On Wed, Mar 11, 2015 at 2:38 AM, Kalevi Suominen <[email protected]> wrote:
>
>
> 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.

Oh, that's a bit high. Is it possible to do the stabilization automatically?

Aaron Meurer

>
>>
>> On Sun, Feb 8, 2015 at 2:57 PM, Kalevi Suominen <[email protected]> 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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