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ELE3340 Analog and Digital Communications 2. Amplitude Modulation

ELE3340

ANALOG AND DIGITAL

COMMUNICATIONS

2. AMPLITUDE MODULATION

Wing-Kin Ma, Dept. Electronic Eng., The Chinese University of Hong Kong 1


QUICK REVIEW OF SOME BASIC CONCEPTS

Fourier transform of a signal x(t):

X() =

x(t)ejtdt

and its inverse

x(t) = 1

2

X()ejtd

Note that is in rad/sec.

We use the notation

x(t) X()

to mean that X() is the Fourier transform ofx(t), & thatx(t) is the inverse Fourier

transform ofX().

In addition, F[x(t)] stands for the Fourier transform ofx(t).


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Some Properties:

Frequency translation:

x(t)ejct X( c)

x(t) cos(ct) 1

2

X( c) + X(+ c)

x(t) sin(ct) 1

2j

X( c) X(+ c)

Convolution: let be the convolution operator.

x(t) y(t) X()Y()

x(t)y(t) X() Y()



Dirac delta: let (t) be the Dirac delta function.

x(t) (t ) =x(t )

(t) 1

1 2()


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DOUBLE-SIDEBAND SUPPRESSED CARRIER

Let m(t) be a (baseband) message signal that is real-valued, such as voice (in more

advanced study we may have complex-valued message signals).

In double-sideband suppressed carrier (DSB-SC) modulation, the transmitted

signal or the modulated signal is

DSBSC(t) =m(t) cos(ct)

The signal cos(ct) is called the carrier signal, & c the carrier frequency. The

message signal m(t) is also called the modulating signal.

The modulation process is depicted as follows:

ttmc

cos)()(tm

cosct

(Modulating signal) (Modulated signal)

(Carrier)



DSB-SC modulation simply shifts the freq. contents ofm(t) to the carrier frequency.

Specifically, ifm(t) M() then

m(t) cos(ct) 1

2[M(+ c) + M( c)]

( )M

DSB-SC( )

c

c2 B2 B

4 B

Suppose that m(t) has a (baseband) bandwidth ofBHz (or 2Brad/sec). Then

DSB-SC modulation occupies a bandwidth of2BHz.


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The portion of the spectrum that lies above c is called the upper sideband (USB).

The portion of the spectrum that lies below c is called the lower sideband (LSB).

DSB-SC( )

c

USBLSB

c

LSBUSB



Demodulation

Conceptually, this can be done by frequency-shifting DSBSC(t) back to the baseband.

( )e tDSB-SC

( )t

cosct

Low-pass filter

1( )

2m t

Since

e(t) =DSBSC(t)cos(ct) =m(t)cos2(ct) =

1

2m(t)[1 + cos(2ct)]

we have

E() =

1

2 M() +

1

4 [M(+ 2c) + M( 2c)]Using a low-pass filter with cutoff freq. at BHz, we may keep 1

2M() while eliminating

14

[M(+ 2c) + M( 2c)].Wing-Kin Ma, Dept. Electronic Eng., The Chinese University of Hong Kong 8

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DSB-SC( )

cc

( )E

2 c2 c 2 B2 B



The method of recovering m(t) described above is called coherent demodulation or

synchronous demodulation.

It requires that the local carrier at the receiver has exact frequency and phase

synchronism with respect to that at the transmitter.

Ex 2.1 Suppose that the local carrier generated at the receiver is cos(ct + ), where

is a phase error due to asynchronism. Derive the resulting coherent demodulator

output. What happens if= /2?

The problem of frequency & phase synchronization can be handled by carrier

acquisition methods.

Alternatively, we can consider modulation that enables noncoherent demodulation.


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Ex 2.2 Suppose that the message signal is a cosine wave

m(t) = cos(mt).

Sketch the time-domain waveform & the spectrum of the modulated signal.



Ex 2.3 (Switching modulator) We consider an alternate implementation of modulation,

in which the carrier multiplication operation is replaced by a simpler switching

operation:

( )e t( )m t

( )p t

Band-pass filterDSB-SC

( )t

Here, m(t) is multiplied by a periodic pulse train with period Tc= 2c

:

p(t) =

1, 0 |t| < Tc4

0, Tc4 |t| < Tc

2

andp(t) =p(t + nTc) for all n.

Determine the Fourier transform ofe(t). Explain why and how a bandpass filterapplying to e(t) can generate the DSB-SC modulated signal.


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AMPLITUDE MODULATION

Amplitude modulation (AM) is a scheme that enables noncoherent demodulation.

The AM modulated signal is given by

AM(t) = (A + m(t)) cos(ct)

where A is chosen such that A + m(t)> 0 for allt.

Demodulation can be achieved by an envelope detector (noncoherent).



The spectrum ofAM(t)

AM() =1

2[M(+ c) + M( c)] + A[(+ c) + ( c)]

is similar to that of DSB-SC except for the presence of carrier.

