0

3.9kviews

For a 30 km long fiber attenuation 0.8 dB /km at 1300 nm. If a 200-watt power is launched into the fiber, find the output power.

written 2.3 years ago by prajapatijaimin • 3.3k

•   modified 2.3 years ago

optical fiber communication

ADD COMMENT FOLLOW SHARE EDIT

1 Answer

0

1.3kviews

written 2.3 years ago by prajapatijaimin • 3.3k

Solution:

z = 30 km

P(0) = 200 µW

Attenuation in optical fiber is given by,

$\begin{aligned} 0.8=10 \times \frac{1}{30} \log \left[\frac{200 \mu \mathrm{W}}{\mathrm{P}(\mathrm{z})}\right] \\\\ 2.4=10 \times \log \left[\frac{200 \mu \mathrm{W}}{\mathrm{P}(\mathrm{z})}\right] \end{aligned}$

ADD COMMENT SHARE EDIT