Week11 PartialResponseSignals...
Transcript of Week11 PartialResponseSignals...
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EE4601Communication Systems
Week 11
Partial Response Signals
Linear Zero Forcing Equalization
0 c©2011, Georgia Institute of Technology (lect11 1)
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Objective
Objective: Signals with a baud rate of 2W symbols/sec in a bandwidth of W Hzwith realizable filters.
{a }k
ka ε {−1,+1}+ h(t)
kTg(t) c(t)
w(t)
Assume c(t) = δ(t) (ideal channel)
Then h(t) = g(T − t)
p(t) = g(t) ∗ g(−t)
P (f) = |G(f)|2
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Partial response signaling
Assume that P (f) has the following form
a (t-nT)δnΣn
a(t)=
{a }k
(f)HN+
T
c(t)
T
{c }k
HN(f) =1
2Wrect
(
f
2W
)
=
1 |f | < W0 else
where W = 1/2T i.e., the baud rate is R = 1/T = 2W symbols/sec
P (f) = (1 + e−j2πfT )HN(f)
= 2e−jπfT
ejπfT + e−jπfT
2
HN(f)
= 2 cos(πfT )e−jπfTHN(f)
= 2 cos(πfT )e−jπfT 1
2Wrect
(
f
2W
)
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Duobinary
P (f) =
2T cos(πfT )e−jπfT |f | < W0 else
π/2
−π/2
-W=-1/2T
|P(f)| arg(P(f))
2
W=1/2T
f f
To get p(t) we write{
P (f) = HN(f) +HN(f)ej2πfT
}
↔
{
p(t) = sinc
(
t
T
)
+ sinc
(
t− T
T
)}
p(t)
−3T −2T −T
0
T 2T 3T
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Duobinary
T
a (t-nT)δnΣn
a(t)=
{a }k
c(t)
T
{c }kp(t)
c(t) =∑
nanp(t− nT )
ck = c(kT ) =∑
nanp((k − n)T ) =
∑
nanpk−n
But pj = p(jT ) =
1 j = 0, 1
0 j 6= 0, 1
Therefore, ck = ak + ak−1
Since ak ∈ {−1,+1} ck ∈ {−2, 0, 2} (3-level)
We can recover {ak} from {ck} by ak = ck − ak−1 assuming an initial value, e.g.a0 = −1 or + 1. This is called decision feedback detection.
Problem : Errors due to noise propagate, i.e., ak = ck − ˆak−1.If ˆak−1 is in error then ak is likely to be in error.
0 c©2011, Georgia Institute of Technology (lect11 5)
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Precoding
+ k a = 2d
k -1
{b }k
bk {0,1}ε
(f)HN+
{c }k
T
c(t)
T
a(t)
Same as before
P(f)
Impulsegenerator
D
a(t)dk {0,1}ε
{a }k
logicD-flip flop
0 - -11 - +1
dk = bk ⊕ dk−1 = bk + dk−1(mod 2)
Example:
{bk} 0 0 1 1 1 0 1 0 0 0
{dk} 1 1 1 0 1 0 0 1 1 1 1{ak} +1 +1 +1 −1 +1 −1 −1 +1 +1 +1 +1
{ck} +2 +2 0 0 0 −2 0 +2 +2 +2
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Modified Duobinary
Note that ck =
±2 if bk = 0
0 if bk = 1Therefore {bk} can be recovered from {ak} by using symbol by symbol detection.
+ k a = 2d
k -1
{b }k
(f)HN+
{c }k
Impulsegenerator
2D
a(t)dk {0,1}ε
{a }k
0 - -11 - +1
2T
c(t)
kT
a(t)
Same as before
P(f)
a(t) =∑
n anδ(t− nT ), dk = bk ⊕ dk−2, ck = ak − ak−2.
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Modified Duobinary
P (f) = (1− e−j4πfT )HN(f)
= j2e−j2πfT
ej2πfT − e−j2πfT
j2
HN(f)
= j2HN(f) sin 2πfTe−j2πfT
=
2T sin 2πfTej(π/2−2πfT ) |f | < 1/2T0 |f | > 1/2T
π/2
−π/2
-W=-1/2T
|P(f)| arg(P(f))
W=1/2T
f f
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Modified Duobinary Pulse
P (f) = HN(f)− e−j4πfTHN(f)
p(t) = sinc(t/T )− sinc((t− 2T )/T )
−3T −2T −T 0 T 2T 3T 4T 5T 6T
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Modified Duobinary with Precoding
Example:
{bk} 0 0 1 1 1 0 1 0 0 0
{dk} 1 1 1 1 0 0 1 0 0 0 0 0{ak} +1 +1 +1 +1 −1 −1 +1 −1 −1 −1 −1 −1
{ck} 0 0 −2 −2 2 0 −2 0 0 0
Note: ck =
±2 if bk = 10 if bk = 0
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Error Probability
Here we consider the error probability of precoded duobinary signaling withsymbol-by-symbol detection.
