chapter 5 am, fm and digital modulation systems

7/30/2019 Chapter 5 AM, FM and Digital Modulation Systems

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Chapter 5 AM, FM and Digital Modulation Systems

CHAPTEROBJECTIVES:

Amplitude modulation and single sideband; Frequency and phase modulation;

Digitally modulated signals (OOK, BPSK, FSK, MPSK, MPSK, QAM, QPSK,p/4QPSK, and OFDM);

Spread system and CDMA systems.

This chapter is concerned with the bandpass techniques of amplitude modulation (AM),single-sideband (SSB), phase modulation (PM), and frequency modulation (FM); and

with the digital modulation techniques of on-off keying (OOK), binary phase-shift keying

(BPSK), frequency-shift keying (FSK), quadrature phase-shift keying (QPSK),quadrature amplitude modulation (QAM) and orthogonal frequency-division

multiplexing (OFDM). All of these bandpass signaling techniques consist of modulating

an analog or digital baseband signal on to a carrier. In particular, the modulated bandpass

signal can be described by

} (5-1))(Re{)(tj Ctgts

=

Where cc f2= and is the carrier frequency. The desired type of modulated signal,

, is obtained by selecting the appropriate modulation mapping function

cf

( )ts ( )[ ]tmg ,where is the analog or digital baseband signal.( )tmThe spectrum of the bandpass waveform is

( ) ([ cc ffGffgfV += *2

1)( )] (5-2a)

and the PSD of the waveform is

( ) ([ cgcgv ffPffPfP +=4

1)( )]

]

(5-2b)

where G and is the PSD of g(t)( ) ( )[ tgFf = ( )fPg

The goals of this chapter are:

Studyg(t) ands(t) for various analog and digital modulations; Evaluate the spectrum for these modulations; Study some adopted standards; Learn about spread spectrum sytems.

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5-1 AMPLITUDE MODULATION

The complex envelope of an AM signal is given by

(5-3))](1[)( tmAtg c +=

Where the constant has been included to specify the power level and is the

modulating signal (which may be analog or digital). These equations reduce to the

following representation of the AM signal:

cA )(tm

ttmAts cc cos)](1[)( += (5-4)

A waveform illustrating the AM signal, as seen on an oscilloscope, is shown in fig 1. For

convenience, it is assumed that the modulating signal m is a sinusoid.

corresponds to the in-phase component

)(t )](1[ tmAc +

( )tx of the complex envelope; it also correspondsto the real envelope )(tg when 1)( tm (the usual case).

Fig. 5-1 AM signal waveform.

If has a peak positive value of +1 and a peak negative value of 1, the AM signal is

said to be 100% modulated.

)(tm

Definition. The percentage of positive modulation on an AM signal is

% Positive modulation = [ ] 100)(max100max =

tmA

AA

c

c (5-5a)

and the percentage of negative modulation is

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% negative modulation = [ ] 100)(min100min =

tmA

AA

c

c (5-5b)

The overall modulation percentage is

% modulation =[ ] [ ]

1002

)(min)(max100

2

minmax

= tmtm

A

AA

c

(5-6)

where is the maximum value ofmaxA )](1[ tmAc + , is the minimum value, and is

the level of the AM envelope in the absence of modulation [i.e., =0 ].

minA cA

)(tm

Eq. (5-6) may be obtained by averaging the positive and negative modulation as given by

Eqs. (5-5a) and (5-5b). , and are illustrated in Fig. 5-1b, where, in this

example, and

maxA

Amin

minA

5.0

cA

cAA 5.1

max=

cA= , so that the percentages of positive and negative

modulation are both 50% and the overall modulation is 50%.

The percentage of modulation can be over 100% ( will have a negative value),

provided that a four-quadrant multiplier is used to generate the product of

and

minA

)](1[ tmAc +

tccos

1[Ac +

so that the true AM waveform, as given by Eq (5-4), is obtained. However,

if the transmitter uses a two-quadrant multiplier that produces a zero output

when is negative, the output signal will be)](tm

[ ]

+=

.0

,cos)(1

)(

ttmA

ts

cc

1)(

1)(

, where

is the carrier frequency.

cf

There are numerous ways in which the modulation ( )tm may be mapped into thecomplex envelope such that an SSB signal will be obtained. SSB-AM is by far the

most popular type. It is widely used by the military and by radio amateurs in high-

frequency (HF) communication systems. It is popular because the bandwidth is the sameas that of the modulating signal (which is half the bandwidth of an AM or DSB-SC

signal). For these reasons, we will concentrate on this type of SSB signal. In the usualapplication, the term SSB refers to the SSB-AM type of signal, unless otherwise denoted.

