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Beliebt Rechnen >

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

Lösung

inverse laplace transformation

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

Lösung

H (t) - e^{- \frac{t}{2}} cos (\frac{3 t}{2}) - \frac{e ^{- \frac{t}{2}} sin ( \frac{3 t}{2} )}{3}

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

= L^{- 1} {\frac{1}{s} + \frac{- s - 1}{s ^{2} + s + 1}}

Schreibe

\frac{- s - 1}{s ^{2} + s + 1}

um:

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

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

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)}

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

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

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

Fasse

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

zusammen:

H (t) - e^{- \frac{t}{2}} cos (\frac{3 t}{2}) - \frac{e ^{- \frac{t}{2}} sin ( \frac{3 t}{2} )}{3}

= H (t) - e^{- \frac{t}{2}} cos (\frac{3 t}{2}) - \frac{e ^{- \frac{t}{2}} sin ( \frac{3 t}{2} )}{3}

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