ADC

PCM signal

Sample

Quantize

Analog Input

Signal

Encode

Line Code

xs(t)

xQ(t)

xk(t)

x(t)

Analog to Digital Conversion

1

1. Sampling of Continuous Signal

First step of analog to digital conversion is the periodic sampling

of continuous analog signal. A sampled signal is considered at some

distinct instant t = nT; n = 0, 1, 2, 3,… for duration τ.

If some one asks what is the amplitude of the sampled signal

between two consecutive sample, the answer is ‘undefined’ same is

true for digital signal.

τ

xs(t)

t

Sampled signal

2

Let an analog signal x(t), band limited to B Hz shown in fig.1, is

sampled at a rate fc = 1/T samples/sec. If the width of each sample is τ

then sampled signal xs(t) can be expressed as the product of x(t) and a

unit amplitude pulse train s(t) of period T and width of each pulse of τ

shown in fig.2. Sampled form xs(t) of x(t) is shown in fig.3.

T

τ

s(t)

tτ

xs(t)

t

x(t)

-B B

f

X(f)

Fig. 1 (a) Continuous base band signal(b) Spectrum of x(t)

Fig.2 Unit amplitude pulse train Fig. 3 Sampled signal

3

Lt us express s(t) in terms of Fourier co-efficient,

1

0 2sin

2cos

2s(t)

nnn

T

ntb

T

nta

TT

a

ndT

nc

T

n

T

n

T

nt

n

T

tdT

nt

tdT

nttsa

T

Tn

sinsin

/sin

2sin

2

2cos.1

2cos)(

2/

2/

2/

2/

2/

2/

Where

;Where d = τ/T, known as duty cycle.

T

τ

s(t)

t

1.sin000

T

ncLtaLta nnn

4

0

2sin.1

2sin)(

2/

2/

2/

2/

tdT

nt

tdT

nttsb

T

Tn

1

1

2cossin2

2cossin

2s(t )

n

n

T

ntndcdd

T

ntndc

TT

T

τ

s(t)

t

5

1

2cossin)(2)()().()(n

cs tnfndctdxtdxtstxtx

1

cossin)(2)(n

ctnndctxdtdx

Now

)()( Xtx Let

)()(2

1)cos()( ccc nXnXtntx

11

11

1

sin)(sin)()(

sin)(sin)()(

sin)()(2

12)()()(

nc

nc

nc

nc

nccss

ndcnXdndcnXddX

ndcnXdndcnXddX

ndcnXnXddXXtx

n

cnn

c ndcnXdndcnXddX sin)(sin)()(0,

6

...sin)()(sin)(2sin)2(...)( dcdXdXdcdXdcdXX cccs

n

cnn

c ndcnXdndcnXddX sin)(sin)()(0,

ffc+Bfc-B-fc

-B B fc

Xs(f)

7

Plot of above equation is shown in fig.4.

To extract x(t)↔X(ω) from xs(t)↔Xs(ω) a low pass filter with cutoff

frequency between B and fc-B can be used. There must not any

overlapping between X(ω) and X(ω- ωc) for proper filtering, therefore

fc-B≥B fc≥2B because of practical characteristics of low pass filter.

Therefore sampling rate of a signal must be greater than or equal to the

maximum frequency of the original/ base band signal.

ffc+Bfc-B-fc

-B B fc

Xs(f)

Fig. 4 Spectrum of Sampled signal

The condition fc≥2B is called Nyquist sampling rate.8

Example:

An analog signal of highest frequency of 3.4 KHz

sampled maintaining a guard band of 300Hz, find

sampling rate.

fc = G.B+2B = 0.3+2×3.4 =7.1KHz

9

2. Quantization of Sampled Signal

Second step of digitations of an analog signal is quantization of

sampled signal into some known discrete amplitude levels. Now

both the axis of the signal becomes discrete hence a quantized

signal is actually a digital signal.

Due to quantization the signal losses its originality and an error

is introduced with the signal known as quantization error. Actually

amplitude of a sample is adjusted with its nearest quantization

level. The spacing between adjacent levels are called quantization

interval.

10

If ‘a’ be the quantization interval then mean square error of a sample

of amplitude Sj+є in the vicinity of jth level Sj, in the interval, of

fig. 5 is expressed like,

Sj

Sj+a/2

Sj+a/2

Sj+є

Fig. 5 Quantization error

12

1)(

22/

2/

22 ad

aE

a

a 22

aaWhere

11

Let the signal at transmitting end is quantized into total N discrete levels with

maximum plus-minus signal excursion of P volts, maximum excursion in positive or

negative direction is V and the range of quantized samples is A .

