Chinese researchers achieved a new milestone in fiber-optic technology, achieving ultra-fast data transmission over long distances without signal repeaters.
And I actually need to correct myself. When I woke up this night I suddenly remembered that I did the one thing that dBm is not great for in the example, and did it wrong.
If you want to add up the power of five lasers you cannot add up their individual dBm numbers! Because that is already an addition operation when you do it on the absolute value scale! So adding them in the log scale would be like multiplying their powers, which makes no physical sense.
So, again, the correct way:
5 lasers of 4 mW or 6 dBm each, added up, actually comes out to 20 mW and then convert 10 * log10(20mW/1mW) = 13 dBm.
Or alternatively we convert the factor of 5 into dB first, 10 * log10(5) = 7 dB, and the add that one to the 6 dBm to also arrive at 13 dBm.
Then the rest of the example goes on like this:
13 dBm - 12 dB + 15 dB = 16 dBm into the fiber
16 dBm - 30 dB = -14 dBm at the end of the fiber
-14 dBm + 20 dB - 12 dB = -6 dBm after the WSS
and now split it for the 5 receivers, by
either subtracting the log of the factor 5 from above, -6 dBm - 7 dBm = -13 dBm per receiver, which is 10^(-13/10) = 0.05 mW.
or converting -6 dBm to 10^(-6/10) = 0.25 mW and then divide that absolute power by 5 to also arrive at 0.05 mW per receiver.
That was very interesting to read and it perfectly explained my question. Thank you very much.
My pleasure!
And I actually need to correct myself. When I woke up this night I suddenly remembered that I did the one thing that dBm is not great for in the example, and did it wrong.
If you want to add up the power of five lasers you cannot add up their individual dBm numbers! Because that is already an addition operation when you do it on the absolute value scale! So adding them in the log scale would be like multiplying their powers, which makes no physical sense.
So, again, the correct way:
5 lasers of 4 mW or 6 dBm each, added up, actually comes out to 20 mW and then convert 10 * log10(20mW/1mW) = 13 dBm.
Or alternatively we convert the factor of 5 into dB first, 10 * log10(5) = 7 dB, and the add that one to the 6 dBm to also arrive at 13 dBm.
Then the rest of the example goes on like this:
13 dBm - 12 dB + 15 dB = 16 dBm into the fiber
16 dBm - 30 dB = -14 dBm at the end of the fiber
-14 dBm + 20 dB - 12 dB = -6 dBm after the WSS
and now split it for the 5 receivers, by
either subtracting the log of the factor 5 from above, -6 dBm - 7 dBm = -13 dBm per receiver, which is 10^(-13/10) = 0.05 mW.
or converting -6 dBm to 10^(-6/10) = 0.25 mW and then divide that absolute power by 5 to also arrive at 0.05 mW per receiver.