this post was submitted on 28 Jun 2026
323 points (98.5% liked)
Technology
85921 readers
4382 users here now
This is a most excellent place for technology news and articles.
Our Rules
- Follow the lemmy.world rules.
- Only tech related news or articles.
- Be excellent to each other!
- Mod approved content bots can post up to 10 articles per day.
- Threads asking for personal tech support may be deleted.
- Politics threads may be removed.
- No memes allowed as posts, OK to post as comments.
- Only approved bots from the list below, this includes using AI responses and summaries. To ask if your bot can be added please contact a mod.
- Check for duplicates before posting, duplicates may be removed
- Accounts 7 days and younger will have their posts automatically removed.
Approved Bots
founded 3 years ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
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.