Page 472 - Fiber Optic Communications Fund

Nonlinear Effects in Fibers 453

is calculated by averaging over the random phases. The FWM field acts as noise on the channel, leading
to performance degradations. The FWM impairment becomes smaller as the dispersion increases, since the
phase matching becomes more difficult, and it increases as the channel spacing decreases. Therefore, FWM
is one of the dominant impairments in OFDM systems in which the subcarriers are closely spaced [25, 26].
There are a number of approximations in this FWM model. Firstly, the modulation of the channels is ignored.
When the channel modulation is included, the signal field q is not independent of time and, therefore, the
n
second term on the left-hand side of Eq. (10.140) can not be ignored. Owing to dispersion, different channels
propagate at different speeds, which is not taken into account in this simple model. The FWM model may be
modified taking into account pump depletion [27], pump modulation [28–30], pump modulation and walk-off
[31–33]. Experimental validation of the CW FWM model can be found in Ref.[34].

Example 10.5
A WDM system consists of three channels centered at −Δf,0,and Δf, with Δf = 50 GHz. The fiber loss
coefficient = 0.046 km −1 and the fiber length L = 40 km. Calculate the efficiency of the non-degenerate
2
2
FWM tone at −2Δf if (a) =−4ps /km, (b) = 0ps /km. Ignore .
3
2
2
Solution:
From Eq. (10.90), we have
1 − exp (−L) 1 − exp (−0.046 × 40)
L eff = =
0.046
= 18.28 km. (10.233)
From Eq. (10.225), the FWM efficiency is

2
2
+ 4e −L sin (Δ jkln L∕2)∕L 2
= eff . (10.234)
jkln 2 2
+(Δ jkln )
Let j =−1, k = 0, and l = 1sothat n = j + k − l =−2 corresponding to the FWM tone at −2Δf.FromEq.
(10.216), we have
Δ = [Ω Ω −Ω Ω ]
jkln 2 l n j k
2
= (2Δf) [1 ⋅ (−2)−(−1) ⋅ 0]. (10.235)
2
2
(a) =−4ps /km:
2
9 2
Δ =−4 × 10 −27 ×(2 × 50 × 10 ) ×(−2) m −1
−101−2
−1
= 7.89 × 10 −4 m ; (10.236)
2
3 2
−3 2
(0.046 × 10 ) + 4exp (−0.046 × 40)sin (7.89 × 10 −1 × 40∕2)∕(18.28 × 10 )
−101−2 =
(0.046 × 10 ) +(7.89 × 10 )
−3 2
−4 2
−3
= 3.4 × 10 . (10.237)
2
(b) = 0ps /km:
2
−1
Δ −101−2 = 0m . (10.238)
Now, from Eq. (10.225), we have = 1.