The transmit and receiver filters are implemented as the root-duobinary pulse,such that
|G(f)| = |H(f)| =√
|P (f)|
We note that the noise process at the output of the receiver filter has powerspectral density
Φnn(f) =No
2|H(f)|2 =
No
2P (f)
Since p(t) is not a Nyquist pulse, pk = p(kT ) = δk0, the noise samples are
correlated, and, hence the symbol-by-symbol detector is suboptimal. This losscan be recovered by using a sequence detector, but we will not discuss here.
The Gaussian noise samples at the output of the receiver matched filter H(f)are zero mean and have variance
σ2n =
No
2
∫ ∞
−∞|P (f)|df =
No
2
∫ 1/2T
−1/2T2 cos(πfT )df =
2No
π
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Modified Duobinary with Precoding
Note that the sampled outputs of the matched filter have the Gaussian densityfunction
yk ∼
N(±2, 2No/π) , xk = 0
N(0, 2No/π) , xk = 1
where the means ±2 each occur with probability 1/4 and the mean 0 occurs with
probability 1/2.Assuming that the receiver makes decisions according to
bk =
1, |yk| < 10, |yk| > 1
we have the probability of error
Pb =1
4·Q+
1
4·Q+
1
2· 2Q =
3
2Q
where
Q = Q
(
1
σ
)
= Q
√√√√
π
2No
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Modified Duobinary with Precoding
The energy per bit is
Eb =∫ ∞
−∞|G(f)|2df =
∫ ∞
−∞P (f)df =
4
π
Hence, π4Eb = 1 and
Q = Q
√√√√
π
2No·π
4Eb
= Q
√√√√
(
π
4
)2 2Eb
No
and
Pb =3
2Q
√√√√
(
π
4
)2 2Eb
No
When compared to binary antipodal signaling with
Pb = Q
√√√√2Eb
No
the loss in Eb/No performance is −10log10(π/4)2 = 2.1 dB.
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EE4601Communication Systems
Week 11
Linear Zero Forcing Equalization
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Equalization
• The cascade of the transmit filter g(t), channel c(t), receiver filter h(t) yields
the overall pulsep(t) = g(t) ∗ c(t) ∗ h(t)
• The signal at the output of the matched filter is
y(t) =∑
k
akp(t− kT ) + n(t)
and the sampled output is
yn = y(nT ) =∑
k
akpn−k + nn
=∑
k
pkan−k + nn
• Assume a causal, finite-length, channel such that p(t) = 0 for t < 0 and
t > LT .
• The discrete-time channel pn = p(nT ), can be represented by the vector
p = (p0, p1, . . . , pL)
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Equalization
• An equalizer is a digital filter that is used to mitigate the effects of inter-
symbol interference that is introduced by a time dispersive channel.
• The tap co-efficients of the equalizer are denoted by the vector
w = (w0, w1, · · · , wN−1)T
where N is the number of equalizer taps.
• If the equalizer is used to process the sampled outputs of the receivermatched filter, then the output of the equalizer is
xn =N−1∑
j=0
wjyn−j
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Linear Transversal Equalizer
0w 1w
^nxnx~
εn
T T T T
wN wN-2 -1
ny
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Overall Discrete-time Model
• The overall channel and equalizer can be represented by a overall digital
filter with impulse response
q = (q0, q1, . . . , qN+L−1)T
where
qn =N−1∑
j=0
wjpn−j
= wTp(n)
withp(n) = (pn, pn−1, pn−2, . . . , pn−N+1)
T
and pi = 0, i < 0, i > L. That is, q is the discrete convolution of p and w.
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Perfect Equalization
• Let the component of p of greatest magnitude be denoted by pd1. Note that
we may have d1 6= 0.
• Let the number of equalizer taps be equal to N = 2d2 + 1 where d2 is aninteger.
• Perfect equalization means that
q = ed = (0, 0, . . . , 0︸ ︷︷ ︸
d zeroes
, 1, 0, . . . , 0, 0)T
where d zeroes precede the “1” and d is an integer representing the overalldelay, a parameter to be optimized.
• Unfortunately, perfect equalization is difficult to achieve and does not always
yield the best performance.
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Zero Forcing Equalizer
• With a zero-forcing (ZF) equalizer, the tap coefficients w are chosen to
minimize the peak distortion of the equalized channel, defined as
Dp =1
|qd|
N+L−1∑
n=0
n6=d
|qn − qn|
where q = (q0, . . . , qN+L−1)T is the desired equalized channel and the delay
d is a positive integer chosen to have the value d = d1 + d2.
• Lucky showed that if the initial distortion without equalization is less than
unity, i.e.,
D =1
|pd1|
L∑
n=0
n6=d1
|pn| < 1 ,
then Dp is minimized by those N tap values which simultaneously causeqj = qj for d − d2 ≤ j ≤ d + d2. However, if the initial distortion before
equalization is greater than unity, the ZF criterion is not guaranteed tominimize the peak distortion.
