dc digital communication module iii part1

8/7/2019 DC Digital Communication MODULE III PART1

1/112

DIGITAL

MODULATION

TECHNIQUES

MODULE 3

PART I


2/112

DIGITAL COMMUNICATION

Module IIIModule III-- Part IPart I

Digital Modulation Techniques: Digital Modulation Techniques: Digital

modulation formats Coherent binary modulation

techniques - PSK, FSK, QPSK, MSK, Non-

coherent binary modulation techniques DPSK -Comparison of binary and quarternary modulation

techniques - M-ary modulation techniques PSK ,

QAM , FSK ( Block level treatment only)


3/112

Binary phase shift keying In binary phase shift keying the transmitted signal is a sinusoid of

fixed frequency. It has one fixed phase when the data is at one level and phase is

changed by 1800 when the data is at another level. If the sinusoid is of amplitude A, it has a power ps = (A/2)2 = A2/2and so A = (2PS) .The transmitted signal is either

In BPSK the data b(t) is a stream of binary digits with voltage levelswhich can be taken as +1 volt for logic high and -1 volt for logic zero.Hence the modulated signal can be represented as

tPtv sBPSK 0cos2)( =

)cos(2)( 0 += tPtv sBPSK tPs 0cos2 =

tPtbtv sBPSK 0cos2)()( =


4/112

Binary phase shift keying

10 1 0 1 0 0 1


5/112

Generation Of BPSK Signal

BPSK signal may be generated by applying a carrier signal to abalanced modulator.

The base band signal b(t) is applied as a modulating signal to thebalanced modulator. A NRZ level encoder converts the binary data

sequence in to bipolar NRZ signal.

Balanced

Modulator

Bipolar NRZ

Level Encoder

Carrier signal

BPSK

Signald(t)b(t)


6/112

Reception Of BPSK Signal The received signal has the form

Where is a nominally fixed phase shift corresponding to the timedelay /0 which depends on the length of path from transmitter toreceiver and the phase shift produced by the amplifiers in the front-end of the receiver preceding the demodulator.

The demodulation technique employed is synchronousdemodulation for which the carrier signal should be reproduced atthe receiver.

tPtbtv sBPSK 0cos2)()( =

)cos(2)(0

+= tPtbs

)/(cos2)(00

+= tPtbs


7/112

Reception Of BPSK Signal

SQUARE LAW

DEVICEBANDPASS

FILTERFREQUENCY

DIVIDER

SYNCHRONOUS

DEMODULATORINTEGRATOR

BIT

SYNCHRONIZER

)cos(2)( 0 +tPtb s )(cos 02

+t )(2cos 0 +t )cos( 0 +t

Recovered

Carrier

)cos(2)( 0 +tPtb s

)(cos2)( 02 +tPtb s

bs TP

tb 2)(Sc

Ss


8/112

Reception Of BPSK Signal

The received signal is squared to generate the signal

The dc component is removed by the band pass filter whose passband is centered around 2f0. At the output we obtain cos2(0 t+ ).

A frequency divider is used to obtain the original carrier cos (0 t+ ).

The received carrier is multiplied with the received signal togenerate which is equal to

It is then applied to an integrator. A bit synchronizer is a device that can recognize precisely the

moment which corresponds to the end of the time interval allocatedto one bit and the beginning of the next.

)(2cos2

1

2

1)(cos 00

2 ++=+ tt

)(cos2)( 02 +tPtb s

++ )(2cos

2

1

2

12)( 0 tPtb s


9/112

Reception Of BPSK Signal At the end of the bit interval it closes switch Sc very briefly to discharge

(dump) the integrator capacitor and leaves the switch Sc open duringthe entire course of the next bit interval.

After this it closes Sc briefly. The output signal at the end of the bit

interval from the integrator (and before closing Sc) is of interest to us. This output signal is made available by switch Ss which samples the

output voltage just prior to dumping of the capacitor. If bit interval Tb is equal to an integral number n of cycles of the carrier

frequency fo such that

the output voltage at the end of a bit interval extending from

time (k-1)Tb to kTb is

bb TnorTn 00

22

==

)(0 bkTv

dttPkTbdtPkTbkTvb

b

b

b

kT

Tk

sb

kT

Tk

sbb )(2cos2

12)(

2

12)()( 0

)1()1(

0 ++=


10/112

Reception Of BPSK Signal The second integral is zero since the integral of a sinusoid over a whole

number of cycles has the value zero.

Thus the system reproduces the bit stream b(t).

bsbb TPkTbkTv2

12)()(0 = b

sb T

PkTb

2

)(=


11/112

Spectrum of BPSK Signal The waveform b(t) is a NRZ binary waveform. The waveform

consists of rectangular pulses of amplitude Vb and each pulse isTb/2 around its centre.

The fourier transform of the single pulse is given

For large number of such positive and negative pulses the powerspectral density S(f) is expressed as S(f) = X(f)2 /Ts where X(f)2denotes the average value of x(f) due to all the pulses in b(t).Ts isthe symbol duration.

b

bbb

fT

fTTVfX

)sin()( =

bV

bV2bT

2bT


12/112

Spectrum of BPSK Signal

In BPSK only one bit is transmitted at a time and so bit duration Tband symbol duration Ts are the same. Tb=Ts

This equation gives the PSD of base band signal b(t). The BPSK signal is generated by modulating a carrier by the base

band signal b(t). Due to modulation by the carrier frequency fc the spectral

components are translated to fc+f and fc-f and the magnitudes aredivided by half.

222)sin(

)(

=

b

b

s

bb

fT

fT

T

TVfS

)1.(..........

)sin(

)(

2

2

= bb

bbfT

fT

TVfS


13/112

Spectrum of BPSK Signal

If the voltage levels of the binary data are vb, it can be expressedin terms of power as Vb= Ps ie, b(t) =Ps . Using this

Equation (1) may be re written as:

The spectrum of NRZ base band signal and the correspondingBPSK signal expressed by equations (2) and (3) are plotted below.

++

+

=22

2

)(

)(sin

2

1

)(

)(sin

2

1)(

bc

bc

bc

bcbBPSK

Tff

Tff

Tff

TffTVfS

)2.........()(

)(sin

)(

)(sin

2)(

22

++

+

=

bc

bc

bc

bcbsBPSK

Tff

Tff

Tff

TffTPfS

)3.(..........)sin(

)(

2

=

b

bbs

fT

fTTPfS


14/112

Spectrum of BPSK Signal)( fSbsTP

2

bsTP

fbf bf2bf-bf2-

cf cf bc ff +bc ff bc ff +bc ff

The spectrum of the BPSK signalis centred on fc and extends fromfc-fb+ fc+fb. So the BW of BPSKsignal is 2fb.


