7080 Chapter9 Slides

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Chapter 9: Signaling Over Bandlimited Channels

A First Course in Digital CommunicationsHa H. Nguyen and E. Shwedyk

February 2009

A First Course in Digital Communications 1/36
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Introduction

Have considered only the detection of signals transmitted over

channels of infinite bandwidth.Bandlimited channels are common: telephone channel, or evenfiber optics, etc.

Bandlimitation depends not only on the channel media butalso on the source, specifically the source rate, R

s(symbols/sec).

Band limitation can also be imposed on a communicationsystem by regulatory requirements.

The general effect of band limitation on a transmitted signal

of finite time duration is to disperse(or spread) it outSignal transmitted in a particular time slot interferes withsignals in other time slots inter-symbol interference(ISI).Shall consider the demodulation of signals which are not onlycorrupted by additive, white Gaussian noisebutalsobyISI.

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Major Approaches to Deal with ISI

1 Force the ISI effect to zeroNyquists first criterion.2 Allow some ISI but in a controlled manner

Partial response

signaling.

3 Live with the presence of ISI and design the bestdemodulation for the situationMaximum likelihoodsequence estimation (Viterbi algorithm).

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Communication System Model

)(fHC)(fHT)Rate( br

( )

(WGN)

tw

)(fHR

bkTt =

)(fHC)(fHT )(fHR

bkTt =

bT0

bT2

1 1 0

( )tw

( )tr ( )ty

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ISI Example

C

R

Lowpass Filter

1( ) expC

th tRC RC

=

)(tsAV

bT0t

)(tsB

bT0t

/1(1 e )b

T RCV

RC

)(tsA

V

bT0t

bT2

bT3 bT4 bT5

)( bA Tts

)2( bA Tts

)3( bA Tts

)4( bA Tts

)(tsB

bT0t

bT2 bT3

bT4

bT5

)( bB Tts

)2( bB Tts

)3( bB Tts

)4( bB Tts

0 1 2 3

During this interval, ( ) ( ) ( ) ( 2 ) ( 3 ) ( )B B b B b B b

t s t s t T s t T s t T t= + + + +r b b b b w

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Ch 9 Si li O B dli i d Ch l


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Time-Domain Nyquists Criterion for Zero ISI

The samples ofsR(t) due to an impulse should be 1 at t= 0 and

zero at all other sampling times kTb (k= 0).

0)()()( fHfHfH RCT t )(tsR

)()( tsfS RR

0at

appliedpulseIm

=t

0t

)(tsR

1

bT2 bT2bTbT bT3 bT4

)(ofSamples tsR


Ch te 9 Si li O e B dli ited Ch els
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Frequency-Domain Nyquists Criterion for Zero ISI

f

0

bT2

1

bT

1

bT

2

bT2

3

bT2

1

bT

1

)(fSR

b

RT

fS 1

b

RT

fS 2

+

b

RT

fS1

f0

bT2

1

bT2

1

bT

Rk b

k

S f T

=

+

IfW < 12Tb ISI terms cannot be made zero.IfW = 1

2Tb SR(f) =Tb over f | 12Tb |, sR(t) =

sin(t/Tb)(t/Tb)

.

IfW > 12Tb

Infinite number ofSR(f) to achieve zero ISI.


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Pulse Shaping when W = 12Tb

f0

bT2

1

bT2

1

bT

)(fSR

3 2 1 0 1 2 30.4

0.2

0

0.2

0.4

0.6

0.8

1

t/Tb

sR(

t)

sR(t) decays as 1/t

if the sampler is

not perfectly synchronized in time, con-siderable ISI can be encountered.


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0 1 2 3 4 5 6 7 8 9 10 11

1

0.5

0

0.5

1

t/Tb

x(t)/V

Input to the impulse modulator

0 1 2 3 4 5 6 7 8 9 10 112

1

0

1

2

t/T

b

y(t)

/V

Output of the receive filter


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Raised Cosine Pulse Shaping

SR(f) =SRC(f) =

Tb, |f| 12Tb

Tbcos2 Tb2 |f| 12Tb , 12Tb |f| 1+2Tb

0, |f| 1+2Tb

.

sR(t) =sRC(t) = sin(t/Tb)

(t/Tb)

cos(t/Tb)

1 42t2/T2b=sinc(t/Tb)

cos(t/Tb)

1 42t2/T2b.

