What is the Laplace Transform of the output power spectrum if the input signal is a white noise?

Question

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?

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:

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