Okay, I'll bite, Rich. I take the L-network, parallel the L and C, and then
measure the resonant frequency using a GDO. I then re-configure the
network back into an L and connect it to the load. If I understand what
you have claimed (correct me if I am wrong), the input impedance of
the L-network at the resonant frequency (determined above) will be
a short cicuit even when the load is attached. If this is what you are
claiming, then I respectfully disagree. If this is not what you are claiming,
then please elaborate.
73 de Mike, W4EF.........
----- Original Message -----
From: "measures" <2@vc.net>
To: "Michael Tope" <W4EF@dellroy.com>; "AMPS" <amps@contesting.com>
Sent: Sunday, April 22, 2001 8:55 PM
Subject: Re: [AMPS] basic question about dipping the plate
> >
> >Hi Rich,
> >
> >Since a dipmeter doesn't measure impedance, how could I use one to prove that
> >the input impedance of an L-network at resonance doesn't look like a dead
> >short circuit?
> >
> ? The series L and C must be paralled to measure resonance.
>
> cheers, Mike.
> >
> >> >
> >> >I believe that is only true if there is no load connected across the
> >> >shunt
> >> >side of the L-network.
> >> >In that case the L-network simply becomes a hi-q series resonant trap to
> >> >ground (e.g. a near
> >> >dead short). Connect a finite resistive load across the shunt element and
> >> >that is no longer true.
> >>
> >> ? Try proving this with a dipmeter, Mike.
> >>
> >
> >
> >
> >--
> >FAQ on WWW: http://www.contesting.com/FAQ/amps
> >Submissions: amps@contesting.com
> >Administrative requests: amps-REQUEST@contesting.com
> >Problems: owner-amps@contesting.com
> >
> >
>
>
> - Rich..., 805.386.3734, www.vcnet.com/measures.
> end
>
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