Fourier transform definition conventions and formulas

There are several slightly different ways to define a Fourier transform. This means that when you look up a theorem about the Fourier transform you have to ask yourself which convention the source is using. All the common conventions can be summarized in the following definition

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

where m is either 1 or 2π, σ is +1 or -1, and q is 2π or 1. This means there are eight potential definitions, one for each choice of m, σ, and q, though only six of these are widely used. Still, that’s six definitions! The differences are small, but they are annoying when you just want to quickly look something up.

We will refer to each definition by its choice of σ, q, and m. To make the notation slightly simpler, we will use τ = 2π. The eight possible Fourier transforms are then

$F_{+11},\, F_{+1\tau}, \,F_{+\tau1}, \,..., \,F_{-\tau\tau}.$

F_-τ1 may be the most common definition. It is the convention used by the classic text by Stein and Weiss and Wikipedia.

$({\cal F}_{-\tau 1} f)(\omega) = \int_{-\infty}^\infty \exp(-2\pi i \omega x) f(x) \, dx$

Other definitions are widely used as well. For example, Mathematica uses F_+1τ and probabilists use F₊₁₁ for “characteristic functions”, what they call a Fourier transform.

These notes are a quick reference for translating between conventions. They are divided into three parts:

1. Converting between definitions
2. Comparison of results, organized by theorem
3. Comparison of results, organized by convention

In first section shows, for example, how to convert between the ordinary frequency Fourier transform (q = 2π) and the angular frequency Fourier transform (q = 1). The second and third sections have the same information, organized differently. The second section will take one theorem at a time and discuss how it varies according to the various conventions. The third section will take one convention at a time and restate all the theorems.

Converting between definitions

For a function f(x), let F(f)(ω) be its Fourier transform. You can convert between the eight possible definitions by applying three equations. Here a * stands for any particular choice of a parameter, as long as the same choice is applied on both sides of the equation:

$({\cal F}_{+**} f)(\omega) &=& ({\cal F}_{-**} f)(-\omega) \\ ({\cal F}_{*\tau*} f)(\omega) &=& ({\cal F}_{*1*} f)(2\pi\omega) \\ ({\cal F}_{**\tau} f)(\omega) &=& \frac{1}{\sqrt{2\pi}}({\cal F}_{**1} f)(\omega)$

Another way to convert between conventions is to compare each to a single convention. Based on the assumption that F_-τ1 is most common, I’ll show how each of the others relates to it.

$\begin{eqnarray*} ({\cal F}_{-11} f)(\omega) &=& ({\cal F}_{-\tau 1} f)\left(\frac{\omega}{2\pi}\right) \\ ({\cal F}_{-1\tau} f)(\omega) &=&\frac{1}{\sqrt{2\pi}} ({\cal F}_{-\tau 1 } f)\left(\frac{\omega}{2\pi}\right) \\ ({\cal F}_{-\tau\tau} f)(\omega) &=&\frac{1}{\sqrt{2\pi}} ({\cal F}_{-\tau 1} f)(\omega) \\ ({\cal F}_{+\tau1} f)(\omega) &=& ({\cal F}_{-\tau 1} f)(-\omega) \\ ({\cal F}_{+11} f)(\omega) &=& ({\cal F}_{-\tau 1} f)\left(-\frac{\omega}{2\pi}\right) \\ ({\cal F}_{+1\tau} f)(\omega) &=&\frac{1}{\sqrt{2\pi}} ({\cal F}_{-\tau 1} f)\left(-\frac{\omega}{2\pi}\right) \\ ({\cal F}_{+\tau\tau} f)(\omega) &=&\frac{1}{\sqrt{2\pi}} ({\cal F}_{-\tau 1} f)(-\omega) \end{eqnarray*}$

Comparisons organized by theorem

Integration

The integral of a function is its Fourier transform evaluated at 0. This is true for all conventions with m = 1. When m = τ an extra term is needed.

