Wiener Filter Additive Noise Uncorrelated

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

H(jω)=S_yx(ω)/S_xx(ω)

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 :(

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

$$x[n]=d[n]+v[n]\tag{1}$$

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:

$$R_{dv}[n]=E\{d^*[k]v[k+n]\}=0\tag{2}$$

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

\begin{align}R_{dx}[n]&=E\{d^*[k]x[k+n]\}\\&=E\{d^*[k]\left(d[k+n]+v[k+n]\right)\}\\&=E\{d^*[k]d[k+n]\}+E\{d^*[k]v[k+n]\}\\&=E\{d^*[k]d[k+n]\}=R_{d}[n]\tag{3}\end{align}

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

$$S_{dx}(\omega)=S_{d}(\omega)\tag{4}$$

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

$$H(\omega)=\frac{S_{dx}(\omega)}{S_x(\omega)}=\frac{S_d(\omega)}{S_d(\omega)+S_v(\omega)}\tag{5}$$

• 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... – Adrian Guerra Mar 4 '18 at 11:15
• @AdrianGuerra: It totally depends on the application, but in many cases the noise or the signals (such as speech) have zero mean. – Matt L. Mar 4 '18 at 11:34