1 0.5 0 0.5 1

0

0.2

0.4

0.6

0.8

1

fTb

SR

(f)

=0

=0.5

=1.0

3 2 1 0 1 2 30.4

0.2

0

0.2

0.4

0.6

0.8

1

t/Tb

sR(t)

=0

=0.5

=1


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0 1 2 3 4 5 6 7 8 9 10 112

1

0

1

2

t/Tb

y(t)/V

With the rectangular spectrum

0 1 2 3 4 5 6 7 8 9 10 112

1

0

1

2

t/Tb

y(t)/V

With a raised cosine spectrum


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p g g

Eye Diagrams

To observe and measure (qualitatively) the effect of ISI.

0 1.0 2.02.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

t/Tb

y(t)/V

0 1.0 2.02.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

t/Tb

y(t)/V

Left: Ideal lowpass filter, Right: A raised-cosine filter with = 0.35.


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Eye Diagrams with SNR=V2/2w = 20dB

0 1.0 2.02.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

t/Tb

y(t)/V

0 1.0 2.02.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

t/Tb

y(t)/V

Left: Ideal lowpass filter, Right: A raised-cosine filter with = 0.35.


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Design of Transmitting and Receiving Filters

Have shown how to design SR(f) =HT(f)HC(f)HR(f) to

achieve zero ISI.When HC(f) is fixed, one still has flexibility in the design ofHT(f) and HR(f).

Shall design the filters to minimize the probability of error.

bkTt=

( ) ( )b

k

t V t kT

=

=

x

( )ty( )tr ( )bkTy

( )

Gaussian noise, zero-mean

PSD ( ) watts/Hz

t

S fw

w

)(fHC)(fHT )(fHR

Noise is assumed to be Gaussian (as usual) but does notnecessarily have to be white.


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bkTt=

( ) ( )b

k

t V t kT

=

=

x

( )ty( )tr ( )bkTy

( )


PSD ( ) watts/Hz

t

S fw

w

)(fHC)(fHT )(fHR

y(mTb) = V + wo(mTb),where wo(mTb) N(0, 2w), with 2w =

Sw(f)|HR(f)|2df.

V V0

2

w

( ( ) | )bf y mT V( ( ) | )bf y mT V

( )b

y mT

Since P[error] =Q VwNeed to maximize

V2

2w

.


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bkTt=

( ) ( )b

k

t V t kT

=

=

x

( )ty( )tr ( )bkTy

( )


PSD ( ) watts/Hz

t

S fw

w

)(fHC)(fHT )(fHR

Given the transmitted powerPT, the channels frequency responseHC(f) and the additive noises power spectral densitySw(f)

chooseHT(f) andHR(f) so that the zero ISI criterion is satisfied

and theSNR = V2

2w

is maximized.


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1 Compute the average transmitted power:

PT = V2

Tb

|HT(f)|2df (watts).

2

Write the inverse of the SNR as2w

V2 =

1

PTTb

|HT(f)|2df

Sw(f)|HR(f)|2df

= 1

PTTb

|SR(f)|2

|HC(f)

|2

|HR(f)

|2

df

Sw(f)|HR(f)|2df .3 Apply the Cauchy-Schwartz inequality:

A(f)B(f)df2

|A(f)|2df

|B(f)|2df

, which

holds with equality if and only ifA(f) =K B(f). Identify

|A(f)| = Sw(f)|HR(f)|,|B(f)| = |SR(f)||HC(f)||HR(f)| . Then|HR(f)|2 = K|SR(f)|

Sw(f)|HC(f)|, |HT(f)|2 =|SR(f)|

Sw(f)

K|HC(f)| .

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Design Under WhiteGaussian Noise and Ideal Channel

If the channel is ideal, i.e., HC(f) = 1 for|f| W andK1=K2 then|HT(f)| = |HR(f)| =|SR(f)|/K1.

IfSR(f) is a raised-cosine spectrum then both HT(f) andHR(f) are square-root raised-cosine(SRRC) spectrum:

HT(f) =HR(f) =Tb, |f|

1

2TbTbcos

Tb2

|f| 12Tb

, 12Tb |f|

1+2Tb

0, |f| 1+2Tb.

hT(t) = hR(t)

= sSRRC(t) =(4t/Tb)cos[(1 + )t/Tb] + sin[(1 )t/Tb]

(t/Tb)[1 (4t/Tb)2] .


