Consider a random process with auto-correlation function: $$r_{\rm dd} [k] = \beta^{\lvert k \rvert}\quad\text{where}\quad 0 < \beta < 1. $$ Suppose also that the observation is: $$ x[n] = d[n] + v[n] $$ where $v[n]$ is uncorrelated white noise with variance $\sigma^2$.

Design a first order Wiener–Hopf filter to reduce the noise in $x[n]$ of the form

$$W(z) = w(0) + w(1)z^{-1}$$

  • $\begingroup$ Is this homework? Can you show your attempt at solving this? $\endgroup$
    – Atul Ingle
    Jan 3 '17 at 16:43
  • $\begingroup$ the w=R^(-1) * p , which R is the correlation matrix of X[n] and p is the cross-correlation vector between d[n] and X[n]. i should compute the inverse of R and p to solve the function to get the W. is that right? it is my opinion. $\endgroup$
    – Haoming Li
    Jan 3 '17 at 22:27
  • $\begingroup$ @HaomingLi , your question should have the homework tag. $\endgroup$
    – Gilles
    Jan 3 '17 at 22:49
  • $\begingroup$ -1 for removing the homework tag. $\endgroup$
    – msm
    Jan 3 '17 at 22:59
  • $\begingroup$ Sorry about that, i am fresh here and i update it now, my solution above is correct or not plz guide me 0.0 $\endgroup$
    – Haoming Li
    Jan 3 '17 at 23:20

I refer to the notation from the Wikipedia article.

Your received signal is $x[n]$, and its Autocorrelation is given by $R_x[n]=R_d[n]+R_v[n]$ when noise and signal are uncorrelated. Hence,

$$ R_x[n]=\beta^{|k|}+\sigma^2\delta[n] $$

The cross-correlation between the received signal and the signal of interest is


under the assumption again that signal and noise are uncorrelated.

Now, the Wiener-Hopf equation gets you

$$ \begin{pmatrix}R_x[0] & R_x[1]\\ R_x[1] &R_x[0]\end{pmatrix}\begin{pmatrix}w[0]\\w[1]\end{pmatrix}=\begin{pmatrix}R_{xd}[0]\\R_{xd}[1]\end{pmatrix} $$

Filling in the variables, we get

$$ \begin{pmatrix}1+\sigma^2 & \beta\\ \beta &1+\sigma^2\end{pmatrix}\begin{pmatrix}w[0]\\w[1]\end{pmatrix}=\begin{pmatrix}1\\\beta\end{pmatrix} $$

Now, you just need to solve for $w[0],w[1]$ to get your filter coefficients.


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