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)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net*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 >

**References**:**Fourier Transform of a "step" function***From:*etaghtron@gmail.com