4
$\begingroup$

Background: I have a variety of BPSK which uses a raised cosine transmitted pulse shape:

$$ h(t) = {1 \over 2}\, (1+\cos(\pi t))\, \Pi(t/2) $$

After matched filtering in the receiver, the end result has ISI. Decimated to 1 sample per symbol, the pulse shape works out to [1/6, 1, 1/6].

So, I might filter and decimate this signal to 1 sample per symbol, then apply an equalizing filter to mitigate that ISI.

My question: modulo the potential consequences to computational complexity, is this any different from combining the matched filtering and equalization into a single filter?

I know filtering by two successive FIR filters is equivalent to one filter with the impulse response of the two convolved together. But with the decimation step between, I'm less sure.

$\endgroup$
1
$\begingroup$

Ignoring the higher computational costs, it's possible to do what you want. Let's consider the case without ISI, where the channel response $C(f)=1$. You transmit the sequence $$s(t)=\sum_{k=0}^N a_k h(t-kT),$$ where $h(t)$ is your raised cosine pulse shape and $T$ is the pulse rate. Ignoring noise, the output of the matched filter at times $t_k=kT$ is the data sequence $a_k$.

If the channel response $C(f)$ is time invariant but not flat, then the received signal (before the matched filter) is $$r(t) = s(t) \ast c(t),$$ where $c(t)$ is the channel impulse response. Now you need to design a matched filter + equalizer combination filter $g(t)$ such that $r(t) \ast g(t)$, sampled at times $t_k$, is the data sequence $a_k$.

Note that all I have done is simply to delay the decimation until after the combination filter. You say that

I know filtering by two successive FIR filters is equivalent to one filter with the impulse response of the two convolved together.

This is true when the filters are time-invariant. A decimator is time-variant, and it can't be trivially combined with other LTI filters.

$\endgroup$
  • $\begingroup$ can't be trivially combined – but if we'd do a Polyphase decomp of the decimation filter, pull it over to the low-rate side of the decimation, then do the same decomp on the second filter, shouldn't we be able to merge them? $\endgroup$ – Marcus Müller Jul 12 '17 at 13:51
  • $\begingroup$ @MarcusMüller I think it should be possible, but I've never tried it myself. Harder to do might be to make it adaptive (as an equalizer should be). $\endgroup$ – MBaz Jul 12 '17 at 14:40
  • $\begingroup$ To be clear, in my case, the channel is C(f) = 1, but there's still ISI because h(t)*h(t) isn't a nyquist pulse. So, the output of the matched filter isn't the data sequence. So I'm not looking for an adaptive equalizer: I'm looking for an equalizer to counteract the known ISI introduced by the modem designer's poor selection of transmit filtering. In any case, I'm still confused since you say "all I have done is simply to delay the decimation until after the combination filter" but then "A decimator is time-variant, and it can't be trivially combined with other LTI filters." $\endgroup$ – Phil Frost Jul 12 '17 at 19:02
  • $\begingroup$ What I meant with that comment is: say you have a set of filters $h_i(t)$ connected serially. If they're all LTI, the overall impulse response is the convolution $h_1(t) \ast h_2(t) \ldots$. However, if one of the filters is a decimator, you can no longer find the overall response in that way. In that sense, it can't be easily "combined" with other LTI filters into an "overall" filter. $\endgroup$ – MBaz Jul 12 '17 at 19:38
0
$\begingroup$

I think that the cascaded equivalent will be $$ H_1(z) H_2(z^M) $$ Where $M$ is the decimation factor.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.