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Chapter 9: Signaling Over Bandlimited Channels
A First Course in Digital CommunicationsHa H. Nguyen and E. Shwedyk
February 2009
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Chapter 9: Signaling Over Bandlimited Channels
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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Chapter 9: Signaling Over Bandlimited Channels
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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Chapter 9: Signaling Over Bandlimited Channels
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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Chapter 9: Signaling Over Bandlimited Channels
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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Chapter 9: Signaling Over Bandlimited Channels
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
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Chapter 9: Signaling Over Bandlimited Channels
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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Chapter 9: Signaling Over Bandlimited Channels
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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Chapter 9: Signaling Over Bandlimited Channels
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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Chapter 9: Signaling Over Bandlimited Channels
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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Chapter 9: Signaling Over Bandlimited Channels
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
( )
Gaussian noise, zero-mean
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
( )
Gaussian noise, zero-mean
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
and choose the largest.
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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
and choose the largest.
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 .
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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.
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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 =
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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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