# 6.2 Derivación de la ecuación de coeficientes de fourier

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Este modulo establece los pasos para derivar la ecuación de coeficientes de Fourier de la forma general de las series de Fourier.

## IntroducciÓN

Usted debería estar familiarizado con la existencia de la ecuación general de las series de Fourier que es la siguiente:

$f(t)=\sum$ c n ω 0 n t
De lo que estamos interesados aquíes el como combinar los coeficientes de Fourier, ${c}_{n}$ , dado a una función $f(t)$ . En la siguiente explicación lo llevaremos paso por paso por la derivación de la ecuación generar para los coeficientes de Fourier dado a una función.

## Derivación

Para resolver la ecuación para ${c}_{n}$ , tenemos que hacer una pequeña manipulación algebraica. Primero que todo, tenemos que multiplicar los dos lados de por $e^{-(i{\omega }_{0}kt)}$ , donde $k\in \mathbb{Z}$ .

$f(t)e^{-(i{\omega }_{0}kt)}=\sum$ c n ω 0 n t ω 0 k t
Ahora integraremos los dos lados sobre el periodo, $T$ :
$\int_{0}^{T} f(t)e^{-(i{\omega }_{0}kt)}\,d t=\int_{0}^{T} \sum \,d t$ c n ω 0 n t ω 0 k t
En el lado derecho podemos intercambiar la sumatoria y el integral y sacar la constante fuera del integral.
$\int_{0}^{T} f(t)e^{-(i{\omega }_{0}kt)}\,d t=\sum$ c n t T 0 ω 0 n k t
Ahora que hemos hecho esto lo que al parecer es mas complicado, nos enfocaremos en tan solo el integral, $\int_{0}^{T} e^{i{\omega }_{0}(n-k)t}\,d t$ , que se encuentra en el lado derecho de la ecuación. Para este integral debemos considerar solo dos casos: $n=k$ y $n\neq k$ . Para $n=k$ tenemos:
$\forall n, n=k\colon \int_{0}^{T} e^{i{\omega }_{0}(n-k)t}\,d t=T$
Para $n\neq k$ , tenemos:
$\forall n, n\neq k\colon \int_{0}^{T} e^{i{\omega }_{0}(n-k)t}\,d t=\int_{0}^{T} \cos ({\omega }_{0}(n-k)t)\,d t+i\int_{0}^{T} \sin ({\omega }_{0}(n-k)t)\,d t$
Pero $\cos ({\omega }_{0}(n-k)t)$ tiene periodos con números enteros para, $n-k$ , entre $0$ y $T$ . Imagine la grafica de un coseno; por que tiene periodos con números enteros, hayáreas de igual valor debajo y arriba del eje de las ordenadas en la grafica. Este hecho es verdadero para $\sin ({\omega }_{0}(n-k)t)$ también. Lo que significa
$\int_{0}^{T} \cos ({\omega }_{0}(n-k)t)\,d t=0$
También para el integral de una función de seno. Por eso, podemos concluir lo siguiente sobre nuestro integral:
$\int_{0}^{T} e^{i{\omega }_{0}(n-k)t}\,d t=\begin{cases}T & \text{if n=k}\\ 0 & \text{otherwise}\end{cases}$
Regresemos a nuestra complicada ecuación, , para ver si podemos encontrar una ecuación para nuestros coeficientes de Fourier. Usando los hechos que ya hemos probado, podemos ver queúnica vez que la ecuación tiene valores de no cero como resultado es cuando $k$ y $n$ son iguales:
$\forall n, n=k\colon \int_{0}^{T} f(t)e^{-(i{\omega }_{0}nt)}\,d t=T{c}_{n}$
Finalmente nuestra ecuación general para los coeficientes de Fourier es:
${c}_{n}=\frac{1}{T}\int_{0}^{T} f(t)e^{-(i{\omega }_{0}nt)}\,d t$

## Pasos para encontrar los coeficientes de fourier

Para encontrar los coeficientes de Fourier de una $f(t)$ periódica:

• Por alguna $k$ , multiplique $f(t)$ por $e^{-(i{\omega }_{0}kt)}$ , saque elárea por debajo de la curva (dividiendo por $T$ ).
• Repita el paso (1) para todo $k\in \mathbb{Z}$ .

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