Question

1. You have 25 km of optical cable with an attenuation coefficient of 0.27 dB/km. What is the total attenuation due to the cable?

3. Considering further question 1. If the power transmitted is 1 mW and the receiver sensitivity is -30 dBm what is the maximum distance of optical fiber that can be used before amplification is needed? You can assume only cable losses.

Answer 1

a) The attenuation inside the optical fibre follows the law:

Where I= intensity at any point 'x' and is initial intensity, i.e. at x=0.

Attenuation coefficient

So, attenuation coefficient in normal units can be written as:

So, the total attenuation due to cable will be:

Thus the ratio of incident intensity (or power) to the intensity (or power) at output is 25.684.

b) Initial power=1mW

Receiver sensitivity refers to the minimum power that can be received, given as: -30dBm

So, minimum power received:

The length for which this amount of power can be transmitted:

This is the maximum distance up to which power can be transmitted without amplification.