490
12 Nonlinear Processes in Optical Fibers
Fig. 12.9 Two optical waves at frequencies ν 1 and ν 2 mix to generate two third-order sidebands
M =
N
2
2
(N − 1)
(12.12)
If the channels are equally spaced, a number of the new waves will have the same
frequencies as the injected signals. Thus the resultant crosstalk interference plus the
depletion of the original signal waves can severely degrade multichannel system
performance unless steps are taken to diminish it.
Example 12.3 Consider the case of three DWDM signals propagating in a fiber at
optical frequencies ν 1 , ν 2 , and ν 3 . What frequency components result from four-wave
mixing?
Solution: From Eq. (12.12) for N = 3 channels there are M = 9 mixing products
generated. These are as follows:
ν 123 = ν 1 + ν 2 − ν 3 (identical to ν 213 = ν 2 + ν 1 − ν 3 )
ν 321 = ν 3 + ν 2 − ν 1 (identical to ν 231 = ν 2 + ν 3 − ν 1 )
ν 312 = ν 3 + ν 1 − ν 2 (identical to ν 132 = ν 1 + ν 3 − ν 2 )
ν 112 = 2ν 1 − ν 2
ν 113 = 2ν 1 − ν 3
ν 221 = 2ν 2 − ν 1
ν 223 = 2ν 2 − ν 3
ν 331 = 2ν 3 − ν 1
12 Nonlinear Processes in Optical Fibers
Fig. 12.9 Two optical waves at frequencies ν 1 and ν 2 mix to generate two third-order sidebands
M =
N
2
2
(N − 1)
(12.12)
If the channels are equally spaced, a number of the new waves will have the same
frequencies as the injected signals. Thus the resultant crosstalk interference plus the
depletion of the original signal waves can severely degrade multichannel system
performance unless steps are taken to diminish it.
Example 12.3 Consider the case of three DWDM signals propagating in a fiber at
optical frequencies ν 1 , ν 2 , and ν 3 . What frequency components result from four-wave
mixing?
Solution: From Eq. (12.12) for N = 3 channels there are M = 9 mixing products
generated. These are as follows:
ν 123 = ν 1 + ν 2 − ν 3 (identical to ν 213 = ν 2 + ν 1 − ν 3 )
ν 321 = ν 3 + ν 2 − ν 1 (identical to ν 231 = ν 2 + ν 3 − ν 1 )
ν 312 = ν 3 + ν 1 − ν 2 (identical to ν 132 = ν 1 + ν 3 − ν 2 )
ν 112 = 2ν 1 − ν 2
ν 113 = 2ν 1 − ν 3
ν 221 = 2ν 2 − ν 1
ν 223 = 2ν 2 − ν 3
ν 331 = 2ν 3 − ν 1
