inverselaplace s/(s^4+4)

Lösung

inverse laplace transformation

\frac{s}{s ^{4} + 4}

- \frac{1}{4} e^{- t} sin (t) + \frac{1}{4} e^{t} sin (t)

Schritte zur Lösung

L^{- 1} {\frac{s}{s ^{4} + 4}}

Ermittle den Partialbruch von

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

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

Schreibe

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

um:

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

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

Schreibe

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

um:

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

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

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

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

= - \frac{1}{4} e^{- t} sin (t) + \frac{1}{4} e^{t} sin (t)

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