15/112

Geometrical representation of BPSK signal

The BPSK signal can be expressed as

Let cos2fct be the carrier signal represented by

Now bit energy Eb is defined as Eb= Ps Tb and So

tfPtbtv csBPSK 2cos2)()( =

tfTP

PTPtb c

bs

s

bs 2cos2

)(=

tfT

c

b

2cos2 )(1 t

)()()( 1 tTPtbtvThen bsBPSK =

bbs ETP =

tfT

TPtb c

b

bs 2cos2

)(=

)()( 1 tEtvThen bBPSK =


16/112

Geometrical representation of BPSK signal

At the receiver and the point at represents symbol 1 andrepresents symbol 0.

The separation between these two points represents the isolation ofsymbols 1 and 0 from each other. This separation is called the distance d.

In the case of BPSK the distance d is given by

)( 1 tonEb +bE

2 bbb EEEd =+=

2 bEd=

bE bE+

2Distance bE=

0


17/112

Differential Phase Shift Keying

In coherent BPSK, we require expensive circuits toregenerate the carrier at the receiver side.

DPSK is a modification of BPSK which eliminatethe need for a synchronous carrier.

The block diagram of the system that generates

DPSK is shown below.


18/112


DELAY Tb

BALANCEDMODULATOR

)(tb)(td

)( bTtb

tPs 0cos2

tPtbtv sBPSK 0cos2)()( =tPOr s 0cos2


19/112


-101111

11-1011

1111-10-10-10-10

VoltageLogiclevel

VoltageLogiclevel

VoltageLogiclevel

b(t)b(t-Tb)d(t)

DELAY Tb

BALANCED

MODULATOR

)(tb)(td

)( bTtb tPs 0cos2

tPtbtvsBPSK 0

cos2)()( =

tPOr s 0cos2

TRUTH TABLE OF EX-OR GATE


20/112


d(t)

b(t-Tb)

b(t)

0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 1 0 0 1 1 0 0 1 1 1 1 0

0


21/112


The input to the EXOR gate is the input bit stream d(t)and b(t) delayed by one bit interval given by b (t-Tb).

Because of the feed back involved in the system there is

the some difficulty involved in determining the logiclevels at the start of the waveform We can eliminate this difficulty by assuming that b(t)=0 at

the start of the interval 0. We can also just as well

assume b(t)=1 but ultimately there is no change in thedifferential nature of the wave form.

In the figure waveform of d(t) the bit stream b(t) and b(t-

Tb) is shown. Assuming that b(t)=0 we can complete thewaveform of DPSK by taking b(t) = d(t) b(t-Tb).


22/112


The response of b(t) to d(t) is that b(t) changes level atthe beginning of each interval in which d(t)=1 and b(t)does not change level when d(t)=0.There is no

correspondence between the levels of d(t) and b(t) asb(t)=1 some times when d(t)=0 and sometimes whend(t)=1

The only invariant feature of the system is that a changein b(t)[up or down] occurs whenever d(t)=1 and nochange in b(t) occurs when d(t)=0


23/112


Even if we assume that b(t)=1 at the start of the intervalthis particular invariant feature of the system will holdgood.The levels of b(t) will get inverted, but a change in

level occurs whenever b(t)=1 and no change in leveloccurs when b(t)=0 Now b(t) is applied to a balanced modulator to which the

other input is 2Ps Cos0t. The modulator output whichis the transmitted signal is

Altogether when d(t)=0 the phase of the carrier does notchange at the beginning of the bit interval while whend(t)=1 there is a phase change of magnitude 1800

tPtbtv sDPSK 0cos2)()( = tPs 0cos2 =


24/112

Demodulation of DPSK

In the receiver shown above the received signal and the receivedsignal delayed by the bit time Tb are applied to a multiplexer. Themultiplexer output is

))(cos()cos(2)()( 00 ++ bsb TttPTtbtb

)1......(22

2coscos)()( 00

+

+= bbsbT

tTPTtbtb

SYNCHRONOUS

DEMODULATOR

(MULTIPLIER)

DELAY Tb

)cos(2)( 0 +tPtb s

))(cos(2)( 0 + bsb TtPTtb

To Integrator andbit synchronizer

nTb 20 =


25/112


This signal is applied to a bit synchronizer and integrator. If the bit interval Tb is equal to an integral number n of

cycles of the carrier frequency f0 such that n.2/ 0 =Tb

or 2n= 0Tb, the output voltage v0(kTb) at the end of abit interval extending from (k-1) Tb to kTb is obtained byintegrating the equation(1).

When 2n = 0Tb the second part of the eq (1) isreduced to zero after integration.

The first part b(t) b(t-Tb) Ps cos 2n is actually a constantmultiplied by b(t) b (t-T

b).

Now we can recover the original data sequence d(t) fromthe product b(t) b(t-Tb),


26/112


If d(t)=0 then there was no phase change andb(t) =b(t-Tb), both being +1 volt or -1 volt. In this caseb(t) b(t-Tb)= +1.

If however d(t)=1 then there was a phase change andeither b(t) = 1 or b(t-Tb) = -1volt or vice versa. In eithercase b(t) b (t-Tb)=-1 volt.

( ) ( ) 1 ( ) 0bb t b t T d t = + =

( ) ( ) 1 ( ) 1bb t b t T d t = =


27/112

Advantages and disadvantages

The main advantage of DPSK is that we can avoid theexpensive circuit needed to generate a local carrier atthe receiver .

There is no ambiguity about the polarity of thetransmitted bit sequence.

In DPSK a bit determination is made on the basis of the

signal received in two successive bit intervals. Hence noise in one bit interval may cause errors to two

bit determination. The error rate is DPSK is greater than in PSK. Also there is tendency for bit errors to occur in pairs.


28/112

Quadrature Phase Shift Keying

In the case of BPSK the channel bandwidth is 2fb sinceone bit constitutes a symbol in this case.

But if we can combine two or more bits to form a symbol

and then modulate it the bit rate will be reduced andhence bw requirement is reduced.

In QPSK we bunch together two bits to form a symbol

and then modulate these symbols on quadraturecarriers.


