On Oct 19, 4:12 am, Roy Lewallen wrote:
Wimpie wrote:

So you will end up with a high Z transmission line with significantly
lower propagation speed. A difficulty is: where is the return
conductor? You can assume the return conductor about 0.125 lambda
away when the influence of the ferrite is not that large. Very close
to the feedpoint, the impedance of the ferrite loaded line is strongly
dependent on the distance from the feedpoint.
. . .

The second conductor of the transmission line is the ground plane.

When you would take a long thick ferrite tube, the field will even not
come out of the ferrite because of the attenuation in combination with
the low EM wave propagation speed inside the ferrite. Ferrite
absorbing tiles also make use of both the epsilon.r and permeability.

So finally I did not answer your question about how to determine the
feedpoint impedance. Probably you will need full 3D EM software that
can handle 3D structures of different mu and eps. As far as I know,
all momentum based EM software cannot handle this.

I don't believe that's necessary. As you say, the field doesn't escape
the ferrite. This means that the E field between transmission line
conductors (the whip and ground plane) is zero. Therefore, the impedance
of the line (E/H) is infinite. The loss of the ferrite is adequate to
suppress any reflections from the end of the line. A reflectionless,
infinite impedance line will have an infinite input impedance. This is,
as it should be, the same result I got from a somewhat different
perspective.

Roy Lewallen, W7EL

Thanks for the replies..

so I think you are saying that the 19" wire inside a 19" tube of
ferrite will present a Hi Z a the base and will therefore consume
little power i.e. it looks like an open circuit...

Does that seem resonable to you? It's the the same result as if the
ferrite were copper. i.e a cavity resonantor. I know the answer is
correct for copper but copper is not lossy.

From the other perspective, the ferrite is lossy so it will be
absorbing energy and dissipating heat and therefore the base must look
like SOME impedance that has an R comonent.

So I think the more fundamenal nature of my questoin is about how a
wire radiates in concert with the return current to the ground plane.
What if the wire is surrounded by a lossy medium such that the field
is greatly attenuated before it can get to the ground. How does the
return current flow and how does energy get absorbed by the lossy
medium?

Mark