Communication Systems, 5e

Chapter 4: Linear CW Modulation

A. Bruce CarlsonPaul B. Crilly

© 2010 The McGraw-Hill Companies

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

3

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

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

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

6

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

7

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(a) Spectrum; (b) Waveform

Bandpass signal

8

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

9

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(a) Rotating phasor;

(b) Phasor diagram with rotation suppressed

Envelope and phase Quadrature Carriers

CW Phasor Diagrams

jtfjtAtv ccomplex 2exp

jtAtvcomplex exp

10

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)

11

Quadrature to Envelope-Phase

• Conversion is as expected

sintAtvq

costAtvi 2q

2i tvtvtA

tvtv

i

qarctan

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

13

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

14

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

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

16

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

17

Conventional AM

• Baseband

• Bandpass

• Fourier Domain

tm1AtA c

tf2costmAtf2cosA

tf2costm1Atv

cccc

cc

ccc

ccc ffMffM

2Affff

2AfV

carrier message

18

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

19

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

20

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

21

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

22

DSB (with Suppressed Carrier)

• Baseband

• Bandpass

• Fourier Domain

tmAtA c

tf2costmAtv cc

ccc ffMffM

2AfV

Message only no carrier !

23

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

24

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

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 !

26

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

27

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.

28

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

29

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

30

Power in AM Waveform tf2costm1Ats cc

2cc

2sT tf2costm1AtsPS

2c222

cs tf2costmtm21AP

2c222

c

2c

2c

2c

2cs

tf2costmA

tf2costm2A

tf2cosAP

zero mean

message

31

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

32

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

33

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

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

35

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

36

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

37

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