# Cross-correlation of filtered random processes

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?

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.

• Thanks! Yes from equation (1) it is straightforward to derive the conditions. – CyberRob Sep 3 '19 at 11:06

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.

• Thanks for your fast answer and the remarks on 𝑋(𝜔). I will re-edit the question to correct the misconceptions. But how did you get to the equation above? So, by definition $𝑅_{𝑦_1,𝑦_2}(\tau)=𝐸[𝑦_1(𝑡)𝑦_2(𝑡+\tau)]=𝐸[(ℎ_1\star 𝑥)(\tau)⋅(ℎ_2\star 𝑥)(t+\tau)]$. So why is this equal to $(𝑅_𝑥\star ℎ_1\star \tilde ℎ _2)(\tau)$? – CyberRob Sep 2 '19 at 17:18