Denis,
The 1_2_1 current distribution comes for the characteristics of a
binomial polynomial.
4 elements would be 1 3 3 1
and phase would be 0, -90, -180, -270.
A 5 element array is: 1 4 6 4 1
and phase is 0, -90, -180, -270, -360
A way to figure out the current distribution is to take the polynomical
coefficients (antenna currents) shift them to the right and add it to
the original coefficients.
For example, to get a 4 element distribution:
original 3 element currents: 1 2 1
original 3 element currents shifted 1 2
1
-------------------------------
4 element excitation currents 1 3 3
1
Hope this helps! I believe ON4UN's antenna book has some ways to creat
the appropriate excitation currents and phases for at least the 3
element case.
Keep us informed if you decide to build something! There are ways to
obtain higher gains and narrower bandwidths with the same number of
elements. However you then have to come up with more difficult ways to
create the proper phasing and current. I also suspect that the binomial
current distribution is "relativly" kind to minor differences to
theoretical currents vs. actual currents.
73
chris, n4vi
Denis Coolican wrote:
>
> I noticed in the books that an in line 3 vertical assenbly would have
> the following current magnitudes at a particular phase angle. My
> question is why is the middle vertical at an magnitude of 2 whilst the
> the end ones are 1
>
> ie Element 1 1 @ 0 degrees
> Element 2 2 @ -90 degrees
> Element 3 1 @ -180 degrees
>
> The modelled pattern works out with good front to back ratio.
>
> What would the magnitudes be for a 4 in line vertical array?
>
> Regards
>
> Denis Ve6AQ
>
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