29/112

QPSK Transmitter

TOGGLE

Flipflop

D Flipflop

D Flipflop

Adder

Even

Clock

Odd

Clock

s 0P cos t

s 0P sin t

Clock se(t)

So(t)

be(t)

bo(t)

)( tvQPSK

b(t)


30/112

QPSK Transmitter Waveforms

1 2 3 4 5 6 7 8 9

0 1 0 1 1 0 0 1 0

1 1 10

0 0 01

Clock

Odd

Clock

Even

Clock

be(t)

bo(t)

b(t)


31/112

QPSK Transmitter

The toggle FF is driven by a clock waveform whose period is the bittime Tb.

The toggle FF generates an odd clock waveform and an even clockwaveform.

These clocks have periods 2Tb.The active edge of one of the clocksand the active edge of the other are separated by bit time Tb.

The bit stream b(t) is applied as the data input to both D flip flops,

one driven by the odd clock and the other driven by the even clock. The FFs register alternate bits in this stream b(t) and hold each

such registered bits for two bit intervals, that is for a time 2Tb.

The bit stream bo(t) registers bits 1,3,5 etc and the bit stream be(t)

registers bits 2,4,6 etc. The bit stream be(t) and bo(t) is either +1volt or -1volt depending on

the data is 1 or 0.


32/112

QPSK Transmitter

The stream be(t) is superimposed on a carrier and bitstream bo(t) is super imposed on a carrier by the use oftwo multipliers to generate the signals se(t) and so(t).

These signals are then added to generate the transmitter output

signal vm(t) which is

When a bit stream with bit time Tb multiplies a carrier, the generated

signal has a bandwidth of 2fb. When b0(t) and be(t) are both of bit duration 2Tb,the frequency is

1/2Tb = 0.5fb so the bw required is 20.5 fb = fb. BW required is halfthat for BPSK

When b0(t) = 1 S0(t) = (Ps sin 0t) and - (Ps sin 0t). When be(t) = 1 Se(t) = (Ps cos 0t) and - (Ps cos 0t)

m s o 0 s e 0v (t) P b (t) sin t P b (t) cos t = +

s 0P cos ts 0P sin t


33/112

QPSK Transmitter

These 4 signals can be represented by phasors as shown below.The phasors are in phase quadrature.

s 0P cos ts 0P cos t

s 0P sin t

s 0P sin t

Q d Ph Shif K i


34/112


The four possible output signals can berepresented as below

)()()( tststv eom +=


s 0P sin t

s 0P sin t

1

1

+=

=

e

o

b

b

1

1

+=

+=

e

o

b

b

1

1

=

+=

e

o

b

b

1

1

=

=

e

o

b

b

{ }1,0

{ }1,1

{ }0,0

{ }0,1

Q d Ph Shif K i


35/112


-451 1

-1351 0

+450 1+1350 0

PHASE

SHIFT

BINARY

INPUT

+135 -45 +45 -135

0 0 1 1 0 1 1 0Dibit input

Phase change


s 0P sin t

s 0P sin t

{ }1,0

{ }1,1

{ }0,0

{ }0,1

( )45cos0

+t( )135cos 0 +t

( )135cos 0 t ( )45cos 0 t

Q d Ph Shif K i


36/112


The QPSK we lump two bits together to form a symbol . The symbol can have any one of four possible values corresponding to

the two bit sequences 00, 01, 10 and11.

We use four distinct signals to represent these symbols.

At the receiver each signal represents one symbol and correspondinglytwo bits .

QPSK Receiver


37/112

QPSK Receiver

Raise input

To fourth

power

Bandpass

Filter

4f0

Frequency

Divider

(by 4)

LATCH

+

b

b

Tk

Tk dt)12(

)12(

+ b

b

Tk

kT d

)22(

2

ttbPttbP esos 00 cos)(sin)(s(t) +=

)(4 ts

t04cos

t0cos

t0sin

tts 0sin)(

tts 0cos)(

g timeat samplinbPT oss

g timeat samplinbPT ess

SamplingSwitch)(tb

QPSK R i


38/112

QPSK Receiver

In QPSK receiver synchronous detection is used and hence it isrequired to locally regenerate the carriers cos0t and sin0t. Carrierregeneration is similar that employed for regeneration . The stepsinvolved in carrier regeneration are :

1. Incoming signal is raised to 4th power.

2. This signal is passed through a bandpass filter of passband 4f0which removes all the frequencies generated during the first step

except 4f0.

3. The signal having frequency 4f0 is passed through a frequencydivider which divides the frequency by 4. The sin40t and

cos40t terms generated in the first step are converted to sin 0tand cos 0t .

QPSK R i


39/112

QPSK Receiver

The regenerated carriers are applied to two synchronousdemodulators, the other input of which is the input signal.

These multipliers or balanced modulators multiplies the incomingsignal with carriers sin 0t and cos 0t

The output of the multipliers are

The first term when expanded gives

) tttbPttbPi esos 000 sincos)(sin)()( +

) tttbPttbPii esos 000 coscos)(sin)()( +

tttbPttbP esos 0002 cossin)(sin)( +

tt

tbP

ttbPes

os 000 cossin22

)(

2cos2

1

2

1

)( +

=

ttbP

ttbPtbP esosos 00 2sin2

)(2cos)(

2

1)(

2

1 +=

QPSK R i


40/112

QPSK Receiver

This signal is then applied to an integrator and bit synchronizer. If the bit duration Tb is an integral number of cycles of the input

signal integration of eq (3) gives zero value for the 2nd and thirdterms.

The integration is performed for Ts =2Tb duration and so the firstterm gives the output Ps b0(t).2Tb = Tb Ps b0(t).

The odd bit stream is thus recovered. Similarly at the output of the

second integrator we get the signal Tb Ps be(t). A bit synchronizer is required to establish the beginnings and endsof the bit intervals of each bit stream so that the time for integration,dumping and sampling can be synchronized .

At the end of each integration time for each individual integratorand just before the accumulation is damped the integrator output issampled.

QPSK R i r


41/112

QPSK Receiver

Samples are taken alternately from one and the other integratoroutput at the end of each bit time Tb and these samples are held inthe latch for the bit time Tb.

Each individual integrator output is sampled at intervals 2Tb. Thelatch output is the recovered bit stream b(t).

