---
title: "Fourier transform"
firstLetter: "F"
publishDate: 2021-02-22
categories:
- Mathematics
- Physics
- Optics

date: 2021-02-22T21:35:54+01:00
draft: false
markup: pandoc
---

# Fourier transform

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):

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

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

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

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 interesting 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:

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

Derivatives in the frequency domain have an analogous property:

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