Fig.6 shows a quantization technique of 6 levels. Here quantized amplitude would be

±a/2, ±3a/, ±5a/2 and ±7a/2. Now V = 4a , A = 7a and P = 8a. Therefore for N levels

quantization, V = Na/2 , A = (N-1)a and P = Na.

A P

V

0

a/2

3a/2

5a/2

-5a/2

-3a/2

-a/2

4a

-4a

7a/2

-7a/2

Now peak signal to quantization ratio

(SQR) of quantization system will be,

NN

a

aN

E

VSQR

log208.43log20

12/

4/log10

)(log10

2

22

2

2

Fig. 6 quantization technique of 8 levels 12

Example-2

A sinusoidal wave of peak amplitude A has mean square amplitude of

A2/2. According to equation of previous slide SQR of the sine wave will

be, SQR(dB)=10log{(A2/2)/(a2/12)} =7.78+20log(A/a)

0 2 4 6 8 10 125

10

15

20

25

30

A/a

SQ

R(d

B)

Fig. 7 SQR of a sinusoidal wave

13

SQR increases with increase in amplitude (loudness of voice signal) of

baseband signal. Our aim is to keep SQR constant irrespective of amplitude of

baseband signal

3. Pulse Code Modulation

Pulse Code Modulation (PCM) is essentially final step of analog to

digital conversion where each quantized sample in converted to

equivalent binary pulses, resembles to decimal to binary

conversion.

Fig.8 show 8 levels PCM technique. In PCM gray code is

preferable compare to ordinary binary code to combat variable

number of bit error for adjacent levels like 3 and 4, 7 and 8 etc

when sampled pulses are transmitted directly. Because in gray

code there is only one bit difference between adjacent decimal

numbers hence there is only possibility of single bit error instead

of variable number of bit error during reception of each sample.

14

TT

6

34

t

Fig. 8(a) Quantized sample of 8 levels quantization

1 1 0

TT

t

1 0 0 0 1 1

Fig.8(b) PCM data of fig. 8(a)15

Exerceise-1

A voice signal of highest frequency of 3.4KHz sampled

maintaining a guard band of 200Hz. After sampling the signal is

quantized into 256 levels. Determine sampling rate and bit rate of

PCM.

16

Companding

From the analysis of previous section we can say, if the quantization

interval is kept directly proportional to amplitude of signal then SQR

will remain constant over the entire dynamic range.

Let a = kA; where A is the amplitude a signal at a particular instant

and k is a constant.

kkAk

A

a

AdBSQR log2079.10

12log10

12/log10

12/log10)(

222

2

2

2

17

Therefore SQR(dB) is independent of signal amplitude. Above

phenomenon can be implemented by compressing the signal

nonlinearly (larger the sample amplitude, the more it is

compressed) then providing uniform quantization is provided on it

like fig.9.

Compressed output signal

Input signal

Fig. 9 Non linear compression of signal

18

At receiving end signal is recovered by reverse operation

(nonlinear expansion). The process of compression and expansion

of signal is called companding shown in fig10.

Input Output

CompressionPCM

D/A Expansion

Fig. 10 Companding technique

Input signal x is compressed in North America and Japan according

to following relation known as μ-law companding.

)1ln(

)1ln()sgn()(

xxxF ;-1 ≤ x ≤ 1

19

where sgn(x) is the polarity of x and μ is a parameter used to define

amount of compression. Practical compander uses μ = 255.

1)1(1

)sgn()(1 yyyF

The reverse operation or expansion of signal is expressed as,

; where y is the compressed signal.

20

Another companding technique prevalent in Europe and most of

the region of the world is known as A-law companding

expressed as,

)Aln(1

xA)xsgn()x(FA

0 ≤ |x| ≤ 1/A

Inverse or expansion of the signal is expressed as,

A

)Aln(1y)ysgn()y(F 1

A

)Aln(1

1y0

Where

Practical compander uses A = 100.

21

22

PAM, PWM and PPM

In all of above pulse modulation techniques unit amplitude pulse

train is used as carrier wave.

In pulse amplitude modulation (PAM) the amplitude of the

carrier pulses are proportional to instantaneous amplitude of

continuous base band signal x(t) shown in fig.11.

