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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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Model of a Binary Digital CommunicationSystem
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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Model of a Binary Digital CommunicationSystem
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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Model of a Binary Digital CommunicationSystem
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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Model of a Binary Digital CommunicationSystem
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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Geometric Representation of Signals
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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Geometric Representation of Signals
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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Geometric Representation of Signals
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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Geometric Representation of Signals
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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Geometric Representation of Signals
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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Geometric Representation of Signals
( 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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Optimum Receiver Design
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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Optimum Receiver Design
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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Optimum Receiver Design
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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Optimum Receiver Design
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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Optimum Receiver Design
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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Optimum Receiver Design
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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Optimum Receiver Design
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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Optimum Receiver Implementation
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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Optimum Receiver Implementation
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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Optimum Receiver Implementation
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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Optimum Receiver Implementation
Matched Filter Receiver
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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Optimum Receiver Implementation
Matched Filter Receiver
( )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