blog, fourier

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 T} a_0 \left[-\frac{1}{n \omega_0} \cos{n \omega_0 (\frac{T}{2})} + \frac{1}{n \omega_0} \cos{n \omega_0 (-\frac{T}{2})} \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} = - \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) }


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