Determinando la serie de Fourier de una función “f(t)”. Fourier.

A continuación, se muestra un bloque de problemas resueltos que ayudan a determinar la serie de Fourier de una función $f(t)$ .

Problemas resueltos

Problema 1. Encontrar la serie de Fourier para la función $f(t)$ definida por

$\displaystyle f(t) =\left\{ \begin{matrix} -1 \, , \quad \quad -\frac{T}{2} < t < 0 \\ 1, \, \quad \quad \quad 0 < t < \frac{T}{2} \end{matrix} \right.$

y $f(t + T) = f(t)$ .

Solución. De la serie de Fourier

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

Se va a determinar el valor de $a_0$ . Entonces

$\displaystyle a_0 = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \, dt}$

$\displaystyle a_0 = \frac{2}{T} \int_{-T/2}^{0}{(-1) \, dt} + \frac{2}{T} \int_{0}^{T/2}{(1) \, dt}$

$\displaystyle a_0 = - \frac{2}{T} \int_{-T/2}^{0}{dt} + \frac{2}{T} \int_{0}^{T/2}{dt} = -\frac{2}{T} [t + C]_{-T/2}^{0} + \frac{2}{T} [t + C]_{0}^{T/2}$

$\displaystyle a_0 = -\frac{2}{T} \left[0 - (-\frac{T}{2}) \right] + \frac{2}{T} \left(\frac{T}{2} - 0\right) = -\frac{2}{T} \left(\frac{T}{2} \right) + \frac{2}{T} \left(\frac{T}{2}\right)$

$a_0 = - 1 + 1$

$\displaystyle a_0 = 0$

Después, se calcula el valor de $a_n$

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

$\displaystyle a_n = \frac{2}{T} \int_{-T/2}^{0}{(-1) \cdot \cos{n \omega_0 t} \, dt} + \frac{2}{T} \int_{0}^{T/2}{(1) \cdot \cos{n \omega_0 t} \, dt}$

$\displaystyle a_n = - \frac{2}{T} \int_{-T/2}^{0}{\cos{n \omega_0 t} \, dt} + \frac{2}{T} \int_{0}^{T/2}{\cos{n \omega_0 t} \, dt}$

$\displaystyle a_n = - \frac{2}{T} \left[\frac{1}{n \omega_0} \sin{n \omega_0 t} + C \right]_{-T/2}^{0} + \frac{2}{T} \left[\frac{1}{n \omega_0} \sin{n \omega_0 t} + C \right]_{0}^{T/2}$

$\displaystyle a_n = - \frac{2}{T} \left[\frac{1}{n \omega_0} \sin{(n \omega_0 \cdot 0)} - \frac{1}{n \omega_0} \sin{(n \omega_0 \cdot (-\frac{T}{2}))} \right] + \frac{2}{T} \left[\frac{1}{n \omega_0} \sin{(n \omega_0 (\frac{T}{2}))} - \frac{1}{n \omega_0} \sin{(n \omega_0 \cdot 0)} \right]$

$\displaystyle a_n = - \frac{2}{T} \left[- \frac{1}{n \omega_0} \sin{(n \omega_0 \cdot (-\frac{T}{2}))} \right] + \frac{2}{T} \left[\frac{1}{n \omega_0} \sin{(n \omega_0 (\frac{T}{2}))} \right]$

Sabiendo que $\displaystyle \omega_0 = \frac{2\pi}{T}$

$\displaystyle a_n = - \frac{2}{T} \left[- \frac{1}{n \cdot \frac{2\pi}{T}} \sin{(n \frac{2\pi}{T} \cdot (-\frac{T}{2}))} \right] + \frac{2}{T} \left[\frac{1}{n \cdot \frac{2\pi}{T}} \sin{(n \frac{2\pi}{T} (\frac{T}{2}))} \right]$

$\displaystyle a_n = - \frac{2}{T} \left[- \frac{T}{2n\pi} \sin{(-n \pi)} \right] + \frac{2}{T} \left[\frac{T}{2n \pi} \sin{(n \pi)} \right]$

$\displaystyle = - \frac{1}{n \pi} \sin{(n \pi)} + \frac{1}{n \pi} \sin{(n \pi)} = 0$

$a_n = 0$

Luego, se determina el valor de $b_n$

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

$\displaystyle b_n = \frac{2}{T} \int_{-T/2}^{0}{(-1) \cdot \sin{n \omega_0 t} \, dt} + \frac{2}{T} \int_{0}^{T/2}{(1) \cdot \sin{n \omega_0 t} \, dt}$

$\displaystyle b_n = - \frac{2}{T} \int_{-T/2}^{0}{\sin{n \omega_0 t} \, dt} + \frac{2}{T} \int_{0}^{T/2}{\sin{n \omega_0 t} \, dt}$

$\displaystyle b_n = - \frac{2}{T} \left[-\frac{1}{n \omega_0} \cos{n \omega_0 t} \right]_{-T/2}^{0} + \frac{2}{T} \left[-\frac{1}{n \omega_0} \cos{n \omega_0 t} \right]_{0}^{T/2}$

$\displaystyle b_n = - \frac{2}{T} \left[-\frac{1}{n \omega_0} \cos{(n \omega_0 (0))} - \left(-\frac{1}{n \omega_0} \cos{(n \omega_0 (-\frac{T}{2}))} \right) \right] + \frac{2}{T} \left[-\frac{1}{n \omega_0} \cos{(n \omega_0 (\frac{T}{2}))} - \left(-\frac{1}{n \omega_0} \cos{(n \omega_0 (0))}\right) \right]$

$\displaystyle b_n = - \frac{2}{T} \left[-\frac{1}{n \omega_0} (1) - \left(-\frac{1}{n \omega_0} \cos{(n \frac{2\pi}{T} (-\frac{T}{2}))} \right) \right] + \frac{2}{T} \left[-\frac{1}{n \omega_0} \cos{(n \frac{2\pi}{T} (\frac{T}{2}))} - \left(-\frac{1}{n \omega_0 } (1) \right) \right]$

$\displaystyle b_n = - \frac{2}{T} \left[-\frac{1}{n \omega_0}- \left(-\frac{1}{n \omega_0} \cos{(-n \pi)} \right) \right] + \frac{2}{T} \left[-\frac{1}{n \omega_0} \cos{(n \pi)} - \left(-\frac{1}{n \omega_0 } \right) \right]$

$\displaystyle b_n = - \frac{2}{T} \left[-\frac{1}{n \omega_0} + \frac{1}{n \omega_0} \cos{(-n \pi)} \right] + \frac{2}{T} \left[-\frac{1}{n \omega_0} \cos{(n \pi)} + \frac{1}{n \omega_0 } \right]$

$\displaystyle b_n = - \frac{2}{T} \left(-\frac{1}{n \omega_0} + \frac{1}{n \omega_0} \cos{n \pi} \right) + \frac{2}{T} \left(-\frac{1}{n \omega_0} \cos{n \pi} + \frac{1}{n \omega_0 } \right)$

