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Working on my communication class right now, and I'm having a bit of a problem with the matched filters.

As I see it the matched filters "eliminate" some of the noise. We project the received signal on our orthonormal base and everything that can't be generated by it is considered to be noise.

Now let X(t) be our only signal we send, time-limited to [0;T]. The matched filter at the receiver's end would have an impulse response of h(t) = X(T-t). After our received signal R(t) passes through this filter we sample at ts=T and we get our Y(t).

After all that we can make our final decision based on Y(t) and the expected X(t), to see if the transmitter sent X(t) or zero.

Now my question is why do we use a shifted time-reversed function, instead of just X(t)? Someone told me that it's because having h(t)=X(t) is too costly in the real world on the hardware level. But I'm not really sure it's this. Or maybe I'm completely wrong and and with h(t)=X(t) it would work or be less effective.

Thank in advance for any answers.

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    $\begingroup$ The short answer is because you want to implement a correlation, which is equivalent to convolution (i.e., filtering) with a time reverse of the function you want to correlate with. Since the time reverse of the transmitted pulse is non-causal, you shift it to the right to make it causal, hence $X(T-t)$ instead of $X(-t)$. $\endgroup$ – Matt L. Jun 19 '15 at 16:02
  • $\begingroup$ Also note that, in most (if not all) practical systems, $X(t)$ is symmetric, so the matched filter's response is indeed $X(t)$ (since $X(t)=X(T-t)$). $\endgroup$ – MBaz Jun 19 '15 at 16:09
  • $\begingroup$ Try reading this answer to understand why time-reversal is necessary. And avoid talking anymore (if possible) with the person who told you that time-reversal is too costly in the real world on the hardware level. $\endgroup$ – Dilip Sarwate Jun 19 '15 at 18:46
  • $\begingroup$ The convolutional process is done to maximize signal to noise ratio - in other word solving for the function maximize signal to noise ratio in presence of white noise it turns out the you need to convolve the expected signal with the input signal. Typical implementations in the frequency domain are just a vector multiplication for both type of processes - not an issue of cost! $\endgroup$ – Moti Jun 21 '15 at 6:06

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