Amplitude_Modulation

8/9/2019 Amplitude_Modulation

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ap er

Am litude Modulation


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Baseband vs Passband Transmission

ase an s gna s: Voice (0-4kHz)

TV 0-6 MHz

A signal may be sent in

its baseband formatwhen a dedicated wiredchannel is available.

erw se, mus econverted to passband.


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Modulation: What and Why?

passband range is calledModulation.

baseband frequency range is called

Demodulation. 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

,(generally a sinusoidal wave) known as thecarrieris chan ed based on the informationsignal that we wish to transmit (modulating

signal). That could be the amplitude, phase, or frequency,

which result in Amplitude modulation (AM),

,modulation (FM). The last two are combined asAngle Modulation


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Types of Amplitude Modulation (AM)

This is the most widely used type of AM modulation.In fact, all radio channels in the AM band use this type.

Double Sideband Suppressed Carrier (DSBSC):

This is the same as the AM modulation above but. Single Sideband (SSB): In this modulation, only half

of 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)

bandwidth 2B rad/s (orB Hz). m(t) M().

et c t e a carr er s gna , c t = cosct , c >>

gDSBSC(t) = m(t)cos(ct)

(1/2) [ (c) + (+ c)].


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Time and Frequency Representation of DSBSCModulation Process


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DSBSC Demodulation

gDSBSC(t)

e(tHLPF( )

BW = 2 B f

(t)

e (t)=gDSBSC(t)cos(ct)

= m(t)cos2(ct)

c(t)

DSBSC Demodulator (receiver)

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

c c . The output signalf(t) of the LPF will be

t = 1/2 m t 1/2 .


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Time and Frequency Representation of DSBSCDemodulation Process


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Modulator Circuits

as ca y we are a er mu p y ng a s gna wa carrier.

ere are t ree rea zat ons o t s operat on:

Multiplier Circuits

Non-L near C rcu ts

Switching Circuits


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Non-Linear Devices (NLD)

- -

linear. One such example is the diode (iD=evD/vT). The output of a NLD can be expressed as a power

series of the input, that isy(t) = ax(t) + bx2(t) + cx3(t) +

, ,and the output can be approximated by the first twoterms.

When the inputx(t) is the sum of two signal, m(t)+c(t),x2(t) will have the product term m(t)c(t)


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Non-Linear Modulatorsx (t)

+ z(t)

y1(t)

y2(t)

on- near ev cea( . )+b( . )2

HBPF( )Cntr Freq. = C

BW = 4 B

m(t)

q(t)

x (t)

DSBSC modulation using non-linear device

c(t)Non-Linear Device

a( . )+b( . )2+

2)()cos()()()(1 tmttmtctx C

Undesired

C

UndesiredUndesired

C

Desired

C

UndesiredUndesired

CCC

CC

tbb

tattbmtbmtam

tbttbmtbmtamta

)2cos(22

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

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

2

22

1

)()cos()()()(1 tmttmtctx C

CCC

CCC

CC

tbb

tattbmtbmtam

tbttbmtbmtamta

tmtbtmtaty

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

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

)()cos()()cos()(

2

22

22

Desired

C

Undesired

ttbmtam

tytytz

)cos()(4)(2

)()()( 21

UndesiredUndesiredUndesiredDesiredUndesiredUndesired


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Switching Modulators

ny per o c unc on can e expresse as aseries of cosines (Fourier Series).

e n ormat on s gna , m t , can t ere ore e,

equivalently, multiplied by any periodicunc on, an o owe y .

Let this periodic function be a train of pulses.

Multiplication by a train of pulses can berealized by simple switching.


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Switching Modulator: Diode Bridge


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Demodulation of DSBSC

e mo u a or c rcu s can e use or emo u a on, u

replacing the BPF by a LPF of bandwidthB Hz.

and frequency synchronization with the incoming carrier.

This t e of demodulation is therefore called coherentdemodulation (or detection).


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From DSBSC to DSBWC (AM)

,

the operation of coherent demodulation, areso histicated and could be uite costl .

If we can let m(t) be the envelope of the

modulated si nal, then a much sim ler circuit,the envelope detector, can be used for demodulation (non-coherent demodulation).

How can we makem(t) be the envelope of themodulated signal?