QPSK Signal Space Representation


42/112


The QPSK signal can be represented as

Let and be the carriers

ttbPttbP oses 00 sin)(cos)( +

ttbT

TPttbT

TP ob

bse

b

bs 00 sin)(1

cos)(1

+=

0 0

2 2( ) cos ( ) sinb e b o

s s

E b t t E b t t T T

= +

)(1tu

)(2tu

tTtu s01 cos

2

)( =t

Ttu

s02

sin2

)( =

)()()()()( 21 tutbEtutbEts obeb +=

)()( 21 tuEtuE bb =



43/112


)(1 tu

bE

1

1

+=

=

e

o

b

b

11

+=+=

e

o

bb

11

=+=

e

o

bb

1

1

=

=

e

o

b

b

)(2 tu

bE

bE

)()( 21 tuEtuE bb +

)()( 21 tuEtuE bb )()( 21 tuEtuE bb

)()( 21 tuEtuE bb +

sE

sE

sE

sE

bEd 2=



44/112


The distance of a signal point from origin is

is the symbol energy

Thus = The distance between signal point is given by

Points which differ in a single bit are separated by the distance

The distance specifies the ability of the system to distinguishbetween two signals without error.

bsTP2

bsTP2 sE

bsTP2 sE

bEd 2=

bE2

Spectrum of QPSK Signal


45/112


The input sequence b(t) is of bit duration Tb.It is a NRZ bipolarwaveform. It consists of pulses of amplitude vb and each pulses is Tb/2 around its center .

The Fourier transform of such a pulse is given by

For a large number of such positive and negative pulses the powerspectral density is expressed as

In the case of b(t),Ts = Tb so

b

bbb

fT

fTTVfX

)sin()( =

sT

fXfS

2)(

)( =

222)sin(

)(

=

b

b

s

bb

fT

fT

T

TVfS

2

2 )sin()(

=

b

bbbfT

fTTVfS



46/112

Spectrum of QPSK Signal Putting

The signal b(t) is divided into be(t) and bo(t) each of bit duration 2Tb = Ts.

If the symbols 1 and 0 are equally likely then we can write powerspectral densities of be(t) and bo(t) as

The base band PSD of QPSK signal equals the sum of individual PSDs

of be(t) and bo(t)

sb PV =2

)(()(

=

s

sbs

fT

fTSinTPfS

2

)(()(

=

s

ssse

fT

fTSinTPfS

2

)(()(

=

s

ssso

fT

fTSinTPfS

)()()( 0 fSfSfS eB +=2

)(2)(

=

s

sssB

fT

fTSinTPfS



47/112


When this signal is multiplied by a carrier having frequency fo itsspectrum is shifted to f0 and divided by two.

+++

=

2

0

0

2

0

0

)())((

)())(()(

s

s

s

sssQ

TffTffSin

TffTffSinTPfS



48/112


)( fSssTP2

ssTP

sT1

0f 0f 20bff +

2

bc

ff

20

bff +2

0bff

sT2sT1sT2 22

11 b

bs

f

TT

==

2bf2bf

M ary PSK


49/112

M-ary PSK In BPSK we transmit each bit individually. Depending on whether

b(t) is logic 0 or 1 we transmit one or another of a sinusoid for bittime Tb, the sinusoids differing in phase by 2/2=1800 .

In QPSK we group together two bits and depending on which of thefour two bit word develops we transmit one or another of foursinusoids of duration 2Tb, the sinusoids differing in phase by

2/=900

.

This scheme can be extended . Let as group together N bits so thatin this N bit symbol extending over NTb there are 2N =M possible

symbols. We can represent the symbols by sinusoids of durationNTb = Ts which differ from one another by phase 2/M.

M-ary PSK


50/112

M ary PSK

In M-ary PSK the wave forms used to identify symbols are

The waveforms are represented by dots in at signal space in whichcoordinate axes are the ortho normal waveforms

( )

mwhere

MmtCosPtv

m

msm

)12(

)1,.....1,0(2)( 0

+=

=+=

sin2

)( 02 tT

tus

=cos2

)( 01 tT

tus

=

[ ]ttptv mmsm 00 sinsincoscos2)( =

tPstP mms 00 sinsin2coscos2 =

M-ary PSK


51/112

M-ary PSK

mso

mse

oe

Pp

Pp

bypandpDefining

sin2

cos2

=

=

tptptv oem 00 sincos)( =

1( )u t

2( )u t

3

5

7

3

2/M

2/M

2/M

M-ary PSK


52/112

M ary PSK

SERIAL

TO

PARALLEL

CONVERTER

DIGITAL

TO

ANALOG

CONVERTER

..

......

SINUSOIDAL

SIGNAL

SOURCE

PHASE

CONTROLLED

BY

v(sm)

)(tb

0

12

1N

)( msv output

M-ary PSK


53/112

M ary PSK The bit stream b(t) is applied to a serial and parallel converter. This converter has facility for storing N bits of a symbol.

When the N bits are presented serially these bits are assembledtogether and is presented all at once of N output lines of theconverter parallelly.

The converter output remains unchanged for the duration NTb of asymbol during which the converter is assembling a new group of N

bits The converter output is applied to a D/A converter. D/A converter

produces an output voltage which assumes one of 2N = M differentvalues in a one to one correspondence to M possible symbols

applied to its input. The D/A output is a voltage v(sm) which depends on the symbol

sm(m =0,1. M-1).

M-ary PSK


54/112

M ary PSK Finally v(sm) is applied as a control input to a special type of constant

amplitude sinusoidal signal source whose phase m is determinedby v(sm).

The output is a fixed amplitude sinusoidal waveform whose phasehas one to one correspondence to the assembled N bit symbol.

M-ary PSK Receiver


55/112

M ary PSK Receiver

Raise to

Mth

Power

BPF

Frequency

Divider

(by M) t0sin

t0cos

bT

dt0

bT

dt0

ADC

ttvm 0cos)(

ttvm 0sin)(

espT

ospT

tptptv oem 00 sincos)( =

Symbol

M-ary PSK Receiver


56/112

M ary PSK Receiver The carrier recovery circuit requires the following operations

(i) The input signal is raised to the Mth power(ii) The signal is passed through a BP filter of center frequency Mf0(iii) The output of BPF is divided by M.

The recovered carriers are multiplied with the input signal and isapplied to integrators which works with the help of bit synchronizers.

The integrator eliminates all sinusoidal terms and produces theoutputs T

sP

eand T

sP

o.

These signals are applied to a device which reconstructs the digital Nbit symbol which constitutes the transmitted signal .

If serial bit stream output is needed a parallel to serial converter maybe used at the output.

Problems Associated With QPSK


57/112

The BW of QPSK signal is found to be fb as the main lobe of thespectrum has a width of fb.

The main lobe contains 90% of the total signal energy.

Still a small amount of power lies outside this main lobe in the formof side lobes.

When QPSK is used for multi channel communication on adjacentcarriers this side lobe power poses many problems.

The wide spectrum of QPSK signal is due to the nature of the baseband signal.

