4. nyquist criterion for distortionless baseband binary ...ece411/slides_files/topic4-2.pdf · @g....

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4. Nyquist Criterion for DistortionlessBaseband Binary Transmission

Objective: To design under the following two conditions:

)( and )( thth dT

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

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

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Design of Bandlimited Signals for Zero ISI - 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) such that it is ISI free is

⎩⎨⎧

≠== 0 0

1 1)( nnnTp (1)

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

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

=−∑ )(

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

Proof.

Proof. When we sample at ,

we have the following pulses L,2 ,1 ,0 , ±±== kkTt

∑ −≡k

kTttptp )()()( δδ

∑ −=k

kTtkTp )()( δ

The Fourier transform of is given by

)(tpδ

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

kTtkTpF

))(()(

∑ −=k

fkTjkTp )2exp()( π

On the other hand

∑ −=k

kTtFkTpfP ))(()()( δδ

)(kTp( is constant for t.)

∑ −=k T

)(1 ( 3)

= 1 ( from (1) ) ( 4)

From (3) and (4), ISI free ⇔

1)(1=−∑

k TkfP

which gives the result in (2).

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Investigate possible pulses which satisfy the Nyquist criterion

Since , we have

WffP >= ||for 0)(

)()()()( fHfHfHfP dcT=

and distinguish the following three cases:

∑ −=n

TnfPfZ )/()(We write

WffHc >= ||for 0)(

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)

1/T-1/T

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

W1/T-W 1/T-W

-1/T+W-1/T

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

<1. , or (i.e., bit rate > 2W, impossible!) No choices for P(f) such that Z(f) = 0.

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

In this case, if we choose

⎩⎨⎧ ≤

=otherwise 0|

)(Wf|T

fP i.e., ⎟⎠⎞

⎜⎝⎛⋅=

WfrectTfP

which results in ⎟⎠⎞

⎜⎝⎛=

Ttctp sin)(

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

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

Nyquist channel.

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:channelNyquist Ideal

TTRRBWT

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 the tails of p(t) decay as 1/|t|.

Hence, a small mistiming error in sampling the output of the matched filter at the demodulator results in an infinite series of ISI components.

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

The raised cosine frequency characteristic is given by

⎪⎪⎪⎪

+<≤−⎥⎦

⎤⎢⎣

⎡ −−+

−<≤

)1()1( 2

))1(|(|cos141

)1(0 21

ααα

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

20RB =

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

0000 )2()2()( BfBTBfPBfPfP ≤≤−=++−+

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

0000 3 )2()2()( BfBTBfPBfPfP ≤≤=++−+

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tp ∝

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

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

tBtp 02sinc)( = 220

161cos2

απα

− (5)

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∑∞+

−∞=

==−n T

RTnRfP 1 , )(

CriteriaNyquist

Ideal Nyquist ChannelRaised Cosine Spectrum

Summary:

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

⎪⎩

⎪⎨

⎧≤≤−⎟⎟

⎞⎜⎜⎝

⎛−

=−−

otherwise 0

10)Hz(10 if 10

1. From the shape of P(f), explain whether this pulse satisfies the Nyquist 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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Hz 10 6=R

Solution:

610−

610610− 0

f (Hz)

Figure 1Figure 2

610−

bR−=− 610

TfZ =)(

TRTnRfPfZn

/1 ,)()( ==−= ∑∞

−∞=

1. The Nyquist criterion is

If we choose , then p(t) satisfies Nyquist criterion for 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 ,1 where0)( and ,1)0( ±±=== nnTppConsequently,

sion. transmisfree-ISI , i.e. other,each with interferenot will

).... ,2 ,1 ,0( )( pulses the,...,2 , ,0at sampled is signal received theif Therefore,

±±=−±±= nnTtpTTt

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1Hz 105.05.0 and 0 601 =⇒×=== αRBf

01 ) 1( Bf α−=0)1( BW α+=22

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 the symbol rate 2W are called correlative coding or partial response signaling 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

is ISI zerofor condition The

1⎩⎨⎧

(2) otherwise ,0

1 and 0 ,1)(

example,for )( samples in the valuenonzero additional

one allow toi.e., instant, timeoneat ISI controlled have to)( signal limited-band design the that weSuppose

2⎩⎨⎧ ==

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).( usingby increased is efficiency Spectral ; )]([)(an domain th frequency on

bandwidthsmaller a has )]([)();(han duration t melarger ti a has )(

tptpFfP

tpFfPtptp

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

(A) Duobinary Signaling (Class I partial response)

The prefix “duo” implies doubling of the transmission capacity of a straight binary system. Figure 1 shows a duobinary encoder.

