path: root/source/know/concept/fourier-transform/index.md

---
title: "Fourier transform"
sort_title: "Fourier transform"
date: 2021-02-22
categories:
- Mathematics
- Physics
- Optics
layout: "concept"
---

The **Fourier transform** (FT) is an integral transform which converts a
function $$f(x)$$ into its frequency representation $$\tilde{f}(k)$$.
Great volumes have already been written about this subject,
so let us focus on the aspects that are useful to physicists.

The **forward** FT is defined as follows, where $$A$$, $$B$$, and $$s$$ are unspecified constants
(for now):

$$\begin{aligned}
    \boxed{
        \tilde{f}(k)
        \equiv \hat{\mathcal{F}}\{f(x)\}
        \equiv A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x}
    }
\end{aligned}$$

The **inverse Fourier transform** (iFT) undoes the forward FT operation:

$$\begin{aligned}
    \boxed{
        f(x)
        \equiv \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}
        \equiv B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k x) \dd{k}
    }
\end{aligned}$$

Clearly, the inverse FT of the forward FT of $$f(x)$$ must equal $$f(x)$$
again. Let us verify this, by rearranging the integrals to get the
[Dirac delta function](/know/concept/dirac-delta-function/) $$\delta(x)$$:

$$\begin{aligned}
    \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{f(x)\}\}
    &= A B \int_{-\infty}^\infty \exp(-i s k x) \int_{-\infty}^\infty f(x') \exp(i s k x') \dd{x'} \dd{k}
    \\
    &= 2 \pi A B \int_{-\infty}^\infty f(x') \Big(\frac{1}{2\pi} \int_{-\infty}^\infty \exp(i s k (x' - x)) \dd{k} \Big) \dd{x'}
    \\
    &= 2 \pi A B \int_{-\infty}^\infty f(x') \: \delta(s(x' - x)) \dd{x'}
    = \frac{2 \pi A B}{|s|} f(x)
\end{aligned}$$

Therefore, the constants $$A$$, $$B$$, and $$s$$ are subject to the following
constraint:

$$\begin{aligned}
    \boxed{\frac{2\pi A B}{|s|} = 1}
\end{aligned}$$

But that still gives a lot of freedom. The exact choices of $$A$$ and $$B$$
are generally motivated by the [convolution theorem](/know/concept/convolution-theorem/)
and [Parseval's theorem](/know/concept/parsevals-theorem/).

The choice of $$|s|$$ depends on whether the frequency variable $$k$$
represents the angular ($$|s| = 1$$) or the physical ($$|s| = 2\pi$$)
frequency. The sign of $$s$$ is not so important, but is generally based
on whether the analysis is for forward ($$s > 0$$) or backward-propagating
($$s < 0$$) waves.



## Derivatives

The FT of a derivative has a very useful property.
Below, after integrating by parts, we remove the boundary term by
assuming that $$f(x)$$ is localized, i.e. $$f(x) \to 0$$ for $$x \to \pm \infty$$:

$$\begin{aligned}
    \hat{\mathcal{F}}\{f'(x)\}
    &= A \int_{-\infty}^\infty f'(x) \exp(i s k x) \dd{x}
    \\
    &= A \big[ f(x) \exp(i s k x) \big]_{-\infty}^\infty - i s k A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x}
    \\
    &= (- i s k) \tilde{f}(k)
\end{aligned}$$

Therefore, as long as $$f(x)$$ is localized, the FT eliminates derivatives
of the transformed variable, which makes it useful against PDEs:

$$\begin{aligned}
    \boxed{
        \hat{\mathcal{F}}\{f'(x)\} = (- i s k) \tilde{f}(k)
    }
\end{aligned}$$

This generalizes to higher-order derivatives, as long as these
derivatives are also localized in the $$x$$-domain, which is practically
guaranteed if $$f(x)$$ itself is localized:

$$\begin{aligned}
    \boxed{
        \hat{\mathcal{F}} \Big\{ \dvn{n}{f}{x} \Big\}
        = (- i s k)^n \tilde{f}(k)
    }
\end{aligned}$$

Derivatives in the frequency domain have an analogous property:

$$\begin{aligned}
    \dvn{n}{\tilde{f}}{k}
    &= A \dvn{n}{}{k}\int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x}
    \\
    &= A \int_{-\infty}^\infty (i s x)^n f(x) \exp(i s k x) \dd{x}
    = \hat{\mathcal{F}}\{ (i s x)^n f(x) \}
\end{aligned}$$



## Multiple dimensions

The Fourier transform is straightforward to generalize to $$N$$ dimensions.
Given a scalar field $$f(\vb{x})$$ with $$\vb{x} = (x_1, ..., x_N)$$,
its FT $$\tilde{f}(\vb{k})$$ is defined as follows:

$$\begin{aligned}
    \boxed{
        \tilde{f}(\vb{k})
        \equiv \hat{\mathcal{F}}\{f(\vb{x})\}
        \equiv A \int_{-\infty}^\infty f(\vb{x}) \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
    }
\end{aligned}$$

