Transformada seno de Fourier

Si f(t) está definida sólo para 0 < t < \infty, f(t) se puede representar por

\displaystyle f (t) = \frac{2}{\pi} \int_{0}^{\infty}{F_s (\omega) \sin{\omega t} \, d\omega}

donde F_s (\omega) está dado por

\displaystyle F_s(\omega) = \mathcal{F}_s [f(t)] = \int_{0}^{\infty}{f(t) \sin{\omega t} \, dt}

y F_s(\omega) se denomina transformada seno de Fourier de f(t). Su inversa es

\displaystyle f(t) = \mathcal{F}_s^{-1} [F_s (\omega)] = \frac{2}{\pi} \int_{0}^{\infty}{F_s (\omega) \sin{\omega t} \, d\omega}

como se observó en un principio.

Transformada coseno de Fourier

Si f(t) está definida sólo para 0 < t < \infty, f(t) se puede representar por

\displaystyle f (t) = \frac{2}{\pi} \int_{0}^{\infty}{F_c(\omega) \cos{\omega t} \, d\omega}

donde F_c (\omega) está dado por

\displaystyle F_c(\omega) = \mathcal{F}_c [f(t)] = \int_{0}^{\infty}{f(t) \cos{\omega t} \, dt}

y F_c(\omega) se denomina transformada coseno de Fourier de f(t). Su inversa es

\displaystyle f(t) = \mathcal{F}_c^{-1} [F_c (\omega)] = \frac{2}{\pi} \int_{0}^{\infty}{F_c (\omega) \cos{\omega t} \, d\omega}

como se observó en un principio.

Problemas resueltos

Problema 1. Encontrar \mathcal{F}_c [e^{-\alpha t}] y \mathcal{F}_s [e^{-\alpha t}] para t>0, \alpha > 0

Solución. Para la transformada coseno de Fourier

\displaystyle F_c(\omega) = \mathcal{F}_c [f(t)]

\displaystyle\mathcal{F}_c [f(t)] = \int_{0}^{\infty}{f(t) \cos{\omega t} \, dt}

\displaystyle \mathcal{F}_c [e^{-\alpha t}] = \int_{0}^{\infty}{e^{-\alpha t} \cos{\omega t} \, dt}

Resolviendo esta integral como indefinida se observa que

\displaystyle \int{e^{-\alpha t} \cos{\omega t} \, dt} = \cos{\omega t}  \left(-\frac{1}{\alpha} e^{-\alpha t} \right) - \int{\left(-\frac{1}{\alpha} e^{-\alpha t} \right) (-\omega \sin{\omega t}) \, dt}

\displaystyle \int{e^{-\alpha t} \cos{\omega t} \, dt} = -\frac{1}{\alpha} e^{-\alpha t} \cos{\omega t} - \frac{\omega}{\alpha} \int{e^{-\alpha t} \sin{\omega t} \, dt}

\displaystyle \int{e^{-\alpha t} \cos{\omega t} \, dt} = -\frac{1}{\alpha} e^{-\alpha t} \cos{\omega t} - \frac{\omega}{\alpha} \left[\sin{\omega t} \left( -\frac{1}{\alpha} e^{-\alpha t} \right) - \int{\left( -\frac{1}{\alpha} e^{-\alpha t} \right) (\omega \cos{\omega t}) \, dt} \right]

\displaystyle \int{e^{-\alpha t} \cos{\omega t} \, dt} = -\frac{1}{\alpha} e^{-\alpha t} \cos{\omega t} - \frac{\omega}{\alpha} \left[-\frac{1}{\alpha} e^{-\alpha t} \sin{\omega t} + \frac{\omega}{\alpha} \int{e^{-\alpha t} \cos{\omega t} \, dt} \right]

\displaystyle \int{e^{-\alpha t} \cos{\omega t} \, dt} = -\frac{1}{\alpha} e^{-\alpha t} \cos{\omega t} + \frac{\omega}{\alpha^2} e^{-\alpha t} \sin{\omega t} - \frac{\omega^2}{\alpha^2} \int{e^{-\alpha t} \cos{\omega t} \, dt}

\displaystyle \int{e^{-\alpha t} \cos{\omega t} \, dt} + \frac{\omega^2}{\alpha^2} \int{e^{-\alpha t} \cos{\omega t} \, dt} = -\frac{1}{\alpha} e^{-\alpha t} \cos{\omega t} + \frac{\omega}{\alpha^2} e^{-\alpha t} \sin{\omega t} + C

\displaystyle \left(1 + \frac{\omega^2}{\alpha^2} \right) \int{e^{-\alpha t} \cos{\omega t} \, dt} = -\frac{1}{\alpha} e^{-\alpha t} \cos{\omega t} + \frac{\omega}{\alpha^2} e^{-\alpha t} \sin{\omega t} + C

