Reverse engineering Fourier conventions

The most annoying thing about Fourier analysis is that there are many slightly different definitions of the Fourier transform. One time I got sufficiently annoyed that I creates a sort of Rosetta Stone of Fourier theorems under eight different conventions. Later I discovered that Mathematica supports these same eight definitions, but with slightly different notation.

When I created my Rosetta Stone I wanted to have a set of notes that answered the question “What are the basic Fourier theorems under this convention?” Recently I was reading a reference and wanted to answer the opposite question “Given the theorems this book is stating, what convention must they be using?”

The eight definitions correspond to

\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 +1 or -1, and q is 2π or 1.

I’m posting these notes for my future reference and for anyone else who may need to do the same sleuthing.

Notation

For the rest of the post, let F and G be the Fourier transforms of f and g respectively. We write

f(x) \leftrightarrow F(\omega)

for the pair of a function and its Fourier transform.

Define the inner product of f and g as

\langle f, g \rangle = \int_{-\infty}^\infty f(x)\, g(x) \, dx

if f and g are real-value. If the functions are complex-valued, replace g with the complex conjugate if g.

We will sometimes denote the Fourier transform of a function by putting a hat on top of it.

Convolution theorem

The convolution theorem gives a quick way to determine the parameter m. Suppose convolution is defined by

(f*g)(x) = \int_{-\infty}^\infty f(x-y)\, g(y) \, dy

Then

f(x)*g(x) \leftrightarrow \sqrt{m} F(\omega)*G(\omega)

and so you can find m immediately. If f*g = F*G with no extra factor out front, m = 1. Otherwise if there’s a factor of √2π out front, then m = 2π. If there’s any other factor, you’ve got an arcane definition of Fourier transform that isn’t one of the eight considered here.

Some authors, like Walter Rudin, include a scaling factor in the definition of convolution, in which casethe argument of this section doesn’t hold.

Parseval and Plancherel

Parseval’s theorem says that the inner product of f and g is proportional to the inner product of F and G. The proportionality constant depends on the definition of Fourier transform, specifically on m and q, and so you can determine m or q from the form of Parseval’s theorem.

\langle f, g \rangle = k \langle F, G \rangle

If k = 1, then either q = 2π and m = 1 or q = 1 and m = 2π. If you know m, say from the statement of the convolution theorem, then Parseval’s theorem tells you q.

Plancherel’s theorem is the special case of Parseval’s theorem with f = g. It can be used the same way to solve for m or q.

If k = 2π, then q = m = 1. If k= 1/2π, then q = m = 1.

Double transform

Theorems about taking the Fourier transform twice carry the same information as Parseval’s and Plancherel’s theorems, i.e. they also let you determine m or q.

We have

\hat{\hat{f}}(x) = k f(-x)

with the same conclusions based on k as above:

  • If k = 1, then either q = 2π and m = 1 or q = 1 and m = 2π.
  • If k = 2π, then q = m = 1.
  • If k= 1/2π, then q = m = 1.

Shift and differentiation

So far we none of our theorems have allowed us to reverse engineer σ. Either the shift or differentiation theorem will let is find σq.

The shift theorem says

f(x-h) \leftrightarrow \exp(ikh\omega) F(\omega)

where k = σq. Since σ = ±1 and q = 1 or 2π, the product σq determines both σ and q.

Similarly, the differentiation theorem says the Fourier transform of the derivative of f transforms as

f'(x) \leftrightarrow ik\omega F(\omega)

and again k = σq.

2 thoughts on “Reverse engineering Fourier conventions

  1. Nice! You wrote f * g = sqrt(m) F * G, but I don’t think you meant that. You mean something like “the Fourier transformation of the convolution of f and g is the product of their Fourier transforms, up to a constant factor” – right?

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