Verified answer

Hola

En forma lineal

F(s) = (5s + 1) / (s(3s + 2))

llevamos las constantes afuera

F(s) = (5/3) (s + (1/5)) / ( s (s + (2/3) )

Simplificamos un poco

F(s) = (5/3) (s + (2/3) + (1/5) - (2/3)) / ( s (s + (2/3) )

F(s) = (5/3) (s + (2/3) + (3/15) - (10/15)) / ( s (s + (2/3) )

F(s) = (5/3) (s + (2/3) - (7/15)) / ( s (s + (2/3) )

F(s) = (5/3) [ (s + (2/3)) / ( s (s + (2/3) ) -

- (7/15) / ( s (s + (2/3) )]

F(s) = (5/3) [ (1/s) - ( (7/15) / ( s (s + (2/3) ) )]

Ahora trabajamos con fracciones parciales.

En este caso, con 2 divisores, es sencillo

(1/s) - (1/(s + (2/3)) = (s + (2/3) - s)/ ( s (s + (2/3) )

(1/s) - (1/(s + (2/3)) = (2/3)/ ( s (s + (2/3) )

Multiplicamos todo por (3/2)

1/ ( s (s + (2/3) ) = (3/2) (1/s) - (3/2) (1/(s + (2/3))

remplazamos en F(s)

F(s) = (5/3) [ (1/s) - (7/15) {(3/2) (1/s) - (3/2) (1/(s + (2/3))} ]

F(s) = (5/3) [ (1/s) - (7/10) (1/s) + (7/10) (1/(s + (2/3)) ]

F(s) = (5/3) [ (3/10) (1/s) + (7/10) (1/(s + (2/3)) ]

F(s) = (1/2) (1/s) + (7/6) (1/(s + (2/3))

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Ahora es fácil,

empleamos antitransformadas estándares.

L(1) = 1/s

L(e^(-at)) = 1/(s + a)

Nos queda

f(t) = (1/2) + (7/6) e^(-2/3) t

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Saludos

5s+1

--------

s(3s+2)

6s

------

s(5s)

6s

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5s2