0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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]. $\endgroup$
    – Juancho
    Commented Feb 4, 2017 at 16:52
  • $\begingroup$ I went from $R_{WW}[n_{1},n_{2}]=E[W[n_{1}]W[n_{2}]]$ then used linearity of expectation to get that $R_{WW}[n_{1},n_{2}]=R_{UU}[n_{1},n_{2}]*h[n_{1}]*h[n_{2}]$ $\endgroup$
    – Austin
    Commented Feb 4, 2017 at 16:57
  • $\begingroup$ 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. $\endgroup$
    – Juancho
    Commented Feb 4, 2017 at 17:03

1 Answer 1

4
$\begingroup$

It's basically just about using the definitions and doing the math:

$$\begin{align}R_{WW}[n]&=E\{W^*[k]W[k+n]\}\\ &=E\left\{\sum_mU^*[k-m]h^*[m]\sum_lU[k+n-l]h[l]\right\}\\ &=\sum_m\sum_lh^*[m]h[l]E\left\{U^*[k-m]U[k+n-l]\right\}\\ &=\sum_m\sum_lh^*[m]h[l]R_{UU}[n+m-l]\\ &=\sum_mh^*[m]\sum_lh[l]R_{UU}[n+m-l]\\ &=\sum_mh^*[m]\left(h\star R_{UU}\right)[n+m]\\ &=\sum_mh^*[-m]\left(h\star R_{UU}\right)[n-m]\\ &=h^*[-n]\star h[n]\star R_{UU}[n]\end{align}$$

where $*$ denotes complex conjugation, and $\star$ denotes convolution.

$\endgroup$
5
  • 2
    $\begingroup$ 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 $$ H(e^{j\omega}) = \mathscr{F}\{h[n]\} $$ $\endgroup$ Commented Feb 4, 2017 at 21:49
  • 1
    $\begingroup$ @robertbristow-johnson: It's an eternal dilemma: in many textbooks random processes have upper case letters to distinguish them from deterministic signals. $\endgroup$
    – Matt L.
    Commented Feb 4, 2017 at 22:03
  • $\begingroup$ 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 $R_{WW}[n]=E[W^{*}[k]W[k+n]]$ $\endgroup$
    – Austin
    Commented Feb 4, 2017 at 22:21
  • $\begingroup$ i like the textbooks that use a boldface non-italics for the random process. problem is that notation can be confused with a vector. $\endgroup$ Commented Feb 4, 2017 at 22:57
  • $\begingroup$ @Jake: The definition with the complex conjugate is the general definition for complex-valued processes. You can ignore it for real-valued processes. $\endgroup$
    – Matt L.
    Commented Feb 5, 2017 at 12:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.