Fourier transforms in Mathematica

Unfortunately there are many slightly different ways to define the Fourier transform. So the first two questions when using Mathematica (or any other software) to compute Fourier transforms are what definition of Fourier transform does it use, and what to do if you want to use a different definition.

The answer to the first question is that Mathematica defines the Fourier transform of f as

\hat{f}(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp(i\omega x)\, f(x) \, dx

The answer to the second question is that Mathematica defines a parameterized Fourier transform by

\hat{f}(\omega) = \sqrt{\frac{|b|}{(2\pi)^{1-a}}} \, \int_{-\infty}^\infty \exp(ib\omega x)\, f(x) \, dx

where a defaults to 0 and b defaults to 1.


For example, if φ(x) = exp(-x²/2), then we can compute Mathematica’s default Fourier transform with

    φ[x_] := Exp[-x^2/2]
    FourierTransform[φ[x], x, ω]

This computes

 \hat{\varphi}(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp(i\omega x)\, \varphi(x) \, dx = \exp(-\omega^2/2)

But if we set (a, b) to (0, 2π) with

    FourierTransform[φ[x], x, ω FourierParameters -> {0, 2 Pi}]

we compute

\hat{\varphi}(\omega) = \int_{-\infty}^\infty \exp(2\pi i\omega x)\, \varphi(x) \, dx = \sqrt{2\pi} \exp(-2\pi^2\omega^2/2)

Theorems under various conventions

Several years ago I got frustrated enough with the multitude of Fourier transform conventions that I made a sort of Rosetta stone, giving several of the most basic theorems under each of eight definitions. I used the parameterization

\hat{f}(\omega) = \frac{1}{\sqrt{m}} \int_{-\infty}^\infty \exp(\sigma q i\omega x)\, f(x) \, dx

where m is either 1 or 2π, σ is either 1 or -1, and q is either 1 or 2π.

My σ and q are the sign and absolute value of Mathematica’s b. The relation between my m and Mathematica’s a is slightly more complicated and given in the table below.

\begin{tabular}{lllll} \hline \(\sigma\) & m & q & a & b\\ \hline \textpm{} 1 & 1 & 1 & 1 & \textpm{} 1\\ \textpm{} 1 & 1 & 2\(\pi\) & 0 & \textpm{} 2\(\pi\)\\ \textpm{} 1 & 2\(\pi\) & 1 & 0 & \textpm{} 1\\ \textpm{} 1 & 2\(\pi\) & 2\(\pi\) & -1 & \textpm{} 2\(\pi\)\\ \hline \end{tabular}

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