Import standard modules:
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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:
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pass
The Fourier transform is one of the fundamental mathematical operations that is made use of in signal processing and interferomtry. It is introduced here.
Consider the class of integrable functions $f$. Then the Fourier transform $\mathscr{F}$ is defined as:
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
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:
For any function $f$ define
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
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
$$ \mathscr{F}: [\mathbb{R}^n \rightarrow \mathbb{C}] \rightarrow [\mathbb{R}^n \rightarrow \mathbb{C}] \,, \quad n\in \mathbb{N}\\ \forall f:\,\mathbb{R}^n \rightarrow \mathbb{C}, \int_{-\infty}^{+\infty} \lvert f(x) \rvert \,d^nx = \int_{-\infty}^{+\infty} \ldots \int_{-\infty}^{+\infty}\lvert f(x) \rvert \,dx_1\ldots dx_n\in \mathbb{R}\\ \begin{align} \mathscr{F}\{f\}(s_1, \ldots, s_n) \,&=\,\mathscr{F}\{f\}({\bf s})\\ &=\,\int_{-\infty}^{+\infty}f(x)\,e^{-\imath 2\pi( x_1s_1+\ldots+x_ns_n)}d^nx\\ &=\,\int_{-\infty}^{+\infty}f(x)\,e^{-\imath 2\pi( {\bf x}\cdot{\bf s})}d^nx\\ \mathscr{F}^{-1}\{F\}(x_1, \ldots, x_n) \,&=\,\mathscr{F}^{-1}\{F\}({\bf x})\\ &=\, \int_{-\infty}^{+\infty}F(s)\,e^{\imath 2\pi (x_1s_1+\ldots+x_ns_n)}d^ns\\ &=\, \int_{-\infty}^{+\infty}F(s)\,e^{\imath 2\pi ({\bf x}\cdot {\bf s})}d^ns\\ \end{align} $$
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
The properties of the delta function $\delta(x)$ are discussed in $\S$ 2.2➞ , its Fourier transform is given by:
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
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 ➞ )
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
With that, we've already walked a long way into signal processing.
A complex-valued function $f: \mathbb{R} \rightarrow \mathbb{C}$ is a Hermitian function, if, and only if
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.
The Fourier transform of a complex conjugate of the function is the complex conjugate of the reversed Fourier transform of the function.
because
$$ \begin{align} \mathscr{F}\left\{f^*\right\}(s) \,&= \,\int_{-\infty}^{+\infty}f^*(x)\,e^{-\imath 2\pi xs}\,dx\\ &=\,\left[\int_{-\infty}^{+\infty}f(x)\,e^{\imath 2\pi xs}\,dx\right]^*\\ &=\,\left[\int_{-\infty}^{+\infty}f(x)\,e^{-\imath 2\pi x(-s)}\,dx\right]^*\\ &=\,\left(\mathscr{F}\left\{f\right\}\right)^*(-s)\\ &=\,\left(\mathscr{F}\left\{f\right\}\right)_-^*(s)\end{align} $$