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Definition of AM

such thatA+m(t) 0. OrAmpeak

)cos()()cos(

cos

ttmtA

mg

CC

CAM

Called DSBWC. Here will refer toit as Full AM, or simply AM

p . 0 1


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Spectrum of AM

)()(21)()()( CCCCAM MMAtg


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Efficiency of AM transmission

=

Assume that the message signal is cos(mt) (what

where 0 1 (i.e., a fractionof the amplitude

modulation index)), or( ) cos( ) cos( )m Cz t A t t

cos ( ) cos ( )2

C m C mt t

A

cos cos2 2

C m C m


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e power o s s gna s e sum o e wo powersof the two sinusoids (because they have different

2 2

22 2

A A

A

The power of the carrier term in the modulated signal

2 2 2z

is ( ) cos( )Cw t A t 2

2w

AP


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ere ore, e e c ency o e ransm ss on

becomes2

2

2 22

2

2z

A

P

P P A A

Since 0 1 to avoid the touching of the upper and2 2

,efficiency of the AM signal is max

10.333 33.3%

1 2

, ,of AM modulation to more than one third, or stated in otherwords, at least 2/3 of the ower of the AM si nal is wasted.


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The Buy and Price of AM

uy: mp c y n emo u a on.

Price: Waste in Power

gAM(t) =Acosct + m(t) cosct

arr er ower c = carr es no n orma on

Sideband Power Ps = Pm/2 (useful)

Power efficiency = = Ps/(Pc + Ps)= Pm/(A2 +Pm)


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Tone Modulation

m = cosm g(t)=[A+Bcos(mt)] cosct=A[1+ cos(mt)] cosct

= =

Under best conditions, =1 max =1/3 =33%

. , .

For practical signals, < 25%

? Would you use AM or DSBSC?


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Generation of AM

s gna s can e genera e y anymodulator, by usingA+m(t) as input instead of

.

In fact, the presence of the carrier term canma e even s mp er. e can use or switching instead of generating a local carrier.

e sw c ng ac on can e ma e y a s ng ediode instead of a diode bridge.


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AM Generator

(to ensure switching

at every period). A

vR= cosct+m t + cosct- cos ct+ =(1/2)cosct+(2/m(t) cosct+ other terms (suppressed by BPF)

v t = 1/2 cost+ 2/m t cost


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AM Modulation Process (Frequency)


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AM Demodulation: Rectifier Detector

ecause o e presence o a carr er erm n e

received signal, switching can be performed in.


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Rectifier Detector: Time Domain


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Rectifier Detector

e s gna s app e o a o e an a

resistor circuit the negative part of the AM.

resistor is a half wave rectified version of the

. ,multiplied by w(t).

31


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Rectifier Detector

e vo age across e res s or w e

VR={[A+m(t)]cosct}w(t)

= [A+m(t)]cosct[cosct-1/3cos3ct +1/5cos5ct-.]

= 1/[A + m(t)]+ other higher terms.

32


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Rectifier Detector

e g er erms can e suppresse y pass ng

VRthrough a low pass filter with cutoff.

be suppressed by a capacitor to give the desired

.using a full wave rectifier.

33


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Rectifier Detector

s n eres ng o no e a rec er e ec on s

in effect synchronous detection without using a.

ensures that its zero crossings are periodic and

of the carrier at the transmitter is built into theAM si nal itself.

34


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Rectifier Detector (Frequency Domain)


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Envelope Detection

careful selection of =RC IfRCis too large, discharging will be

slow and the circuit cannot follow adecreasing envelope.

WhenRCis too small the ri les willbe high.

1/(2B)


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Envelope Detection

os ve cyc e o mo u a e s gna ma es o e

to conduct and capacitor c charges to maximum.

As the input falls below this peak diode stopscon uc ng s nce capac or vo age s near yequal to the peak voltage of input signal.

38


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Envelope Detection

39


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Envelope Detection

e capac or now sc arges roug res s or

at a slow rate (with a time constant RC). The

envelope of modulated signal. Capacitor

ripple signal of frequency c in the output.

40


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Envelope Detection

s r pp es n e ou pu can e re uce y

increasing the time constant RC so that

positive peaks (RC>>1/c).

owever can can e ncrease eyon1/2B ,where B is the highest frequency in

,capacitor to follow envelope of m(t).