$\int_{-\infty}^\infty f(x)\, dx =({\cal F}_{**1} f)(0) = \sqrt{2\pi} ({\cal F}_{**\tau} f) (0)$

Shifting

Shifting the argument of a function rotates its Fourier transform. The amount of rotation does not depend on the choice of scaling factor m, but does depend on the sign convention σ and the frequency convention q:

$({\cal F}_{\sigma qm} f(x - h))(\omega) = \exp(i \sigma qh\omega) ({\cal F}_{\sigma qm} f(x))(\omega)$

Inversion

The inversion formulas are simplest when the frequency conventions q is 2π and the scaling factor is 1, or vice versa. F_+τ1 and F_-τ1 are inverses of each other, as are F_+1τ and F_-1τ. The other conventions involve extra factors of 2π.

${\cal F}_{+\tau 1}( {\cal F}_{-\tau 1} f) &=& f \\ {\cal F}_{+1 \tau}( {\cal F}_{-1 \tau} f) &=& f \\ {\cal F}_{+ 11}( {\cal F}_{-11} f) &=& 2\pi f \\ {\cal F}_{+\tau\tau}({\cal F}_{-\tau\tau} f) &=& \frac{1}{2\pi} f$

Parseval

Define the inner product of two functions to be the integral of their product over the real line. (Take the complex conjugate of the latter function if the functions are complex-valued.)

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

Then Parseval’s theorem says that the inner product of two functions equals the inner product of their Fourier transforms. This is true for definitions F_+τ1, F_-τ1, F_+1τ, and F_-1τ.

$\langle f, g \rangle = \langle {\cal F}_{\pm 1\tau} f, {\cal F}_{\pm 1\tau} g \rangle = \langle {\cal F}_{\pm \tau 1} f, {\cal F}_{\pm \tau 1} g \rangle$

For definitions F₊₁₁ and F_-11 the inner product of the Fourier transforms is larger by a factor of 2π. For definitions F_+ττ and F_-ττ the inner product of the Fourier transforms is smaller by a factor of 2π.

$\langle f, g \rangle = \frac{1}{2\pi} \langle {\cal F}_{\pm 11} f, {\cal F}_{\pm 11} g \rangle = 2\pi \langle {\cal F}_{\pm \tau \tau} f, {\cal F}_{\pm \tau \tau} g \rangle$

Plancherel

Plancherel’s formula is Parseval’s formula with g = f. This says a function and its Fourier transform have the same L² form for definitions F_+τ1, F_-τ1, F_+1τ, and F_-1τ. For definitions F₊₁₁ and F_-11 the norm of the Fourier transforms is larger by a factor of √2π. For definitions F_+ττ and F_-ττ the inner product of the Fourier transforms is smaller by a factor of √(2π).

Differentiation

The Fourier transform of the derivative of a function is a multiple of the Fourier transform of the original function. The multiplier is -σqi where σ is the sign convention and q is the angle convention. The scale convention m does not matter.

Convolution

The convolution of two functions is defined by

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

Fourier transform turns convolutions into products:

${\cal F}(f*g) = \sqrt{m} \, {\cal F}( f ) {\cal F}( g )$

So for conventions with m = 1, the Fourier transform of the convolution is the product of the Fourier transforms. For the conventions with m = 2π there is an extra factor of √(2π).

Theorems organized by convention

(-, τ 1)

$\begin{align*} \hat{f}(\omega) &=& \int_{-\infty}^\infty \exp(-2\pi i \omega x) f(x) \, dx \\ f(x) &=& \int_{-\infty}^\infty \exp(2\pi i \omega x) f(\omega) \, d\omega \\ \widehat{f(x-h)} &=& \exp(-2\pi i h \omega) \hat{f}(\omega) \\ \hat{f}(0) &=& \int_{-\infty}^\infty f(x) \, dx \\ \int_{-\infty}^\infty f(x) \bar{g}(x) \, dx &=& \int_{-\infty}^\infty \hat{f}(\omega) \bar{\hat{g}}(\omega) \, d\omega \\ \int_{-\infty}^\infty | f(x) |^2 \, dx &=& \int_{-\infty}^\infty |\hat{f}(\omega)|^2 \, d\omega \\ \widehat{f'(x)} (\omega) &=& 2\pi i \omega \hat{f}(\omega) \\ \widehat{(f*g)}(\omega) &=& \hat{f}(\omega) \hat{g}(\omega) \end{align*}$

