00d Matched Filter

8/3/2019 00d Matched Filter

1/39

D.1

Baseband Data Transmission I

After this lecture, you will be able to

describe the components of a digital transmission system

Information source, transmitter, channel, receiver and

destination calculate the signaling rate and bit rate of a system

design the matched filter of a receiver derive the condition for maximum signal-to-noise ratio at

the receiver

determine the error rate Error rate versus received signal energy per bit per hertz

of thermal noise


2/39

D.2

Reference

Reference

Chapter 4.1 - 4.3, S. Haykin, Communication Systems, Wiley.


3/39

D.3

Introduction

Digital communication system

Information source produces a message (or a sequence of symbol) to be

transmitted to the destination.

Example 1Analog signal (voice signal): sampling, quantizing and

encoding are used to convert it into digital form

Informationsource Transmitter Receiver

Destination

NoiseTransmittedmessage

channel

Receivedmessage


4/39D.4

Introduction (1)

Sampling and quantizing

Digits

0

1

2

3

4

5

67

8

Quantizationnoise

t

3

2

1

0

7

65

4


5/39D.5

Introduction (2)

encodingDigits

0

2

3

4

5

6

7

8

Binary code

000

001

010

011

100

101

110

111

Return-to-zero

0

1

2

3

4

5

6

7


6/39D.6

Introduction (3)

Example 2

digital source from a digital computer

Transmitter operates on the message to produce a signal suitable for

transmission over the channel.


7/39D.7

Introduction (4)

Channel

medium used to transmit the signal from transmitter to thereceiver

Attenuation and delay distortions

Noise

Receiver performs the reverse function of the transmitter

determine the symbol from the received signal

Example: 1 or 0 for a binary system

Destination

the person or device for which the message is intended.


8/39D.8

Signaling Rate

Digital message

An ordered sequence of symbols drawn from an alphabet offinite size .

Example

Binary source: =2 for alphabet 0,1 where 0 and 1 aresymbols

A 4 level signal has 4 symbols in its alphabet such as 1, 3

Signaling Rate

The symbols are suitably shaped by a shaping filter into a sequenceof signal-elements. Each signal-element has the same duration ofT

second and is transmitted immediately one after another, so that the

signal-element rate (signaling rate) is 1/T elements per second(bauds).


9/39

D.9

Bit Rate

Symbols: 1,2,3 and 4

Binary digit: 0 and 1


10/39

D.10

Bit Rate

Bit Rate

The bit rate is the product of signaling rate and no ofbit/symbol.

Example

A 4-level PAM with a signaling rate = 2400 bauds/s. Bit rate (Data rate) =2400 X log2(4) = 4800 bits/s (bps)


11/39

D.11

Matched Filter (1)

A basic problem that often arises in the study ofcommunication systems is that of detecting a pulsetransmitted over a channel that is corrupted by channel noise

t t

Square pulse Signal at the receiving end


12/39

D.12

Matched Filter (2)

A matched filter is a linear filter designed to provide themaximum signal-to-noise power ratio at its output. This isvery often used at the receiver.

Consider that the filter inputx(t) consists of a pulse signal

g(t) corrupted by additive noise w(t). It is assumed that thereceiver has knowledge of the waveform of the pulse signalg(t). The source of uncertainty lies in the noise w(t). Thefunction of receiver is to detect the pulse signalg(t) in anoptimum manner, given the received signal x(t).

)()()( twtgtx += )()()( tntgty o +=


13/39

D.13

Matched Filter (3)

The purpose of the circuit is to design an impulse responseh(t) of the filter such that the output signal-to-noise ratio ismaximized.

Let g(f) and h(f) denoted the Fourier Transform ofg(t) and

h(t).

= dfftjfGfHtg )2exp()()()(0

The signal power =22

0 )2exp()()()(

= dfftjfGfHtg

Signal Power


14/39

D.14

Matched Filter (4)

Noise Power Since w(t) is white with a power spectral density , the

spectral density function of Noise is

The noise power =

2

0N

20 )(2

)( fHNfSN =

dffHN

tnE2

02 )(2)]([

=


15/39

D.15

Matched Filter (5)

S/N Ratio Thus the signal to noise ratio become

....(1)

(the output is observed at )

2

20 )(

2

)2exp()()(

=

dffHN

dffTjfGfH

Tt=


16/39

D.16

Matched Filter (6)

Our problem is to find, for a given G(f), the particular form of thetransfer functionH(f) of the filter that makes at maximum.

Schwarzs inequality:

If


17/39

D.17

Matched Filter (7)

Applying the schwarzs inequality to the numerator of

equation (1), we have

2)2exp()()( dffTjfGfH dffH

2)( dffG

2)(

(2)

(Note: 12 =fTje )


18/39

D.18

Matched Filter (8)

Substituting (2) into (1) ,

The S/Nratio0

2N

dffG

2)(

or

0

2

N

E (3)

where the energy E= dffG

2)( is the input signal energy


19/39

D.19

Matched Filter (9)

otice that the S/N ratio does not depend on the transferunction H(f) of the filter but only on the signal energy.

The optimum value of H(f) is then obtained as

)2exp()()( * fTjfkGfH =


20/39

D.20

Matched Filter (10)

Taking the inverse Fourier transform ofH(f) we have

h(t)=k dftTfjfG )](2exp[)(*

and )()(* fGfG = for real signal )(tg

h(t)=k dftTfjfG )](2exp[)(

h(t)=kg(T-t) ..(4)

Equation (4) shown that the impulse response of the filter is

the time-reversed and delayed version of the input signal(t). Matched with the input signal


21/39

D.21

Matched Filter (11)

Example:

The signal is a rectangular pulse.

