Integración de la serie de Fourier. Fourier.

Se observa que la integración término por término de una serie trigonométrica

$\displaystyle \frac{1}{2} a_0 + \sum_{n=1}^{\infty}{a_n \cos{\omega_0 t} + b_n \sin{n\omega_0 t}}$

divide los coeficientes $a_n$ y $b_n$ por $\pm n omega_0$ y provocará que la integración tienda a aumentar la convergencia.

Teorema de integración de la serie deFourier

Sea $f(t)$ continua por tramos en el intervalo $-T/2 < t < T/2$ y sea $f(t + T)$ . De la serie de Fourier

$\displaystyle f(t) = \frac{1}{2} a_0 + \sum_{n=1}^{\infty}{(a_n \cos{n \omega_0 t} + b_n \sin{n \omega_0 t})}$

se puede integrar término por término para obtener

$\displaystyle \int_{t_1}^{t_2}{f(t) \, dt} = \frac{1}{2} a_0 (t_2 - t_1) + \sum_{n=1}^{\infty}{\frac{1}{n \omega_0} [-b_n(\cos{n\omega_0 t_2} - \cos{n \omega_0 t_1}) + a_n (\sin{n\omega_0 t_2} - \sin{n \omega_0 t_1})]}$

Demostración.

Como $F(t+T) = F(t)$ y $f(t)$ es continua por tramos, la función $F(t)$ esta definida como

$\displaystyle F(t) = \int_{0}^t{f(\tau) \, d\tau} - \frac{1}{2} a_0 t$

(donde $\displaystyle a_0 = \frac{1}{2} \int_{-T/2}^{T/2}{f(t) \, dt}$ ) es continua y períodica con período $T$ . Puesto que

$\displaystyle F'(t) = f(t) - \frac{1}{2} a_0$

se sigue que $F'(t)$ también sea continua. Sea la expansión de $F(t)$ en serie de Fourier

$\displaystyle F(t) = \frac{1}{2} \alpha_0 + \sum_{n=1}^{\infty}{(\alpha_n \cos{n \omega_0 t} + \beta_n \sin{n \omega_0 t})}$

Entonces, para $n \ge 1$ , $\alpha_n$ tiene el siguiente resultado (utilizando el método de integración por partes)

$\displaystyle \alpha_n = \frac{2}{T} \int_{-T/2}^{T/2}{F(t) \cos{n\omega_0 t} \, dt}$

$\displaystyle \alpha_n = \frac{2}{T} \left\{ \left[\frac{1}{n \omega_0} F(t) \sin{n \omega_0 t} \right]_{-T/2}^{T/2} - \int_{-T/2}^{T/2}{F'(t) \cdot \frac{1}{n \omega_0}\sin{n \omega_0 t} \, dt} \right\}$

$\displaystyle \alpha_n = \frac{2}{T} \left\{ \left[\frac{1}{n \omega_0} F( \frac{T}{2}) \sin{n \omega_0 (\frac{T}{2})} - \frac{1}{n \omega_0} F(-\frac{T}{2}) \sin{n \omega_0 (-\frac{T}{2})} \right] - \frac{1}{n \omega_0} \int_{-T/2}^{T/2}{F'(t) \sin{n \omega_0 t} \, dt} \right\}$

$\displaystyle \alpha_n = \frac{2}{T} \left\{ \frac{1}{n \omega_0} \sin{n \omega_0 (\frac{T}{2})} \left[ F( \frac{T}{2}) - F(-\frac{T}{2}) \right] - \frac{1}{n \omega_0} \int_{-T/2}^{T/2}{F'(t) \sin{n \omega_0 t} \, dt} \right\}$

$\displaystyle \alpha_n = \frac{2}{T} \left\{- \frac{1}{n \omega_0} \int_{-T/2}^{T/2}{F'(t) \sin{n \omega_0 t} \, dt} \right\}$

$\displaystyle \alpha_n = - \frac{1}{n \omega_0} \cdot \frac{2}{T} \int_{-T/2}^{T/2}{F'(t) \sin{n \omega_0 t} \, dt}$

