baseband data transmission i after this lecture, you will ...em/dtss05pdf/00d matched filter.pdf ·...

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

D.2

Reference

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

D.3

Introduction

Digital communication system

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

transmitted to the destination.– Example 1

• Analog signal (voice signal): sampling, quantizing and encoding are used to convert it into digital form

Informationsource Transmitter Receiver Destination

NoiseTransmittedmessage

channel

Receivedmessage

D.4

Introduction (1)

– Sampling and quantizing

Digits

0

1

2

3

45678

Quantizationnoise

t

3

2

1

0

7

6

5

4

D.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

D.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.

D.7

Introduction (4)

Channel– medium used to transmit the signal from transmitter to the

receiver– Attenuation and delay distortions – Noise

Receiver– performs the reverse function of the transmitter

• determine the symbol from the received signalExample: 1 or 0 for a binary system

Destination– the person or device for which the message is intended.

D.8

Signaling Rate

Digital message– An ordered sequence of symbols drawn from an alphabet of

finite size µ.• Example

Binary source: µ=2 for alphabet 0,1 where 0 and 1 are symbolsA 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 sequence

of signal-elements. Each signal-element has the same duration of Tsecond and is transmitted immediately one after another, so that the signal-element rate (signaling rate) is 1/T elements per second (bauds).

D.9

Bit Rate

Symbols: 1,2,3 and 4

Binary digit: 0 and 1

D.10

Bit Rate

Bit Rate– The bit rate is the product of signaling rate and no of

bit/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)

D.11

Matched Filter (1)

– A basic problem that often arises in the study of communication systems is that of detecting a pulse transmitted over a channel that is corrupted by channel noise

t t Square pulse Signal at the receiving end

D.12

Matched Filter (2)

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

– Consider that the filter input x(t) consists of a pulse signal g(t) corrupted by additive noise w(t). It is assumed that the receiver has knowledge of the waveform of the pulse signal g(t). The source of uncertainty lies in the noise w(t). The function of receiver is to detect the pulse signal g(t) in an optimum manner, given the received signal x(t).

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

D.13

Matched Filter (3)

– The purpose of the circuit is to design an impulse response h(t) of the filter such that the output signal-to-noise ratio is maximized.

Let g(f) and h(f) denoted the Fourier Transform of g(t) andh(t).

∫∞

∞−

= dfftjfGfHtg )2exp()()()(0 π

The signal power =22

0 )2exp()()()( ∫∞

∞−

= dfftjfGfHtg π

Signal Power

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 =

20N

20 )(2

)( fHNfSN =

dffHNtnE 202 )(2

)]([ ∫∞

∞−

=

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()()(

∫

∫∞

∞−

∞

∞−=dffH

N

dffTjfGfH πη

Tt =

D.16

Matched Filter (6)

– Our problem is to find, for a given G(f), the particular form of the transfer function H(f) of the filter that makes η at maximum.

Schwarz’s inequality:

If ∞<∫∞

∞−

dxx 21 )(φ and ∞<∫

∞

∞−

dxx 22 )(φ ,

2

21 )()(∫∞

∞−

dxxx φφ ≤ dxx∫∞

∞−

21 )(φ dxx∫

∞

∞−

22 )(φ

This equality holds, if and only if, we have )()( *21 xkx φφ =

where k is an arbitrary constant, and * denotes complexconjugation.

D.17

Matched Filter (7)

Applying the schwarz’s inequality to the numerator ofequation (1), we have

≤∫∞

∞−

2)2exp()()( dffTjfGfH π dffH∫∞

∞−

2)( dffG∫∞

∞−

2)(

……(2)

(Note: 12 =fTje π )

D.18

Matched Filter (8)

Substituting (2) into (1) ,

The S/N ratio0

2N

≤η dffG∫∞

∞−

2)(

or

0

2N

E≤η ……(3)

where the energy E= dffG∫∞

∞−

2)( is the input signal energy

D.19

Matched Filter (9)

Notice that the S/N ratio does not depend on the transferfunction H(f) of the filter but only on the signal energy.The optimum value of H(f) is then obtained as

)2exp()()( * fTjfkGfH π−=

D.20

Matched Filter (10)Taking the inverse Fourier transform of H(f) we have

h(t)=k dftTfjfG )](2exp[)(* −−∫∞

∞−

π

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

h(t)=k dftTfjfG )](2exp[)( −−−∫∞

∞−

π

h(t)=kg(T-t) …..(4)

Equation (4) shown that the impulse response of the filter isthe time-reversed and delayed version of the input signalg(t). ”Matched with the input signal”

D.21

Matched Filter (11)

Example:The signal is a rectangular pulse.

