Demodulation and Detection Techniques
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Transcript of Demodulation and Detection Techniques
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EEGR 453Communications Theory
Module #3 (Lessons 14-17):Baseband Demodulation
and Detection (Sklar: Chapt 3)
Dr. Yacob AstatkeFall 2011
(Original Lesson done by: Sorour Falahati)
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Introduction
In Module 2- chapter 2 we saw that the basebandsignals will be sent using waveforms that are shapedlike pulses that will be affected by the channelscharacteristics and by noise (AWGN). This effect is
known as Inter-symbol Interference or (ISI).In this chapter, we will discuss the design ofdemodulators that are capable of recoveringbaseband pulses with the best possible SNR(using a receiving filter) and free of ISI (using anEqualizing filter).
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Learning Objectives
After completing this lesson the student willbe able to: Understand the vector representation of signals and
noise in a signal space. Understand the design of a typical receiver structure
using a two step process:Step#1: Demodulation and samplingStep#2: Detection
Understand the design of an important receiver circuitcalled the optimum receiving filter also known as thematched filter or correlator .
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Last time we talked about:
Transforming the information source to aform compatible with a digital system Sampling
Aliasing Quantization
Uniform and non-uniform
Baseband modulationBinary pulse modulationM-ary pulse modulation
M-PAM (M-ay Pulse amplitude modulation)
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Formatting and transmission of basebandsignal
Information (data) rate: Symbol rate :
For real time transmission:
7/21/2011 Lecture 14-18 5
[bits/sec] /1 bb T R ec][symbols/s /1 T R
mR Rb
Sampling at rate
(sampling time=Ts)
Quantizing each sampled
value to one of the L levels in quantizer.
Encoding each q. value to bits
(Data bit duration Tb=Ts/l )
EncodePulse
modulateSample Quantize
Pulse waveforms(baseband signals)
Bit stream(Data bits)Format
Digital info.
Textualinfo.
Analoginfo.
source
Mapping every data bits to a
symbol out of M symbols and transmittinga baseband waveform with duration T
s s T f /1 Ll 2log
M m 2log
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Overview of Chapter 3In chapter 2 we saw that the baseband signalswill be sent using waveforms that are shaped likepulses.The baseband pulses will be affected by the
channels characteristics (filtering) and by noise(electrical and thermal AWGN).Therefore, the received baseband pulses will nothave an ideal pulse shape, and will not exactlyoccupy their original symbol interval. This effectis known as : Inter-symbol Interference or(ISI).
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Overview of Chapter 3
7/21/2011 Lecture 14-18 7
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Overview of Chapter 3The goal of the demodulator is to recover abaseband pulse with the best possible SNR(using a receiving filter) and free of ISI (usingan Equalizing filter) (see Fig 3.1).The optimum receiving filter is called thematched filter or correlator .In some cases the receiving and equalizingfilters are grouped together to form areceiving and equalizing filter.
7/21/2011 Lecture 14-18 8
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Receiver Structure :a two step process
Step#1-Demodulation and sampling: Waveform recovery and preparing the
received signal for detection:
Improving (SNR): the signal power to the noisepower using matched filterReducing ISI using equalizerSampling the recovered waveform
Step#2-Detection: Estimate the transmitted symbol based on
the received sample
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Steps in designing the receiverFind optimum solution for receiver designwith the following goals:
1. Maximize SNR2. Minimize ISISteps in design:
Model the received signal Find separate solutions for each of the goals.
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Receiver structure
7/21/2011 Lecture 14-18 11
Step 1 waveform to sample transformation Step 2 decision making
Frequencydown-conversion
Receivingfilter
Equalizingfilter
Threshold
comparison
For bandpass signals Compensation forchannel inducedISI
Baseband pulse(possibly distorted)
Sample
(teststatistic)
Baseband pulseReceived waveform
)(t r )(T z
im
Demodulate & Sample Detect
z(T) has a voltage value that is directly proportional to the energy of thereceived symbol and inversely proportional to the noise .
