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@w00zie: Hi: I'm putting this here just because I find it easier to write in here. This is not an adequate answer from a DSP point of view (and I would like to see one also) but let me just it explain more clearly since I babbled above and it's still not that clear. Take the more standard AR(1): $y_{t} = a y_{t-1} + \epsilon_{t}$. ( denote this as first ...


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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-\...


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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 $\{y_2(t)\}$ ...


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