In PWM the width of a pulse is proportional to the instantaneous

amplitude of continuous base band signal. In PPM the position of

the pulse i.e. distance from a reference instant is proportional to the

instantaneous amplitude of continuous base band signal.

23Fig. 11 PAM, PWM and PPMwave

0 Ts 2Ts 3Ts 4Ts 5Ts 6Ts

x(t),xp(t)

t

τ

Ts

τ

s(t)

t

t

xpo(t)

PWM

PPM

xd(t)

t

PAM

24

PAM wave xp(t) can be generated using simple electronic circuit of fig.12 where

pulses at G1 and G2 of FETs are shown in fig.13 and fig.14 respectively. PAM signal

actually resembles to flat top sampled signal of period Ts. Since the width of each PAM

pulse is τ therefore pulses at G2 stars at the position of τ unit delayed from kTs to

discharge the capacitor C.

CG2

G1

x(t) xp(t)

G1

0 Ts 2Ts 3Ts 4Ts 5Ts 6Ts

t

G2

0 τ Ts 2Ts 3Ts 4Ts 5Ts 6Ts

t

Fig. 12 Circuit to convert continuous signal to PAM wave

Fig. 13 Pulses at G1 of above circuit

Fig. 14 Pulses at G2 of above circuit

PAM (Pulse Amplitude Modulation)

25

25

CG2

G1

x(t) xp(t)

G1

0 Ts 2Ts 3Ts 4Ts 5Ts 6Ts

t

G2

0 τ Ts 2Ts 3Ts 4Ts 5Ts 6Ts

t

0 Ts 2Ts 3Ts 4Ts 5Ts 6Ts

x(t),xp(t)

t PAM

τ

26

In pulse width modulation (PWM) width of a carrier pulse is proportional to instantaneous

amplitude of the baseband signal of course the amplitude of each pulse is equal.

PWM can be generated simply from a comparator circuit of fig.15 where input at non-

inverting terminal is the baseband signal x(t) and a sawtooth wave at inverting terminal.

Resulting wave is shown in fig.16 where starting position of each pulse is at kTs but terminating

points depend on width of each pulse, hence amplitude of baseband signal.

Fig. 15 PWM and PPM generator

Sawtooth

wave

x(t)

PWM wave

Monostable

Multivibrator

PPM

wave

Comparator

PWM (Pulse Width Modulation)

xd(t)

t

27

xd(t)

t

t

xpo(t)

PWM

PPM

Fig. 16 Output waves of PWM and PPM generator

In pulse position modulation (PPM),

amplitude and width of each pulse is

kept same but starting position of each

pulse i.e. the distance between instant,

t = kTs and stating point of kth pulse

depends on instantaneous amplitude of

base band signal.

PPM wave can be generated simply

applying PWM pulses at the input of a

monostable multivibrator.

Sawtooth

wave

x(t)

PWM wave

Monostable

Multivibrator

PPM

wave

Comparator

PPM (Pulse Position Modulation)

28

Let us now consider the case of conversion of PWM signal to

PAM. PWM pulses of fig. 17 (a) is passed through a ramp generator

with large capacitor at output to hold the final amplitude of the

signal, the resulting wave is shown in fig. 17(b).

A pulse train of period Ts is locally generated and added with the

signal of fig 17(b) such that pulses fall on the constant part of the

wave shown in fig. 17(c). Finally pulses of fig.17(c) is passed

through a clipper circuit of threshold voltage shown by doted line of

fig. 17.(c).

Clipper circuit actually clipped the signal below the threshold

level, therefore the resulting wave would be like fig.17(d) i.e. the

equivalent PAM signal. Baseband signal can be regenerated by

passing the PAM wave through a low pass filter.

Conversion of PWM to PAM

29

PWM

Fig. 17(a) PWM wave

Fig.17 (b) Output of Ramp Generator

Fig. 17(c) Locally generated pulses are added with ramp output

Fig. 17(d) Out put of clipper circuit

30

It is found that average amplitude of sample to sample difference

is less than that of original sampled wave of voice signal therefore

less number bit is required for PCM, maintaining same SQR. If

PCM is done on consecutive pulse difference instead of individual

pulse known as differential pulse code modulation (DPCM).

When sample to sample difference is expressed by a single bit

then the modulation scheme is called delta modulation is

considered as an special case of DPCM.