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Receiver structure
#1)(Hypotesis )( then)( 1 H T z T z if
#1)(Hypotesis )( then)( 1 H T z T z if
7/21/2011 Lecture 14-18 12
#2)(Hypotesis )( then)( 2 H T z T z if
T = Symbol duration
mi noerrorcorrectionusedui error
correction
used
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Topics covered in Chapter 3Vector representation of signals and noise(signal space), an important tool tofacilitate Signals presentations, receiver structures
Detection operationsReceiver structure Demodulation (and sampling) DetectionFirst step for designing the receiver Matched filter receiver
Correlator receiver
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Signals and Noise:Sources of Performance Degradation
Major sources of errors: Thermal noise (AWGN)
disturbs the signal in an additive fashion (Additive)
has flat spectral density for all frequencies of interest (White)is modeled by Gaussian random process (Gaussian Noise)
Inter-Symbol Interference (ISI)Due to the filtering effect of transmitter, channel and receiver,symbols are smeared.
7/21/2011 Lecture 14-18 14
FormatPulse
modulate
Bandpass
modulate
Format DetectDemod.
& sample
)(t si)(t g iim
im
)(t r )(T z
channel)(t hc
)(t n
transmitted symbol
estimated symbol
M i ,,1M-ary modulation
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Signals and Noise:Example: Impact of the channel
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Signals and Noise:Example: Channel impact
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)75.0(5.0)()( T t t t hc
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A Vectorial View of Signals and NoiseWhat is a signal space? Vector representations of signals in an N-dimensionalorthogonal spaceWhy do we need a signal space? It is a means to convert signals to vectors and vice versa.
It makes it easy to calculate signals energy andEuclidean distances between signals .Why are we interested in Euclidean distancesbetween signals? For detection purposes: The received signal is transformed
to a received vector. The reference signal which has theminimum distance to the received signal is estimated asthe transmitted signal.
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Vectorial view of Signals and NoiseAny arbitrary finite set of waveformswhere each member of the set is of duration T ,
can be expressed as a linear combination of Northogonal waveforms where
The basis functions must satisfy the following:
7/21/2011 Lecture 14-18 18
M i
i t s1
)(
N j j
t 1
)( M N
N1,...,k j, Tt0 )()(0
jk j
T
ji K dt t t
otherwise 0
k jfor1 jk Kronecker Delta Function
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Vectorial view of Signals and Noise
When the K j constants are non zero , thesignal space is called orthogonal .When the K j constants are normalizedsuch that K
j= 1, the signal space is called
orthonormal .
7/21/2011 Lecture 14-18 19
1 because k j if )()(0
jT
ji K dt t t
0 because k j if 0)()(0
dt t t T
ji
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Example of an orthonormal basis functionsExample: 2-dimensional orthonormal signal space
Example: 1-dimensional orthonornal signal space
7/21/2011 Lecture 14-18 20
1)()(
0)()()(),(
0)/2sin(2
)(
0)/2cos(2)(
21
20
121
2
1
t t
dt t t t t
T t T t T
t
T t T t T
t
T
T t
)(1 t
T 1
0
)(1 t
)(2 t
0
1)(1 t )(1 t 0
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Vectorial view of Signals and Noise
),,()()()()( 321332211 mmmmmmmm aaat at at at s s
7/21/2011 Lecture 14-18 21
The relationship between
si(t), i(t), and coefficientsaij(t) is given by equations3.10 and 3.11.
Review example 3.1 on how tofind the coefficients a
ij(t).
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Vectorial view of Signals and Noise
Why should we use this approach ? It allows the receiver/detector to comparethe received signal with the knownprototypes or reference signals that belong
to the set of M waveforms {s i(t)}. It will decide which of the prototypes (i.e.
reference signals s i(t)) is closest in distance tothe received vector r .
See Fig 3.4 for more information
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Vectorial view of Signals and Noise
7/21/2011 Lecture 14-18 23
Reference signals s i
Received signal r
Detector decision: is r close in resemblance to reference signal s j or s k ?
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A signal space with 3 reference waveforms
7/21/2011 Lecture 14-18 24
),()()()(
),()()()(
),()()()(
),()()()(
212211
323132321313
222122221212
121112121111
z z t z t z t z
aat at at s
aat at at s
aat at at s
z
s
s
s
)(1 t
)(2 t
),( 12111 aas
),( 22212 aas
),( 32313 aas
),( 21 z z z
Transmitted signalalternatives
Received signalat
matched filteroutput
You choose thecorrect waveform s i(t)after you computethe distances
between the receivedsignal z , and thestored waveforms:s1, s2, s3 .
Choose the signalwith the minimumdistance.
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Signal Space and Basis FunctionsWhy should we use this approach ? In the previous example, both the transmitter
and receiver must be able to generate the threesignal waveforms s 1(t), s 2(t), s 3(t)
The receiver needs to generate the threewaveforms because it has to use them asreference signals to make its decision.
