mohatta2015.pdf
signal processing from power amplifier operation control point of view
signal processing from power amplifier operation control point of view
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MORE DETAILS 75<br />
The first term is the signal term and has average power<br />
S = w 2 100. (4.21)<br />
For the MMSE solution, w 2 = -0.0397 and S is 0.16. Notice that in (4.20), the<br />
coefficient in front of s 2 is — Ww 2 — 0.397 φ 1. Thus, the MMSE solution gives a<br />
biased estimate of the symbol.<br />
The remaining terms are interference and noise, which we call impairment. The<br />
impairment has power<br />
I + N = (ωι(-10)+ω 2 9) 2 + ω 2 81 + (ω 2 + ω 2 )100. (4.22)<br />
Notice that this is a general expression, for any weight values. Substituting the<br />
MMSE weights gives an SINR of 0.657, as expected.<br />
Like the ZF DFE, if we only use r\ and r 2 to detect s 2 we will not do as well<br />
as MF at low SNR, as MF would also collect the copy of s 2 in r 3 . With partial<br />
ZF linear equalization, we avoided r;j to avoid ISI from future symbols. With the<br />
MMSE strategy, we don't need to avoid Γ3 or other future received samples. So,<br />
we should use as many future received values as we can. As you might suspect, at<br />
low SNR, MMSE linear equalization behaves like matched filtering, as the weights<br />
tend to be a scaled version of the MF weights.<br />
4.2.2 Maximum SINR solution<br />
We have seen that error performance, at least when noise is the only impairment, is<br />
related to output SINR. Thus, another reasonable strategy is to maximize output<br />
SINR. To do this, we need to minimize the sum of the ISI and noise powers.<br />
Consider using r\ and r 2 to detect s 2 . Like the ZF solution, we can weight r 2<br />
by -1/10, so that<br />
z 2 = wm - O.lr-2, (4.23)<br />
where w\ is the weight for r\ to be optimized. Substituting the models for n and'<br />
r 2 into (4.23), we can model z 2 using (4.6), except now<br />
«2 = (wi(-10) - 0.9)si + wi9s 0 + «Ί"ι - 0.1n 2 . (4.24)<br />
Now we need to find w\.<br />
From (4.6), notice that the signal power is 1, independent of w\. Thus, to<br />
maximize SINR, we simply need to minimize the power in u 2 , denoted U 2 . This<br />
power is given by<br />
U 2 = (ωι (-10) - 0.9) 2 + 81w 2 + (ω 2 +0.01)100<br />
= 281ω 2 + 18w x + 1.81. (4.25)<br />
This is plotted in Fig. 4.4. Observe that it is minimized at Wi = —0.032028, which<br />
results in U 2 = 1.5217. Thus, the SINR is 1/1.5217 or 0.657. By comparison, the<br />
partial ZF solution has a higher U 2 = 2.4661, which gives a lower SINR of 0.406.<br />
Observe that the unity-gain maximum SINR approach (SINR = 0.657) performs<br />
the same as the MMSE approach described earlier (SINR = 0.657). Coincidence?<br />
No. Let's revisit the MMSE weight solution. If we divide each weight by 10|wi |, we<br />
get the same solution as the unity-gain max SINR solution. As scaling the weights<br />
doesn't affect SINR, we discover that the MMSE solution also maximizes SINR!