\displaystyle \left(\frac{\alpha^2+ \omega^2}{\alpha^2} \right) \int{e^{-\alpha t} \cos{\omega t} \, dt} = -\frac{1}{\alpha} e^{-\alpha t} \cos{\omega t} + \frac{\omega}{\alpha^2} e^{-\alpha t} \sin{\omega t} + C

\displaystyle \int{e^{-\alpha t} \cos{\omega t} \, dt} = \left(\frac{\alpha^2}{\alpha^2+ \omega^2} \right) \left[ -\frac{1}{\alpha} e^{-\alpha t} \cos{\omega t} + \frac{\omega}{\alpha^2} e^{-\alpha t} \sin{\omega t} + C \right]

\displaystyle \int{e^{-\alpha t} \cos{\omega t} \, dt} = - \left(\frac{\alpha^2}{\alpha^2+ \omega^2} \right) \frac{1}{\alpha} e^{-\alpha t} \cos{\omega t} + \left(\frac{\alpha^2}{\alpha^2+ \omega^2} \right)\frac{\omega}{\alpha^2} e^{-\alpha t} \sin{\omega t} + C

\displaystyle \int{e^{-\alpha t} \cos{\omega t} \, dt} = - \left(\frac{\alpha}{\alpha^2+ \omega^2} \right) e^{-\alpha t} \cos{\omega t} + \left(\frac{\omega}{\alpha^2+ \omega^2}\right) e^{-\alpha t} \sin{\omega t} + C

Agregando los límites (para poder resolverlo como integral definida)

\displaystyle \int_{0}^{\infty}{e^{-\alpha t} \cos{\omega t} \, dt} = \left[- \left(\frac{\alpha}{\alpha^2+ \omega^2} \right) e^{-\alpha t} \cos{\omega t} + \left(\frac{\omega}{\alpha^2+ \omega^2}\right) e^{-\alpha t} \sin{\omega t} + C \right]_{0}^{\infty}

\displaystyle = \left[- \left(\frac{\alpha}{\alpha^2+ \omega^2} \right) e^{-\alpha (\infty)} \cos{\omega (\infty)} + \left(\frac{\omega}{\alpha^2+ \omega^2}\right) e^{-\alpha (\infty)} \sin{\omega (\infty)}\right] - \left[- \left(\frac{\alpha}{\alpha^2+ \omega^2} \right) e^{-\alpha (0)} \cos{\omega (0)} + \left(\frac{\omega}{\alpha^2+ \omega^2}\right) e^{-\alpha (0)} \sin{\omega (0)}\right]

\displaystyle = \left[- (0) + (0) \right] - \left[- \left(\frac{\alpha}{\alpha^2+ \omega^2} \right) (1) + \left(\frac{\omega}{\alpha^2+ \omega^2}\right) (0) \right] = \frac{\alpha}{\alpha^2+ \omega^2}

\displaystyle \mathcal{F}_c [e^{-\alpha t}] = \frac{\alpha}{\alpha^2+ \omega^2}

Para la transformada seno de Fourier

\displaystyle F_s(\omega) = \mathcal{F}_s [f(t)]

\displaystyle \mathcal{F}_s [f(t)] = \int_{0}^{\infty}{f(t) \sin{\omega t} \, dt}

\displaystyle \mathcal{F}_s [e^{-\alpha t}] = \int_{0}^{\infty}{e^{-\alpha t} \sin{\omega t} \, dt}

Resolviendo esta integral como indefinida se observa que

\displaystyle \int{e^{-\alpha t} \sin{\omega t} \, dt} = \sin{\omega t}  \left(-\frac{1}{\alpha} e^{-\alpha t} \right) - \int{\left(-\frac{1}{\alpha} e^{-\alpha t} \right) (\omega \cos{\omega t}) \, dt}

\displaystyle \int{e^{-\alpha t} \sin{\omega t} \, dt} = -\frac{1}{\alpha} e^{-\alpha t} \sin{\omega t} + \frac{\omega}{\alpha} \int{e^{-\alpha t} \cos{\omega t} \, dt}

\displaystyle \int{e^{-\alpha t} \sin{\omega t} \, dt} = -\frac{1}{\alpha} e^{-\alpha t} \sin{\omega t} + \frac{\omega}{\alpha} \left[\cos{\omega t} \left( -\frac{1}{\alpha} e^{-\alpha t} \right) - \int{\left( -\frac{1}{\alpha} e^{-\alpha t} \right) (-\omega \sin{\omega t}) \, dt} \right]

