1

Digital Communication SystemsLecture-3, Prof. Dr. Habibullah JamalUnder Graduate, Spring 2008

2

3

4

5

Baseband Demodulation/Detection In case of baseband signaling, the received signal sis already in pulse-like form. Why is then is demodulator

required?

Arriving baseband pulses are not in the form of ideal pulse shapes, each one occupying its own symbol interval.

The channel (as well as any filtering at the transmitter) causes intersymbol interference (ISI).

Channel noise is another reason that may cause bit error is channel noise.

6

Effect of Channel

Figure 1.16 (a) Ideal pulse. (b) Magnitude spectrum of the ideal pulse.

7

Figure 1.17 Three examples of filtering an ideal pulse. (a) Example 1: Good-fidelity output. (b) Example 2: Good-recognition output. (c) Example3: Poor-recognition output.

8

0 2 4 6 8 10 12 14 16 18 20-2

-1

0

1

2

0 2 4 6 8 10 12 14 16 18 20-2

-1

0

1

2

0 2 4 6 8 10 12 14 16 18 20-2

-1

0

1

2

Effect of Noise

9

FREQUENCYDOWN

CONVERSIONTRANSMITTEDWAVEFORM

AWGN

RECEIVEDWAVEFORM RECEIVING

FILTEREQUALIZING

FILTER THRESHOLDCOMPARISON

FORBANDPASSSIGNALS

COMPENSATIONFOR CHANNELINDUCED ISI

DEMODULATE & SAMPLESAMPLEat t = T

DETECT

MESSAGESYMBOL

ORCHANNELSYMBOL

ESSENTIAL

OPTIONAL

Figure 3.1: Two basic steps in the demodulation/detection of digital signals

3.1.2 Demodulation and Detection

The digital receiver performs two basic functions:

Demodulation, to recover a waveform to be sampled at t = nT.

Detection, decision-making process of selecting possible digital symbol

10

3.2 Detection of Binary Signal in Gaussian Noise For any binary channel, the transmitted signal over a symbol interval

(0,T) is:

The received signal r(t) degraded by noise n(t) and possibly degraded by the impulse response of the channel hc(t), is

(3.1)Where n(t) is assumed to be zero mean AWGN process For ideal distortionless channel where hc(t) is an impulse function and

convolution with hc(t) produces no degradation, r(t) can be represented as:

(3.2)

1

2

( ) 0 1( )

( ) 0 0i

s t t T for a binarys t

s t t T for a binary

2,1)()(*)()( itnthtstr ci

Ttitntstr i 02,1)()()(

11

3.2 Detection of Binary Signal in Gaussian Noise

The recovery of signal at the receiver consist of two parts Filter

Reduces the effect of noise (as well as Tx induced ISI) The output of the filter is sampled at t=T.This reduces the received

signal to a single variable z(T) called the test statistics Detector (or decision circuit)

Compares the z(T) to some threshold level 0 , i.e.,

where H1 and H2 are the two possible binary hypothesis

1

2

0( )

H

H

z T

12

Receiver Functionality

The recovery of signal at the receiver consist of two parts:1. Waveform-to-sample transformation (Blue Block)

Demodulator followed by a sampler At the end of each symbol duration T, predetection point yields a

sample z(T), called test statistic

(3.3)

Where ai(T) is the desired signal component,

and no(T) is the noise component

2. Detection of symbol Assume that input noise is a Gaussian random process and

receiving filter is linear

(3.4)

0( ) ( ) ( ) 1,2iz T a T n T i

2

0

0

0

0 2

1exp

2

1)(

n

np

13

Then output is another Gaussian random process

Where 0 2 is the noise variance

The ratio of instantaneous signal power to average noise power , (S/N)T, at a time t=T, out of the sampler is:

(3.45)

Need to achieve maximum (S/N)T

2

11

00

1 1( | ) exp

22

z ap z s

2

22

00

1 1( | ) exp

22

z ap z s

20

2

i

T

a

N

S

14

3.2.2 The Matched Filter

Objective: To maximizes (S/N)T

Expressing signal ai(t) at filter output in terms of filter transfer function H(f) (Inverse Fourier transform of the product H(f)S(f)).

