       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
>
>
>

```

• Prev by Date: Re: implicit function
• Next by Date: Re: newbie question about the symbol E
• Previous by thread: Re: Laplace Transforms of piecewise continuous functions
• Next by thread: Re: How to convert a list of functions in a list valued function