@G. Gong 1

4. Nyquist Criterion for DistortionlessBaseband Binary Transmission

4. Nyquist Criterion for DistortionlessBaseband Binary Transmission

Objective: To design under thefollowing two conditions:

)(and)( thth dT

(a). There is no ISI at the sampling instants (Nyquistcriterion, this section ).

(b). A controlled amount of ISI is allowed (correlativecoding, next section)

@G. Gong 2

Design of Bandlimited Signals for ZeroISI - Nyquist criterion

Recall the output of the receiving filter, sampled at t = kT,is given by

)(kTy kbµ= )(kTno+∑≠

−+kn

n nTkTpb )(µ

Thus, in time domain, a sufficient condition for µp(t) suchthat it is ISI free is

≠== 00

11)( nnnTp (1)

Question. What is the condition for P(f) in order for p(t) tosatisfy (1) (Nyquist, 1928)?

@G. Gong 3

Theorem. (Nyquist) A necessary and and sufficient condition forp(t) to satisfy (1) is that the Fourier transform P(f) satisfies

TTnfP

n

=−∑ )(

This is known as the Nyquist pulse-shaping criterion orNyquist condition for zero ISI.

Proof.

(2)

Proof. When we sample at,

we have the following pulsesL,2,1,0, ±±== kkTt

)(tp

∑ −≡k

kTttptp )()()( δδ

∑ −=k

kTtkTp )()( δ

The Fourier transform ofis given by

)(tpδ

−=

=

∑k

kTtkTpF

tpFfP

)()(

))(()(

δ

δδ

∑ −=k

fkTjkTp )2exp()( π

On the other hand

∑ −=k

kTtFkTpfP ))(()()( δδ

)(kTp( is constant for t.)

∑ −=k T

kfPT

)(1 ( 3)

= 1 ( from (1) ) ( 4)

From (3) and (4), ISI free ⇔

1)(1 =−∑k T

kfPT

which gives the result in (2).

@G. Gong 5

Investigate possible pulses which satisfy theNyquist criterion

Since , we have

WffP >= ||for0)(

)()()()( fHfHfHfP dcT=

and distinguish the following three cases:

∑ −=n

TnfPfZ )/()(We write

WffHc >= ||for0)(

Suppose that the channel has a bandwidth of W, then

W 1/T-W 1/T 1/T+W-W-1/T+W-1/T-1/T-W

Fig. 4.1 Z(f) for the case T < 1/(2W)

f

Z(f)

1/T-1/T

Fig. 4.2 Z(f) for the case T = 1/(2W)

f

Z(f)

TW

21=

W1/T-W 1/T-W

-1/T+W-1/T

Fig. 4.3 Z(f) for the case T > 1/(2W)

f

Z(f)

WT

21<1. , or (i.e., bit rate > 2W, impossible!) No

choices for P(f) such that Z(f) = 0.W

T21 >

2. , i.e., (the Nyquist rate)W

T21=

TW

21=

In this case, if we choose

≤

=otherwise0|

)(Wf|T

fP i.e.,

⋅=

WfrectTfP

2)(

which results in

=

Ttctp sin)(

This means that the smallest value of T for which thetransmission with zero ISI is possible is

WT

21= W

TR 21 =≡( , bit rate ) This is called the ideal

Nyquist channel.

@G. Gong 8

),(

12

:channelNyquistIdeal

bbo

o

TTRRBWT

BR

===

==

In other words,

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Disadvantages:

(a) an ideal LPF is not physically realizable.

(a) Note that

||1sin)(tT

tctp ∝

=

Thus, the rate of convergence to zero is slow since thetails of p(t) decay as 1/|t|.

Hence, a small mistiming error in sampling the outputof the matched filter at the demodulator results in aninfinite series of ISI components.

@G. Gong 10

3. For , i.e, , in this case, there existsnumerous choices for P(f) such that Z(f) = T. Theimportant one is so called the raised cosine spectrum.

