1. You have 25 km of optical cable with an attenuation coefficient of 0.27 dB/km. What...

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.

Solutions

Expert Solution

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.

genius_generous answered 1 year ago

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