NEC-LIST: Modeling short dipoles, chapter 2

Hi, again,

If you know the measured GAIN in dbi of the antenna which would
include the efficiency and directivity, then you're done after you
calculate the mismatch loss. But if you know the radiation resistance
and copper losses separately you need to figure how much power gets to
the radiation resistance compared to how much power is available from
the source. Again, assume Rs = source = 50ohms; Rrad = 3 ohms; Rc = 2
ohms (copper losses). It is helpful to describe the source as an
a.c. voltage source of Vs volts RMS in series with Rs; then the
available power in the matched case would be [(Vs/2)^2]/Rs =
Vs^2/(4*Rs). With all the reactances tuned out what you have as an
actual load is Rrad+Rc and the voltage across Rrad is
Vs*Rrad/(Rs+Rrad+Rc), and the power dissipated in it is Prad =
Vs^2*Rrad/((Rs+Rc+Rrad)^2). Compare this to Pavail: Gc = 'circuit'
gain of antenna = Prad/Pavail = 4*Rrad*Rs/((Rs+Rrad+Rc)^2) after some
algebra. Note that this reduces to Gc = 1 = 0 dB for Rrad = Rs and Rc
= 0.

73, Paul Gili, AA1LL, Greenville, NH
paul.e.gili_at_lmco.com
Received on Wed Nov 26 1997 - 09:41:58 EST