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Beliebt Rechnen >

inverselaplace 1/((s^3+1)(s^2+1))

Lösung

inverse laplace transformation

\frac{1}{( s ^{3} + 1 ) ( s ^{2} + 1 )}

Lösung

\frac{1}{6} e^{- t} + \frac{1}{3} - 2 e^{\frac{t}{2}} cos (\frac{3 t}{2}) + \frac{1}{2} cos (t) + \frac{1}{2} sin (t)

Schritte zur Lösung

L^{- 1} {\frac{1}{( s ^{3} + 1 ) ( s ^{2} + 1 )}}

Ermittle den Partialbruch von

\frac{1}{( s ^{3} + 1 ) ( s ^{2} + 1 )} : \frac{1}{6 ( s + 1 )} + \frac{- 2 s + 1}{3 ( s ^{2} - s + 1 )} + \frac{s + 1}{2 ( s ^{2} + 1 )}

= L^{- 1} {\frac{1}{6 ( s + 1 )} + \frac{- 2 s + 1}{3 ( s ^{2} - s + 1 )} + \frac{s + 1}{2 ( s ^{2} + 1 )}}

Schreibe

\frac{- 2 s + 1}{3 ( s ^{2} - s + 1 )}

um:

\frac{1}{3} \cdot \frac{- 2 ( s - \frac{1}{2} )}{( s - \frac{1}{2} ) ^{2} + \frac{3}{4}}

= L^{- 1} {\frac{1}{6 ( s + 1 )} + \frac{1}{3} \cdot \frac{- 2 ( s - \frac{1}{2} )}{( s - \frac{1}{2} ) ^{2} + \frac{3}{4}} + \frac{s + 1}{2 ( s ^{2} + 1 )}}

Schreibe um

= L^{- 1} {\frac{s}{2 ( s ^{2} + 1 )} + \frac{1}{6 ( s + 1 )} + \frac{1}{3} \cdot \frac{- 2 ( s - \frac{1}{2} )}{( s - \frac{1}{2} ) ^{2} + \frac{3}{4}} + \frac{1}{2 ( s ^{2} + 1 )}}

Wende die lineare Eigenschaftder inversen Laplace-Transformation an:

Für Funktionen

f (s), g (s)

und Konstanten

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= \frac{1}{6} L^{- 1} {\frac{1}{s + 1}} + \frac{1}{3} - 2 L^{- 1} {\frac{s - \frac{1}{2}}{( s - \frac{1}{2} ) ^{2} + \frac{3}{4}}} + \frac{1}{2} L^{- 1} {\frac{s}{s ^{2} + 1}} + \frac{1}{2} L^{- 1} {\frac{1}{s ^{2} + 1}}

L^{- 1} {\frac{1}{s + 1}} : e^{- t}

L^{- 1} {\frac{s - \frac{1}{2}}{( s - \frac{1}{2} ) ^{2} + \frac{3}{4}}} : e^{\frac{t}{2}} cos (\frac{3 t}{2})

L^{- 1} {\frac{s}{s ^{2} + 1}} : cos (t)

L^{- 1} {\frac{1}{s ^{2} + 1}} : sin (t)

= \frac{1}{6} e^{- t} + \frac{1}{3} - 2 e^{\frac{t}{2}} cos (\frac{3 t}{2}) + \frac{1}{2} cos (t) + \frac{1}{2} sin (t)

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