Differential Pulse Code Modulation (DPCM)

31

Major components of transmitter and receiver of delta modulation are sampler,

predictor, quantizer, adder and smoothing filter shown in fig. 18. In transmitter the

analog baseband signal x(t) is sampled and a difference signal, is generated. Here

is jth sampled pulse of x(t) is xj and gj is jth predicted pulse.

x(t)

gj

xj +

_

Sampler

Predictor

Quantizer± k´

+

gj

± k´ +

_

Predictor

Smoothing

filter

Fig. 18(a) Delta modulator

Fig. 18(b) Delta demodulator

j

j

32

Predicted pulse is determined from linear combination of some previous sampled

pulses shown in fig. 19 is expressed as,

k

1s

sjsj xhg

where is the jth estimated sample determined as, g ± k´

Based on difference signal a pulse of amplitude k´ is generated by a quantizer

like,

jx

j

negativeisifk

positiveisifkP

j

j

j

gj

......Z-1 Z-1 Z-1 Z-1

h1 h2 hk

Σ

Fig. 19 Predictor circuit

33

Let us demonstrate delta modulation transmitter and receiver with

the graph of fig.20. Here predicted sample g1 at 1st sampling instant is

determined from linear combination of some fixed previous samples.

Now g1 has to be compared with 1st sample x1 and it is found from

fig.20 that x1 > g1, therefore transmitter will generate g1+ k´ and 1st

estimated sample, = g1+ k´ will be detected at receiver.

At 2nd sampling instant g2 is determined from the similar relation

and compared with x2 where x2 > g2, therefore transmitter will

generate g2 + k´ and 2nd estimated sample = g2+ k´ will be detected

at receiver. Same condition is also found at 3rd sampling instant.

1x

2x

34

At 4th sampling instant x4 < g4 therefore transmitter will generate

-k´ and 4th estimated sample, = g4- k´ will be detected at

receiver and so on.

Estimated samples are discrete points at the vicinity of the

baseband wave but some times points are above the curve and

some times fall below the curve depending on the profile of the

baseband curve. These discrete points are made a continuous using

a smoothing filter.

4x

35

Fig. 20 Comparison of and xj)(ˆ tx j

x(t)

gj

xj +

_

Sampler

Predictor

Quantizer± k´

+

k´

36

Since the estimated sampled points are little away from the baseband wave

hence a quantization noise is incorporated with delta modulation like fig. 21. If the

step size k´ is lowered then the separation between and xj will be reduced and

the zigzag curve of points would be smoother.1x

Fig. 21 Output of the demodulator for larger step size

t

x(t)

37

When step size k´ is very small then the zigzag curve of estimated points are simply

unable to follow the baseband curve when the curve changes rapidly like fig.22, known as

slope overload distortion.

Maximum slope supported by the zigzag curve is k´/Ts ; where Ts is the sampling period

Therefore a tradeoff has to be made with quantization noise and slope overload distortion

taking optimum step size kopt. Output of the demodulator for optimum step size is shown in

fig. 23.

K´

Ts

Fig. 22 Output of the demodulator for

smaller step size

Fig. 23 Output of the demodulator for

optimum step size

38

Example-1

Determine optimum step size for the signal x(t)= 5sin(2π20t) considering

sampling frequency of 8KHz.

Ans. The slope of the signal x(t),

The maximum slope is obtained taking =1.

The maximum slope of the signal mmax

= = 200π . From the trajectory of zigzag curve of fig. 23, the maximum slope

supported by it,

= kfs=k8000

Therefore 200 = 8000k

kopt

= 200π/8000 = π/40 volt

)202cos( t

2025

)202cos(2025)(

tdt

tdx

sT

k)tan(

39

Line codingBinary data (logic 0 or 1 of PCM) can be transmitted using a number

of different types of serial pulses. The choice of a particular pair of

pulses to represent the symbols 1 and 0 is called Line Coding. Line

coding can be represented by the diagram of fig. below.

ADC

PCM signal

Sample

Quantize

Analog Input

Signal

Encode

Line Code

xs(t)

xQ(t)

xk(t)

x(t)

Fig. Line coding

40

Each line code is described by a ‘symbol mapping function’ ak and a

‘pulse shape’ p(t):

k

bk kTtpatx )()(

Line code can be classified based on symbol mapping functions (ak) like,

Unipolar: In unipolar signalling binary symbol 0 is represented by the absence

of a pulse called space and the other binary symbol 1 is represented by the

presence of a pulse called mark. It is also called on-off keying.

