# LTI filtering for wide-sense stationary process

Why is it that if $U[n]$ is wide-sense stationary and it is convolved with $h[n]$ to produce $W[n]$, the autocorrelation becomes $R_{WW}[n] = R_{UU}[n]*h[n]*h[-n]$?

I know that in general $R_{WW}[n_{1},n_{2}]=R_{UU}[n_{1},n_{2}]*h[n_{1}]*h[n_{2}]$ and that wide-sense stationary means $m_{U}[n] = m_{U}$ along with $R_{UU}[n_{1},n_{2}]=R_{UU}[n_{1}-n_{2},0]$, but I can't get to the above relation from these facts.

• The expression for Rww[n] results from directly applying the definition of autocorrelation to an LTI system. Please give a reference to how you obtained Rxx[n1,n2]. Feb 4 '17 at 16:52
• I went from $R_{WW}[n_{1},n_{2}]=E[W[n_{1}]W[n_{2}]]$ then used linearity of expectation to get that $R_{WW}[n_{1},n_{2}]=R_{UU}[n_{1},n_{2}]*h[n_{1}]*h[n_{2}]$ Feb 4 '17 at 16:57
• You must have a wrong sign somewhere. Check for example this deduction (pg. 4). The operations for discrete time and continuous time are very similar. Feb 4 '17 at 17:03

\begin{align}R_{WW}[n]&=E\{W^*[k]W[k+n]\}\\ &=E\left\{\sum_mU^*[k-m]h^*[m]\sum_lU[k+n-l]h[l]\right\}\\ &=\sum_m\sum_lh^*[m]h[l]E\left\{U^*[k-m]U[k+n-l]\right\}\\ &=\sum_m\sum_lh^*[m]h[l]R_{UU}[n+m-l]\\ &=\sum_mh^*[m]\sum_lh[l]R_{UU}[n+m-l]\\ &=\sum_mh^*[m]\left(h\star R_{UU}\right)[n+m]\\ &=\sum_mh^*[-m]\left(h\star R_{UU}\right)[n-m]\\ &=h^*[-n]\star h[n]\star R_{UU}[n]\end{align}
where $*$ denotes complex conjugation, and $\star$ denotes convolution.
• too bad they didn't use the convention of lower case for "time-domain" functions leaving the upper case for "frequency-domain". using whatever notational convention remaining for the Fourier transform they might see that $$\mathscr{F}\{R_{WW}[n]\} = |H(e^{j\omega})|^2 \mathscr{F}\{R_{UU}[n]\}$$ where $$H(e^{j\omega}) = \mathscr{F}\{h[n]\}$$ Feb 4 '17 at 21:49
• Thanks for your answer. I'm actually a bit confused about the starting point. Where does the complex conjugate come from? I know that $R_{WW}[n_{1},n_{2}]=E[W[n_{1}]W[n_{2}]]$, but haven't seen $R_{WW}[n]=E[W^{*}[k]W[k+n]]$ Feb 4 '17 at 22:21