How to find the impulse response of filter that outputs the autocorrelation of the input? - Signal Processing Stack Exchange most recent 30 from dsp.stackexchange.com 2019-07-22T21:48:50Z https://dsp.stackexchange.com/feeds/question/54535 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://dsp.stackexchange.com/q/54535 0 How to find the impulse response of filter that outputs the autocorrelation of the input? AJ_Kauchy https://dsp.stackexchange.com/users/39807 2019-01-01T01:42:06Z 2019-01-01T10:55:52Z <p>I am trying to solve question 9.61 in the attached image. </p> <p><a href="https://i.stack.imgur.com/23WY7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/23WY7.jpg" alt="enter image description here"></a></p> <p>I guess, by evaluating the autocorrelation for a specific instance, say <span class="math-container">$t = 0$</span> , that the impulse response, <span class="math-container">$h(t)$</span>, should be : <span class="math-container">$$h (t) = x(-t)$$</span>.</p> <p>But can I get this from equating the convolution integral to the given output integral, i.e.</p> <p><span class="math-container">$$y(t) = \int_{-\infty}^{\infty} x(\tau)x(t + \tau) d\tau = \int_{-\infty}^{\infty} x(\tau)h(t - \tau) d\tau$$</span> .</p> <p>Is it then right to equate as below?</p> <p><span class="math-container">$$x(t + \tau) = h (t - \tau)$$</span></p> <p>Thank you!</p> https://dsp.stackexchange.com/questions/54535/-/54538#54538 1 Answer by Matt L. for How to find the impulse response of filter that outputs the autocorrelation of the input? Matt L. https://dsp.stackexchange.com/users/4298 2019-01-01T10:55:52Z 2019-01-01T10:55:52Z <p><strong>HINT:</strong></p> <p>Note that</p> <p><span class="math-container">$$\int_{-\infty}^{\infty}x(\tau)x(t+\tau)d\tau=\int_{-\infty}^{\infty}x(\tau-t)x(\tau)d\tau$$</span></p>