So perhaps you would have better results if you started with a Float
with higher precision, like Float("0.07071", 100).

Aaron Meurer

On Sun, Feb 8, 2015 at 2:57 PM, Kalevi Suominen <jks...@gmail.com> 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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