Week11 PartialResponseSignals...

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EE4601 Communication Systems Week 11 Partial Response Signals Linear Zero Forcing Equalization 0 c 2011, Georgia Institute of Technology (lect11 1)

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.

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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.

0 c©2011, Georgia Institute of Technology (lect11 10)

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EE4601Communication Systems

Week 11

Linear Zero Forcing Equalization

0 c©2010, Georgia Institute of Technology (lect8 1)

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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.

0 c©2010, Georgia Institute of Technology (lect8 7)

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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).

0 c©2010, Georgia Institute of Technology (lect8 8)

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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 .

0 c©2010, Georgia Institute of Technology (lect8 9)

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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 .

0 c©2010, Georgia Institute of Technology (lect8 10)

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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.

0 c©2011, Georgia Institute of Technology (lect11b 12)

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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)