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RC and SRRC Waveforms ( = 0.5)

3 2 1 0 1 2 30.2

0

0.2

0.4

0.6

0.8

1

1.2

t/Tb

sR(t)

Raised cosine (RC)

Squareroot RC


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Example: Transmission raterb = 3600 bits/sec, P[bit error] 104. Channel model:

HC(f) = 102 for|f| 2400 Hz and HC(f) = 0 for |f| >2400 Hz. Noise model:

Sw(f) = 1014 watts/Hz, f (white noise).(a) Since rb = 3600 bits/sec and W= 2400 Hz, choose a raised-cosine spectrum

with rb2 = 600 or= 13 .

SR(f) =

1

3600, |f|


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Duobinary Modulation

If bandwidth is very limited and one cannot afford to use more

than 12Tb Hz, the only way to achieve zero ISI is to have flatSR(f), which is practically difficult to implement.

An alternative is to allow a certain amount of ISI but in acontrolled manner Duobinarymodulation.

Shall restrict the ISI to only one term, namely that due to theprevious symbol.

0t

(1)

bT2 bT2bTbT bT3 bT4

)(ofSamples tsR

3b

T

Interfering term

(1)

[sampled]( )Rs t


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Overall System Response of Duobinary Modulation

SR(f) = 2Tbcos(f Tb), 12Tb f 12Tb

0, elsewhereF sR(t) = cos(t/Tb)

14 ( tTb )2

.

0.4 0.2 0 0.2 0.4 0.60

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

fTb

SR

(f)/Tb

3 2 1 0 1 2 30.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

t/Tb

sR(t)


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Using Precoder in Duobinary Modulation

y(tk) = Vk+ Vk1+ wo(tk), where tk =kTb

Tb

2

, k = 0,

1,

2, . . .

=

2V + wo(tk), if bitsk and (k 1) are both 10 + wo(tk), if bitsk and (k 1) are different

2V + wo(tk), if bitsk and (k 1) are both 0.

Precoder

Output data bits

1,k kb b 1k k k= d b d

1 out

1 out

0 then ( ) 2 + ( )

1 then ( ) 0 + ( )

k k k k k

k k k k k

t V t

t t

= = =

= = =

b d d y w

b d d y w

2V 2V0

D1D0 D0

( )ky t

( ( ) | 0 )k T

f y t ( ( ) | 0 )k Tf y t

( ( ) |1 )k Tf y t


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P[y(tk) = 2V] =P[y(tk) = 2V] = 1

4; P[y(tk) = 0] =

1

2.

P[error] 1

4area4+

1

2[area1+area2] +

1

4area3 =

3

2Q

V

w

,

LetHC(f) = 1over |f| 1

2Tb and consider AWGN with PSD N02 . Then

V2

2w

max

=PTTb

N0/2 12Tb

12Tb

2Tbcos(f Tb)df

2 = (PTTb) 2

N0

4

2

P[error]duobinary =

3

2 Q 42PTTbN0 .For binary PAM with zero ISI

V2

2w

max

is

PTTb

|SR(f)|

Sw(f)

|HC(f)|df

2=PTTb

N0/2 1

2Tb

1

2Tb

|SR(f)|df

2

=PTTb 2

N0P[error]binary =Q

2PTTb

N0

.

Duobinary modulation requires an addition of4

2or2.1 dB to achieve the

same error probability as the zero-ISI modulation.


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Maximum Likelihood Sequence Estimation (MLSE)

Do not attempt to eliminate the ISI but rather takes it into

account in the demodulator.The criterion is to minimize the sequence error probability.

Consider a sequence ofNequally likely bits transmitted overa bandlimited channel where the transmission begins at t= 0and ends at t=N Tb, with Tb the bit interval.

h(t) is the impulse response of the overall chain:modulator/transmitter filter/channel, assumed to be nonzeroover [0, LTb]The number of ISI terms is L.

Impulse

Modulator ?)(th

Modulator/Transmitter Filter/Channel

{ }kb

1

0

( ) ( )N

k b

k

t t kT

=

= b b ( )i

s t ( )tr

( )tw


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The receiver sees one of the M= 2N possible signals,si(t) =

N1k=0 bi,kh(t kTb), i= 1, 2, . . . , M = 2N, corrupted by

w(t): This as a humongousM-ary demodulation problem in AWGN.