$\displaystyle b_n = - \frac{2}{T} \left(-\frac{1}{n \cdot \frac{2\pi}{T}} + \frac{1}{n \cdot \frac{2\pi}{T}} \cos{n \pi} \right) + \frac{2}{T} \left(-\frac{1}{n \cdot \frac{2\pi}{T}} \cos{n \pi} + \frac{1}{n \cdot \frac{2\pi}{T}} \right)$

$\displaystyle b_n = - \frac{2}{T} \left(-\frac{T}{2 \pi n} + \frac{T}{2\pi n} \cos{n \pi} \right) + \frac{2}{T} \left(-\frac{T}{2\pi n} \cos{n \pi} + \frac{T}{2 \pi n} \right)$

$\displaystyle b_n = \frac{1}{\pi n} - \frac{1}{\pi n} \cos{n \pi} - \frac{1}{\pi n} \cos{n \pi} + \frac{1}{\pi n} = \frac{2}{\pi n} - \frac{2}{\pi n} \cos{n \pi}$

Si $n = 1, 3, 5, 7, \cdots$

$\displaystyle \frac{2}{\pi n} - \frac{2}{\pi n} \cos{n \pi} = \frac{2}{\pi n} - \frac{2}{\pi n} (-1) = \frac{2}{\pi n} + \frac{2}{\pi n} = \frac{4}{\pi n}$

Si $n = 2, 4, 6, 8, \cdots$

$\displaystyle \frac{2}{\pi n} - \frac{2}{\pi n} \cos{n \pi} = \frac{2}{\pi n} - \frac{2}{\pi n} (1) = \frac{2}{\pi n} - \frac{2}{\pi n} = 0$

Entonces

$\displaystyle b_n = \frac{4}{\pi n}$

donde $n=1, 3, 5, 7, \cdots$ .

Finalmente, sustituyendo los valores de cada coeficiente en la serie de Fourier

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

$\displaystyle f(t) = \frac{1}{2} (0) + \sum_{n=impar}^{\infty}{\left[ (0) \cos{\omega_0 t} + (\frac{4}{\pi n}) \sin{\omega_0 t} \right]}$

$\displaystyle \therefore f(t) = \frac{4}{\pi} \sum_{n=impar}^{\infty}{\frac{1}{n} \sin{\omega_0 t}}$

Problema 2. Obtener la serie de Fourier de la siguiente figura

Solución. Analizando la gráfica, en $\displaystyle -\frac{T}{2} < t \le 0$ , $f(t)$ es

$\displaystyle f(t) - f(t_0) = \frac{f(t_1) - f(t_0)}{t_1 - t_0} (t - t_0)$

$\displaystyle f(t) - (-1) = \frac{(1) - (-1)}{0 - (-\frac{T}{2})} (t - (-\frac{T}{2}))$

$\displaystyle f(t) + 1 = \frac{1 + 1}{\frac{T}{2}} (t + \frac{T}{2})$

$\displaystyle f(t) + 1 = \frac{2}{\frac{T}{2}} (t + \frac{T}{2})$

$\displaystyle f(t) + 1 = \frac{4}{T} (t + \frac{T}{2})$

$\displaystyle f(t) + 1 = \frac{4}{T} t + 2$

$\displaystyle f(t) = \frac{4}{T} t + 1$

Y en $\displaystyle 0 \le t < \frac{T}{2}$ , $f(t)$ es

$\displaystyle f(t) - f(t_0) = \frac{f(t_1) - f(t_0)}{t_1 - t_0} (t - t_0)$

$\displaystyle f(t) - (1) = \frac{(-1) - (1)}{\frac{T}{2} - 0} (t - 0)$

$\displaystyle f(t) - 1 = \frac{-1 - 1}{\frac{T}{2}} (t)$

$\displaystyle f(t) - 1 = \frac{-2}{\frac{T}{2}} t$

$\displaystyle f(t) - 1 = -\frac{4}{T} t$

$\displaystyle f(t) - 1 = - \frac{4}{T} t$

$\displaystyle f(t) = - \frac{4}{T} t + 1$

Por esa parte, $f(t)$ esta definida de la siguiente manera

$\displaystyle f(t) = \left\{ \begin{matrix} \frac{4}{T} t + 1 \, , \quad \quad -\frac{T}{2} < t \le 0 \\ - \frac{4}{T} t + 1, \, \quad \quad 0 \le t < \frac{T}{2} \end{matrix} \right.$

Depués, recordando la serie de Fourier

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

Calculando $a_0$

$\displaystyle a_0 = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \, dt}$

$\displaystyle a_0 = \frac{2}{T} \int_{-T/2}^{0}{\left( \frac{4}{T} t + 1 \right) \, dt} + \frac{2}{T} \int_{0}^{T/2}{\left(-\frac{4}{T} t + 1 \right) \, dt}$

$\displaystyle a_0 = \frac{2}{T} \cdot \frac{4}{T} \int_{-T/2}^{0}{t \, dt} + \frac{2}{T} \int_{-T/2}^{0}{dt} - \frac{2}{T} \cdot \frac{4}{T} \int_{0}^{T/2}{t \, dt} + \frac{2}{T} \int_{0}^{T/2}{dt}$

$\displaystyle a_0 = \frac{8}{T^2} \int_{-T/2}^{0}{t \, dt} + \frac{2}{T} \int_{-T/2}^{0}{dt} - \frac{8}{T^2} \int_{0}^{T/2}{t \, dt} + \frac{2}{T} \int_{0}^{T/2}{dt}$

$\displaystyle a_0 = \frac{8}{T^2} \int_{-T/2}^{0}{t \, dt} - \frac{8}{T^2} \int_{0}^{T/2}{t \, dt} + \frac{2}{T} \int_{-T/2}^{0}{dt} + \frac{2}{T} \int_{0}^{T/2}{dt}$

$\displaystyle a_0 = \frac{8}{T^2} \int_{-T/2}^{0}{t \, dt} - \frac{8}{T^2} \int_{0}^{T/2}{t \, dt} + \frac{2}{T} \int_{-T/2}^{T/2}{dt}$

$\displaystyle a_0 = \frac{8}{T^2} \left[ \frac{1}{2} t^2 + C \right]_{-T/2}^{0} - \frac{8}{T^2} \left[ \frac{1}{2} t^2 + C \right]_{0}^{T/2} + \frac{2}{T} [t + C]_{-T/2}^{T/2}$

$\displaystyle a_0 = \frac{8}{T^2} \left[\frac{1}{2} (0)^2 - \frac{1}{2} (-\frac{T}{2})^2\right] - \frac{8}{T^2} \left[ \frac{1}{2} (\frac{T}{2})^2 - \frac{1}{2} (0)^2 \right] + \frac{2}{T} \left[\frac{T}{2} - (-\frac{T}{2}) \right]$

