Page 200 - Fiber Optic Communications Fund

Optical Modulators and Modulation Schemes 181

where m is an integer, m ∈[−M∕2, M∕2 − 1], and M is even. Assume that the occurrence of any of these
symbols is equally probable. Find the mean optical power.
Solution:
Since the M symbols could occur with equal probability, the mean power is
M∕2−1
P 0 ∑
2
P ̄ out = (2m + 1) . (4.118)
M
m=−M∕2
Let n = m + M∕2 + 1. Now, Eq. (4.118) becomes
M
0
̄
P = P ∑ [2n −(M + 1)] 2
out
M
n=1
M
P ∑ 2 2
0
= [4n +(M + 1) − 4n(M + 1)]. (4.119)
M
n=1
Using the following relations,
M
∑ M(M + 1)
n = ,
2
n=1
M
∑ 2 M(M + 1)(2M + 1)
n = .
6
n=1
Eq. (4.119) is simplified as
[ 2 ]
P 0 4M(M + 1)(2M + 1) 2 4M(M + 1)
̄
P = + M(M + 1) −
out
M 6 2
2
P (M − 1)
0
= . (4.120)
3
2
Note that the mean power scales as M for M ≫ 1.
Example 4.4
Repeat Example 4.3 for an M-ary rectangular QAM signal.

Solution:
Let X and Y be the number of levels of in-phase and quadrature components with M = XY. The QAM signal
in a symbol interval 0 < t < T may be written as (see Eq. (4.113))
s
A (t)= m (t) cos (2f t)+ m (t) sin (2f t), (4.121)
I
c
Q
out
c