( )M

AM( )

c

c2 B2 B

4 B


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Let mp be the peak amplitude (+ve or -ve) ofm(t); i.e., |m(t)| mp for all t.

The modulation index is defined as

=mp

A

In order to enable envelope detection, we need mp A and as a result

0 1



Power efficiency

The advantage of envelope detection in AM has its price. Some power has been spent

on the carrier which contains no information.

AM(t) =A cos(ct) c(t)

+ m(t) cos(ct) s(t)

Letx2(t) = limT1T

T /2T/2

x2(t)dtdenote the average power of the given signal x(t).

The power efficiency of AM is defined as

=useful power

total power =

s2(t)

c2(t) + s2(t)

= m2(t)

A2

+ m2

(t)


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Ex 2.4 Consider a sinusoidal message signal m(t) = cos(mt). Show, in this special

case, that the best possible power efficiency of AM can only be 33%.



QUADRATURE AMPLITUDE MODULATION

In DSB-SC and AM, the transmission bandwidth is 2BHz, twice of the message signal.

In quadrature amplitude modulation (QAM), two DSB signals are transmitted overthe same carrier frequency:

QAM(t) =m1(t) cos(ct) + m2(t)sin(ct)

where m1(t) & m2(t) denote the two message signals.

The message signals m1(t) &m2(t) arein-phase &quadrature-phase components of

QAM(t).

QAM generally requires coherent demodulation.


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Modulation and demodulation (coherent) of QAM:

Low-passfilter

Low-passfilter

2/ 2/

)(1

tm

)(2

tm

)(1

tx

)(2 tx

)(1

tm

)(2 tm

tc

cos

tcsin

tc

cos2

tcsin2

)(QAM t

Ex 2.5 Show how the coherent demodulator in the above fig. can recover m1(t) &

m2(t).



SINGLE SIDEBAND MODULATION

The idea is to transmit either the USB or LSB.

SSB-USB( )

cc

2 B

SSB-LSB( )

cc

2 B


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SSB may be seen as a special case of QAM.

Let

M+() =M()u(), M() =M()u()

be the +ve freq. & -ve freq. portions ofM(), respectively. (here u()is the unit step

function)

Let us focus on SSB modulation using USB

USB() =M+( c) + M(+ c)

Taking inverse transform ofUSB() yields

USB(t) =m+(t)ejct + m(t)e

jct

= [m+(t) + m(t)] m(t)

cos(ct) j[m(t) m+(t)] mh(t)

sin(ct)



Hilbert transform

Let

H() = jsgn() =

j =ej/2, >0

j =ej/2,

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( )h t

t

| ( ) |H

( )h

2

2



In SSB, the quadrature-phase component mh(t) is the Hilbert transform of

m(t). To see this,

Mh() = F[jm+(t) +jm(t)]

= jM+() +jM()

= jsgn()M()


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Ex 2.6 Show that for the LSB counterpart, the SSB-modulated signal can be expressed

as

LSB(t) =m(t) cos(ct) + mh(t) sin(ct)

Ex 2.7 Consider again a sinusoidal message signal m(t) = cos(mt). Assuming USB

transmission, sketch the spectrum and time-domain waveform of the modulated signal.



Modulation

There are two ways of performing SSB modulation.

Selective filtering method: apply a band-pass filter to the DSB-SC signal, eliminating

the unwanted sideband.

c

c

This necessitates a very sharp cutoff band in the filter design, which is not too easy to

achieve in practice (at least compared to AM).


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VESTIGIAL SIDEBAND MODULATION

The generation of SSB signals is rather difficult in practice.

The generation of DSB signals is simple, but DSB signals require twice the signal

bandwidth of SSB.

Vestigial sideband (VSB) modulation was designed to provide a compromise

between DSB and SSB.



The modulation process is similar to the selective filtering method in SSB.

VSB( )t

2cosct

Band-pass filter( )m t

( )iH

Instead of eliminating one sideband completely (as in SSB), we allow the band-pass

filter to have a gradual cutoff of one sideband.

This results in some increase in transmission bandwidth (say, 20%), but it also makes

the band-pass filter easier to realize.


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c

c

cc

( )iH DSB-SC

( )

VSB( )



The (coherent) demodulation process is similar to that in DSB-SC and SSB.

( )e tVSB ( )t

2cosct

Low-pass filter ( )m t

( )o

H

However, to ensure perfect recovery ofm(t), the low-pass filter Ho() is specially

designed.


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The VSB signal spectrum is given by

VSB() = [M(+ c) + M( c)]Hi()

At the demodulator, the signal e(t) has its Fourier transform given by

E() = VSB(+ c) + VSB( c)

=M()[Hi( c) + Hi(+ c)]

+ M(+ 2c)Hi(+ c) + M( 2c)Hi( c)

Now, if the demodulating low-pass filter Ho()is such that

Ho() = 1

Hi(+ c) + Hi( c), || 2B

andHo() = 0 for || >2B , then

M() =E()Ho() =M()



c

c

cc

( )iH

( )o

H


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