Digital Signal Processing Reference
In-Depth Information
Assume we have a 2-bit PAM system, and let σ s = 1. The average error
probability calculated using Eq. (11.46) is then 0 . 0236 . If we equalize these
errors using a unitary matrix, then the resulting error vector is
y =0 . 0222[11111]
and the average error probability computed using Eq.
(11.55) becomes
0 . 0020 . The gain in using the unitary matrix is therefore
G
=0 . 0236 / 0 . 0020 = 11 . 8 .
Thus the use of a unitary matrix to equalize the error components can be
very powerful. Several examples of minimum error probability systems will
be presented in Chaps. 17 and 18.
11.5.2.A How large should the SNR be?
For QAM symbols, since A is as in (11.42), the convexity condition y k <A 2 / 3
canbewrittenas
σ s
(2 b
y k <
(11 . 56)
1)
that is, the SNR σ s /y k at the input of the k th detector should be
SNR k = σ s
y k (2 b
1) .
(11 . 57)
Notice that as the number of bits b is increased the threshold (2 b
1) also
increases. For example if b = 6 (64-QAM) then
SNR k
63 ,
(11 . 58)
which is about 18 dB. The reader may wonder whether this is an unusually high
SNR. Observe however that this is the SNR at the detector input . For 64-QAM
this SNR corresponds to a symbol error probability of 0.14. Assuming we have
a Gray coded system, the bit error probabilty is approximately 0 . 14 / 6=0 . 023.
Except in situations where this error is acceptable, we can assume that the
SNR is large enough to justify the convexity requirement on
P e ( k ) . As another
example, let b = 2 (4-QAM, or QPSK). The requirement now is
SNR k
3
(11 . 59)
or about 4 . 77 dB, which corresponds to a symbol error probability of 0 . 0816 .
Again, in situations where the error probabilties are smaller than this, the con-
vexity condition is satisfied.
Summary . For any chosen set of multipliers
in Fig. 11.7(b)
satisfying the zero-forcing condition, the unitary U that minimizes the average
error probability is any unitary matrix such that
{
α k }
and
{
β k }
=1 / M.
|
U km |
 
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