Re: Problem with LaplaceTransform and InverseLaplaceTransform
- To: mathgroup at smc.vnet.net
- Subject: [mg67597] Re: Problem with LaplaceTransform and InverseLaplaceTransform
- From: bghiggins at ucdavis.edu
- Date: Sat, 1 Jul 2006 05:12:56 -0400 (EDT)
- References: <e7td5h$3kh$1@smc.vnet.net><e82nnu$r2v$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Maxim,
When you include the zero in the integration range you do not get the
same result. The result is off by a factor of 1/2. Is this a bug in
Mathematica or am I missing something?
Integrate[DiracDelta[Ï?], {Ï?, 0, t}, Assumptions -> t â??
Reals] + 2*Sum[Integrate[(-1)^k*DiracDelta[Ï? - Pi*
k], {Ï?, 0, t}, Assumptions -> {t
â?? Reals}], {k, Infinity}] // Simplify
Piecewise[{{1/2, t < Pi}}, -(1/2) + (-1)^Floor[t/Pi]]
Your result is
Piecewise[{{1, Inequality[0, LessEqual, t, Less, Pi]},
{(-1)^Floor[t/Pi], t >= Pi}}]
These calculations were done on a Mac OSX (10.4.6) with Mathematica
Version 5.2
Cheers,
Brian
ab_def at prontomail.com wrote:
> aXi wrote:
> > Can someone point me to site that would help me in using Mathematica
> > 5.2for purpose of calculations related to Systhems theory. I'm having
> > trouble with several topics... for example doing
> > InverseLaplaceTransform of function:
> >
> > (1/s) * (TanH[pi*s/2])
> >
> > Thanks in advance!
>
> If you want the inverse Laplace transform of Tanh[Pi*s/2]/s, it can be
> found as follows. First take the transform of Tanh[Pi*s/2]:
>
> In[1]:= fs = Tanh[Pi*s/2]/s;
>
> In[2]:= InverseLaplaceTransform[s*fs, s, t]
>
> Out[2]= DiracDelta[t] + 2*Sum[(-1)^K$40*DiracDelta[(-K$40)*Pi + t],
> {K$40, 1, Infinity}]
>
> Then integrate the result:
>
> In[3]:= Integrate[DiracDelta[t], t] +
> 2*Sum[Integrate[(-1)^k*DiracDelta[t - Pi*k], t], {k, Infinity}]
>
> Out[3]= UnitStep[t] + (-1 + (-1)^Floor[t/Pi])*UnitStep[-1 + t/Pi]
>
> Strictly speaking, we need to evaluate Integrate[f[tau], {tau, 0-, t}],
> 'including' the zero in the integration range.
>
> It is easy to see that the result is equal to (-1)^Floor[t/Pi] or
> Sign[Sin[t]]. Here's a check:
>
> In[4]:= LaplaceTransform[Sign[Sin[t]], t, s] - fs // Simplify
>
> Out[4]= 0
>
> Maxim Rytin
> m.r at inbox.ru