Optical power launched into the fiber at the transmitter end is 150 W. The power at the end of the 10 km length of the link working in the first windows is -38.2d Bm. Another system of the same length working in the second window is 47.5 W. The same length system working in the third window has 50% launched power. Calculate fiber attenuation for each case and mention the wavelength of operation.

Solution:

$ \begin{aligned} & P(0)=150 \mu \mathrm{W} \\\\ & z=10 \mathrm{~km} \\\\ & P(z)=-38.2 \mathrm{dBm} \Rightarrow\left\{\begin{array}{l}\\ -38.2=10 \log \frac{\mathrm{P}(\mathrm{z})}{1 \mathrm{~mW}} \\\\ \mathrm{P}(\mathrm{z})=0.151 \mu \mathrm{W} \\ \end{array}\right.\\ \end{aligned} $

$ \alpha=10 \times \frac{1}{z} \log \left[\frac{P(0)}{P(z)}\right\rfloor $

Attenuation in 1st window:

$ \alpha_1=10 \times \frac{1}{10} \log \left[\frac{150}{0.151}\right]\\ $

$ \alpha_1=2.99 \mathrm{~dB} / …