$\displaystyle a_0 = \frac{8}{T^2} \left[- \frac{1}{2} \left( \frac{T^2}{4} \right) \right] - \frac{8}{T^2} \left[ \frac{1}{2} \left(\frac{T^2}{4} \right) \right] + \frac{2}{T} \left(\frac{T}{2} + \frac{T}{2} \right)$

$\displaystyle a_0 = \frac{8}{T^2} \left(- \frac{T^2}{8} \right) - \frac{8}{T^2} \left(\frac{T^2}{8} \right) + \frac{2}{T} (T) = -1 -1 +2$

$a_0 = 0$

Calculando $a_n$ , resulta

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

$\displaystyle a_n = \frac{2}{T} \int_{-T/2}^{0}{\left(\frac{4}{T} t + 1 \right) \cos{n \omega_0 t} \, dt} + \frac{2}{T} \int_{0}^{T/2}{\left(-\frac{4}{T} t + 1 \right) \cos{n \omega_0 t} \, dt}$

$\displaystyle a_n = \frac{2}{T} \cdot \frac{4}{T} \int_{-T/2}^{0}{t \cos{n \omega_0 t} \, dt} + \frac{2}{T} \int_{-T/2}^{0}{\cos{n \omega_0 t} \, dt} - \frac{2}{T} \cdot \frac{4}{T} \int_{0}^{T/2}{t \cos{n \omega_0 t} \, dt} + \frac{2}{T} \int_{0}^{T/2}{\cos{n \omega_0 t} \, dt}$

$\displaystyle a_n = \frac{8}{T^2} \int_{-T/2}^{0}{t \cos{n \omega_0 t} \, dt} + \frac{2}{T} \int_{-T/2}^{0}{\cos{n \omega_0 t} \, dt} - \frac{8}{T^2} \int_{0}^{T/2}{t \cos{n \omega_0 t} \, dt} + \frac{2}{T} \int_{0}^{T/2}{\cos{n \omega_0 t} \, dt}$

$\displaystyle a_n = \frac{8}{T^2} \int_{-T/2}^{0}{t \cos{n \omega_0 t} \, dt} - \frac{8}{T^2} \int_{0}^{T/2}{t \cos{n \omega_0 t} \, dt} + \frac{2}{T} \int_{-T/2}^{0}{\cos{n \omega_0 t} \, dt} + \frac{2}{T} \int_{0}^{T/2}{\cos{n \omega_0 t} \, dt}$

$\displaystyle a_n = \frac{8}{T^2} \int_{-T/2}^{0}{t \cos{n \omega_0 t} \, dt} - \frac{8}{T^2} \int_{0}^{T/2}{t \cos{n \omega_0 t} \, dt} + \frac{2}{T} \int_{-T/2}^{T/2}{\cos{n \omega_0 t} \, dt}$

De la primera y segunda integral

$\displaystyle \int{t \cos{n \omega_0 t} \, dt}$

se resuelve por el método de integración por partes (donde $u=t$ , $du=dt$ , $dv=\cos{n \omega_0 t}$ y $v = \frac{1}{n \omega_0} \sin{n \omega_0 t}$ ). Entonces

$\displaystyle \int{t \cos{n \omega_0 t} \, dt} = (t) \left( \frac{1}{n \omega_0} \sin{n \omega_0 t} \right) - \int{\left(\frac{1}{n \omega_0} \sin{n \omega_0 t} \right)dt}$

$\displaystyle \int{t \cos{n \omega_0 t} \, dt} = \frac{1}{n \omega_0} t \sin{n \omega_0 t} - \frac{1}{n \omega_0} \int{\sin{n \omega_0 t} \, dt}$

$\displaystyle \int{t \cos{n \omega_0 t} \, dt} = \frac{1}{n \omega_0} t \sin{n \omega_0 t} + \frac{1}{n^2 \omega_0^2} \cos{n \omega_0 t} + C$

Para la tercera integral, se toma la propiedad de ortogonalidad

$\displaystyle \frac{2}{T} \int_{-T/2}^{T/2}{\cos{n \omega_0 t} \, dt} = 0$

Regresando y remplazando la variable $t$ por sus respectivos límites interior y superior, resulta lo siguiente

$\displaystyle a_n = \frac{8}{T^2} \int_{-T/2}^{0}{t \cos{n \omega_0 t} \, dt} - \frac{8}{T^2} \int_{0}^{T/2}{t \cos{n \omega_0 t} \, dt} + \frac{2}{T} \int_{-T/2}^{T/2}{\cos{n \omega_0 t} \, dt}$

$\displaystyle a_n = \frac{8}{T^2} \left[ \frac{1}{n \omega_0} t \sin{n \omega_0 t} + \frac{1}{n^2 \omega_0^2} \cos{n \omega_0 t} + C \right]_{-T/2}^{0} - \frac{8}{T^2} \left[ \frac{1}{n \omega_0} t \sin{n \omega_0 t} + \frac{1}{n^2 \omega_0^2} \cos{n \omega_0 t} + C \right]_{0}^{T/2} + \frac{2}{T}(0)$

$\displaystyle a_n = \frac{8}{T^2} \left[ \left(\frac{1}{n \omega_0} (0) \sin{n \omega_0 (0)} + \frac{1}{n^2 \omega_0^2} \cos{n \omega_0 (0)} \right) - \left(\frac{1}{n \omega_0} (-\frac{T}{2}) \sin{(n \omega_0 (-\frac{T}{2}) )} + \frac{1}{n^2 \omega_0^2} \cos{n \omega_0 (-\frac{T}{2}))} \right) \right] - \frac{8}{T^2} \left[ \left(\frac{1}{n \omega_0} (\frac{T}{2}) \sin{(n \omega_0 (\frac{T}{2}) )} + \frac{1}{n^2 \omega_0^2} \cos{n \omega_0 (\frac{T}{2}))} \right) - \left(\frac{1}{n \omega_0} (0) \sin{n \omega_0 (0)} - \frac{1}{n^2 \omega_0^2} \cos{n \omega_0 (0)} \right) \right] + 0$

$\displaystyle a_n = \frac{8}{T^2} \left[\frac{1}{n^2 \omega_0^2} (1) - \left(\frac{1}{n \omega_0} (-\frac{T}{2}) \sin{(n \omega_0 (-\frac{T}{2}) )} + \frac{1}{n^2 \omega_0^2} \cos{n \omega_0 (-\frac{T}{2}))} \right) \right] - \frac{8}{T^2} \left[ \left(\frac{1}{n \omega_0} (\frac{T}{2}) \sin{(n \omega_0 (\frac{T}{2}) )} + \frac{1}{n^2 \omega_0^2} \cos{n \omega_0 (\frac{T}{2}))} \right) - \frac{1}{n^2 \omega_0^2} (1) \right]$

