Serie de Fourier

La serie de Fourier nos permite representar cualquier función periódica mediante una suma de senos y cosenos.

Función periódica:

$\begin{array}{cc}& f(t+T)=f(t)\\ & f(t+kT)=f(t),\mathrm{\forall }k\in \mathbb{Z}\\ \end{array}$

Representación trigonométrica

$\begin{array}{cc}& f(t)=\frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cdot \cos \left(\frac{2\pi n}{T}t\right)+\sum _{n=1}^{\mathrm{\infty }}{b}_n\cdot \sin \left(\frac{2\pi n}{T}t\right)\\ & \omega =\frac{2\pi }{T}\\ & a_n=\frac{2}{T}\underset{-T/2}{\overset{T/2}{\int }}f(t)\cos \left(\frac{2\pi n}{T}t\right)\partial t\\ & b_n=\frac{2}{T}\underset{-T/2}{\overset{T/2}{\int }}f(t)\sin \left(\frac{2\pi n}{T}t\right)\partial t\\ & a_0=\frac{2}{T}\underset{-T/2}{\overset{T/2}{\int }}f(t)\partial t\\ \end{array}$

También tiene una REPRESENTACION EXPONENCIAL como se verá mas adelante

Teorema de Dirichlet

$f(t+T)=f(t)=\{\begin{array}{cc}& \text{continua trozos }I=[-{}^T╱{}_2,{}^T╱{}_2]\\ & \mathrm{\forall }{t}_0\in I,\underset{t\to t_0}{\lim }f(t)\text{ finito}\\ \end{array}$

Por lo que la serie de Fourier correspondiente a f es convergente y su suma vale:

$\begin{array}{cc}& f(t),\text{ en punto continuo}\\ & \frac{f(t^{+})+f(t^{-})}{2},\text{ en punto discontinuo}\\ \end{array}$

Observaciones:

$\begin{array}{cc}& f(t)\text{ par }f(t)=f(-t)\\ & \underset{-b}{\overset{b}{\int }}f(t)\partial t=2\underset{0}{\overset{b}{\int }}f(t)\partial t\\ & f(t)\text{ impar }f(t)=-f(-t)\\ & \underset{-b}{\overset{b}{\int }}f(t)\partial t=0\\ \end{array}$

f(t)	g(t)	f(t)xg(t) o f(t)/g(t)
Par	Par	Par
Impar	Impar	Par
Par	Impar	Impar

Forma exponencial o compleja de la serie de Fourier

$\begin{array}{cc}& f(t)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{C}_k\cdot e^{j\frac{2\pi }{T}kt}\\ & C_k=\frac{1}{T}\underset{{}^{-T}╱{}_2}{\overset{{}^T╱{}_2}{\int }}f(t)\cdot e^{-j\frac{2\pi }{T}kt}\partial t\\ \end{array}$

Demostración parcial:

$\begin{array}{cc}& \left\{\begin{array}{cc}& a_n\\ & b_n\\ \end{array}\right\}=\frac{2}{T}\underset{-T/2}{\overset{T/2}{\int }}f(t)\left\{\begin{array}{cc}& \cos \left(\omega nt\right)\\ & \sin \left(\omega nt\right)\\ \end{array}\right\}\partial t\\ & f(t)=\frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cdot \underset{\frac{e^{j\omega \cdot nt}+e^{-j\omega \cdot nt}}{2}}{\underbrace{\cos \left(\omega nt\right)}}+\sum _{n=1}^{\mathrm{\infty }}{b}_n\cdot \underset{\frac{e^{j\omega \cdot nt}-e^{-j\omega \cdot nt}}{2j}}{\underbrace{\sin \left(\omega nt\right)}}\omega \text{=}\frac{2\pi }{T}\\ & f(t)=\frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}{e}^{j\omega nt}(\frac{a_n}{2}-j\frac{b_n}{2})+e^{-j\omega nt}(\frac{a_n}{2}+j\frac{b_n}{2})\\ & \frac{a_n}{2}\pm j\frac{b_n}{2}=\frac{1}{T}\underset{-T/2}{\overset{T/2}{\int }}f(t)[\cos \left(\omega nt\right)\pm j\sin \left(\omega nt\right)]\partial t=\frac{1}{T}\underset{-T/2}{\overset{T/2}{\int }}f(t)e^{\pm j\omega nt}\partial t\\ \end{array}$

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Ejercicio de ejemplo:(FALTA DIBUJO DE MATHEMATICA)

