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Digital Modulation Technique

Coherent Non - Coherent

Binary

(m) = 2

M - ary Hybrid

* ASK M-ary ASK M-ary APK* FSK M-ary FSK M-ary QAM* PSK M-ary PSK

(QPSK)

Binary

(m) = 2

M - ary

* ASK M-ary ASK* FSK M-ary FSK* DPSK M-ary DPSK

Digital modulation techniques



Amplitude Shift Keying (ASK)

l Pulse shaping can be employed to remove spectral spreading.

l ASK demonstrates poor performance, as it is heavily affectedby noise and interference.

l Long string of zeros causes synchronization loss.

BasebandData

ASK modulatedsignal

A cos wct A cos wct0 0



Phase Shift Keying (PSK)

l Binary Phase Shift Keying (BPSK) demonstrates better performance

than ASK and FSK.l PSK can be expanded to a M-ary scheme, employing multiple phases

and amplitudes as different states.

l Filtering can be employed to avoid spectral spreading.

BasebandData

Binary PSK modulated

signal

s1 s1s0 s0

where s0 = -A cos wct and s1 = A cos wct





Binary Phase Shift Keying (BPSK)



Signal constellation for binary PSK modulation.



Probability of error using dmin

For binary problems detection in Gaussiannoise, Q(x) function can be used.





P(ε) = Q(d2min /2No)

Since σ2 = No /2



Prob. of error for BPSK



Simple scheme for generating BPSK modulated signal.



A scheme for coherent demodulation of BPSK modulated signal

following concept of optimum correlation receiver



Power Spectrum for BPSK Modulated Signal



Quadrature/Quarternary phase shift keying (QPSK)

QPSK is effectively two independent BPSK systems (I and Q), andtherefore exhibits the same performance but twice the bandwidth

efficiency.QPSK can be filtered using raised cosine filters to achieve excellent outof band suppression.

Large envelope variations occur during phase transitions, thus requiring

linear amplification.

Cos Wc t

900

Odd Data

Even Data

(NRZ)

(NRZ)

QPSK

Q-Channel

I-Channel

Q

(-1,-1)

(-1,1)

I

(1,1)

(1,-1)

Wc = Carrier Frequency, I = In phase channel, Q = Quadrature channel



QPSK- Features

• Very important for developing concepts of two

dimensional I-Q modulations as well as for its

practical relevance.

• QPSK is an expanded version from binary PSK where

in a symbol consists of two bits and two

orthonormal basis functions are used.

• A group of two bits is often called a ‘dibit’. So,

four dibits are possible. Each symbol carries same

energy.











410)2sin(]4 / )12sin[(2)2cos(]4 / )12[(2)(

410]4 / )12(2[2

)(

toiT t t f iT E t f iCos

T E t s

toiT t it f CosT

E t s

cci

ci

114

72cos

2)(

014

52cos

2)(

004

32cos2)(

104

2cos2

)(

3

2

1

dibit input for t T

E t

dibit input f or t T

E t

dibit input f or t T

E t

bit diinput f or t T

E t

f S

f S

f S

f S

ct

c

c

c

In QPSK the carrier is given by

The transmitted signals are given by

T t t t

T t t t

f T

f T

cb

cb

02sin2

)(

02cos2

)(

2

1

The basic functions

Th f i d h i d i l d fi d b



4,3,2,1

412sin

412cos

i

i E

i E

Si

There are four message points and the associated signal vectors are defined by

Input

dibit

Phase of QPSK

signal(radians)

Coordinates of message

points

Si1 Si2

10

00

01

11

4

2

E 2

E

4

3 2 E

2 E

4

5 2

E 2

E

4

7

2 E 2

E

QPSK W f



QPSK Waveform



QPSK Transmitter

QPSK Receiver

Probabilit of error in QPSK(coherent)



Probability of error in QPSK(coherent)

► A QPSK system is equivalent to two coherent binary PSK systems working in parallel and

using carriers that are in-phase and quadrature.

►The in-phase channel output x1 and the Q-channel output x2 may be viewed as the

individual outputs of the two coherent binary PSK systems.