WT

21> W

T21 <

The raised cosine frequency characteristic is given by

+≥

+<≤−

−−+

−<≤

=

)1(0

)1()1(2

))1(|(|cos141

)1(021

)(

0

000

0

0

00

Bf

BfBB

BfB

BfB

fP

α

ααα

απ

α

]1,0[∈αwhere is called the rolloff factor and( i.e., ) .

20RB =

TB

21

0 =

0)1( Bα+0)1( Bα−0)1( Bα−−0)1( Bα+−0B0B−

021B

f

0000 )2()2()( BfBTBfPBfPfP ≤≤−=++−+

Z(f) = T by the following sum of three terms at anyinterval of length 2Bo:

0000 3)2()2()( BfBTBfPBfPfP ≤≤=++−+

…

@G. Gong 12

1)(3t

tp ∝

This function has much better convergence property thanthe ideal Nyquist channel. The first factor in (5) isassociated with the ideal filter, and the second factor thatdecreases as 1/|t|2 for large |t|. Thus

The time response p(t), the inverse Fourier transform ofP(f), is given by

tBtp 02sinc)( = 220

20

161cos2

tBtB

απα

− (5)

@G. Gong 13

@G. Gong 14

∑∞+

−∞=

==−n T

RTnRfP 1,)(

CriteriaNyquist

Bo-Bo

Ideal Nyquist ChannelRaised Cosine Spectrum

Summary:

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Example 1The Fourier transform P(f) of the basis pulse p(t) employed in acertain binary communication system is given by

≤≤−

−

=−−

otherwise0

10)Hz(10if10

110)(

666

6 ff

fP

1. From the shape of P(f), explain whether this pulse satisfies theNyquist criterion for ISI free transmission.

2. Determine p(t) and verify your result in part 1.3. If the pulse does satisfy the Nyquist criterion. What is the

transmission rate ( in bits/sec.) and what is the roll-off factor?

@G. Gong 16

Hz106=R

Solution:

610−

610610− 0

P(f)

f (Hz)

Figure 1Figure 2

610−

bR=

610

bR−=− 610

TfZ =)(

TRTnRfPfZn

/1,)()( ==−= ∑∞

−∞=

1. The Nyquist criterion is

If we choose , then p(t) satisfies Nyquist criterionfor ISI free transmission, shown as Figure 2.

@G. Gong 17

)(10sinc])([)( 621 tfPFtp == −2. We have

)10(sinc)( 62 ttp =

)( sTt µ

0 2-1-2-3 31

....,2,1where0)(and,1)0( ±±=== nnTppConsequently,

sion.transmisfree-ISI,i.e.other,eachwithinterferenotwill

)....,2,1,0()(pulsesthe,...,2,,0atsampledissignalreceivedtheifTherefore,

±±=−±±= nnTtpTTt

@G. Gong 18

1Hz105.05.0and0 601 =⇒×=== αRBf

01 )1( Bf α−=0)1( BW α+=22

10

RT

B ==

In this case, we have,

610=Rwhere the transmission rate (bits/s).

3. The relationship between the bandwidth and the roll-off factor is

@G. Gong 19

5. Correlative Coding and Equalization5. Correlative Coding and Equalization

A. Correlative Coding

The schemes which allow a controlled amount of ISI to achieve thesymbol rate 2W are called correlative coding or partial responsesignaling schemes.

• For zero ISI, the symbol rate R = 1/T < 2W, the Nyquist rate.• We may relax the condition of zero ISI in order to achieve

R = 2W.

@G. Gong 20

(1)0,00,1

)(

isISIzeroforconditionThe

1

≠=

=nn

nTp

{ }

(2)otherwise,0

1and0,1)(

example,for)(samplesin thevaluenonzeroadditional

oneallowtoi.e.,instant,timeoneatISIcontrolledhaveto)(signallimited-banddesign thethat weSuppose

2 ==

=nn

nTp

nTp

tp

@G. Gong 21

).(usingbyincreasedisefficiencySpectral;)]([)(andomain thfrequencyon

bandwidthsmallerahas)]([)();(handuration tmelarger tiahas)(

2

11

22

12

tptpFfP

tpFfPtptp

⇒=

=⇒

Note. The ISI we introdece by using p2(t) is deterministic or“controlled” and, hence, its effect on signal detection at thereceiver can be removed, as discussed below.

