Propiedad de linealidad en la transformada inversa de Laplace. Laplace.

Introducción

Si $c_1$ y $c_2$ son constantes arbitrarias y $F_1 (s)$ y $F_2 (s)$ son las transformadas de Laplace de $f_1 (t)$ y $f_2 (t)$ , entonces,

$\mathcal{L}^{-1} [F_1 (s) + F_2 (s)] = c_1 \cdot \mathcal{L}^{-1}[F_1 (s)] + c_2 \cdot \mathcal{L}^{-1}[F_2 (s)]$

$\mathcal{L}^{-1} [F_1 (s) + F_2 (s)] = c_1 \cdot f_1 (t) + c_2 \cdot f_2 (t)$

Tambien se aplica para mas de dos funciones.

Problemas resueltos

Problema 1. Hallar $f(t)$ para $\displaystyle F(s) = \frac{5s+4}{s^3} - \frac{2s-18}{s^2+9} + \frac{24-30\sqrt{s}}{s^4}$ .

Solución. De la función brindada por el problema, se utiliza el símbolo $\mathcal{L}^{-1}$ para aplicar la transformada inversa de Laplace.

$\displaystyle F(s) = \frac{5s+4}{s^3} - \frac{2s-18}{s^2+9} + \frac{24-30\sqrt{s}}{s^4}$

$\displaystyle \mathcal{L}^{-1} [F(s)] = \mathcal{L}^{-1} \left[\frac{5s+4}{s^3} - \frac{2s-18}{s^2+9} + \frac{24-30\sqrt{s}}{s^4} \right]$

$\displaystyle f(t) = \mathcal{L}^{-1} \left[\frac{5s}{s^3} + \frac{4}{s^3} - \frac{2s}{s^2+9} + \frac{18}{s^2+9} + \frac{24}{s^4} - \frac{30\sqrt{s}}{s^4} \right]$

$\displaystyle f(t) = \mathcal{L}^{-1} \left[\frac{5}{s^2} + \frac{4}{s^3} - \frac{2s}{s^2+9} + \frac{18}{s^2+9} + \frac{24}{s^4} - \frac{30{s}^{1/2}}{s^4} \right]$

$\displaystyle f(t) = \mathcal{L}^{-1} \left[\frac{5}{s^2} + \frac{4}{s^3} - \frac{2s}{s^2+9} + \frac{18}{s^2+9} +\frac{24}{s^4} - \frac{30}{s^{7/2}} \right]$

Aplicando la propiedad de linealidad, resulta

$\displaystyle f(t) = \mathcal{L}^{-1} \left[\frac{5}{s^2} \right] + \mathcal{L}^{-1} \left[\frac{4}{s^3}\right] - \mathcal{L}^{-1} \left[\frac{2s}{s^2+9}\right] + \mathcal{L}^{-1} \left[\frac{18}{s^2+9} \right] +\mathcal{L}^{-1} \left[\frac{24}{s^4} \right] - \mathcal{L}^{-1} \left[\frac{30}{s^{7/2}} \right]$

$\displaystyle f(t) = 5\ \mathcal{L}^{-1} \left[\frac{1}{s^2} \right] + 4\ \mathcal{L}^{-1} \left[\frac{1}{s^3}\right] - 2\ \mathcal{L}^{-1} \left[\frac{s}{s^2+9}\right] + 18 \ \mathcal{L}^{-1} \left[\frac{1}{s^2+9} \right] +24\ \mathcal{L}^{-1} \left[\frac{1}{s^4} \right] - 30 \ \mathcal{L}^{-1} \left[\frac{1}{s^{7/2}} \right]$

$\displaystyle f(t) = 5 \left(\frac{t}{1!} \right) + 4 \left(\frac{t^2}{2!}\right) - 2 \left(\cos{3t} \right) + 18 \left(\frac{1}{3} \sin{3t} \right) +24 \left(\frac{t^3}{3!} \right) - 30 \left(\frac{t^{5/2}}{\frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi}} \right)$

$\displaystyle f(t) = 5 \left(\frac{t}{1} \right) + 4 \left(\frac{t^2}{2}\right) - 2 \left(\cos{3t} \right) + 18 \left(\frac{1}{3} \sin{3t} \right) +24 \left(\frac{t^3}{6} \right) - 30 \left(\frac{t^{5/2}}{\frac{15}{8} \sqrt{\pi}} \right)$

$\displaystyle f(t) = 5t + 2t^2 - 2 \cos{3t} + \frac{18}{3} \sin{3t} + \frac{t^3}{6} - \frac{30 t^{5/2}}{\frac{15}{8} \sqrt{\pi}}$

$\displaystyle f(t) = 5t + 2t^2 - 2 \cos{3t} + 6 \sin{3t} + \frac{t^3}{6} - \frac{16 t^{5/2}}{\sqrt{\pi}}$

$\displaystyle f(t) = 5t + 2t^2 - 2 \cos{3t} + 6 \sin{3t} + \frac{t^3}{6} - \frac{16 t \cdot t^{1/2}}{\sqrt{\pi}}$

$\displaystyle f(t) = 5t + 2t^2 - 2 \cos{3t} + 6 \sin{3t} + \frac{t^3}{6} - 6t \cdot \frac{\sqrt{t}}{\sqrt{\pi}}$

Finalmente

$\displaystyle \therefore f(t) = 5t + 2t^2 - 2 \cos{3t} + 6 \sin{3t} + \frac{t^3}{6} - 6t \sqrt{\frac{t}{\pi}}$

Problema 2. Hallar $f(t)$ para $\displaystyle F(s) = \frac{6}{2s-3} - \frac{3+4s}{9s^2-16} + \frac{8-6s}{16s^2+9}$ .

