Import standard modules:


In [ ]:
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
from IPython.display import HTML 
HTML('../style/course.css') #apply general CSS

Import section specific modules:


In [ ]:
pass

2.4 The Fourier Transform

The Fourier transform is one of the fundamental mathematical operations that is made use of in signal processing and interferomtry. It is introduced here.

2.4.1 Definition of the Fourier transform and its inversion

Consider the class of integrable functions $f$. Then the Fourier transform $\mathscr{F}$ is defined as:

$$ \mathscr{F}: [\mathbb{R} \rightarrow \mathbb{C}] \rightarrow [\mathbb{R} \rightarrow \mathbb{C}]\\ \forall f:\,\mathbb{R} \rightarrow \mathbb{C}, \int_{-\infty}^{+\infty} \lvert f(x) \rvert \,dx \in \mathbb{R}\\ \mathscr{F}\{f\}(s) = \int_{-\infty}^{+\infty}f(x)\,e^{-\imath 2\pi xs}dx $$

The Fourier transform is invertible with the inverse Fourier transform $\mathscr{F}^{-1}$ defined as

$$ \forall F:\,\mathbb{R} \rightarrow \mathbb{C}, \int_{-\infty}^{+\infty} \lvert F(s) \rvert \,ds \in \mathbb{R}\\ \mathscr{F}^{-1}\{F\}(x) = \int_{-\infty}^{+\infty}F(s)\,e^{\imath 2\pi xs}ds\\ \Rightarrow \mathscr{F}^{-1}\{\mathscr{F}{f}\}(x) = f(x) \land \mathscr{F}\{\mathscr{F}^{-1}{F}\}(s) = F(s) $$

For the proof of the above Fourier inversion therem we refer to e.g. this Wikipedia article.

Often the Fourier transform operator (and, sloppily, also its inverse) is abbrevated by a tilde

$$ \begin{align} \mathscr{F}\{f\}(s) \,&=\, \tilde{f}(s)\\ \mathscr{F}^{-1}\{F\}(x) \,&= \tilde{F}(x) \end{align} $$

and the two domains are separated by using lower case letters in the original domain and capital letters in the Fourier domain. A capitalized letter denotes the Fourier transform of its associated lower case letter. A Fourier pair is indicated by the $\rightleftharpoons$ symbol:

$$ \mathscr{F}\{f\}(s) = F(s) \,\Rightarrow\, f \rightleftharpoons F $$

For any function $f$ define

$$ f_{-}(x)=f(-x) $$

Then

$$ \begin{align} \mathscr{F}\{f\}(s)\,&=\, \int_{-\infty}^{+\infty} f(x) e^{-\imath xs}dx\\ &\underset{x^\prime = -x}{=}\,\int_{+\infty}^{-\infty} f(-x^{\prime}) e^{\imath x^\prime s} \frac{dx}{dx^{\prime}}\,dx^\prime\\ & = \, \int_{+\infty}^{-\infty} f(-x^{\prime}) e^{\imath x^\prime s}\,dx^\prime\\ & = \mathscr{F}^{-1}\{f_-\}(s) \end{align} $$

and

$$ \begin{align} \mathscr{F}^{-1}\{F\}(x)\,&=\, \int_{-\infty}^{+\infty} F(s) e^{\imath sx}ds\\ &\underset{s^\prime = -s}{=}\,\int_{+\infty}^{-\infty} F(-s^{\prime}) e^{-\imath s^\prime x} \frac{ds}{ds^{\prime}}\,ds^\prime\\ & = \, \int_{+\infty}^{-\infty} F(-s^{\prime}) e^{\imath s^\prime x}\,ds^\prime\\ & = \mathscr{F}\{F_-\}(s) \end{align} $$

Therefore, a repeated application of the same transformation results in the reverse function

$$ \begin{align} \mathscr{F}\{\mathscr{F}\{f\}\}(x) \,&=\, f_-(x) \,&=\, f(-x)\\ \mathscr{F}^{-1}\{\mathscr{F}^{-1}\{F\}\}(x) \,&=\, F_-(x) \,&=\, F(-x)\\ \end{align}\qquad $$

Equation 2.4.1

One can hence define the inverse Fourier transform as a triple forward Fourier transform. A corrolary is that an even function with $f_- = f$ the Fourier transformation has the same result as an inverse Fourier transform. Further, for an odd function $f(x) = -f(-x)$ the result of a Fourier transformation is the negative of the reverse transformation.

