-----Original Message-----
From: Peter Chadwick <Peter_Chadwick@mitel.com>
To: Steve Thompson <amps@txrx.demon.co.uk>;
amps@contesting.com <amps@contesting.com>; 'Rich Measures'
<measures@vcnet.com>
To: <amps@contesting.com>
Date: 11 August 1999 08:47
Subject: RE: [AMPS] Pi-Net math
>It's really back to how you define resonance. My definition
would be the
>frequency where the input impedance is purely resistive,
and I'd measure it
>that way, using an R + jX bridge. Whether or not that will
correspond to
>'resonance' as shown on a GDO, I don't know. Under those
circumstances,
>whether the circuit is doubly loaded or not is immaterial.
>
In this case I don't think it matters how you define or
measure resonance. Without external resistances the circuit
has a resonant frequency. If you put a resistor across one
of the capacitors, the resonant frequency will lower as the
resistor provides a path without phase shift for some of the
current (consider an extreme example of a .001 ohm resistor
across one of the caps). If you put the DESIGNED load
resistor values across both ends, they carry the same amount
of non-phase shifted current, and have no overall effect. I
explain what I mean much faster on a blackboard...
Draw the pi network as a parallel tuned circuit with two
capacitors in series across the coil. If you shunt one of
the capacitors with a resistor, the resonant frequency
changes. If you shunt each capacitor with a resistor where
the same amount of current flows in each resistor, you can
redraw it as a single resistor across the coil. The resonant
frequency hasn't changed but the Q is lower.
Steve
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