(3.46)

where S(f) is the Fourier transform of input signal s(t) Output noise power can be expressed as:

(3.47) Expressing (S/N)T as:

(3.48)

dfefSfHta ftji

2)()()(

dffHN 202

0 |)(|2

dffHN

dfefSfH

N

SfTj

T 20

22

|)(|2

)()(

15

Now according to Schwarz’s Inequality:

(3.49)

Equality holds if f1(x) = k f*2(x) where k is arbitrary constant and * indicates complex conjugate

Associate H(f) with f1(x) and S(f) ej2 fT with f2(x) to get:

(3.50)

Substitute in eq-3.48 to yield:

(3.51)

dffSdffHdfefSfH fTj222

2 )()()()(

dxxfdxxfdxxfxf2

2

2

1

2

21 )()()()(

dffSNN

S

T

2

0

)(2

16

Or and energy E of the input signal s(t):

Thus (S/N)T depends on input signal energy

and power spectral density of noise and

NOT on the particular shape of the waveform

Equality for holds for optimum filter transfer function H0(f)

such that: (3.54)

(3.55)

For real valued s(t):

(3.56)

0

2max

N

E

N

S

T

dffSE2

)(

0

2max

N

E

N

S

T

fTjefkSfHfH 20 )(*)()(

fTjefkSth 21 )(*)(

( ) 0( )

0

ks T t t Th t

else where

17

The impulse response of a filter producing maximum output signal-to-noise ratio is the mirror image of message signal s(t), delayed by symbol time duration T.

The filter designed is called a MATCHED FILTER

Defined as:

a linear filter designed to provide the maximum

signal-to-noise power ratio at its output for a given

transmitted symbol waveform

( ) 0( )

0

ks T t t Th t

else where

18

3.2.3 Correlation realization of Matched filter

A filter that is matched to the waveform s(t), has an impulse response

h(t) is a delayed version of the mirror image of the original signal waveform

( ) 0( )

0

ks T t t Th t

else where

Signal Waveform Mirror image of signal waveform

Impulse response of matched filter

Figure 3.7

19

This is a causal system Recall that a system is causal if before an excitation is applied at time

t = T, the response is zero for - < t < T The signal waveform at the output of the matched filter is

(3.57)

Substituting h(t) to yield:

(3.58) When t=T,

(3.59)

dthrthtrtzt

)()()(*)()(0

dtTsr

dtTsrtz

t

t

0

0

)(

)()()(

dsrtzT

)()()(0

20

The function of the correlator and matched filter are the same

Compare (a) and (b) From (a)

Tdttstrtz

0)()()(

T

TtdsrTztz

0)()()()(

21

From (b)

But

At the sampling instant t = T, we have

This is the same result obtained in (a)

Hence

0'( ) ( )* ( ) ( ) ( ) ( ) ( )

tz t r t h t r h t d r h t d

)()]([)()()( tTstTsthtTsth

t

dtTsrtz0

)()()('

0 0'( ) '( ) ( ) ( ) ( ) ( )

T T

t Tz t z T r s T T d r s d

)(')( TzTz

T

dsrTz0

)()()('

22

Detection Matched filter reduces the received signal to a single variable z(T), after

which the detection of symbol is carried out The concept of maximum likelihood detector is based on Statistical

Decision Theory It allows us to

formulate the decision rule that operates on the data optimize the detection criterion

1

2

0( )

H

H

z T

23

P[s1], P[s2] a priori probabilities These probabilities are known before transmission

P[z] probability of the received sample

p(z|s1), p(z|s2)

conditional pdf of received signal z, conditioned on the class s i

P[s1|z], P[s2|z] a posteriori probabilities After examining the sample, we make a refinement of our previous

knowledge P[s1|s2], P[s2|s1]

wrong decision (error) P[s1|s1], P[s2|s2]

correct decision

Probabilities Review

24

Maximum Likelihood Ratio test and Maximum a posteriori (MAP) criterion:

If

else

Problem is that a posteriori probabilities are not known. Solution: Use Bay’s theorem:

1 2 1( | ) ( | )p s z p s z H

1 1

2 2

1 1 2 21 1 2 2

)( | ) ( ) ( | ) (( | ) ( ) ( | ) ( )( ) ( )

H H

H H

p z s P s p z s P sp z s P s p z s P sP z P z

How to Choose the threshold?

