B. P. Lathi, Zhi Ding - Modern Digital and Analog Communication Systems-Oxford University Press (2009)

10.6 Optimum Receiver for White Gaussian Noise Channels 557
Hence,
and d 2 = 0.4£. Moreover, for M = 16, each symbol carries the information of log2 16 = 4
bits. Hence, the energy per bit Eh is
and
Hence, for large Eb/N
Eb E 5d 2
N 4N SN
P e M =3Q()
=3Q () (10. 104)
A comparison of this with binary PSK [Eq. 10.33)] shows that 16-point QAM requires almost
2.5 times as much power as does binary PSK; but the rate of transmission is increased by a
factor of log2 M = 4. This comparison does not take into account the fact that P b , the BER, is
somewhat smaller than P e M.
In terms of receiver implementation, because N = 2 and M = 16, the receiver in
Fig. 10. 19 is preferable. Such a receiver is shown in Fig. 10.24c. Note that because all signals
are equiprobable,
Ei
a; =-- 2
PSK is a special case of QAM with all signal points lying on a circle. Hence, the same analytical
approach applies. However, the analysis may be more convenient if a polar coordinate
is selected. We use the following example to illustrate the two different approaches.
Example 10.3 MPSK
Determine the error probability of the optimum receiver for equiprobable MPSK signals, each
with energy E.
I
Figure 10.25a shows the MPSK signal configuration for M = 8. Because all the signals are
equiprobable, the decision regions are conical, as shown. The message m1 is transmitted
by a signal s1 (t) represented by the vector s 1 = (s1 , 0). If the projection in the signal
space of the received signal r is q = (q1 , q2), and the noise is n = (n1 , n2), then
.._,_,__.,,...
qi q2
q = (s1 + n1 , n2) = (✓ E + n1 , n2 )