Where the wavevector $$\vb{k} = (k_1, ..., k_N)$$.
Likewise, the inverse FT is given by:

$$\begin{aligned}
    \boxed{
        f(\vb{x})
        \equiv \hat{\mathcal{F}}^{-1}\{\tilde{f}(\vb{k})\}
        \equiv B \int_{-\infty}^\infty \tilde{f}(\vb{k}) \exp(- i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{k}}
    }
\end{aligned}$$

In practice, in $$N$$D, there is not as much disagreement about
the constants $$A$$, $$B$$ and $$s$$ as in 1D:
typically $$A = 1$$ and $$B = 1 / (2 \pi)^N$$, with $$s = \pm 1$$.
Any choice will do, as long as:

$$\begin{aligned}
    \boxed{
        A B
        = \bigg( \frac{|s|}{2 \pi} \bigg)^{\!N}
    }
\end{aligned}$$


{% include proof/start.html id="proof-constants-ndim" -%}
The inverse FT of the forward FT of $$f(\vb{x})$$ must be equal to $$f(\vb{x})$$ again, so:

$$\begin{aligned}
    \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\}
    &= A B \int \exp(- i s \vb{k} \cdot \vb{x})
    \int f(\vb{x}') \exp(i s \vb{k} \cdot \vb{x}') \ddn{N}{\vb{x}'} \ddn{N}{\vb{k}}
    \\
    &= (2 \pi)^N A B \int f(\vb{x}')
    \Big( \frac{1}{(2 \pi)^N} \int \exp(i s \vb{k} \cdot (\vb{x}' - \vb{x})) \ddn{N}{\vb{k}} \Big) \ddn{N}{\vb{x}'}
    \\
    &= (2 \pi)^N A B \int f(\vb{x}')
    \Big( \prod_{n = 1}^N \frac{1}{2 \pi} \int \exp(i s k_n (x_n' - x_n)) \dd{k_n} \Big) \ddn{N}{\vb{x}'}
\end{aligned}$$

Here, we recognize the definition of the Dirac delta function again,
leading to:

$$\begin{aligned}
    \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\}
    &= (2 \pi)^N A B \int f(\vb{x}')
    \Big( \prod_{n = 1}^N \delta(s(x_n' - x_n)) \Big) \ddn{N}{\vb{x}'}
    \\
    &= \frac{(2 \pi)^N A B}{|s|^N} \int f(\vb{x}') \: \delta(\vb{x}' - \vb{x}) \ddn{N}{\vb{x}'}
    = \frac{(2 \pi)^N A B}{|s|^N} f(\vb{x})
\end{aligned}$$
{% include proof/end.html id="proof-constants-ndim" %}


Differentiation is more complicated for $$N > 1$$,
but the FT is still useful,
notably for the Laplacian $$\nabla^2 \equiv \idv{ {}^2}{x_1^2} + ... + \idv{ {}^2}{x_N^2}$$.
Let $$|\vb{k}|$$ be the norm of $$\vb{k}$$,
then for a localized $$f$$:

$$\begin{aligned}
    \boxed{
        \hat{\mathcal{F}}\{\nabla^2 f(\vb{x})\}
        = - s^2 |\vb{k}|^2 \tilde{f}(\vb{k})
    }
\end{aligned}$$


{% include proof/start.html id="proof-laplacian" -%}
We insert $$\nabla^2 f$$ into the FT,
decompose the exponential and the Laplacian,
and then integrate by parts (limits $$\pm \infty$$ omitted):

$$\begin{aligned}
    \hat{\mathcal{F}}\{\nabla^2 f\}
    &= A \int \big( \nabla^2 f \big) \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
    \\
    &= A \int \Big( \sum_{n = 1}^N \pdv{ {}^2 f}{x_n^2} \Big) \Big( \prod_{m = 1}^N \exp(i s k_m x_m) \Big) \ddn{N}{\vb{x}}
    \\
    &= A \sum_{n = 1}^N \bigg[ \pdv{f}{x_n} \exp(i s \vb{k} \cdot \vb{x}) \bigg]
    - A \sum_{n = 1}^N i s k_n \int \pdv{f}{x_n} \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
\end{aligned}$$

Just like in 1D, we get rid of the boundary term
by assuming that all derivatives $$\idv{f}{x_n}$$ are nicely localized.
To proceed, we then integrate by parts again:

$$\begin{aligned}
    \hat{\mathcal{F}}\{\nabla^2 f\}
    &= - A \sum_{n = 1}^N i s k_n \int \pdv{f}{x_n} \Big( \prod_{m = 1}^N \exp(i s k_m x_m) \Big) \ddn{N}{\vb{x}}
    \\
    &= - A \sum_{n = 1}^N i s k_n \bigg[ f \exp(i s \vb{k} \cdot \vb{x}) \bigg]
    + A \sum_{n = 1}^N (i s k_n)^2 \int f \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
\end{aligned}$$

Once again, we remove the boundary term
by assuming that $$f$$ is localized, yielding:

$$\begin{aligned}
    \hat{\mathcal{F}}\{\nabla^2 f\}
    &= - A s^2 \sum_{n = 1}^N k_n^2 \int f \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
    = - s^2 \sum_{n = 1}^N k_n^2 \tilde{f}
\end{aligned}$$
{% include proof/end.html id="proof-laplacian" %}



## References
1.  O. Bang,
    *Applied mathematics for physicists: lecture notes*, 2019,
    unpublished.