inverselaplace 1/((s^2+4)(s^2+2s+2))

Lösung

inverse laplace transformation

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

- \frac{1}{10} cos (2 t) - \frac{1}{20} sin (2 t) + \frac{1}{10} e^{- t} cos (t) + \frac{1}{5} e^{- t} sin (t)

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{( s ^{2} + 4 ) ( s ^{2} + 2 s + 2 )} : \frac{- s - 1}{10 ( s ^{2} + 4 )} + \frac{s + 3}{10 ( s ^{2} + 2 s + 2 )}

= L^{- 1} {\frac{- s - 1}{10 ( s ^{2} + 4 )} + \frac{s + 3}{10 ( s ^{2} + 2 s + 2 )}}

Schreibe

\frac{s + 3}{10 ( s ^{2} + 2 s + 2 )}

um:

\frac{1}{10} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 1} + \frac{1}{5} \cdot \frac{1}{( s + 1 ) ^{2} + 1}

= L^{- 1} {\frac{- s - 1}{10 ( s ^{2} + 4 )} + \frac{1}{10} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 1} + \frac{1}{5} \cdot \frac{1}{( s + 1 ) ^{2} + 1}}

Schreibe um

= L^{- 1} {- \frac{s}{10 ( s ^{2} + 4 )} + \frac{1}{10} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 1} + \frac{1}{5} \cdot \frac{1}{( s + 1 ) ^{2} + 1} - \frac{1}{10 ( s ^{2} + 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)}

= - \frac{1}{10} L^{- 1} {\frac{s}{s ^{2} + 4}} - \frac{1}{10} L^{- 1} {\frac{1}{s ^{2} + 4}} + \frac{1}{10} L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 1}} + \frac{1}{5} L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 1}}

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

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

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

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

= - \frac{1}{10} cos (2 t) - \frac{1}{10} \cdot \frac{1}{2} sin (2 t) + \frac{1}{10} e^{- t} cos (t) + \frac{1}{5} e^{- t} sin (t)

Fasse

- \frac{1}{10} cos (2 t) - \frac{1}{10} \frac{1}{2} sin (2 t) + \frac{1}{10} e^{- t} cos (t) + \frac{1}{5} e^{- t} sin (t)

zusammen:

- \frac{1}{10} cos (2 t) - \frac{1}{20} sin (2 t) + \frac{1}{10} e^{- t} cos (t) + \frac{1}{5} e^{- t} sin (t)

= - \frac{1}{10} cos (2 t) - \frac{1}{20} sin (2 t) + \frac{1}{10} e^{- t} cos (t) + \frac{1}{5} e^{- t} sin (t)