# 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.

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

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. – Dilip Sarwate 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. – Tendero 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. – Alalalala 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? – Alalalala Apr 26 '18 at 21:57