(A) Duobinary Signaling (Class I partial response)

The prefix “duo” implies doubling of the transmission capacityof a straight binary system. Figure 1 shows a duobinaryencoder.

Figure 1. Block diagram of duobinary encoder

{ }na

{ }nb

1/T{ }ˆny { }ˆ

nb { }ˆnapre-coder

DelayT

+Ideal channel

)( fHNyquist

Post-coder

duobinarydecoder

Duobinary encoder

{ }ny

overall channel

{ }

=−=+

=.0if

1ifconverter)level(or

precodermemorylesstheofoutputthepolar,NRZthe,2)

n

nn

n

adad

b

b

{ } { }.isdurationsymbol)or(bitsampleThe

.0fortindependenbeingandwith1,0,1).T

kaaaa knnnn ≠∈ +

Legend in Figure 1:

3) The frequency response of the duobinary encode is given by)()()( 1 fHfHfH NyquistI =

which is a cascaded two filters, since we have

DelayT

+= )(1 fH

)(tδ )()()(1 Tttth −+= δδ

Figure 2. Transfer function of delay operator

@G. Gong 24

The effect of (3): , two level {d, -d}, uncorrelated,three level {-2d, 0, 2d}, correlated.

}{ nb}{ ny⇒

Notice that

==−≠

===+=

−

−

−

−

0if2if0

1if2

1

1

1

1

nn

nn

nn

nnn

aadaaaad

bby (4)

(3),valuepreviousitsandpulseinputpresenttheofsumby thedrepresentebecanoutputfilterThe4)

1

1

−

−

+= nnn

nn

n

bbybb

y

Note that here we consider noiseless channel and µ = 1which is from the following formula:

)()()( kTnnTtpbbkTYy okn

nkk +−+== ∑≠

µµ (3)-a

@G. Gong 25

)]()([)()()()( 1 TttFfHfHfHfH NyquistNyquistI −+== δδ

>

≤=

=

2/1if,0

2/1if,1)(

functiontransferhas2/1bandwidthofchannelNyquistidealanSince 0

Tf

TffH

TB

Nyquist

4) On the frequency domain, the transfer function of theduobinary encoder is:

>

≤−=

.2/1if,0

2/1if)exp()cos(2)(

Tf

TffTjfTfHI

ππ

)]2exp(1)[( fTjfHNyquist π−+=

)]exp())[exp(exp()( fTjfTjfTjfHNyquist πππ −+−=

)cos()exp()(2 fTfTjfHNyquist ππ−=

Then

@G. Gong 26

2π

2π

−

2/R2/R− 0f

)]([arg fH I

)( fH I

f2/R− 2/R0

0.2 ratebit1 ==T

R

Figure 3. The frequency response of the duobinary encoder

)]()([sinc1 TttTt

T−+∗

= δδ

)( fHI5) The impulse response corresponding to consists of twosinc (Nyquist) pulses that are time-displaced by T seconds withrespect to each other, which can be derived as follows.

])([])([])([)( 1111 tHFtHFtHFth NyquistII

−−− ∗==

)()/sin(

)()/sin()/sin(

tTtTtT

TtTt

tTt

−=

−−=

ππ

ππ

ππ

−+

=

TTt

Tt

Tsincsinc1

−−+=

TTtTTt

TtTt

T /)()/)(sin(

/)/sin(1

ππ

ππ

@G. Gong 28

signaling.duobinaryby thereducediserrorationsynchronizbittodueISItheTherefore,cahnnel.Nyquistidealin thedencountere/1is

n thedecay thaofratefasteriswhich,/1asdecay)(oftailsThe:Note 2

t

tthI

t3T 4T2TT0-T-2T

)(thI

0.1

Figure 4.

