Hmm, the formula is a bit off, but... 30 x 32.785 = 983.55. I'll also
bet length is expressed in feet.
Looks eerily like someone wants you to cut a one-wavelength piece of
coax cut at the mean repeater frequency.
Just a guess.
73, Russ WB8ZCC
On 8/13/2010 11:38 AM, Sid wrote:
I have a note in my file that I do not recall where it came from
relative to cable length between the duplexer and the TX or between
the duplexer and additional filter. Length = (30)(32.785)(vf/freq).
30 is for 30 degrees, vf is velocity factor, freq is the average of
the pass and reject frequencies. If too short add 180 degrees. Don't
know if this is good info or not. The article would be appreciated. Sid.
--- In Repeater-Builder@yahoogroups.com
<mailto:Repeater-Builder%40yahoogroups.com>, Nate Duehr <n...@...> wrote:
>
>
> On Aug 5, 2010, at 11:20 AM, Kevin Custer wrote:
>
> > Allan Crites and I are currently in discussion which will be used
as the basis of a RB web article that will explain exactly what is
happening, why it happens, and why an 'optimized' cable length can be
used to transfer power ending up with the stated loss of the duplexer
and have little reflected power toward the transmitter - so long as
the duplexer is tuned properly and exhibits good return loss on the
frequency it's designed to pass.
>
> There's already a great book on that topic, it's called the ARRL
Antenna Handbook, and the chapter on transmission lines covers it in
more detail than anyone will ever need to know in the real-world,
who's not a practicing RF Engineer.
>
> That book if read cover-to-cover, is also damn good for insomnia. Or
at least it'll keep you distracted while you can't sleep! :-)
>
> --
> Nate Duehr
> n...@...
>
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>
