# First order Wiener–Hopf filter design

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

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

• Is this homework? Can you show your attempt at solving this? – Atul Ingle Jan 3 '17 at 16:43
• 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. – Haoming Li Jan 3 '17 at 22:27
• @HaomingLi , your question should have the homework tag. – Gilles Jan 3 '17 at 22:49
• -1 for removing the homework tag. – msm Jan 3 '17 at 22:59
• Sorry about that, i am fresh here and i update it now, my solution above is correct or not plz guide me 0.0 – 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

$$R_{xd}[n]=E[d[n'](d[n'+n]+v[n'+n])]=R_d[n]$$

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.