This signal consists of abrupt changes and abrupt changes give riseto spectral components at high frequencies.

When this base band spectra is translated to a high carrierfrequency as a result of multiplication it retains its shape.



58/112

Q

One solution is to pass the base band signal through a LPF toeliminate the abrupt changes and to band limit it.

But the band limiting give rise to a pulse that is not time limitedand hence inter symbol interference results.

Another alternative is to pass the modulated signal through aband pass filter to suppress the side lobes.

When a signal like QPSK where there are abrupt phase changes

is passed through a band pass filter substantial changes in theamplitude of the signal occurs at times of abrupt phase changesdue to the effect of filters.

Such amplitude variations can cause problems in QPSK that

employ repeaters.



59/112

Q So a method of modulation called Minimum Shift Keying isdeveloped where the major changes are:

(i) The base band wave form that multiplies the quadrature

carriers are much smoother than the abrupt rectangular wave

form of QPSK.(ii) The wave form of MSK exhibits phase continuity. There are

no abrupt phase changes as in QPSK. So inter symbol

interference is also avoided.

Minimum Shift Keying


60/112

y g In MSK also we start with the original bit stream b(t). It is divided in to an odd and even bit stream bo(t) and be(t) as in fig b and

fig c . The odd stream consists of alternate bits b1,b3,b5 etc and the even

stream consists of b2,b4,b6 etc.

Each bit in both streams is held for two bit intervals 2Tb=Ts, the symboltime.

The MSK transmitter also generates two sinusoidal waveformssin (2t/4Tb) and cos (2t/4Tb).

It is essential that sin (2t/4Tb) passes through zero precisely at the endof symbol time be(t) and cos (2t/4Tb) passes through zero at the end ofthe symbol time bo(t).

Now the products of be(t)sin(2t/4Tb) and bo(t) cos (2t/4Tb) is generatedwhich are shown in fig e and fig f.

In MSK the transmitted signal is

)1.....(sin4

2cos)(2cos

4

2sin)(2)( 00 t

T

ttbPt

T

ttbPtv

b

os

b

esMSK

+

=



61/112

y g

In the case of QPSK the quadrature carries are multiplied by therectangular abruptly changing odd and even bit streams. But in MSKthe carriers are multiplied by the smoother waveforms shown in fig eand f.

The side lobes generated by this smoother waveforms will be smallerthan those associated with the rectangular waveform and have easierto suppress.

Equation(1) can be re-written as

+

+

+= tTtT

tbPtv

bb

esMSK 00

4

2sin4

2sin2

)(2)(

+

+ t

Tt

T

tbP

bb

os

4

2sin

4

2sin

2

)(2 00

+

+

+= t

T

tbtbt

T

tbtbPtv

b

eo

b

oesMSK

4

2sin

2

)()(

4

2sin

2

)()(2)( 00



62/112

y g

Now bo = 1 and, be = 1

If bo = be than CL = 0 and CH =1If bo = -be than CH = 0 and CL =1

( ) ( )

++

+= t

tbtbt

tbtbPtv eooesMSK 00 sin

2

)()(sin

2

)()(2)(

2

2 eoL

eoH

bbCandbbCDefining =+=

=+= 00 LH and

)2.......(sin)(2sin)(2)( ttCPttCPtv LLsHHsMSK +=



63/112

y g Thus depending on the value of bits be and bo in each bit interval the

transmitted signal is at angular frequency H or at L as in the case ofFSK and the magnitude is always. sP2

ShiftedH-1-1

0L-11

Shifted

L1-1

0H11

PhaseTransmittedFrequency

bebo



64/112

y g In MSK the two frequencies fH and fL are chosen to ensure that the two

possible signals are orthogonal over the bit interval Tb.

0sinsin

0

= tt LT

H

b

( ) ( ) 0coscos2

1..

0

=+ ttei LHT

LH

b

( ) ( ) 02cos2cos21

..0

=+ tfftffei LHT

LH

b

( )

( )

( )

( )

0

2

2sin

2

2sin

2

1..

0

=

+

+

bT

LH

LH

LH

LH

ff

tff

ff

tffei

( )( )

( )( )

)3......(02

2sin

2

2sin

2

1.. =

++

LH

bLH

LH

bLH

ff

Tff

ff

Tffei



65/112

Equation(3) will be satisfied if

Now we know that

)

( )

2(4)

2

H L b

H L b

f f T n

f f T m

=

+ =

422

42

4

2000

bb

b

H

ff

f

T+=+=+=

( )54

.. 0 +=b

Hfffei

4

22

4

2

4

2 000

bb

b

L

ff

f

T

Similarly ===

( )64

.. 0 =b

L

fffei



66/112

Substituting (5) and (6) in equations (4)

Similarly

Altogether we have

Since n=1 fH and fL are close together as possible for orthogonalityto prevail.

So this system is called Minimum shift keying.

nTf

ff

f bbb =

+

442 00 nT

fi.e. b

b =2

2 1 =ni.e.

mTf

ff

f bbb =

++

442 00 mTfi.e. bo =22

mffi.e. bo =

14 m

ffi.e. bo 4 =

1=n mf

f bo4

=



67/112

The carrier frequency f0 should be an integral multiple of fb/4

4)1(

444 0

bbbbH

fm

ffm

fff +=+=+=

4)1(

444 0 bbbbL fmffmfff ===

Minimum Shift Keying Transmitter


68/112

BPF

BPF

ADD

SUB

ADD

)( 0

+

+

+

-

+

+

t0sin

tcos

t)sin(21

0 +

t)sin(21

0

)(2 tbP os

)(2 tbP es

tttbP os 0sincos)(2

tttbP es 0cossin)(2

tt )sin()sin( 00 + ttei 0cossin..

tt )sin()sin( 00 ++ ttei 0sincos..

)(tvMSK

)(0

+

Minimum Shift Keying Transmitter


69/112

First and are multiplied resulting in the productwhich is equal to

Two band pass filters are used to separate the two terms of the

above equation and These components are then added to produce

and i.e.

