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. -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at http://groups.google.com/group/sympy. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/83a765a9-c984-4fa2-bbc2-00010e02ed83%40googlegroups.com. For more options, visit https://groups.google.com/d/optout.
