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
Fourier theorems will serve as an important tool box for the rest of the course. While they are presented without context at the beginning of our introduction, we hope that the reader will eventually realise their tremendous impact not only in radio interferometry but in general. In fact, had Fourier lived to see the first dynamite explosion, he would probably have received 2 or 3 Nobel prizes.
For all functions $ f:\,\mathbb{R} \rightarrow \mathbb{C} $ we assume that they are (Lebesgue-) integrable (i.e. $\int_{-\infty}^{+\infty} \lvert f(x) \rvert \,dx \in \mathbb{R}$) so that their Fourier transform $\mathscr{F}\{f\} = F$ exists. The Fourier transform has been introduced in $\S$ 2.4 . Functions in the "Fourier domain" are denoted by capital letters, funcions in the "original domain" by lower case letters.
The Fourier transform is linear with respect to:
Proof:
Proof:
The same is true for the inverse Fourier transform and for the multi-dimensional Fourier transform. We leave the proof to the reader.
The Fourier transform of a scaled function $\mathscr{F}\{f_{\rm c}\}$, where $f_{\rm c}(x) \,= \,f(ax)$, is given by:
Proof:
$$ \begin{align} \mathscr{F}\{f_{\rm c}\}(s) \,&= \,\int_{-\infty}^{+\infty}f_{\rm c}(x)\,e^{-\imath 2\pi xs}\,dx\\ &= \,\int_{-\infty}^{+\infty}f(ax)\,e^{-\imath 2\pi xs}\,dx\\ &\underset{x^\prime = ax}{=}\,\int_{-\text{sign}(a)\infty}^{+\text{sign}(a)\infty}f(x^{\prime})\,e^{-\imath 2\pi \frac{x^\prime}{a} s}\frac{dx}{dx^\prime}\,dx^\prime\\ &=\,\frac{1}{\lvert a \rvert}\int_{-\infty}^{+\infty}f(x^\prime)\,e^{-\imath 2\pi x^\prime \frac{s}{a}}\,dx\\ &=\, \frac{1}{\lvert a \rvert}\mathscr{F}\{f\}(\frac{s}{a}) \end{align}\qquad $$
Note that the sign of the scaling factor determines the signs on the limits of the integral. This theorem holds true for the inverse Fourier transform and for the multi-dimensional Fourier transform. We leave the proof to the reader.
The Fourier transform of a shifted function $\mathscr{F}\{f_{\rm t}\}$, where $f_{\rm t}(x) \,= \,f(x-a)$, is given by
Proof:
$$ \begin{align} \mathscr{F}\{f_{\rm t}\}(s) \,&= \,\int_{-\infty}^{+\infty}f_{\rm t}(x)\,e^{-\imath 2\pi xs}\,dx\\ &= \,\int_{-\infty}^{+\infty}f(x-a)\,e^{-\imath 2\pi xs}\,dx\\ &\underset{x^\prime = x-a}{=}\,\int_{-\infty}^{+\infty}f(x^{\prime})\,e^{-\imath 2\pi (x^\prime + a) s}\frac{dx}{dx^\prime}\,dx^\prime\\ &=\,e^{-\imath 2\pi a s}\int_{-\infty}^{+\infty}f(x^\prime)\,e^{-\imath 2\pi x^\prime s}\,dx\\ &=\, e^{-\imath 2\pi a s}\,\mathscr{F}\{f\}(s) \end{align}\qquad $$
The same is true for the inverse Fourier transform but with a change of sign, i.e.
where $F_{\rm t}(s) \,= \,\mathscr{F}\{f\}(s-a)$. This also holds for the multi-dimensional Fourier transform. We leave the proof to the reader. Thus, if $g(x) \,= \,e^{\imath 2\pi a x}$, we also have that
Proof:
which again is also valid for the inverse Fourier transform
where $G(s) \,= \,e^{-\imath 2\pi a s}$.
The Fourier transform of the convolution of two functions is the product of the Fourier transforms of the functions, i.e.