[ ]mg

Theorem. An SSB signal (i.e., SSB-AM type) is obtained by using the complexenvelope

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( ) ( ) ([ tmjtmAtg c = )] (5-15)

Which results in the SSB signal waveform

( ) ( ) ( )[ ttmttmAts ccc ]sincos = (5-16)

where the upper (-) sign is used for USSB and the lower (+) sign is used for LSSB. ( )tm denotes the Hilbert transform of ( )tm , which is given by

(5-17)( ) ( ) ( )thtmtm =where

( )t

th

1= (5-18)

and corresponding to a phase shift network:( ) ( )[ thfH = ] 090

(5-19)( )

=j

jfH

,

0

0

f

f

Fig. 5-4 Spectrum for USSB signal.

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Fig. 5-4 illustrates this theorem. Assume that m(t) has a magnitude spectrum that is of

triangular shape, as shown in Fig. 5-4a. Then, for the case of USSB (upper signs), thespectrum ofg(t) is zero for negative frequencies, as illustrated in Fig. 5-4b, and s(t) has

the USSB spectrum shown in Fig. 5-4c. This result is proved as follows:

Proof. We need to show that the spectrum of is zero on the appropriate sideband(depending on the sign chosen). Taking the Fourier transform of (5-15), we get

)(ts

( ) ( ) ( )[{ tmjfMAfG c = ]}

]

(5-20)

And, using (5-17), we find that the equation becomes

(5-21)( ) ( ) ( )[ fjHfMAfG c = 1To prove the result for the USSB case, choose the upper sign. Then, from (5-19), (5-21)

becomes

(USSB case) (5-22)( )( )

=

0,0

0,2

f

ffMAfG

c

Hence the band pass signal is:

(5-23)( )( )

( )

+

=

cc

c

c

c

cc

cffffM

ffA

ff

ffffMAfS

,

,0

,0

,

This is indeed a USSB signal (see Fig. 5-4)

If the lower signs of (5-23) were chosen, an LSSB signal would have been obtained.

The normalized average powerof the SSB signal is

( ) ( ) ( )[ ]22222 2

1

2

1tmtmAtgts c +== (5-24)

Since ( ) ( )tmtm 22 = , so that the SSB signal power is

( ) ( )tmAts c222 = (5-25)

which is the power of the modulating signal ( )tm2 multiplied by the power gain factor

.2cA

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The normalized peak envelope power(PEP) is

( ) ( )[ ]2222 2

1max

2

1tmtmAtg c += (5-26)

Fig. 5-5 Generation of SSB

Fig. 5-5 illustrates two techniques for generating the SSB signal. Thephasing methodis

identical to the IQ canonical form discussed earlier as applied to SSB signal generation.

Thefiltering method is a special case in which RF processing (with a sideband filter) isused to form the equivalent , instead of using baseband processing to generate( )tg [ ]mg directly. The filter method is the most popular method because excellent sideband

suppression can be obtained when a crystal filter is used for the sideband filter. Crystal

filters are relatively inexpensive when produced in quantity at standard IF frequencies. In

addition to these two techniques for generating SSB, there is a third technique, calledWeavers method[Weaver, 1956].

SSB signals have both AM and PM. Using (5-15), we have, for the AM component (realenvelope),

( ) ( ) ( ) ( )[ ]22 tmtmAtgtR c +== (5-27)

And for the PM component,

( ) ( )( )

( )

==

tm

tmtgt

tan 1 (5-28)

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SSB signals may be received by using a superheterodyne receiver that incorporates a

product detector with 00 = . Thus, the receiver output is

( ) ( ) ( )tmkAetgktV cj

out == 0Re

(5-29)

where K depends on the gain of the receiver and the loss in the channel. In detecting SSB

signals with audio modulation, the reference phase 0 does not have to be zero,

because the same intelligence is heard regardless of the value of the phase used [although

the waveform will be drastically different, depending on the value of( )tvout 0 ]. Fordigital modulation, the phase has to be exactly correct so that the digital waveshape is

preserved. Furthermore, SSB is a poor modulation technique to use if the modulating

data signal consists of a line code with a rectangular pulse shape. The rectangular shape(zero rise time) causes the value of the SSB-AM waveform to be infinite adjacent to the

switching times of the data because of the Hilbert transform operation. (This result will

be demonstrated in a homework problem.). Thus, and SSB signal with this type ofmodulation cannot be generated by any practical device, since a device can produce only

finite peak value signals. However, if rolled-off pulse shapes are used in the line code,

such as( )

x

xsinpulses, the SSB signal will have a reasonable peak value, and digital data

transmission can then be accommodated via SSB.