Solución. De la función brindada por el problema, se utiliza el símbolo $\mathcal{L}^{-1}$ para aplicar la transformada inversa de Laplace.

$\displaystyle F(s) = \frac{6}{2s-3} - \frac{3+4s}{9s^2-16} + \frac{8-6s}{16s^2+9}$

$\displaystyle \mathcal{L}^{-1} [F(s)] = \mathcal{L}^{-1} \left[\frac{6}{2s-3} - \frac{3+4s}{9s^2-16} + \frac{8-6s}{16s^2+9} \right]$

$\displaystyle f(t) = \mathcal{L}^{-1} \left[\frac{6}{2s-3} - \frac{3}{9s^2-16} - \frac{4s}{9s^2-16} + \frac{8}{16s^2+9} - \frac{6s}{16s^2+9} \right]$

Aplicando la propiedad de linealidad, resulta

$\displaystyle f(t) = \mathcal{L}^{-1} \left[\frac{6}{2s-3} \right] - \mathcal{L}^{-1} \left[\frac{3}{9s^2-16}\right] - \mathcal{L}^{-1} \left[\frac{4s}{9s^2-16} \right] + \mathcal{L}^{-1} \left[\frac{8}{16s^2+9} \right] - \mathcal{L}^{-1} \left[\frac{6s}{16s^2+9} \right]$

$\displaystyle f(t)= 6 \ \mathcal{L}^{-1} \left[\frac{1}{2s-3} \right] - 3 \ \mathcal{L}^{-1} \left[\frac{1}{9s^2-16}\right] - 4 \ \mathcal{L}^{-1} \left[\frac{s}{9s^2-16} \right] + 8 \ \mathcal{L}^{-1} \left[\frac{1}{16s^2+9} \right] - 6\ \mathcal{L}^{-1} \left[\frac{s}{16s^2+9} \right]$

$\displaystyle f(t) = 6 \ \mathcal{L}^{-1} \left[\frac{1}{2(s-\frac{3}{2})} \right] - 3 \ \mathcal{L}^{-1} \left[\frac{1}{9(s^2-\frac{16}{9})}\right] - 4 \ \mathcal{L}^{-1} \left[\frac{s}{9(s^2-\frac{16}{9})} \right] + 8 \ \mathcal{L}^{-1} \left[\frac{1}{16(s^2+\frac{9}{16})} \right] - 6\ \mathcal{L}^{-1} \left[\frac{s}{16(s^2+\frac{9}{16})} \right]$

$\displaystyle f(t) = \frac{6}{2} \ \mathcal{L}^{-1} \left[\frac{1}{s-\frac{3}{2}} \right] - \frac{3}{9} \mathcal{L}^{-1} \left[\frac{1}{s^2-\frac{16}{9}}\right] - \frac{4}{9} \mathcal{L}^{-1} \left[\frac{s}{s^2-\frac{16}{9}} \right] + \frac{8}{16} \mathcal{L}^{-1} \left[\frac{1}{s^2+\frac{9}{16}} \right] - \frac{6}{16} \mathcal{L}^{-1} \left[\frac{s}{s^2+\frac{9}{16}} \right]$

$\displaystyle f(t) =3 \ \mathcal{L}^{-1} \left[\frac{1}{s-\frac{3}{2}} \right] - \frac{1}{3} \mathcal{L}^{-1} \left[\frac{1}{s^2-\frac{16}{9}}\right] - \frac{4}{9} \mathcal{L}^{-1} \left[\frac{s}{s^2-\frac{16}{9}} \right] + \frac{1}{2} \mathcal{L}^{-1} \left[\frac{1}{s^2+\frac{9}{16}} \right] - \frac{3}{8} \mathcal{L}^{-1} \left[\frac{s}{s^2+\frac{9}{16}} \right]$

$\displaystyle f(t) =3 \left(e^{\frac{3}{2}t} \right) - \frac{1}{3} \left(\frac{1}{\frac{4}{3}} \sinh{\frac{4}{3} t} \right) - \frac{4}{9} \left(\cosh{\frac{4}{3}t} \right) + \frac{1}{2} \left(\frac{1}{\frac{3}{4}} \sin{\frac{3}{4}t} \right) - \frac{3}{8} \left(\cos{\frac{3}{4}t} \right)$

$\displaystyle f(t) =3 \left(e^{\frac{3}{2}t} \right) - \frac{1}{3} \left(\frac{3}{4} \sinh{\frac{4}{3} t} \right) - \frac{4}{9} \left(\cosh{\frac{4}{3}t} \right) + \frac{1}{2} \left(\frac{4}{3} \sin{\frac{3}{4}t} \right) - \frac{3}{8} \left(\cos{\frac{3}{4}t} \right)$

Finalmente,

$\displaystyle \therefore f(t) =3 e^{\frac{3}{2}t} - \frac{1}{4} \sinh{\frac{4}{3} t} - \frac{4}{9} \cosh{\frac{4}{3}t} + \frac{2}{3} \sin{\frac{3}{4}t} - \frac{3}{8} \cos{\frac{3}{4}t}$

Propiedad de linealidad en la transformada inversa de Laplace. Laplace.

Introducción

Problemas resueltos

Publicado por Cesar Reyes

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Propiedad de linealidad en la transformada inversa de Laplace. Laplace.

Introducción

Problemas resueltos

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Publicado por Cesar Reyes

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