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

⎩⎨⎧

=−=+

=.0 if

converter) level(or precoder memoryless theofoutput thepolar, NRZ the, 2)

{ } { }. isduration symbol)or (bit sample The

.0for t independen being and with 1,0 , 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

⎪⎩

⎪⎨

==−≠

0 if 2 if0

1 if 2

aadaaaad

bby (4)

(3) , valueprevious its and pulseinput present the

of sum by the drepresente becan output filter The 4)

+= nnn

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

)()( )( kTnnTtpbbkTYy okn

nkk +−+== ∑≠

μμ (3)-a

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

⎪⎩

⎪⎨⎧

2/1 if ,0

2/1 if ,1)(

function transfer has 2/1 bandwidth of channelNyquist idealan Since 0

Nyquist

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

⎪⎩

⎪⎨⎧

≤−=

.2/1 if ,0

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

TffTjfTfHI

)]2exp(1)[( fTjfHNyquist π−+=

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

)cos()exp()(2 fTfTjfHNyquist ππ−=

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2/R2/R− 0f

)]([ arg fH I

)( fH I

f2/R− 2/R0

0.2 ratebit 1==

Figure 3. The frequency response of the duobinary encoder

)]()([sinc1 TttTt

T−+∗⎟

⎠⎞

⎜⎝⎛= δδ

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

] )([] )([] )([)( 1111 tHFtHFtHFth NyquistII−−− ∗==

)()/sin(

)()/sin()/sin(

tTtTtT

−−=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −

+⎟⎠⎞

⎜⎝⎛=

Tsincsinc1

⎥⎦

⎤⎢⎣

⎡−−

TtT /)(

)/)(sin(/

)/sin(1ππ

@G. Gong 28

signaling.duobinary by the reduced iserror ation synchronizbit todue ISI theTherefore, cahnnel.Nyquist ideal in the dencountere /1 is

n thedecay tha of ratefaster is which ,/1 asdecay )( of tailsThe :Note 2

t3T 4T2TT0-T-2T

Figure 4.

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(B). Decoding of the Duobinary Signaling

{ }{ }

get we, from ˆ estimate

previous thegsubtractin Then, . at timereceiver theby received as pulse original theof estimate the

represent ˆlet ly,Specifical . (3) Eq.on baseddecoder

feedback a using sequence coded-duobinary thefromdetected bemay sequence level- twooriginal The

DelayT

Figure 5Decision feedback

Drawback: error propagation

DelayT

Levelconverter

Duobinaryencoder

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

Precoder

Figure 6

Duobinary Scheme with Precoder

To uniquely determine the source bit in the kth signaling interval, 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) Then before. as duration

sample with where, sequence level- twoingcorrespond aproducing converter, level a toapplied is sequencebinary The

)8( 0 if 2

1 if 0 ⎩⎨⎧

⎪⎩

⎪⎨⎧

(6) 1. if ,

0 if ,

bygiven is sequence precoded The

where the bar represents the complement of the symbol.

⎪⎩

⎪⎨

==−≠

0 if 2 if0

1 if 2

bbdbbbbd

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

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{ } { }

guessingrandomly , if

0 is symbol then , if

1 is symbol then , if: from sequence

binary original thedetectingfor rule decision following thededuce we(8) From

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detector.in theoccur cannot n propagatioerror Hence, required. is

onepresent n theother tha sampleinput any ofknowledge no that isdetector thisof feature usefulA

{ }ny { }ny { }na

threshold d

Rectifier Decisiondevice

Figure 7

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Duobinary Scheme with Precoding

{ } na { } nb{ } ˆny { } ˆ

nb { } ˆna

+Ideal channel

)( fHNyquist

Post-coder

duobinarydecoder

{ } ny

overall channelD

+ LC{ } *

Summary: Correlative coding can achieve a transmission rate of 2W symbols per second by using the duobinary scheme together with the precoding.

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Example. Precoding with memory and duobinary coding.

Consider the binary data sequence 0010110. To proceed with the precoding of this sequence, which involves feeding the precoderoutput back to the input, we add and extra (initialization) bit to the precoding 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 as shown in row 2 of the following table. The polar formart {bn} of the sequence {b*

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

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

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Binary Sequence

0 0 1 0 1 1 0

Precoded sequence

1 1 1 0 0 1 0 0

Two-level sequence

+d +d +d -d -d +d -d -d Duobinary Encoder output

+2d +2d 0 -2d 0 0 -2d

Detected binary sequence

0 0 1 0 1 1 0

{ }*nb

Example 1. Duobinary coding with precoding.