The outputs of adder and subtractor are then multiplied by the oddand even bit streams and resulting in the

products and

When these products are added we get

which is the required MSK wave form.

t0sin tcostt cossin 0 ( )tt )sin()sin(2

100 ++

t)sin(2

10 + t)sin(2

10

( )tt )sin()sin( 00 ++

tt )sin()sin( 00 + tt sincos 0

)(2 tbP os )(2 tbP estttbP os 0sincos)(2 tttbP es 0cossin)(2

0 02 ( ) sin cos 2 ( ) cos sin s e s oP b t t t P b t t t +

MSK Waveforms


70/112

b1=1 b2=1 b3=0 b4=0 b5=0 b6=0 b7=0 b8=0

b1

b3

b5

b7

b2 b4 b6 b8

b(t)

bo

(t)

be(t)

bT

t

4

2cos

bT

t

4

2sin

b

eT

ttb 4

2sin)(

b

oT

ttb

4

2cos)(

)(a

)(b

)(c

)(d

)(e

)( f

2b 4b 6b

8b

1b 3b

5b 7b

MSK Waveforms


71/112

b

eT

ttb

4

2sin)(

b

oT

ttb42cos)(

2b 4b 6b

8b

1b 3b5b 7b

(1,1) (-1,1) (-1,-1) (-1,-1) (-1,1) (1,1) (1,1)

bH ff 5.1=

bL ff 1=

)(tvMSK

eo bb = eo bb = eo bb = eo bb = eo bb = eo bb = eo bb =

)(g

)(h

)(i

Minimum Shift Keying Receiver


72/112

Storesample

Store

sample

+

b

b

Tk

Tkdt

)12(

)12(

+ b

b

Tk

kT dt

)22(

2

tt = cossinx(t) 0

tt = sincosy(t) 0

bTkatSample )12(t +=

bTkatSample )22(t +=

bkTatSwitch =t

)(tvMSK

Minimum Shift Keying Receiver2222


73/112

ADD

SUB

)(tvMSKSquaring

Circuit

BPF

BPF

Filter &Amp

H2

L2

2

2

tP Hs 2sinsin)(sin)(

2222

ttCttC LLHH +

tP Ls 2sin

tHsin2

1

tLsin2

1

tt = cossinx(t) 0

tt = sincosy(t) 0

tk scos

+

+

+

_



74/112

Synchronous detection is used in MSK receiver. First the carrier is recovered by a carrier recovery circuit as in fig(2).

The input signal

is first squared and the output of the squarer consists , among otherterms and which consists of double

frequency terms and .

When it is passed through two BPFs of central frequencies and

and assuming phase modification the output signal is

and .

The frequency is divided by 2 to obtain and

The signals are then added to obtain which is

equal to

sin)(2sin)(2 ttCPttCP LLsHHs +

sin)( 22

ttC LL ttC HH 22 sin)(

tH2cos tL2cos

H2

L2 tP Hs 2sin

tP Ls 2sin

tH

sin tL

sin

tt LH sinsin +tt LHLH

+

2cos

2sin



75/112

So the expression becomes Similarly the signals are subtracted to obtain

The subtracted signal is thus Thus the carriers are recovered.

Also the signals and are multiplied and passed

through a LP to generate signal at symbol frequency

When passed through a LPF the output is

Once the carriers are regenerated the incoming MSK signal ismultiplied by these carriers x(t) and y(t).

tt = cossinx(t) 0

tttt LHLHLH

+

=2

cos2

sinsinsin

tt = sincosy(t) 0

tHsin tLsin

s( ) ( )tttttz LHLHLH ++== coscossinsin)(( ) ttLH = 2coscos

t

fb

= 422cos tfb

= 22cos



76/112

The outputs of the multipliers are passed to an integrate and dumpcircuit.

The integrator integrates staggered overlapping intervals of symboltime Ts=2Tb.

The first integrator integrates over (2k-1)Tb to (2k+1)Tb and thesecond one over 2kTb to (2k+2)Tb .

The sinusoidal terms are eliminated by the integrator.

At the end of each integration time the integrator output is storedand then it is dumped.

The switch at the output swings back and forth at the bit rate so thatthe final output waveform is the original data bit stream.

Signal space representation of MSK


77/112

2( ) sinL Ls

u t tT

=

tT

tu Hs

H sin2)( =0

1

=

+=

L

H

C

C

1

0

=

=

L

H

C

C

0

1

==

L

H

C

C

1

0

+=

=

L

H

C

C

sss ETP =2 sd E=



78/112

Here CH(t) and CL(t) can assume 1.

When CH(t) = 1, CL(t)=0 and CL(t) = 1, CH(t)=0

The ortho normal unit vectors are given by

sin)(2

sin)(2

ttCT

TPttCT

TP LLs

ssHH

s

ss +=

)()()()( tutCEtutCE LLsHHs +=

sin

2

)( tTtuwhere HsH = sin

2

)(nd tTtua LsL =

tT

tu cs

L sin2

)( =tT

tu Hs

H sin2

)( =

sin)(2sin)(2)( ttCPttCPtv LLsHHsMSK +=



79/112

The end points of 4 possible signal vectors are indicated by dots. The smallest distance between signal points is bs EEd 42 ==

Binary Frequency Shift Keying


80/112

In BFSK two frequencies are used to represent binary data 0 and 1 The binary data waveform d(t) generates a signal

where d(t) is +1 or -1 corresponding to the logic levels 1 and 0 of the

data waveform. The transmitted signal is of amplitude and is either

Where 0 is the nominal carrier frequency and d(t) is represented bya constant offset (0+) or (0-) from the carrier frequency.

(0+) is the higher frequency and (0-) is the low frequency and isrepresented by H and L

[ ]ttdtPtv sBFSK += )(cos2)( 0

sP2

tPts sH )cos(2)( 0 +=

tPtsor sL )cos(2)( 0 =

Binary Frequency Shift Keying


81/112

A scheme for generating BFSK is shown below. Two balanced modulators are used, one with carrier frequency H

and the other with L Voltage values of pH(t) and pL(t) are related to the voltage values of

d(t) as given below.

When pH(t)=1, pL(t)=0 generated signal is

When pH(t)=0, pL(t)=1 generated signal is At the output of the adder, corresponding to d(t) one of these

waveforms occur, but not both

+1V0V0-1V

0V+1V1+1V

pL(t)pH(t)b(t)d(t)

tP Hs cos2

tP Ls cos2

BFSK-Complete Block Diagram2


82/112

ADDER

)(tpTP Hbs

)(tpTP Lbs

tTH

bcos

tT

L

b

cos2

)(tvBFSK

LEVEL

SHIFTER

LEVEL

SHIFTERINVERTER

POLAR TONONPOLAR

CONVERTOR

)(td

)(tb

)(tpH

)(tpL

BFSK- Transmitter


83/112

The input signal b(t) is either 1 or 0. The upper level shifter produces corresponding to 1 and no

signal corresponding to 0.