42


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Envelope Detection

e enve ope etector output sc

t =A+m(t) with ripple frequency of c.

The dc term A can be blocked by a low

.

43


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44


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NOTE

o ec er e ec or an nve ope e ec or

seems to be equivalent but they are distinct and.

46


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Difference

e rec er e ec or s as ca y a sync ronous

detector while envelope detector is non linear .

rectifier detector is designed to filter m(t) from terms

c

. However in case of envelope detector the timeconstant for low ass filter does de ends on .

47


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Quadrature Amplitude Modulation (QAM)

n or e mo u a e s gna

occupies double the bandwidth of the baseband.

It is possible to send two signals over the samean , one mo u a e w a cos ne an one

with sine.

n eres ng enoug , e wo s gna s can ereceived separately after demodulation.


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EFFECT OF LACK OF PHASE


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-

SYNCHRONISATION IN DSBSC

ccmSCDSB Attsts cos)()(if is unknown

ttstv cSCDSB cos)()(

ttsA

cmc

ccmc

2coscos)(2

Output of LPF cos)()( tsA

tv mc

o


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Contd

)(2)( tstv mc

o Due to lack of phase synchronization, we will see that

the wanted signal at the output of LPF will be

.In other words, phase error causes an attenuation ofthe output signal proportional to the cosine of the

phase error.The worst scenario is when

=

/2, which will give rise

.



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Su ose that the local oscillator is not stable at f but

SYNCHRONISATION IN DSBSC

at fc+ f, then

ttstv cSCDSB cos)()(

tttsA

s

c

ccmc

2coscos

coscos

cm2

A

Output of LPF sv

mo cos

2Thus, the recovered baseband information signal willvary sinusoidal according to cos t


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Solution

synchronization circuitry which is required to detect

and t and by providing the carrier signal to therece ver.A synchronizer is introduced to curb the

synchronization problem exhibited in a coherentsystem.

Let the baseband si nal be tts cos

Received DSB-SC signal ttsAts cos


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( )2 PLL BPF 2

Mathematical analysis of the synchronizer is shownbelow:

ttAA

ttAAts

mc

cmmc

2cos12cos1

coscos)(

22

22222

ttttAA

cmcmmc 2cos2cos2cos2cos1

4

422

ttttAA mcmccmmc 2cos

212cos

212cos2cos1

4

22

Output of BPF

t

AA

c

mc

2cos4

22


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SYNCHRONISER

tk ccoswhere k is a constant of proportionality.

The frequency and phase of the local oscillator signal.

It requires additional circuitry such as synchronizer


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Single-Side Band (SSB) Modulation

as we as occup es ou e e

bandwidth of the baseband signal, although the twosides carr the same information.

Why not send only one side, the upper or the lower?

Modulation: similar to DSBSC. Onl chan e thesettings of the BPF (center frequency, bandwidth).

Demodulation: similar to DSBSC (coherent)


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Complex receivers Require more expensive receivers

because envelope detection cannot be used

Tuning Difficulties - More difficult to tune thanconventional AM receivers. Receivers need a precise

.

59


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SSB Representation

represent the SSB signalin the time domain?

gUSB(t) = ?

gLSB(t) = ?


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Time-Domain Representation of SSB (1/2)

+ -

Letm+(t)

M+() and m-(t)

M-()Then:m(t) = m+(t) + m-(t) [linearity]

BecauseM+(),M-() are not even

m+(t), m-(t) are complex.

Since their sum is real the must beconjugates.

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

m-t = [m t- mht ]

What is mh(t) ?


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Time-Domain Representation of SSB (2/2)

+ -

M+() =M()u(M-() =M()u(-sgn()=2u() -1 u()= + sgn(); u(-) = - sgn()M+() = [M() +M()sgn()]

M-() = [M() - M()sgn()]

ompar ng o:

m+(t) = [m(t) +j mh(t)] [M() +j Mh()]

m t = m t - m t - -

We findMh() = - j M()sgn() where mh(t)Mh()


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Hilbert Transform

h .

The transfer function of this transform is given by:H() = -j sgn()

It is basically a /2 phase shifter


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Time-Domain Operation for Hilbert


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Time Domain Operation for Hilbert

Transformation

or er rans orma on = - sgn .

What is h(t)?

