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

Question

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.

Answer 1

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.

Answer 2

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