Fourier series (periodic signals)
The considered framework is $L^2[0,T]$, that is, $T$-periodic functions with finite energy on a period.
for all $f,g\in L^2[0,T]$, we have the inner product $$ \langle f,g \rangle = \frac{1}{T}\int_0^T f(t) \bar{g}(t)\mathrm{d} t $$ and the induced norm $$ ||f|| = \left(\frac{1}{T}\int_0^T |f(t)|^2 d t\right)^{\frac{1}{2}} $$
Fourier Coefficients and Fourier series: the Fourier transform of periodic functions and its inverse
Fourier analysis for periodic function relies on the following theorem.
Bases of periodic functions The family of the trignometric functions $ \lbrace t \rightarrow e^{i\frac{2\pi}{T}n t} = e_{n}(t)\rbrace_{n\in\mathbb{Z}}$ is an orthogonal basis of $L^2([0,T])$.
**Fourier Coefficients (Fourier transform of periodic functions)** Let $f\in L^2([0,T])$. The Fourier complex coefficients of $f$ are: $$ \hat f[n] = c_n(f) = \frac{1}{T}\int_0^T e^{-i\frac{2\pi}{T}n t}\mathrm{d} t\quad \forall n\in\mathbb{Z}\ . $$
Spectrum of a periodic signal The spectrum is the set of the squared modulus of the fourier coefficients: $$ \text{spectrum}(f)=\lbrace |\hat f[n]|^2 \rbrace $$ sometimes, one can display the modulus instead of the squared modulus.
Some usefull properties:
let $f,g\in L^2([0,T])$. Then
- Linéarity: $\widehat{(\lambda f+g)}[n] = \lambda \hat f[n] + \hat g[n]$
- Conjugation: $\widehat{\bar{f}}[n] = \overline{\hat f[n]}$
- Time reversal: Let $g:t\mapsto f(-t)$, then $\hat g[n] = \hat f[-n]$
- Time shift: Let $a\in\mathbb{R}$ and $f_a:t\mapsto f(t+a)$, then $\hat f_{a}[n] = e^{i\frac{2\pi}{T}na}\hat f[n]$
- Derivation: If $f$ is $k$ time differentiable, then $\hat{f^{(k)}} = (i\frac{2\pi}{T}n)^k \hat f[n]$ The conjugation properties implies that the spectrum is symmetric with respect to the frequency 0.
Thanks to the Fourier coefficients, we can now define the Fourier serie of $f$
Fourier Serie We call Fourier series of $f\in L^2([0,T])$, denoted by $S(f)$ the series $$ S(f)(t) = \sum_{n=-\infty}^{+\infty} \hat f[n] e^{i\frac{2\pi}{T}n t} $$
The family $\lbrace e_n\rbrace_{n\in\mathbb{Z}}$ being an ortonormal basis of $L^2([0,T])$, we know that $$ \lim_{N\rightarrow +\infty} \left\Vert f-\sum_{-N}^{N}\hat f[n] e^{i\frac{2\pi}{T}n}\right\Vert=0 $$
Moreover, one has the energy conservation and the angle conservation properties, known as the Plancherel-Parseval theorem
Let $f,g \in L^2([0,T])$ and $\lbrace \hat f[n]\rbrace_{n\in\mathbb{Z}}$ and $\lbrace \hat g[n]\rbrace_{n\in\mathbb{Z}}$ their respective Fourier coefficients. Then $$ \Vert f\Vert^2 = \frac{1}{T}\int_{-T/2}^{T/2} |f(t)|^2 \mathrm{d} t = \sum_{n=-\infty}^{+\infty} |\hat f[n]|^2\ . $$ and $$ \langle f,g \rangle = \frac{1}{T} \int_{-T/2}^{T/2} f(t)\bar{g}(t) \mathrm{d} t = \sum_{n=-\infty}^{+\infty}\hat f[n] \bar{\hat g}_n $$
We can recover any periodic functions from its Fourier coefficients in an “$L^2$ sense”. This implies the following properties of the Fourier coefficients:
Let $f\in L^2([0,T])]$, then $$ \lim_{|n|\rightarrow +\infty} \hat f[n] = 0 $$
Hence, the coefficients are attenuated with high frequencies. More generally, the ore regular the function, the faster its Fourier coefficients converge to $0$ with high frequencies.