Polar: In polar signalling a binary 1 is represented by a pulse p(t) and a binary

0 by the opposite (or antipodal) pulse –p(t).

Bipolar: Bipolar Signalling is also called ‘alternate mark inversion’ which uses

three voltage levels (+V, 0, -V) to represent binary symbols. Zeros, as in

unipolar, are represented by the absence of a pulse and ones (or marks) are

represented by alternating voltage levels of +V and –V.

41

Line code can again be classified based on pulse shapes p(t) like,

NRZ (Nonreturn-to-zero): The pulse occupies the full duration of

a symbol.

RZ (Return to Zero): The pulse occupies the half of duration of a

symbol.

Manchester (split phase): In Manchester encoding, the duration of

the bit is divided into two halves. The voltage remains at one level

during the first half and moves to the other level during the second

half. Binary logic 1 is +ve in 1st half and -ve in 2nd half. Binary

logic 0 is -ve in 1st half and +ve in 2nd half.

42

BINARY DATA

A

-A0(b) Polar NRZ

A

0(c) Unipolar RZ

A

-A0

(d) Bipolar RZ

1 1 0 1 0 0 1

Mark (hole)

Mark (hole)

Mark (hole)

Mark (hole)

space space space

A

-A

0(e) Manchester NRZ

Binary Signaling Formats

VoltsA

Time

0(a) Unipolar NRZ

Tb

According to above classification different types of line coding is shown

in fig. below.

Fig. Line coding

Multiplexing Techniques

43

44

45

46

47

Time division MultiplexingFig. below shows the frame of 30-channel system where out of 32 time slot (TS) the

0th TS is for frame alignment, 16th TS is for signaling and the rest 30 TS (1 to 15 and

17 to 31) for speech. Each TS contains 8 PCM bits, equivalent to each quantized

pulse of 256 level system. Such frame is called E1 frame flowed by European.

0 1 to 15 16 17 to 30 31

Frame Alignment Signaling

b1 b2 b3 b4 b5 b6 b7 b8

Fig.E1 frame of 30 Channels

48

The bit rate of the 30-channel PCM frame is, 8 kHz (sampling rate) × 8 (bits per sample

or TS) × 32 (number of TS/frame) = 2.048 Mbps. The total number of bite/frame =

8×32 = 256bits. We know the sampling rate of voice signal is 8000 samples/sec,

therefore sampling period, Ts = (1/8000) sec. The 8 PCM bits of each sample is placed

in a TS, therefore the length of a frame, Tf = (1/8000) sec 125 μs. The width of each bit,

d = 125/256 = 0.488 μs.

Multiplexing hierarchy of European system is shown in fig.3.

E1

M

U

X

30 Voice

channels,

each of 64

kbps

.

..

E1 = 2.048 Mbps

( 30 voice channel)

E2

M

U

X

E1

E2 = 8.448 Mbps

( 120 voice channels)

E3

M

U

X

E3 = 34.368 Mbps

( 480 voice channels)

E4

M

U

X

E2

E3

E4 = 139.264 Mbps

( 1920 voice channels)E5

M

U

X

E5= 564.992 Mbps

( 7680 voice channels)E4

Fig.3 European synchronous digital hierarchy

49

North American TDM frame contains 24 TS where each TS contains 8

bits of a user (equivalent of 8 PCM bits/sample) and the 0th TS

contains a single bit for frame alignment called T1 frame like fig.4.

0 Channel-1

Frame Alignmentb1 b2 b3 b4 b5 b6 b7 b8

Channel-2 … … … Channel-24

Fig.4 T1 frame of 30 Channels

The total number of bits/frame = 8×24+1 = 193 bits. The 8 PCM bits of each sample is

placed in a TS, therefore the length of a frame, Tf = (1/8000) sec 125 μs. The bit rate of T1

frame = 193/125 =1.544 Mbps.

50

The objective of digital modulation is to convert the rectangular

digital pulses to smooth sinusoidal wave hence considerable

reduction in bandwidth is achieved. The bandwidth reduction is

essential to cope the transmitted wave with the transmission

medium of lower bandwidth.

For example MODEM is connected between PC and telephone

line to convert the rectangular data pulse from the computer

(infinite bandwidth) into continuous FSK wave of lower bandwidth

to cope with the allowed bandwidth of the telephone line

(transmission medium).

