Those impedances seem pretty low for a folded monopole, unless the
conductor diameter is large.

When modeling two parallel wires like a folded monopole or dipole with
any NEC-2 based program, it's essential that the segment junctions be
aligned. For the folded dipole or monopole, simply make the wires the
same lengths and give them the same number of segments.

Roy Lewallen, W7EL

John wrote:

"Roy Lewallen" wrote in message
...

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



Okay! I did it!

I didn't separate out the transmission line from the antenna. Instead, I
just modeled the folded monopole. I plotted on a Smith chart the resultant
terminal impedance as the vertical element varied from .23 wavelengths to
.245 wavelengths. It went from (73.31 - J 40.81) through (85.71 + J 0.7908)
to (93.59 + J 22.56). Easy!

I found that a vertical element of .2325 wavelengths gave me 76.05 - J 30.27
ohms. That's a sweet spot for this particular antenna in that feeding it
with a .143 wavelength piece of 75 Ohm coax will match to a 50 Ohm source.

To further my education, I also checked the anti-resonance point you
mentioned.

Thanks!

John

"Roy Lewallen" wrote in message
...
Those impedances seem pretty low for a folded monopole, unless the
conductor diameter is large.

When modeling two parallel wires like a folded monopole or dipole with
any NEC-2 based program, it's essential that the segment junctions be
aligned. For the folded dipole or monopole, simply make the wires the
same lengths and give them the same number of segments.

Roy Lewallen, W7EL


You are correct again. I had different numbers of segments of the parallel
wires. When the segments are about the same number, the terminal impedance
goes up. I am also trying to run as many segments as practical so the
accuracy is greatest.

John