Steve.
I have watched the video, more than once actually.
If you would say, "For a given gamma (shunt) wire size and spacing there is a
point on the tower that will reflect a 50 +jx impedance at the feedpoint", it
would be more clear.
Also your instance that the tower must "be inductive", i.e. greater that 90
degrees tall is simply wrong.
Wes
On 9/6/2019 1:58 PM, VE6WZ_Steve wrote:
Hi Wes,
To be clear, I am referring only to matching a shunt-fed tower using either
the traditional Gamma match (a single series capacitor) or the Omega match.
If you currently have a shunt-fed tower with a single capacitor Gamma
capacitor in series with the coax at the feed-point, and you have a 1:1 match,
if you move that tap point higher or lower on the tower, you will never get a
1:1 match. That is because the Real R measured at the feed point will be
greater or lower than 50 ohms.
This is not “theory”. It is how it actually works.
Perhaps if you were to watch my YouTube video this will be clear. I spent a
fair bit of time actually showing these real measurements in the field on my
tower. In the video I show how the real resistance changes as the shunt wire
is moved higher or lower on the tower. I also show exactly what happens when
you move the wire closer or further from the tower. A movie is worth a
million words.
Start the video at 2:10 if you don't want to watch it all.
https://youtu.be/cHlc5MTGTFM
Wes…..PLEASE….After you have watched the video let me know if there is
something wrong with the “fifty-ohm” point.
Steve VE6WZ
On Sep 6, 2019, at 1:51 PM, Wes <wes_n7ws@triconet.org
<mailto:wes_n7ws@triconet.org>> wrote:
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
<mailto: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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