AndyS wrote:
Andy asks:
I have a quick question about the number of microwatts that an
antenna will extract from the air and put into a known load.

Assume a 70 ohm perfect antenna used for receiving a signal....

Preceived = Pwr density ( watts/sq mtr) X Capture Area (sq
mtrs)

Is this the power that the 70 ohm antenna will put into a 70
ohm resistor connected to the antenna (as the load), OR is half
this power dissipated in the antenna and the other half sent to
the load ( considering the antenna as a Thevenin generator)....?????

Thanks for any discussion on this. I know it is simple, but I
would like to hear what others have to say about it....

Andy W4OAH

The short answer is that the product of impinging wave power density and
capture area equals the power dissipated by the load, assuming a
lossless antenna and matched load.

A lossless antenna won't dissipate any power applied to it. So it
certainly won't dissipate half, or any, of the power from the wave.

You won't find "capture area" in many texts, but when you do, it'll be
another term for the much more common "aperture". Among variations of
this are "aperture", "effective aperture", "maximum effective aperture",
"scattering aperture", "loss aperture", "collecting aperture", and
"physical aperture". The aperture is the ratio of power delivered to the
load, to the power density of the impinging wave. If the antenna is
lossless, terminated in the complex conjugate of its transmitting
feedpoint impedance, and oriented for maximum response to the wave's
polarization, the effective aperture, maximum effective aperture, and
scattering aperture are all equal, and represent the cross section of
the power in the impinging wave delivered to the load.

Kraus's _Antennas_ is a good source of information about this topic.

Roy Lewallen, W7EL