jallem's formulas are right.Of course ,ypu can calculate more easily.First, you need computate the power with the voltage,Vrms(V)]^2/R. Attention here ,the impedance 's value is 50 Ω or the charateristic impedance of your system. Then you can convert the power from Watt to mW.jallem said:P=(Vrms)^2/R
P(mW) = ([Vrms(V)]^2)10^3/R = {[Vrms(mV)1000]^2}1000/R
10log[P(mW)]=20log[Vrms(mV)]+90-10log(R) = P(dBm)
P(dBm)=20log[Vrms(mV)]+90-27.77(R=600 ohm)
P(dBm)=20log[Vrms(mV)]+90-17(R=50 ohm)
P(dBm)=20log[Vrms(mV)]+62.21dB ......(R=600 ohm)
P(dBm)=20log[Vrms(mV)]+77dB ..........(R=50 ohm)
Then what is the solution, know-it-all?Jallem's only two first formulas are right.
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