From: Paul
Date: Sun, 24 Mar 2002 20:33:13 +0000
Subject: Re: Ready :-)) - Re: [TSSP] short H/D and stuff
Terry,
I've been working out how to calculate the current profile using
just 11 pings of the coil.
For the ringdown, for each frequency component,
E_loss/Es = 2*pi/Q
where E_loss is energy lost per cycle and Es is the stored energy,
(both functions of time).
Consider E_loss to be the sum of the energy lost in the coil (Ec)
plus the energy lost in the extra series resistance (Ex). Then we
can say
Ec/Es = 2*pi/Qs ...(1)
where Qs is the Q measured without the extra resistor. With the
resistor added at point x = 1..10 in the coil, we have
(Ec + Ex)/Es = 2*pi/Qx ...(2)
where Qx is the associated observed Q. We have 10 of these
equations.
Substituting (1) into (2) gives
Ex/Es = 2*pi/(1/Qx - 1/Qs) ...(3)
Now at any cycle of the ringdown,
Es = Lee * Ib^2/2
where Ib is the peak base current, and the ten equations
Ex = R * Ix^2/(2*f) x = 1..10
in which Ix is the peak current at the resistor point x,
With these, (3) becomes
Ix^2/Ib^2 = 2*pi*f*Lee/R * (1/Qx - 1/Qs)
which is independent of time now that we have eliminated all the
energy terms. Noting that Ib = Ix when x = 1, we have
1 = 2*pi*f*Lee/R * (1/Q1 - 1/Qs)
and the other nine equations
Ix^2/Ib^2 = 2*pi*f*Lee/R * (1/Qx - 1/Qs) ; x = 2..10
The first equation gives
2*pi*f*Lee/R = Q1 * Qs/(Qs - Q1)
and using this to eliminate all the unknowns in the remaining
nine equations gives
Ix^2/Ib^2 = (1/Qx - 1/Qs) * Q1 * Qs/(Qs - Q1) ; x = 2..10
Therefore 10 pings, one for each link, plus a ping of the coil
without the resistor, can be used to determine the current
profile Ix/Ib through the formula
Ix/Ib = 1 ; x = 1
Ix/Ib = sqrt{ (Qs - Qx)/(Qs - Q1) * Q1/Qx } ; x = 2..10
and this recipe can be applied to all the frequency components
in the ping waveform.
The analysis code is just about ready. I'll send across a copy of
the source for you to try to compile.
--
Paul Nicholson,
--
Maintainer Paul Nicholson, paul@abelian.demon.co.uk.