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Chapter: Digital

Modulation Techniques

1

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Introduction Digital ModulationDigital data needs to be carried on an

analog signal.A carrier signal (frequency f c)

performs the function of transporting

the digital data in an analogwaeform.

The analog carrier signal is

manipulated to uniquely identify thedigital data being carried.

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! Mechanisms for Modulating Digital Data

into Analog "ignal is done by certaintechniques.

Introduction Digital Modulation

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Binary Phase Shift Keying (BPSK)! #n $%"& the transmitted signal is a sinusoid

of 'ed amplitude.! #t has one 'ed phase when data is at one

leel and when the data is at the otherleel the phase is dierent by *+,-.

! The transmitted signal is

! #n $%"& the data b(t) is a stream of binaydigit. Then transmitted $%"& signal is gien

as

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5

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$inary %hase "hift &eying($%"&)! The receied signal has a

! The output oltage v o(kT b ) at the end of

a bit interal etending from time (k-1)T b to KT b is

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Spectrum of BPSK! The waeform b(t) is a /01 binary

waeform whose power spectral densityma2es an ecursion between

and we hae

The %ower "pectral density of the $%"&signal is

s P +

s P −

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8

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Geometrical Representation ofBPSK ! A $%"& signal can be represented in terms of

one orthonormal signal as

The $%"& signal can be drawn as

3ig. 4eometrical representation of $%"&"ignal

The distance 5d6 between the signals

7here 8b9 %s Tb is the energy contained in a bit

duration.

( ) t CosT t ub 01

/2 ω −=

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Dierential Phase Shift Keying! D%"& is the modi'cation of $%"&.

! D%"& eliminates the ambiguity about whetherthe demodulated data is inerted or not.

! #t also aoids the need to proide thesynchronous carrier required at the

demodulator for detecting a $%"& signal.

! 3ig. 4enerating a D%"& "ignal

! The data stream to be transmitted is d(t).

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11

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Dierential Phase Shift KeyingCont.! b(t) is applied to a balanced modulator

to which is applied the carrier The modulator output which is thetransmitted signal is

! 7hen d(t) 9, the phase of the carrierdoes not change at the beginning of thebit interal.

! 7hile when d(t) 9* there is a phasechange of magnitude .

t Cos P s 02 ω

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Dierentially Encoded PhaseShift Keying! D%"& demodulator required a deice which

operates at the carrier frequency and proides adelay Tb.

! D8%"& eliminates the need for such a piece ofhardware.

! #n D8%"& synchronous demodulation recoers thesignal b(t) and the decoding of b(t) to generated(t) is done at baseband.

! The transmitter of D8%"& is identical with D%"&.

3ig. $aseband decoder to obtain d(t) from b(t)

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153ig. 8rrors in Dierentially 8ncoded %"& occurs

(a)

(b)

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!adrat!re Phase Shift Keying(PSK)! %"& that uses phase shifts of ;,-9 ?! Dierent signals generated each

representing = bits.

! adantage: higher data rate than in %"&(= bits per bit interal) while bandwidthoccupancy remains the same.

! ?@%"& can easily be etended to +@%"&

i.e. n@%"& ! higher rate %"& schemes are limited by

the ability of equipment to distinguish

small dierences in phase.

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!adrat!re Phase Shift Keying(Cont.)

! 3ig. Type D 3lip 3lop symbol

3ig. 3lip 3lop

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3ig. An oset %"&

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!adrat!re Phase Shift Keying(Cont.)! The transmitted output signal is gien

by

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22

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23

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24

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d t Ph Shift K i

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!adrat!re Phase Shift Keying

3ig. %"& 0eceier

PSK Si l S

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PSK Signal SpaceReprsentation (Cont.)! The four quadrature signal can be represented

as

! These signals were represented in terms of

two orthonormal signals

! The %"& signal v m(t) can be gien as

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PSK Si l S

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PSK Signal SpaceReprsentation (Cont.)

" t d t Ph

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"on#oset !adrat!re PhaseShift Keying! An additional Bip@Bop is placed either

before een or odd Bip@Bop.! "o in each transition time Tb for %"&

and =Tb for %"&.

! ne bit for %"& and two bit for %"&change for * to @*.

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($SK)

#n A"& the two binary alues are represented byto dierent amplitudes of the carrier frequency.

The resulting modulated signal for one bit time is

"usceptible to noise.

A"& is also called n@ &eying.

The simplest and most common form of operate

as a switch.Application: A"& is used to transmit digital data

oer optical 'ber.

=

0,0

1),2cos()(

binary

binaryt f At s

cπ

A lit d Shift K i (C t )

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Amplitude Shift Keying (Cont.)

Nbaud = baud rate

f c = carrier frequency

A lit d Shift K i (C t )

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! The bandwidth $ of A"& is proportional

to the signal rate ".$ 9 (*Ed)"

! 5d6 is due to modulation and 'ltering

lies between , and *.

