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 > >