Let us consider a random wide-sense stationary process $n(t)$, which passes through a filter $h(t, \tau)$. Its autocorrelation function is $$R_{n^\prime}(t)=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} R_{n}(t-\tau+\tau^\prime) h(\tau) h(\tau^\prime) d \tau d \tau^\prime$$ For white noise input, $R_{n}(t)=\delta(t)$ and $$R_{n^\prime}(t)=\int_{-\infty}^{\infty} h(\tau-t) h(\tau) d \tau $$ The power spectrum of the output is $$S_{n^\prime}(s)=\int_{-\infty}^{\infty} R_{n^\prime}(t) e^{-s t} d t=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(\tau-t) h(\tau) e^{-s t} d t d \tau$$ If we integrate with respect to $t$: \begin{equation} \int_{-\infty}^{\infty} h(\tau-t) e^{-s t} d t=-e^{-s \tau} \int_{-\infty}^{\infty} h(\gamma) e^{s \gamma} d \gamma=-e^{-s \tau} H(-s) \end{equation} and then with respect to $\tau$, we obtain $$S_{n^\prime}(s)=-H(-s)\int_{-\infty}^{\infty} h(\tau) e^{-s \tau} d \tau=-H(-s)H(s)$$ I think that the minus sign in the last equation is a mistake and that the true formula is: $$S_{n^\prime}(s)=H(-s)H(s)$$ Can anyone tell me if my steps are correct, please?


1 Answer 1


The minus sign is indeed a mistake; you probably forgot that the integration limits also change sign, and thus compensate the minus sign:

$$\begin{align}\int_{-\infty}^{\infty}h(\tau-t)e^{-st}dt\Big|_{\gamma=\tau-t}&=-e^{-s\tau}\int_{\infty}^{-\infty}h(\gamma)e^{\gamma s}d\gamma\\&=e^{-s\tau}\int_{-\infty}^{\infty}h(\gamma)e^{\gamma s}d\gamma\\&=e^{-s\tau}H(-s)\end{align}$$

  • $\begingroup$ Aaargh! I am really heedless, I forgot. Thank you! $\endgroup$
    – Mark
    Jul 9, 2022 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.