mohatta2015.pdf

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