Re: dB and dBm
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...