I'll once again separate the "antenna" from the "transmission line" to
make it easier to see what's happening.

If you're dealing with an air-dielectric folded dipole, the transmission
line stub is nearly a quarter wavelength long. So at resonance, its
impedance is high and it doesn't have much effect on the feedpoint
impedance. As you lower the frequency or shorten the antenna, the
resistance of the antenna (as opposed to the transmission line) drops
fairly slowly, and the reactance becomes negative relatively quickly.
This is in parallel with the transmission line, whose reactance becomes
more positive as the line gets electrically shorter. If you look at the
net result of this parallel combination, you get a feedpoint impedance
that has a rising resistance as frequency drops or the antenna shortens,
and a reactance that gets more negative.

At some frequency below resonance, the increasing positive reactance of
the transmission line equals the negative reactance of the antenna,
creating a parallel resonant (sometimes called anti-resonant) circuit.
Just before this happens, the resistance skyrockets and the feedpoint
reactance heads positive. Exactly at parallel resonance, the reactance
is zero (by definition of resonance) and the resistance is very high.
And just below that frequency, the reactance heads rapidly to a high
positive value, then begins decreasing as the frequency drops below
that. The frequency or length where you hit anti-resonance depends on
the impedance of the transmission line. I fished up a model of a 17.56
foot high folded monopole with #12 conductors spaced 6 inches apart
which I had lying around. It's resonant at about 13.25 MHz., where its
feedpoint impedance is 143 ohms. It hits anti-resonance at about 8.5
MHz, where its feedpoint resistance is about 15k ohms. Below that, the
feedpoint reactance is positive, and decreases as the frequency is lowered.

If you want to model a folded monopole as a separate unfolded monopole
and transmission line (which is a way to model one made from twinlead,
since you can separately adjust the transmission line length to account
for the reduced velocity factor of the transmission line mode), here's
what you have to do.

First, make the unfolded monopole from two wires, connected in parallel
at the bottom and top, or from a single wire of equivalent diameter.
Next, choose the impedance of the transmission line to be 1/4 the
impedance of the actual line. You have to use a transmission line model
for this, not a transmission line made from wires. Make sure it's in
parallel, not series, with the source at the base of the monopole. In
EZNEC, a transmission line is connected in parallel with a source if
they're on the same segment. Finally, multiply the reported feedpoint
impedance by four to find the Z of the actual folded monopole.

Roy Lewallen, W7EL

John wrote:
"Roy Lewallen" wrote in message
...

John wrote:

. . .
I thought it was supposed to be backwards from the usual unfolded

monopole

such that it would go up in resistance and become inductive.?.

Why would it do that?

Roy Lewallen, W7EL



Well, you said earlier that the folded monopole could be modeled as an
unfolded monopole with a shorted transmission line in parallel. I thought I
understood. When I modeled the unfolded monopole, I saw it do as usual when
the element was varied in length. But when I included the shorted section of
transmission line and varied it directly with the element, I thought I saw
the terminal reactance go inductive as the length was decreased below
1/4-wave resonance and I thought the terminal resistance went up. So, I was
expecting the same from EZNEC by modeling the folded version.

I guess I'm really lost here.

John