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#### David83

Hi,

Why digital communication books (at least ones I read) labeling the equalizer coefficients from -N to N, and not from 0 to 2N? Does this signify the fact that inter-symbol interference (ISI) occurs from previous and future symbols?

suppose that 'I' is the information sequence, g(t) is the Tx filter impulse response, c(t) is the channel impulse response and h(t) is the Rx filter impulse response. So 'y' is the received signal at the receiver.

y = I*g(t)*c(t)*h(t) = I*x(t) ---> Y(z) = I(z)X(z) ---> I(z) = Y(z)/X(z) ---> best equalizer will be 1/X(z)

When the channel is modeled as a digital filter the received signal is given by y = sum(I(k)x(n-k)) (convolution)

Now because X(z) is two sided ( having both z^+ and z^- ) so is it's reciprocal 1/X(z). So yes, ISI comes from previous and next symbols. But I'm a bit confused about the physical justification of this.

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If we interpret the convolution physically, it will be as the following: when sampling at time k, inputs to the equalizer will be from the received signals at time k-N up to K+N, i.e.: the desired symbol is in the mid tap.

the desired symbol is in the mid tap

of course. but when sampling at time k, the (k+N)th sample isn't received yet. that's the "?"

Yes, I agree.

By the way, 1/X(z) is not the "best" equalizer. ZF equalizer is optimum in the sense that it inverts the channel and eliminates ISI completely. However, in terms of BER performance it does not behave well, especially in channels with spectral nulls, because it enhances the noise, too. In general a MSE equalizer performs better.

Regards

• hunter555persia

### hunter555persia

Points: 2
Actually the best would be the MLSE which is base on Viterbi algorithm. ZF and MSE equalizers are sub-optimum linear equalizers and DFE equalizer is a sub-optimum nonlinear equalizer that performs much better than linear ones.

• David83

### David83

Points: 2
You are right.

I have another question: why when do whitening filtering before equalization we always choose $G^*(1/z^*)$ and not $G(z)$ which are the factorization of the cascaded transfer functions?

X(z) = G(z)G*(1/z*)

G*(1/z*) corresponds to the anti-causal part of X(z). We choose 1/G*(1/z*) as the whitening filter so the whole cascaded transfer function i.e. X(z)/G*(1/z*) won't have any anti-causal part. But I think (I'm NOT sure) this whitening filter is only mathematically important, since it whitens the noise and error rate calculations are simplified. Sometimes the whitening filter is absorbed into the equalizer, which would yield a causal overall system.

So? what is the problem with anti-causal part? Why would we eliminate it?

In this model X(z) must physically exist. But It has an anti-causal part, so the same confusion in post#2 arises. I can't justify this and don't know the answer. Please let me know if you find any justification for this.

I will. Thanks

What is peak distortion? and why it is defined the way it is defined? and what does this have to do with ZF equalization?

What is peak distortion? and why it is defined the way it is defined? and what does this have to do with ZF equalization?
Imagine a binary sistem. Peak distortion is the ratio between the maximum absolute value of ISI and the signal alone. It is equal to the maximum closure of the eye in the eye pattern.
For a given support of the equalizer (span of the filter before and after the "main sample"), the equalizer that achieves the minimum peak distortion is the zero-forcing equalizer.
Regards

Z

zorro , do you have any explanation for post#2 please ??

Imagine a binary sistem. Peak distortion is the ratio between the maximum absolute value of ISI and the signal alone. It is equal to the maximum closure of the eye in the eye pattern.
For a given support of the equalizer (span of the filter before and after the "main sample"), the equalizer that achieves the minimum peak distortion is the zero-forcing equalizer.
Regards

Z

But as I understand, ZF forces the samples at the output of the concatenation of the transmit filter, channel, matched filter, sampler, noise whitening filter and the equalizer to be zero everywhere except at the sampling time of the current symbol. How to interpret peak distortion as one equivalent to this? For example Proakis says that we need to minimize the following metric:

$\mathcal{D}=\displaystyle\sum_{n=-\infty \atop n\neq 0}^{\infty}|q_n|$

where $\{q_n\}$ are the coefficients of the effective model of the concatenation mentioned above.

