[TowerTalk] Shunt Feeding a Vertical

[Joe Subich, W4TV]

On 2019-09-06 4:58 PM, VE6WZ_Steve wrote:

 > Wes…..PLEASE….After you have watched the video let me know if there is
 > something wrong with the “fifty-ohm” point.

Steve, the tap point on the tower is not 50 Ohms.  The true impedance of
the tap point is *transformed to 50 Ohms* by the (complex) impedance and
length of the transmission line formed by the tower and "gamma rod".
That assumes you have a single capacitor gamma match (shunt feed).

For an omega match (two capacitor), the impedance at the tap point is
first transformed by the impedance and length of the "gamma rod" and
is *transformed again* by the capacitive L network (two capacitors)
before it reaches 50 Ohms.

73,

    ... Joe, W4TV


On 2019-09-06 4:58 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 <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 at 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
>>
>>