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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156 ADVANCED TOPICS<br />
Let's look at the Alice and Bob example. From Table 1.1, we see that there are<br />
three messages encoded with two binary symbols. The combination si = — 1 and<br />
«2 = +1 is not a valid message. Thus, with MLPD, we would not consider that<br />
combination when forming metrics. In Table 6.1, we would not look at rows 3 and<br />
7. It turns out that in this case, we wouldn't have selected rows 3 or 7 anyway,<br />
because their metrics are too large. However, in the general case, there are times<br />
when the noise would cause the metrics in rows 3 or 7 to be the best. By using<br />
knowledge that these rows cannot occur, we would avoid making a packet error at<br />
such times.<br />
In general, MLPD is fairly complex, as there are usually a lot of possible packet<br />
values. As a result, a lower-complexity iterative approach has been developed, in<br />
which equalization and decoding are performed more than once, with each process<br />
feeding information to the other. This approach is referred to as turbo equalization.<br />
7.2 MORE DETAILS<br />
7.2.1 MAP symbol detection<br />
Notice that when we compute each metric, there is one term in the summation that<br />
is larger than the rest. To reduce complexity, we can approximate the summation<br />
by its largest term, i.e.,<br />
L(s 2 = +1) « e ( - 4,l/lll0) = 0.8187 (7.5)<br />
L(s 2 = -1) « e ( - r,4/1(,l,) = 0.7261. (7.6)<br />
While the approximation is not that accurate in this example, it does lead to the<br />
same detected value. This approximation is referred to as the dual maxima approximation.<br />
The name comes from the fact that there are two (dual) summations, and<br />
we keep the maximum exponent (closest to zero) in each case. The approximation<br />
works best at high SNR (low σ 2 ).<br />
There is a numerical issue with MAPSD. At high SNR, σ 2 becomes small, making<br />
the exponents negative numbers with large magnitudes. This makes the result<br />
close to zero, which may end up being represented by zero in a machine, such as a<br />
calculator or computer. This is sometimes called the underflow problem.<br />
How do we solve this? If we scale both symbol metrics by the same positive<br />
number, we don't change which one is bigger. In the example above, what if we<br />
scaled each metric by exp(40/100), where we've used the notation exp(a:) = e x .<br />
This would cause the largest term in the summation to be 1 and all other terms to<br />
be less than one. This would guarantee that at least one term would not underflow,<br />
which is enough to determine a detected value with MAPSD.<br />
However, if we perform the scaling after computing the individual terms, it would<br />
be too late. So, we use the fact that<br />
exp(o) exp(6) = exp(o + b). (7.7)<br />
Since we will ultimately negate the sequence metric, we need to subtract 40 from<br />
each sequence metric before forming the exponentials. In general, we would find the