@G. Gong 29

(B). Decoding of the Duobinary Signaling

{ }{ }

1

1

ˆˆgetwe,fromˆestimate

previousthegsubtractinThen,.at timereceiverthebyreceivedaspulseoriginaltheofestimatethe

representˆletly,Specifical.(3)Eq.onbaseddecoder

feedbackausingsequencecoded-duobinarythefromdetectedbemaysequencelevel-twooriginalThe

−

−

−=

=

nnn

nn

n

n

n

n

byb

yb

nTtb

b

yb

DelayT

++

-

ny nb

Figure 5Decision feedback

Drawback: errorpropagation

⊕

DelayT

Levelconverter

Duobinaryencoder

{ }na { }ny1/T{ }nb{ }*nb

Precoder

Figure 6

Duobinary Scheme with Precoder

To uniquely determine the source bit in the kth signalinginterval, even if an error is made on the (k-1)th bit, the kthsource bit, we introduce the precoding:

(5)*1

*−⊕= nnn bab where ⊕ is modulo 2 operation.

@G. Gong 31

{ }{ }

(7)Thenbefore.asduration

samplewithwhere,sequencelevel-twoingcorrespondaproducingconverter,levelatoappliedissequencebinaryThe

1

*

−+=

±=

nnn

nn

n

bbyT

dbbb

)8(0if2

1if0

=±=

=⇔n

nn ad

ay

{ }

=

==

−

−

(6)1.if,

0if,

bygivenissequenceprecodedThe

*1

*1*

*

nn

nnn

n

ab

abb

b

where the bar represents the complement of the symbol.

==−≠

===

−

−

−

0if2if0

1if2

1**

1**

1**

nn

nn

nn

n

bbdbbbbd

y

From (4), we have Combining with (6),

@G. Gong 32

{ } { }

nn

nn

nn

nn

ady

ady

adyya

guessingrandomly,if

0issymbolthen,if

1issymbolthen,if:fromsequence

binaryoriginalthedetectingforruledecisionfollowingthededucewe(8)From

=

>

<(A)

@G. Gong 33

detector.in theoccurcannotnpropagatioerrorHence,required.is

onepresentn theother thasampleinputanyofknowledgenothatisdetectorthisoffeatureusefulA

{ }ny { }ny { }na

threshold d

Rectifier Decisiondevice

Figure 7

@G. Gong 34

Duobinary Scheme with Precoding

{ }na { }nb{ }ˆny { }ˆ

nb { }ˆna

D

+Ideal channel

)( fHNyquist

Post-coder

duobinarydecoder

{ }ny

overall channelD

+ LC{ }*

nb

Summary: Correlative coding can achieve atransmission rate of 2W symbols per secondby using the duobinary scheme together withthe precoding.

@G. Gong 35

Example. Precoding with memory and duobinary coding.

Consider the binary data sequence 0010110. To proceed with theprecoding of this sequence, which involves feeding the precoderoutput back to the input, we add and extra (initialization) bit to theprecoding output. This extra bit is chosen arbitrarily to be 1. Hence,using (4), we find that the sequence {b*

n} at the precoder output is asshown in row 2 of the following table. The polar formart {bn} of thesequence {b*

n} is shown in row 3 of the table. Finally, using (7), wefind the duobinary encoder output has the amplitude levels given inrow 4 of the table. To detect the original binary sequence, we applythe decision rule, given by (A), so, obtain the binary sequence givenin row 5 in the table.

The last row shows that, in absence of noise, the original binarysequence is detected correctly.

@G. Gong 36

BinarySequence 0 0 1 0 1 1 0

Precodedsequence 1 1 1 0 0 1 0 0

Two-levelsequence +d +d +d -d -d +d -d -dDuobinaryEncoderoutput +2d +2d 0 -2d 0 0 -2d

Detectedbinarysequence 0 0 1 0 1 1 0

{ }na

{ }*nb

{ }nb

{ }ny

{ }na

Example 1. Duobinary coding with precoding.

@G. Gong 37

Example 2. Find the error probability of the duobinary signaling inAWGN where the symbols are equally likely.

=+±=

=0if)(2

1if)(

ko

kok akTnd

akTny

Solution.

