Re: Fourier Transform of a "step" function

*To*: mathgroup at smc.vnet.net*Subject*: [mg128510] Re: Fourier Transform of a "step" function*From*: Murray Eisenberg <murray at math.umass.edu>*Date*: Thu, 25 Oct 2012 23:35:07 -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>

On Oct 25, 2012, at 1:43 AM, etaghtron at gmail.com wrote: > 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. You may construct the function in Mathematica as a Piecewise expression: f[x_] := Piecewise[{{a + b x + c x^2, 0 < x < x0}, {0, x0 < x}}] Then: FourierTransform[f[x], x, y] If you have specific numeric values for the coefficients or the dividing point x0, you may of course insert those directly -- or, instead, in the definition of f[x_] just append an expression such as: /. {a - >2.1, b -> -1, c -> 3.2, x0 ->1.5} In this example you could also use an If expression to define the function -- f[x_] := If[0 < x < x0, a + b x + c x^2, 0] -- but typically a piecewise-defined function will be treated more consistently by other Mathematica operations than one defined using If. --- Murray Eisenberg murray at math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2838 (W) 710 North Pleasant Street fax 413 545-1801 Amherst, MA 01003-9305

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