Re: a question about the UnitStep function

• To: mathgroup at smc.vnet.net
• Subject: [mg58422] Re: a question about the UnitStep function
• From: dh <dh at metrohm.ch>
• Date: Sat, 2 Jul 2005 04:06:19 -0400 (EDT)
• References: <da2msl\$944\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

Hi Zhou,
Your example is clearly wrong. Note also that
integrand=f[z] f[z-x]
where the argument of the second f[] (symmetric function!) has been
negated, gives a different, only partly evaluated result. This result,
if numerically evaluated, is correct.
I think this is something for Wolfram.  I will beg Daniel Lichtblau to
take  note.

sincerely, Daniel

Zhou Jiang wrote:
> Dear Mathgroup,
> I want to let Mathematica compute the convolution of two sqare waves. I did as follows
>
> f[x_]:=(UnitStep[x+1]-UnitStep[x-1])/2;
>
> integrand=f[z] f[x-z];
>
> Assuming[Element[x, Reals], Integrate[integrand, {z, -Infinity, Infinity}]]
>
> Mathematica gave me the result as follows,
> ((-1 + x) UnitStep[-1 + x] - x UnitStep[x] + (2 + x) UnitStep[2 + x])/4
>
> I plot the result to check
>
> Plot[%,{x,-10,10}, PlotRange->All];
>
> It is clear wrong since the convolution of two square waves should be convergent. Can anyone give me some help with the subtlties about the UnitStep function? Any thoughts are appriciable.
>
>

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