(-, 1, 1)

$\begin{align*} \hat{f}(\omega) &=& \int_{-\infty}^\infty \exp(- i \omega x) f(x) \, dx \\ f(x) &=& \frac{1}{2\pi} \int_{-\infty}^\infty \exp( i \omega x) f(\omega) \, d\omega \\ \widehat{f(x-h)} &=& \exp(- i h \omega) \hat{f}(\omega) \\ \hat{f}(0) &=& \int_{-\infty}^\infty f(x) \, dx \\ \int_{-\infty}^\infty f(x) \bar{g}(x) \, dx &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \hat{f}(\omega) \bar{\hat{g}}(\omega) \, d\omega \\ \int_{-\infty}^\infty | f(x) |^2 \, dx &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty |\hat{f}(\omega)|^2 \, d\omega \\ \widehat{f'(x)} (\omega) &=& i \omega \hat{f}(\omega) \\ \widehat{(f*g)}(\omega) &=& \hat{f}(\omega) \hat{g}(\omega) \end{align*}$

(-, 1, τ)

$\begin{align*} \hat{f}(\omega) &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp(- i \omega x) f(x) \, dx \\ f(x) &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp( i \omega x) f(\omega) \, d\omega \\ \widehat{f(x-h)} &=& \exp(- i h \omega) \hat{f}(\omega) \\ \hat{f}(0) &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(x) \, dx \\ \int_{-\infty}^\infty f(x) \bar{g}(x) \, dx &=& \int_{-\infty}^\infty \hat{f}(\omega) \bar{\hat{g}}(\omega) \, d\omega \\ \int_{-\infty}^\infty | f(x) |^2 \, dx &=& \int_{-\infty}^\infty |\hat{f}(\omega)|^2 \, d\omega \\ \widehat{f'(x)} (\omega) &=& i \omega \hat{f}(\omega) \\ \widehat{(f*g)}(\omega) &=& \sqrt{2\pi} \hat{f}(\omega) \hat{g}(\omega) \end{align*}$

(-, τ τ)

$\begin{align*} \hat{f}(\omega) &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp(-2\pi i \omega x) f(x) \, dx \\ f(x) &=& 2\pi \int_{-\infty}^\infty \exp(2\pi i \omega x) f(\omega) \, d\omega \\ \widehat{f(x-h)} &=& \exp(-2\pi i h \omega) \hat{f}(\omega) \\ \hat{f}(0) &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(x) \, dx \\ \int_{-\infty}^\infty f(x) \bar{g}(x) \, dx &=& 2\pi \int_{-\infty}^\infty \hat{f}(\omega) \bar{\hat{g}}(\omega) \, d\omega \\ \int_{-\infty}^\infty | f(x) |^2 \, dx &=& 2\pi \int_{-\infty}^\infty |\hat{f}(\omega)|^2 \, d\omega \\ \widehat{f'(x)} (\omega) &=& 2\pi i \omega \hat{f}(\omega) \\ \widehat{(f*g)}(\omega) &=& \sqrt{2\pi} \hat{f}(\omega) \hat{g}(\omega) \end{align*}$

(+, τ 1)

