Owen Duffy wrote:


An alternative is to calculate the power collected by a lossless,
matched receiver as Pr=S*A.

In this case, S=0.3**2/(120*pi)

Kraus derives A (the effective apperture) for a short dipole to be
3/8/pi*wavelength**2.

This gives the power collected by the receiver as 6.4mW. If the
antenna and receiver were disected by the ground plane, wouldn't there
be 3.2mW developed in each half of the receiver load?

I believe that's correct. Note that Kraus uses a plain wire dipole for
his aperture correction, but a dipole with end hats (and therefore
uniform current) for input impedance calculations. See my other recent
posting for further comments.

Roy Lewallen, W7EL