For a multi-dimensional function, the Fourier transform and its inverse are defined as

$$ \mathscr{F}: [\mathbb{R}^n \rightarrow \mathbb{C}] \rightarrow [\mathbb{R}^n \rightarrow \mathbb{C}] \,, \quad n\in \mathbb{N}\\ \forall f:\,\mathbb{R}^n \rightarrow \mathbb{C}, \int_{-\infty}^{+\infty} \lvert f(x) \rvert \,d^nx = \int_{-\infty}^{+\infty} \ldots \int_{-\infty}^{+\infty}\lvert f(x) \rvert \,dx_1\ldots dx_n\in \mathbb{R}\\ \begin{align} \mathscr{F}\{f\}(s_1, \ldots, s_n) \,&=\,\mathscr{F}\{f\}({\bf s})\\ &=\,\int_{-\infty}^{+\infty}f(x)\,e^{-\imath 2\pi( x_1s_1+\ldots+x_ns_n)}d^nx\\ &=\,\int_{-\infty}^{+\infty}f(x)\,e^{-\imath 2\pi( {\bf x}\cdot{\bf s})}d^nx\\ \mathscr{F}^{-1}\{F\}(x_1, \ldots, x_n) \,&=\,\mathscr{F}^{-1}\{F\}({\bf x})\\ &=\, \int_{-\infty}^{+\infty}F(s)\,e^{\imath 2\pi (x_1s_1+\ldots+x_ns_n)}d^ns\\ &=\, \int_{-\infty}^{+\infty}F(s)\,e^{\imath 2\pi ({\bf x}\cdot {\bf s})}d^ns\\ \end{align} $$

2.4.2 Fourier transform of the Gaussian

Remember the Gaussian $f(x) = a e^{-\frac{(x-\mu)^2}{2\sigma^2}}$ as defined in $\S$ 2.2 ➞ , its Fourier transform is

$$ \begin{align} \mathscr{F}\{f\}(s) \,&=\, \int_{-\infty}^{+\infty} a e^{-\frac{(x-\mu)^2}{2\sigma^2}} e^{-\imath 2\pi x s} dx\\ &=\int_{-\infty}^{+\infty} a e^{-\frac{(x-\mu)^2}{2\sigma^2}} e^{-\imath 2\pi (x-\mu+\mu) s} dx\\ &\underset{x^{\prime} = x-\mu}{=}\,\int_{-\infty}^{+\infty} a e^{-\imath 2\pi \mu s} e^{-\frac{{x^{\prime}}^2}{2\sigma^2}} e^{-\imath 2\pi x^\prime s} dx^{\prime}\\ &=\, a e^{-\imath 2\pi \mu s} \int_{-\infty}^{+\infty} e^{-\left[\left( \frac{x^{\prime}}{\sqrt{2}\sigma}+\imath\pi s\sqrt{2}\sigma\right)^2+2\pi^2s^2\sigma^2\right]}dx^{\prime}\\ &=\, a e^{-\imath 2\pi \mu s} e^{-2\pi^2s^2\sigma^2}\int_{-\infty}^{+\infty} e^{-\left( \frac{x^{\prime}}{\sqrt{2}\sigma}+\imath\pi s\sqrt{2}\sigma\right)^2}dx^{\prime}\\ &\underset{x^{\prime\prime} = \,\frac{x^{\prime}}{\sqrt{2}\sigma}+\imath\pi s\sqrt{2}\sigma}{=}a e^{-\imath 2\pi \mu s} e^{-2\pi^2s^2\sigma^2}\int_{-\infty}^{+\infty}\sqrt{2}\sigma \,e^{-{x^{\prime\prime}}^2}dx^{\prime\prime}\\ &= e^{-\imath 2\pi \mu s}\, a\sqrt{2\pi}\sigma\,e^{-2\pi^2s^2\sigma^2}. \end{align} $$

