# Besides computational costs, is filtering, decimating, then filtering different from just filtering?

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.

## 2 Answers

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.

• 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? – Marcus Müller Jul 12 '17 at 13:51
• @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). – MBaz Jul 12 '17 at 14:40
• 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." – Phil Frost Jul 12 '17 at 19:02
• 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. – MBaz Jul 12 '17 at 19:38

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