2/(jt) 2 sgn(-) [symmetry]

1/( t) -j sgn()SinceMh() = - j M()sgn() =H() M()

Then tmtmh )(*1

)(

d

t

m )(1


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na y

tjtjLSB

tjtj

USB

CC

CC

etmetmtg

etmetmtg

)()()(

)()()(

)(

1

)(

1

)(2

)(2

)(

etjmetm

etjmetmtg

tj

h

tj

tj

h

tj

USB

CC

CC

)(1

)(1

)(

)sin()()cos()(

etjmetmtg

ttmttm

tj

h

tj

LSB

ChC

CC

)(21)(

21 etjmetm tjh

tj CC

)()()( CCUSB MMG

ChC


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Generation of SSB

e ec ve er ng e o

Realization based on spectrum analysisase- t et o

Realization based on time-domain expression

o e mo u a e s gna


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Selective Filtering


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Phase Shiftingsincos ttmttmt

)sin()()cos()()( ttmttmtg ChCLSB

Phase-shifting Method:


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g

Frequency-Domain Illustration


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SSB Demodulation (Coherent)

)2sin()(21)]2cos(1)[(

21)cos()(

)sin()()cos()()(

ttmttmttg

ttmttmtg

ChCCSSB

ChCSSB

)(2

1OutputLPF tm


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-

structure (low cost) - reliable

information)

72

DSBFC is wasteful of Power


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2/3 of total transmitted power taken up by carrier.

ower

Pc= 1000W

lsb = usb =

lsb usbc

The total power being transmitted is (1000).(1 + 0.8 ) = 1320W2

In transmittin 1320W of the total ower the carrier contains

1000W and does not contain any information being transmitted.The side freq each have 160W and each carries a copy of thesame info signal.

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So, 1320W is being used in order to transmit only 160W.

DSB is wasteful of Bandwidth


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DSB has a Wide Bandwidth

wasteful BW usage i.e info in USB = info in LSB

If so much of the transmitted wave is not required, then

1. No need to send 2 co ies of the same info. Waste of ower andBW2. Eg. Within a band of say, 100kHz, we can transmit only 5signals that occupy 20kHz, but 10 stations if they agree to limit

74

their transmitted BW to 10kHz.


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Sin le Side Band SSB S stem


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Motivation: Both DSBFC and DSBSC occupy a bandwidth of 2B.How can we reduce the bandwidth requirements?

Due the symmetric condition (info in USB = info in lsb), one ofthe sidebands is sufficient to rovide the com lete informationin the original signal.

Power

Pu sb = 1 6 0 W

N o C a rr ie rN o l s f

F r e q u e n c yflsb fu s b fc

76



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AM Sin le Sideband Full Carrier SSBFC Carrier is transmitted at full power with only one of the sidebands.

Half as much bandwidth will be required (BWSSBFC=1/2BWDSBFC).2

RPc

itude Modulating

si nal, m

2m

PP clsb

2m

PP cusb

2cPm

PP

DSBFC

Amp

f

2

ude Modulating

0P2m 2 cPm

SSBFC

R

Pc

77Ampli , m 4

cus4

ct

f



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The carrier is totally removed together with one of the

sidebands.

a as muc an wi t is require SSBFC= DSBFC .

ude Modulating

22

cPm

SSBSC

0cP

Am

plit s gna , m 0lsbP

4

PP cusb 4t

f

78



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AM Sin le Sideband Reduced Carrier SSBRC

Conserve BW and considerably power

One sideband is totally removed and carrier voltage is reduced to

reduced to approx. 1% of its unmodulated power

The carrier is totally suppressed during modulation and to be

SSBRC

2

mplitud

e

Modulatingsignal,fm 0lsbP 4

2mPP cusb 4

01.0

2

cct

PmPP

cc .

79f



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AM Inde endent Sideband ISB

It is a form of DSB transmission in which the transmitter

consists of two independent SSBSC modulators.

different information, with suppressed carrier.

It conserves both power and BW as two info sources are

.

2

ISB

RVP cc /)1.0(2

Mo u at ng

signal,fm 4

2mPP cusb 2

01.0 cctm

PP

4

2mPP clsb

Ch A Ch B

80

f



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The carrier and one complete SB are transmitted, but only part (a

vestige) of the second SB is transmitted. The carrier is transmitted.

