Lec8 Optimum Receiver

8/11/2019 Lec8 Optimum Receiver

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TE312: Introduction toDigital Telecommunications

PART II

BASEBAND DIGITALTRANSMISSION

Lecture #8Optimum Digital Receivers


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Introduction

Points to be discussed in this lecture

Model of a Binary Digital CommunicationSystem.

Geometric Representation of Signals.

Optimum Receiver Design

Implementation of Optimum Receivers.


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Introduction

Reading Assignment

Simon Haykin, Digital Communications,

John Wiley & Sons, Inc., 1988, Chapter 3,Sec. 3.1~3.8.

Simon Haykin, Communication Systems, 4thEd., John Wiley & Sons, Inc., 2001, Chapter5.


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Model of a Binary Digital CommunicationSystem

{ }1 2,b b

{ }1 2s ,s

ModulatorVector

Transmitter

DataSource

( ) ( ){ }1 2,s t s t

Transmitter

( )r t

VectorReceiverDetector

Estimater

WaveformChannel b

Noise

( )w t

Receiver


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Binary Data Source:

Data source emits a bit , 1,2ib i= at everyseconds where

bT

1 2bit 0, bit 1b b= = .

is the bit duration (in Sec.) and bTbT 1/bR = is thetransmission bit rate (in bits per sec.).

Data source is characterized by the a prioriprobability forip , 1, 2.ib i=

[ ]( ) [ ]( )1 1 2 2 1 2bit 0 , bit 1 , 1.0p P b p P b p p= = + =


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Transmitter

Modulator maps the bit into one of two distinct

real-valued signalsi

b

1( )s t and 2 ( )s t of durationwith finite energy

bT

1Eand

2E , respectively.

21 1

0

22 2

0

0 ( )

0 ( )

b

b

T

T

E s t dt

E s t dt

< = <

< = <


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Channel

The channel is linear and distortionless withbandwidth much larger than the message signal.

The signal ( ), 1,2i

s t i= is perturbed by zero-mean, stationary, additive white Gaussian noise

(AWGN) process ( )W t with sample function ( ).w t

Received signal is expressed as( )r t

( ) ( ) ( )ir t s t w t = + 1,2k= , 0 bt T .


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Receiver

The receiver estimates the transmitted bit inthe bit interval based on the observation it

makes on the received signal

b

( ).r t

Since the received signal is corrupted with

noise, the estimated bit will be in error leading to

average probability of bit error

( )r t

ep

( )e i

p P b b=


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Geometric Representation of Signals

A set of energy signals ( )is t , 1,2i= can be

represented as a linear combination of a set of

2N orthonormal basis functions { ( )}j t

1

( ) ( )N

i ij j

j

s t s t =

= 1,2i= 0 bt T

Orthonormality of ( )j

t implies that

01( ) ( )0

bTm j

m jt t dt m j

==


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Coefficients of expansion ijs are given by

0( ) ( )

bT

ij i js s t t dt = 1,21,2

ij

==

( )is t is generated from1,2i= ijs 1,2j= by a bankof multipliers followed by a summer.

ij

s , are generated from1,2j= ( )is t , 1,2i= by a

bank of correlators (multiplication followed byintegration).

2


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Geometric Representation of Signals( ) 1 t

0

bT

dt 1is( )is t

0

bT

dt 2is

( )2 t

2is

1is

( )is t

( )1 t

( )2 t


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Gram-Schmidt Orthogonalization Procedure

A set of orthonormal basis functions }{ ( )j t 1,2j= is obtained as follows:

Define1

( )s t (first signal is arbitrarily selected)as

( ) ( )1 11 1 12 2 12( ) where 0s t s t s t s = + = ,0 bt T

Obtain 1( )t by squaring and integrating1

1

1

(( )

)s tt

E = 1Eis the energy of 1( )s t


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Define 2 ( )s t as

2 21 1 22 2

2 21 1 22 2

( ) ( ) ( )

( ) ( ) ( )

s t s t s t

s t s t s t

= + =

,0b

t T

Obtain 2 ( )t by squaring both sides andintegrating from 0to

bT

2 21 12 2

2 21

( ) ( )( ) s t s t tE s

=

2E is the energy of 2( )s t


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Define the correlation coefficient as

1 201 2

1 ( ) ( )bT

s t s t dt E E

= 21 2s E =

Thus

( )2 1

22

2 1

1 ( ) ( )( )

1

s t s t t

E E

=


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The coefficients of expansion 11 12 21 22, , ,s s s s aregiven by

11 1s E= , 12 0s =

21 2s E= ,2

22 21s E=

Each signal in the set { ( )is t , 1,2i= , can beuniquely determined by thesignal vector

is

1

2

i

i

s

s

=

is


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( Energy of the signal )is t is equal to the squared-length of its vector

22 2 2

01

( )bT

ij i i

j

s s t dt E =

= = = is (Prove)

Energy of 1 2( ) ( )s t s t is equal to the square of theEuclidian distance between their vectors

( )2 222 2

12 1 2 1 2 1 201

[ ( ) ( )]bT

j j

j

d s s s t s t dt =

= = = s s

(Prove)


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Signal point associated with the signal vector

represents the transmitted signal ( )is

is t .

