Page 454 - Fiber Optic Communications Fund

Nonlinear Effects in Fibers 435

should not change as a function of Z and hence, we set k to be a constant. To solve Eq. (10.110), we multiply
Eq. (10.110) by dg∕dT and integrate from −∞ to T to obtain
T dg T d g dg T dg
2
−k g dT − 2 dT + g 3 dT = C, (10.111)
∫ 2 ∫ 2 ∫
−∞ dT −∞ dT dT −∞ dT
( ) 2 4
2 dg g
2
−kg − + = C, (10.112)
4 dT 4
where C is the constant of integration. To obtain Eq. (10.112), we have assumed that g(±∞) = 0. When
< 0, Eq. (10.112) can be rewritten as
2
[ 4 ] 1∕2
dg 2 2 g
= √ C + kg − (10.113)
dT − 2 4
or
g T
dg 2
∫ [ ] 1∕2 = √ ∫ dT, (10.114)
C + kg −
g 0 2 g 4 − 2 0
4
where g = g(0). Using the table of integrals [11], Eq. (10.114) can be solved to give
0
( )
T
g(T)= √ sech √ , (10.115)
− 2
√
where = 2k. Therefore, the total solution is
( )
T 2
q = √ sech √ exp (i Z∕2). (10.116)
−
2
The above solution represents a fundamental soliton that propagates without any change in pulse shape. It
acquires a phase shift due to propagation that is proportional to the square of the amplitude.

Example 10.3
2
The FWHM of a fundamental soliton is 50 ps. Fiber dispersion coefficient =−21 ps /km, and nonlinear
2
−1
coefficient = 1.1W −1 km . Calculate the peak power required to form a soliton. Ignore fiber loss.
Solution:
From Eq. (10.116), we have
( )
2 2 T
2
P(t)= |q| = sech √ . (10.117)
− 2
Let
2
= P peak , (10.118)

√
− 2
= T , (10.119)
0