The lower level shifter on the other hand produces

corresponding to 0 and no signal corresponding to 1. When these signals are multiplied and added the output is

corresponding to b(t)=1 and corresponding to b(t)=0

bsTP

bsTP

tP Hs cos2

tP Ls cos2

BFSK Receiver


84/112

ENVELOPE

DETECTOR

ENVELOPE

DETECTOR

BPF

BPF

COMPARATOR

[ ]ttdtPs + )(cos2 0

)(td

bf2

bf2

bH fff += 0

bL fff = 0

BFSK Receiver

BPSK signal is applied to two bandpass filters with centre


85/112

g pp pfrequencies f0+fb and f0-fb and with bandwidths 2fb.

Also fH-fL=2fb. The filter frequency ranges selected do not overlapand each filter has a pass band wide enough to pass the main lobein the spectrum.

Hence the filter will pass nearly all energy in the transmission at fHwhere as it will be blocked by the other filter.

The second filter will pass nearly all energy in the transmission at fLwhere as it will be blocked by the first filter.

The filter outputs are applied to two envelope detectors and theoutputs of envelope detectors are applied to a comparator.

If the transmitted signal is at fH the output of the upper envelopedetector will produce a higher voltage when compared to the lower

detector. The comparator compares the envelope detector outputs and a

decision is made in favour of high.

BFSK Receiver


86/112

If the transmitted signal is at fL the output of the lower envelopedetector will produce a higher voltage when compared to the upperdetector.

The comparator compares the envelope detector outputs and a

decision is made in favour of low.

BFSK Receiver (Synchronous detection)BFSK SIGNAL


87/112

COMPARATOR

CARRIERRECOVERY

CIRCUITtLcos

tHcos

OUTPUT

bT

dt0

bT

dt0

[ ]ttdtPs + )(cos2 0

BFSK Receiver


88/112

In the presence of noise and other interference the demodulationmethod using envelope detection produces wrong results.

So synchronous detection method is employed in thesecircumstances.

The carrier recovery circuit regenerates the two carriers cosHt andcosLt .

The incoming signal is multiplied by these two carriers and ispassed through an integrate and dump circuit synchronized by bit

synchronizers. The sinusoidal terms are eliminated and a term containing the bit

information is obtained at the output.

This information is applied to a comparator circuit which comparesthe voltage levels of the integrator outputs and makes a decisionregarding the transmitted bit i.e. high or low.

This type of demodulation is called coherent detection.

Spectrum of BFSK Signal


89/112

The BFSK signal may be written as

The expression for BPSK signal is given by

When we compare equations (1) and (2) we can see that the two

terms of eq(1) are identical to eq(2) with the difference that b(t) is abipolar waveform assuming +1 and -1 where as pH(t) and pL(t) arenon polar signals assuming values of +1 and 0.

If we can convert eq(1) in the same form as eq(2) we can easily

obtain the spectrum by comparing with the spectrum of BPSKsignal.

)1......(cos2)(cos2)()( tPtptPtpts LsLHsH +=

)2......(cos2)()( 0tPtbts sBPSK =


)('

11

)( tt HH 1)(1)(' tth HH


90/112

Substituting in eq(1)

The first term of eq(3) has a spectrum that is a single impulse

located at fH and the second term of has a spectrum that is a singleimpulse located at fL.

)('22)( tptpHH +=

)('2

1

2

1)( tptp LL +=

1)(1)(' =+= tptpwhen HH0)(1)(' == tptpwhen HH1)(1)(' +=+= tptpwhen LL

0)(1)(' == tptpwhen LL

cos)('2

1

2

12cos)('

2

1

2

12)( ttpPttpPts LLsHHs

++

+=

cos ' ( )cos cos ( ) ' ( )cos .........( 3 )2 2 2 2

s s s s H H H L L L

P P P P t p t t t t p t t = + + +


PSD f bit t i i b)(' tPs

2

)sin()( bfTTPfS


91/112

PSD of bit stream is given by

When it is multiplied by its spectrum is translated to fHand divided by 2.

Similarly the spectrum of is given by

Complete spectra is obtained by adding all the 4 terms.

)('2

tpP Hs )sin()(

=

b

bbs

fTfTTPfS

tHcos

++

+

=22

)(

)(sin

)(

)(sin

2)(

bH

bH

bH

bHbsq

Tff

Tff

Tff

TffTPfS

ttpP LLs cos)('2

++

+

=22

)(

)(sin

)(

)(sin

2)(

bL

bL

bL

bLbsp

Tff

Tff

Tff

TffTPfS


[ ])()(1)( fffffS


92/112

The spectra is plotted for fH-fL=2fb With such a selection the BW may be made minimum without

interference between spectra of the signals.

[ ] )()(21)( LHBPSK fffffS +

=

+

+

22

)(

)(sin

)(

)(sin

2 bL

bL

bH

bHbs

Tff

Tff

Tff

TffTP



93/112

HfHf Lf Lf

bL ff bL ff + bH ff bH ff +

HfLfHf Lf

bf2

bf4

Geometrical Representation of BFSK

In the general case a modulated signal can be represented in terms


94/112

In the general case a modulated signal can be represented in termsof ortho normal vectors u1(t) and u2(t) in signal space.

In the case of PSK the orthogonality of the vectors u1(t) and u2(t)

results from their phase quadrature. But in the case of BFSK we should select the frequencies carefully

to ensure orthogonality.

Let m and n be integers. Then

)()()( 2211 tuCtuCts +=

1 2( ) cos Hs

u t tT

=

1

2( ) cos 2 b

b

u t m f t T

= 22

( ) cos 2 bb

u t n f t T

=

bitsymbolforTT bs == 1 2( ) cos Ls

u t tT

=


The vectors u1(t) and u2(t) are the mth

and nth

harmonics of thef f f


95/112

The vectors u1(t) and u2(t) are the m and n harmonics of thefundamental frequency fb.

Different harmonics are orthogonal over the interval of thefundamental period.

If the frequencies fH and fL in a BFSK system is selected to be

the corresponding signal vectors are

bH mff =

L b f n f =nmwhere >

1

2( ) cos 2 ( ) H s b b b

b

s t P T m f t E u t T

= =

2

2( ) cos 2 ( ) L s b b b

b

s t P T n f t E u t T

= =


Now sH(t) and sL(t) are orthogonal and can be represented in signali b l


96/112

Now sH(t) and sL(t) are orthogonal and can be represented in signalspace as given below.

The distance between signal points is which is less thanthat of BPSK signals.

