Consider a bare step-index fiber with $n_1=1.46, n_2=1.0$ and $a=50$ $\mu \mathrm{m}$. Show that the pulse dispersion is about $2240 \mathrm{~ns} / \mathrm{km}$.

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written 17 months ago by

prajapatijaimin • 3.1k

• modified 17 months ago

optical fiber communication

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written 17 months ago by

prajapatijaimin • 3.1k

Solution:

The pulse dispersion in the case of step index fiber is given by,

$ \Delta T=\frac{n_1 L}{c}\left(\frac{n_1}{n_2}-1\right)=\frac{1.46 \times 10^3}{3 \times 10^8}(1.46-1)\\ $

$ =0.2238 \times 10^{-5}\\ $

$ =2238 \times 10^{-9} \mathrm{sec} $

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