On Wednesday, March 11, 2015 at 6:59:33 AM UTC+2, Aaron Meurer wrote:
>
> So perhaps you would have better results if you started with a Float
> with higher precision, like Float("0.07071", 100).
>
The floats are of the order 10**(-19000) in both the numerator and the
denominator before rounding.
> On Sun, Feb 8, 2015 at 2:57 PM, Kalevi Suominen <[email protected]
> <javascript:>> wrote:
> >
> >
> > On Sunday, February 8, 2015 at 12:06:46 AM UTC+2, Sten Sogaard wrote
> >>
> >>
> >> H = 1/(0.07071*s**2 + 258.0*s + 1100000.)
> >> inverse_laplace_transform(1/s*H,s,t)
> >>
> >>
> >> Any ideas?
> >
> >
> > The current implementation leads to floating point numbers with very
> large
> > positive and negative exponents. After the small numbers are rounded to
> > zero, nan results.
> >
> > However, the computation succeeds with suitably normalized coefficients:
> > ```
> > In [38]: H1 = H.subs(s, 10000.*s)
> > In [40]: print(inverse_laplace_transform(1/s*H1, s, t))
> > -1.0*(-9.09090909090909e-7*exp(0.182435299109037*t) +
> > 4.74279450417556e-7*sin(0.349688926583877*t) +
> >
> 9.09090909090909e-7*cos(0.349688926583877*t))*exp(-0.182435299109037*t)*Heaviside(t)
>
>
> > ```
> > (Of course, the result has to be normalized again.)
> >
> > For a more satisfactory solution, cancellation of exponents in the
> > intermediate results would be required.
> >
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