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Re: Fourier Transform of a "step" function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg128507] Re: Fourier Transform of a "step" function
  • From: Bob Hanlon <hanlonr357 at gmail.com>
  • Date: Thu, 25 Oct 2012 23:34:07 -0400 (EDT)
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  • References: <20121025054335.0BFEB68B5@smc.vnet.net>

f[x_] = Piecewise[
   {{a + b x + c x^2, 0 < x < 20}}, 0];

g[w_] = FourierTransform[f[x], x, w] //
  FullSimplify

-((1/(Sqrt[2*Pi]*w^3))*
      (I*(2*c - w*(I*b + a*w) +
            E^(20*I*w)*(-2*c +
                 I*(b + 40*c)*w +
                 (a + 20*(b + 20*c))*w^2))))

f2[x_] = InverseFourierTransform[g[w], w, x]

(1/2)*(a + x*(b + c*x))*
   (Sign[20 - x] + Sign[x])

f2 differs from the original function only on the step transitions

Simplify[f2[x], #] & /@
 {x < 0, x == 0, 0 < x < 20, x == 20, x > 20}

{0, a/2, a + x*(b + c*x),
   (1/2)*(a + 20*(b + 20*c)), 0}

f2 has the same Fourier transform as the original function

g[w] == FourierTransform[f2[x], x, w]

True


Bob Hanlon


On Thu, Oct 25, 2012 at 1:43 AM,  <etaghtron at gmail.com> wrote:
> Dear Group,
>
> I am new to Mathematica. So question might be naive.
>
> I have a function, which is defined in this form
>
> F(x) = a + b x + c x^2(0 < x < x0)
> F(x) = 0 (x > x0)
>
> This function is simple. However, its Fourier transform is not straightforward due to the limits of the function. I probably can do it manually, but I have a few other functions that are more complex than this one, which also requires Fourier transform. So I'd like to know if Mathematica can handle this.
>
> If you could please shed some light on this, I would be really grateful.
>
> Best Regards,
> Frank
>



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