SSB has many advantages, such as a superior detected signal-to-noise ration compared to

that of AM and the fact that SSB has one-half the bandwidth of AM or DSB-SC signals.

5-6 PHASE MODULATION AND FREQUENCY MODULATION:

Representation of PM and FM Signals

Phase modulation (PM) andFrequency modulation (FM) are special cases of angle

modulated signaling. In this kind of signaling the complex envelope is

(5-33)( ) ( )tjceAtg=

where the real envelope, ( ) ( ) cAtgtR == , is a constant and phase ( )t is a linear

function of modulating signal ( )tm . However, is nonlinear function of themodulation. The resulting angle modulated signalis given by:

)(tg

( ) ( )[ ttAts cc ]+= cos (5-34)

For PM the phase is directly proportional to the modulating signal. i.e.;

( ) ( )tmDt p= (5-35)

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Where is the Phase sensitivity of the phase modulator, having units of radians/volt.pD

For FM, the phase is proportional to the integral of ( )tm so that

( ) ( )

=t

f dmDt (5-36)

where the frequency deviation constant has units of radians/volt-sec.fD

Comparing the last two equations , we see that if we have a PM signal modulated by

, there is also FM on the signal, corresponding to a different modulation wave

shape that is given by

( )tmp

( ) ( )

=

dttdm

DDtm

p

f

p

f (5-37)

Wherefandp denote frequency and phase respectively.

Similarly if we have a FM signal modulated by ( )tmf , the corresponding phasemodulation on this signal is

( ) ( )

=t

f

p

f

p dmD

Dtm (5-38)

Fig. 5-7 Generation of FM from PM or vice versa.

A PM circuit can be used to synthesize an FM circuit by inserting an integrator in cascadewith phase modulator input.

Direct PM circuits are realized by passing a un modulated sinusoidal signal though a time

varying circuit which introduces a phase shift that varies with the applied modulating

voltage. is the gain of a PM circuit (rad/V).Similarly a direct FM circuit is obtained

by varying the tuning of an oscillator tank (resonant) circuit according to the modulating

voltage.

pD

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Definition:

If a band pass signal is represented by

( ) ( ) ( )ttRts cos=

Where ( ) ( )ttt c += then the instantaneous frequency(hertz) of ( )ts is

( ) ( )( )

==

dt

tdttf ii

2

1

2

1

or

( ) ( )+= dttdftf ci

21 (5-39)

For the case of FM, using (5-36), the instantaneous frequency:

( ) ( )tmDftf fci2

1+= (5-40)

This is the reason for calling this type of signaling frequency modulation.

The instantaneous frequency is the frequency that is present over a particular instance of

time.

The frequency deviation of a carrier frequency is

( ) ( )( )

+=

dt

tdftftf cid

2

1(5-41)

And the peak frequency deviation is

( )

=

dt

tdF

2

1max (5-42)

In applications such as the unipolar digital modulation, the peak-to-peak deviation isused. This is defined by

( ) ( )

=

dt

td

dt

tdFpp

2

1min

2

1max (5-43)

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For FM signal the Peak frequency deviation is related to peak modulating voltage by

pfVDF2

1= (5-44)

The peak phase deviation may be defined by

( )[ t ] max= (5-45)

which, for PM, is related to peak modulating voltage by

ppVD= (5-46)

where V .( )[ ]tmp max=

Definition.

The phase modulation index is given by

=p (5-47)

Where is the peak phase deviation.

The frequency modulation index is given by

B

Ff

= (5-48)

where is the peak frequency deviation and B is the bandwidth of the modulating

signal, which for case of sinusoidal modulation, is , the frequency of sinusoid.

F

mf

For case of PM or FM signaling with sinusoidalmodulation such that the PM and FM

signals have the same peak frequency deviation, p if identical to f .

Spectra of Angle Modulated SignalsUsing (4-12), we find that the spectrum of angle modulated signal is given by

( ) ( ) ([ cc ffGffGfS += 2

1)] (5-49)

where

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( ) ( )[ ] ( )tjceAtgfG== (5-50)

Example 2

Spectrum of a PM or FM Signal With Sinusoidal Modulation

Assume that the modulation on the PM signal is

( ) tAtm mmp sin= (5-51)

Then

( ) tt m sin= (5-52)

where == mpp AD is the phase modulation index.

