0.12 Linear equalization  (Page 7/17)

where ${\overline{S}}_{\ell }$ is the $\ell$ th column of $\overline{S}$ . Thus, the set of these $J_{LS}^{\min ,\ell }$ are all along the diagonal of

Hence, the minimum value on the diagonal of $\Phi$ (e.g., at the $(j,j)$ th entry) corresponds to the optimum $\delta$ .

Consider the low-order example with $n=1$ (so $F$ has two parameters), $\alpha =2$ (so $\alpha >n$ ), and $p=5$ (so $p>n+\alpha$ ). Thus,

and

For the example, assume that the true channel is

A two-tap equalizer $F={\left[f_0{f}_1\right]}^T$ can provide perfect equalization for $\delta =1$ with $f_0=1/b$ , $f_1=-a/b$ , since

Consider

which results in

With $a=0.6$ , $b=1$ , and $r\left[1\right]=0.8$ ,

and

The effect of channel noise will be simulated by rounding these values for $r$ in composing

Thus, from [link] ,

and from [link] ,

Since the second diagonal term in $\Phi$ is the smallest diagonal term, $\delta =1$ is the optimum setting (as expected) and the second column of ${\overline{F}}^{†}$ is the minimum summed squared delayed recovery error solution(i.e.,  $f_0=0.9548$ ( $\approx 1/b=1$ ) and $f_1=-0.5884$ ( $\approx -a/b=-0.6$ )).

With a “better” received signal measurement, for instance,

the diagonal of $\Phi$ is $[1.3572,0.0000,0.1657]$ and the optimum delay is again $\delta =1$ , and the optimum equalizer settings are $0.9960$ and $-0.6009$ , which is a better fit to the ideal noise-free answer.Infinite precision in $\overline{R}$ (measured without channel noise or other interferers) produces a perfect fit to the “true” $f_0$ and $f_1$ and a zeroed delayed sourcerecovery error.

Summary of least-squares equalizer design

The steps of the linear FIR equalizer design strategy are as follows:

1. Select the order $n$ for the FIR equalizer in [link] .
2. Select maximum of candidate delays $\alpha$ ( $>n$ ) used in [link] and [link] .
3. Utilize set of $p$ training signal samples $\left\{s\right[1],s[2],...,s[p\left]\right\}$ with $p>n+\alpha$ .
4. Obtain corresponding set of $p$ received signal samples $\left\{r\right[1],r[2],...,r[p\left]\right\}$ .
5. Compose $\overline{S}$ in [link] .
6. Compose $\overline{R}$ in [link] .
7. Check if ${\overline{R}}^T\overline{R}$ has poor conditioning induced by any (near) zero eigenvalues.M atlab will return a warning (or an error) if the matrix is too close to singular. The condition number (= maximum eigenvalue / minimum eigenvalue)of ${\overline{R}}^T\overline{R}$ should be checked. If the condition number is extremely large,start over with different $\left\{s\right[·\left]\right\}$ . If all choices of $\left\{s\right[·\left]\right\}$ result in poorly conditioned ${\overline{R}}^T\overline{R}$ , then most likely the channel has deep nulls that prohibit the successful application ofa $T$ -spaced linear equalizer.
8. Compute ${\overline{F}}^{†}$ from [link] .
9. Compute $\Phi$ by substituting ${\overline{F}}^{†}$ into [link] , rewritten as
   $$\Phi ={\overline{S}}^T[\overline{S}-\overline{R}{\overline{F}}^{†}].$$
10. Find the minimum value on the diagonal of $\Phi$ . This index is $\delta +1$ . The associated diagonal element of $\Phi$ is the minimum achievable summed squared delayed source recovery error ${\sum }_{i=\alpha +1}^p{e}^2\left[i\right]$ over the available data record.
11. Extract the $(\delta +1)$ th column of the previously computed ${\overline{F}}^{†}$ . This is the impulse response of the optimum equalizer.
12. Test the design. Test it on synthetic data, and then on measured data (if available).If inadequate, repeat design, perhaps increasing $n$ or twiddling some other designer-selected quantity.