The BW is typically 25% greater than that of SSBSC.

R

VPc

2

de Modulating 2m 2

VSB

Am

plit signal,fm usblsb PP 4cusb

lsb

cct P

mPP

4

f

81

DSBFC AM a e


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82

SSBFC AM a e


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ONE SIDEBAND)

100% modulated SSBFC wave with a sin le fre uenc modulatin wave

83

SSBSC AM a e


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THE WAVEFORM IS NOT AN ENVELOPE; IT IS A SINE WAVE AT A SINGLEFREQUENCY EQUAL TO THE CARRIER FREQUENCY PLUS/MINUS THEMODULATING SIGNAL FREQUENCY

84

ISB AM a e


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THAT OF THE MODULATING SIGNAL FREQUENCY

It is a form of DSB transmission in which the transmitter consists of twoindependent SSBSC modulators.

Out ut consists of two totall inde endent sidebands each of different

85

information, with suppressed carrier.


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Multi lexin methods of transmittin more than one si nal alon a

single transmission path/stream i.e many to one.

Demultiplexing separate the stream back into its componenttransmission i.e one to man .

Path refers to the physical link.

Channel refers to a portion that carries a transmission between a

. .

87

MULTIPLEXING


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MULTIPLEXING

Multiplexing(FDM) and Time Division Multiplexing(TDM).

Advantages:

ncrease num er o c anne s so a more n o can e ransm e

Save cost by using one channel to send many info signals

88

F Di i i M lti l i (FDM)


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Frequency Division Multiplexing (FDM)

FDM multi le sources that ori inall occu ied the same fre uenc

spectrum are each converted to different frequency band and

transmitted simultaneously. FDM is an analo techni ue the information enterin an FDM

system must be analog. If the source is digital, it must be

converted to analog before being frequency-division multiplexed.

S lit the total channel bandwidth into several smaller channels ofdifferent frequencies.

Different signal travel over the medium concurrently.

interfering with one another. Modulation is used to lift the centre freq of the baseband signal up

89

.

FDM


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In communication systems, Frequency Division Multiplexing (FDM)is a method in which each signal (channel) is allocated a frequencyslot within the overall line/transmission bandwidth.

In other words the total available frequency bandwidth on the

information signal occupies one of these channels The signal will have exclusive use of this frequency slot all the time

(i.e. each subscriber occupies his/her own slot).


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The bandwidth of a voice signalis taken to be 4kHz, with aneffective spectrum of 300 to

3400Hz.

If this signal is used to amplitudemodulate a 64 kHz carrier, thespectrum becomes the modulated

kHz, extending from 60 to 68kHz.

To make efficient use of bandwidth,.

If three voice signals are used tomodulate carriers at 64, 68, and 72kHz, the spectrum output is asshown.


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With FDM, eachnarrowband channels are

another in the frequency


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another in the frequencydomain.

FDM system wherefour 5kHz channels arefrequency-division

multiplexed into asingle 20kHz combinedchannel.


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- .

Each voice band channel is bandlimited with an antialising filter prior tomodulating the channel carrier.

Figure (b) shows the output spectrum.


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Figure shows first stages of telephone mux

Group multiplexer takes 12 voice ch and puts them on subcarriers at 64,, z us ng .

The resulting spectrum extends 48kHz starting at 60kHz. Five such carriers are combined by LSB on subcarriers at 420, 468,

612kHz to produce a supergroup from 312 to 552kHz an contains 60

voice channels. 96


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If the BW of the tx medium ermits 10 su er rou s can becombined to form a mastergroup.

For a wider BW applications such as satellite links, it is

97

FDM in Tele hone S stem


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.

The voice is used to modulate subcarrier. Each subcarrier is on different

frequency.

Balance

Modulator

Voice

fv

fv+fc

fv-fc fv-fc

fc=60kHz

Ch

12 BPF

Ch

11 BPF

56 64kHz60 64kHz0 -4

kHz

0 -4kHz Linear

fc=

64kHz

104 108kHz

Mixer

fc=

104kHz

Ch

1 BPF

SelectsUSB

100 108kHz

DSBSC

SSBSC0 -4kHz


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Voice si nal am litude modulates 1 of 12 chs in the 60 to 108kHz ran e.

The carrier freq begin at 60kHz with a spacing of 4kHz.(slightly higher

than the highest typical freq of voice) .

