I am optimizing a filter $f$ using mean squared error criterion.

$$MSE = f^T*(R_{xx}-R_{yx}^T*R_{yy}^{-1}*R_{yx})*f \tag{1}$$


  • $x\in \mathbb{R}^n$ input column vector
  • $y\in \mathbb{R}^m$ output column vector and $y=Hx+AWGN$. $H\in \mathbb{R}^{m\times n}$ is ISI channel response convolution matrix.
  • $R_{xx}=E[x^{}x^T]$, $R_{yx}=E[y^{}x^T]$, $R_{yy}=E[y^{}y^T]$

I know that $f$ can be found by imposing some meaningful constraints on (1).

My question is about finding the right $n$ for estimating $R_{xx}, R_{yx}, R_{yy}$. I know that as I increase $n$ (the dimension of my input vector $x$), these estimates improve. Therefore, if I fix $m$ (dimension of $y$) and plot $MSE$ on (y-axis) vs $n$ on (x-axis), I have a decreasing curve from left to right.

What I don't know is how to find the right value of $n$ when memory of ISI channel is $\mu$.

  • $\begingroup$ are you running this filter online or offline? $\endgroup$ – ruoho ruotsi Dec 14 '16 at 17:24
  • $\begingroup$ I am running this filter offline. Therefore my filter is not adaptive. $\endgroup$ – NAASI Dec 14 '16 at 17:34
  • $\begingroup$ what is $f$ in your equations and from which model do you start deriving your MMSE estimator? As you say, MSE becomes less, the higher $n$ is. Then what is the "right" value for you? $\endgroup$ – Maximilian Matthé Dec 14 '16 at 19:06
  • $\begingroup$ $f$ is the filter that I am optimizing. As you see in (1), if I add unit energy constraint, then $f$ turns out to be the eigen vector corresponding to the smallest eigen value of matrix $(R_{xx}-R_{yx}^{T}*R_{yy}^{-1}*R_{yx})$. So finding $f$ is not a problem here. The estimate of these correlation matrices is dependent on $n$. The longer the $n$, the better the estimate. Better the estimate, smaller the MSE. $\endgroup$ – NAASI Dec 14 '16 at 19:11
  • $\begingroup$ What I am asking is if there is some way of deciding $n$ so that for $n^{'}>n$, change in MSE is not greater than $\epsilon$ $\endgroup$ – NAASI Dec 14 '16 at 19:13

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