Re: convolution involving UnitStep

• To: mathgroup at smc.vnet.net
• Subject: [mg126125] Re: convolution involving UnitStep
• From: Ulf-Dietrich Braumann <braumann at uni-leipzig.de>
• Date: Fri, 20 Apr 2012 07:42:32 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <jmogf6\$46k\$1@smc.vnet.net>

```Hi,

while I was also ending up with a ConditionalExpression for t>1, then I
tried to flip the functions under the integral due to assumend
commutativity, but still got the same ConditionalExpression. For both
cases, however, Mathematica 8 could plot in the interval t \[Element] [0,1]. Later
I launched Mathematica 6, below you find the results:

In[133]:= y[t_] := Integrate[h[t - s] g[s], {s, 0, t}];
y[t] // FullSimplify

Out[134]= (-1 + Cos[1 - t]) UnitStep[-1 + t] -
2 (-1 + Cos[t]) UnitStep[t]

In[135]:= y[t_] := Integrate[h[s] g[t - s], {s, 0, t}];
y[t] // FullSimplify

Out[136]= \[Piecewise] {
{1 + Cos[1 - t] - 2 Cos[t], t > 1},
{2 - 2 Cos[t], 0 < t <= 1}
}

No idea how to get Mathematica 8 to provide a piecewise functional solution for 0
< t <= 1 as well. Mathematica 7 btw. behaves also strange, even though the
output is not restricted to t>1, the plots seem to be wrong.

I also have tried to define g as g[t_]:= 2UnitBox[t-1/2]+UnitStep[t-1] but
then the output of Integrate no longer can be simplified and given as
\[Piecewise] defined function.

Greetings - Ulf-Dietrich

On Thu, 19 Apr 2012, J Davis wrote:

> h[t_] = Sin[t];
> g[t_] = 2 UnitStep[t] - UnitStep[t - 1];
> y[t_] = Integrate[h[t-s]g[s],{s,0,t}]
>
> results in a conditional expression requiring t>1, but I want to
> evaluate and plot t values from [0,1] as well as t>1.
>
> I tried HeavisideTheta as well as := in the definition of y to no
> avail. Thanks for any help...
>
>

```

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