Is there a way of minimizing the number of
waveforms that need to be generated by boththe transmitter and receiver ? Yes, see example 3.1
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Orthogonal Representation of Waveforms
7/21/2011 Lecture 14-18 26
0 because k j if 0)()(0
dt t st sT
k j
1 because k j if )()(0
j
T
k j K dt t st s
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Orthogonal Representation of Waveforms
7/21/2011 Lecture 14-18 27
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Orthogonal Representation of Waveforms
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Orthogonal Representation of Waveforms
eq#3.11 )()(
1
0
dt t t s K
aT
ji j
ij
2))5.0(33()5.0()1)(3()1)(1()()(1
5.0
5.0
002112
dt dt dt t t saT
1))5.0(33()5.0()1)(3()1)(1()()(1
5.0
5.0
001111
dt dt dt t t saT
)()2()()1()()()( 212121111 t t t at at s
7/21/2011 Lecture 14-18 30
A better method of finding the coefficients a ij for thebasis functions is to use equation 3.11.For our case Kj = 1 (orthonormal function)
)()()( 2221212 t at at s )()()( 2321313 t at at s
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Orthogonal Representation of Waveforms
eq#3.11 )()(
1
0
dt t t s K
aT
ji j
ij
) () ( )()(1
5.0
5.0
001221 dt dt dt t t sa
T
)() ()() ()()()( 212221212 t t t at at s
7/21/2011 Lecture 14-18 31
A better method of finding the coefficients a ij for the
basis functions is to use equation 3.11.For our case Kj = 1 (orthonormal function)
) () ( )()(
1
5.0
5.0
002222 dt dt dt t t sa
T
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Orthogonal Representation of Waveforms
eq#3.11 )()(
1
0
dt t t s K
aT
ji j
ij
) () ( )()(1
5.0
5.0
001331 dt dt dt t t sa
T
)() ()() ()()()( 212321313 t t t at at s
7/21/2011 Lecture 14-18 32
A better method of finding the coefficients a ij for the
basis functions is to use equation 3.11.For our case Kj = 1 (orthonormal function)
) () ( )()(
1
5.0
5.0
002332 dt dt dt t t sa
T
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Orthogonal Representation of Waveforms
7/21/2011 Lecture 14-18 33
By using orthogonal basis functions, thethree waveforms can be recreatedusing only two basis functions (seeexample 3.1)The Gram-Schmidtorthogonalization procedure allowsthe user to find the appropriate set oforthogonal basis functions { j(t)} forany given signal set {s i(t)} (see slidesat the end of the lecture ).
L11 Begin
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Example of distances in signal space
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The Euclidean distance between signals z(t) and s i(t):
3,2,1
)()()()( 2222
11,
i
z a z at z t sd iii z si
)(1 t
)(2 t ),( 12111 aas
),( 22212 aas
),( 32313 aas
),( 21 z z z
z sd ,1
z sd ,2 z sd ,3
1 E
3 E
2 E
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SNR parameter for DCSRelationship between SNR and (Eb/N o ) :
Why do we use ( Eb/N o) for DCS instead ofSNR that is used for Analog Comm Systems ?
The answer lies in the types of signals that arecommonly used in analog and digital communicationsystems (note: Eb/No is a unit-less ratio like SNR).
sec*sec*
Watt/HzJoules
unitWatt Watt
N S
N E
o
b
7/21/2011 Lecture 14-18 37
RW
N S
N E
W N RS
W N T S
N E
o
bbb
o
b *
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SNR parameter for DCSMost analog signals can be classified aspower signals they have a finiteaverage power but have infinite energy .Thats because most analog waveforms haveinfinite duration (ex: cos function).Since we cannot compute the energy ofanalog waveforms, it is better to use theirpower (rate of delivering energy ) toevaluate their performance.
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SNR parameter for DCSMost digital signals can be classified as energysignals they have a finite energy but havezero average power .Thats because most digital waveforms have a
finite duration because they are transmitted andreceived within a time frame or window known asthe symbol time Ts .(see Pb1.1(b))The average power of a symbol over a symboltime Ts is zero . Therefore, it is better to use theenergy of a symbol (integral of power over Ts)to evaluate the performance of a DCS .
