John Popelish wrote:
You might. What is the feed point impedance of a dipole that has its
ends terminated in the complex conjugate of their local impedance, so
that no energy reflects at the ends?

It's roughly approximately the same as a terminated
Rhombic antenna, i.e. hundreds of ohms.

Let's make some rough assumptions. The SWR on the
1/2WL dipole is 20:1 which makes rho roughly
(20-1)/(20+1)= 0.9 That makes the power reflection
coefficient (rho^2) roughly equal to 0.8

If we are supplying 100 watts to the antenna then
Pfor - Pref = 100W and we know that Pref/Pfor = 0.8
So we can solve for Pfor = 500W and Pref = 400W.

If we assume the Z0 of the dipole is 600 ohms, that
makes Vfor = 548 volts and Ifor = 0.91 amps. Also
Vref = 490 volts and Iref = 0.81.

So (Vfor-Vref)/(Ifor/Iref) =
(548-490)/(0.91+0.81) = 58V/1.72A = 34 ohms.

But that is just half of the dipole's impedance so we
have to double it to get a feedpoint impedance in the
ballpark of 68 ohms. However, please note that 34
ohms is roughly the feedpoint impedance of a 1/4
wavelength monopole, i.e. half a dipole.

These are obviously rough ballpark assumptions but
you can observe the concepts involved. For the dipole
feedpoint impedance to be low, the voltages have to
subtract and the currents have to add. This agrees
with the extra 180 degree phase shift that happens
when the current is reflected at the ends of the dipole.
--
73, Cecil http://www.w5dxp.com