inverselaplace 1/((s-1)(s^2+9))

Lösung

inverse laplace transformation

\frac{1}{( s - 1 ) ( s ^{2} + 9 )}

\frac{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{30} sin (3 t)

Schritte zur Lösung

L^{- 1} {\frac{1}{( s - 1 ) ( s ^{2} + 9 )}}

Ermittle den Partialbruch von

\frac{1}{( s - 1 ) ( s ^{2} + 9 )} : \frac{1}{10 ( s - 1 )} + \frac{- s - 1}{10 ( s ^{2} + 9 )}

= L^{- 1} {\frac{1}{10 ( s - 1 )} + \frac{- s - 1}{10 ( s ^{2} + 9 )}}

Schreibe um

= L^{- 1} {- \frac{s}{10 ( s ^{2} + 9 )} + \frac{1}{10 ( s - 1 )} - \frac{1}{10 ( s ^{2} + 9 )}}

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}{10} L^{- 1} {\frac{1}{s - 1}} - \frac{1}{10} L^{- 1} {\frac{s}{s ^{2} + 9}} - \frac{1}{10} L^{- 1} {\frac{1}{s ^{2} + 9}}

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

L^{- 1} {\frac{s}{s ^{2} + 9}} : cos (3 t)

L^{- 1} {\frac{1}{s ^{2} + 9}} : \frac{1}{3} sin (3 t)

= \frac{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{10} \cdot \frac{1}{3} sin (3 t)

Fasse

\frac{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{10} \frac{1}{3} sin (3 t)

zusammen:

\frac{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{30} sin (3 t)

= \frac{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{30} sin (3 t)