The pointwise convergence of Fourier series is given by the Dirichlet’s Theorem
Dirichlet Let $f\in L^2([0,T])$, such that $f$ is piecewise $C^1$. For all $t_0\in [0,T]$, we have $$ \lim_{N\rightarrow +\infty} \sum_{n=-N}^{N} \hat f[n] e^{i\frac{2\pi}{T}nt_0} = \frac{f(t_0^+) + f(t_0^-)}{2}\ . $$ In particular, if $f$ is continuous in $t_0$, then we have $$ f(t_0) = \sum_{n=-\infty}^{+\infty} \hat f[n] e^{i\frac{2\pi}{T}nt_0} $$
Trigonometric Fourier coefficients
When $f$ has some simitrical properties, it can be usefull to use the trigonometric Fourier coefficients instead of the complex coefficients.
Coefficients de Fourier trigonométriques Let $f\in L^2([0,T])$. The trigonometric Fourier coefficients of $f$ are given by: $$ \begin{aligned} a_0(f) & = \frac{1}{T}\int_{0}^{T} f(t) \mathrm{d} t\\ a_n(f) & = \frac{2}{T}\int_{0}^{T} f(t) \cos\left(\frac{2\pi}{T}nt\right) \mathrm{d} t\quad \forall n \geq 1\\ b_n(f) & = \frac{2}{T}\int_{0}^{T} f(t) \sin\left(\frac{2\pi}{T}nt\right) \mathrm{d} t\quad \forall n \geq 0\ . \end{aligned} $$
On peut bien sûr déduire les coefficients trigonométriques à partir de coefficients complexes et vice-versa, qu’on résume dans la proposition suivante (dont la preuve est obtenue par simples calculs)
Soit $f\in L^2([0,T])$ $$\hat f_0 = a_0(f)\quad \hat f_n = \frac{a_n(f)- i b_n(f)}{2}\quad \hat f_{-n} = \frac{a_n(f) + i b_n(f)}{2}$$ $$a_n(f) = \hat f_n + \hat f_{-n}\quad b_n(f) = i(\hat f_n - \hat f_{-n})$$
Trigonometric Fourier coefficients are particularly interesting when $f$ has some properties of symmetry
Let $f\in L^2([0,T])$.
- If $f$ is even, then $\forall n\geq 1$ $b_n(f) = 0$ and $a_n(f) = \frac{4}{T}\int\limits_{0}^{T/2} f(t) \cos(\frac{2\pi}{T}nt)\mathrm{d} t$
- If $f$ is odd, then $\forall n\geq 0$ $a_n(f) = 0$ and $\forall n\geq 1$ $b_n(f) =\frac{4}{T}\int\limits_{0}^{T/2} f(t) \sin(\frac{2\pi}{T}nt)\mathrm{d} t$
The Fourier series then reads $$ S(f)(t) = a_0(f) + \sum_{n=1}^{+\infty} a_n(f) \cos\left(\frac{2\pi}{T}n t\right) + \sum_{n=1}^{+\infty} b_n(f) \sin\left(\frac{2\pi}{T}n t\right)\ , $$ and the Plancherel-Parseval’s Theorem $$ |f|^2 = |a_0(f)|^2 + \frac{1}{2}\sum_{n=1}^{+\infty} |a_n(f)|^2 + \frac{1}{2}\sum_{n=1}^{+\infty} |b_n(f)|^2\ . $$
Gibbs phenomenon
Fourier series allows one to decompose any periodic function as an infinite linear combination of trigonometric functions. In practice, one can only use a finite linear combination. We describe here the behavior of truncated Fourier series.
When the function is continuously differentiable, the Fourier series converges pointwise, rapidly, to the function. Problems appear when there is a singularity. Take for example the square wave function: let $f\in L^2([-\pi,\pi])$ such that $$ \begin{cases} f(t) = 1 & \text{ if } 0\leq t < \pi \\ f(t) = -1 & \text{ if } -\pi\leq t < 0 \end{cases} $$ The function $f$ is odd, we calculate the trigonometric coefficients $$ b_n(f) = \frac{2}{\pi}\int_{0}^{\pi} \sin(nt)\mathrm{d} t\ . $$ which lead to $$ S(f)(t) = \frac{4}{\pi}\sum_{n=-\infty}^{+\infty} \frac{1}{2n+1} \sin( (2n+1)t) \ . $$
Next figures illustrate the behavior of the truncated sum of the Fourier series of the square wave function.