From (3)-a and (8), we have -2d -d 2dd0

00 1

Since for {ak}, 0 and 1 are equallylikely, the output levels ±2d eachoccur with ¼ and the output level 0occurs with prob. ½ assuming nonoise. If the thresholds are set at ±d,errors occurs as follows:If ak = 0, then

dkTndkTnddkTndkTnd

oo

oo

>−>+−−<<+

)(or)(2(ii))(or)(2(i)

Thus, error occurs when

1ifor

0if

=>−<

=<<−

kkk

kk

adydyadyd

We write N = no(kT).Consequently,

0ifor =>−< kadNdN

@G. Gong 38

}]1|{}1|{[41}0|{

21)( dyPdyPdydPeP kkk >+−<+<<−=

}]{}{[41}]{}{[

21 dNPdNPdNPdNP >+−<+−<+>=

}]{}{[43 dNPdNP >+−<=

=

2/23

0NdQ

Remark. If the factor of 3/2 is ignored, the fraction of F = (4/π)2

amounts to a degradation in signal-to-noise ratio of 2.1 dB ofduobinary over direct binary. That is, to achieve the same errorprobability, the transmission power for duobinary must be 2.1 dBgreater than that for direct binary, assuming ideal channel filteringand AWGN. This is the sacrifice that paid for the smallerbandwidth required by duobinary signaling. .

@G. Gong 39

Eye pattern is an experimental tool to evaluate thecombined effect of receiver noise and ISI on overall systemperformance in an operational environment.

The eye pattern deserves its name from the fact that itresembles the human eye for binary waves. The interiorregion of the eye pattern is called the eye opening.

It is defined as the synchronized superposition of all possiblerealizations of the signal of interest (e.g. received signal,receiver output) viewed within a particular signaling interval.

B. Eye Pattern

5. Correlative Coding and Equalization (Cont.)

Fig. 8 (a) Distorted binary wave with noisy, but no ISI

Binarydata

0 0 1 0 1 1 1 0 0

Tt

0

t

T

Fig. 8 (b)Eye pattern

Binarydata

0 0 1 0 1 1 1 0 0

Tt

0

Fig 9. (a) Distorted binary wave with noisy and ISI

Fig. 9 (b)Eye pattern

t

T

@G. Gong 41

Figure 10

Remark 1. An eye pattern provides a great deal of useful informationabout the performance of a data transmission system, as describedin Figure 10. Specifically, we may make the following statements:

1. The width of the eye opening defines the time interval overwhich the received signal can be sampled without error from ISI.It is apparent that the preferred time for sampling is the instant

of time at which the eye is open the widest.

2. The sensitivity of the system to timing errors is determined by therate of closure of the eye as the sampling time is varied.

3. The height of the eye opening, at a specified sampling time,defines the noise margin of the system.

4. When the effect of ISI is severe , traces from the upper portionof the eye pattern cross traces from the lower portion, with the resultthat the eye is completely closed. In such a situation, it is impossible to

avoid errors due to the combined presence of ISI and noise in the system.

@G. Gong 43

Remark 2. In the case of an M-ary system, then eyepattern contains (M - 1) eye openings stacked upvertically one on the other, where M is the numberof discrete amplitude levels used to construct thetransmitted signal.

In a strictly linear system with truly random data, all these eyeopenings would be identical. Figures 11 and 12 show the eyediagrams for a baseband PAM transmission system usingM = 2 and M = 4 respectively, under the idealized conditions:no channel nose and no bandwidth limitation (i.e., noiselessand zero ISI), and Figures 13 show the eye diagrams with abandwidth limitation.

Note. For how to generate eye diagrams, see Handout 3.

