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.
Is it worth the effort and complexity for 2 dB or less? Can the builder
construct an almost lossless matching network? Thats up to the
individual IMO. In my case I work hard for every dB or fraction I can
get.
Perhaps someone here will model this and report one way or the other. I'm
just going with what I have thought was the correct assumption.
73 Carl KM1H
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