modulation techniques - wayne state universitywebpages.eng.wayne.edu/ece4700/lecture...

ERG2310A-II p. II-1

Modulation Techniques

Modulation: translates an information-bearing signal (message signal) to a new spectral location (frequency domain)

Frequency (f)

a(t) ↔ A(f)

-fw 0 +fw

Baseband signal

A( f-fc )

(fc –fw) fc (fc+fw) Frequency (f)0

Bandpass signal

• Communication channels bandpass transfer (frequency) response

⇒ translates the message signal to be within the channel transfer response

1013Hz5x1014 HzOptical

2GHz100GHzMillimeterwave

100MHz5GHzMicrowave

2MHz100MHzVHF

100kHz5MHzShortwave Radio

2kHz100kHzLongwave Radio

BandwidthCarrier frequencyFrequency Band

Selected frequency bands

BW

f0f

H(f)

fc: carrier frequency

• Facilitates antenna reception

ERG2310A-II p. II-2

Modulation Techniques

If more than one message signal utilizes a channel

modulation allows translation of different signals to different spectral locations

multiplexing allows two or more message signals to be transmitted by a single transmitter (frequency division multiplexing)

desired modulated signal can be selected by a receiver

f

A1(f)

f

A2(f)

f

A3(f)

modulator

modulator

modulator

f

A1c(f)

f

A3c(f)

f

A2c(f)

f

demodulatorf

channel

f

A2c(f)

f

Low-pass filter

demodulator

Low-pass filter

f

A3c(f)

fc1

fc3

fc2

fc2

fc3

fc3

fc2

fc1

fc3fc2fc1

ERG2310A-II p. II-3

Frequency Translation

Recall: If )()( 2c

tfj ffAeta c −↔π)()( fAta ↔ , then (Fourier transform pairs)

Consider a message signal x(t) , which is bandlimited to the frequency range 0 to Wand has its Fourier transform is X(f) , is multiplied by cos (2π fc t) .

{ } [ ])()(21)2cos()( ccc ffXffXtftx −++=ℑ π

Baseband signal Frequency translated signal

ERG2310A-II p. II-4

Recovery of Baseband Signal

To recovery the baseband signal, we can simply multiply the translated signal with cos (2π fc t).

[ ] )4cos(2)(

2)()2(cos)()2cos()2cos()( 2 tftxtxtftxtftftx cccc ππππ +==

We obtain the baseband signal x(t) and a signal whose spectral range extends from (2fc-W) to (2fc+W). As fc >> W, the extra signal is removed by a low-pass filter.

Frequency translated signal

Baseband signal

[ ])2()2(21)(

21

cc ffXffXfX ++−+

ERG2310A-II p. II-5

Analog (Continuous-Wave) Modulation

A parameter of a high-frequency sinusoidal carrier is varied proportionallyto the message signal x(t) .

[ ])(cos)()( tttAts c φω +=General modulated signal:

ωc : carrier frequency

A(t) : instantaneous amplitude

φ(t) : instantaneous phase deviation

When A(t) is linearly related to the modulating (message) signal

Amplitude modulation (AM)

When φ(t) is linearly related to the modulating signal

Phase modulation (PM)

When time derivative of φ(t) is linearly related to the modulating signal

Frequency modulation (FM)

[ ] )(cos)()( txttAts oc ∝+= A(t) where φω

[ ] )()(cos)( txttAts cc ∝+= (t) where φφω

[ ] )()(cos)( txttAts cc ∝+=dt

(t)d where φφω

*FM & PM are commonly called Angle Modulation

ERG2310A-II p. II-6


Unmodulated carrier frequency

Message signal

Amplitude-modulated signal

Angle-modulated signal (frequency-modulated)

ERG2310A-II p. II-7


Message signal

Unmodulated carrier

Phase-modulated signal

Frequency-modulated signal

ERG2310A-II p. II-8

Amplitude Modulation

The envelope of the modulated carrier has the same shape as the message signal.

The amplitude of the carrier wave [ Ac cos(2π fct) ] varies linearly with the baseband message signal x(t).