$\displaystyle \alpha_n = - \frac{1}{n \omega_0} \cdot \frac{2}{T} \int_{-T/2}^{T/2}{\left[f(t) - \frac{1}{2} a_0 \right] \sin{n \omega_0 t} \, dt}$

$\displaystyle \alpha_n = - \frac{1}{n \omega_0} \cdot \frac{2}{T} \int_{-T/2}^{T/2}{\left[f(t) \sin{n \omega_0 t} - \frac{1}{2} a_0 \sin{n \omega_0 t} \right] \, dt}$

$\displaystyle \alpha_n = - \frac{1}{n \omega_0} \cdot \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} + \frac{1}{n \omega_0} \cdot \frac{2}{T} \int_{-T/2}^{T/2}{\frac{1}{2} a_0 \sin{n \omega_0 t} \, dt}$

$\displaystyle \alpha_n = - \frac{1}{n \omega_0} \cdot \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} + \frac{1}{2n \omega_0} a_0 \cdot \frac{2}{T} \int_{-T/2}^{T/2}{\sin{n \omega_0 t} \, dt}$

$\displaystyle \alpha_n = - \frac{1}{n \omega_0} \cdot \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} + \frac{1}{n \omega_0 T} a_0 \left[-\frac{1}{n \omega_0} \cos{n \omega_0 t} + C \right]_{-T/2}^{T/2}$

$\displaystyle \alpha_n = - \frac{1}{n \omega_0} \cdot \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n \omega_0 t} \, dt} = - \frac{1}{n \omega_0} b_n$

Y $\beta_n$ tiene el siguiente resultado (utilizando el método de integración por partes)

$\displaystyle \beta_n = \frac{2}{T} \int_{-T/2}^{T/2}{F(t) \sin{n\omega_0 t} \, dt}$

$\displaystyle \beta_n = \frac{2}{T} \left\{ \left[-\frac{1}{n \omega_0} F(t) \cos{n \omega_0 t} +C \right]_{-T/2}^{T/2} - \int_{-T/2}^{T/2}{F'(t) \cdot (-\frac{1}{n \omega_0} \sin{n\omega_0 t}) \, dt} \right\}$

$\displaystyle \beta_n = \frac{2}{T} \left\{ \left[-\frac{1}{n \omega_0} F(\frac{T}{2}) \cos{n \omega_0 (\frac{T}{2})} + \frac{1}{n \omega_0} F(-\frac{T}{2}) \cos{n \omega (-\frac{T}{2})} \right] + \frac{1}{n \omega_0} \int_{-T/2}^{T/2}{F'(t) \sin{n\omega_0 t} \, dt} \right\}$

$\displaystyle \beta_n = \frac{2}{T} \left\{ \left[-\frac{1}{n \omega_0} F(\frac{T}{2}) \cos{n \omega_0 (\frac{T}{2})} + \frac{1}{n \omega_0} F(\frac{T}{2}) \cos{n \omega (\frac{T}{2})} \right] + \frac{1}{n \omega_0} \int_{-T/2}^{T/2}{F'(t) \sin{n\omega_0 t} \, dt} \right\}$

$\displaystyle \beta_n = \frac{2}{T} \left\{ \frac{1}{n \omega_0} \int_{-T/2}^{T/2}{F'(t) \sin{n\omega_0 t} \, dt} \right\}$

$\displaystyle \beta_n = \frac{1}{n \omega_0} \cdot \frac{2}{T} \int_{-T/2}^{T/2}{F'(t) \sin{n\omega_0 t} \, dt} = \frac{1}{n \omega_0} a_n$

Regresando a la expansión de $F(t)$

$\displaystyle F(t) = \frac{1}{2} \alpha_0 + \sum_{n=1}^{\infty}{(\alpha_n \cos{n \omega_0 t} + \beta_n \sin{n \omega_0 t})}$

$\displaystyle F(t) = \frac{1}{2} \alpha_0 + \sum_{n=1}^{\infty}{\left[\left( - \frac{1}{n \omega_0} b_n \right) \cos{n \omega_0 t} + \left( \frac{1}{n \omega_0} a_n \right) \sin{n \omega_0 t} \right]}$

