Re: Mode Splitting

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

Original poster: "Antonio Carlos M. de Queiroz" <acmdq-at-uol-dot-com.br> 

Tesla list wrote:
 >
 > Original poster: "Bob (R.A.) Jones" <a1accounting-at-bellsouth-dot-net>

 > 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.

The splitting always occurs (I am assuming a system with two capacitors
and
two inductors). In an SSTC where the rise time is fast and the losses
small, the output voltage rising transient has three frequencies on it.
The two natural oscillations of the system and the driving frequency.
If the driving frequency coincides with one of the natural frequencies,
the rising transient is a sum of the two natural oscillations with an
oscillatory term that rises linearly with the time. But this combination
may not be the one that produces the fastest rise until breakout occurs.

 > 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.

Ok for impedances in Laplace transform ("s"). Steady-state sinusoidal
analysis ("jw") can't be directly used in these systems.

 >  >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.

Yes. The transformer can be eliminated to simplify the analysis, or
included back after a simplified synthesis.

 > 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.

What you call 1/4 wave mode I assume that is the natural oscillation
frequency of the secondary system alone.

 > 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.

I don't see how one of the modes can disappear. Any combination of two
capacitors and two inductors (assuming that they form a 4th-order
system)
must resonate at two different frequencies. The frequencies can't be
equal,
because this would generate waveforms that grow linearly with the time,
without limit, violating energy conservation. This explains why mode
splitting must occur.

Antonio Carlos M. de Queiroz