The standard form of an amplitude-modulated (AM) signal is given by:

[ ] )2cos()(~1)( tftxmAts cac π+=

Where is the normalized message signal and ma is called the modulation index.

x(t) s(t) envelope

Ac

)(~ tx

ERG2310A-II p. II-9

Amplitude Modulation: DSB-LC

[ ]

)cos()(~)cos()2cos()(~)2cos(

)2cos()(~1)(

ttxmAtAtftxmAtfA

tftxmAts

caccc

caccc

cac

ωωππ

π

+=+=

+=AM signal:

cc fπω 2= where

Double-sideband –large carrier (DSB-LC)

[ ] [ ])()(2

)()(2

)( ccac

ccc ffXffX

mAffff

AfS −+++−++= δδ

ERG2310A-II p. II-10

Amplitude Modulation: DSB-LC[ ] )2cos()(~1)( tftxmAts cac π+=

Ac

Distorted signal !

So, Ac has to be large enough or we have to control the modulation index ma.


ma < 1 ma = 1 ma > 1Ac(1+ma)

Ac(1-ma)Ac

Amplitude Modulation: DSB-LC[ ] )2cos()(~1)( tftxmAts cac π+=

x(t)

+DSB-LC

Effect of modulation index ma :

Define modulation depth =c

c

AA−maxA

For a sinusoidal message signal, Amax=Ac(1+ma), thus the modulation depth is ma .



Product modulator:

[ ] )2cos()(~1)( tftxmAts cac π+=Generation of DSB-LC signal:

Square-law modulator: + filter

vin vout

x(t)

cosωct

Nonlinear device

s(t)221 ininout vavav +=ttxv cin ωcos)( +=

4444 34444 21)(

1

21

22

221 cos)(21cos)()(

ts

ccout ttxaaatatxatxav ωω

++++= where Ac=a1 and ma=2a2/a1

Chopper/rectifier modulator:

+

-

+

-

Bandpass filter at ωc

vo(t)x(t)

Accosωct

ωc+

-

+

-

Bandpass filter at ωc

vo(t)x(t)

Accosωct



The chopper or rectifier can generate a periodic waveform whose fundamental frequency is ωc rad/sec.

The periodic signal p (t) can be represented as

∑∞

−∞=

=n

tjnn

cePtp ω)(

.)()()( ∑∞

−∞=

=n

tjnn

cetfPtptf ω

Applying the frequency translation property of the Fourier transform, we get

.)()}()({ ∑∞

−∞=

−=ℑn

cn nFPtptf ωω

Chopper/rectifier modulator:

Consider p(t)f(t):

Let f(t) = Accosωct + x(t)



Demodulation of DSB-LC signal:

By envelope detector: the diode cuts off the negative part of the DSB-LC signal while RC acts as a lowpass filter to retrieve the envelope.

cfRCW /1/1 >>>> where W is the message signal bandwidth


Amplitude Modulation: DSB-LC[ ] )2cos()(~1)( tftxmAts cac π+=DSB-LC signal:

Consider the average power of s(t) :

[ ]

[ ]

[ ] [ ]

[ ][ ]

sbc

xxac

cac

caaaac

caac

cac

s

PP

txPPmA

ttxmA

ttxmtxmtxmtxmA

ttxmtxmA

ttxmA

tsP

2

)(~12

0)2cos(0~)(~12

)2cos()(~2)(~1)(~2)(~12

)2cos(1(21)(~2)(~1

)(cos)(~1

)(

222

222

22222

222

222

2

+=

=+=

==+=

+++++=

+++=

+=

=

where

as and (t)x if

ω

ω

ω

ω

Thuscxaxacsbcc PPmPmAPAP 2222

21

41

21 === ;

where Psb : average power per sideband

ssbsccsbxaa PPPPPPPmtxm212

2111)( 2 ≥−=⇒≤⇒≤⇒≤ For

At least 50% of total transmitted power resides in the carrier term which conveys no information wasteful of power



Fraction of total transmitted power contained in the sidebands is:

xa

xa

cxac

cxa

sbc

sb

PmPm

PPmPPPm

PPP

2

2

2

2

122

+=

+=

+=µ

If x(t) is a single sinusoid, i.e. cosωmt, then

Thus, and is known as the transmission efficiency of DSB-LC AM system.

Example:

21)(2 == txPx

2

2

2 a

a

mm+

=µ

A given AM (DSB-LC) broadcast station transmits an average carrier power output of 40kW and uses a modulation index of 0.707 for sine-wave modulation. Calculate (a) the total average power output; (b) the transmission efficiency; and(c) the peak amplitude of the output if the antenna is represented by a 50-ohm resistive load.