A

)(tg

T t

The impulse response of the matched filter has exactlythe same waveform as the signal.

kA

)(th

T t

D.22

Matched Filter (12)

The output signal of the matched filter has a triangularwaveform.

)(tgo

T t

TkA2

D.23

Matched Filter (13)

In this special case, the matched filter may beimplemented using a circuit known as integrate-and-dump circuit.

∫T

0)(tr

Sample at t = T

D.24

Realization of the Matched filter (1)

Assuming the output of y(t) = r(t)⊗h(t)

= ∫ −t

dthr0

)()( τττ ...(5)

Substitute (4) into (5) we have

∫ −−=t

dtTgrty0

)]([)()( τττ

When t = T

y(T)= τττ dgrT

)()(0∫

D.25

Realization of the Matched filter (2)

∫T

0)(tr

)(tg

Correlator

D.26

Error Rate of Binary PAM (1)Signaling

– Consider a non-return-to-zero (NRZ) signaling (sometime called bipolar). Symbol 1 and 0 are represented by positive and 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 density No/2. In the signaling interval , the received signal is

• A is the transmitted pulse amplitude• Tb is the bit duration

+−++

=sent was0 symbol)(sent was1 symbol)(

)(twAtwA

tx

bTt ≤≤0

D.27

Error Rate of Binary PAM (2)

Receiver

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

– Given the noisy signal x(t), the receiver is required to make a decision in each signaling interval

∫T

0)(tx

Sample at t = Tb

Decisiondevice

λ

λλ

<>

yy

if 0 if 1y

D.28


– In actual transmission, a decision device is used to determine the 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 is

x(t) = -A + n(t)If the signal is input to a bandlimited low pass filter(matched filter implemented by the integrate-and-dumpcircuit), the output y(t) is obtained as:

y(t)= ∫∫ +−=bTt

b

dttnT

Adttx00

)(1)(

D.29


As the noise )(tn is white and Gaussian, we maycharacterize )(ty as follows:

• )(ty is Gaussian distributed with a mean of –A

• the variance of y(t) can be shown as b

y TN2

02 =σ

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

D.30


The Probability density function of aGaussian distributed signal is given as

)2

)(exp(2

1)0( 2

2

yyy

yyyfσσπ−

−=

)/

)(exp(/

1)0(0

2

0 bby TN

AyTN

yf +−=∴

π

where is the conditional probability density function of the random variable y, given that 0 was sent

)0|(yf y

D.31


– Let p10 denote the conditional probability of error, given that symbol 0 was sent• This probability is defined by the shaded area under the

curve of from the threshold λ to infinity, which corresponds to the range of values assumed by y for a decision in favor of symbol 1

)0|(yf y

D.32


The probability of error, conditional on sending symbol 0 isdefined by

)sent was0 Symbol(10 λ>= yPP

∫∞

=λ

dyyf y )0(

dyTNAy

TN bb

)/

)(exp(/

1

0

2

0

+−= ∫

∞

λπ

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 and A for symbol 1. The threshold λis set to be 0.

D.34


Define a new variable z = bo TN

Ay/

+

and then dy =bT

N0 dz.

We have dzzPob NE

)exp(1 2

/10 −= ∫

∞

π

where bE is the transmitted signal energy per bit, defined by bb TAE 2=

D.35


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

∫∞

−=u

dzzu )exp(2)erfc( 2

π

Therefore, the conditional probability of error

)erfc(21

010 N

EP b=

( Note: ∫ −=u

dzzu0

2 )exp(2)erf(π

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

D.36


In some literature, Q function is used instead of erfcfunction.

duuxx

)2

exp(21)Q(

2

∫∞

−=π

)2

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

D.37


Case IISimilary, the conditional probability density function of Ygiven that symbol 1 was sent, is

)/

)(exp(/

1)1(0

2

0 bby TN

AyTN

yf −−=

π

dyTNAy

TNP

bb

)/

)(exp(/

1

0

2

001

−−= ∫

∞−

λ

π

D.38


By setting λ =0 and putting

zTNAy

b

−=−

/0

we find that 1001 PP =

The average probability of symbol error eP is obtained as

011100 PPPPPe +=

If the probability of 0 and 1 are equal and equal to ½

)erfc(21

0NEP b

e =

D.39


baseband data transmission i after this lecture, you will ...em/dtss05pdf/00d matched filter.pdf ·...

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