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SNR parameter for DCSInsert Pb 1.1 (a) and (b) solutions
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Example of distances in signal space
),()()()(
),()()()(
),()()()(
),()()()(
212211
323132321313
222122221212
121112121111
z z t z t z t z
aat at at s
aat at at s
aat at at s
z
s
s
s
7/21/2011 Lecture 14-18 41
)(1 t
)(2 t
),( 12111 aas
),(22212
aas
),( 32313 aas
),( 21 z zz
z sd ,1
z sd ,2 z sd ,3
1 E
3 E
2 E
Does expressing s i(t) andr(t)=z(t) in terms of theorthogonal basisfunctions help us in
computing the energyof the signals ?Yes !!!!
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Waveform EnergyThe energy Ei of a waveform can beexpressed as a function of its orthogonalcomponents (see equation 3.17 for derivation):
If orthonormal basis functions are used (i.e. Kj=1)the normalized energy over a symbol duration Tis (see equation 3.17):
7/21/2011 Lecture 14-18 42
M ii t s 1)(
2
1
ij
N
j
ji a K E Waveform energy
N
jiji a E
1
2
Normalized Waveform energy
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Pb versus (Eb/N o) Curves for DCSThe performance of a DCS can becharacterized by looking at the biterror probability P b versus ( Eb/N o)curves as shown in Figure 3.6.The curves indicate the minimumrequired ( Eb/N o) in order to meet thebit error rate P b required for properdetection of the received signals.
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Pb versus (E b/N o) Curves for DCS
7/21/2011 Lecture 14-18 44
For a given P b= Po, select
the minimum required(E b/No)= x o for accuratedetection of the receivedsignal.
For a given P b= P o, the smaller the required (E b/No) the more efficientthe detection of the received signal.
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Detection of Binary Signals in AWGNMaximum Likelihoodreceiver
The received signal is:z(T)= a i(T) + n o(T)
ai(T) desired signal
no(T) noise signal If we assume that the noise
signal is AWGN, then z(T) will
also be a Gaussian randomvariable with a mean of eithera1 (s1=1 was sent) or a 2 (s2=0was sent).
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Detection of Binary Signals in AWGNConditional Probability Density Functions (pdf)
7/21/2011 Lecture 14-18 47
z(T) is a voltage signal that is proportional to the energy ofthe received symbol .The detection decision is based on the energy and not the shapeof the received signal .
p (z /s i)ConditionalProbabilities
Shaded area is prob of errors1 sent but classified as s2
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Detection of Binary Signals in AWGNMaximum Likelihood receiver The detection piece of the
receiver is described as :
The main question is how tochoose the threshold ( ) suchthat the probability of erroris minimized ?
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Detection of Binary Signals in AWGNMaximum Likelihood receiver The likelihood ratio test can be rewritten as:
If the product on the left is greater than theone on the right, then the detector choosesH1 it assumes signal s1(t) was sent.
If not, the detector chooses H2 it assumessignal s2(t) was sent.
)(*)( )(*)( 22112
1
s P s z p s P s z p H
H
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Detection of Binary Signals in AWGNMaximum Likelihood receiver If P(s1)= P(s2) , we obtain the optimum
threshold ( ) that will minimize theprobability of making an incorrect decision
The detector will choose the hypothesis(H) of the signal with the maximumlikelihood max likelihood detector .
)( )( 212
1
s z p s z p H
H
o
7/21/2011 Lecture 14-18 50
2
210
aa
f
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Detection of Binary Signals in AWGN
Error Probability An error of a binary detector can occur
under two conditions: Case 1=P(e/s 1): Signal s1(t) is sent, but the
channel noise forces the receiver output tochoose s 2(t) because : z (T)
o
o
7/21/2011 Lecture 14-18 51
o
f l
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Detection of Binary Signals in AWGNError Probability PB
The probability P B represents the area under the tail ofthe likelihood pdfs p(z/s 1) or p(z/s 2) that falls on theincorrect side of the threshold.
