1. A multimode step index fiber has a relative refractive index difference of 2% and core refractive index of 1.48. The maximum optical bandwidth that may be obtained with a particular source on a 5km link is 3Mhz. Determine the rms pulse broadening per kilometre resulting from chromatic dispersion mechanisms. Assuming waveguide dispersion to be negligible, estimate the rms spectral width of the source used, if the material dispersion parameter for the fiber at the operating wavelength is 80 ps nm-1 km-1.
2. A 1km length multimode step index fiber has a spectral width of 50nm with intermodal and intramodal dispersion of 10 ns/km and 100 ps/km respectively. Calculate the total dispersion and maximum bit rate.
(1) Relative refractive index ∆ = 2 % = 0.02
Core refractive index n1 = 1.48
Link length L = 5 km = 5000 m
Maximum optical bandwidth
Material dispersion parameter M = 80 ps nm-1 km-1
Spectral width due to waveguide dispersion
The expression of the rms pulse broadening due to chromatic dispersion τ:
NA is the numerical aperture, n1 is core refractive index
c is the velocity of light
The expression for numerical aperture NA:
∆ is the relative refractive index
The expression to calculate the rms pulse broadening per km
The expression to calculate spectral width of source:
Answer: 98.6 ns/km, 0.4 ns/km
(2) Total dispersion
Intermodal or modal dispersion
Intramodal or chromatic dispersion
Number of bits per second can be determined as
Answer: 11.18 ns, 100 Mbit/s km
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