The phase function ( )t , as given by (5-52), could also be obtained if FM were used,where

( ) tAtm mmf cos= (5-53)

andm

mf

f

AD

== . The peak frequency deviation would be

2

mf ADF= .

The complex envelope is

(5-54)( ) ( ) tjctj

cmeAeAtg

sin==

which is periodic with periodm

mf

1=T . Consequently, ( )tg could be represented by a

Fourier series that is valid over all the time (


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( ) ( )

nc

nj

cn JAeAc =

=

sin

2

1(5-56)

The integral known as the Bessel function of the first kind of the nth order, ( )nJ -cannotbe evaluated in closed form, but it can be evaluated numerically.

Examining the integral shows that

(5-58)( ) ( ) ( ) nn

n JJ 1=

Taking the Fourier transform of the equation (5-55), we obtain

(5-59)( ) ( )=

=

=n

n

mn nffcfG

or

(5-60)( ) ( ) ( )=

=

=n

n

mnc nffJAfG

Using this result in (5-49) we get the spectrum of angle modulated signal.

5-9 BINARY MODULATED BANDPASS SIGNALING

The most common binary bandpass signaling techniques are as follows:

On Off keying (OOK), also called amplitude shift keying (ASK), which consistsof keying (switching) a carrier sinusoid on and off with a uni-polar binary signal.

Morse code radio transmission is an example of this technique. OOK was one ofthe first modulation techniques to be used and precedes analog communication

systems.

Binary Phase-Shift Keying (BPSK), which consists of shifting the phase of a

sinusoidal carrier or 180 with a unipolar binary signal. BPSK is equivalent

to PM signaling with a digital waveform.

00 0

Frequency-Shift Keying (FSK), which consists of shifting the frequency of a

sinusoidal carrier from a mark frequency to a space frequency, according to thebaseband digital signal. FSK is identical to modulating an FM carrier with a

binary digital signal.

On Off keying (OOK)

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The OOK signal is represented by

( ) ( ) ttmAts cc cos= (5-70)

Where is a unipolar baseband data signal. The complex envelope is( )tm

( ) ( )tmAtg c= (5-71)

and the PSD of this complex envelope is proportional to that for a uni polar signal. ThePSD is

( ) ( )

+=

22 sin

2 b

b

b

c

gfT

fTTf

Af

(5-72)

where has the peak value of( )tm 2=A so that ( )ts has an average normalized power

of2

2

cA .

The null-to null bandwidth is RBT 2= and the absolute bandwidth is .=TB

The transmission bandwidth of the OOK signal is BBT 2= where B is baseband

bandwidth, since OOK is AM type signaling.

If filtered, the absolute baseband bandwidth is ( )RrB += 12

1 where r is the roll off factor

of the filter.

This gives an absolute transmission bandwidth of ( )RrBT += 1

for OOK with raised cosine roll off filtering.

Binary Phase Shift Keying (BPSK)

The BPSK signal is represented by

( ) ( )tmDtAts pcc += cos (5-75a)

where is polar baseband data signal. Let has peak value of and a

rectangular pulse shape.

( )tm ( ) 1=tm

We show that BPSK is also a form of AM type signaling.

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Expanding (5-75a) we get

( ) ( )( ) ( )( ) ttmDAttmDAts cpccpc sinsincoscos =

The representation of BPSK reduces to

( ) ) ) ( ) ttmDAtDAts cpccpc sinsincoscos = (5-75b)

The first term of above equation is called the pilot carrier term and the second term is

called the data term. The level of pilot carrier term is set by the value of peak deviation.

For digital angle-modulated signals, the digital modulation index h is defined by

=

2h (5-76)

where 2

bs T=

is the maximum peak to peak phase deviation (radians) during the time

required to send one symbol . For binary signaling the symbol time is equal to bit

time.(T ).

sT

By maximizing power in the data term by letting2

900

=== pD radians which

corresponds to a digital modulation index of 1=h . For this optimum case of , theBPSK signal becomes

1=h

( ) ( ) ttmAtscc

sin= (5-77)

The complex envelop for this BPSK signal is

(5-78)( ) ( )tmjAtg c=

The PSD of complex envelop for BPSK is

( )

=

b

b

bcgfT

fTTAf

sin2 (5-79)

The null to null bandwidth for BPSK is also 2R, the same as that found for OOK.

