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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.
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written 17 months ago by
prajapatijaimin
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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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