Page 502 - Fiber Optic Communications Fund

Nonlinear Effects in Fibers 483

Eqs. (10.439) and (10.440) represent the evolution of the complex amplitudes of the pump and the Stokes
waves. To obtain the expression for the energy exchange between the pump and the Stokes waves, we multiply
∗
Eq. (10.439) by A and subtract its complex conjugate to obtain
p
dP
p
=−gP P − P , (10.441)
p
p s
dZ
2
where P = |A | and g = 2Ω. Similar operations on Eq. (10.440) leads to
p
p
dP
s
= gP P − P . (10.442)
p s
s
dZ
From Eqs. (10.441) and (10.442), we see that the gain coefficients at the pump and the Stokes frequencies are
identical, but they are not the same in Eqs. (10.406) and (10.407). This is because of our linear approximation
(first term in the Taylor expansion) to the Raman time response function.
SRS has a number of applications. If an intense Raman pump is launched to the fiber, it can amplify a weak
signal if the frequency difference lies within the bandwidth of the Raman gain spectrum. SRS can also be used
to construct Raman fiber lasers which can be tuned over a wide frequency range (∼10 THz) [84–88]. In other
types of fiber amplifiers, the SRS process can be detrimental since the pump energy is used to amplify the
range of wavelengths over which amplification is not desired. In WDM systems, a channel of higher frequency
transfers energy to a channel of lower frequency, leading to Raman cross-talk and performance degradations
[89–91].

10.12 Additional Examples

Example 10.9

When the bandwidth of the WDM signal and/or the dispersion slope are large, can not be ignored. In this
3
case, show that the phase mismatch factor given by Eq. (10.216) should be modified as
[ ]
3
Δ =(Ω Ω −Ω Ω ) + (Ω +Ω ) . (10.443)
jkln l n j k 2 j k
2
Solution:
2 2 3 3
= + Ω + Ω + Ω , (10.444)
1
j
j
0
j
j
2 6
Ω =Ω +Ω −Ω . (10.445)
n j k l
Consider the contribution to the phase mismatch Δ due to the last term of Eq. (10.444),
jkln
3 3 3 3 3 3 3 3 3 3
[Ω +Ω −Ω −Ω ]= [Ω +Ω −Ω −(Ω +Ω −Ω ) ]. (10.446)
j
l
n
k
6 j k l 6 j k l
Using the formula
3
3
3
2
2
3
2
2
(a + b + c) = a + b + 3a b + 3ab + c + 3(a + b) c + 3(a + b)c , (10.447)