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I recognize the filter in the problem is a matched filter. But I don't understand how this filter actually work, and I am not quite sure about how to calculate the expectations and the auto-correlation in this case. Any help is greatly appropriated.


marked as duplicate by Dilip Sarwate, MBaz, A_A, lennon310, Stanley Pawlukiewicz May 4 '18 at 2:01

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First you should find $Y(t)$. In this case, it can be calculated as:

$$Y(t)=X(t)*h(t)=\int_{-\infty}^{\infty}h(t)X(t-\tau)\ \mathrm{d}\tau$$

But you have more information about $X(t)$ and $h(t)$:

$$Y(t)=\int_{-\infty}^{\infty}S(-t)[S(t-\tau)+N(t-\tau)]\ \mathrm{d}\tau$$

If we assume that the integral converges (which makes sense, because if not $Y(t)$ would not exist), then we can do the following:

$$\begin{align} \mathbb{E}[Y(t)]&= \mathbb{E}\left[\int_{-\infty}^{\infty}S(-t)[S(t-\tau)+N(t-\tau)]\ \mathrm{d}\tau\right]\\ &=\int_{-\infty}^{\infty}\mathbb{E}\left[S(-t)S(t-\tau)+S(-t)N(t-\tau)\right]\ \mathrm{d}\tau \end{align} $$

But $S(t)$ is deterministic, so:

$$\mathbb{E}[Y(t)]= \int_{-\infty}^{\infty}S(-t)S(t-\tau)+S(-t)\mathbb{E}[N(t-\tau)]\ \mathrm{d}\tau$$

The problem states that the mean of the noise process is zero, thus:

$$\mathbb{E}[Y(t)]= \int_{-\infty}^{\infty}S(-t)S(t-\tau)\ \mathrm{d}\tau$$

So, the expectation of $Y(t)$ is the convolution between $S(t)$ and its reversed version (also known as autocorrelation of $S(t)$):


Because $S(t)$ is known, then $\mathbb{E}[Y(t)]$ is, too.

To find the autocorrelation, you have to do something similar to this procedure. I'll leave that to you.

  • $\begingroup$ Ah, but finding the autocorrelation function of the output of the matched filter is the hard part. $\endgroup$ – Dilip Sarwate Apr 26 '18 at 14:32
  • $\begingroup$ @DilipSarwate I may have washed my hands of the matter, but given that the OP stated that he had no idea how to begin, I supposed that showing how to find the expectation would be a good start for the OP to come around a solution for the autocorrelation, or at least show an attempt by his own. $\endgroup$ – Tendero Apr 26 '18 at 14:37
  • $\begingroup$ Thanks for the answer, Tendero. And Dilip, very informative link you provided. So the expected output of the filter is just the correlation of the signal, very interesting result. $\endgroup$ – Alalalala Apr 26 '18 at 21:17
  • $\begingroup$ When I calculated the autocorrelation, the result came out to be [Rs(t)]^2, is this so? Or did I drop some cross-terms? $\endgroup$ – Alalalala Apr 26 '18 at 21:57

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