Signal point associated with observation vectorrrepresents the received signal ( ).r t

, 1,2.i= + =ir s w

rand are samples of random vectors andw R W

1 11

2 2 2

i

i

s wr

r s w

+ = = +

r


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Each 2r is a sample value of a Gaussian

random variable 2R R , respectively.1 andr

and1

Mean values of R are1 2andR

[ ] [ ]1 21 1 2 2, , 1,2R i R i

m E R s m E R s i= = = = =

Variances of R are1 2andR

1 2

2 2 0

2R R

N = = (Prove)


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1 2R are mutually uncorrelated, hence are

statistically independent

andR

[ ] ( )( )1 2 1 1 2 2Cov 0i iR R E R s R s = = (Prove)

andR

Conditional pdfs of 2R given1 ( )is t

( ) ( )1

2

1 1 1

00

1 1| ( ) exp

R i if r s t r s

NN

=

( ) ( )2

2

2 2 2

00

1 1| ( ) exp

R i if r s t r s

NN

=


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Conditional probability density function of given( )

R

is t (likelihood function)

( )( ) ( )( ) ( ))1 21 2

| | | t i R i R if s t f r s t f r s=R r 1,2i=

( )2

2

1j

j

r s=

( ) 100

1exp j iNN

=

Detection Problem:

Perform mapping from to an estimate of , with

minimum probability of bit error .

r ib b

ep


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Probability of bit error

( ) ( ), 1 sent |e i ip b P b= r r

( )sent |iP b r is the a posteriori probability of the

binary integer .ibMaximum a posteriori (MAP) probability optimumdecision rule is

1

1 2

set if

( sent| ) ( sent| )

b b

P b P b

=

r r


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Using Bayes Rule

( ) ( )( )|sent | i ii p f bP b

f=R

R

rrr

The MAP decision rule becomes

( ) ( )

1

1 1 2 2

set if

| sent | sent

b b

p f b p f b

=

R Rr r


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Substituting for ( ) ( )1 2| sent and | sentf b f bR Rr r andsimplifying yield

( ) ( )

( ) ( )

2 2

1 11 2 12

2 2

1 21 2 22r s

+

1

2

1

set if

1

exp

1exp

o

o

b b

p N

p

N

r s r s

r s

=

+


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Optimum Receiver Implementation

Determine the received signal vector elements.1 2andr r

Compute the decision rule based on the receivedsignal vector elements 2andr r and vectorelements of two signals ,

1

1,2; 1,2ij

s i j= = .

1 1 2 2 ln2 2

i oi i i

E Nr s r s p+ + , 1,2.i=

Choose the largest.


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Correlation Receiver

( )1 t

1r

2r

bt T=

bt T=

( ) 222 2lnoN Ep

i

r s Choose

thelargest

b

( ) 112 2lnoN Ep

( )r t

( )2 t

0

bT

dt

0

bTdt


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Example: Consider the signal set 1 2( ), ( )s t s t of

orthogonal signals below. Design an optimumreceiver for this signal set assuming an AWGNchannel and (a) 1 20.4, 0.6p p= = (b) 1 2p p= .

A

tbT

A

t

0 0 b

T

( )1s t ( )2s t


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Matched Filter Receiver

Output ( )jy t of a linear filter with impulse response( )and inputjh t ( ) ( ) ( )ir t s t w t = + is given by

( ) ( ) ( ) ( ) ( )j j jy t r t h t r h t d

= =

Substitute ( )j bT t for ( )jh t ,output ( )jy t becomes

( ) ( ) ( ) ( ) ( )j j b j by t r t T t r T t d

= = +


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Output ( )jy t at bt T= is given by

( ) ( ) ( )j b j jy T r d r

= = The fil ter with the impulse response ( ) ( )j j bh t T t =

is called a matched filter.

A receiver that uses matched filters in place ofcorrelators is called a matched filter receiver.


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( )1 bT t ( )r t1r

2r

( ) 222 2lnoN Ep

bt T=

bt T=

i

r s

( )2 bT t

Choosethe

largest

b

( ) 112 2lnoN Ep

Lec8 Optimum Receiver

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