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I have a wide-sense-stationary (WSS) process $\{x(t)\}$ and two linear filters with impulse functions $h_1$ and $h_2$.

Let $\delta(\omega)$ be the power spectrum of $\{x(t)\}$ and $$H_1:\omega\mapsto H_1(\omega)$$ and $$H_2:\omega\mapsto H_2(\omega)$$ the transfer functions of the filters. The outputs of the filters are denoted $$y_1(t)=(x \star h_1)(t)$$ and $$y_2(t)=(x \star h_2)(t),$$ where $\star$ denotes the convolution.

How can we compute the correlation of $\{y_1(t)\}$ and $\{y_2(t)\}$ and when are these two random variable uncorrelated?

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As an addition to Dilip's answer I'll show you how to derive that result:

$$\begin{align}R_{y_1,y_2}(\tau)&=E[y_1(t+\tau)y_2(t)]\\&=E\left[\int_{-\infty}^{\infty}x(\alpha)h_1(t+\tau-\alpha)d\alpha\int_{-\infty}^{\infty}x(\beta)h_2(t-\beta)d\beta\right]\\&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}E[x(\alpha)x(\beta)]h_1(t+\tau-\alpha)h_2(t-\beta)d\alpha d\beta\\&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}R_x(\alpha-\beta)h_1(t+\tau-\alpha)h_2(t-\beta)d\alpha d\beta\\&\stackrel{\gamma=\alpha-\beta}{=}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}R_x(\gamma)h_1(t+\tau-\alpha)h_2(t-\alpha+\gamma)d\alpha d\gamma\\&\stackrel{\zeta=\alpha-t}{=}\int_{-\infty}^{\infty}\underbrace{\int_{-\infty}^{\infty}R_x(\gamma)h_2(\gamma-\zeta)d\gamma}_{(R_x\star h_2^-)(\zeta)}\; h_1(\tau-\zeta) d\zeta\\&=(R_x\star h_1\star h_2^-)(\tau)\qquad\qquad\qquad (1)\end{align}$$

with $h_2^-(t)=h_2(-t)$.

The cross-spectral density $S_{y_1,y_2}(\omega)$ is the Fourier transform of the cross-correlation function:

$$S_{y_1,y_2}(j\omega)=S_x(j\omega)H_1(j\omega)H_2^*(j\omega)\tag{2}$$

where $S_x(j\omega)$ is the power spectral density of $x(t)$, and $H_1(j\omega)$ and $H_2(j\omega)$ are the frequency responses of the two filters.

Using $(2)$ it is straightforward to define a condition on $H_1(j\omega)$ and $H_2(j\omega)$ such that the cross-spectral density, and, consequently, the cross-correlation function become zero.

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  • $\begingroup$ Thanks! Yes from equation (1) it is straightforward to derive the conditions. $\endgroup$ – CyberRob Sep 3 '19 at 11:06
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There are a lot of misconceptions in the way that the problem has been posed (in particular, $X(\omega)$ as defined by the OP in the first version of his question -- he has since then deleted the definition -- ) has nothing to do with the matter), but when $\{x(t)\}$ is a wide-sense-stationary (WSS) process, then the processes $\{y_1(t)\}$ and $\{y_2(t)\}$ are jointly WSS and their crosscorrelation function is given by $$R_{y_1, y_2}(\tau) = E[y_1(t)y_2(t+\tau)] = R_x\star h_1\star \tilde{h}_2\big|_{\tau}$$ where $R_x$ denotes the autocorrelation function of $\{x(t)\}$, $\star$ denotes convolution, $h_1$ is the impulse response of one filter, while $\tilde{h}_2$ is the time-reversed impulse response of the other filter, that is, $\tilde{h}_2(t) = h_2(-t)$.

I repeat again that $X(\omega)$ as defined by the OP has nothing to do with the matter. In particular, $R_x$ is not the inverse Fourier transform of $|X(\omega)|$ or $|X(\omega)|^2$; these are random variables whereas $R_x$ is a deterministic function.

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  • $\begingroup$ Thanks for your fast answer and the remarks on 𝑋(πœ”). I will re-edit the question to correct the misconceptions. But how did you get to the equation above? So, by definition $𝑅_{𝑦_1,𝑦_2}(\tau)=𝐸[𝑦_1(𝑑)𝑦_2(𝑑+\tau)]=𝐸[(β„Ž_1\star π‘₯)(\tau)β‹…(β„Ž_2\star π‘₯)(t+\tau)]$. So why is this equal to $(𝑅_π‘₯\star β„Ž_1\star \tilde β„Ž _2)(\tau)$? $\endgroup$ – CyberRob Sep 2 '19 at 17:18

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