A

)(tg

T t

The impulse response of the matched filter has exactly

the same waveform as the signal.

kA

)(th

T t


22/39

D.22

Matched Filter (12)

The output signal of the matched filter has a triangular

waveform.

)(tgo

T t

TkA2


23/39

D.23

Matched Filter (13)

In this special case, the matched filter may be

implemented using a circuit known as integrate-and-

dump circuit.

T

0)(tr

Sample at t= T


24/39

D.24

Realization of the Matched filter (1)

Assuming the output ofy(t) = r(t)h(t)=

t

dthr0

)()( ...(5)

Substitute (4) into (5) we have

=t

dtTgrty0

)]([)()(

When t= T

y(T)= dgr

T

)()(0

R li i f h M h d fil (2)


25/39

D.25

Realization of the Matched filter (2)

T

0)(tr

)(tg

Correlator

E R t f Bi PAM (1)


26/39

D.26

Error Rate of Binary PAM (1)

Signaling

Consider a non-return-to-zero (NRZ) signaling (sometimecalled bipolar). Symbol 1 and 0 are represented by positiveand negative rectangular pulses of equal amplitude and

equal duration.Noise

The channel noise is modeled as additive white Gaussian

noise of zero mean and power spectral densityNo/2. In thesignaling interval , the received signal is

A is the transmitted pulse amplitude

Tb is the bit duration

+

++= sentwas0symbol)(

sentwas1symbol)()(

twA

twAtx

bTt0

E R t f Bi PAM (2)


27/39

D.27


Receiver

It is assumed that the receiver has prior knowledge of thepulse shape, but not its polarity.

Given the noisy signalx(t), the receiver is required to make adecision in each signaling interval

T

0)(tx

Sample at t= Tb

Decisiondevice

y

y

if0

if1y

E o Rate of Bina PAM (3)


28/39

D.28


In actual transmission, a decision device is used to determinethe received signal. There are two types of error

Symbol 1 is chosen when a 0 was actually transmitted

Symbol 0 is chosen when a 1 was actually transmitted

Case I

Suppose that a symbol 0 is sent then the received signal isx(t) = -A + n(t)

If the signal is input to a bandlimited low pass filter

(matched filter implemented by the integrate-and-dump

circuit), the outputy(t) is obtained as:

y(t)= +=bTt

b

dttnT

Adttx00

)(1)(



29/39

D.29


As the noise )(tn is white and Gaussian, we may

characterize )(ty as follows:

)(ty is Gaussian distributed with a mean of A the variance ofy(t) can be shown as

b

yT

N

2

02 =

(Proof refers to p.254, S. Haykin, CommunicationSystems)



30/39

D.30


The Probability density function of a

Gaussian distributed signal is given as

)2

)(exp(

2

1)0(

2

2

yy

y

yyyf

=

)/

)(exp(

/

1)0(

0

2

0 bb

yTN

Ay

TNyf

+=

where is the conditional probability density

function of the random variable y, given that 0 was sent

)0|(yfy



31/39

D.31


Letp10 denote the conditional probability of error, given thatsymbol 0 was sent

This probability is defined by the shaded area under thecurve of from the threshold to infinity, whichcorresponds to the range of values assumed by y for adecision in favor of symbol 1

)0|(yfy



32/39

D.32


The probability of error, conditional on sending symbol 0 is

defined by

)sentwas0Symbol(10 >= yPP

=

dyyfy )0(

dyTN

Ay

TNbb

)/

)(exp(

/

1

0

2

0

+=



33/39

D.33


Assuming that symbols 0 and 1 occur with equalprobability, i.e.2/110 == PP

If no noise, the output at the matched filter will be A for symbol 0 andA for symbol 1. The threshold is set to be 0.



34/39

D.34


Define a new variablez=bo TN

Ay/

+

and then dy =bT

N0 dz.

We havedzzP

ob

NE

)exp(1 2

/

10 =

where bE is the transmitted signal energy per bit,

defined by bb TAE2=



35/39

D.35


At this point we find it convenient to introduce the definiteintegration called complementary error function.

=u

dzzu )exp(2

)erfc( 2

Therefore, the conditional probability of error

)erfc(2

1

0

10N

EP b=

( Note: =u

dzzu0

2 )exp(2

)erf(

and erfc(u)=1-erf(u) )



36/39

D.36


In some literature, Q function is used instead of erfc

function.

duu

xx

)2

exp(2

1)Q(

2

=

)2

erfc(21)Q( xx = and )2Q(2)erfc( xx =



37/39

D.37


Case II

Similary, the conditional probability density function of

given that symbol 1 was sent, is

)/

)(exp(/

1)1(0

2

0 bb

yTN

AyTN

yf =

dyTN

Ay

TNP

bb

)/

)(exp(

/

1

0

2

0

01

=



38/39

D.38

o o y ( )

By setting =0 and putting

zTN

Ay

b

=

/0

we find that 1001 PP =

The average probability of symbol error eP is obtained as011100 PPPPPe +=

If the probability of 0 and 1 are equal and equal to

)erfc(2

1

0N

EP be =



39/39

D.39

y ( )

Documents

00d Matched Filter