# Expectation and Auto-correlation of an Output from a Matched Filter. [duplicate] [see the attached image]

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.

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 $S(t)$):

$$\mathbb{E}[Y(t)]=S(t)*S(-t)$$

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.

• Ah, but finding the autocorrelation function of the output of the matched filter is the hard part. Apr 26 '18 at 14:32
• @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. Apr 26 '18 at 14:37
• 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. Apr 26 '18 at 21:17
• When I calculated the autocorrelation, the result came out to be [Rs(t)]^2, is this so? Or did I drop some cross-terms? Apr 26 '18 at 21:57