A continuous 12 km long optical fiber link has a 1.5 dB/km loss. What is the minimum visual power level that must be launched into the fiber to maintain an optical power level of $0.3 \mu \mathrm{W}$ at the receiving end? What is the required input power if the fiber has a loss of $2.5 \mathrm{~dB} / \mathrm{km}$?

Solution:

Given:

z = 12 km= 1.5 dB/km

P(0) = 0.3 µW

i) Attenuation in optical fiber is given by,

$ \begin{aligned}\\ & \alpha=10 \times \frac{1}{\mathrm{z}} \log \left(\frac{\mathrm{P}(0)}{\mathrm{P}(\mathrm{z})}\right) \\\\ & 1.5=10 \times \frac{1}{12} \log \left(\frac{0.3 \mu \mathrm{W}}{\mathrm{P}(\mathrm{z})}\right) \\\\ & \log \left(\frac{0.3 \mu \mathrm{W}}{\mathrm{P}(\mathrm{z})}\right)=\frac{1.5}{0.833}\\ \end{aligned} $

$ \begin{aligned} & \left(\frac{0.3 \mu \mathrm{W}}{\mathrm{P}(\mathrm{z})}\right)=10^{1.8} \\\\ & \mathrm{P}(\mathrm{z})=\left(\frac{0.3 \mu \mathrm{W}}{10^{1.8}}\right)=\frac{0.3}{63.0} \\\\ & \mathrm{P}(\mathrm{z})=4.76 \times 10^{-9} \mathrm{~W}\\ \end{aligned}\\ $

Optical power output= $4.76 \times 10^{-9} \mathrm{~W}$

ii) Input power:

α = 2.5 dB/km

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

$ \begin{gathered}\\ 2.5=10 \times \frac{1}{\mathrm{z}} \log \left(\frac{\mathrm{P}(0)}{4.76 \times 10^{-9}}\right) \\\\ \log \left(\frac{\mathrm{P}(0)}{4.76 \times 10^{-9}}\right)=\frac{2.5}{0.833}=3 \\\\ \frac{\mathrm{P}(0)}{4.76 \times 10^{-9}}=10^3=1000 \\\\ \mathrm{P}(0)=4.76 \mu \mathrm{W}\\ \end{gathered} $

Input power= 4.76 µW