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

⎩⎨⎧

= 0 if )( 2

1 if )(

kok akTnd

Solution.

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

Since for {ak}, 0 and 1 are equally likely, the output levels ±2d each occur with ¼ and the output level 0 occurs with prob. ½ assuming no noise. If the thresholds are set at ±d, errors occurs as follows:If ak = 0, then

dkTndkTnd d kTndkTnd

>−>+−−<<+

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

Thus, error occurs when

1 if or

=>−<

=<<−

We write N = no(kT). Consequently,

0 if or =>−< kadNdN

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

21)( dyPdyPdydPeP kkk >+−<+<<−=

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

21 dNPdNPdNPdNP >+−<+−<+>=

}]{}{[43 dNPdNP >+−<=

⎟⎟⎠

⎞⎜⎜⎝

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 of duobinary over direct binary. That is, to achieve the same error probability, the transmission power for duobinary must be 2.1 dB greater than that for direct binary, assuming ideal channel filtering and AWGN. This is the sacrifice that paid for the smaller bandwidth required by duobinary signaling. .

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Eye pattern is an experimental tool to evaluate the combined effect of receiver noise and ISI on overall system performance in an operational environment.

The eye pattern deserves its name from the fact that it resembles the human eye for binary waves. The interior region of the eye pattern is called the eye opening.

It is defined as the synchronized superposition of all possible realizations 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

Binary data

0 0 1 0 1 1 1 0 0

Fig. 8 (b)Eye pattern

Binary data

0 0 1 0 1 1 1 0 0

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

Fig. 9 (b)Eye pattern

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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 instantof 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 eye pattern contains (M - 1) eye openings stacked up vertically one on the other, where M is the number of discrete amplitude levels used to construct the transmitted signal.

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

−0.5

deEye Diagram

Sample instance

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−1.5

−0.5

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

−0.5

Eye Diagram

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−1.5

−0.5

Eye Diagram

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−1.5

−0.5

Eye Diagram

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2

−1.5

−0.5

Eye Diagram

Figure 13. Band-width Limitation

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Equalization

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

In practice we often encounter channels whose frequency response characteristics are either unknown or change with time. The methodology to overcome this problem is to employ channel equalizers.

Channel equalizers: To compensate for the channel distortion, a linear filter with adjustable parameters may be employed. The filter parameters are adjusted on the basis of measurements of the channel characteristics. These adjustable filters are called channel equalizers or, simply, equalizers. (Figure 14)

EffectiveChannel p(t)

Equalizerw(t)

Figure 15

Figure 14

DelayT

Nw− 1−w1+−Nw

DelayT

1w Nw1−Nw

DelayT…

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

∑−=

Nkk kTtwtw )()( δ

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Recall that the output of the overall filter may be sampled periodically 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.

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

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

Zero-forcing equalizer

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

∑−=

−=∗=N

Nnn nTtpwtwtptg )()()()(

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

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

⎩⎨⎧

≠== 0 if 0

0 if 1 kkgk

From (8), we may force the conditions

(9) ..., ,2 ,1for 00for 1

⎩⎨⎧

±±±=== Nk

(8) ∑−=

Nnnknk pwg

At the time instance t = kT,

)(nTppn =where and )(kTggk =

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

(10) ..., ,2 ,1for 00for 1

⎩⎨⎧

±±±===∑

−=− Nk

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0:010:0

......:::::

......

......:::::

......

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢⎢

−−

−−−−−

−−−−+−

pppppppppp

Equivalently, we have the following matrix form

A tapped-delay-line equalizer described by Eq. (10) or (11) is referred to as a zero-forcing equalizer. Such an equalizer is optimum in the sense 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+

r(t)x(t)

Figure 16

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

Figure 17

matched filter

noise path)()()( fHfHfH CT=

+ equalizer

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

Tx filter physicalchannel

Signal path

Signal path: the “effective” channel has 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 pass through the equalizer which is not equalizing the matched filter⇒ equalizing will enhance residual additive noise.

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

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

02|)(| ftjefH π−

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Preset equalizers: On channels whose frequency-response characteristics are unknown, but time-invariant, we may measure the channel characteristics, adjust the parameters of the equalizer, and once adjusted, the parameters remain fixed during the transmission of data.

Two types of equalizers

Adaptive equalizers: update their parameters on a periodic basis during the transmission of data.