@G. Gong 44

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−1.5

−1

−0.5

0

0.5

1

1.5

Time

Am

plitu

deEye Diagram

Sample instance

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−1.5

−1

−0.5

0

0.5

1

1.5

Time

Am

plitu

de

Eye Diagram

Sample instance

Figure 11. M = 2

Noiseless and zero ISI

Figure 12. M = 4

@G. Gong 45

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Time

Am

plitu

de

Eye Diagram

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−1.5

−1

−0.5

0

0.5

1

1.5

Time

Am

plitu

de

Eye Diagram

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−1.5

−1

−0.5

0

0.5

1

1.5

Time

Am

plitu

de

Eye Diagram

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Time

Am

plitu

de

Eye Diagram

Figure 13. Band-width Limitation

@G. Gong 46

Equalization

In the preceding sections, we discussed that if a band-limited channelHc(f) is known, then it is possible to achieve ISI-free transmission byusing a suitable pair of Tx and Rx.

In practice we often encounterchannels whose frequencyresponse characteristics areeither unknown or change withtime. The methodology toovercome this problem is toemploy channel equalizers.

Channel equalizers: To compensate for thechannel distortion, a linear filter withadjustable parameters may be employed.The filter parameters are adjusted on thebasis of measurements of the channelcharacteristics. These adjustable filters arecalled channel equalizers or, simply,equalizers. (Figure 14)

EffectiveChannel p(t)

Equalizerw(t)

Figure 15

Figure 14

+

y(kT)

DelayT

Nw− 1−w1+−Nw

…

…

DelayT

0w

DelayT

1w Nw1−Nw

DelayT…

…

p(kT+NT) p(kT+T) p(kT-T) p(kT-NT)p(kT)

∑−=

−=N

Nkk kTtwtw )()( δ

@G. Gong 48

Recall that the output of the overall filter may be sampledperiodically to produce the sequence

kk by µ= okn+∑≠

−+kn

nkn pbµ

)(nTppn =)(kTyyk =where )(kTnn ook =and

The middle term of the equation (1) represents the ISI.

(1)

In the practical system, it is reasonable to assume that the ISIaffects a finite number of symbols. Hence the ISI observed atthe output of the receiving filter may be viewed as beinggenerated by passing the data sequence though a linear filter.

Suppose that the equalizer is connectedin cascade with the “effective” channel(which consists of the Tx filter,physical channels and Rx filter), asshown in Figure 15.

Zero-forcing equalizer

Let g(t) denote the impulse response ofthe equalized systems, then

∑−=

−=∗=N

Nnn nTtpwtwtptg )()()()(

{ }ngNote that any term in the sequenceis the weighted sum of consecutive 2N+1 terms of .{ }np

To eliminate the ISI, according to theNyquist criterion for the distortionlesstransmission, we should satisfy

≠== 0if0

0if1kkgk

From (8), we may force the conditions

(9)...,,2,1for00for1

±±±=== Nk

kgk

(8)∑−=

−=N

Nnnknk pwg

At the time instance t = kT,

)(nTppn =where and )(kTggk =

From (8) and (9), we obtain a set oflinear equations:

(10)...,,2,1for00for1

±±±===∑

−=− Nk

kpwN

Nnnkn

@G. Gong 50

(11)

0:010:0

:

:

......:::::

......

......

......:::::

......

1

0

1

0112

10121

101

12101

2110

=

−

−

−+

+−+

−−

−−−−−

−−−−+−

N

N

NNNN

NN

NN

NN

NNNN

w

www

w

ppppp

pppppppppp

ppppp

ppppp

Equivalently, we have the following matrix form

A tapped-delay-line equalizer described by Eq. (10) or (11) is referredto as a zero-forcing equalizer. Such an equalizer is optimum in thesense that it minimizes the peak distortion (ISI).

@G. Gong 51

In summary ,(i) in presence of additive white Gaussian noise, a matched

filter is the optimum detector; and(ii) in the presence of ISI , an equalizer is the desired

structure to mitigate ISI.

Front-end

receiver

Matched filter equalizer+

n(t)

r(t)x(t)

Figure 16

Intuitively, the optimum receiver should consist of a matchedfilter and an equalizer in tandem, as shown in Figure 16.

