given dispersion parameter, d = 22 ps/km-nm for a given fiber at wavelength of 1550 nm. What is the pulse spreading in ns/km due to material dispersion when led operating at 1550 nm having a spectral width of 75-nm is used to launch light into the system?
Answers
Answer:
Vp = c/n = 3 x 108/1.4504 = 2.068 x 108 m/s
ng = n - l (dn/dl ). From the graph dn/dl = -.01/um at l = 1 um.
Therefore, ng = 1.4605.
vg = c/ng = 2.054 x 108 m/s.
2- A glass fibre has material dispersion given by: |l^2 (d^2n/dl^2| =0.025. Determine the material dispersion parameter at l = 850 nm. A laser diode at l = 850 nm with a relative spectral width of 0.0012 l is used at the transmitter to launch a light pulse into the fibre . Estimate the root mean square pulse spreading per km.
Solution:
Material dispersion coefficient can be written as:
= 0.025/(3 x 108 x 850 nm) = 98.1 ps/nm km.
Since laser spectral width dl = 0.0012 l , therefore the material dispersion is = (98.1 ps/nm.km).(1.02 nm) = 100 ps/km.
3- (a) Discuss the effects of fibre dispersion and attenuation on digital communication and how they can be minimised or prevented.
Solution: See lecture notes
(b) Consider a fibre with a core and cladding refractive indices of 1.458 and 1.44, respectively. An optical pulse propagated through the fibre will be affected by the its dispersion property. We assume that both material and waveguide dispersion are very small. Calculate the pulse dispersion per kilometre for the maximum pulse repetition rate and bandwidth-length product for:
i) multimode step index fibre
ii) multimode graded index fibre.