$\displaystyle F(t) = \frac{1}{2} \alpha_0 + \sum_{n=1}^{\infty}{\frac{1}{n \omega_0} \left(- b_n \cos{n \omega_0 t} + a_n \sin{n \omega_0 t} \right)}$

Mencionando una vez más lo anterior

$\displaystyle F(t) = \int_{0}^t{f(\tau) \, d\tau} - \frac{1}{2} a_0 t$

se observa que los límites de integración fueron $t=0$ y $t=t$ . Si se cambian por $t=t_1$ y $t=t_2$ , la expresión anterior tiene el siguiente resultado

$\displaystyle F(t_2) - F(t_1) = \int_{t_1}^{t_2}{f(\tau) \, d\tau} - \frac{1}{2} a_0 (t_2 - t_1)$

Despejando $\int_{t_1}^{t_2}{f(\tau) \, d\tau}$ , resulta

$\displaystyle \int_{t_1}^{t_2}{f(\tau) \, d\tau} =F(t_2) - F(t_1) + \frac{1}{2} a_0 (t_2 - t_1)$

Continuando

$\displaystyle \int_{t_1}^{t_2}{f(\tau) \, d\tau} = \sum_{n=1}^{\infty}{\frac{1}{n \omega_0} \left(- b_n \cos{n \omega_0 t_2} + a_n \sin{n \omega_0 t_2} \right)} - \sum_{n=1}^{\infty}{\frac{1}{n \omega_0} \left(- b_n \cos{n \omega_0 t_1} + a_n \sin{n \omega_0 t_1} \right)} + \frac{1}{2} a_0 (t_2 - t_1)$

$\displaystyle \int_{t_1}^{t_2}{f(\tau) \, d\tau} = \sum_{n=1}^{\infty}{\left[\frac{1}{n \omega_0} \left(- b_n \cos{n \omega_0 t_2} + a_n \sin{n \omega_0 t_2} \right) - \frac{1}{n \omega_0} \left(- b_n \cos{n \omega_0 t_1} + a_n \sin{n \omega_0 t_1} \right) \right]} + \frac{1}{2} a_0 (t_2 - t_1)$

$\displaystyle \int_{t_1}^{t_2}{f(\tau) \, d\tau} = \sum_{n=1}^{\infty}{\frac{1}{n \omega_0} \left[ \left(- b_n \cos{n \omega_0 t_2} + a_n \sin{n \omega_0 t_2} \right) - \frac{1}{n \omega_0} \left(- b_n \cos{n \omega_0 t_1} + a_n \sin{n \omega_0 t_1} \right) \right]} + \frac{1}{2} a_0 (t_2 - t_1)$

$\displaystyle \int_{t_1}^{t_2}{f(\tau) \, d\tau} = \sum_{n=1}^{\infty}{\frac{1}{n \omega_0} \left(- b_n \cos{n \omega_0 t_2} + a_n \sin{n \omega_0 t_2} + b_n \cos{n \omega_0 t_1} - a_n \sin{n \omega_0 t_1} \right)} + \frac{1}{2} a_0 (t_2 - t_1)$

$\displaystyle \int_{t_1}^{t_2}{f(\tau) \, d\tau} = \sum_{n=1}^{\infty}{\frac{1}{n \omega_0} \left[b_n (- \cos{n \omega_0 t_2} + \cos{n \omega_0 t_1}) + a_n (\sin{n \omega_0 t_2} - \sin{n \omega_0 t_1}) \right]} + \frac{1}{2} a_0 (t_2 - t_1)$

$\displaystyle \int_{t_1}^{t_2}{f(\tau) \, d\tau} = \frac{1}{2} a_0 (t_2 - t_1) + \sum_{n=1}^{\infty}{\frac{1}{n \omega_0} \left[b_n (- \cos{n \omega_0 t_2} + \cos{n \omega_0 t_1}) + a_n (\sin{n \omega_0 t_2} - \sin{n \omega_0 t_1}) \right]}$

Intercambiando $\tau$ por $t$ (variable comodín), se tiene el resultado esperado

$\displaystyle \therefore \int_{t_1}^{t_2}{f(t) \, dt} = \frac{1}{2} a_0 (t_2 - t_1) + \sum_{n=1}^{\infty}{\frac{1}{n \omega_0} \left[b_n (- \cos{n \omega_0 t_2} + \cos{n \omega_0 t_1}) + a_n (\sin{n \omega_0 t_2} - \sin{n \omega_0 t_1}) \right]}$

