Re: Mode Splitting

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

Original poster: "Malcolm Watts" <m.j.watts-at-massey.ac.nz> 

Hi Bob,

On 20 Aug 2004, at 13:01, Tesla list wrote:

 > Original poster: "Bob (R.A.) Jones" <a1accounting-at-bellsouth-dot-net>
 >
 >
 > ----- Original Message -----
 > From: "Tesla list" <tesla-at-pupman-dot-com>
 > To: <tesla-at-pupman-dot-com>
 > Sent: Thursday, August 19, 2004 5:28 PM
 > Subject: Re: Mode Splitting
 >
 >
 >  > Original poster: "Antonio Carlos M. de Queiroz" <acmdq-at-uol-dot-com.br>
 >
 >  >
 >  > Better to say: With k=0 the primary and the secondary systems
 >  resonate > (oscillate) at the same frequency. When the coils become
 >  coupled, this > single frequency splits in two, one above and the
 >  other below the > original > frequency, and both systems oscillate at
 >  both frequencies > simultaneously.
 >
 > Perhaps very clumsily I was trying state that the two modes are
 > independent. Yes its true that in the usual impulsive system they are
 > excited simultaneously but in a master oscillator SSTC or using a
 > signal generator either mode can be driven independently to the extent
 > of their Q and separation. It has been incorrectly stated that mode
 > splitting does not occur in an SSTC as if some how its a property of
 > the drive signal as opposed to a property of the system.

Agreed. It is easy to show mathematically that for any waveform other
than constant amplitude sinewave, other frequencies are present.

 >  >  > That probably needs expansion. The
 >  >  > reflected impedance of the primary is either inductive or
 >  capacitive
 > hence
 >  >  > the wave of one mode is shortened and the other is lengthened.
 >  The >  > shortened one will have the same polarity at both end while
 >  the
 > lengthened
 >  >  > one will have the opposite polarity at its ends with one null
 >  near the >  > primary end. Of cause the real effect is distributed
 >  along the coil
 > with
 >  >  > the distributed inductive coupling from the primary.  Incedently
 >  I
 > don't
 >  >  > think the higher order modes of the secondary split because at
 >  those
 > higher
 >  >  > frequencies the reflected impedance of primary is always
 >  inductive so
 > they
 >  >  > are just shifted. In the case of a top load coil all modes are
 > truncated at
 >  >  > the top.
 >  >
 >  > I don't see much use in considering steady state impedances in this
 >  > case, > where there are two frequencies involved and the waveforms
 >  are all > transient.
 >
 > I think we would agree that the system is linear  (assuming a closed
 > spark gap) just a collection of Ls Cs and Rs so it can be completely
 > characterized by it complex impedances which I assume you refer to as
 > steady state impedances. Hence in the usual complex circuit analysis
 > whether we get a transient or not is a function of the excitation
 > signal.
 >
 >  >From the perspective of the secondary we can replace coupled primary
 >  circuit
 > with a parallel tuned circuit in series with primary end of the
 > secondary using the relationship ((Lm.s)^2)/Zp were Lm is the mutual
 > inductance and Zp is the series impedance of the primary. Similar to
 > the way you eliminate the coupling in several of your papers. The
 > referred impedance is equivalent to a parallel tuned circuit which is
 > high impedance (assuming both are at the same frequency) at the
 > frequency of the 1/4 wave mode of the secondary. But that mode
 > requires a low impedance so even though the impedance is real
 > (something that's bugged me for a long time) the mode can not be
 > supported. Either side of the 1/4 wave frequency the impedance is low
 > and either inductive or capacitive hence it can support a 1/4 plus a
 > bit or a truncated 1/4 wave mode.
 >
 > Using this description its also easy to visualize what happens if you
 > vary the primary and secondary frequencies.  As the separation
 > frequency increase one mode move down the resonance curve of the
 > referred primary and one moves up to the peak. At the peak the
 > impedance is too high to support that mode and it disappears leaving
 > only the other mode  and the uncoupled resonance of the primary that
 > has very little feed thru to the secondary. Well easy for me to
 > visualize. Apart from finding the roots of the transfer function.  I
 > have not read any explanation of the mode splitting that even come
 > close to holding water.

My take on it is that the incrementing/decrementing waveforms *must*
produce sidebands because the change in amplitude is altering the
slope of the sinusoidal waveform continuously.

Malcolm

 > One surprise, for me anyway, was that one of the split modes has null.
 >   But in fact that's the only way the two orthogonal modes can sum to
 > a maximum at the primary while summing to zero at the top end and be
 > almost zero along the secondary. Then (if the frequencies and phase
 > are right) after a time interval sum to zero at the primary and a max
 > at the top of the coil.  There was  minor problem because two
 > orthogonal modes or for that matter three or any finite number can not
 > initially sum to zero over the length of secondary. To do that you
 > need you need a contribution from all the higher order modes.
 >
 > An other intersting point is that the closer the two spilt modes are
 > in frequency the better they cancel along the secondary so less
 > contribution is required from the higher order modes. Putting this an
 > other way. In an impulsive system the tighter the coupling the more
 > energy is wasted in the higher order modes. This ignores the
 > distributed effects of the coupling so it may only be partially true.
 > In any case I do not mean to suggest that the wasted energy is
 > necessarily significant relative to other losses only that its
 > inevitable.
 >
 > Bob
 >
 >
 >