Minimum mean-square error (MSE) equalizer: :The tap weights are chosen to minimize MSE of all the ISI terms plus the noise power at the output of the equalizer.

Remark. Most high-speed telephone line modems use an MSE weight criterion, because it is superior to a zero-forcing criterion, and it is more robust in the present of noise and large ISI.

Remark on equalization of digital data transmission

The zero-forcing equations (10) or (11) in Sec. 4. 5 do not account for the effect of noise. In addition, a finite-length filter equalizer can minimize worst-case ISI only if the peak distortion (i.e., the magnitude of the difference between the channel output and desired signal ) is sufficiently small.

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

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

Transmission Rate

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

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

The bandwidth dilemma

∞<|| f

@G. Gong 56

(c)(d)

(e) 35 dB

General shape of psd

(e) 50 dB

TffcTfH cX )(sin)( 2 −= π

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

)( fH X

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

, where PX is the total signal power over all frequencies.

)(/ cXXN fHPW =

(c ) Null-to-null bandwidth. It is defined as the width of the main spectral lobe, where the most of the signal power is contained (the most popular measure of bandwidth. )

(d) Fractional power containment bandwidth. Federal Communication Commission (FCC Rules and Regulations Section 2.202). It states that the occupied bandwidth is the band that exactly .5% of the signal power above the upper band limit and exactly 0.5% of the signal power below the lower band limit. Thus 99% of the signal power is inside the occupied band.

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

must have fallen at least to a certain stated level below that found at the band center. Typical attenuation levels might be 35 or 50 dB.

)( fH X

(f) Absolute bandwidth. This is the interval between frequencies, outside of which the spectrum is zero. (Useful abstraction. For all realizable waveforms, this is infinite.)

Example. Digital Telephone Circuits.

Compare the system bandwidth requirements for a terrestrial 3-kHz analog telephone 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 is 8000 samples/s and each voice sample is quantized to one of 256 levels. The bit stream is then transmitted using a PAM waveform and received with zero ISI.

Solution. The resulting of the sampling and quantization process yields PCM words such that each word has one of L = 256 levels. If each sample were sent as a 256-ary PAM pulse (symbol). Thus the required system bandwidth without ISI for sending Rs symbols/s would be . Since each PCM word is converted to 8 bits. Thus, the system bandwidth required using PCM is

2/sRW ≥

.kHz 32)symbols/s 8000)(lbits/symbo 8(21

=≥PCMW

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

A Note on Relation between Channel bandwidth 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 physical quantities. 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)()()()( channel Effective

μfHfHfHfP RCT

)( of bandwidth theas same theis signal tted transmi theof bandwidth the

)( signal ted transmit theof psd The 2

@G. Gong 61

accuracymission high trans 3)efficiencyn utilizatiopower mittedhigh trans 2)

efficiencyn utilizatio spectrumhigh 1)achieve order toin

way thatasuch in designed be should system The)( of bandwidth : )( of bandwidth :

signal) ed transmitt theof bandwidth ( )( of bandwidth :Let

fHBfHB

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

@G. Gong 62

ssion. transmifree-ISI ofconstraint under the 2/1 as small as becan interval

symbol theas , ISI without2 rate aat sequenceninformatiobinary asmit can tran we, "0" symbolfor )2sinc(2 and "1" symbolfor )2sinc(2 Using

)()2sinc(2 :pairFourier thehave We

WtWWtW

fHWtW T

channelNyquist ideal The

,1)()()()(

⎪⎩

⎪⎨⎧

WffPfHfHfH RCT

4) Consider a special case where

@G. Gong 63

Summary of Chapter 4 (Chapter 6 in the textbook)

∑∞

−∞=

BTX nfTjnRfHT

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

signals PAMbinary ofdensity spectrumpower and signals 1.PAM2 π

⎩⎨⎧

∑∞

∞−

12 Spectrum Cosine Raised

2 ChannelNyquist Ideal

0 ,00 ,1

sion transmisfree-ISIfor criteriaNyquist 3.

TRTnRfP

2. ISI due to bandlimited channel

4. Correlative coding and equalization

Duobinary signaling: achieving the maximum transmission rate 2W 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-forcing equalizer.

Modified Duobinary Scheme with Precoding

{ } na { } nb{ } ˆny { } ˆ

nb { } ˆna

+Ideal channel

)( fHNyquist

Post-coder

duobinarydecoder

{ } ny

overall channel2D

+ CL{ } *

4. nyquist criterion for distortionless baseband binary ...ece411/slides_files/topic4-2.pdf · @g....

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