Figure 17

{ }na

matchedfilter

noise path)()()( fHfHfH CT=

{ }nb

+ equalizer

)(tn

precoder )( fHT )( fH C02|)(| ftjefH π−

Tx filter physicalchannel

Signal path

Signal path: the “effective” channelhas the impulse response:

⇒ a further stretching of the “pulse”⇒ the matched filter accentuates ISI⇒ the equalizer needs to work harder.

Noise path: the filter noise must passthrough the equalizer which is notequalizing the matched filter⇒ equalizing will enhance residualadditive noise.

Question: What should the front-end of the receiver match to ?

It should match to the Tx and the physicalchannel: (Fig. 17).)()()( fHfHfH CT=

02|)(| ftjefH π−

@G. Gong 53

Preset equalizers: On channelswhose frequency-responsecharacteristics are unknown, buttime-invariant, we may measure thechannel characteristics, adjust theparameters of the equalizer, andonce adjusted, the parametersremain fixed during thetransmission of data.

Two types of equalizers

Adaptive equalizers: update theirparameters on a periodic basisduring the transmission of data.

Minimum mean-square error(MSE) equalizer: :The tap weightsare chosen to minimize MSE of allthe ISI terms plus the noise powerat the output of the equalizer.

Remark. Most high-speedtelephone line modems use an MSEweight criterion, because it issuperior to a zero-forcing criterion,and it is more robust in the presentof noise and large ISI.

Remark on equalization of digital data transmission

The zero-forcing equations (10) or (11) in Sec. 4. 5 do notaccount for the effect of noise. In addition, a finite-lengthfilter equalizer can minimize worst-case ISI only if the peakdistortion (i.e., the magnitude of the difference between thechannel output and desired signal ) is sufficiently small.

The sequence {wk} , in Figure 11 in Sec. 4. 5, can be chosen in away such that one can minimize the mean-square error (MSE) ofall the ISI terms plus the noise power at the output of theequalizer. This is called minimum MSE equalizer. (Here, MSE isdefined as the expected value of the squared difference betweenthe desired data symbol and the estimated data symbol . )

@G. Gong 55

A Few Remarks on the Definition of bandwidthand Relation between Channel Bandwidth and

Transmission Rate

For all bandlimited spectra, the waveforms are not realizable, andfor all realizable waveforms, the absolute bandwidth is infinite.The mathematical description of a real signal does not permit thesignal to be strictly duration limited and strictly bandlimited.

All bandwidth criteria have in common that attempt to specify ameasure of the width, W, of a nonegative real-valued psd definedfor all frequencies .

The bandwidth dilemma

∞<|| f

@G. Gong 56

(a)

(b)

Tfc

1−

Tfc

1+

(c)(d)

(e) 35 dB

(e) 50 dB

General shape ofpsd

TffcTfH cX )(sin)( 2 −= π

fc

(a) Half-power bandwidth. This isthe interval between frequencies atwhich has dropped to half-power, or 3 dB below the peak value.

)( fHX

(b) Equivalent rectangular or noiseequivalent bandwidth. It is defined by

, where PX is the totalsignal power over all frequencies.

)(/ cXXN fHPW =

(c ) Null-to-null bandwidth. It isdefined as the width of the mainspectral lobe, where the most of thesignal power is contained (the mostpopular measure of bandwidth. )

(d) Fractional power containmentbandwidth. Federal CommunicationCommission (FCC Rules andRegulations Section 2.202). It statesthat the occupied bandwidth is theband that exactly .5% of the signalpower above the upper band limit andexactly 0.5% of the signal powerbelow the lower band limit. Thus 99%of the signal power is inside theoccupied band.

(e) Bounded power spectral density.Everywhere outside the specified band,

must have fallen at least to acertain stated level below that found atthe band center. Typical attenuationlevels might be 35 or 50 dB.

)( fHX

(f) Absolute bandwidth. This is theinterval between frequencies, outsideof which the spectrum is zero. (Usefulabstraction. For all realizablewaveforms, this is infinite.)

Example. Digital Telephone Circuits.

Compare the system bandwidth requirements for a terrestrial 3-kHz analogtelephone voice channel with that of a digital one. For the digital channel,the voice is formatted as a PCM bit stream, where the sampling rate is8000 samples/s and each voice sample is quantized to one of 256 levels.The bit stream is then transmitted using a PAM waveform and receivedwith zero ISI.