)(1 tu

)(2 tu

)( tsL

)( tsH

bE

bE

2 bd E=

2 bd E=

1cos2)( fortPts HsH =

0cos2)( fortPts LsL =

M-ary FSK Transmitter


97/112

SERIAL

TO

PARALLEL

CONVERTER

N- BIT

D/A

CONVERTER

FREQUENCY

MODULATOR.

)(td

0d

1Nd

M-ary FSKoutput

M-ary FSK Receiver


98/112

BPF(f0)

BPF(f1)

BPF(fM-1)

ENVELOPE

DETECTOR

ENVELOPE

DETECTOR

ENVELOPE

DETECTOR

LARGEST

SIGNAL

OUTPUT

N-BIT

ADC

M-aryFSK

M-ary FSK At the transmitter a serial to parallel converter converts the serial bit

stream to parallel formS ti th d t i l d i ll l f d d t


99/112

stream to parallel form. Sometimes the data is already in parallel form and we need not use

a serial to parallel converter. The converter output is applied to a frequency converter that

generates a carrier waveform whose frequency is determined bythe modulating waveform. The transmitted signal for the duration of the symbol interval is of

frequency f0, f1, .,fM-1 where M=2N.

At the receiver the incoming signal is applied to M parallel BPFseach followed by an envelope detector. The BPFs have centre frequencies f0, f1, .,fM-1 The envelope detector apply their outputs to a device which

determines which of the detector outputs is the largest. Largest output is applied to an N-bit DAC whose output is theoriginal transmitted signal. This parallel data may be converted toserial form if required.

M-ary FSK

The probability of error is minimized by selecting frequencies f0

, f1

,fM 1 mutually orthogonal


100/112

.,fM-1 mutually orthogonal. For this the carrier frequency is selected to be successive even

harmonics of the symbol frequency fs=1/Ts

Thus the lowest frequency is f0=kfs, f1=(k+2)fs, f3=(k+4)fs, . In this case the spectral density patterns of the individual possible

transmitted signals overlap as shown below.

In order to pass M-ary FSK the required BW is B=2Mfs.

Since fs=fb/N and M=2N, B=2N+1 fb/N

M-ary FSK

kff fkf )( fkf )4(


101/112

2fs

BW = 2Mfs

skff =1 sfkf )2(2 += sfkf )4(3 +=

Quadrature Amplitude Shift Keying (QASK)

QASK is an amplitude and phase shift keying system in which thephase as well as amplitude of the carrier is varied according to the


102/112

p p gmodulating bit stream.

In the case of BPSK,QPSK and M-ary PSK we transmit in any

symbol interval one signal or another which are distinguished fromone another in phase but are all of the same amplitude.

In each of these systems the end points of the signal vectors insignal space falls on the circumference of a circle.

The ability to distinguish one signal vector from another in thepresence of noise will depend on the distance between vector endpoints.

So the noise immunity can be improved if the signal vectors are

varied in phase amplitude. Such a type of modulation system is called QAPSK or QASK.

QASK Example


103/112

Quadrature Amplitude Shift Keying (QASK)

Consider a system in which a symbol is transmitted for every 4 bitcombinations. There are thn 24=16 possible symbols.


104/112

p y Now we have to generate 16 distinguishable signals.

One possible geometrical representation of these 16 signals is

shown in figure(1). Each signal point is equally distant from its neighbouring points, the

distance being d=2a.

When all the 16 signals are equally likely the average energy

associated with a signal is

( ) ( ) ( ) ( )[ ] 222222222 1099994

1aaaaaaaaaEs =+++++++=

s

Ea 1.0=

sEad 1.022 ==

QASK Each symbol represents 4 bits and the normalised symbol energy

Es=4Eb where Eb is the normalised bit energy.


105/112

s b b gy

A typical signal can be represented by

where k1 and k2 are each equal to 1 or 3

bs EEa 1.01.0 == bEd 4.02=

)1().........()()( 2211 tauktauktvQASK +=

tT

tus

01 cos2

)( =

tTtu s02 sin

2

)( =

sEa 1.0=)3(


106/112

QASK Constellation

a3D(a, 3a) C(3a, 3a)

2a )(2 tu


107/112

a

a2

a

a2

a3

a a2 a3aa2a3

A(a, a) B(3a, a)

O )(1 tu

QASK Generator

D Qkb )(tAe

tPs 0cos


108/112

D Q

D Q

D Q

DAC

DAC

ADDER

CLOCK GENERATOR (TS)

1+kb

2+kb

3+kb

)(tAo

tPs 0sin

QASK Generator The 4 bit symbol bk+3 bk+2 bk+1 bk is stored in the 4 bit register made

up of 4 flip flops. A new symbol is presented once per interval T =4T and the


109/112

A new symbol is presented once per interval Ts=4Tb and thecontents are updated at each active edge of the clock which has aperiod Ts.

Two bits are presented to one D/A converter and two bits are to thesecond converter.

The converter output Ae(t) modulates the balanced modulator whoseinput carrier is the even function and Ao(t) modulatesthe modulator with odd function carrier.

The transmitted signal is then

Comparing eq(3) with the general equation

tPs 0cos

)3.......(sin)(cos)()( 00 ttAPttAPtv osesQASK +=

tPktPktv ssQASK 0201 sin2.0cos2.0)( += 2.032.0)(),( = ortAtA oe

QASK Receiver

T

)(tAe

ttAPttAPoses 00

sin)(cos)( +

0b


110/112

Raise to

4th

Power

BPF(4f0)

FrequencyDivider

(by 4) t0sin

t0cos

sT

dt0

sT

dt0

ADC

ADC

)(tAo

0

1b

2b

3b

QASK Receiver

As in the case of QPSK carriers are regenerated at the receiver.

The incoming signal is raised to the 4th power.


111/112

A band pass filter extracts the component at 4f0.

It is divided by 4 to get

Neglecting all terms which are not in the frequency range 4f0

( )40024 sin)(cos)()( ttAttAPtv oesQASK +=

ttAtAtAtA

Ptv oeoesQASK 04444

44cos

8

)()(6)()()(

+=

t

tAtAtAtA

Poeoe

s 0

22

4sin2

))()()(()(

+

QASK Receiver

When the signal is divided by 4 we obtain the quadrature carriers

andt0sin t0cos


112/112

Two balanced modulators are used to multiply the incoming signalwith the retrieved carriers.

After that it is applied to an integrator circuit operating insynchronism with a symbol time synchronizer.

At the output of the integrator all sinusoidal terms are reduced tozero and we get Ae(t) and Ao(t).

These voltage levels are applied to a D/A converter which producesthe original bit stream.

dc digital communication module iii part1

Documents