PB can be computed by integrating p(z/s 1) from - to
( ) or p(z/s 2) from ( ) to + o o
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i f i Si l i G
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Detection of Binary Signals in AWGN
Error Probability The probability of errors P(e/s 1) and P(e/s 2)
are written as follows:
dz s z p s H P se P 0
2212 )()(
dz s z p s H P se P 0
1121 )()(
7/21/2011 Lecture 14-18 53
D i f Bi Si l i AWGN
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Detection of Binary Signals in AWGN
Error Probability The total probability of bit error P B is the
sum of probabilities of all the ways that anerror can occur. For the binary case:
)()(),(2
1
2
1i
ii
ii B s P se P se P P
)()()()(2211
s P se P s P se p P B
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D i f Bi Si l i AWGN
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Detection of Binary Signals in AWGN
Error Probability PB can be computed as follows:
We obtain the solution:
Q(x) is called the complementary error
function or co-error function (Appendix B). 20 is the variance of the output noise
(average noise power ).
dz s z p s H P P aa
B 2
221
210
)(
7/21/2011 Lecture 14-18 56
0
21
2 aaQ P B
D i f Bi Si l i AWGN
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Detection of Binary Signals in AWGNError Probability Q(x) can be approximated for values of
x>3 using equation 3.44
At this stage, we have chosen the value ofthe threshold ( ) that optimizes thesystem by minimizing P B.
o
7/21/2011 Lecture 14-18 57
2exp
2
1 2 x
x xQ
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Receiver structure
7/21/2011 Lecture 14-18 59
T = Symbol duration
D t ti f Bi Si l i AWGN
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Detection of Binary Signals in AWGN
The Matched Filter The output of the sampler in step 1 is:
ai(t): is the signal at the filter output 20 : is the average noise power . Our goal is to find the transfer function
Ho(f) of a filter that maximizes the aboveequation (3.45).
3.45Equ20
2
i
T
a N S
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D t ti f Bi Si l i AWGN
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Detection of Binary Signals in AWGN
The Matched Filter The maximum output (S/N) T depends on
the input signal energy and the powerspectral density of the noise.
The max (S/N) T does not depend on theparticular shape of the waveform.
3.52Equ2max0 N
E N S
T
3.53Equdf S(f) 2-
E
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D t ti f Bi Sig l i AWGN
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Detection of Binary Signals in AWGN
The Matched Filter The maximum output (S/N) T equation holds
true iff we employ a filter with the optimumtransfer function shown below:
The optimum filter transfer function is amirror image of the message signal s(t),delayed by the symbol time duration T .
Eq3.56 elsewhere 0
Tt0 t)-ks(T (t)h 0
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Matched filter receiver
Problem: Design the receiver filter such that the SNR is
maximized at the sampling time whenis transmitted.
Solution: The optimum filter, is the Matched filter, given by
which is the time-reversed and delayed version of the conjugate ofthe transmitted signal
7/21/2011
Lecture 14-18 63
)(t h
)()()( * t T st ht h iopt
)2exp()()()( * fT j f S f H f H iopt
M it si ,...,1),(
T 0 t
)(t si
T 0 t
)()( t ht h opt
Detection of Binar Signals in AWGN
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Detection of Binary Signals in AWGN
The Matched Filter The output z(T) of step 1 is therefore the
convolution of the received input waveform r(t)with the impulse response of the filter h(t).
If we choose t = T, and k=1 the equation is:
d t hr t ht r t z t
0)()()(
3.59Eq# )(0
d sr T z T
7/21/2011 Lecture 14-18 64
d t T sr d t T sr t z t t 00
)(
Detection of Binary Signals in AWGN
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Detection of Binary Signals in AWGN
The Matched Filter = The Correlator Equ # 3.59 indicates that z(T) can be
obtained by computing the cross -correlation of r(t) and s(t).
Therefore, the receiver filter can beimplemented as follows:The received signal r(t) is correlated with eachprototype signal s i(t) (i=1 M) using a bank of M
matched filters or M correlators.The correct signal s i(t) will be chosen as the onethat yields the maximum output z i(t).
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Implementation of matched filter receiver
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Implementation of matched filter receiver
7/21/2011 Lecture 14-18 66
Bank of M matched filters
Matched filter output:Observation
vector
)()( t T st r z ii
M i ,...,1),...,,())(),...,(),(( 2121 M M z z zT zT zT z z
M z
z 1
z)(t r
)(1 T z )(*1 t T s
)(* t T s M )(T z M
z
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Example of matched filter receivers using basic
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Example of matched filter receivers using basicfunctions
Number of matched filters (or correlators) is reduced by 1 compared tousing matched filters (correlators) to the transmitted signal.
7/21/2011 Lecture 14-18 68
T t
)(2 t s
T t
)(1 t
T 1
0
1 z z)(t r z
1 matched filter
T t
)(1 t
T 1
0
1 z
T A
0T t
)(1 t s
T A
0
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Optimizing Error PerformanceThe Matched Filter The matched filter is therefore the
optimum receiving filter that maximizes theargument of Q(x) . Thats because P B
decreases as x increases.