Digital Modulation Techniques

51

In wired communication the digital modulation is necessary for,

To reduce the bandwidth of the baseband signal according the

capacity of the transmission medium and receiver circuits.

If the wavelength of the signal is considerable with the length of

the wire then the wire acts an antenna and radiates most of the

signal energy. The remedial measure of the situation is to increase

the frequency of the signal i.e. modulation with high frequency

carrier.

52

In wireless communications the reasons of digital modulation are,

In wireless communication digital modulation is necessary to avoid

interferences of surrounding users with choose of appropriate carrier.

Noisy immunity is increased for a high frequency modulated wave.

Size of an antenna depends on wavelength of a signal. After

modulation (using high frequency carrier) wavelength of the

modulated wave is reduced hence size of the antenna is reduced.

A modulated wave can propagate far longer way than the baseband

signal.

53

Frequency Shift Keying (FSK) In binary FSK two sinusoidal waves of frequency f1 and f2 over a

period [0, T] are used to represent binary logic 1 and 0 respectively.

The waves are like,

for logic 1

for logic 0

; where θ is the initial phase of sinusoidal wave i.e. at the stating

point symbol wave.

)2cos()( 11 tfAtS TktkT )1(

;)2cos()( 22 tfAtS

;

TktkT )1(

54

Input binary bits

as a selector

FSK wave

Oscillator-1

Oscillator-2

M

U

X

Fig. 1 FSK modulator

FSK wave can be generated using two oscillators of frequency f1

and f2 connected to a multiplexer like fig.1. The multiplexer is

switched between the oscillators by binary input bits.

)2cos()( 11 tfAtS

)2cos()( 22 tfAtS

55

Frequencies f1 and f2 are chosen such that S1(t) and S2(t) are mutually orthogonal.

In that case at receiving end S1(t) and S2(t) can be detected using cross

correlation called coherent demodulation.

Tk

kTdttStS

)1(

21 0)()(

To satisfy above condition,

and ; where m and n are integers. The difference

between frequencies,

For m = 2 the separation becomes 1/T which ensure phase

continuity at bit transitions called Sunde’s FSK shown in fig.2. In

case of discontinuity of phase at each bit transition point like fig. 3,

spectral width becomes wider.

nTff 2)(2 21 mTff 2)(2 21

T

mff

2)( 21

56

S(t)1 0 1 0

t

S(t)1 0 1 0

t

Fig.2 Continuous FSK

Fig.3 Discontinuous FSK

57

The received signal contaminated by awgn is expressed as,

r(t)=Si(t)+n(t); where i=1,2 and n(t) is the additive white

Gaussian noise (awgn). The coherent detector of orthogonal FSK

is shown in fig.4.

Decision

r(t)

1

0

Tk

kTdt

)1(

tftf 21 2cos2cos

Fig. 4 Coherent demodulator of FSK

58

Binary Phase Shift Keying (BPSK)

In BPSK binary logic 1 and 0 are presented by two sinusoidal waves

of same frequency and peak amplitude over the period [0, T] but

initial phases are 0 and π respectively. BPSK wave is expressed like,

)2cos()02cos()(1 tfAtfAtS cc TktkT )1( for logic 1

)2cos()2cos()(2 tfAtfAtS cc TktkT )1( for logic 0

Wave shape of BSK is shown in fig.5.

1 0 1 0

Fig. 5 BPSK waveform

;

;

59

PSK signals are represented graphically in two dimensional co-ordinate

system called signal constellation. Here x axis is represented by a

function,

tfT

t c 2cos2

)(1 Tt 0

and y axis by,

tfT

t c 2sin2

)(2 Tt 0

;

;

Orthogonal basis function Unit vector

and and

Correlation Dot product

and . = 0 and . = 1

Component along x-axis from

is given as,

Component along x-axis

from is given as,

tfT

t c 2cos2

)(1 tfT

t c 2sin2

)(2

0)()(0

21 T

tt 1)()(0

11 T

tt

jbiaS ˆˆ

j ii

)()( 21 tbtaS

atST

0 1 )(

i j

i

aiS ˆ.

