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Transformada de laplace de integrales, por WikiMatematica.org
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Transformada de laplace de integrales

De por WikiMatematica.org



    \mathcal{L}\left\{ \int_0^t f(\tau)\, d\tau \right\} = {1 \over s} \mathcal{L}\{f\}

La Transformada Inversa de una integral esta dada por la forma


\mathcal{L}^{-1}\left\{\frac{1}{S} F(s)\right\} = \int_0^t f(\tau)\, d\tau

Contenido

Ejemplo 1


 \mathfrak{L}\left\{{\frac{1}{s(s+1)}}\right\}
 = \mathfrak{L^{-1}}\left\{{{ \frac{\frac{1}{s}+ 1}{s}}}\right\}
 = \int_{0}^{t}e^{-\tau}d\tau
 = -e^{-\tau}\mid _{0}^{t}\textrm{}
 = -e^{-t} + 1

Ejemplo 2


Resolver la ecuacion integral f(t)= 3t^{2} -e^{-t} - \int_{0}^{t}f(\tau)e^{t-\tau}d\tau


 \mathfrak{L}\left\{{f(t)}\right\} =\mathfrak{L}\left\{{{3t^{2} -e^{-t} - \int_{0}^{t}f(\tau)e^{t-\tau}d\tau}}\right\}


 F(s)= \frac{6}{s^{3}} - \frac{1}{s+1} - \mathfrak{L^{-1}}\left\{{{f(t)*e^{t}}}\right\}


 F(s)=\frac{6}{s^{3}} - \frac{1}{s+1} - \frac{F(s)}{(s-1)}


 F(s) + \frac{F(s)}{s-1} = \frac{6}{s^{3}} - \frac{1}{(s+1)} entonces  F(s)(1 + \frac{1}{s-1})= \frac{6}{s^{3}} - \frac{1}{s+1}



F(s)= \frac{6}{s^{3}} - \frac{6}{s^{4}} - \frac{1}{s+1} + \frac{1}{s(s+1)}




f(t)= 3t^{2}- t^{3} -e^{-t} + \int_{0}^{t}e^{-\tau}d\tau



f(t)=3t^{2}- t^{3} -e^{-t} + 1 -e^{-t}


 f(t)= 3t^{2}- t^{3} -2e^{-t} + 1

Ejemplo 3

Ejemplo2.jpg


Voltajes en los componentes :
inductor: L\frac{\mathrm{di} }{\mathrm{d} t}
resistor: RI
capacitor: \frac{1}{c} \int_{0}^{t}i(\tau)d\tau la ecuacion correspondiente para el voltaje seria


E(t)=L\frac{\mathrm{di} }{\mathrm{d} t} + RI + \frac{1}{c} \int_{0}^{t}i(\tau)d\tau


E(t)=\begin{Bmatrix}
120t,  0\leq t\leq 1 & \\ 
0, t> 1 & 
\end{Bmatrix}


 E(t)=120t -120t\upsilon (t-1)


\mathfrak{L}\left\{{{ E(t)}}\right\}=\mathfrak{L}\left\{{{ 120t -120\upsilon (t-1)}}\right\}


 \mathfrak{L}\left\{{{ E(t)}}\right\}=\mathfrak{L}\left\{{{ 120t}}\right\} - 120\mathfrak{L}\left\{{{\upsilon (t-1) }}\right\}



\frac{120}{s^{2}} - 120\mathfrak{L}\left\{{{ (t-1+1)\upsilon (t-1)}}\right\}


\frac{120}{s^{2}} - 120\mathfrak{L}\left\{{{(t-1)\upsilon (t-1) + \upsilon (t-1)}}\right\}


 \mathfrak{L}\left\{{E(t)}\right\} = \frac{120}{s^{2}} - 120[\frac{1}{s^{2}}e^{-s} + \frac{1}{s}e^{-s}]


 \mathfrak{L}\left\{{E(t)}\right\} = \frac{120}{s^{2}} - \frac{120}{s^{2}}e^{-s} - \frac{120}{s}e^{-s}


para el inductor tenemos: \mathfrak{L}\left\{{L\frac{\mathrm{d} i}{\mathrm{d} t}}\right\} = L\mathfrak{L}\left\{{\frac{\mathrm{d} i}{\mathrm{d} t}}\right\}=L[SI(s)]=LSI(s)


para la resistencia tenemos: \mathfrak{L}\left\{{RI}}\right\} = RI(s)


y para el capacitor tenemos que \mathfrak{L}\left\{{\frac{1}{c} \int_{0}^{t}i(\tau)d\tau}\right\} = \frac{1}{cs}I(s)


0.1SI(s) + 20I(s) + \frac{10}{s}I(s) = \frac{120}{s^{2}} - \frac{120}{s^{2}}e^{-s} - \frac{120}{s}e^{-s}


 [0.1S + 20 + \frac{10^{3}}{s}]IS = \frac{120}{s^{2}} - \frac{120}{s^{2}}e^{-s} - \frac{120}{s}e^{-s}


I(s) = \frac{1200}{s(s+100)^{2}} - \frac{1200}{s(s+100)^{2}}e^{-s} - \frac{1200}{(s+100)^{2}}e^{-s}


hacemos fracciones parciales para : \frac{1200}{s(s+100)}=-\frac{3}{25(s+100)} -\frac{12}{(s+100)^{2}} + \frac{3}{25s}



 I(s)= -\frac{3}{25(s+100)} -\frac{12}{(s+100)^{2}} + \frac{3}{25s} + [\frac{3}{25(s+100)} +\frac{12}{(s+100)^{2}} - \frac{3}{25s}]e^{-s} + \frac{1200}{(s + 100)^{2}}e^{-s}




i(t) = \frac{-3}{25}e^{-100t} - 12te^{100t}+ \frac{3}{25} + \frac{3}{25}e^{-100(t-1)}\upsilon (t-1) + 12(t-1)\upsilon (t-1)e^{100(t-1)} - \frac{3}{25}\upsilon (t-1)-1200(t-1)e^{100(t-1)}\upsilon (t-1)


i(t)=\frac{3}{25}[1-\upsilon (t-1)]-\frac{3}{25}[e^{-100t}-e^{-100(t-1)}\upsilon (t-1)]-12[te^{100t}-99(t-1)\upsilon (t-1)e^{100(t-1)}]

Ejemplo4

\pounds {\int_{0}^{t}e^{-\tau }sen(t-\tau )d\tau }

f(\tau)= e^{-\tau } f(t)=e^{-t} g(t-\tau)=sen(t-\tau) g(t)=sen(t)

\pounds{f(t)}\pounds{g(t)}=\frac{1}{s+1}\frac{1}{s^2+1}=\frac{1}{(s+1)(s^2+1)}

Ejemplo 5:

Evaluar: \mathcal{L}^{-1}\left\{\frac{1}{S(S^2 + 1)}\right\}

Identificamos F(s) = \frac{1}{S^2 + 1} , por lo tanto f(t) = sin(t)


\int_0^t sin(\tau)\, d\tau = \left [- cos(\tau)\right]_{0}^{t} = -cos(t) -(-1) , y el resultado es 1 - cos(t)

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