mohatta2015.pdf

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The first term is the signal term and has average power
S = w 2 100. (4.21)
For the MMSE solution, w 2 = -0.0397 and S is 0.16. Notice that in (4.20), the
coefficient in front of s 2 is — Ww 2 — 0.397 φ 1. Thus, the MMSE solution gives a
biased estimate of the symbol.
The remaining terms are interference and noise, which we call impairment. The
impairment has power
I + N = (ωι(-10)+ω 2 9) 2 + ω 2 81 + (ω 2 + ω 2 )100. (4.22)
Notice that this is a general expression, for any weight values. Substituting the
MMSE weights gives an SINR of 0.657, as expected.
Like the ZF DFE, if we only use r\ and r 2 to detect s 2 we will not do as well
as MF at low SNR, as MF would also collect the copy of s 2 in r 3 . With partial
ZF linear equalization, we avoided r;j to avoid ISI from future symbols. With the
MMSE strategy, we don't need to avoid Γ3 or other future received samples. So,
we should use as many future received values as we can. As you might suspect, at
low SNR, MMSE linear equalization behaves like matched filtering, as the weights
tend to be a scaled version of the MF weights.
4.2.2 Maximum SINR solution
We have seen that error performance, at least when noise is the only impairment, is
related to output SINR. Thus, another reasonable strategy is to maximize output
SINR. To do this, we need to minimize the sum of the ISI and noise powers.
Consider using r\ and r 2 to detect s 2 . Like the ZF solution, we can weight r 2
by -1/10, so that
z 2 = wm - O.lr-2, (4.23)
where w\ is the weight for r\ to be optimized. Substituting the models for n and'
r 2 into (4.23), we can model z 2 using (4.6), except now
«2 = (wi(-10) - 0.9)si + wi9s 0 + «Ί"ι - 0.1n 2 . (4.24)
Now we need to find w\.
From (4.6), notice that the signal power is 1, independent of w\. Thus, to
maximize SINR, we simply need to minimize the power in u 2 , denoted U 2 . This
power is given by
U 2 = (ωι (-10) - 0.9) 2 + 81w 2 + (ω 2 +0.01)100
= 281ω 2 + 18w x + 1.81. (4.25)
This is plotted in Fig. 4.4. Observe that it is minimized at Wi = —0.032028, which
results in U 2 = 1.5217. Thus, the SINR is 1/1.5217 or 0.657. By comparison, the
partial ZF solution has a higher U 2 = 2.4661, which gives a lower SINR of 0.406.
Observe that the unity-gain maximum SINR approach (SINR = 0.657) performs
the same as the MMSE approach described earlier (SINR = 0.657). Coincidence?
No. Let's revisit the MMSE weight solution. If we divide each weight by 10|wi |, we
get the same solution as the unity-gain max SINR solution. As scaling the weights
doesn't affect SINR, we discover that the MMSE solution also maximizes SINR!