EE/RF pedant here (there had to be one, right?).
However, I doubt that e.g. subtracting 3dBm is sufficient, say
Without going into detail, it needs to be said that dB is a relative measurement while dBm is absolute. Thus, one would state that 3 dB is subtracted from X dBm in order to represent half power (which is what you were getting at with the 200 -> 100 mW issue). Simple example: 15 dBm = 31.623 mW, 12 dBm = 15.829 mW (indeed, 3 dB down is half power on a linear scale). 3 dBm, however is 1.995 mW. Subtracting 3 dBm from 15 dBm would then give you 14.72 dBm (29.628 mW). Not obvious? Yes. Important? Definitely.
While I'm at it an OT(?) question: Does somebody know how to _simply_ (using a multimeter or an old 20MHz scope) measure the power output of a wireless NIC? Just a rough (+-10mW) estimate would suffice. The antennae are external so I have access to the SMA. Then I could measure the mapping myself.
To simply answer, "no." You cannot absolutely measure transmitted RF power without a calibrated receiver of some sort (e.g. a commercial RF power meter... see Agilent), two antennas with known directivity patterns and known efficiencies, and/or a 2D motorized az/el stage such that you can easily rotate one of the two antennas and integrate the received power. If you have, however, two known antennas with known gains (say, Hyperlink patch antennas), and you know - or can estimate - the insertion loss in the cables and coax connectors, and a second wireless NIC with software you believe is giving you approximate values of received power, you can use the Friis equation to find the transmitted power. This will put you within 10 dBm/mW easily. http://en.wikipedia.org/wiki/Friis_Transmission_Equation Any other questions related to this I can answer off-list. Cheers, Charles