$\begin{align*} \hat{f}(\omega) &=& \int_{-\infty}^\infty \exp(2\pi i \omega x) f(x) \, dx \\ f(x) &=& \int_{-\infty}^\infty \exp(-2\pi i \omega x) f(\omega) \, d\omega \\ \widehat{f(x-h)} &=& \exp(2\pi i h \omega) \hat{f}(\omega) \\ \hat{f}(0) &=& \int_{-\infty}^\infty f(x) \, dx \\ \int_{-\infty}^\infty f(x) \bar{g}(x) \, dx &=& \int_{-\infty}^\infty \hat{f}(\omega) \bar{\hat{g}}(\omega) \, d\omega \\ \int_{-\infty}^\infty | f(x) |^2 \, dx &=& \int_{-\infty}^\infty |\hat{f}(\omega)|^2 \, d\omega \\ \widehat{f'(x)} (\omega) &=& -2\pi i \omega \hat{f}(\omega) \\ \widehat{(f*g)}(\omega) &=& \hat{f}(\omega) \hat{g}(\omega) \end{align*}$

(+, 1, 1)

$\begin{align*} \hat{f}(\omega) &=& \int_{-\infty}^\infty \exp( i \omega x) f(x) \, dx \\ f(x) &=& \frac{1}{2\pi} \int_{-\infty}^\infty \exp(- i \omega x) f(\omega) \, d\omega \\ \widehat{f(x-h)} &=& \exp( i h \omega) \hat{f}(\omega) \\ \hat{f}(0) &=& \int_{-\infty}^\infty f(x) \, dx \\ \int_{-\infty}^\infty f(x) \bar{g}(x) \, dx &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \hat{f}(\omega) \bar{\hat{g}}(\omega) \, d\omega \\ \int_{-\infty}^\infty | f(x) |^2 \, dx &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty |\hat{f}(\omega)|^2 \, d\omega \\ \widehat{f'(x)} (\omega) &=& - i \omega \hat{f}(\omega) \\ \widehat{(f*g)}(\omega) &=& \hat{f}(\omega) \hat{g}(\omega) \end{align*}$

(+, 1, τ)

$\begin{align*} \hat{f}(\omega) &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp(i \omega x) f(x) \, dx \\ f(x) &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp(-i \omega x) f(\omega) \, d\omega \\ \widehat{f(x-h)} &=& \exp(i h \omega) \hat{f}(\omega) \\ \hat{f}(0) &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(x) \, dx \\ \int_{-\infty}^\infty f(x) \bar{g}(x) \, dx &=& \int_{-\infty}^\infty \hat{f}(\omega) \bar{\hat{g}}(\omega) \, d\omega \\ \int_{-\infty}^\infty | f(x) |^2 \, dx &=& \int_{-\infty}^\infty |\hat{f}(\omega)|^2 \, d\omega \\ \widehat{f'(x)} (\omega) &=& -i \omega \hat{f}(\omega) \\ \widehat{(f*g)}(\omega) &=& \sqrt{2\pi} \hat{f}(\omega) \hat{g}(\omega) \end{align*}$

(+, τ τ)

$\begin{align*} \hat{f}(\omega) &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp(2\pi i \omega x) f(x) \, dx \\ f(x) &=& 2\pi \int_{-\infty}^\infty \exp(-2\pi i \omega x) f(\omega) \, d\omega \\ \widehat{f(x-h)} &=& \exp(2\pi i h \omega) \hat{f}(\omega) \\ \hat{f}(0) &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(x) \, dx \\ \int_{-\infty}^\infty f(x) \bar{g}(x) \, dx &=& 2\pi \int_{-\infty}^\infty \hat{f}(\omega) \bar{\hat{g}}(\omega) \, d\omega \\ \int_{-\infty}^\infty | f(x) |^2 \, dx &=& 2\pi \int_{-\infty}^\infty |\hat{f}(\omega)|^2 \, d\omega \\ \widehat{f'(x)} (\omega) &=& -2\pi i \omega \hat{f}(\omega) \\ \widehat{(f*g)}(\omega) &=& \sqrt{2\pi} \hat{f}(\omega) \hat{g}(\omega) \end{align*}$

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