Thanks

David83,

ZF equalizer minimizes the metric D that you said, that is the peak distortion. But it can not reduce it to zero, unless the equalizer has infinite length.
Suppose that the equalizer has Nb taps before the curren symbol and Na taps after it. The set of coefficients that minimizes the metric D (with the constraint q0=1) is the set that makes that the response has Nb zeros before the current symbol and Na zeros after it. (Here is the reason of the name "zero forcing".)
The equalizer has Na+Nb+1 degrees of freedom: it can satisfy Na+Nb+1 conditions. They are:
qn=0 for -Nb<=n<=1
q0=1
qn=0 for 1<=n<=Na

Doing so, the sampes of the impuse response increase beyond the above mentioned range, i.e. for n<-Nb and n>Na. Nevertheless, the residual peak distortion is less than the original one (I don't remember exactly, but I think that this last statement is true with the condition that initially the eye is not completely closed, i.e. D<1).

Regards

Z

I came up with an idea for confusion of post#2 (future samples):
If the equalizer has N1 taps for future samples and N2 taps for previous samples, the receiver produces any output with delay of N1 samples. This delay would be enough to get all the needed future samples. So as a whole the system will produce equalized output with a delay David83,

ZF equalizer minimizes the metric D that you said, that is the peak distortion. But it can not reduce it to zero, unless the equalizer has infinite length.
Suppose that the equalizer has Nb taps before the curren symbol and Na taps after it. The set of coefficients that minimizes the metric D (with the constraint q0=1) is the set that makes that the response has Nb zeros before the current symbol and Na zeros after it. (Here is the reason of the name "zero forcing".)
The equalizer has Na+Nb+1 degrees of freedom: it can satisfy Na+Nb+1 conditions. They are:
qn=0 for -Nb<=n<=1
q0=1
qn=0 for 1<=n<=Na

Doing so, the sampes of the impuse response increase beyond the above mentioned range, i.e. for n<-Nb and n>Na. Nevertheless, the residual peak distortion is less than the original one (I don't remember exactly, but I think that this last statement is true with the condition that initially the eye is not completely closed, i.e. D<1).

Regards

Z

Correct. But I still confused about the relationship between $\mathcal{D}$ and ZF. Why we use the absolute value in the minimization?

Thanks

---------- Post added at 18:31 ---------- Previous post was at 18:19 ----------

I came up with an idea for confusion of post#2 (future samples):
If the equalizer has N1 taps for future samples and N2 taps for previous samples, the receiver produces any output with delay of N1 samples. This delay would be enough to get all the needed future samples. So as a whole the system will produce equalized output with a delay This makes sense. Now I am confused about another thing: the concatenation of the transmit filter, channel, matched filter, and sampler has an equivalent discrete-time model with coefficients from say $-N_1$ up to $N_2$. The noise output of this model is colored, and hence we do noise whitening. The concatenation of the above time-discrete model and the noise whitening filter results in an equivalent discrete-time model whose coefficients start from 0 up to $N_2$. This means now we do not have interference from future symbols. How to connect between the original model and the white noise model?

Regards

Well now the whitening filter has future samples in its implementation.

---------- Post added at 18:56 ---------- Previous post was at 18:44 ----------

Correct. But I still confused about the relationship between $\mathcal{D}$ and ZF. Why we use the absolute value in the minimization?

Well some ISI components are positive while others may be negative. Now positive or negative information stream will yield different values for ISI. When we minimize the sum of absolute value of 'q' , the ISI is minimized regardless of information symbol sign. If minimization is carried out on sum of 'q' (NOT absolute) the sign of information stream may result in large ISI.

How do you type those formulas ??

Well now the whitening filter has future samples in its implementation.

So, future samples are absorbed by the whitening filter.

Returning to the whitening filter: why we use $1/G^*(1/z^*)$ as the whitening filter again?

Another thing, in DSP books it is said that z-transform reduces to discrete-time Fourier Transform by setting $z=e^{jw}$. However, in the digital communication books, when evaluating the power spectrum from the z-transform, it is set $z=e^{jwT}$, where T is the symbol duration. Why?

About typing: just write [ tex ] in the left and [ /tex ] in the right, and put Latex commands inside. Just eliminate the spaces inside [].

Regards

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