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Derivación de la Ecuación de Coeficientes de Fourier

Module by: Michael Haag. E-mail the authorTranslated By: Fara Meza, Erika Jackson

Based on: Derivation of Fourier Coefficients Equation by Michael Haag

Summary: 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:

ft= n = c n ej ω 0 nt f t n c n ω 0 n t
(1)
De lo que estamos interesados aquí es el como combinar los coeficientes de Fourier, c n c n , dado a una función ft 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 ecuación 1 para c n c n , tenemos que hacer una pequeña manipulación algebraica. Primero que todo, tenemos que multiplicar los dos lados de ecuación 1 por e(j ω 0 kt) ω 0 k t , donde kZ k .

fte(j ω 0 kt)= n = c n ej ω 0 nte(j ω 0 kt) f t ω 0 k t n c n ω 0 n t ω 0 k t
(2)
Ahora integraremos los dos lados sobre el periodo, TT:
0Tfte(j ω 0 kt)d t =0T n = c n ej ω 0 nte(j ω 0 kt)d t t T 0 f t ω 0 k t t T 0 n c n ω 0 n t ω 0 k t
(3)
En el lado derecho podemos intercambiar la sumatoria y el integral y sacar la constante fuera del integral.
0Tfte(j ω 0 kt)d t = n = c n 0Tej ω 0 (nk)td t t T 0 f t ω 0 k t n c n t T 0 ω 0 n k t
(4)
Ahora que hemos hecho esto lo que al parecer es mas complicado, nos enfocaremos en tan solo el integral, 0Tej ω 0 (nk)td t t T 0 ω 0 n k t , que se encuentra en el lado derecho de la ecuación. Para este integral debemos considerar solo dos casos: n=k n k y nk n k . Para n=k n k tenemos:
0Tej ω 0 (nk)td t =T  ,   n=k    n n k t T 0 ω 0 n k t T
(5)
Para nk n k , tenemos:
0Tej ω 0 (nk)td t =0Tcos ω 0 (nk)td t +j0Tsin ω 0 (nk)td t   ,   nk    n n k t T 0 ω 0 n k t t T 0 ω 0 n k t t T 0 ω 0 n k t
(6)
Pero cos ω 0 (nk)t ω 0 n k t tiene periodos con números enteros para, nk n k , entre 00 y TT. Imagine la grafica de un coseno; por que tiene periodos con números enteros, hay áreas de igual valor debajo y arriba del eje de las ordenadas en la grafica. Este hecho es verdadero para sin ω 0 (nk)t ω 0 n k t también. Lo que significa
0Tcos ω 0 (nk)td t =0 t T 0 ω 0 n k t 0
(7)
También para el integral de una función de seno. Por eso, podemos concluir lo siguiente sobre nuestro integral:
0Tej ω 0 (nk)td t ={T  if  n=k0  otherwise   t T 0 ω 0 n k t T n k 0
(8)
Regresemos a nuestra complicada ecuación, ecuación 4, para ver si podemos encontrar una ecuación para nuestros coeficientes de Fourier. Usando los hechos que ya hemos probado, podemos ver que única vez que la ecuación ecuación 4tiene valores de no cero como resultado es cuando kk y nn son iguales:
0Tfte(j ω 0 nt)d t =T c n   ,   n=k    n n k t T 0 f t ω 0 n t T c n
(9)
Finalmente nuestra ecuación general para los coeficientes de Fourier es:
c n =1T0Tfte(j ω 0 nt)d t c n 1 T t T 0 f t ω 0 n t
(10)

Pasos para Encontrar los Coeficientes de Fourier

Para encontrar los coeficientes de Fourier de una ft f t periódica:

  1. Por alguna kk, multiplique ft f t por e(j ω 0 kt) ω 0 k t , saque el área por debajo de la curva (dividiendo por TT).
  2. Repita el paso (1) para todo kZ k .

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