2
$\begingroup$

I have the following problem.

enter image description here

Am trying to understand how the channel block affects the input signal s(n). I know that x(n), the input signal to the filter is basically: \begin{equation} x\left(n\right)\:=\:s\left(n\right)* h\left(n\right)+v\left(n\right) \end{equation} However, the end goal is to find the Wiener filter tap weights: w0, w1, and w2.

So, in order to find that, I need to find the correlation matrix R of the filter tap inputs and the cross-correlation vector p relating the input of the filter to the output

My process to solving this problem was first by decomposing the channel block using partial fractions. As in the following:

\begin{equation} H\left(z\right)\:=\:\frac{1}{1-1.2z^{-1}+0.35z^{-2}}=\frac{1}{\left(1-0.5z^{-1}\right)\left(1-0.7z^{-1}\right)}=-\frac{2.5}{1-0.5z^{-1}}+\frac{3.5}{1-0.7z^{-1}} \end{equation} This in turn, using the inverse z transform, would turn into the impulse response: \begin{equation} h\left(n\right)\:=\:-2.5\left(0.5\right)^n+3.5\left(0.7\right)^n \end{equation} Am stuck at this point because I don't know what to do next given the concept of convolution stated above. The essence later is to find the correlation matrix R which should be in this form: \begin{equation} \left[R\right]\:=E\left[\begin{pmatrix}x\left(n\right)x\left(n\right)&x\left(n\right)x\left(n-1\right)&x\left(n\right)x\left(n-2\right)\\ \:x\left(n-1\right)x\left(n\right)&x\left(n-1\right)x\left(n-1\right)&x\left(n-1\right)x\left(n-2\right)\\ \:x\left(n-2\right)x\left(n\right)&x\left(n-2\right)x\left(n-1\right)&x\left(n-2\right)x\left(n-2\right)\end{pmatrix}\right]\: \end{equation} and also, to find the coss-correlation matrix p, which is in this form: \begin{equation} \left[p\right]\:=\:\begin{pmatrix}E\left(x\left(n\right)s\left(n\right)\right)\\ E\left(x\left(n-1\right)s\left(n\right)\right)\\ E\left(x\left(n-2\right)s\left(n\right)\right)\end{pmatrix} \end{equation} Please help me move on the right path on this problem. Thank you in advance.

$\endgroup$
  • $\begingroup$ given s(n) and y(n) the solution to this problem for determining the Wiener coefficients is detailed here: dsp.stackexchange.com/questions/31318/… $\endgroup$ – Dan Boschen Sep 13 at 21:20
  • $\begingroup$ Would you please post an answer using analysis? Am really stuck and confused on this. I tried to follow the example you provided but still very confused. The essence of this problem is to find the wiener filter taps w_i $\endgroup$ – Raykh Sep 19 at 22:39
  • $\begingroup$ This appears to be a homework or quiz problem, so we avoid providing the full solution for you. But I'll provide further hints as an answer that will hopefully help you-- show your work down that path and we can help see where you are stuck. $\endgroup$ – Dan Boschen Sep 20 at 14:25
2
$\begingroup$

This appears to be a homework problem so I will avoid providing the complete solution here, but the following should suggest a path toward a solution:

Typically this problem is done when the channel is NOT known and results in a least squared approach given the effects of the added noise $v(n)$. Solving for the equalizer coefficients when the channel is a known quantity results in a similar process to solving for inverse convolution using a matrix equation approach as is linked here: Compensating Loudspeaker frequency response in an audio signal with a convolution equation of the form:

$$h[n] * w[n] = \delta[0]$$

Where $h[n]$ represents the non-causal channel response and $w[n]$ represents the non-causal equalizer coefficients. I use "non-causal" to mean each is centered about $n=0$ which simplifies the equation to remove any delay in the result, as in $\delta[0]$ vs $\delta[n-m]$. The result in turn for the causal solution simply changes the indexing back such that the first sample starts at $n=0$ rather than the middle sample.

In basic English this is saying knowing $h[n]$, solve for $w[n]$ that would result in an impulse response when the two systems are cascaded, since the output and input of a system whose impulse response is an impulse would be identical.

This would result in $y[n] = s[n]$ when the noise is not added, and since the noise is white, this would also be the solution that any least squares approach using actual signals $s[n]$ and $x[n]$ would converge to. As constructed I do not see this as a least squares problem but simply an inverse convolution problem. This is similar to the use of least-squares versus zero-forcing equalizers and has similar construction: the least-squares solution converges to the zero-forcing solution as SNR increases.

In this case we solve for $w[n]$ using inverse convolution.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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