We made use of the fact that $\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}$, see $\S$ 2.2 ➞ . For the normalised Gaussian this means

$$ f(x) \, = \,\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{x^2}{2\sigma^2}}\\ \Rightarrow\\ \begin{align} \mathscr{F}\{f\}(s) \,&=\, e^{-2\pi^2\sigma^2s^2}\\ &\underset{\sigma_s\,=\,\frac{1}{\sqrt{2\pi\sigma}}}{=}\, e^{-\frac{s^2}{2\sigma_s^2}} .\\ \end{align} $$

2.4.3 Fourier transform of Dirac's delta function

The properties of the delta function $\delta(x)$ are discussed in $\S$ 2.2➞ , its Fourier transform is given by:

$$ \begin{align} \mathscr{F}\{\delta\}(s) \,&=\, \int_{-\infty}^{+\infty} \delta(x)\, e^{-\imath 2\pi x s} dx ,\\ &=\, e^{0},\\ &=\, 1. \end{align} $$

The result is a constant, which means that the (inverse) Fourier transform of a constant is a (scaled) delta-function. The shifted delta function has the Fourier transform

$$ \begin{align} \mathscr{F}\{\delta_{x_0}\}(s) \,&=\, \int_{-\infty}^{+\infty} \delta(x-x_0)\, e^{-\imath 2\pi x s} dx\\ &=\, e^{-\imath 2\pi x_0 s} \end{align} \qquad . $$

2.4.4 Fourier transform of the comb function

Recall the definition of the Shah function $III_{T^{-1}}(s)\,=III\left(\frac{s}{T}\right)\,=\sum_{m=-\infty}^{+\infty} T \delta\left(s-m T\right)$ and its Fourier series $III_{T^{-1}}(s)\,=\sum_{m=-\infty}^{+\infty} e^{\imath \frac{2\pi m}{T}s}$ ($\S$ 2.2➞ ). The Fourier transform of the Shah function can be easily computed as

$$ \begin{align} \mathscr{F}\{III_T\}(s) \,&=\, \int_{-\infty}^{+\infty} \left(\sum_{m=-\infty}^{+\infty} \frac{1}{T} \delta\left(x-\frac{m}{T}\right)\right)\, e^{-\imath 2\pi x s} dx\\ &=\, \frac{1}{T}\sum_{m=-\infty}^{+\infty} e^{-\imath 2\pi \frac{m}{T} s}\\ &\underset{m^{\prime} = -m}{=}\, \frac{1}{T}\sum_{m^{\prime}=-\infty}^{+\infty} e^{\imath 2\pi \frac{m^{\prime}}{T} s}\\ &= \frac{1}{T} III_{T^{-1}}(s) \end{align} \qquad $$

The Fourier transform of the Shah function is simply a scaled Shah function. Since $III_T$ is an even function, we also have (see Eq 2.4.1 ➞ )

$$ \mathscr{F}^{-1}\{III_T\}(x) \,=\,\frac{1}{T} III_{T^{-1}}(x)\qquad $$

2.4.5 Fourier transform of the rectangle and the sinc function

Remember the rectangle function ➞ $\Pi(x) = \left\{ \begin{array}{lll} 0 & {\rm for} & x < -\frac{1}{2} \\ 1 & {\rm for} & -\frac{1}{2} \le x \le \frac{1}{2} \\ 0 & {\rm for} & x > \frac{1}{2} \\ \end{array}\right. $. Then, also remembering Euler's formula ➞,

$$ \begin{align} \mathscr{F}\{\sqcap\}(s) \,&=\, \int_{-\frac{1}{2}}^{+\frac{1}{2}} e^{-\imath 2\pi x s}\,dx\\ &\underset{s \neq 0}{=}\, \left[\frac{-1}{2\pi\imath s}e^{-\imath 2\pi x s}\right]_{-\frac{1}{2}}^{+\frac{1}{2}}\\ &=\, \frac{-1}{2\pi\imath s}\left\{\left[ \cos{(\pi s)}-\imath\sin{(\pi s)}\right] - \left[\cos{(\pi s)}+\imath\sin{(\pi s)}\right] \right\}\\ &=\,\frac{\sin{(\pi s)}}{\pi s}\\ \mathscr{F}\{\sqcap\}(s) \,&=\, \int_{-\frac{1}{2}}^{+\frac{1}{2}} e^{-\imath 2\pi x s}\,dx\\ &\underset{s = 0}{=}\, 1\\ &\Rightarrow\\ \mathscr{F}\{\sqcap\}(s) \,&=\,sinc(s) \end{align} \qquad $$