$\displaystyle a_n = \frac{8}{T^2} \left[\frac{1}{n^2 \omega_0^2} - \left(-\frac{T}{2n \omega_0} \sin{(n (\frac{2\pi}{T}) (-\frac{T}{2}) )} + \frac{1}{n^2 \omega_0^2} \cos{(n (\frac{2\pi}{T}) (-\frac{T}{2}))} \right) \right] - \frac{8}{T^2} \left[ \left(\frac{T}{2n \omega_0} \sin{(n (\frac{2\pi}{T}) (\frac{T}{2}) )} + \frac{1}{n^2 \omega_0^2} \cos{(n (\frac{2\pi}{T}) (\frac{T}{2}))} \right) - \frac{1}{n^2 \omega_0^2} \right]$

$\displaystyle a_n = \frac{8}{T^2} \left[\frac{1}{n^2 \omega_0^2} - \left(-\frac{T}{2n \omega_0} \sin{(-n \pi)} + \frac{1}{n^2 \omega_0^2} \cos{(-n\pi)} \right) \right] - \frac{8}{T^2} \left[ \left(\frac{T}{2n \omega_0} \sin{(n \pi)} + \frac{1}{n^2 \omega_0^2} \cos{n \pi)} \right) - \frac{1}{n^2 \omega_0^2} \right]$

$\displaystyle a_n = \frac{8}{T^2} \left[\frac{1}{n^2 \omega_0^2} + \frac{T}{2n \omega_0} \sin{(-n \pi)} - \frac{1}{n^2 \omega_0^2} \cos{(-n\pi)} \right] - \frac{8}{T^2} \left[ \frac{T}{2n \omega_0} \sin{(n \pi)} + \frac{1}{n^2 \omega_0^2} \cos{(n \pi)} - \frac{1}{n^2 \omega_0^2} \right]$

$\displaystyle a_n = \frac{8}{T^2} \left[\frac{T^2}{n^2 \cdot 4\pi^2} + \frac{T^2}{2n \cdot 2\pi} \sin{(-n \pi)} - \frac{T^2}{n^2 \cdot 4\pi^2} \cos{(-n\pi)} \right] - \frac{8}{T^2} \left[ \frac{T^2}{2n \cdot 2\pi} \sin{(n \pi)} + \frac{T^2}{n^2 \cdot 4\pi^2} \cos{(n \pi)} - \frac{T^2}{n^2 \cdot 4\pi^2} \right]$

$\displaystyle a_n = \frac{8}{T^2} \left[\frac{T^2}{4n^2\pi^2} + \frac{T^2}{4\pi n} \sin{(-n \pi)} - \frac{T^2}{4 n^2 \pi^2} \cos{(-n\pi)} \right] - \frac{8}{T^2} \left[ \frac{T^2}{4n \pi} \sin{(n \pi)} + \frac{T^2}{4 \pi^2 n^2} \cos{(n \pi)} - \frac{T^2}{4 \pi^2 n^2} \right]$

$\displaystyle a_n = \frac{8}{T^2} \frac{T^2}{4n^2\pi^2} + \frac{8}{T^2} \frac{T^2}{4\pi n} \sin{(-n \pi)} - \frac{8}{T^2} \cdot \frac{T^2}{4 n^2 \pi^2} \cos{(-n\pi)} - \frac{8}{T^2} \cdot \frac{T^2}{4n \pi} \sin{(n \pi)} - \frac{8}{T^2} \cdot \frac{T^2}{4 \pi^2 n^2} \cos{(n \pi)} + \frac{8}{T^2} \cdot \frac{T^2}{4 \pi^2 n^2}$

$\displaystyle a_n = \frac{2}{n^2\pi^2} + \frac{2}{\pi n} \sin{(-n \pi)} - \frac{2}{n^2 \pi^2} \cos{(-n\pi)} - \frac{2}{n \pi} \sin{(n \pi)} - \frac{2}{\pi^2 n^2} \cos{(n \pi)} + \frac{2}{\pi^2 n^2}$

$\displaystyle a_n = \frac{2}{n^2\pi^2} - \frac{2}{\pi n} \sin{(n \pi)} - \frac{2}{n^2 \pi^2} \cos{(n\pi)} - \frac{2}{n \pi} \sin{(n \pi)} - \frac{2}{\pi^2 n^2} \cos{(n \pi)} + \frac{2}{\pi^2 n^2}$

$\displaystyle a_n = \frac{4}{n^2\pi^2} - \frac{4}{\pi n} \sin{(n \pi)} - \frac{4}{n^2 \pi^2} \cos{(n\pi)}$

Si $n=1,3,5,7,\cdots$

$\displaystyle \frac{4}{n^2\pi^2} - \frac{4}{\pi n} \sin{(n \pi)} - \frac{4}{n^2 \pi^2} \cos{(n\pi)} = \frac{4}{n^2\pi^2} - \frac{4}{n^2 \pi^2}(-1)$

$\displaystyle = \frac{4}{n^2\pi^2} + \frac{4}{n^2 \pi^2} = \frac{8}{n^2 \pi^2}$

Si $n=2,4,6,8,\cdots$

$\displaystyle \frac{4}{n^2\pi^2} - \frac{4}{\pi n} \sin{(n \pi)} - \frac{4}{n^2 \pi^2} \cos{(n\pi)} = \frac{4}{n^2\pi^2} - \frac{4}{n^2 \pi^2}(1) = 0$

Por tanto, cuando $n$ es impar, el valor de $a_n$ es

$\displaystyle a_n = \frac{8}{n^2 \pi^2}$

Calculando el valor de $b_n$

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

$\displaystyle b_n = \frac{2}{T} \int_{-T/2}^{0}{\left( \frac{4}{T}t + 1\right) \sin{n \omega_0 t} \, dt} + \frac{2}{T} \int_{0}^{T/2}{\left( -\frac{4}{T}t + 1\right)\sin{n \omega_0 t} \, dt}$

$\displaystyle b_n = \frac{2}{T} \cdot \frac{4}{T} \int_{-T/2}^{0}{t \sin{n \omega_0 t} \, dt} + \frac{2}{T} \int_{-T/2}^{0}{\sin{n \omega_0 t} \, dt} - \frac{2}{T} \cdot \frac{4}{T} \int_{0}^{T/2}{t \sin{n \omega_0 t} \, dt} + \frac{2}{T} \int_{0}^{T/2}{\sin{n \omega_0 t} \, dt}$

$\displaystyle b_n = \frac{8}{T^2} \int_{-T/2}^{0}{t \sin{n \omega_0 t} \, dt} + \frac{2}{T} \int_{-T/2}^{0}{\sin{n \omega_0 t} \, dt} - \frac{8}{T^2} \int_{0}^{T/2}{t \sin{n \omega_0 t} \, dt} + \frac{2}{T} \int_{0}^{T/2}{\sin{n \omega_0 t} \, dt}$

$\displaystyle b_n = \frac{8}{T^2} \int_{-T/2}^{0}{t \sin{n \omega_0 t} \, dt} - \frac{8}{T^2} \int_{0}^{T/2}{t \sin{n \omega_0 t} \, dt} + \frac{2}{T} \int_{-T/2}^{0}{\sin{n \omega_0 t} \, dt} + \frac{2}{T} \int_{0}^{T/2}{\sin{n \omega_0 t} \, dt}$