2 1 2( | ) ( | )p s z p s z H

( | ) ( )( | )

( )p z s P si ip s zi p z

25

MAP criterion:1

2

1 2

2 1

( )( | ) ( )( )( | ) ( )

H

H

likelihood ratio test LRTp z s P sL zp z s P s

When the two signals, s1(t) and s2(t), are equally likely, i.e., P(s2) = P(s1) = 0.5, then the decision rule becomes

This is known as maximum likelihood ratio test because we are selecting the hypothesis that corresponds to the signal with the maximum likelihood.

In terms of the Bayes criterion, it implies that the cost of both types of error is the same

1

2

1

2

max( | )( ) 1( | )

H

H

likelihood ratio testp z sL zp z s

26

Substituting the pdfs2

11 1

00

1 1: ( | ) exp

22

z aH p z s

2

22 2

00

1 1: ( | ) exp

22

z aH p z s

2

11 1

0012

22

2 200

1 1exp

22( | )( ) 1 1

( | ) 1 1exp

22

z aH H

p z sL z

p z s z aH H

27

Taking the log of both sides will give

1

2 21 2 1 2

2 20 0

2

( ) ( )ln{ ( )} 0

2

H

z a a a aL z

H

1

2 21 2 1 2 1 2 1 2

2 2 20 0 0

2

( ) ( ) ( )( )

2 2

H

z a a a a a a a a

H

1

2 21 2 1 2

2 20 0

2

( ) ( )exp 1

2

H

z a a a a

H

Hence:

28

Hence

where z is the minimum error criterion and 0 is optimum threshold

For antipodal signal, s1(t) = - s2 (t) a1 = - a2

1

20 1 2 1 2

20 1 2

2

( )( )

2 ( )

H

a a a az

a a

H

1

1 20

2

( )

2

H

a az

H

1

2

0

H

z

H

29

This means that if received signal was positive, s1 (t) was sent, else s2(t) was sent

30

Detection of Binary Signal in Gaussian Noise

The output of the filtered sampled at T is a Gaussian random process

31

The impulse response of a filter producing maximum output signal-to-noise ratio is the mirror image of message signal s(t), delayed by symbol time duration T.

The filter designed is called a MATCHED FILTER and is given by:

( ) 0( )

0

ks T t t Th t

else where

Matched Filter and Correlation

32

Hence

where z is the minimum error criterion and 0 is optimum threshold

For antipodal signal, s1(t) = - s2 (t) a1 = - a2

1

1 20

2

( )

2

H

a az

H

1

2

0

H

z

H

Bay’s Decision Criterion and Maximum Likelihood Detector

33

Probability of Error

Error will occur if s1 is sent s2 is received

s2 is sent s1 is received

The total probability of error is the sum of the errors

0

2 1 1

1 1

( | ) ( | )

( | ) ( | )

P H s P e s

P e s p z s dz

0

1 2 2

2 2

( | ) ( | )

( | ) ( | )

P H s P e s

P e s p z s dz

2

1 1 2 21

2 1 1 1 2 2

( , ) ( | ) ( ) ( | ) ( )

( | ) ( ) ( | ) ( )

B ii

P P e s P e s P s P e s P s

P H s P s P H s P s

34

If signals are equally probable

Numerically, PB is the area under the tail of either of the conditional distributions p(z|s1) or p(z|s2) and is given by:

2 1 1 1 2 2

2 1 1 2

( | ) ( ) ( | ) ( )

1( | ) ( | )

2

BP P H s P s P H s P s

P H s P H s

2 1 1 2 1 2

1( | ) ( | ) ( | )

2by Symmetry

BP P H s P H s P H s

0 0

0

1 2 2

2

2

00

( | ) ( | )

1 1exp

22

BP P H s dz p z s dz

z adz

35

The above equation cannot be evaluated in closed form (Q-function) Hence,

0

1 2

0

2

2

00

2

0

2

( )

2

1 1exp

22

( )

