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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)
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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)?
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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).
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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.
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),(
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.
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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 ≤≤=++−+
…
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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)
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∑∞+
−∞=
==−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?
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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.
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)(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
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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
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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.
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(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
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).(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
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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
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)]()([)()()()( 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
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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
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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.
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{ }{ }
(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),
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{ } { }
nn
nn
nn
nn
ady
ady
adyya
guessingrandomly,if
0issymbolthen,if
1issymbolthen,if:fromsequence
binaryoriginalthedetectingforruledecisionfollowingthededucewe(8)From
=
>
<(A)
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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.
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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.
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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.
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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
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}]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. .
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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.
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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.
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−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
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−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
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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
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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
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(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).
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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−
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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 . )
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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
⇒
∝
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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.
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ssion.transmifree-ISIofconstraintunder the2/1assmallasbecaninterval
symboltheas,ISIwithout2rateaatsequenceninformatiobinaryasmitcan tranwe,"0"symbolfor)2sinc(2and"1"symbolfor)2sinc(2Using
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channelNyquistidealThe
,0
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>
<====
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WffPfHfHfH RCT
4) Consider a special case where
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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
-