Solution: (a) The total average power output is ).2/1(2 2acsbcs mPPPP +=+=

For ma = 0.707, .50)4/11(40 kW=+=sP

(b) The transmission efficiency is %.205.25.0

)707.0(2)707.0(

2

2

==+

=µ

(c) Consider .1042,2

622

×==⇒= cc RPARAP

The peak amplitude of the output is V.3414)1( =+ Ama


Amplitude Modulation: DSB-SCThe “wasted” carrier power in DSB-LC can be eliminated by setting ma=1 and suppressing the carrier.

Thus the modulated signals becomes

In this case, the carrier frequency component is suppressed, thus it is called double-sideband suppressed-carrier modulation (DSB-SC) .

Its spectral density is:

)2cos()(~)( tftxAts cc π=

[ ])()(2

)( ccc ffXffXA

fS −++=

Average Power of the modulated signal:

xc

xxc

cc

s

PA

txPPA

ttxA

tsP

2

22

222

2

41

)(~21

)(cos)(~)(

=∴

=

==

=

=

sb

sb

P

2P

where

ω


Amplitude Modulation: DSB-SC

A signal spectrum can be translated an amount ± ωc rad/sec in frequency by multiplying the signal with any periodic waveform whose fundamental frequency is ωc rad/sec.

The periodic signal p (t) can be represented as

∑∞

−∞=

=n

tjnn

cePtp ω)(

.)()()( ∑∞

−∞=

=n

tjnn

cetxPtptx ω

Applying the frequency translation property of the Fourier transform, we get

.)()}()({ ∑∞

−∞=

−=ℑn

cn nXPtptx ωω

Consider p(t)x(t):



Example:A periodic signal consists of the exponentially decreasing waveform e-at, 0 ≤ t < T, repeated every T seconds. A given signal f(t) is multiplied by this periodic signal. Determine an expression describing the spectrum and the time waveform of the resulting amplitude-modulated signal if all components except those centered at ±ωc, ωc = 2π/T, are discarded.

Solution: The Fourier series for the given periodic signal can be written as

∑∞

−∞=

=n

tjnn

cePtp ,)( ω where ∫ +−==

−−−T

c

aTtjnat

n jnae

Tdtee

TP c

0.111

ωω

The spectrum of the product p(t)f(t) is

.)(1)(1)1(1

ω+ω

ω−+ω−ω

ω+− −

cc

cc

aT Fja

Fja

eT

The corresponding terms in the Fourier series are

.)(1)(1)1(1

ω−

+ω+

− ω−ω− tj

c

tj

c

aT cc etfja

etfja

eT

Combining yields the time waveform

),cos()(12022

θ+ωω+

− −

ttfa

eT c

c

aT

where ),/(tan 10 acω−=θ − ./2 Tc π=ω



Generation of DSB-SC signal by balanced modulator:

AM Modulator

AM Modulator

+

)(21 tx

)(21 tx−

ttxA cc ωcos)(211

+

ttxA cc ωcos)(211

−

+

-tAtx cc ωcos)(

tA cc ωcos



Demodulation of DSB-SC signal:Assuming that the transmitted signal is

ttxts cωcos)()( =

To demodulate the signal, we have

ttxtxttxtts

c

cc

ωωω

2cos)()(cos)(cos)(

21

21

2

+==

Taking the Fourier transform of both sides, we get

)2()2(

)(}cos)({

414121

c

c

c

XXXtts

ωωωω

ωω

−+++

=ℑ


Amplitude Modulation: DSB-SCConsider a small frequency error, ∆ω, and a phase error, θ0, are introducedin the locally generated carrier signal at the receiver. The signal at the receiver becomes

].)2cos[()(])cos[()(

])cos[(cos)(])cos[()(

021

021

00

θωωθω

θωωωθωω

+∆+++∆=

+∆+=+∆+

ttxttx

tttxtts

c

ccc

After passing via the low-pass filter, the output is

].)cos[()()( 021 θω +∆= ttxteo

Phase error and frequency error results in undesirable distortion. In somecases, they vary randomly, resulting in unacceptable performance.

Remedy:Using a synchronized oscillator to recover the original signal f(t) from themodulated signal φ(t). (Synchronous detection, or coherent detection)



The original signal x(t) can be recovered from the modulated signal s(t) by multiplyings(t) by cosωct (i.e. synchronous detection).

The same circuits as those used for modulation can be used for demodulation withthe following minor differences.

1. Since the desired output spectrum is centered about ω=0 and therefore a low-passfilter is needed at the output.

2. The oscillator in the demodulator must be synchronized to the oscillator in the demodulator to achieve proper demodulation.

