That is what I think I am seeing.
And Vo = 10 volts, I assume.
The graphs are so poor that I have been able to make little sense of them.
I also get a peak value of 8.626 at t=1.478, found by solving for when
the derivative i(t) = 0 and applying that time to the i(t) formula,
but 5.345 from the i peak formula. At least one of them is wrong.
If I solve produce a formula for peak current, based on the i(t)
formula, I get:
ipeak=(Vo/(L*w0))*exp((-R/(2*L^2*w0))*atan(2*L^2*w/R))*...
sin(atan(2*L^2*w)/R)
and this gives a value of 8.626, which just verifies that I did the
differentiation and simplification right, since the i(t) it was based
on looks like it has the same peak value. It doesn't prove that i(t)
was right to start with.
(snip)
Yes. I used values of Vo=1, C=1, R=1 and L=R^2*C/4 and a very
slightly higher value (1.0001 times) for L, for the under damped case.
Using Mathcad to evaluate the given expressions for i(t).
I experimented with the over damped case, earlier, (deriving the i
peak formula I pasted to A.B.S.E) to solve for i peak) and verified
that i peak based on the the given over damped i(t) converged to i
peak given for the critical damped case as critical damping was
approached from the over damped side.
So I think the over damped i(t) formula is at least consistent with
the critical i peak formula. But I haven't yet derived the critical i
peak formula from the critical damped i(t) formula.
I've lost track.
So does this lead you to suspect the formula for i(t) for the under
damped case is in error?
Yes, indeed. You get the same numerical value from his web page formula
as I do for the peak current in the underdamped case, and I think this
formula, at least as I am able to read it, is wrong.
The formulas I posted mostly came from an application note published by
Sprague concerning their energy storage capacitors.
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I also solved the problem by writing an expression for the current as:
I(s) = (Vo/L)/(s^2 + s*R/L + 1/L/C)
and looking up the appropriate Laplace transform. The result is:
i(t) = (Vo/L/(a-b))*(Exp(a*t)-Exp(b*t))]
where a and b are the roots of the denominator of the I(s) expression.
This can be whipped into the form of the expressions for the under and
overdamped cases from the Sprague app note that I posted yesterday. The
critically damped case requires taking limits or explicitly making the
roots equal and then transforming.
tpeak can be found as: tp = (Ln(a/b) - 2*Pi*j*n)/(b-a), j=SQRT(-1)
What is neat about that expression is that if you have a calculator or PC
program that can deal with
complex numbers (including taking logs and
exponentials of complex arguments), you succesively set the variable n to
0,1,2,... and this expression will give the times for the successive peaks
(positive and negative) in the underdamped case; and with n=0, it will give
the peak time for the overdamped case. Then you just substitute those
times back into the expression for i(t) and you get the peak currents. So
you can get numerical results without having to manipulate the expression
for i(t) and possibly making a mistake in the algebra.
I've done all this and verified that the formulas from the Sprague app
note that I posted yesterday are correct (or made minor corrections as
necessary). The numerical results that you have gotten so far are the same
as mine, and I'm confident the formulas I posted yesterday are correct.