dbm math contradiction question.

[yefj]

Hello ,i am trying to do dbm math without the logarithmic expression. I want to convert 15 dbm so its 10 dbm +5 dbm 5dbm=4dbm+1dbm=(10-3-3)dbm+(10-3-3-3)dbm = 10/2*2*2mW *10/2*2*2mW=2.5*1.25mW=3.125mW so (10+5)dbm=31.25dBm but on the other hand 15dBm=3+3+3+3+3dbm=2^5mW where did i go wrong in my first calculation?

 

[kaz1]

Hello ,i am trying to do dbm math without the logarithmic expression. I want to convert 15 dbm so its 10 dbm +5 dbm 5dbm=4dbm+1dbm=(10-3-3)dbm+(10-3-3-3)dbm = 10/2*2*2mW *10/2*2*2mW=2.5*1.25mW=3.125mW so (10+5)dbm=31.25dBm but on the other hand 15dBm=3+3+3+3+3dbm=2^5mW where did i go wrong in my first calculation?

for 15 dBm: 10^(15/10) = 31.623 mW

Reverse: 10*log10(31.623) = 15 dBm

Why do you convert the math into a puzzle that goes wrong?

 

[FvM]

1. You can't add dBm.
2. You can add dBm and dB, but it's a power multiplication rather than summation. Example you have a signal of 10 dBm (10 mW). If you send it through a 6dB amplifier you get 16 dBm signal (40 mW)
3. If you add two 10 dBm signals with a lossless power combiner, you get 13 dBm (power x2 = +3 dB).