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

Lösung

inverse laplace transformation

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

\frac{1}{10} e^{- t} cos (t) - \frac{3}{10} e^{- t} sin (t) - \frac{1}{10} cos (2 t) + \frac{sin ( 2 t )}{5}

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

Schreibe

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

um:

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

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

Schreibe um

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

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)

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

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

= \frac{1}{10} e^{- t} cos (t) - \frac{3}{10} e^{- t} sin (t) - \frac{1}{10} cos (2 t) + \frac{4}{10} \cdot \frac{1}{2} sin (2 t)

Fasse

\frac{1}{10} e^{- t} cos (t) - \frac{3}{10} e^{- t} sin (t) - \frac{1}{10} cos (2 t) + \frac{4}{10} \frac{1}{2} sin (2 t)

zusammen:

\frac{1}{10} e^{- t} cos (t) - \frac{3}{10} e^{- t} sin (t) - \frac{1}{10} cos (2 t) + \frac{sin ( 2 t )}{5}

= \frac{1}{10} e^{- t} cos (t) - \frac{3}{10} e^{- t} sin (t) - \frac{1}{10} cos (2 t) + \frac{sin ( 2 t )}{5}

\int 8 (x + 1) d x

\int \frac{( x ^{5} - 2 x ^{2} )}{( 1 + x ^{3} ) ^{\frac{3}{2}}} d x

\frac{d}{d u} (2 uv)

\frac{d}{d x} (ln ∣ coth (5 x) ∣)

\frac{\partial}{\partial u ( v )} (\frac{e ^{v}}{u ( v ) + v ^{3}})