Re: Laplace Transforms of piecewise continuous functions
- To: mathgroup at smc.vnet.net
- Subject: [mg28316] Re: [mg28292] Laplace Transforms of piecewise continuous functions
- From: Jack Goldberg <jackgold at math.lsa.umich.edu>
- Date: Wed, 11 Apr 2001 02:00:56 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Michael;
If a function f is defined over an interval, say [a,b], then
f*(UnitStep[x-a]-UnitStep[x-b])
is zero outside [a,b] and f inside. So you can piece together functions f
and g as follows:
f*(UnitStep[x-a]-UnitStep[x-b])+g*(UnitStep[x-c]-UnitStep[x-d])
Besides the ease in contructing piecewise functions this way, Mathematica knows
how to integrate f[x]*UnitStep[x-a]. One warning: Expand before trying
to integrate so that Mathematica is faced with sums like f[x]*UnitStep[x-a].
Also, you need version 3.xx or higher otherwise you have to download a
package.
Jack Goldberg
On Sat, 7 Apr 2001, Michael A. Powers wrote:
> Hi,
>
> The documentation isn't very clear on how to compute a Laplace Transform of
> a piecewise continuous function f(t). Say I have a function f(t) such that:
>
> f(t) = {3 over 0<=t<2, -2 over 2<=t<3, 0 over 3 <=t}
>
> how can I use the LaplaceTransform[] function to compute this easily?
> (aside from separately integrating the pieces, and adding)
>
> Thanks,
>
> -Mike
>
>
>