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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 23 months ago by prajapatijaimin • 3.2k

•   modified 23 months ago

optical fiber communication

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written 23 months ago by prajapatijaimin • 3.2k

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} $

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