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Digital Communication SystemsLecture-3, Prof. Dr. Habibullah JamalUnder Graduate, Spring 2008
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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.
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Effect of Channel
Figure 1.16 (a) Ideal pulse. (b) Magnitude spectrum of the ideal pulse.
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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.
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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
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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
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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)()()(
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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
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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
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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
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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
)()(
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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
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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
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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
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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
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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
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The function of the correlator and matched filter are the same
Compare (a) and (b) From (a)
Tdttstrtz
0)()()(
T
TtdsrTztz
0)()()()(
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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
)()()('
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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
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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
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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
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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
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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
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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:
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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
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This means that if received signal was positive, s1 (t) was sent, else s2(t) was sent
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Detection of Binary Signal in Gaussian Noise
The output of the filtered sampled at T is a Gaussian random process
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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
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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
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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
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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
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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
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Table for computing of Q-Functions
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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
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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
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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
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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?
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Examples
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Examples (continued)
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Generalized One Dimensional Signals
One Dimensional Signal Constellation
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Binary Baseband Orthogonal Signals Binary Antipodal Signals
Binary orthogonal Signals
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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
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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)
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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
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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:
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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
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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
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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
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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