I disagree too. This idea that there is a "fifty-ohm" point somewhere on the
tower and your assignment is to find it is simply not true.
Consider the gamma rod (shunt wire or whatever you want to call it) as one
conductor of a parallel wire transmission line, with the other conductor being
the adjacent length of tower. This line is short circuited at the top and
bottom ends with a source inserted in one leg, Since the whole object of this
is the fact that the source can't be inserted in the tower leg, we use the
gamma/shunt leg.
If we make these two conductors the same diameter and length, and adjust them
appropriately for resonance we have a folded-monopole or half a folded dipole.
It's commonly accepted that a symmetrical folded antenna has a feedpoint Z of
four times a single wire, so a resonant quarter-wave folded monopole over
perfect ground would have a feedpoint of ~140 ohm.
I don't know anyone who would argue that the top of a resonant quarter-wave
vertical has an impedance of 140 ohm and the feed wire is simply bringing that
down to earth.
Wes N7WS
On 9/5/2019 5:27 PM, K9MA wrote:
I must disagree. I’ve been shunt feeding a 21 meter tower on 160 for years,
with a simple gamma match. At most, it is electrically 3/16 wavelength.
See ON4UN’s book.
73,
Scott K9MA
----------
Scott Ellington. K9MA
--- via iPhone
On Sep 5, 2019, at 5:40 PM, VE6WZ_Steve <ve6wz@shaw.ca> wrote:
Your tower is too short for a shunt-feed. It needs to be inductive at the QRG
of interest, 90 degrees or more (1/4 WL).
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