Therefore, the matched filter provides themaximum distance between the twoprototype signals s 1(t)=a 1 and s 2(t)=a 2 at thereceiver.
0
21
2 aa
Q P B
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Optimizing Error Performance
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Optimizing Error PerformanceChoosing the right signals/waveforms If = 90 s1(t) and s 2(t) are orthogonal to
each other = cos(90)= 0
If = 180 s1(t) and s 2(t) are antipodalsignals = cos(180)= -1
Eq#3.71 0
N
E Q P b B
Eq#3.70 20
N E Q P b B
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Optimizing Error PerformanceChoosing the right signals/waveforms
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Note: Antipodal signals are easier to detect because there ismore distance between the two signals less chance of error.
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Optimizing Error PerformanceMatched Filter detection of Antipodal Signals
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Optimizing Error Performance
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Optimizing Error Performance
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Optimizing Error PerformanceError Probability for Binary Signaling
The bipolar signals are easier to detectunder AWGN compared to unipolar signals.
Eq#3.71 0
N E
Q P b B
Eq#3.70 20
N E Q P b B
7/21/2011 Lecture 14-18 75
Eq#3.7 1 0
N
E Q P b B
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Optimizing Error PerformanceGiven the same
(Eb/No)=10 dB the P bfor Bipolar is 100times better than thatof Unipolar
Pb = 10 -3 for Unipo
Pb = 10 -5 for BipoGiven the same P b=10-4 Bipolar requires3dB less power thanUnipolar
(Eb/No)= 11 dB forUnipo
(Eb/No)= 8 dB for Bipo
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Intersymbol Interference (ISI)
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Intersymbol Interference (ISI)ISI can be reduced using an equalizing filter
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Intersymbol Interference (ISI)
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Intersymbol Interference (ISI)The best ISI filter is the Nyquist filter H(f)
Minimum bandwidth required in order to detectsignals arriving at Rs symbols/sec without ISI is Rs/2Hertz.
The Ideal Nyquist filter is the sync function shown in
Fig 3.16 h(t) = sinc (t/T) To avoid ISI, the sampling timing must be done at each
time T. At t=0 sampler detects only h(t) since h(t-T)=0 At t=T sampler detects only h(t-T) since h(t)= 0
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Intersymbol Interference (ISI)
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Intersymbol Interference (ISI)The best ISI filter is the Nyquist filter H(f) Minimum bandwidth required in order to detect
signals arriving at Rs symbols/sec without ISI is Rs/2Hertz.
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BW=1/2T
Intersymbol Interference (ISI)
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Intersymbol Interference (ISI)Practical Filter is a Raised Cosine filter H(f)
Insert Figure 3.17
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Required Bandwidthfor baseband signalsW = 0.5(1+ r)Rs
Required Bandwidthfor bandpass signalsW = (1+ r)Rs
See Example 3.3 formore information.
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Intersymbol Interference (ISI)Bandwidth requirements
Insert Figure 3.17
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Required Bandwidthfor baseband signalsW = 0.5(1+ r)Rs
Required Bandwidthfor bandpass signalsW = (1+ r)Rs
Signal space contd
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Signal space contdTo find an orthonormal basis functions for a given
set of signals, Gram-Schmidt procedure can beused.Gram-Schmidt procedure:
Given a signal set , compute an orthonormal
basis1. Define2. For compute
If letIf do not assign any basis function.
3. Renumber the basis functions such that basis is
This is only necessary if for any i in step 2.
Note that 7/21/2011 Lecture 14-18 82
M ii t s 1)(
N
j j t 1)( )(/)(/)()( 11111 t st s E t st
M i ,...,2 1
1
)()(),()()(i
j j jiii t t t st st d
0)( t d i)(/)()( t d t d t iii 0)( t d i
)(),...,(),( 21 t t t N
0)( t d i
M N
Example of Gram-Schmidt procedure
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p p
Find the basis functions and plot the signal space for
the following transmitted signals:
Using Gram-Schmidt procedure:
T t
)(1 t s
T t
)(2 t s
)( )(
)()(
)()(
21
12
11
A A
t At s
t At s
ss
)(1 t1
s2s
T A
T A
0
0
T t
)(1 t
T
1
0
0)()()()(
)()()(),(
/)(/)()(
)(
122
0 1212
1111
0
22
11
t At st d
Adt t t st t s
At s E t st
Adt t s E
T
T
1
2