When the mean optical power launched into an 8 km fiber length is 12 µW, the mean optical power at the fiber output is 3 µW. Determine, Overall signal attenuation in dB, The overall signal attenuation for a $10 \mathrm{~km}$ optical link using the same fiber with splices at $1 \mathrm{~km}$ intervals, each giving an attenuation of $1 \mathrm{~dB}$.

Solution:

Given:

z = 8 km

P(0) = 120 µW

P(z) = 3 µW

1) Overall attenuation is given by,

$ \begin{aligned} & \alpha=10 . \log \left[\frac{P(0)}{P(z)}\right] \\\\ & \alpha=10 . \log \left[\frac{120}{3}\right] \\\\ & \alpha=16.02 \mathrm{~dB}\\ \end{aligned} $

2) Overall attenuation for 10 km,

Attenuation per km,

$ \propto_{\mathrm{dB}}=\frac{16.02}{z}=\frac{16.02}{8}=2.00 \mathrm{~dB} / \mathrm{km} $

Attenuation in 10 km link = 2.00 x 10 = 20 dB

In the 10 km link, there will be 9 splices at 1 km intervals. Each splice introduces attenuation of 1 dB.

Total attenuation = 20 dB + 9 dB = 29 dB