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Re: [TowerTalk] quad questions

To: towertalk@contesting.com
Subject: Re: [TowerTalk] quad questions
From: jim Jarvis <jimjarvis@optonline.net>
Date: Wed, 03 Dec 2008 05:14:39 -0500
List-post: <towertalk@contesting.com">mailto:towertalk@contesting.com>
Very interesting discussion.

When I put my lightning bolt quad up, I was concerned with their  
system of
simply tying all loops together at a common feedpoint.   I went so  
far as to
talk with Cebik about this, and wound up with a simple analysis.

Suppose you had a 1 wavelength loop and a 2 wavelength loop, in  
parallel,
and in the same spatial plane.   e.g.  a 20m and a 10m quad DE.   What
happens?

First, the 2 wavelength pattern is precisely opposite that of the 1  
wavelength pattern.
In the case of the 20/10m example,  the 20m element would fill in the  
nulls in
the 10m pattern.    That null filling would take place proportional  
to the current in
the respective loops.   And that current would be proportional to the  
respective
impedance of the loops, at the feedpoint.

The 2 lambda loop (20m element operating at 10m) would have a very  
high impedance,
on the order of 2k ohms.   The 1 lambda loop (10m element on 10m)  
would be around
100 ohms.   100/2000 = 1/20 = 5%.    So, your side nulls would be  
filled by around 5%.
Not such a big deal, if you can accept the f/s ratio of maybe 13dB   
Note, it's not the f/b
of the array -- that's established by the parasitic element.

What has me confused is the opposite case...  what happens when the  
10m element is
operated at 14 mHz, in parallel with the 20m element?   The system of  
tying the elements
together for feeding is being used, so it must work.   What's  
happening?   The suggestion has
been made that the 1/2 lambda loop can be considered a quarter wave  
parallel transmission line,
which is shorted at the far end.   The far end short  is thus  
transformed to an open at the feed point.
I'm clearly missing something, here.  Your thoughts?

In the case where each element is fed through a stub of 72 ohm coax,  
to transform the
loop impedance to 50 ohms, and a switch is used to select the desired  
DE, the unused
elements should be left floating, not tied to common shield.   You'll  
still have some parasitic
coupling to the unused elements, but the amount of induced current is  
much less than in
the parallel feed situation.   Probably on the order of 0.5%-1.0%,  
according to Cebik's
model, at the time, getting you back to the 20dB+  f/s ratio.

N2EA
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