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Actualy I think it depends on what do you refer to, power or voltage. For power it is as the_risk_master said, but for voltage it is:
20*log(Vout/Vin) for dB, and 20*log(V/0.77) for dBm, not sure for 0.77!
And another thing, xdB = 10 log (P) is not true for one thing, P is in watts so it has to be devided by watts so we could use log()... It should be 10*log(Pout/Pin)...
For the second question, I am not sure what is your question, but I can help you this way:
10*log(P/1mW)=16 => P=1mw*10^(1.6)≈39.8mW this may mean that your signal must be at least 39mW...
I have probably written it that way (dBm +-dB) a thousand times and never realized that it was possibly not technically accurate :!:
Maybe 16dBm +-3dBm :?:
13dBm ~= 20mW
19dBm ~= 79mW
You can convert that into E or I based on the load
dB is an expression of the power ratio between any two signals.
dBm ( sometimes called dBmW )has one of the signals to reference built in: 1 milliwatt and is therefore a measure of signal level. So 0dBm = 1mW.
2mW = 10xlog10(2) ~ 3dBm.
If you work it backwards you can prove that there is a square relationship between power and voltage or current which is equal to 2 x log of a number. Therefore if the ratio is in voltage or current then the formula is 20 x log(ratio).
A 'Bell' unit was too large ( like a Farad ) so they used 1/10 of a Bell or a decibel. If I remember right, it was Alexander Bell of the Telephone fame that described the exponential relationship between the power in an audio signal and it's apparent loudness.