33

Amplitude Shift Keying (Cont.)

Example 3

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We have an available bandwidth of 100 kHz which spans

from 00 to !00 kHz" What are the carrier fre#uency and

the bit rate if we modulated our data by usin$ A%& with d

' 1(

Solution

The middle of the bandwidth is located at 250 kHz. Thismeans that our carrier frequency can be at fc = 250 kHz.

We can use the formula for bandwidth to find the bit rate

(with d = 1 and r = 1).

Example 3

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! The digital data stream changes the frequencyof the carrier signal f c.

! frequency of carrier signal is aried to representbinary * or ,

! Amplitude and phase is not changeable.

! Adant: 3"& is less susceptible to errors thanA"& Jspeci'c frequency changes oer a numberof interals so oltage (noise) spi2es can beignored

! Disadantage: 3"& spectrum is = A"&spectrum.

! application: oer oice lines in high@freq. radio

transmission etc. 36

(%SK)

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3

$3"& 4enera

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3!

0eceier for a $3"& si

" #n B$SK the %inary &ata 'aeform &(t) generates a

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" #n B$SK the %inary &ata 'aeform &(t) generates a

%inary signa*

" +ere &(t) , -1 or 1 correspon&s to the ogic 1 an& /

of the &ata 'aeform.

BFSK Spectrum

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BFSK Spectrum

" #n terms of the aria%e p+ an& p0* he B$SK Signa

is*

" #n the BPSK case %(t) is %ipoar i.e. it aternates

%et'een -1 an& 1. p+ an& p0 as a sum of constant

an& a %ipoar aria%e* that is*

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eome r ca epresen a on o

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eome r ca epresen a on o'rthogonal %SK

An orthogonal $3"& can be generatedwith the suitable selection of thefrequencies of the unit ector with m andn integers.

The ector u* and u= are the mth and nth

harmonics of the fundamental frequenciesf b.

The frequency f G and f K in the $3"& are

selected to be with (m L n)

The corresponding signal ectors are

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The corresponding signal ectors are

The distance between the signal endpoints is

Signal space representation of


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eome r ca epresen a on onon#orthogonal %SK

7hen two 3"& signals sH(t) and sL(t)

are [email protected] us represent the higher frequency

signal sH(t) as

/ow represent the lower frequencysignal sL(t) as

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3ig. "ignal "pace representation forS

H(t) and S

L(t) are not orthogonal


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The distance separating sH(t) and sL(t)

is,

when the two signals are not

orthogonal we hae to ealuate S11,S12 and S22.


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7e are using the preious eq. getting

The distance separating sH(t) and sL(t)

is,

/ow simplifying the equation we get

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/ow simplifying the equation we get

The 'nal result is then

#f then the optimumdistance d opt

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! #t uses 5two@dimensional6 signaling.

! riginal information stream is split into two

sequences that consist of odd and een symbolse.g. $2 and A2

! A2 sequence (in@phase comp.) is modulated by

Cos(=f ct) $2 sequence (quadrature@phasecomp.) is modulated by "in(=f ct).

! Composite signal is sent through the channel A2

Cos(=f ct)E $2 "in(=f ct).5/

(AM)

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! Ad: data rate 9 = bit per interal

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Modulation (AM)

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Modulation (AM)

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16 QAM Constellation

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16-QAM Constellation

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5!

Duobinary 8ncoding

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! #t is also 2nown as correlatie coding andpartial response signalling.

! #t basically introduce controlled inter@symbol interference (#"#) in data stream.

! "o encoding a binary bit stream byduobinary enoding eects a reduction ofma. freq. than ma. req. of unencodeddata stream.

! "o bandwidth reduces by usingduobinary signalling.

6/

Duobinary 8ncoding

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(C l ti di )

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! The waeform D(2) is therefore

! The inerter output is Thedierential encoder (called precoder )

output is

! The input #= 9 b(2@*). "o that the inerter

output d(2) is

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(Correlatie coding)

"pectrum of Duobinary 8ncoding

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63

p y g

Example 5

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ind the minimum *W for an %& si$nal transmittin$ at

000 bps" The transmission mode is half/duple and the

carriers must be separated by !+000 Hz

Solution:

$7 9 baud rate E (f c* Jf c,)

The baud rate is the same as the bitrate

$7 9 =,,, E I,,, 9F,,, GH

p

Example 6

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ind the ma bit rates for an %& si$nal if the *W of the

medium is 1+000 Hz and the difference between the

carriers must be at least 000 Hz" Transmission is in full/duple mode"

Solution:

$7 9 baud rate E (f c* Jf c,) The $7 for each direction is N,,, GH$aud rate 9 N,,, J=,,, 9 ?,,,

$aud rate 9 bit rate$it rate 9 ?,,, bps

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