Page 461 - Fiber Optic Communications Fund

442 Fiber Optic Communications

Substituting Eq. (10.151) in Eq. (10.150), we find
( )
A p p ( )
i − A p = −i − A 2 p A . (10.152)
p
Z ′ Z ′ 2
Separating the real and imaginary parts, we obtain
′
′
A (T , Z )
p ′ ′
=− A (T , Z ) (10.153)
p
Z ′ 2
or
′
′
′
′
A (T , Z )= A (T , 0)e −Z ∕2 , (10.154)
p
p
′
′
(T , Z ) ′
p
′
2
= A (T , 0)e −Z . (10.155)
p
Z ′
′
Integrating Eq. (10.155) from 0 to Z , we find
Z ′ ′
′
′
2
′
′
(T , Z )− (T , 0)= A (T , 0) ∫ e −Z dZ ′
p
p
p
0
′
2
′
= A (T , 0)L (, Z ) (10.156)
p EFF
where
1 − e −x
L (, x)= . (10.157)
EFF

Substituting Eqs. (10.154) and (10.156) in Eq. (10.151), we find
′
2
′
′
′
′
′
′
e
q (T , Z )= A (T , 0)e −Z ∕2 i[ p (T ,0)+A (T ,0)L EFF (,Z )]
p
p p
′
′
2
′
′
= q (T , 0)e iA (T ,0)L EFF (,Z )−Z ∕2. (10.158)
p
p
Using Eqs. (10.148) and (10.149), Eq. (10.158) can be rewritten as
2
′
′
e
q (T, Z) ≡ q (T , Z )= q (T − d Z, 0)e −Z∕2 i|q p (T−d p Z,0)| L EFF (,Z). (10.159)
p
p
p
p
Similarly, Eq. (10.145) can be solved by setting
q = A e i s . (10.160)
s s
Substituting Eq. (10.160) in Eq. (10.145) and proceeding as before, we find
A (T, Z)= A (T, 0)e −Z∕2, (10.161)
s
s
d
s 2
p
= 2|q (T, Z)|
dZ
2 −Z,
= 2|q (T − d Z, 0)| e (10.162)
p
p
Z
2 −Z
(T, Z)= (T, 0)+ 2 ∫ |q (T − d Z, 0)| e dZ, (10.163)
p
s
p
s
0
q (T, Z)= q (T, 0)e −Z∕2+i XPM (T,Z), (10.164)
s
s