60

Let us represent S1(t) and S2(t) in terms of and )(1 t )(2 t

)()(

)(sin2

)(cos2

22sin

2sin2cos

2

2cos

2sinsin2coscos

)2cos()(

2111

21

tStS

tT

AtT

A

Ttf

TAtf

T

TA

tfAtfA

tfAtS

ii

cici

cici

ici

2/f2cos)( 2

0

122

0

21 TAdttAdttSE

TT

2/f2s)( 2

0

122

0

22 TAdttinAdttSE

TT

Energy of S1(t) or S2(t)

)(sin)(cos

)()()(

21

2111

tEtE

tStStS

ii

i

61

The abscissa of the constellation point, The ordinate of the constellation point,

In BPSK θi = 0 or π

i

T

T

ii

ES

dtttStS

dtttSS

cos

)()()(

)()(

1

012111

011

i

T

T

ii

ES

dtttStS

dtttSS

sin

)()()(

)()(

2

022111

022

EEES i orcos1

0sin2 iES

62

0,),( 121 SSS

The co-ordinate of the constellation points,

In Cartesian co-ordinate system shown in fig.6.

Fig.6 BPSK Constellation

S2=-√E S1=√E

Ψ1(t)

Ψ2(t)

0,cos iE )0,( E )0,( Eor

63

Now BPSK modulator and its coherent demodulator can be

implemented like fig. 3.

Local

Oscillator

×Polar NRZ data pulse

sequence b(t)

Ab(t)cos(2πfct)

Acos(2πfct)

Fig.3 BPSK (a) Modulator (b) Demodulator

CR

×r(t)

Acos(2πfct)

1

01 or 0

L Tk

kT

dt)1(

k

k kTtpbtb )()(

Where bk ε {1, -1} and

T

Tttp

2/)(

64

Here output of the correlator of demodulator assuming S1(t)

transmitted is,

2)2cos()(0

2

)2cos()()4cos(12

)2cos()()2(cos

)2cos()()2cos(

)2cos()()(

)2cos()(

)1(

)1()1(

)1(

2

)1(

)1(

1

)1(

ATdttftn

AT

dttftndttfA

dttftntfA

dttftntfA

dttftntS

dttftrL

Tk

kT

c

Tk

kT

c

Tk

kT

c

Tk

kT

cc

Tk

kT

cc

Tk

kT

c

Tk

kT

c

If S2(t) is transmitted then, 2

ATL

CR

×r(t)

Acos(2πfct)

1

01 or 0

L Tk

kT

dt)1(

65

M-ARY PSKIn BPSK bit rate and modulation symbol rate are equal. In MPSK n = log2M bits are

represented by a modulation symbol hence bit rate is n times higher than symbol rate.

Therefore n times more information can be transmitted by MPSK (for the same

baud/symbol rate) scheme compared to that of BPSK at the expense of bit error rate.

MPSK wave is expressed as,

tfAtfA

tfAtS

cici

ici

2sinsin2coscos

)2cos()(

)()( 2211 tStS ii

Tt 0Where i = 1, 2, 3,…M, and

are orthogonal basis functions. M

ii

)12(

)(),( 21 tt

66

The abscissa of the constellation points,

i

T

ii EdtttSS cos)()(0

11

The ordinate of the constellation point,

i

T

ii EdtttSS sin)()(0

22

The co-ordinates of the constellation points in Cartesian co-ordinate

system are,

iiii EESS sin,cos),( 21

In polar co-ordinate system amplitude is,

EEE ii 22

sincos

The phase angle,2

21tani

ii

S

S

67

Therefore the polar co-ordinates of the signal constellation are ),( iE

For M = 8, initial phases are 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2 and 7π/4

taking i = 0, 1, 2 ,..., 7. Constellation of 8-ARY PSK is shown in fig.1

below with its decision range.

Ψ1(t)

Ψ2(t)

Fig.1 8-ARY PSK Constellation

68

QPSKQuadrature PSK is a special case of M-ary PSK (MPSK) where M = 4; Therefore the

initial phases, ; i = 1, 2, 3, 4 provides π/4, 3π/4, 5π/4 and 7π/4. Bit rate of

QPSK scheme is lower than that of M-ARY PSK for M > 4 (for same baud rate) but

provide better performance in context of bit error rate. Each modulation symbol

carries two binary bits correspond to a point on constellation shown in fig. 2. In any

constellation gray code is used to ensues single bit discrepancy between adjacent

points.

4

)12(

ii

01 11

00 10

θi

Ψ1(t)

Ψ2(t)

Fig. 2 QPSK Constellation