$\begin{array}{cc}& f(t)=\{\begin{array}{cc}& 1,-1\le t\le 0\\ & -1,0<t<1\\ \end{array}\text{, }f(t+2)=f(t),\text{ funcion impar}\\ & \omega n=\frac{2\pi n}{T}=\pi n\\ & a_0=0\text{ (por ser impar)}\\ & a_n=\frac{2}{T}\underset{-T/2}{\overset{T/2}{\int }}\underset{\text{impar}}{\underbrace{\underset{\text{impar}}{\underbrace{f(t)}}\underset{\text{par}}{\underbrace{\cos \left(\pi nt\right)}}}}\partial t=0\\ & f(t)=\sum _{n=1}^{\mathrm{\infty }}{b}_n\cdot \sin \left(\pi nt\right)\\ & b_n=\frac{2}{T}\underset{-T/2}{\overset{T/2}{\int }}\underset{\text{par}}{\underbrace{\underset{\text{impar}}{\underbrace{f(t)}}\underset{\text{impar}}{\underbrace{\sin \left(\frac{2\pi n}{T}t\right)}}}}\partial t=2\cdot \frac{2}{T}{\underset{0}{\overset{T/2}{\int }}f(t)\sin \left(\frac{2\pi n}{T}t\right)\partial t|}_{T=2}=2\underset{0}{\overset{1}{\int }}\underset{-1}{\underbrace{f(t)}}\sin \left(\pi nt\right)\partial t\\ & =-2\underset{0}{\overset{1}{\int }}\sin \left(\pi nt\right)\partial t=-2{\left(\frac{-\cos \left(\pi nt\right)}{\pi n}\right)|}_0^1=\frac{2}{\pi n}(\underset{{(-1)}^n=\pm 1}{\underbrace{\cos (\pi n)}}-1)=\{\begin{array}{cc}& 0,n=2k,k\in \mathbb{N}\\ & \frac{-4}{\pi n},n=1,3,5...=2k+1,k\in \mathbb{N}\\ \end{array}\\ & f(t)=\sum _{n=1}^{\mathrm{\infty }}\left(\frac{-4}{\pi n}\right)\cdot \sin (\pi nt)\text{ para n impar }=\sum _{k=0}^{\mathrm{\infty }}\left(\frac{-4}{\pi (2k+1)}\right)\cdot \sin (\pi (2k+1)t)\\ & \\ \end{array}$

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$\begin{array}{cc}& Ej:\cos (t+1);T=2\pi \\ & f(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{C}_n\cdot e^{j\frac{2\pi }{T}nt}\\ & C_n=\frac{1}{T}\underset{{}^{-T}╱{}_2}{\overset{{}^T╱{}_2}{\int }}f(t)\cdot e^{-j\frac{2\pi }{T}nt}\partial t\\ & \omega \cdot n=\frac{2\pi }{T}\cdot n=\frac{2\pi }{2\pi }\cdot n=n\text{; }f(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{C}_n\cdot e^{jnt}\\ & f(t)=\cos (t+1)=\frac{e^{j(t+1)}+e^{-j(t+1)}}{2}=\frac{1}{2}{e}^j\underset{n=1}{\underbrace{e^{jt}}}+\frac{1}{2}{e}^{-j}\underset{n=-1}{\underbrace{e^{-jt}}}\\ & C_1=\frac{1}{2}{e}^j={\tau }_{-1}\text{ (}{\tau }_n=C_n^{*})\\ & C_{-1}=\frac{1}{2}{e}^{-j}\\ & C_n=0,\mathrm{\forall }n\ne \pm 1\\ \end{array}$

(FALTA DIBUJO DE MATHEMATICA)

