📂Fourier Analysis

Various Meanings of the Fourier Transform

The Fourier transform is widely treated across various fields such as mathematics, physics, and engineering, and thus it comes to have different meanings depending on the perspective from which it is viewed. Here, its meanings in the context of mathematics, quantum mechanics, and signal processing are introduced. Let’s first define the Fourier transform and inverse transform, as they are defined in various forms in this document.

$$\hat{f}(\xi) := \int_{-\infty}^{\infty} f(t) e^{-i\xi x}dx$$

In Mathematics

In mathematics, the Fourier transform is fundamentally an integral transform with a kernel of $e^{-i\xi x}$. Since the inner product in function spaces is defined by integration, the Fourier transform of $f$ can be thought of as the inner product of $f$ and $e^{i\xi x}$.

$$\begin{equation} \hat{f}(\xi) = \langle f, e^{i\xi x} \rangle \end{equation}$$

As will be explained below, the reason why we can know the frequency $\xi$ contained in signal $f$ when obtaining the Fourier transform of $f$ is due to this meaning. Moreover, a bit more meaning can be given to kernel $e^{i \xi x}$. Consider the Laplacian as a linear operator.

$$\Delta \phi = \nabla ^{2} \phi = \lambda \phi$$

The $\lambda$ that satisfies the above equation is called the eigenvalue of the Laplacian, and $\phi$ is the eigenvector corresponding to $\lambda$. Functions that yield themselves upon being differentiated twice are exponential functions $e^{i\xi x}$. Hence, the following equation holds.

$$\nabla ^{2} e^{i\xi x} = (-\xi^{2}) e^{i\xi x}$$

Therefore, $e^{i\xi x}$ is an eigenvector of the Laplacian, and $(1)$ can be described as follows.

Fourier transform of $f$ $=$ Inner product of $f$ and the eigenvector of the Laplacian

In graph signal processing, the graph Fourier transform is defined based on the above interpretation.

In Signal Processing

The Fourier series of $f$ refers to the expansion of $f$ as a series of exponential functions as follows.

$$f(t) = \sum \limits_{\omega=-\infty}^{\infty} c_{\omega}e^{i\omega t}$$

Then, due to the [Euler’s formula], $e^{i \omega t} = \cos (\omega t) + i \sin (\omega t)$ holds, so $t$ represents time, and $\omega$ represents the frequency of the wave (which is the same as the oscillation frequency). However, as explained above, the Fourier transform of $f$ is the same as taking the inner product of $f$ and $e^{i\omega t}$. Since exponential functions with different frequencies $\omega$ and $\omega^{\prime}$ are orthogonal to each other, taking the inner product of $f$ with $e^{i \omega t}$ makes all terms with different frequencies $e^{-i \omega^{\prime} t}$ become $0$, leaving only the coefficient of $e^{i \omega t}$.

$$\begin{equation} \hat{f}(\omega) = \langle f, e^{i\xi x} \rangle = \left\langle \sum \limits_{\omega=-\infty}^{\infty} c_{\xi}e^{i\omega t} \right\rangle = | c_{\omega} |^{2} \label{signalprocess} \end{equation}$$

Therefore, if we calculate the Fourier transform of $f$ $\hat{f}(\omega)$ and find that $\hat{f}(\omega) \ne 0$ is $\omega$, this means that $\omega$ is exactly one of the signals contained in signal $f$. For instance, let’s say signal $f$ is as follows.

$$f(t) = 2\sin (2\pi 50t) + 1.7\sin (2\pi 100t) + 0.3 \cos (2\pi 200t) + 4\cos(2\pi 300 t)$$

If you calculate $\hat{f}$, you will see below picture, and you can confirm that the values of the functions at the frequencies that make up $f$ are not $0$.

Therefore, the Fourier transform can be thought of as a tool that allows us to view a certain signal $f$ in both time and frequency domains.

In Quantum Mechanics

In quantum mechanics, the motion of small particles is described by the Schrödinger equation and wave functions. A wave function with a wave number of $k$ and an angular frequency of $\omega$ is expressed as follows depending on the position and time.

$$\psi (x,t) = e^{i kx -\omega t}$$

However, according to the de Broglie relations, the wave number and momentum satisfy $k = \dfrac{p}{\hbar}$, and the energy and angular frequency satisfy $\omega = \dfrac{E}{\hbar}$, so the wave function is as follows.

$$\psi (x,t) = e^{\frac{i}{\hbar} (px - Et)}$$

Therefore, applying the above explanations, the wave function can move back and forth between momentum-position domains and energy-time domains by the Fourier transform. In other words, the Fourier transform is a tool that allows us to view the wave function from the two perspectives of momentum and position (energy and time).