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Frequency Split

Hi everyone,
             As promised, here is the list of frequency splits to
achieve the "sweet values" of k (the coupling constant). This work
follows on from presentations by Mark Barton and others. Also,
Abramyans paper on a transformer-type accelerator was referenced.
     The split may be expressed as F(hi) = yF(lo) where y is a
multiplying factor given in the table below and n = No of 1/2 cycles 
to effect complete energy transfer from primary to secondary, and
F(hi) and F(lo) are the high and low split frequencies respectively.

   k     |   n   |   y
-------------------------
  0.6    |   2   |  1.63
  0.385  |   3   |  1.5
  0.28   |   4   |  1.33
  0.22   |   5   |  1.25
  0.18   |   6   |  1.2
  0.153  |   7   |  1.167
  0.133  |   8   |  1.14
  0.117  |   9   |  1.125
  0.105  |  10   |  1.11
  
     Beyond 10  1/2 cycles, the difference between the frequencies is
becoming smaller and smaller.

     To calculate F(lo), choose the k you want to set the system 
at then using the corresponding value of y from the table, 
use F(lo) = 2 x fr/(1+y)   where fr = secondary resonant frequency.

This formula assumes the frequency split to be equal about fr and
also assumes that the secondary is tuned to the same frequency as the 
primary. Then use F(hi) = yF(lo)  to find F(hi).

Derivations :
***********
   y = SQRT( (1+k)/(1-k) )    derived by rearranging the
   formula presented by Mark : k = 1 - 2/(1 + (F(hi)^2 / F(lo)^2) )

   F(lo) = 2 x fr/(1+y) derived from :  fr = ( F(lo) + F(hi) )/2
   and :  F(hi) = yF(lo)

One further point : k = (F(hi) - F(lo))/fr  becomes a good 
approximation for values of k below 0.15 or so.

Example : Let fr = 120kHz  and aim for k = 0.22   (n = 5)
*******

Then  F(lo) = 2 x fr/(1 + y)  = 2 x 120kHz/( 1 + 1.25)
                              = 106.67kHz

and  F(hi) = yF(lo)  =>  F(hi) = 106.67kHz x 1.25
                               = 133.33kHz

Comments and corrections welcome as always.

Malcolm