0.12 Linear equalization  (Page 6/17)

The best equalizer is the one corresponding to a delay of 2, since this Jmin is the smallest. In this case, however, any of the last three open the eye.Observe that the number of errors (as reported in err ) is zero when the eye is open.

There is a way to convert the exhaustive search over all the delays $\delta$ in the previous approach into a single matrix operation.Construct the $(p-\alpha )\times (\alpha +1)$ matrix of training data

where $\alpha$ specifies the number of delays $\delta$ that will be searched, from $\delta =0$ to $\delta =\alpha$ . The $(p-\alpha )\times (n+1)$ matrix of received data is

where each column corresponds to one of the possible delays. Note that $\alpha >n$ is required to keep the lowest index of $r[·]$ positive. In the $(n+1)\times (\alpha +1)$ matrix

each column is a set of equalizer parameters, one corresponding to each of the possible delays. The strategy is to use $\overline{S}$ and $\overline{R}$ to find $\overline{F}$ . The column of $\overline{F}$ that results in the smallest value of the cost $J_{LS}$ is then the optimal receiver at the optimal delay.

The $j$ th column of $\overline{F}$ , corresponds tothe equalizer parameter vector choice for $\delta =j-1$ . The product of $\overline{R}$ with this $j$ th column from $\overline{F}$ is intended to approximate the $j$ th column of $\overline{S}$ . The least-squares solution of $\overline{S}\approx \overline{R}\overline{F}$ is

where the number of columns of $\overline{R}$ (i.e.,  $n+1$ ) must be less than or equal to the number of rows of $\overline{R}$ (i.e.,  $p-\alpha$ ) for ${\left({\overline{R}}^T\overline{R}\right)}^{-1}$ to exist. Consequently, $p-\alpha \ge n+1$ implies that $p>n+\alpha$ . If so, the minimum value associated with a particularcolumn of ${\overline{F}}^{†}$ (e.g.,  ${\overline{F}}_{\ell }^{†}$ ) is, from [link] ,