Page 258 - Fiber Optic Communications Fund

Optical Receivers 239

Solution:

P LO = 10 mW,
−1 1
i3∕4
s(t)= e = √ + i √ ,
2 2
P = 0.1585 mW,
r
√
P = P P = 1.259 mW.
0
r LO
From Eq. (5.138), we find the in-phase component of the current as
−1
I = RP Re [s(t)] = 0.9 × 1.259 × √ mA =−0.8012 mA.
I 0
2
From Eq. (5.139), we find

1
I =−RP Im [s(t)] = −0.9 × 1.259 × √ mA =−0.8012 mA.
Q 0
2

5.6.5 Polarization Effects

So far we have ignored the state of polarization of light. In this section, we consider the optical signal modu-
lated in orthogonal polarizations. Let the transmitted signal be
[ ]
q T,x
q = , (5.144)
T
q T,y
where
A T
q T,x = √ s (t) exp (−i t), (5.145)
c
x
2
A T
q T,y = √ s (t) exp (−i t), (5.146)
y
c
2
s (t) and s (t) are the data in x- and y-polarizations, respectively. In fiber, the propagation constants of x-
x y
and y-polarization components are slightly different due to a possible asymmetry in fiber geometry and, as
a result, these components acquire different amounts of phase shift. Besides, because of the perturbations
during the propagation, there is a power transfer between the x- and y-components. These effects can be taken
into account by a channel matrix (see Chapter 2):
[ ]
M xx M xy
M = . (5.147)
M yx M yy
The output of the fiber-optic link may be written as
[ ]
q r,x
q = = Mq , (5.148)
r q r,y T
q r,x =[M q + M q ]e i c , (5.149)
xy T,y
xx T,x
q =[M q + M q ]e i c , (5.150)
r,y yx T,x yy T,y