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} / \mathrm{km}\\ $

Attenuation in 2nd window:

$ \begin{aligned} & \alpha_2=10 \times \frac{1}{10} \log \left[\frac{150}{47.5}\right]\\ \\ & \alpha_2=\mathbf{0 . 4 9} \mathbf{d B} / \mathrm{km}\\ \end{aligned}\\ $

Attenuation in 3rd window:

$ \begin{aligned} & \alpha_3=10 \times \frac{1}{10} \log \left[\frac{150}{75}\right]\\ \\ & \alpha_3=\mathbf{0 . 3 0} \mathbf{~ d B} / \mathbf{k m}\\ \end{aligned} $

The wavelength in 1st window is 850 nm.

The wavelength in 2nd window is 1300 nm.

The wavelength in 3rd window is 1550 nm.