This is usually accomplished by either a direct connection if the modulator and demodulator are in close proximity or by supplying a sinusoid displaced in frequency but related to the modulator-oscillator frequency. The sinusoid is called a “pilot carrier”.



Pilot Carrier SystemsIt is a common method used in DSB-SC modulation to maintain synchronizationbetween modulator and demodulator.In this case, a sinusoidal tone whose frequency and phase are related to thecarrier frequency is generated and is sent outside the pass-band of the modulatedsignal so it will not alter the frequency response capability of the system.

A tuned circuit in the receiver detects the tone, translate it to the proper frequency,and uses it to correctly demodulate the DSB-SC signal.e.g. Stereo-multiplex system

cosωct÷2 frequency

dividerAtten.

x

∑

∑

∑

L

R

L-R

L+R

38kHz 19kHz

+

+

+

-

+

+

+ To transmitter

L-R (lower sideband)

L-R (upper sideband)

L+R

0 15 19 23 38 53

Audio (mono) DSB-SC

f in kHz

Spectrum used for stereo multiplexing before transmission

Pilo

t ca

rrie

r


Amplitude Modulation: DSB-SCPhase-Locked Loop (PLL)In pilot tone system, phase-locked loop is used to synchronize one sinusoidal to another.

A simplified phase-locked loop stereo demodulator.


Amplitude Modulation: QAM

Quadrature MultiplexingUsing the orthogonality of sines and cosines , it is possible to transmit andreceive two different signals simultaneously on the same frequency.

Thus, each signal can be recovered by synchronous detection of the receivedsignal using carriers of the same frequency but in phase quadrature.

ttfttfts cc ωω sin)(cos)()( 21 +=

ttfttftfttfttftts

cc

cccc

ωωωωωω

2sin)(2cos)()(cossin)(cos)(cos)(

221

121

121

22

1

++=+=

ttftfttfttftttftts

cc

cccc

ωωωωωω

2cos)()(2sin)(sin)(sincos)(sin)(

221

221

121

221

−+=+=

In the low-pass filter, all terms at 2ωcare attenuated, yielding

).()(),()(

221

2

121

1

tftetfte

==

x

xf1(t)

f2(t)

cosωct

sinωct

+

+∑ s(t)

x

x ½ f1(t)

½ f2(t)

cosωct

sinωct

LPF

LPF


Frequency Division Multiplexing (FDM)

Frequency-division multiplexing is the positioning of signal spectra in frequency such that each signal spectrum can be separated out from all the others by filtering.


Frequency Division Multiplexing (FDM)

Example: commercial radio and television receiver


Intermediate Frequency (IF)

Heterodyning means the translating or shifting in frequency.

In the heterodyne receiver the incoming modulated signal is translated in frequency, thus occupying an equal bandwidth centered about a new frequency, known as an intermediate frequency (IF), which is fixed and is not dependent on the received signal center frequency.

The signal is amplified at the IF before demodulation.

If this intermediate frequency is lower than the received carrier frequency but above the final output signal frequency, it is called a superheterodyne receiver.



Advantage:The amplification and filtering is performed at a fixed frequency regardless ofstation selection.

Disadvantage:Image-frequency problem

Two ways to solve this problemi. Choose the intermediate frequency as

high as possible and practical.ii. Attenuate the image frequency before

heterodyning.

The intermediate frequency chosen must be free from other strong transmissionsor otherwise the receivers will amplify these spurious signals as they leak into the high-gain IF stages.



Example:A given radar receiver operating at a frequency of 2.80 GHz and using the super-heterodyne principle has a local oscillator frequency of 2.86 GHz . A second radarreceiver operates at the image frequency of the first and interference results.

(a) Determine the intermediate frequency of the first radar receiver.(b) What is the carrier frequency of the second receiver?(c) If you were to redesign the radar receiver, what is the minimum intermediate

frequency you would choose to prevent image-frequency problems in the 2.80-3.00 GHz radar band?

Solution:

(a) MHz.60GHz80.2GHz86.2 =−=−= cLOIF fff

(b) GHz.92.2GHz12.0GHz80.22 =+=+= IFcIMAGE fff

(c) 100MHz.GHz;20.0GHz80.2GHz00.32 ≥=−=−≥ IFMINMAXIF ffff

modulation techniques - wayne state universitywebpages.eng.wayne.edu/ece4700/lecture...

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