# How to find the impulse response of filter that outputs the autocorrelation of the input?

I am trying to solve question 9.61 in the attached image.

I guess, by evaluating the autocorrelation for a specific instance, say $$t = 0$$ , that the impulse response, $$h(t)$$, should be : $$h (t) = x(-t)$$.

But can I get this from equating the convolution integral to the given output integral, i.e.

$$y(t) = \int_{-\infty}^{\infty} x(\tau)x(t + \tau) d\tau = \int_{-\infty}^{\infty} x(\tau)h(t - \tau) d\tau$$ .

Is it then right to equate as below?

$$x(t + \tau) = h (t - \tau)$$

Thank you!

• Read the answers to “Understanding the matched filter” on this website – Dilip Sarwate Jan 1 at 5:00

HINT:

Note that

$$\int_{-\infty}^{\infty}x(\tau)x(t+\tau)d\tau=\int_{-\infty}^{\infty}x(\tau-t)x(\tau)d\tau$$

• Thank you! I see the next step that needs to be taken to arrive at the solution, but why are the two expressions you listed equal to one another? Secondly, is it right to equate the two terms as indicated in my original question? – AJ_Kauchy Jan 1 at 15:29
• @AJ_Kauchy: The equality in my answer can be shown very easily using variable substitution. The last equation in your question is not correct, but using the equality in my answer you should be able to come up with the correct equation. – Matt L. Jan 1 at 15:37