It's the sinc funtion! Again, because both the sinc and rectangle functions are even, we get

$$ \begin{align} \mathscr{F}\{\sqcap\}(s) \,&=\,sinc(s)\\ \mathscr{F}^{-1}\{\sqcap\}(x) \,&=\,sinc(x)\\ \mathscr{F}\{sinc\}(s) \,&=\,\sqcap(s)\\ \mathscr{F}^{-1}\{sinc\}(x) \,&=\,\sqcap(x)\\ \end{align} \qquad $$

With that, we've already walked a long way into signal processing.

2.4.6 Fourier transforms of real-valued and Hermitian functions

A complex-valued function $f: \mathbb{R} \rightarrow \mathbb{C}$ is a Hermitian function, if, and only if

$$ \forall \,x\in\mathbb{R}\quad f^*(x) = f(-x) \qquad $$

The (inverse) Fourier transform of a real-valued function is a Hermitian function (again making use of Euler's formula ➞ ):

$$ f: \mathbb{R} \rightarrow \mathbb{R}\\ \forall\,x\in\mathbb{R} \quad \left[\,f(x)\in\mathbb{R}\quad\Leftrightarrow \quad f^*(x)=f(x)\,\right]\\ \begin{align} \left(\mathscr{F}\{f\}\right)^*(s)\,&=\, \left(\int_{-\infty}^{+\infty} f(x)\,e^{-\imath 2\pi x s}\,dx\right)^*\\ &=\, \int_{-\infty}^{+\infty} f^*(x)\,\left[\cos{( -2\pi x s)}+\imath\sin{(- 2\pi x s)}\right]^*\,dx\\ &=\, \int_{-\infty}^{+\infty} f^*(x)\,\left[\cos{( 2\pi x s)}-\imath\sin{( 2\pi x s)}\right]^*\,dx\\ &=\, \int_{-\infty}^{+\infty} f(x)\,\left[\cos{( 2\pi x s)}+\imath\sin{( 2\pi x s)}\right]\,dx\\ &=\, \int_{-\infty}^{+\infty} f(x)\,\left[\cos{( 2\pi x (-s))}-\imath\sin{( 2\pi x (-s))}\right]\,dx\\ &=\,\left(\mathscr{F}\{f\}\right)(-s) \end{align} $$

Likewise, the (inverse) Fourier transform of any Hermitian function is a real-valued function. The above also applies for multi-dimensional Fourier transforms.

2.4.7 Fourier transforms of complex conjugate functions

The Fourier transform of a complex conjugate of the function is the complex conjugate of the reversed Fourier transform of the function.

$$ \forall f,F,F_-:\,\mathbb{R} \rightarrow \mathbb{C}\\ F_-(x) = F(-x) \quad \Rightarrow \quad \begin{split} \mathscr{F}\left\{f^*\right\} \,=\, \left(\mathscr{F}\left\{f\right\}\right)_-^* \end{split} $$

because

$$ \begin{align} \mathscr{F}\left\{f^*\right\}(s) \,&= \,\int_{-\infty}^{+\infty}f^*(x)\,e^{-\imath 2\pi xs}\,dx\\ &=\,\left[\int_{-\infty}^{+\infty}f(x)\,e^{\imath 2\pi xs}\,dx\right]^*\\ &=\,\left[\int_{-\infty}^{+\infty}f(x)\,e^{-\imath 2\pi x(-s)}\,dx\right]^*\\ &=\,\left(\mathscr{F}\left\{f\right\}\right)^*(-s)\\ &=\,\left(\mathscr{F}\left\{f\right\}\right)_-^*(s)\end{align} $$

Future Additions:
  • intro: explain the Fourier transform as a decomposition into an orthogonal basis set of sinusoidal waves
  • intro example: composition/decomposition of a 1-d image via summation of sine/cosine waves with phase and amplitude