$\displaystyle b_n = \frac{8}{T^2} \int_{-T/2}^{0}{t \sin{n \omega_0 t} \, dt} - \frac{8}{T^2} \int_{0}^{T/2}{t \sin{n \omega_0 t} \, dt} + \frac{2}{T} \int_{-T/2}^{T/2}{\sin{n \omega_0 t} \, dt}$

De la primera y segunda integral

$\displaystyle \int{t \sin{n \omega_0 t} \, dt}$

se resolverá por el método de integración por partes (donde $u=t$ , $du=dt$ , $dv=\sin{n \omega_0 t}$ y $v=-\frac{1}{n \omega_0}\cos{n \omega_0 t}$ ). Entonces

$\displaystyle \int{t \sin{n \omega_0 t} \, dt} = (t)\left[-\frac{1}{n \omega_0}\cos{n \omega_0 t} \right] - \int{\left( -\frac{1}{n \omega_0}\cos{n \omega_0 t}\right) \, dt}$

$\displaystyle \int{t \sin{n \omega_0 t} \, dt} = -\frac{1}{n \omega_0} t \cos{n \omega_0 t} + \frac{1}{n \omega_0} \int{\cos{n \omega_0 t} \, dt}$

$\displaystyle \int{t \sin{n \omega_0 t} \, dt} = -\frac{1}{n \omega_0} t \cos{n \omega_0 t} + \frac{1}{n^2 \omega_0^2} \sin{n \omega_0 t} + C$

En la tercera integral, por propiedad de ortogonalidad

$\displaystyle \int_{-T/2}^{T/2}{\sin{n \omega_0 t} \, dt} = 0$

Regresando y remplazando la variable $t$ por sus respectivos límites, resulta lo siguiente

$\displaystyle b_n = \frac{8}{T^2} \int_{-T/2}^{0}{t \sin{n \omega_0 t} \, dt} - \frac{8}{T^2} \int_{0}^{T/2}{t \sin{n \omega_0 t} \, dt} + \frac{2}{T} \int_{-T/2}^{T/2}{\sin{n \omega_0 t} \, dt}$

$\displaystyle b_n = \frac{8}{T^2} \left[ -\frac{1}{n \omega_0} t \cos{n \omega_0 t} + \frac{1}{n^2 \omega_0^2} \sin{n \omega_0 t} + C \right]_{-T/2}^{0} - \frac{8}{T^2} \left[ -\frac{1}{n \omega_0} t \cos{n \omega_0 t} + \frac{1}{n^2 \omega_0^2} \sin{n \omega_0 t} + C \right]_{0}^{T/2} + \frac{2}{T} (0)$

$\displaystyle b_n = \frac{8}{T^2} \left[ -\frac{1}{n \omega_0} (0) \cos{(n \omega_0 (0))} + \frac{1}{n^2 \omega_0^2} \sin{(n \omega_0 (0))} - \left( -\frac{1}{n \omega_0} (-\frac{T}{2}) \cos{(n \omega_0 (-\frac{T}{2}))} + \frac{1}{n^2 \omega_0^2} \sin{(n \omega_0 (-\frac{T}{2}))} \right) \right] - \frac{8}{T^2} \left[ -\frac{1}{n \omega_0} (\frac{T}{2}) \cos{(n \omega_0 (\frac{T}{2}))} + \frac{1}{n^2 \omega_0^2} \sin{(n \omega_0 (\frac{T}{2}))} - \left( -\frac{1}{n \omega_0} (0) \cos{(n \omega_0 (0))} + \frac{1}{n^2 \omega_0^2} \sin{(n \omega_0 (0))} \right) \right] + 0$

$\displaystyle b_n = \frac{8}{T^2} \left[- \left( \frac{T}{n \cdot 2\pi } (\frac{T}{2})\cos{(n \frac{2\pi}{T} (-\frac{T}{2}))} + \frac{T^2}{n^2 \cdot 4 \pi^2} \sin{(n (\frac{2\pi}{T}) (-\frac{T}{2}))} \right) \right] - \frac{8}{T^2} \left[ -\frac{T}{n \cdot 2\pi} (\frac{T}{2}) \cos{(n (\frac{2\pi}{T}) (\frac{T}{2}))} + \frac{T^2}{n^2 4 \pi^2} \sin{(n (\frac{2\pi}{T}) (\frac{T}{2}))} - 0 \right]$

$\displaystyle b_n = \frac{8}{T^2} \left[- \left( \frac{T^2}{4 \pi n} \cos{(-n \pi)} + \frac{T^2}{4 n^2 \pi^2} \sin{(-n \pi)} \right) \right] - \frac{8}{T^2} \left[ -\frac{T^2}{4n \pi} \cos{(n \pi)} + \frac{T^2}{4 n^2 \pi^2} \sin{(n \pi)} \right]$

$\displaystyle b_n = - \frac{8}{T^2} \frac{T^2}{4 \pi n} \cos{(-n \pi)} + \frac{8}{T^2} \frac{T^2}{4 n^2 \pi^2} \sin{(-n \pi)} + \frac{8}{T^2}\frac{T^2}{4n \pi} \cos{(n \pi)} - \frac{8}{T^2} \frac{T^2}{4 n^2 \pi^2} \sin{(n \pi)}$

$\displaystyle b_n = - \frac{2}{\pi n} \cos{(-n \pi)} + \frac{2}{n^2 \pi^2} \sin{(-n \pi)} + \frac{2}{n \pi} \cos{(n \pi)} - \frac{2}{n^2 \pi^2} \sin{(n \pi)}$

$\displaystyle b_n = - \frac{2}{\pi n} \cos{(n \pi)} - \frac{2}{n^2 \pi^2} \sin{(n \pi)} + \frac{2}{n \pi} \cos{(n \pi)} - \frac{2}{n^2 \pi^2} \sin{(n \pi)}$

$b_n = 0$

Regresando y sustituyendo los coeficientes de Fourier en la serie de Fourier, el resultado final es

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

$\displaystyle f(t) = \frac{1}{2} (0) + \sum_{n=impar}^{\infty}{(\frac{8}{n^2 \pi^2} \cos{\omega_0 t} + (0) \sin{\omega_0 t})} = \sum_{n=impar}^{\infty}{(\frac{8}{n^2 \pi^2} \cos{\omega_0 t})}$

$\displaystyle \therefore f(t) = \frac{8}{\pi^2} \sum_{n=impar}^{\infty}{ \frac{1}{n^2} \cos{\omega_0 t}}$

Problema 3. Encontrar la serie de Fourier para la función $f(t)$ definida por

$\displaystyle f(t) = \left\{ \begin{matrix} 0 \, , \quad \quad \quad \quad -\frac{T}{2} < t < 0 \\ A \sin{\omega_0 t}, \, \quad \quad 0 < t < \frac{T}{2} \end{matrix} \right.$

y $f(t+T) = f(t)$ , $\omega_0 = \frac{2\pi}{T}$ .

