I have been trying to understand for a while now why a non-causal Wiener filter that has a frequency response of the type


where y(n) = x(n) + n(n), x(n) the input signal, n(n) the noise and both uncorrelated,

can be simplified to

H(jw) = S_xx(jw) / ( S_xx (jw) + S_nn (jw) )

I understand the denominator but why is it that when x and n are uncorrelated

S_yx(jw) = S_xx(jw)

i.e. I don't understand Step 3 in this question but I don't have enough reputation to ask for a clarification/add a comment :(


1 Answer 1


Let me change the notation slightly to make things clear. Let $d[n]$ be the desired signal, $x[n]$ is the noisy input signal, and $v[n]$ is the noise. Furthermore, we assume that


and that the desired signal $d[n]$ and the noise $v[n]$ are uncorrelated and at least one of them has zero mean, which implies that their cross-correlation equals zero:


From $(2)$ it follows that the cross-correlation between the desired signal $d[n]$ and the input signal $x[n]$ is given by


From $(3)$ it follows that the cross-power spectral density of $d[n]$ and $x[n]$ equals the power spectral density of $d[n]$:


With the given assumptions, the frequency response of the non-causal Wiener filter can be written as


  • $\begingroup$ Great, thank you so much! I think "at least one of them has zero mean" was the part I was missing... How come we can make such an assumption? I feel like this isn't specified usually... $\endgroup$ Mar 4, 2018 at 11:15
  • $\begingroup$ @AdrianGuerra: It totally depends on the application, but in many cases the noise or the signals (such as speech) have zero mean. $\endgroup$
    – Matt L.
    Mar 4, 2018 at 11:34

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