A signal with average optical power of 120 W is launched into an 8 km length of fiber. The mean optical power at the fiber output is 3 W. Determine a) the overall signal attenuation in dB and the signal attenuation per kilometer (in dB/km) through the fiber, assuming there are no connectors or splices; b) the overall signal attenuation for a 10 km optical link using the same fiber with splices at 1km intervals, each of the 9 splices adding 0.5 dB of attenuation.

Given Information:

Input Power = Pin = 120 W

Output Power = Pout = 3 W

Length of fiber = 8 km and 10 km

Required Information:

a) Overall signal attenuation in dB and dB/km = ?

b) Overall signal attenuation in dB with 10 km and 9 splices adding 0.5 dB of attenuation  = ?

a)  α ≈ 16 dB and α = 2 dB/km

b)  α = 24.5 dB

Explanation:

a) Overall signal attenuation in dB

The signal attenuation is the log of output and input power ratio

α = 10*log(Pout/Pin)

α = 10*log(3/120)

α = 16.02 ≈ 16 dB

Divide overall attenuation by the entire length to get per km attenuation

α = 16/8 = 2 dB/km

b) Overall signal attenuation in dB with 10 km and 9 splices adding 0.5 dB of attenuation  

Since we know the rate of per km attenuation, we will multiply that with the entire length

2 dB/km*10 km = 20 dB

Now calculate the attenuation due to splices

There are 9 splices and each splice causes 0.5 dB attenuation per km so

9*0.5 dB = 4.5 dB

So the total attenuation is

α = 20 dB + 4.5 dB

α = 24.5 dB

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