Frequency Shift Keying (FSK):

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The FSK signal can be characterized as one of two different types, depending on the

method used to generate it. One type is generated by switching the transmitter output linebetween two different oscillators as shown in Fig 5-23a. This type generates an output

waveform that is discontinuous at the switching times. It is called discontinuous-phase

FSK signal and is represented by

( ) ( )( )[ ]( )

(

+

+=+=

22

11

cos

,coscos

tA

tAttAts

c

c

cc )for binary 1 or 0 is sent.

(5-80)

Fig. 5-23 Generation of FSK.

Where is called the mark (binary 1) frequency and is called the space (binary 0)

frequency.

1f 2f

1 and 2 are startup phases of the two oscillators. The discontinuous phase is

( )

+

+=

tt

ttt

c

c

22

11for binary 1 or 0 interval.

Since FSK transmitter are not usually built this way, we turn into the second type asshown in Fig. 5-23b. The continuous phase FSK signal is generated by feeding the data

signal into a frequency modulator. The FSK signal is represented by

( ) ( )

+=

t

fcc dmDtAts cos

or

( ) ( ) tj cetgts Re= (5-81a)

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where

(5-81b)( ) ( )tjceAtg=

(5-81c)( ) ( )=t

f dmDt

and is a baseband digital signal.( )tm

If the serial data input waveform is binary, such as a polar baseband signal, the resultingFSK signal is called a binary FSK signal. In general the spectra of the FSK signals are

difficult to evaluate since the complex envelop is a non-linear function of .( )tm

5-10 MULTILEVEL MODULATED BANDBASS SIGNALING

Quadrature Phase Shift Keying (QPSK) and M-array Phase shift keying (MPSK)

If the transmitter is a PM transmitter with an M = 4 level digital modulation signal, M-

array phase shift keying(MPSK) is generated at the transmitter output. Since M=4, this

M-array PSK is called Quadrature phase shift keying(QPSK) signaling.

A plot of the complex envelope, ( ) ( )tjceAtg= , one value of g for each of four

multilevel values.

Suppose that the permitted multilevel values at the DAC are 3, -1, +1, and +3 V;

these multilevel values might correspond to PSK phase of 0, 90, 180, and 270degrees, respectively.

MPSK can also be generated by using the two quadrature carriers modulated by x and ycomponents of the complex envelope. In that case ,

( ) ( ) ( ) ( )tjytxeAtg tjc +== (5-91)

where permitted values of x and y are

ici Ax cos= (5-92)

and

ici Ay sin= (5-93)

for the permitted phase angles of i , i=1,2,.M, of the MPSK signal.

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Quadrature Amplitude Modulation (QAM)

Quadrature carrier signaling is called quadrature amplitude modulation. (QAM). The

general QAM signal is

( ) ( ) ( ) ttytyxts cc sincos = (5-94)

where

( ) ( ) ( ) ( ) ( )tjetRtjytxtg =+= (5-95)

with

( )

=

n

nD

nthxtx 1 (5-96)

( )

=

n

nD

nthyty 1 (5-97)

wherel

RD = and ( )nn yx , denotes one of the permitted ( )ii yx , values during the

symbol time that is centered onD

nnTs ==t s.

OQPSK and4QPSK

Offset quadrature phase shift keying(OQPSK) is M=4 PSK in which the alloweddata transition times for the I and Q components are offset by symbol (by 1 bit)

interval.

A4

quadrature phase shift keying (

4

QPSK) signal is generated by altering

between two QPSK constellations that are rotated by4

= with respect to each

other.

045

PSD for MPSK, QAM, QPSK, OQPSK, and4

QPSK

The PSD for MPSK and QAM signals is relatively easy to evaluate for the case of

rectangular bit shape signaling.

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We know that

(5-99)( ) ( )

= sn nTtfctg

where c is complex valued random variable representing the multi level value during

the symbol pulse.

n

thn ( )

=

sT

ttf is the rectangular symbol pulse with symbol

duration T .ssT

D1

= is the symbol(or baud) rate. The rectangular pulse has Fourier

transform

( )

=

=

b

b

b

s

s

sfTl

fTllT

fT

fTTfF

sinsin(5-100)

where T . i.e., there are bits representing each allowed the multilevel value. For

M=16, the mean value of c is

bs lT= l

n

0== nc cm (5-101a)

and the variance is

Cccc nnnc ===

22 (5-101b)

Where C is real positive constant. The PSD for the complex envelope of MPSK or

QAM signals with data modulation of rectangular bit shape is

( )2

sin

=

b

b

gflT

flTkf

for MPSK and QAM (5-102)

where is the number of points in the constellation, and the bit rate islb MClTk 2, ==

bTR

1= . For a total transmitted power of P watts, the value of k is k bplT2= , since

.( ) =dff

s

This PSD for the complex envelope is plotted as shown in Fig. 5-33 below:

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Fig. 5-33 PSD for the complex envelope of MPSK and QAM.

