Re: How do you measure couplin

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

Original poster: Peter Lawrence <Peter.Lawrence-at-Sun.COM> 

Antonio,
         this is great and helps a lot, but I am also looking for the
differential equations of a "losely coupled oscillators" and the
derivation of the equations of its voltage verses time, whereas you
seem to have assumed from the beginning the sum of two sin waves.

One of the reasons is that most descriptions of the frequency splitting
phenomena of "losely coupled oscillators" shows a fourier transform (spectrum
analysis) with two smooth "humps", whereas the fourier transform of two sin
waves is two infinitessimally thin "spikes", and I'm trying to figure out
which is correct...

-Pete Lawrence.


 >
 >Original poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br>
 >
 >Tesla list wrote:
 >
 > > Original poster: Peter Lawrence <Peter.Lawrence-at-Sun.COM>
 >
 > >          is there an exact mathematical formula for the voltage and
 >current in
 > > the primary verses time in a TC (a hypothetical one with no losses) (ie a
 > > losely coupled dual resonator).
 > >
 > > I'm guessing its proportinal to either sin(C*t)+sin(D*t) or
 >sin(C*t)*sin(D*t)
 > > for suitable constants C,D that depend on Fres and K...
 >
 >I prefer to make this analysis by synthesis:
 >Start with the desired mode, a:b, and two of the elements and
 >calculate the required element values making L1*C1=L2*C2.
 >The coupling coefficient is k12=(b^2-a^2)/(b^2+a^2).
 >This coil will oscillate at the frequencies b*w and a*w, where
 >w=(1/(a*b))sqrt((a^2+b^2)/(2*L2*C2)) rad/s.
 >The primary voltage is:
 >v1(t)=v1max*(0.5*cos(a*w*t)+0.5*cos(b*w*t))
 >To find the primary current, just calculate the current in the
 >primary capacitor with this voltage waveform (i1(t)=-C1*dv1/dt):
 >I1(t)=v1max*C1*(0.5*a*w*sin(a*w*t)+0.5*b*w*sin(b*w*t))
 >It's equally easy to find the secondary waveforms. Consider that
 >v2max=v1max*sqrt(C1/C2), by energy conservation, and that the
 >waveform vc2 has the same shape of vc1 (with maximum vc2max), but
 >starts at zero instead of the maximum value.
 >v2(t)=v2max*(0.5*cos(a*w*t)-0.5*cos(b*w*t)).
 >i2(t)=v2max*C2*(0.5*a*w*sin(a*w*t)-0.5*b*w*sin(b*w*t))
 >
 >Ex: C1=10nF, L1=100 uH, L2=100 mH, C2=10 pF, mode 9:10, v1max=10 kV.
 >w=(1/90)*sqrt(181/(2*1e-12))=105.7 krad/s (16823 Hz)
 >A=a*w=951.31e3 (151.5 kHz)
 >B=b*w=1057.0e3 (168.2 kHz)
 >k12=19/181=0.1050
 >v1(t)=5000*(cos(A*t)+cos(B*t)); Maximum=10 kV
 >i1(t)=5.2851(9*sin(A*t)+10*sin(B*t)); Maximum=~100 A
 >v2(t)=158114*(cos(A*t)-cos(B*t)); Maximum=316 kV
 >i2(t)=167.13e-3*(9*sin(A*t)-10*sin(B*t)); Maximum=~3.16 A
 >
 >Looks correct. The output voltage has a negative maximum for a
 >noninverting transformer.
 >I had not observed previously that it was so simple to obtain the
 >secondary waveforms. Interesting.
 >
 >Antonio Carlos M. de Queiroz
 >
 >