Solución. Recordando la serie de Fourier

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

Se calcula el coeficiente $a_0$

$\displaystyle a_0 = \frac{2}{T} \int_{-T/2}^{T/2}{f(t) \, dt}$

$\displaystyle a_0 = \frac{2}{T} \int_{-T/2}^{0}{0 \, dt} + \frac{2}{T} \int_{0}^{T/2}{A \sin{\omega_0 t} \, dt}$

$\displaystyle a_0 = A \cdot \frac{2}{T} \int_{0}^{T/2}{\sin{\omega_0 t} \, dt} = \frac{2A}{T} \int_{0}^{T/2}{\sin{\omega_0 t} \, dt}$

$\displaystyle a_0 = \left[- \frac{2A}{T \omega_0} \cos{\omega_0 t} \right]_{0}^{T/2} = \left[- \frac{2A}{T \omega_0} \cos{(\omega_0 \cdot \frac{T}{2})} \right] - \left[- \frac{2A}{T \omega_0} \cos{(\omega_0 \cdot 0)} \right]$

$\displaystyle a_0 = \left[- \frac{2A}{T \cdot \frac{2\pi}{T}} \cos{(\frac{2\pi}{T} \cdot \frac{T}{2})} \right] - \left[- \frac{2A}{T \cdot \frac{2\pi}{T}} \cos{(0)} \right]$

$\displaystyle a_0 = \left[- \frac{A}{\pi} \cos{\pi} \right] - \left[- \frac{A}{\pi} \cos{0} \right] = \left[- \frac{2A}{\pi} (-1) \right] - \left[- \frac{2A}{\pi} (1) \right]$

$\displaystyle a_0 = \left(\frac{A}{\pi} \right) - \left(- \frac{A}{\pi} \right) = \frac{A}{\pi} + \frac{A}{\pi} = \frac{2A}{\pi}$

$\displaystyle a_0 = \frac{2A}{\pi}$

Calculando $a_n$ , resulta

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

$\displaystyle a_n = \frac{2}{T} \int_{-T/2}^{0}{0 \cdot \cos{n \omega_0 t} \, dt} + \frac{2}{T} \int_{0}^{T/2}{A \sin{\omega_0 t} \cos{n \omega_0 t} \, dt}$

$\displaystyle a_n = \frac{2A}{T} \int_{0}^{T/2}{\sin{\omega_0 t} \cos{n \omega_0 t} \, dt}$

Si $n=1$ , el integrando tiene la siguiente forma

$\displaystyle a_1 = \frac{2A}{T} \int_{0}^{T/2}{\sin{\omega_0 t} \cos{\omega_0 t} \, dt}$

$\displaystyle a_1 = \frac{2A}{T} \left[ \frac{1}{\omega_0} \sin^{2}{\omega_0 t} \right]_{0}^{T/2} = \frac{2A}{T}\left\{ \left[ \frac{1}{\omega_0} \sin^{2}{\omega_0 (\frac{T}{2})} \right] - \left[ \frac{1}{\omega_0} \sin^{2}{\omega_0 (0)} \right] \right\}$

$\displaystyle a_1 = \frac{2A}{T}\left\{ \left[ \frac{1}{\frac{2\pi}{T}} \sin^{2}{(\frac{2\pi}{T}) (\frac{T}{2})} \right] - 0 \right\} = \frac{2A}{T}\left( \frac{T}{2\pi} \sin^{2}{\pi} \right) = 0$

Si $n=2, 3, 4, \cdots$ , el integrando tiene la siguiente forma

$\displaystyle a_n = \frac{2A}{T} \int_{0}^{T/2}{\sin{\omega_0 t} \cos{n \omega_0 t} \, dt}$

Esta integral se resuelve por integración de productos de funciones seno y coseno con diferentes argumentos en la misma variable. Se observa que esta integral de la forma $\displaystyle \int{\sin{mu} \cos{nu} \, du}$ , donde $m=\omega_0$ y $n=\omega_0 n$ , entonces

$\displaystyle \int{\sin{\omega_0 t} \cos{n \omega_0 t} \, dt} = -\frac{1}{2(\omega_0 + n \omega_0)} \cos{(\omega_0 + n\omega_0)t} - \frac{1}{2(\omega_0 + n \omega_0 )}\cos{(\omega_0 - n \omega_0)t}$

$\displaystyle = -\frac{1}{2\omega_0(1 + n)} \cos{\omega_0 (1 + n)t} - \frac{1}{2\omega_0 (1 + n)} \cos{\omega_0(1 - n)t} + C$

Remplazando la variable « $t$ » por sus respectivos límites, resulta lo siguiente

$\displaystyle a_n = \frac{2A}{T} \int_{0}^{T/2}{\sin{\omega_0 t} \cos{n \omega_0 t} \, dt}$

$\displaystyle a_n = \frac{2A}{T} \left\{\left[ -\frac{1}{2\omega_0(1 + n)} \cos{\omega_0 (1 + n)t} - \frac{1}{2\omega_0 (1 - n)} \cos{\omega_0(1 - n)t} \right]_{0}^{T/2} \right\}$

$\displaystyle a_n = \frac{2A}{T} \left\{ \left[ -\frac{1}{2\omega_0(1 + n)} \cos{\omega_0 (1 + n)(\frac{T}{2})} - \frac{1}{2\omega_0 (1 - n)} \cos{\omega_0(1 - n)(\frac{T}{2})} \right] - \left[ -\frac{1}{2\omega_0(1 + n)} \cos{\omega_0 (1 + n)(0)} - \frac{1}{2\omega_0 (1 - n)} \cos{\omega_0(1 - n)(0)} \right] \right\}$

$\displaystyle a_n = \frac{2A}{T}\left\{ \left[ -\frac{1}{2(\frac{2\pi}{T})(1 + n)} \cos{(\frac{2\pi}{T}) (1 + n)(\frac{T}{2})} - \frac{1}{2 (\frac{2\pi}{T}) (1 - n)} \cos{(\frac{2\pi}{T})(1 - n)(\frac{T}{2})} \right] - \left[ -\frac{1}{2 (\frac{2\pi}{T}) (1 + n)} \cos{0} - \frac{1}{2 (\frac{2\pi}{T})(1 - n)} \cos{0} \right] \right\}$

$\displaystyle a_n = \frac{2A}{T} \left\{ \left[ -\frac{T}{4\pi (1 + n)} \cos{\pi(1 + n)} - \frac{T}{4 \pi (1 - n)} \cos{\pi (1 - n)} \right] - \left[ -\frac{T}{4 \pi (1 + n)} (1) - \frac{T}{4 \pi (1 - n)}(1) \right] \right\}$

$\displaystyle a_n = \frac{2A}{T}\left[ -\frac{T}{4\pi (1 + n)} \cos{\pi(1 + n)} - \frac{T}{4 \pi (1 - n)} \cos{\pi (1 - n)} + \frac{T}{4 \pi (1 + n)} + \frac{T}{4 \pi (1 - n)} \right]$

