subbandsP
ower
freqcfmc ff mc ff cfmc ff mc ff
8A 2
c8
A 2c 8
A 2c
8A 2
c
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Comparing AM and DSB
• DSB Power vs. 2 AM Sideband Power– Assume 100% mod index, max m(t)=1 and the
waveform maximum magnitudes are identical– For DSB, Ac=Amax– For AM, Amax=Ac + Ac= 2Ac
44
P2
2A
P2
A
PP
2
m2
2max
m
2max
sb
DSB
4+ times more message power at the same transmitter power
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Comparing AM and DSB (2)
44P
P2
sb
DSB
Why would we use AM instead of DSB?
Pow
er
freqcfmc ff mc ff cfmc ff mc ff
4A 2
c
8A 22
c 8
A 22c
4A 2
c
8A 22
c 8
A 22c
Power
freqcfmc ff mc ff cfmc ff mc ff
8A 2
c8
A 2c 8
A 2c
8A 2
c