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.

4. Considering further Question 3. What is the effect of adding connectors of loss 0.23 dB each?

Answer 1

Answer :-1) For 25 km of cable length, the attenution will be-

= 0.27*25 dB

= 6.75 dB.

Answer :- 3) 1 mW is same as 0 dBm = -30 dB. -30 dBm = -30 -30 dB = -60 dB. Thus for a length of L, we can write-

-60 dB = -30 dB - 0.27*L

=> L = 30/0.27 = 111.11 km.

Answer :- 4) The connector loss wil reduce the total length. If two connectors have been used then we can write-

-60 = -30 - 0.27*L - (0.23*2)

=> L = 109.4 km.