Proof:
$$ \begin{align} \mathscr{F}\{f \circ g\}(s) \,&= \,\int_{-\infty}^{+\infty}(f\circ g)(x)\,e^{-\imath 2\pi xs}\,dx\\ &=\,\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(x-t)g(t)\,dt\,e^{-\imath 2\pi xs}\,dx\\ &=\,\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(x-t)e^{-\imath 2\pi xs}\,dx\,g(t)\,dt\,\\ &=\,\int_{-\infty}^{+\infty}e^{-\imath 2\pi ts}\,\mathscr{F}\{f\}(s)\,g(t)\,dt\,\\ &=\,\mathscr{F}\{f\}(s)\,\int_{-\infty}^{+\infty}g(t)\,e^{-\imath 2\pi ts}\,dt\,\\ &=\, \mathscr{F}\{f\}(s) \mathscr{F}\{g\}(s)\\ &=\, \left(\mathscr{F}\{f\} \mathscr{F}\{g\}\right)(s) \end{align}\qquad $$
where we have used the shift theorem in going from line three to line four. Similarly, we can show that
As usual the result can be extended to the multi-dimensional case. The proof is left as an exercise. It follows that
Proof:
$$ \begin{split} \mathscr{F}\{f\} \circ \mathscr{F}\{g\} \,&=\,\mathscr{F}\left\{ \mathscr{F}^{-1}\left\{ \mathscr{F}\{f\} \circ \mathscr{F}\left\{g\right\}\right\}\right\}\\ &=\, \mathscr{F}\left\{ \mathscr{F}^{-1}\left\{ \mathscr{F}\left\{f\right\}\right\}\mathscr{F}^{-1}\left\{ \mathscr{F}\left\{g\right\}\right\}\right\}\\ &=\, \mathscr{F}\left\{f g\right\} \end{split}\qquad $$
It is also true that
The proof of this result, and for the multi-dimensional case, is left as an exercise.
The Fourier transform of the product $\mathscr{F}\{g f\}(s)$, where $g(x) \,= \,\cos{(2\pi s_0 x)}$, is given by
Proof:
$$ \begin{split} g(x)\,&=\,\cos{(2\pi s_0 x)} \\ &=\, \frac{1}{2}\left(\,e^{\imath 2\pi s_0 x}+e^{-\imath 2\pi s_0 x}\right)\\ &=\,\frac{1}{2}\left(g_1(x)+g_2(x)\right) \end{split}\\ \begin{split} \mathscr{F}\{g\, f\}(s) \,&=\, \frac{1}{2}\left[\mathscr{F}\left\{g_1\, f\right\}+\mathscr{F}\left\{g_2 f\right\}\right]\\ &=\, \frac{1}{2}\left[\mathscr{F}\left\{f\right\}(s-s_0)+\mathscr{F}\left\{f\right\}(s+s_0)\right] \end{split}\qquad $$
where we have used the Euler equations ➞ , linearity ➞ and the shift theorem ➞.
Occasionally, the inverse version of the shift theorem ➞ is also called the modulation theorem. Because of their similarity this is, to some extent, justified. However, we will always refer to the above theorem as the modulation theorem.
The power theorem, also known as Parseval's theorem, states that:
Proof:
$$ \forall F:\,\mathbb{R} \rightarrow \mathbb{C}\\ F_-(x) \underset{\rm def}{=} F(-x) \\ \begin{split} \mathscr{F}\left\{fg^*\right\}(s) \,&=\,\int_{-\infty}^{+\infty}f(x)\,g^*(x)\, e^{-\imath 2\pi s x}\,dx\\ &=\,\mathscr{F}\left\{f\right\}\circ\mathscr{F}\left\{g^*\right\}(s)\\ &=\,\mathscr{F}\left\{f\right\}\circ\left(\mathscr{F}\left\{g\right\}\right)_-^*(s) \end{split}\\ \Rightarrow\\\begin{split} \int_{-\infty}^{+\infty}f(x)\,g^*(x)\,dx \,&=\,\int_{-\infty}^{+\infty}f(x)\,g^*(x)\, e^{-\imath 2\pi \cdot 0\cdot x}\,dx\\ &=\,\mathscr{F}\left\{fg^*\right\}(0)\\ &=\,\mathscr{F}\left\{f\right\}\circ\left(\mathscr{F}\left\{g\right\}\right)_-^*(0)\\ &=\int_{-\infty}^{+\infty}\mathscr{F}\left\{f\right\}(s)\,\left(\mathscr{F}\left\{g\right\}\right)_-^*(0-s)\,ds\\ &=\int_{-\infty}^{+\infty}\mathscr{F}\left\{f\right\}(s)\,\left(\mathscr{F}\left\{g\right\}\right)^*(s)\,ds \end{split} \qquad $$
where we have used the result pertaining to the Fourier transform of complex conjugate functions ➞ and the convolution theorem in going from line two to three.
