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I am given the following problem: I have a filter with impulse response $h(t) = e^{-10t}, t \leq 0$, and autocorrelation function of the input signal, which is WDS and Gaussian with median equal to 0, of: $R_X(\tau) = \frac{\sin 20 \tau}{\tau}$. Amongst other things, I am asked to find the probability that the rate of change of the output is greater than 1V/s, but I am getting confused as of what to do here. I thought of using a relation like: $\frac{dY(t)}{dt} = X(t)$ and compute the impulse response using this, but I get lost. Am I losing something? Any help would be greatly appreciated.

EDIT: I tried using the above fact and I end up with $|H'(w)|^2 = \frac{1}{\omega^2}$, and thus I find that the power is the following integral of: $\sigma^2 = P = \int S_{y'} = \int |H'(\omega)|^2 S_y(f) = \int \frac{1}{\omega^2} S_y(f)df$, and I have found the $S_y = \frac{\pi}{2\pi} (\Pi(\omega + 20) - \Pi(\omega - 20))$. Is something like this correct?

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  • $\begingroup$ sure it's $t\le0$? That makes the impulse response explode! $\endgroup$
    – Marcus Müller
    Commented Jun 24, 2022 at 16:20
  • $\begingroup$ Positive! It's not a mistake, the problem stated $t \leq 0$. $\endgroup$
    – average_discrete_math_enjoyer
    Commented Jun 24, 2022 at 16:22
  • $\begingroup$ What is WDS? ${}{}$ $\endgroup$
    – Dilip Sarwate
    Commented Jun 24, 2022 at 17:56
  • $\begingroup$ Hm, your system produces infinite output for any nonzero input at some time $t$. Far as I can tell, the derivative of this is also unbounded and hence your probability is = 1 $\endgroup$
    – Marcus Müller
    Commented Jun 25, 2022 at 1:02

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