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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