Joel Kolstad wrote:
I'm looking at the form of the equation for the effective area of an antenna
and finding it a little non-intuitive...

We have D/Ae=4*pi/lambda^2, where D is the antenna's directivity and Ae is
the effective area. Hence, after rearranging, we find that Ae is
proportional to lambda^2 or 1/f^2 (f=frequency).

Is this really telling me that if I cut a receiving antenna in the form of a
half-wave dipole at 200MHz, it'll only intercept a quarter as much power as
a half-wave dipole at 100MHz?

Yes. It's half as long, so it won't intercept as much.

For instance... say the incident field has a power density of 1 W/m^2. A
half-wave dipole at resonance is ~73 ohms and its directivity is ~1.64
giving it an effective area of ~.131*lambda^2. At 100MHz, that's 1.18 m^2
and the antenna will produce 1.18 W. At 200MHz, I only get .295W... right?
If I attach a ~73 ohm transmission line to these antennas, equating
V^2/(2*R) to the powers gives me 13.13V (peak to peak) at 100MHz and 6.56V
at 200MHz. This is going to make it twice as hard for me to detect some
signal at 200MHz, won't it??? How does the antenna noise scale with
frequency -- do I still have the same SNR in both cases?

Thermal noise has a uniform frequency distribution, so over a given
bandwidth, the noise power is the same at any frequency. You'll have
better S/N ratio with the 100 MHz dipole than the 200 MHz dipole.
Atomospheric noise is usually more important at HF, and it does vary a
great deal with frequency, but at VHF and above, thermal noise generally
dominates.

Roy Lewallen, W7EL