Page 503 - Fiber Optic Communications Fund

484 Fiber Optic Communications

Eq. (10.446) is simplified as
3 3 3 3 3 3 2 2 3 2 2
{Ω +Ω −Ω −[Ω +Ω + 3Ω Ω + 3Ω Ω −Ω − 3(Ω +Ω ) Ω + 3(Ω +Ω )Ω ]}
k
k
j
l
j
j
k
6 j k l j k j k l l
3
=− {3Ω Ω (Ω +Ω )+ 3(Ω +Ω )Ω (Ω −Ω −Ω )}
k
j
l
j
k
l
j
k
k
j
6
3
=− (Ω +Ω ){Ω Ω −Ω Ω }. (10.448)
l
n
k
j
j
k
2
Adding the contributions due to and , Eq. (10.216) is modified as
3 2
[ ]
3
Δ =[Ω Ω −Ω Ω ] + (Ω +Ω ) . (10.449)
jkln l n j k 2 j k
2
Example 10.10
A WDM system has five channels centered at lΔf, l =−2, −1, 0, 1, 2, Δf = 50 GHz. The launch power
per channel is 3 dBm, and the channels are CW. The transmission fiber has the following parameters:
−1
2
−1
= 0.046 km , L = 20 km, =−4ps /km, and = 1.8W −1 km . The initial phases of the channels are
2
−2 = 0.5 rad, −1 =−0.7 rad, = 1.2 rad, = 0.8 rad, and =−1 rad. Find the FWM power on the
2
1
0
central channel (l = 0). Ignore .
3
Solution:
The FWM tones falling on the central channel should satisfy the condition
(j + k − l)Δf = nΔf = 0. (10.450)
The possible triplets are shown in Table 10.2. Here, ND and D refer to non-degenerate and degenerate FWM
tones, respectively. From Eq. (10.216), we see that Δ jkln is invariant under the exchange of j and k, i.e.,
Δ =Δ . (10.451)
jkln kjln
Hence, the FWM tones corresponding to {j, k, l} and {k, j, l} should be identical. For example, the triplets
{−2, 1, −1} and {1, −2, −1} produce identical FWM tones. So, we need to consider only the tones listed in
Table 10.3. First consider the triplet {−2, 1, −1}. The FWM field for this triplet is given by Eq. (10.221),
L (1 − e − −2,1,−1,0 L )
(−2,1,−1,0) 3∕2 − +iΔ −2,1,−1
(L)= iP e 2 , (10.452)
0 −2,1,−1,0
−2,1,−1,0 = − iΔ −2,1,−1,0
= + i Ω Ω 1
2
−2
−3 −27 9 2 −1
= 0.046 × 10 − i4 × 10 ×(2 × 50 × 10 ) ×(−2) m
−1
−4
= 4.6 × 10 −5 + 7.89 × 10 im , (10.453)