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MathGroup Archive 2001

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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
> 
> 
> 



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