Solution. The resulting of the sampling and quantization process yields PCMwords such that each word has one of L = 256 levels. If each sample were sentas a 256-ary PAM pulse (symbol). Thus the required system bandwidthwithout ISI for sending Rs symbols/s would be . Since each PCMword is converted to 8 bits. Thus, the system bandwidth required using PCM is

2/sRW ≥

.kHz32)symbols/s8000)(lbits/symbo8(21 =≥PCMW

Therefore, the PCM format, using 8-bit quantization and binary signaling withbinary PAM, requests at least eight times the bandwidth required for theanalog channel.

A Note on Relation between Channelbandwidth and transmission rate

Question:In the ideal Nyquist channel, W = R/2 . How can it be possible forthe channel bandwidth W to be smaller than the transmission rate R?

Answer:1) The channel bandwidth W (Hz) and the transmission rate

R (bit per second , or bps) are two different physicalquantities. In general, they are proportional to each other,but it is NOT necessary for them to be equal.

)( fH T )( fH C )( fH R

TransmitterFilter

Physicalchannel

Receiverfilter

Transmittedsignal

Receivedsignal

)1(assuming)()()()(channelEffective

==

µfHfHfHfP RCT

2)

)(ofbandwidththeassametheissignalttedtransmitheofbandwidththe

)(signaltedtransmittheofpsdThe 2

fH

fH

T

T

⇒

∝

@G. Gong 61

accuracymissionhigh trans3)efficiencynutilizatiopowermittedhigh trans2)

efficiencynutilizatiospectrumhigh1)achieveorder toin

way thatasuchindesignedbeshouldsystemThe)(ofbandwidth:)(ofbandwidth:

signal)edtransmitttheofbandwidth()(ofbandwidth:Let

RCT

RR

CC

TT

BBB

fHBfHB

fHB

==

=

3). For the discrete PAM signal formats, the signal bandwidth(e.g., defined as the frequency interval which contains 99% ofthe total power, Definition (d)) may not be equal to thetransmission rate 1/Tb.

@G. Gong 62

ssion.transmifree-ISIofconstraintunder the2/1assmallasbecaninterval

symboltheas,ISIwithout2rateaatsequenceninformatiobinaryasmitcan tranwe,"0"symbolfor)2sinc(2and"1"symbolfor)2sinc(2Using

)()2sinc(2:pairFourierthehaveWe

WWR

WtWWtW

fHWtW T

=

−

↔

channelNyquistidealThe

,0

,1)()()()(

>

<====

Wf

WffPfHfHfH RCT

4) Consider a special case where

@G. Gong 63

Summary of Chapter 4 (Chapter 6 in the textbook)

∑∞

−∞=

−=n

BTX nfTjnRfHT

fS )2exp()()(1)(

signalsPAMbinaryofdensityspectrumpowerandsignals1.PAM2 π

α+=

=

==−

≠=

=

∑∞

∞−

12SpectrumCosineRaised

2ChannelNyquistIdeal

/1)(

0,00,1

)(

siontransmisfree-ISIforcriteriaNyquist3.

WR

WR

TRTnRfP

nn

nTp

n

2. ISI due to bandlimited channel

4. Correlative coding and equalization

Duobinary signaling: achieving the maximum transmission rate2W with zero ISIi) Pre-coder with memory:

encoding and decoding (error propagation)ii) Pre-coder without memory:

encoding and decoding (no error propagation)iii) Psd of the duobinary PAM signals and error probability

nnn yba ,→

nnnn ybba ,,*→

Eye Patterns:

Equalization: to mitigate the effects of ISI, zero-forcingequalizer.

Modified Duobinary Scheme with Precoding

{ }na { }nb{ }ˆny { }ˆ

nb { }ˆna

2D

+Ideal channel

)( fHNyquist

Post-coder

duobinarydecoder

{ }ny

overall channel2D

+ CL{ }*

nb

-