Relación entre el teorema de Parseval y la integración de la serie de Fourier

Sea $f(t)$ una función continua y $f'(t)$ una función continua por tramos en el intervalo $-T/2 < t < T/2$ . Si se multiplica

$\displaystyle f(t) = \frac{1}{2} a_0 + \sum_{n=1}^{\infty}{(a_n \cos{n\omega_0 t} + b_n \sin{n\omega_0 t})}$

por $f(t)$ , integrar término por término, se obtiene el siguiente resultado

$\displaystyle \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} = \frac{1}{4} a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty}{a_n^2 + b_n^2}$

Demostación.

$\displaystyle f(t) = \frac{1}{2} a_0 + \sum_{n=1}^{\infty}{(a_n \cos{n\omega_0 t} + b_n \sin{n\omega_0 t})}$

$\displaystyle f(t) \cdot f(t) = \frac{1}{2} a_0 \cdot f(t) + \sum_{n=1}^{\infty}{(a_n \cos{n\omega_0 t} + b_n \sin{n\omega_0 t}) \cdot f(t)}$

$\displaystyle [f(t)]^2 = \frac{1}{2} a_0 \cdot f(t) + \sum_{n=1}^{\infty}{(a_n f(t) \cos{n\omega_0 t} + b_n f(t) \sin{n\omega_0 t})}$

$\displaystyle \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} = \int_{-T/2}^{T/2}{\frac{1}{2} a_0 \cdot f(t) \, dt} + \int_{-T/2}^{T/2}{\sum_{n=1}^{\infty}{(a_n f(t) \cos{n\omega_0 t} + b_n f(t) \sin{n\omega_0 t})} \, dt}$

$\displaystyle \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} = \frac{1}{2} a_0 \int_{-T/2}^{T/2}{f(t) \, dt} + \sum_{n=1}^{\infty}{\left(\int_{-T/2}^{T/2}{a_n f(t) \cos{n\omega_0 t} \, dt} + \int_{-T/2}^{T/2}{b_n f(t) \sin{n\omega_0 t} \, dt} \right) }$

$\displaystyle \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} = \frac{1}{2} a_0 \cdot \frac{T}{2} \cdot \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \, dt} + \sum_{n=1}^{\infty}{\left(a_n \cdot \frac{T}{2} \cdot \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \cos{n\omega_0 t} \, dt} + b_n \cdot \frac{T}{2} \cdot \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \sin{n\omega_0 t} \, dt} \right) }$

$\displaystyle \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} = \frac{1}{2} a_0 \cdot \frac{T}{2} a_0 + \sum_{n=1}^{\infty}{\left(a_n \cdot \frac{T}{2} a_n + b_n \cdot \frac{T}{2} b_n \right) }$

$\displaystyle \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} = \frac{1}{2} \cdot \frac{T}{2} a_0^2 + \sum_{n=1}^{\infty}{\left(\frac{T}{2} a_n^2 + \frac{T}{2} b_n^2 \right) }$

$\displaystyle \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} = T \cdot \frac{1}{4} a_0^2 + \frac{T}{2} \sum_{n=1}^{\infty}{\left(a_n^2 + b_n^2 \right) }$

$\displaystyle \int_{-T/2}^{T/2}{[f(t)]^2 \, dt} = T \left[\frac{1}{4} a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty}{\left(a_n^2 + b_n^2 \right) } \right]$

$\displaystyle \therefore \frac{1}{T}\int_{-T/2}^{T/2}{[f(t)]^2 \, dt} = \frac{1}{4} a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty}{\left(a_n^2 + b_n^2 \right) }$

Integración de la serie de Fourier. Fourier.

Teorema de integración de la serie deFourier

Relación entre el teorema de Parseval y la integración de la serie de Fourier

Publicado por Cesar Reyes

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Integración de la serie de Fourier. Fourier.

Teorema de integración de la serie deFourier

Relación entre el teorema de Parseval y la integración de la serie de Fourier

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Publicado por Cesar Reyes

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