randyc
2013-04-30
As I understand it, the syntax for producing the power level in dBm to a 50 ohm load for a peak AC voltage of "Vrf" is:
Power = dbm(Vrf.v)
However this expression is always 3 dB too high. Example: using the QUCS 50 ohm power source set to 0 dBm connected to a 50 ohm resistor will yield an output power level of +3 dBm.
Replacing the above expression for Power with the following exact definition:
Power = 10log10(((.707Vrf.v)^2/50)/.001)
Yields the correct power level of 0 dBm
I repeated this with an AC voltage source in series with a 50 ohm resistor and obtained the same 3 dB error.
Is this a bug or am I using the expression incorrectly ?
Cheers,
Randy
Frans
2013-05-01
Dear Randy,
In this document http://qucs.sourceforge.net/docs/functions.pdf on page 5, you can read that all output voltages are peak values. the function dbm however (page 121) takes the rms voltge.
To get the correct value for a sine you could use dbm(Vrf.v/sqrt(2))
randyc
2013-05-01
Thanks, I suspected that was the case :)