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A problem in Statistical and Adaptive Signal Processing (problem 10.15) presents two WSS signals both generated from zero-mean white Gaussian noise with $\sigma_w = 1$. They are described by $v_1(n) = 0.9v_1(n-1) + w(n)$ and $v_2(n) = -0.75v_2(n-1) + w(n)$.

The solution to the problem claims that the cross-correlation $r_{v_1v_2}(l) = h_1(l)*h_2(-l)*r_w(l)$, where $h_1$, $h_2$ are the impulse responses corresponding to the LTI systems generating $v_1$ and $v_2$. I know that for an individual LTI system with WSS input $x$ and output $y$, we have $r_{xy}(l) = h(-l)*r_x(l)$.

I can't find anything in the textbook explaining why we're justified in saying this for signals generated by two different systems.

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  • $\begingroup$ Remember that the Fourier Transform of the cross-correlation of two signals is the product of the Fourier Transforms of the two signals with one of the FTs are complex conjugated. That's maybe how you solve this thing. $\endgroup$ Apr 4 at 1:24

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You have to work through the calculation once and you'll remember forever. I hope you can fill in the details yourself. First, from the definition of cross-correlation (assuming real-valued filters and processes):

$$r_{v_1v_2}[l]=E\big\{v_1[k+l]v_2[k]\big\}\tag{1}$$

We know that

\begin{align*} v_1[k] &= \sum_{m=-\infty}^{\infty}w[m]h_1[k-m]\tag{2}\\ v_2[k] &= \sum_{m=-\infty}^{\infty}w[m]h_2[k-m]\tag{3} \end{align*}

Now you need to plug $(2)$ and $(3)$ into $(1)$, change the indices and swap expectation and summation such that you obtain an expression $E\big\{w[m+n]w[m]\big\}=r_w[n]$ inside the sums. This should get you at least very close to the final result. Edit your answer to show us your work and if necessary we can help you arrive at the final solution.

EDIT:

I just realized that in this answer I've written out the derivation for the continuous-time case. It's exactly the same in discrete time if you exchange integrals with sums.

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  • $\begingroup$ I really like your notational convention, Matt. $\endgroup$ Apr 4 at 16:18
  • $\begingroup$ @robertbristow-johnson: Thx Robert $\endgroup$
    – Matt L.
    Apr 6 at 10:53

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