mohatta2015.pdf

74 LINEAR EQUALIZATION
What if we could use r-.ι, as well? There is a copy of s 2 that would be helpful.
However, there is an even larger copy of s,¡. With the zero-forcing strategy, we
found it best not to use r,-¡ because of the larger copy of s;j. However, with the
MMSE strategy, it turns out that r :i is useful. We will explore this more in the
next chapter.
4.2 MORE DETAILS
So far we have explored partial zero-forcing and MMSE. For partial zero-forcing, a
better approach would be to minimize ISI, accounting for ISI not canceled. This
approach is explored in the Problems section. As for the MMSE solution, here we
provide more details. Then a maximum SINR solution is developed. It is shown
that the MMSE solution also maximizes SINR. Results are then generalized for the
dispersive and MIMO scenarios.
4.2.1 Minimum mean-square error solution
Consider the MMSE solution. In (4.16), we found that the MSE for the dispersive
channel example is given by
E 2 = (w 2 (-10) - l) 2 + (w 2 (9) + wi(-10)) 2 + ( Wl 9) 2 + {w\ + w|)100. (4.17)
We used trial-and-error plots to find the weights that minimized MSE.
There is another way to find such weights. In the plots, we were looking for the
minimum of the MSE. At the minimum, the instantaneous slope of the curve is
zero. From differential calculus, we know that the instantaneous slope is given by
the derivative. Thus, we can take the derivative of E 2 with respect to each weight
and set the derivatives to zero.
For this particular example, it will help to recall the following facts from differential
calculus.
1. The derivative of ax 2 + bx + c with respect to (w.r.t.) x is 2ax + b.
2. The derivative of (ax + d) 2 w.r.t. x is 2a(ax + d).
3. When we take the derivative w.r.t. one variable, we treat the other variable
as a constant.
Using these facts, we can take the derivative of E 2 w.r.t. w\ and ω 2 and set the
results to zero. This gives
0 = 0 + 2(-10)(u) 2 (9)+wi(-10)) + 2(u> 1 9)+2(w 1 )100 (4.18)
0 = 2(-10)(w 2 (-10) - 1) + 2(9)(w 2 (9) + wi(-10)) + 0 + 2(w 2 )100. (4.19)
Solving these two equations gives w\ = —0.0127 and w 2 = —0.0397. Substituting
these results into the MSE equations gives an MSE of 0.603.
As for computing SINR, substituting (4.13) into (4.14) gives
2 2 = W 2 (-10)S2 + (Wl(—10) + W 2 9)si + Wl9S() + Wim + W2H2- (4.20)