SB containing the original voice signal.

All 12 SSB signals are then summed in a linear mixer to produce a single

. Basic group freq spectrum for FDM telephone mux system is shown below.

Channel

4kHz

60 72 76 80

f(kHz)

64 68 84 88 92 96 100 104 108

No.

99

Carrierfrequencies

If more than 12 voice channels are needed, multiple basic groups are

used.


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x amp e ,

carrier frequency is at 60 kHz and the total bandwidth is 96 kHz.

Design a FDM system, given a general rule of 12 channels per.

i. How many basic groups are required?

ii. Draw the circuit diagram of your design

. raw t e requency spectrum o your mu t p exe systemSolution

=. tota 1 channel = 4 kHz,

then 12 channels = 12x4 = 48kHz/basic groupThus 96/48 = 2 basic group

100


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ii. Block Diagram

fc=60kHz

Ch12 BPF

Linear

Mixer

fc=104kHz

Ch1 BPF

Ch 24 BPF

Linear

Mixer

fc=108kHz

Ch

13 BPF

Linear

Mixer

fc=152kHz

48kHz 48kHz

iii. Frequency Spectrum

101 f(kHz)

60 104 108 152

FDM in Telephony


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FDM in Telephony s one n s ages

Reduce number of carrier frequencies

More ractical realization of filters

Group: 12 voice channels 4 kHz = 48 kHzoccupy the band 60-108 kHz

upergroup: groups z = zoccupy the band 312-552

Master rou : 10 S-G 240 kHz = 2400 kHzoccupy the band 564-3084 kHz

FDM Hierarchy


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FDM Hierarchy108 k

10

4

3

11

87

6

1 312 k

Supergroup

543

4

2

1 60 k

0

Vestigial Side Band Modulation (VSB)


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Vestigial Side Band Modulation (VSB)a we wan o genera e us ng se ec ve

filtering but there is no guard band between the two

We will filter-in a vestige of the other band.

an we s recover our message, w ou s or on,after demodulation?

. .

Filtering Condition of VSB


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Filtering Condition of VSBcos2 ttmt

m(t) HVSB( )

(BPF) g

VSB(t)

DSBSC

)()()( CCDSBSC MMG

)()()()( CCVSBVSB MMHG 2cos( ct)

VSB Modulator (transmitter)

C Basebandat

CCVSB MMHX

2

)()2()()(

Cat

C

baseband

CVSB MMH

2

)2()()(

)()()()()( MHHHZ CVSBCVSBLPF

1H

CVSBCVSB HH

VSB Filtering


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VSB Filtering

VSB Filter: Special Case


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VSB Filter: Special Case

1)(LPFH ; || 2 B

If we impose the condition on the filter at the modulator:

VSBc VSB c = ;

Then HLPF= 1 for || 2 B (Ideal LPF)

HVSB() will then have odd symmetry around c over thetransition period.


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AM Broadcasting


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goca e e an z z w

minor variations) z per c anne . z n some countr es

More that 100 stations can be licensed in the

same geograp ca area. Uses AM modulation (DSB + C)

AM station Reception


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p ,

tuning the receiver BPF to its center frequency. Then demodulated.

mpract ca t es:

Requires a BPF with very high Q-factor (Q =fc /B). Particularly difficult if the filter is to be tunable.

Solution: Superheterodyne receiver


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p yep : requency rans a on rom o

Shift the desired station to another fixed pass band (calledIntermediate Fre uenc 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.

ep : emo u a on

Use Envelope Detector


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The Local Oscillator


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a s ou e e requency o e oca

oscillator used for translation from RF to IF?LO c IF -

or fLO =fcfIF (down-conversion)

Tun ng rat o =fLO, max fLO, min Up-Conversion: (1600 + 455) / (530+455) 2

Down-Conversion: (1600455) / (530455) 12

Easier to desi n oscillator with small tunin ratio.

Image Station Problem


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g - , ,

at the same time, down-converting another station to IF

as well. These two stations are called image stations, and they

are spaced by 2x455=910kHz.

Before conversion, use a BPF (at RF) centered atfc ofthe desired station.

e purpose o e er s o ex rac e es re

station, but to suppress its image. Hence, it does nothave to be very sharp.


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Amplitude_Modulation

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