The signal energy per bit is 2 E and the noise spectral density is

2

0 N

The average probability of bit error in each channel of the coherent QPSK

N

N

E erfc

E E

E

erfcP

2 0

0

1

2

1

22

21

No

E erfc

No

E erfc

No

E erfc

PPC

24

1

2

1

22

11

1

2

2

21



No

E erfc

No

E erfc

PP C e

24

1

2

1

2

In the region where 12

o

N

E

We may ignore the second term and so the approximate formula for theaverage probability of symbol error for coherent QPSK system is

No

E erfcPe 2



BFSK

01

b E S

b E S

0

2

b

b

T t t f CosT

t 022

)( 11

b

b

T t t f CosT

t 022

)( 22

for symbol 1

for symbol 0

t f CosT

E t S

b

b

112

2)(

t f CosT

E t S

b

b22 2

2)(



bT

dt t t x x0

11)()(

bT

dt t t x x0

22 )()(

BFSK Tx

BFSK Rx

l = x1 – x2

E



b

b

E

E

x E

x E

l E

0

111

21

Similarly for ‘0’ transmissionb E

l E

0

When symbol ‘1’ was transmitted x1 and x2 has mean value of 0 andb

E respectively.

0

21][][][

N

xVar xVar lVar

20 N

2

0 N

Assuming zero mean additive white Gaussian noise with input PSD with variance

.

dl

N

E l

N

PPb

ee

0 0

2

0

0

2

)(exp

2

1)0 / 1(

02 N

E l Z

b

0

2

2

0

22

1

)exp(1

0

N

E erfc

dz zP

b

N

E

e

b



Using dmin Pe expression,

d2min = E + E = 2E = A2T

Eb

= (E + E) / 2 = E = A2T/2

Similarly

0

122

1

N

E erfcP b

e

The total probability of error = )1 / 0()0 / 1([21 eee PPP

Assuming 1’s & 0’s with equal probabilities

][2

110 ee PP

0

22

1

N

E erfcP b

ePe=

P(ε) = Q(d2min /2No)

P(ε) = Q(A2T /2No) = Q(Eb /N0)



Non- Coherent FSK Demodulation

Input Binary Sequence {bK} 1 0 0 1 0 0 1 1

Differentially Encoded 1 sequence

{dK}

1 0 1 1 0 1 1 1

Transmitted Phase 0 0 Π 0 0 Π 0 0 0

Received Sequence

(Demodulated Sequence)

1 0 0 1 0 0 1 1



Differential Phase Shift Keying [DPSK](Non-coherent PSK)

DPSK Transmitter

bd bd d k k k k k 11

Input Binary Sequence {bK} 1 0 0 1 0 0 1 1

Differentially Encoded 1

sequence {dK}

1 0 1 1 0 1 1 1

Transmitted Phase 0 0 Π 0 0 Π 0 0 0

Received Sequence

(Demodulated Sequence)

1 0 0 1 0 0 1 1



If correlator output is +ve -- A decision is made in favour of symbol ‘1’ If correlator output is -ve --- A decision is made in favour of symbol ‘0’

DPSK Receiver

Relation b/w erfc and Q functions



u

dz zuerf 0

2 )exp(2

)(

u

dz zuerfc )exp(2

)( 2

erfc(u) = 1 – erf(u)

dx x

vQv

2exp

2

1)(

2

221)( verfcvQ

)2(2 uQuerfc

Relation b/w erfc and Q functions

Example 1.



Example 1.Binary data is transmitted over an RF band pass channel with a usable bandwidth of10 MHz at a rate of (4.8) (106) bits/sec using an ASK signaling method. The carrieramplitude at the receiver antenna is 1 mv and the noise power spectral density at thereceiver input is 10-15 watt/Hz. Find the error probability of a coherent and non

coherent receiver.