1exp

22

B

a a

z aP dz

z au

udu

1 2

0

.182B

a aP Q equation B

21( ) exp

22

zQ z

z

36

Table for computing of Q-Functions

37

A vector View of Signals and Noise

N-dimensional orthonormal space characterized by N linearly independent basis function {ψj(t)}, where:

From a geometric point of view, each ψj(t) is mutually perpendicular to each of the other {ψj(t)} for j not equal to k.

ji

jidttt

T

ji if 0

if 1 )()(

0

38

Representation of any set of M energy signals { si(t) } as a linear combinations of N orthogonal basis functions where N M.

where:

N

jjiji Mi

Tttats

1 ,...,2,1

0)()(

Nj

Midtttsa j

T

iij ,...,2,1

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

0

39

Therefore we can represent set of M energy signals {si(t) } as:

Representing (M=3) signals, with (N=2) orthonormal basis functions

Waveform energy:

N

jij

T N

jjij

T

ii tattadttsE1

2

01

2

0

2 )( )]()([ )(

Miaaas iNiii ,...,2,1).......,,( 21

40

Question 1: Why use orthormal functions? In many situations N is much smaller than M. Requiring few

matched filters at the receiver.

Easy to calculate Euclidean distances

Compact representation for both baseband and passband systems.

Gram-Schmidt orthogonalization procedure.

Question 2: How to calculate orthormal functions?

41

Examples

42

Examples (continued)

43

Generalized One Dimensional Signals

One Dimensional Signal Constellation

44

Binary Baseband Orthogonal Signals Binary Antipodal Signals

Binary orthogonal Signals

45

Constellation Diagram Is a method of representing the symbol states of modulated

bandpass signals in terms of their amplitude and phase In other words, it is a geometric representation of signals There are three types of binary signals:

Antipodal Two signals are said to be antipodal if one signal is the

negative of the other The signal have equal energy with signal point on the real

line

ON-OFF Are one dimensional signals either ON or OFF with

signaling points falling on the real line

)()( 01 tsts

46

With OOK, there are just 2 symbol states to map onto the constellation space a(t) = 0 (no carrier amplitude, giving a point at the origin) a(t) = A cos wct (giving a point on the positive horizontal axis

at a distance A from the origin)

Orthogonal Requires a two dimensional geometric representation since

there are two linearly independent functions s1(t) and s0(t)

47

Typically, the horizontal axis is taken as a reference for symbols that are Inphase with the carrier cos wct, and the vertical axis represents the Quadrature carrier component, sin wct

Error Probability of Binary Signals

Recall:

Where we have replaced a2 by a0.

18.2 0

01 Bequationaa

QPB

48

To minimize PB, we need to maximize:

or

We have

Therefore,

02

201 )(

aa

0

01

aa

21 0

20 0 0

( ) 2

/ 2d da a E E

N N

21 0 1 0

20 0 0 0

( ) 21 1

2 2 2 2d da a a a E E

N N

49

TTT

T

d

tstsdttsdtts

dttstsE

0 01

2

0 0

2

0 1

2

0 01

)()(2)()(

)()(

)63.3(2 0

N

EQP d

B

The probability of bit error is given by:

50

The probability of bit error for antipodal signals:

The probability of bit error for orthogonal signals:

The probability of bit error for unipolar signals:

0

2

N

EQP b

B

02N

EQP b

B

0N

EQP b

B

51

Bipolar signals require a factor of 2 increase in energy compared to orthogonal signals

Since 10log102 = 3 dB, we say that bipolar signaling offers a 3 dB better performance than orthogonal

52

Comparing BER Performance

For the same received signal to noise ratio, antipodal provides lower bit error rate than orthogonal

4,

2,

0

10x8.7

10x2.9

10/

antipodalB

orthogonalB

b

P

P

dBNEFor

53

Relation Between SNR (S/N) and Eb/N0

In analog communication the figure of merit used is the average signal power to average noise power ration or SNR.

In the previous few slides we have used the term Eb/N0 in the bit error calculations. How are the two related?

Eb can be written as STb and N0 is N/W. So we have:

Thus Eb/N0 can be thought of as normalized SNR.

Makes more sense when we have multi-level signaling.

Reading: Page 117 and 118.

0 /b b

b

E ST S W

N N W N R