About Adaptive Equalizers

[zorro]

Hi Hunter,

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

Sorry because I din't answer to your question about post #2.
You are right. The response has a main lobe preceded by precursors and followed by a "tail".

Z

 

[zorro]

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?

In the expression \[z=e^{jw}\], \[w\] is the normalized angular frequency, measured in radians/sample.
In \[z=e^{jwT}\] it is the angular frequency, measured in radians/seconds.
Regards

Z

 

[David83]

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

After a second thought: why the receiver waits \[N_1\] samples in the first place? Can we say that the reason for delaying is to get a factorized transfer function to realize the noise whitening filter, for example??

---------- Post added at 04:13 ---------- Previous post was at 04:09 ----------

In the expression \[z=e^{jw}\], \[w\] is the normalized angular frequency, measured in radians/sample.
In \[z=e^{jwT}\] it is the angular frequency, measured in radians/seconds.
Regards

Z

Thanks. About the noise whitening filter transfer function, can we say that we choose \[1/G^*(1/z^*)\] as the transfer function of the noise whitening filter because it will be causal, since \[G^*(1/z^*)\] is anti causal? prof. Proakis says that this filter is causal and stable. Why stable?

 

[hunter555persia]

why the receiver waits \[N_1\] samples in the first place?

Because after waiting for N1 samples those future samples needed for equalization will be received and stored.

 

[David83]

Because after waiting for N1 samples those future samples needed for equalization will be received and stored.

But again, if we did not wait \[N_1\] samples, there will be no interference with them, physically speaking. Right?

 

[hunter555persia]

Yeah I don't understand this. ISI can't come from future samples physically, but in the formulation of text book there are future samples as well.
?????????????????????!!!!!!!!!!!!!!!!!!!

 

[David83]

I think it has to do with the whitening filter, because if we do not have future samples, then we can not factorize the effective channel. It is still confusing!!

 

[zorro]

Hi guys,

About "future samples": as I said before, the pulse has precursors and tail (postcursors). The precursors are the samples of the response before the peak, and they can span over several symbols.
When you are detecting a current symbol, the ISI is composed by the postcursors of the past simbols as well as the precursors of the future ones.
Regards

Z

 

[David83]

Hi guys,

About "future samples": as I said before, the pulse has precursors and tail (postcursors). The precursors are the samples of the response before the peak, and they can span over several symbols.
When you are detecting a current symbol, the ISI is composed by the postcursors of the past simbols as well as the precursors of the future ones.
Regards

Z

When transmitting this future symbols, how they interfere with current symbol? I mean, can the next symbols (or part of them) arrive with the current symbol because the channel is dispersive?

 

[zorro]

... can the next symbols (or part of them) arrive with the current symbol because the channel is dispersive?

Yes. The channel is causal but there is a delay to the peak of the response. Think for example in a relatively slow response (long rise time).
Z

 

[hunter555persia]

zorro is right. Channel causes different signal components to have different delays. So the short delay components of the current symbol are influenced by the long delay components of the next symbol.

 

[bisen_tejeshwari]

-N to N is nothing but 0 to 2N interval

 

[zorro]

-N to N is nothing but 0 to 2N interval

It is not the same.
Normally, when the adaptive equalizer starts, all the coefficients except c0 are 0.
Regards

Z

 

[hunter555persia]

EDIT for post #31: Channel causes different signal components to have different delays. So the LONG delay components of the current symbol are influenced by the SHORT delay components of the next symbol.