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

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

for the pair of a function and its Fourier transform.

Define the inner product of *f* and *g* as

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

Then

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.

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

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 σ q**.

The shift theorem says

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

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

and again *k* = σ*q*.

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?

Thanks. Yes. I meant to put in an arrow indicating Fourier transform pairs.