5-13 SPREAD SPECTRUM SYSTEMS:

We have been concerned primarily with the performance of communication system interms of bandwidth efficiency and energy efficiency (i.e., detected SNR or probability of

bit error rate) with respect to natural noise. In some applications we also need to considerthe multiple access capability, antijam capability, interference rejection, and covert

operation or low probability intercept (LPI) capability. These performance objectives can

be optimized by using spread spectrum techniques.

Multiple access capability is needed in cellular phones and personal communication

applications, where many users share a band of frequencies, because there is not enoughbandwidth to assign a permanent frequency channel to each user. Spread spectrum

techniques can provide a simultaneous use of a wide frequency band via code division

multiple access (CDMA) techniques, an alternative to band sharing.

There are many types of spread spectrum systems (SS). To be considered a spread

spectrum system a system must satisfy two criteria:

The bandwidth of the transmitted signal needs to be much greater than that of themessage.

The relative bandwidth ofs(t) must be caused by an independent modulatingwaveform c(t) called the spreading signal, and this signal must be known by the

receiver in order for the message signal to be detected.

The SS signal is

( ) ( ) tj cetgts Re= (5-120a)

where the complex envelope is a function of both m(t) and c(t). In most cases a product

function is used so that

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(5-120b)( ) ( ) ( )tgtgtg cm=

where and are usual types of modulation complex envelope functions that

generate AM,PM ,FM and so on. The SS signals are classified by the type of mapping

functions that are used for .

( )tgc ( )tgm

( )tgc

The following are some of the most common types of SS signals:

Direct Sequence (DS). Here a DSB-SC (double sideband suppressed carrier) typeof spreading modulation is used and ( )tc is a polar NRZ waveform.

Using a phase modulation.

Frequency Hopping (FH). Here ( )tgc is of the FM type where there arek2=

hop frequencies determined by the k-bit words obtained from the spreading codewaveform .( )tc

A Frequency hopped (FH) signal uses a ( )tgc that is of the FM type, where there

are k2= hop frequencies controlled by a spreading code, in which k chipwords are taken to determine each hop frequency.

Hybrid techniques that include both DS and FH.

5-15 STUDY-AID EXAMPLES

SA5-1 Formula for SSB Power

Prove that the normalized average power for an SSB signal is ( ) ( )tmAts c222 = , as

given by (5-25).

Solution. For SSB

( ) ( ) ([ ]tmjtmAtg c = )

Using (4-17), the normalized average power for an SSB is

( ) ( ) ( ) ( ) ( ) ( )( )tmtmAtgRdffPtsP csss 22222

2

1

2

10 +=====

or

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( ) ( ) ( )( )tmtmAtsP cs 2222 2

1+== (5-137)

But

( ) ( ) ( ) ( )dffPfHdffPtm mm

==2

2

whereH(f) is the transfer function of the Hilbert transform. Using (5-19),

( ) 1=fH

Consequently,

( ) ( ) ( ) ( )tmdffPdffPtm mm2

2 ===

(5-137)

Substituting (5-137) into (5-136), we get

( ) ( )tmAtsP cs222 ==

SA5-2 Evaluation of SSB PowerAn SSB transmitter withAc=100 is being tested by modulating it with a triangular

waveform that is shown Fig. 5-14a, where Vp =0.5 V. The transmitter is connected to a 50

W resistive load. Calculate the actual power dissipated into the load.

Solution. Using (5-25) yields

( ) ( )( )tm

R

A

R

ts

R

VP

L

c

LL

rmss

actual

2222

=== (5-138)

For the waveform shown in Fig. 5-14a,

( ) ( )

3

4

43

14

441

2

4

0

3

24

00

22

p

T

p

m

p

m

m

T

p

m

p

m

T

m

V

VtT

V

T

T

dtVtT

V

Tdttm

Ttm

m

mm

=

=

==

(5-139)

23


24/24

Substituting (5-139) into (5-138)

( ) ( )W

R

VAP

L

pc

actual 67.16503

5.0100

3

2222

=

==

chapter 5 am, fm and digital modulation systems

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