The integral of the square of the absolute value of a function is the integral of the square of the absolute value of its Fourier transform.
Proof:
$$ \begin{split} \int_{-\infty}^{+\infty}\lvert f(x) \rvert^2\,dx \,&=\, \int_{-\infty}^{+\infty}f(x) \,f^*(x) \,dx\\ &=\, \int_{-\infty}^{+\infty}\mathscr{F}\left\{f\right\}(s) \, \left(\mathscr{F}\left\{f\right\}\right)^*(s)\,ds\\ &=\, \int_{-\infty}^{+\infty}\lvert \mathscr{F}\left\{f\right\}(s) \rvert^2\,ds \end{split}\qquad $$
Note that this is just a special case of the power theorem ➞.
The Fourier transform of the cross-correlation ➞ between function $f$ and function $g$ satisfies
Proof:
$$ f_-(x) \underset{\rm def}{=} f(-x) \\ \begin{split} \mathscr{F}\left\{f\star g\right\} \,&=\, \mathscr{F}\left\{{f_-}^* \circ g\right\}\\ &=\,\mathscr{F}\left\{{f_-}^*\right\} \cdot \mathscr{F}\left\{g\right\}\\ &=\,\left(\mathscr{F}\left\{{f}\right\}\right)^* \cdot \mathscr{F}\left\{g\right\} \end{split} $$
where we have used the convolution theorem ➞ and the Fourier transform of complex conjugate functions ➞.
The Fourier transform of the auto-correlation ➞ of a function $f$ satisfies
Proof:
This is just a special case of the cross-correlation theorem.
The Fourier transform $\mathscr{F}\left\{\frac{df}{dx}\right\}(s)$ of the derivative $\frac{df}{dx}$ of a function $f(x)$ satisfies:
Proof:
$$ \begin{split} \mathscr{F}\left\{\ \frac{d}{dx}\mathscr{F}^{-1}\left\{\ \mathscr{F}\left\{\ f \right\} \right\} \right\}(s) \,&=\, \int_{-\infty}^{+\infty} \frac{d}{dx}\left[ \int_{-\infty}^{+\infty} \mathscr{F}\left\{f\right\}(s) e^{\imath 2\pi x s}\,ds \right] e^{-\imath 2\pi x s}\,dx\\ &=\, \int_{-\infty}^{+\infty} \left[ \int_{-\infty}^{+\infty} \imath 2\pi s\mathscr{F}\left\{f\right\}(s) e^{\imath 2\pi x s}\,ds\, \right] e^{-\imath 2\pi x s}\,dx\\ &=\, \int_{-\infty}^{+\infty} \left[ \mathscr{F}^{-1}\left\{\imath 2\pi s\mathscr{F}\left\{f\right\}\right\}\, \right] e^{-\imath 2\pi x s}\,dx\\ &=\, \int_{-\infty}^{+\infty} \left[ \left(\mathscr{F}^{-1}\left\{\imath 2\pi s\right\}\right) \circ f\, \right] e^{-\imath 2\pi x s}\,dx\\ &=\, \left[\mathscr{F}\left\{\mathscr{F}^{-1}\left\{\imath 2\pi s\right\}\right\} \cdot \mathscr{F}\left\{f\right\}(s)\right]\\ &=\,\imath 2\pi s\,\mathscr{F}\left\{f\right\}(s) \end{split}\qquad $$
The result is also valid for the inverse Fourier transform, although with the opposite sign.