TopBand: Folded unipoles

[Chuck (Jack) Hawley]

km1h @ juno.com wrote:

> On Mon, 6 Oct 1997 22:39:19 +0000 w8jitom@postoffice.worldnet.att.net
> writes:
> >> From:          Peter Chadwick <Peter.Chadwick@gpsemi.com>
> LOTS OF SNIPS
> >
> >The only thing that changed is the feedline current is now 1/2 ampere
>
> >at 60 volts, which is STILL 30 watts.
> >
> >We could do the same thing with an L network, or a simple
> >transformer. But make no mistake about it, no matter what method we
> >use ground loss would remain EXACTLY the same.
> >
> >Adding ground loss into this equation makes no difference at all.
> >Since ground CURRENT remains the same, ground loss remains the same
> >in either system. Any thoughts to the contrary are absolutely
> >incorrect.
> >
> >73, Tom W8JI
>
> I'm confused Tom....
> Your first assumption was that the 1/4 wave monopole was over perfect
> ground....correct?
> But in that case of course there is no ground loss and all the math
> works
> just fine.
>
> Take a REAL system with 35 Ohms Z and 12 Ohms of ground loss; not
> uncommon in many residential installations. The VSWR looks real good
> but
> 1/3 of the power is warming the worms( ~ 2dB). No matching section
> involved or required here.
>
> Next fold that wire and we have a 4X impedence transformation so the
> radiation resistance is now 140 Ohms. The ground loss is still 12 Ohms
>
> but that is only 8.6%  of the total. Assuming a quality matching
> network
> the loss in the ground now is only a fraction of a dB.

How come the "impedance transformer" transforms the radiation resistance
and not the ground loss impedance?

Chuck.

--
Charles (Jack) Hawley
Chuck....Ham Radio KE9UW
AKA "Jack" BMW Motorcycles, MOA #224 K100RS
Wife rides...Viki, MOA #18120 K100RS
President IBMWR
c-hawley@uiuc.edu
Sr. Research Engineer Emeritus
University of Illinois, Urbana-Champaign



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