The decision rule is:

Compute:

i = 2

N0

r(t)si(t)dt 1N0

s2i (t)dt, i= 1, 2, . . . , M

and choose the largest.

Substituting si(t) =N1

k=0 bi,kh(t kTb), the decision rule is

Compute:

i = 2

N0

N1k=0

bi,k

r(t)h(t kTb)dt1

N0

N1k=0

N1j=0

bi,kbi,j

h(t kTb)h(t jTb)dt



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Definehkj =

h(t kTb)h(tjTb)dt andrk = r(t)h(t kTb)dt. Then the decision rule is

Compute:

i= 2

N0

N1

k=0bi,krk 1

N0

N1

k=0N1

j=0bi,kbi,jhkj , i= 1, 2, . . . , M


The path metric i can be expressed as

i = 2

N0

N1k=0

bi,krk 1

N0

N1k=0

b2i,kh0 2

N0

N1k=0

bi,k

kj=1

bi,kjhj .



b2 h0 h
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Note that bi,k = 1and bi,kh02 = h02 is a constant. Also sinceh(t) = 0 fort LTb, which means hj = 0 forj L, the pathmetric becomes

i = 2

N0

N1k=0

bi,krk bi,k L1

j=1

bi,kjhj

branch metric

.

Branch metricdepends on: (i) the present output of the matchedfilter, rk; (ii) the present value of the considered bit pattern, bi,k;(iii) the previous L 1 values of the considered bit patternbi,k1, bi,k2, . . . , bi,k(L1).

The system has memory, namely the ISI terms

Can use a finite

state diagram and the trellis to represent the transmitted signals.

States are defined by the previous L 1 bits.Determination of the best path through the trellis can beaccomplished most efficiently with the Viterbi algorithm.


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Example

0bT2bT 2.5 bT

1 2.5b

T

( )h t

t

0bT2bT 2.5 bTbT2 bT2.5 bT

01h =

1h

2h2h

1h

| |1 , | | 2.5

2.5( )0, | | 2.5

b

bh

b

T

TRT

=

>

There are two ISI terms, which are due to h1 = 0.6 and

h2= 0.2L= 3 or the memory is L 1 = 2 bits in length.The branch metric term isbi,krk 0.6bi,kbi,k1 0.2bi,kbi,k2.The inputsbi,k1, bi,k2 represent the system memory andhence are states of the state diagram.



S d T lli Di
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State and Trellis Diagrams

1 2 (defining states)k kb b

denotes present input bit 0 (or 1)k

b =

denotes present input bit 1 (or 1)k

b = +

bT0

bT2 bT3 bT4

1 2k kb b

00

10

01

11

0b

1b

2b

3b

4b

0r

1r

2r

3r

4r

Starting state is chosen to be 00. Before t= 0 everythingiszero.

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C i f B h M i
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Computations of Branch Metrics

bkrk bk2

j=1

bkjhj =bkrk 0.6bkbk1 0.2bkbk2

bk2 bk1 bk Branch Metric

0 (1) 0 (1) 0 (1) rk 0.6 0.2 = rk 0.80 (1) 0 (1) 1 (+1) +rk+ 0.6 + 0.2 = +rk+ 0.80 (1) 1 (+1) 0 (1) rk+ 0.6 0.2 = rk+ 0.40 (1) 1 (+1) 1 (+1) +rk 0.6 + 0.2 = +rk 0.41 (+1) 0 (1) 0 (1) rk 0.6 + 0.2 = rk 0.41 (+1) 0 (1) 1 (+1) +rk+ 0.6 0.2 = +rk+ 0.41 (+1) 1 (+1) 0 (1) rk+ 0.6 + 0.2 = rk+ 0.81 (+1) 1 (+1) 1 (+1) +rk 0.6 0.2 = +rk 0.8


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Branch metrics and (partial) path metrics for all possible paths forthe first 3 bit transmissions.

00

10

01

11

t

1.074

1.074

1.0590.015

1.059 2.133

0.141

0.141

0.933

1.215

0.3160.301

3.017

0.331

1.249

0.884

0.084

1.28

4

bT0

bT2

bT3

01.074r = 1 0.459r = 2 0.484r =



P d T lli U T k 12
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Pruned Trellis Up To k = 12

0k= 1k= 2k= 3k= 4k= 5k= 6k=

6k= 7k= 8k= 9k= 10k= 11k= 12k=

12.305

10.531

11.999

7.825

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