$\displaystyle a_n = \frac{2A}{T} \cdot \frac{T}{4\pi} \left[ -\frac{1}{1 + n} \cos{\pi(1 + n)} - \frac{1}{1 - n} \cos{\pi (1 - n)} + \frac{1}{1 + n} + \frac{1}{1 - n} \right]$

$\displaystyle a_n = \frac{A}{2\pi} \left[ -\frac{1}{1 + n} \cos{\pi(1 + n)} - \frac{1}{1 - n} \cos{\pi (1 - n)} + \frac{2}{(1 + n)(1 - n)} \right]$

$\displaystyle a_n = \frac{A}{2\pi} \left[ \frac{2}{(1 + n)(1 - n)} -\frac{1}{1 + n} \cos{\pi(1 + n)} - \frac{1}{1 - n} \cos{\pi (1 - n)} \right]$

$\displaystyle a_n = \frac{A}{2\pi} \cdot \frac{2}{(1 + n)(1 - n)} - \frac{A}{2\pi} \cdot \frac{1}{1 + n} \cos{\pi(1 + n)} - \frac{A}{2\pi} \cdot \frac{1}{1 - n} \cos{\pi (1 - n)}$

$\displaystyle a_n = \frac{A}{\pi(1 + n)(1 - n)} - \frac{A}{2\pi(1 + n)} \cos{\pi(1 + n)} - \frac{A}{2\pi (1 - n)} \cos{\pi (1 - n)}$

Si $n=2,4,6, \cdots$

$\displaystyle a_n = \frac{A}{\pi(1 + n)(1 - n)} - \frac{A}{2\pi(1 + n)} (-1) - \frac{A}{2\pi (1 - n)} (-1)$

$\displaystyle a_n = \frac{A}{\pi(1 + n)(1 - n)} + \frac{A}{2\pi(1 + n)} + \frac{A}{2\pi (1 - n)}$

$\displaystyle a_n = \frac{A}{\pi(1 + n)(1 - n)} + \frac{A}{\pi(1 + n)(1- n)} = \frac{2A}{\pi(1 + n)(1 - n)}$

Si $n=3,5,7, \cdots$

$\displaystyle a_n = \frac{A}{\pi(1 + n)(1 - n)} - \frac{A}{2\pi(1 + n)} (1) - \frac{A}{2\pi (1 - n)} (1)$

$\displaystyle a_n = \frac{A}{\pi(1 + n)(1 - n)} - \frac{A}{2\pi(1 + n)} - \frac{A}{2\pi (1 - n)}$

$\displaystyle a_n = \frac{A}{\pi(1 + n)(1 - n)} - \frac{A}{\pi(1 + n)(1- n)} = 0$

Finalmente

$\displaystyle a_n = \frac{2A}{\pi(1 + n)(1 - n)}$

Cuando $n=2,4,6, \cdots$

Y calculando $b_n$

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

$\displaystyle b_n = \frac{2}{T} \int_{-T/2}^{0}{(0) \sin{n \omega_0 t} \, dt} + \frac{2}{T} \int_{0}^{T/2}{A \sin{\omega_0 t} \sin{n \omega_0 t} \, dt}$

$\displaystyle b_n = \frac{2A}{T} \int_{0}^{T/2}{\sin{\omega_0 t} \sin{n \omega_0 t} \, dt}$

Si $n=1$ , el integrando tiene la siguiente forma

$\displaystyle b_1 = \frac{2A}{T} \int_{0}^{T/2}{\sin{\omega_0 t} \sin{\omega_0 t} \, dt}$

$\displaystyle b_1 = \frac{2A}{T} \int_{0}^{T/2}{\sin^2{\omega_0 t} \, dt} = \frac{2}{T} \int_{0}^{T/2}{\left(\frac{1}{2} - \frac{1}{2}\cos{2\omega_0 t} \right) \, dt}$

$\displaystyle b_1 = \frac{2A}{T} \left[\frac{1}{2} t - \frac{1}{4 \omega_0} \sin{2\omega_0 t} \right]_{0}^{T/2} = \frac{2A}{T} \left\{ \left[\frac{1}{2} (\frac{T}{2}) - \frac{1}{4 \omega_0} \sin{2\omega_0 (\frac{T}{2})} \right] - \left[\frac{1}{2} (0) - \frac{1}{4 \omega_0} \sin{2\omega_0 (0)} \right] \right\}$

$\displaystyle b_1 = \frac{2A}{T} \left\{ \left[\frac{T}{4} - \frac{1}{4 \omega_0} \sin{\omega_0 T} \right] - 0 \right\} = \frac{2A}{T} \left[\frac{T}{4} - \frac{1}{4 (\frac{2\pi}{T})} \sin{\frac{2\pi}{T} T} \right]$

$\displaystyle b_1 = \frac{2A}{T} \left(\frac{T}{4} - \frac{T}{8 \pi} \sin{2\pi} \right) = \frac{A}{2} - \frac{A}{4 \pi} \sin{2\pi} = \frac{A}{2}$

Si $n=2, 3, 4, \cdots$ , el integrando tiene la siguiente forma

$\displaystyle b_n = \frac{2A}{T} \int_{0}^{T/2}{\sin{\omega_0 t} \sin{n \omega_0 t} \, dt}$

Esta integral se resuelve por integración de productos de funciones seno y coseno con diferentes argumentos en la misma variable. Se observa que esta integral de la forma $\displaystyle \int{\sin{mu} \sin{nu} \, du}$ , donde $m=\omega_0$ y $n=n \omega_0$ , entonces

$\displaystyle \int{\sin{\omega_0 t} \sin{n \omega_0 t} \, dt} = -\frac{1}{2(\omega_0 + n \omega_0)} \sin{(\omega_0 + n \omega_0)t} + \frac{1}{2(\omega_0 - n \omega_0)} \sin{(\omega_0 - n \omega_0)t}$

$\displaystyle = -\frac{1}{2\omega_0 (1 + n)} \sin{\omega_0(1 + n)t} + \frac{1}{2\omega_0(1 - n)} \sin{\omega_0(1 - n)t} + C$

Remplazando la variable « $t$ » por sus respectivos límites, resulta lo siguiente

$\displaystyle = \frac{2A}{T} \int_{0}^{T/2}{\sin{\omega_0 t} \sin{n \omega_0 t} \, dt}$

$\displaystyle = \frac{2A}{T} \left[-\frac{1}{2\omega_0 (1 + n)} \sin{\omega_0(1 + n)t} + \frac{1}{2\omega_0(1 - n)} \sin{\omega_0(1 - n)t} \right]_{0}^{T/2}$