Solution: For Coherent ASK,

).10(226

/ 102 /

8.4 / 10,1;4

7

15

6

2

QPe

Hzwatt

T mv AT A

QP b

b

e

,)16 /(exp2

1 2 be T AP

For Non-Coherent ASK,

= 0.0008



Minimum shift keying

• Continuous Phase Frequency Shift Keying [CPFSK] )

0)0(2cos2

1)0(2cos2

)(

2

1

symbol f or t

symbol f or t

t S

f T E

f T

E

b

b

b

b

)0(2cos2

)( t t S f T

E c

b

b

ratiodeviationtheh

frequencycarrier theWhere

T h

h

h

bygivenares frequencied transmitteThe

t

t

t

f

f f

f f f

T f f

T f f

f f

T T

h

c

b

c

bc

bc

b

b

&

)(

)(2 / 1

&

0)0()(

21

21

2

1

21

2

2

The phase θ(t) is a continuous function of time:

Th i ti f h θ(t) ith ti t f ll th i ti f f



The variation of phase θ(t) with time t follows a path consisting of sequence ofstraight lines, the slope of which represent frequency change.

“Phase tree”plot showing possible paths starting from time t=0.



When h= ½ the frequency deviation equals half the bit rate.

This is the minimum frequency difference (deviation) that allows the two FSK

signals representing symbol ‘1’ & ‘0’.

A CPFSK signal with a deviation ratio of one- half is referred to as “minimum

shift keying”[MSK].

Deviation ratio h is measured with respect to the bit rate 1/Tbat t =Tb

0

1

)0()( Symbol for h

Symbol for h

T b



Phase Trellis, for sequence 1101000



With deviation ratio h=1/2,

b

b

T t t T

t 02

)0()(

)2()]([2

)2()]([2

)( t f Sint SinT

E t f Cost Cos

T

E t s

c

b

bc

b

b

In terms of In phase and Quadrature Components,

+ Sign corresponds to symbol 1- Sign corresponds to symbol 0

In- phase components

bb

bb

b

bb

b

b

b

T t T t T

CosT

E

t T

CosCosT

E

t CosT

E t s

2

2

2])0([

2

])([2

)(1

For the interval of

consists of half cosine pulse

bb T t T

+ Sign corresponds to θ(0) =0- Sign corresponds to θ(0) = п



Quadrature components

For the interval of

consists of half sine pulsebT t 20

b

bb

b

b

b

b

b

b

bQ

T t t T T

E

t T

CosT SinT

E

t SinT

E t s

202

sin2

2])([

2

])([2)(

+ Sign corresponds to θ(Tb) =п /2

- Sign corresponds to θ(Tb) = -п /2



Since the phase states θ(0) and θ(Tb) can each assume one of the two

possible values, any one of the four possibilities can arise

bbc

bb

T t T t f Cost T

CosT

t

)2(

2

2)(1

bc

bb

T t t f Sint T

SinT

t 20)2(2

2)(2

bbc

bb

T t T t f Cost T

CosT

t

)2(

2

2)(1

bc

bb

T t t f Sint T

SinT

t 20)2(2

2)(2

We may express the MSK signal in the form



We may express the MSK signal in the form

bT t t st st s 0)()()(

2211

bbb

T

T

T t T Cos E

dt t t ssb

b

)0(

)()(11

bbb

T

T t T Sin E

dt t t ssb

b

20)(

)()(

2

0

22

Signal spacediagram of MSKsystem

Sequence and waveforms for MSK signal



Sequence and waveforms for MSK signal



For AWGN Channel, x(t) = s(t) + w(t)

bb

T

T

T t T ws

dt t t x xb

b

11

11)()(

The projection of the received signal x(t) onto the reference signal )(1 t is

)(2 t Similarly, the projection of the received signal x(t) onto the reference signal is

b

T

T t ws

dt t t x xb

20

)()(

22

2

0

22

2)(

^ bT

2

)(^ bT

If x2>0, the receiver chooses the estimate

. If, on the other hand, x2<0, it chooses the estimate

.



• To reconstruct the original binary sequence, weinterleave the above two sets of phasedecisions,

If we have the estimates and ,or alternatively if we have the estimates

and , the receiver makes a final decisionin favor of symbol 0.

If we have the estimates and ,

or alternatively if we have the estimates

and , the receiver makes a finaldecision in favor of symbol 1.

0)0(^

2

)(

^ bT

0)0(^

)0(^

2)(

^ bT

2)(

^

bT

0)0(^

2)(

^

bT



T n

b

C

4f c

T b4

1

T b2

1

T b2

1

2

1

2

2

2

2

2

1

h for h

or h

hor

h

R f T

f f

R f

T f f

bC b

c

bC bc

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