inverselaplace 1/(s^3-7s)

Lösung

inverse laplace transformation

\frac{1}{s ^{3} - 7 s}

- \frac{1}{7} H (t) + \frac{1}{14} e^{- 7 t} + \frac{1}{14} e^{7 t}

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{3} - 7 s}}

Ermittle den Partialbruch von

\frac{1}{s ^{3} - 7 s} : - \frac{1}{7 s} + \frac{1}{14 ( s + 7 )} + \frac{1}{14 ( s - 7 )}

= L^{- 1} {- \frac{1}{7 s} + \frac{1}{14 ( s + 7 )} + \frac{1}{14 ( s - 7 )}}

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}{7 s}} + L^{- 1} {\frac{1}{14 ( s + 7 )}} + L^{- 1} {\frac{1}{14 ( s - 7 )}}

L^{- 1} {\frac{1}{7 s}} : \frac{1}{7} H (t)

L^{- 1} {\frac{1}{14 ( s + 7 )}} : \frac{1}{14} e^{- 7 t}

L^{- 1} {\frac{1}{14 ( s - 7 )}} : \frac{1}{14} e^{7 t}

= - \frac{1}{7} H (t) + \frac{1}{14} e^{- 7 t} + \frac{1}{14} e^{7 t}

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