$\displaystyle = \frac{2A}{T} \left\{\left[-\frac{1}{2\omega_0 (1 + n)} \sin{\omega_0(1 + n)(\frac{T}{2})} + \frac{1}{2\omega_0(1 - n)} \sin{\omega_0(1 - n)(\frac{T}{2})} \right] - \left[-\frac{1}{2\omega_0 (1 + n)} \sin{\omega_0(1 + n)(0)} + \frac{1}{2\omega_0(1 - n)} \sin{\omega_0(1 - n)(0)} \right] \right\}$

$\displaystyle = \frac{2A}{T} \left\{\left[-\frac{1}{2 (\frac{2\pi}{T}) (1 + n)} \sin{(\frac{2\pi}{T})(1 + n)(\frac{T}{2})} + \frac{1}{2 (\frac{2\pi}{T}) (1 - n)} \sin{(\frac{2\pi}{T})(1 - n)(\frac{T}{2})} \right] - (-0 + 0) \right\}$

$\displaystyle = \frac{2A}{T} \left[-\frac{T}{4 \pi (1 + n)} \sin{\pi (1 + n)} + \frac{T}{4 \pi (1 - n)} \sin{\pi (1 - n)} \right]$

$\displaystyle = -\frac{A}{2 \pi (1 + n)} \sin{\pi (1 + n)} + \frac{A}{2 \pi (1 - n)} \sin{\pi (1 - n)}$

Si $n=2,4,6, \cdots$

$\displaystyle b_n = -\frac{A}{2 \pi (1 + n)} \sin{\pi (1 + n)} + \frac{A}{2 \pi (1 - n)} \sin{\pi (1 - n)}$

$\displaystyle = -\frac{A}{2 \pi (1 + n)} (0) + \frac{A}{2 \pi (1 - n)} (0) = 0$

Si $n = 3,5,7 \cdots$

$\displaystyle b_n = -\frac{A}{2 \pi (1 + n)} \sin{\pi (1 + n)} + \frac{A}{2 \pi (1 - n)} \sin{\pi (1 - n)}$

$\displaystyle b_n = -\frac{A}{2 \pi (1 + n)} (0) + \frac{A}{2 \pi (1 - n)} (0) = 0$

Finalmente

$b_n = 0$

Regresando y sustituyendo los valores de los coeficientes $a_0$ , $a_n$ y $b_n$ , se tiene el resultado final

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

$\displaystyle f(t) = \frac{1}{2} a_0 + \left(a_1 \cos{\omega_0 t}+b_1 \sin{\omega_0 t}\right) + \sum_{n=2}^{\infty}{(a_n \cos{\omega_0 t}+b_n \sin{\omega_0 t})}$

$\displaystyle f(t) = \frac{1}{2} \cdot \frac{2A}{\pi} + \left[ 0 \cos{\omega_0 t} + \frac{A}{2} \sin{\omega_0 t} \right] + \sum_{n=par}^{\infty}{\left[\frac{2A}{\pi (1+n)(1-n)} \cos{\omega_0 t} + (0) \sin{\omega_0 t} \right]}$

$\displaystyle \therefore f(t) = \frac{A}{\pi} + \frac{A}{2} \sin{\omega_0 t} + \sum_{n=par}^{\infty}{\frac{2A}{\pi (1+n)(1-n)} \cos{\omega_0 t}}$

Problema 4. Desarrollar la función $\sin^5{t}$ en serie de Fourier.

Solución. Utilizando la siguiente identidad

$\displaystyle \sin{t} = \frac{e^{jn t} - e^{-jn t}}{2j}$

Entonces (observando que $n=1$ )

$\displaystyle \sin^5{t} = (\sin{t})^5 = \left( \frac{e^{j t} - e^{-j t}}{2j} \right)^5$

$\displaystyle = \frac{ \left(e^{j t} - e^{-j t} \right)^5}{(2j)^5} = \frac{(e^{j t})^5 + 5 (e^{j t})^4 (-e^{-j t}) + 10 (e^{j t})^3 (-e^{-j t})^2 + 10(e^{j t})^2(-e^{-j t})^3 + 5(e^{j t})(-e^{-j t})^4 + (-e^{-j t})^5}{32j}$

$\displaystyle = \frac{e^{5j t} + 5 (e^{4j t}) (-e^{-j t}) + 10 (e^{3j t}) (e^{-2j t}) + 10(e^{2j t})(-e^{-3j t}) + 5(e^{j t})(e^{-4j t}) + (-e^{-5j t})}{32j}$

$\displaystyle = \frac{e^{5j t} - 5 e^{(4j t - jt)} + 10 e^{(3j t-2j t)} - 10 e^{(2j t-3j t)} + 5 e^{(j t-4j t)} - e^{-5j t}}{32j}$

$\displaystyle = \frac{e^{5j t} - 5 e^{3j t} + 10 e^{j t} - 10 e^{-j t} + 5 e^{-3j t} - e^{-5j t}}{32j}$

$\displaystyle = \frac{e^{5j t} - e^{-5j t} - 5 e^{3j t} + 5 e^{-3j t} + 10 e^{j t} - 10 e^{-j t}}{32j}$

$\displaystyle = \frac{e^{5j t} - e^{-5j t} - 5 (e^{3j t} - e^{-3j t}) + 10 (e^{j t} - e^{-j t})}{32j}$

$\displaystyle = \frac{e^{5j t} - e^{-5j t}}{32j} + \frac{(- 5) (e^{3j t} - e^{-3j t})}{32j} + \frac{(10) (e^{j t} - e^{-j t})}{32j}$

$\displaystyle = \frac{e^{5j t} - e^{-5j t}}{(16)(2j)} + \frac{(- 5) (e^{3j t} - e^{-3j t})}{(16)(2j)} + \frac{(10) (e^{j t} - e^{-j t})}{(16)(2j)}$

$\displaystyle = \frac{1}{16} \left(\frac{e^{5j t} - e^{-5j t}}{2j} \right) + \frac{(- 5)}{16} \left( \frac{e^{3j t} - e^{-3j t}}{2j} \right) + \frac{10}{16} \left(\frac{e^{j t} - e^{-j t}}{2j} \right)$

$\displaystyle = \frac{1}{16} \sin{5 t} - \frac{5}{16}\sin{3t} + \frac{10}{16} \sin{t} = \frac{1}{16} \sin{5 t} - \frac{5}{16}\sin{3t} + \frac{5}{8} \sin{t}$

Finalmente

$\displaystyle \therefore f(t) = \frac{1}{16} \sin{5 t} - \frac{5}{16}\sin{3t} + \frac{5}{8} \sin{t}$

O también

$\displaystyle \therefore f(t) = \frac{5}{8} \sin{t} - \frac{5}{16}\sin{3t} + \frac{1}{16} \sin{5 t}$

Y se concluye que la función $f(t)$ al ser expresado como serie de Fourier, solo consta de tres términos.

Determinando la serie de Fourier de una función “f(t)”. Fourier.

Problemas resueltos

Publicado por Cesar Reyes

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Determinando la serie de Fourier de una función “f(t)”. Fourier.

Problemas resueltos

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Publicado por Cesar Reyes

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