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Zero Forcing Equalizer
• For the case when q = eTd the equalized channel is given by
q = (q0, . . . , qd1−1, 0, . . . , 0, 1, 0, . . . , 0, qd1+N , . . . , qN+L−1)T .
• In this case the equalizer forces zeroes into the equalized channel and, hence,
the name “zero-forcing equalizer.”
• If the ZF equalizer has an infinite number of taps it is possible to selectthe tap weights so that Dp = 0, i.e., q = q. Assuming that qn = δn0 this
condition means thatQ(z) = 1 = C(z)G(z) .
Therefore,
C(z) =1
G(z)
and the ideal ZF equalizer has a discrete transfer function that is simply the
inverse of overall channel G(z).
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Equalizer Tap Solution
• For a known channel impulse response, the tap gains of the ZF equalizer can
be found by the direct solution of a simple set of linear equations. To do so,we form the matrix
P = [p(d1), . . . ,p(d), . . . ,p(N + d1 − 1)]
and the vectorq = (qd1, . . . , qd, . . . , qN+d1−1)
T .
• Then the vector of optimal tap gains, wop, satisfies
wT
opP = qT −→ wop = (P−1)T q .
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Example
• Suppose that a system has the channel vector
p = (0.90,−0.15, 0.20, 0.10,−0.05)T ,
where pi = 0, i < 0, i > 4. The initial distortion before equalization is
D =1
|p0|
4∑
n=1
|pn| = 0.5555
and, therefore, the minimum distortion is achieved with the ZF solution.
• Suppose that we wish to design a 3-tap ZF equalizer. Since p0 is the com-
ponent of p having the largest magnitude, d1 = 0 and the optimal equalizerdelay is d = 1. The desired response is q = eT1 so that q = (0, 1, 0)T .
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Example
• We then construct the matrix
P = [p(0),p(1),p(2)]
=
0.90 −0.15 0.200.00 0.90 −0.15
0.00 0.00 0.90
and obtain the optimal tap solution
wop = (P−1)T q = (0, 1.11111, −0.185185)T .
The overall response of the channel and equalizer is
q = ( 0.0, 1.0, 0.0, 0.194, 0.148, −0.037, −0.009, 0, . . .)T .
• Hence, the distortion after equalization is
Dmin =1
|q0|
6∑
n=1
|qn − qn| = 0.388 .
0 c©2011, Georgia Institute of Technology (lect11b 11)
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Adaptive Solution
• In practice, the channel impulse response is unknown to the receiver and a
known finite length sequence a is used to train the equalizer.
• During this training mode, the equalizer taps can be obtained by using thefollowing steepest-descent recursive algorithm:
wn+1j = wn
j + αǫna∗n−j−d1
, j = 0, . . . , N − 1 , (1)
where
ǫn = an−d − an
= an−d −N−1∑
i=0
wiyn−i (2)
is the error sequence, {wnj } is the set of equalizer tap gains at epoch n.
• α is an adaptation step-size that can be optimized to trade off convergencerate and steady state bit error rate performance.
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Adaptive Solution
• Fact: The adaptation rule in (1) attempts to force the crosscorrelations
ǫna∗n−j−d1 , j = 0, . . . , N − 1, to zero.
• To see that this leads to the desired solution we note
1
2E[ǫna
∗n−j−d1] =
1
2E[an−da
∗n−j−d1]−
1
2
N−1∑
i=0
L∑
ℓ=0
wicℓE[an−i−ℓa∗n−j−d1]
= σ2a
δd2−j −N−1∑
i=0
wipj+d1−i
= σ2a(δd2−j − qj+d1) , j = 0, 1, . . . , N − 1 , (3)
where σ2a =
1
2E[|ak|
2].
• Fact: The conditions 1
2E[ǫna
∗n−j−d1] = 0 are satisfied when qd = 1 and qi = 0
for d − d2 ≤ i < d and d < i ≤ d + d2, which is the zero forcing solution.
Note the ensemble average over the noise and the data symbol alphabet.
0 c©2011, Georgia Institute of Technology (lect11b 13)
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Adaptive Solution
• After training the equalizer, a decision-feedback mechanism is typically em-
ployed where the sequence of symbol decisions a is used to update the tapcoefficients. This mode is called the data mode and allows the equalizer to
track variations in the channel vector c. In the data mode,
wn+1j = wn
j + αǫna∗n−j−d1
, j = 0, . . . , N − 1 ,
where the error term ǫn in (2) becomes
ǫn = an−d −N−1∑
i=0
ciyn−i
and, again, an−d is the decision on the equalizer output an delayed by dsamples.
0 c©2011, Georgia Institute of Technology (lect11b 14)