Re: Measuring self-capacitance directly (Re: flat secondary)

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <twftesla-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br>

Tesla list wrote:
> 
> Original poster: "Paul Nicholson by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <paul-at-abelian.demon.co.uk>
> 
> [Definition of equivalent energy capacitance]
> 
> I used the assumption that any two voltages on the coil at
> resonance differed in phase by an integer multiple of pi radians.

Correct for forced voltages at sinusoidal steady state and a lossless
system. I was thinking about transients or lossy systems.

>... 
> The most noticeable departure from uniform phase of the voltage
> occurs with base-driven CW steady state at resonance.  The
> voltage at the base is in phase with the current, but the phase
> rapidly advances to +90 deg over the lower fraction 1/sqrt(Q) of
> the length of the coil.  The voltage at any point is a mixture of
> the in-phase drive voltage and the quadrature induced voltage due
> to the coil current. By 1/sqrt(Q) of the coil length, the latter
> dominates and for the remainder of the coil the the V is leading I
> by virtually 90 deg.  The currents share a common phase (modulo pi
> for the higher overtones) to within 1 or 2 deg all along the coil.

A lossy system, I imagine.
 
> The free resonance solutions for the isolated resonator approach
> the uniform phase condition much more closely than the forced
> responses, even when coupled to a primary resonator with normal k
> factors, and even when the Q is low.  There's probably a deep
> reason for the fact that neither the non-uniformity of the self
> L and C, nor the longitudinal coupling via the mutual L and C,
> appear to distort the phase uniformity of the normal modes, and so
> far it appears that the standing waves of all the normal modes
> display uniform phase (modulo pi) just like a uniform line.
> This is not something I expected in advance, so the software is
> coded not to assume it.  This feature emerges, but I'm unable to
> supply a proof that it should always be so.

The cause for this may be just linearity. Assuming that the waveforms
obtained from certain initial conditions are periodical, they would
have the same appearance for any value of the initial conditions,
with a scaling factor only.
In a complete cycle, there is no energy exchange among different
sections of the system, and so voltages and currents are always
around the 90 degrees relationship. 
 
> [Measuring the Zbase pole frequencies in grounded base config]

> > This is more complicated to demonstrate than appears to be...
> 
> Ok, I'll try...
> 
> If we relate the impedance and admittance parameters to the
> cascade matrix, defined as usual by
> 
>            base                top
>                  +-----------+
>    I1-->  o------|   coil    |------o  <-- I2
>          V1      +-----------+      V2
>           o-------------------------o
> 
>      | V1 |     | a11    a12| |  V2 |
>      |    |  =  |           | |     |
>      | I1 |     | a21    a22| | -I2 |
> 
> then we have
> 
>    z11 = a11/a21
>    y22 = a11/a12
> 
> For the kind of networks we're talking about here, the elements
> of |a| are rational polynomials in w with denominator unity [*],
> ie the elements a11, a12, a21, a22 have zeros but no poles.
> 
> We see that the zeros of a11 become the shared (and only) zeros of
> z11 and y22.  The poles of z11 are the zeros of a21, and the poles
> of y22 are the zeros of a12.  Therefore, unless the network has the
> property roots(a12) = roots(a21), we cannot expect the poles to
> coincide.

A very elegant demonstration. The idea of using transmission parameters
to avoid the complications with poles in the parameters is very clever.

>... 
> Ok, we can invert the cascade matrix to model this situation. Let
> 
>      | V2 |     | a'11    a'12| |  V1 |
>      |    |  =  |             | |     |
>      | I2 |     | a'21    a'22| | -I1 |
> 
> in which the relevant inverted matrix elements are
> 
>   a'11 = a22/det(a)
>   a'21 = a21/det(a)
> 
> Then we have the top-end impedance (with base open circuit) given
> by
> 
>   z'11 = a'11/a'21 = a22/a21
> 
> We see that the poles of z'11 match those of z11 since they both
> come from the zeros of the polynomial a21, so the method is correct
> for determining z11 poles.  But unless a22 = a11, the zeros will
> differ.  How nice.  I should alter the tssp model to test this
> point.

Nice, but you could just look at the z parameters. z11 and z22 have the 
same poles if there are no branches in series with the ports of the 
network (or other reasons for systematic cancellations of poles by
zeros).

I don't know how many of the members of the list can follow these
theoretical discussions, but they are very beautiful for those who can.

Antonio Carlos M. de Queiroz