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Re: problem
*To*: mathgroup at smc.vnet.net
*Subject*: [mg23715] Re: [mg23664] problem
*From*: "Mark Harder" <harderm at ucs.orst.edu>
*Date*: Mon, 5 Jun 2000 01:09:11 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
Amna,
Since I don't see a Pulse or Square Well function in the Help browser, I
constructed one. Look at UnitStep[x+1]-UnitStep[x-1] with
Plot[UnitStep[x + 1] - UnitStep[x - 1], {x, -2, 2}]
I think this is what you want from your H[x]. Now linearity of the inverse
f.t. implies that
InverseFourierTransform[H[x] ] =InverseFourierTransform[UnitStep[x+1] ] -
InverseFourierTransform[UnitStep[x-1] ]:
So, I get
In[526]:=
Ft1 = InverseFourierTransform[UnitStep[x + 1], x, t]
Out[526]=
I t
I E Pi
-(------------) + Sqrt[--] DiracDelta[t]
Sqrt[2 Pi] t 2
In[527]:=
Ft2 = InverseFourierTransform[UnitStep[x - 1], x, t]
Out[527]=
I Pi
-(-----------------) + Sqrt[--] DiracDelta[t]
I t 2
E Sqrt[2 Pi] t
In[534]:=
FullSimplify[Ft1 - Ft2]
Out[534]=
2
Sqrt[--] Sin[t]
Pi
---------------
t
In other words, Sqrt[2/Pi] * SincFunction[t], which I believe is correct.
-mark harder
-----Original Message-----
From: Amna Latif <hochimin at nexlinx.net.pk>
To: mathgroup at smc.vnet.net
Subject: [mg23715] [mg23664] problem
>I can't figure out the inverse fourier transform of the following function:
>H(w)= 1 |x|<1,
> 0 |x|>1
>Amna
>
>
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