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$\begin{array}{cc}& Ej:\\ & f(t)=\{\begin{array}{cc}& t,-1\le t\le 1\\ & 0,\left|t\right|>1\\ \end{array}\\ & f(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{C}_n\cdot e^{j\omega nt}\omega \text{=}\frac{2\pi }{T=2}=\pi \\ & C_n=\frac{1}{T}\underset{{}^{-T}╱{}_2}{\overset{{}^T╱{}_2}{\int }}f(t)\cdot e^{-j\frac{2\pi }{T}nt}\partial t=\frac{1}{2}\cdot \underset{-1}{\overset{+1}{\int }}t\cdot e^{-j\pi nt}\partial t\to \\ & \left\{\begin{array}{cc}& u=t\partial u=\partial t\\ & \partial v=e^{-j\pi nt}\partial tv=\frac{-1}{j\pi n}{e}^{-j\pi nt}\\ \end{array}\right\}\\ & {\frac{1}{2}\frac{-t}{j\pi n}{e}^{-j\pi nt}|}_{-1}^1-\frac{1}{2}\frac{-1}{j\pi n}\underset{-1}{\overset{+1}{\int }}{e}^{-j\pi nt}\partial t={\frac{1}{2}\frac{-t}{j\pi n}{e}^{-j\pi nt}|}_{-1}^1-\frac{1}{2}\frac{-1}{j\pi n}{\frac{e^{-j\pi nt}}{-j\pi n}|}_{-1}^1=\\ & \frac{-1}{2}\frac{\stackrel{{(-1)}^n}{\overbrace{e^{-j\pi n}}}-(-1)\cdot \stackrel{{(-1)}^n}{\overbrace{e^{+j\pi n}}}}{j\pi n}-\frac{1}{2}{\frac{e^{-j\pi nt}}{{\left(j\pi n\right)}^2}|}_{-1}^1=\frac{-j{(-1)}^n}{\pi n}-\frac{1}{2}\frac{\stackrel{0}{\overbrace{e^{-j\pi n}-e^{-j\pi n}}}}{{\left(j\pi n\right)}^2}=\frac{-j{(-1)}^n}{\pi n}\mathrm{\forall }n\ne 0\\ & C_0=\frac{1}{T}\underset{{}^{-T}╱{}_2}{\overset{{}^T╱{}_2}{\int }}f(t)\partial t=\frac{1}{2}\underset{\text{IMPAR}}{\underbrace{\underset{-1}{\overset{1}{\int }}t\partial t}}=0\\ & f(t)=\{\begin{array}{cc}& \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{-j{(-1)}^n}{\pi n}\cdot e^{j\pi nt},\mathrm{\forall }n\ne 0\\ & 0,n=0\\ \end{array}\\ \end{array}$

Probemos ahora a poner esta misma función en la forma trigonometrica:

$\begin{array}{cc}& f(t)=\{\begin{array}{cc}& \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{-j{(-1)}^n}{\pi n}\cdot e^{j\pi nt},\mathrm{\forall }n\ne 0\\ & 0,n=0\\ \end{array}\to \frac{-j{(-1)}^n}{\pi n}=\frac{2}{2}\cdot \frac{{(-1)}^n}{j\pi n}=\frac{1}{2j}\cdot \frac{2{(-1)}^n}{\pi n}\to \sin \omega t=\frac{e^{j\omega t}-e^{-j\omega t}}{2j}\\ & f(t)=\underset{n=1}{\underbrace{\frac{1}{2j}\frac{-2}{\pi }{e}^{j\pi t}}}+\underset{n=-1}{\underbrace{\frac{1}{2j}\frac{2}{\pi }{e}^{-j\pi nt}}}+\underset{n=2}{\underbrace{\frac{1}{2j}\frac{2}{\pi 2}{e}^{j\pi 2t}}}+\underset{n=-2}{\underbrace{\frac{1}{2j}\cdot \frac{2}{\pi 2}{e}^{-j\pi 2t}}}+\underset{n=3}{\underbrace{\frac{1}{2j}\frac{-2}{\pi 3}{e}^{j\pi 3t}}}+\underset{n=-3}{\underbrace{\frac{1}{2j}\frac{2}{\pi 3}{e}^{-j\pi 3t}}}+...\\ & f(t)=\frac{-2}{\pi }\sin \left(\pi t\right)+\frac{1}{\pi }\sin \left(2\pi t\right)+\frac{-2}{\pi 3}\sin \left(3\pi t\right)+\frac{1}{2\pi }\sin \left(4\pi t\right)+...+\sum _{n=1}^{\mathrm{\infty }}\frac{2{(-1)}^n}{\pi n}\cdot \sin (n\pi t)\\ \end{array}$

Como reglas generales para saber si hemos hecho bien la serie de Fourier, tenemos:

• En la forma trigonometrica (sumas de senoides) todos los coeficientes deben ser números reales (no puede haber j), ya que, como es lógico, nuestra función a representar es real
• Si los coeficientes de mismo valor pero signo opuestos (+-1,+-2,etc..) no son iguales o complejos conjugados es que hemos hecho algo mal

Para este ejercicio en concreto al ser la función impar, el sumatorio estará compuesto por funciones impares, esto es, sin().