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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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

$2240 \mathrm{~ns} / \mathrm{km}$ .

written 2.3 years ago by

prajapatijaimin • 3.3k

• modified 2.3 years ago

optical fiber communication

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written 2.3 years ago by

prajapatijaimin • 3.3k

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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