# Channel equalization affect on input signal

I have the following problem. 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.

• given s(n) and y(n) the solution to this problem for determining the Wiener coefficients is detailed here: dsp.stackexchange.com/questions/31318/… – Dan Boschen Sep 13 '20 at 21:20
• 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 – JordenSH Sep 19 '20 at 22:39
• 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. – Dan Boschen Sep 20 '20 at 14:25

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:

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

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 $$\delta$$ 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.