\displaystyle \int{e^{-\alpha t} \sin{\omega t} \, dt} = -\frac{1}{\alpha} e^{-\alpha t} \sin{\omega t} + \frac{\omega}{\alpha} \left[-\frac{1}{\alpha} e^{-\alpha t} \cos{\omega t} - \frac{\omega}{\alpha} \int{e^{-\alpha t} \sin{\omega t} \, dt} \right]

\displaystyle \int{e^{-\alpha t} \sin{\omega t} \, dt} = - \frac{1}{\alpha} e^{-\alpha t} \sin{\omega t} - \frac{\omega}{\alpha^2} e^{-\alpha t} \cos{\omega t} - \frac{\omega^2}{\alpha^2} \int{e^{-\alpha t} \sin{\omega t} \, dt}

\displaystyle \int{e^{-\alpha t} \sin{\omega t} \, dt} + \frac{\omega^2}{\alpha^2} \int{e^{-\alpha t} \sin{\omega t} \, dt} = -\frac{1}{\alpha} e^{-\alpha t} \sin{\omega t} - \frac{\omega}{\alpha^2} e^{-\alpha t} \cos{\omega t} + C

\displaystyle \left(1 + \frac{\omega^2}{\alpha^2} \right) \int{e^{-\alpha t} \sin{\omega t} \, dt} = -\frac{1}{\alpha} e^{-\alpha t} \sin{\omega t} - \frac{\omega}{\alpha^2} e^{-\alpha t} \cos{\omega t} + C

\displaystyle \left(\frac{\alpha^2 + \omega^2}{\alpha^2} \right) \int{e^{-\alpha t} \sin{\omega t} \, dt} = - \frac{1}{\alpha} e^{-\alpha t} \sin{\omega t} - \frac{\omega}{\alpha^2} e^{-\alpha t} \cos{\omega t} + C

\displaystyle \int{e^{-\alpha t} \sin{\omega t} \, dt} = \left(\frac{\alpha^2}{\alpha^2 +  \omega^2} \right) \left[ -\frac{1}{\alpha} e^{-\alpha t} \sin{\omega t} - \frac{\omega}{\alpha^2} e^{-\alpha t} \cos{\omega t} + C \right]

\displaystyle \int{e^{-\alpha t} \sin{\omega t} \, dt} = - \left(\frac{\alpha^2}{\alpha^2 + \omega^2} \right) \frac{1}{\alpha} e^{-\alpha t} \sin{\omega t} - \left(\frac{\alpha^2}{\alpha^2 + \omega^2} \right)\frac{\omega}{\alpha^2} e^{-\alpha t} \cos{\omega t} + C

\displaystyle \int{e^{-\alpha t} \sin{\omega t} \, dt} = - \left(\frac{\alpha}{\alpha^2 + \omega^2} \right) e^{-\alpha t} \sin{\omega t} + \left(\frac{\omega}{\alpha^2 + \omega^2}\right) e^{-\alpha t} \cos{\omega t} + C

Agregando los límites (para poder resolverlo como integral definida)

\displaystyle \int_{0}^{\infty}{e^{-\alpha t} \sin{\omega t} \, dt} = \left[- \left(\frac{\alpha}{\alpha^2 + \omega^2} \right) e^{-\alpha t} \sin{\omega t} - \left(\frac{\omega}{\alpha^2 + \omega^2}\right) e^{-\alpha t} \cos{\omega t} + C \right]_{0}^{\infty}

\displaystyle = \left[- \left(\frac{\alpha}{\alpha^2 + \omega^2} \right) e^{-\alpha (\infty)} \sin{\omega (\infty)} - \left(\frac{\omega}{\alpha^2 + \omega^2}\right) e^{-\alpha (\infty)} \cos{\omega (\infty)}\right] - \left[- \left(\frac{\alpha}{\alpha^2 + \omega^2} \right) e^{-\alpha (0)} \sin{\omega (0)} - \left(\frac{\omega}{\alpha^2 + \omega^2}\right) e^{-\alpha (0)} \cos{\omega (0)}\right]

\displaystyle = \left[- (0) - (0) \right] - \left[- \left(\frac{\alpha}{\alpha^2 + \omega^2} \right) (0) - \left(\frac{\omega}{\alpha^2 + \omega^2}\right) (1) \right] = \frac{\omega}{\alpha^2 + \omega^2}

\displaystyle \mathcal{F}_s [e^{-\alpha t}] = \frac{\omega}{\alpha^2 + \omega^2}

Finalmente, los resultados esperados son

\displaystyle \mathcal{F}_c [e^{-\alpha t}] = \frac{\alpha}{\alpha^2+ \omega^2}

y

\displaystyle \mathcal{F}_s [e^{-\alpha t}] = \frac{\omega}{\alpha^2 + \omega^2}

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Publicado por Cesar Reyes

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