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Beliebt Rechnen >

inverselaplace (9s)/(5(s^2+5s+11))

Lösung

inverse laplace transformation

\frac{9 s}{5 ( s ^{2} + 5 s + 11 )}

Lösung

\frac{9}{5} e^{- \frac{5 t}{2}} cos (\frac{19 t}{2}) - \frac{9 e ^{- \frac{5 t}{2}} sin ( \frac{19 t}{2} )}{19}

Schritte zur Lösung

L^{- 1} {\frac{9 s}{5 ( s ^{2} + 5 s + 11 )}}

Schreibe

\frac{9 s}{5 ( s ^{2} + 5 s + 11 )}

um:

\frac{9}{5} \cdot \frac{s + \frac{5}{2}}{( s + \frac{5}{2} ) ^{2} + \frac{19}{4}} - \frac{9}{2} \cdot \frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{19}{4}}

= L^{- 1} {\frac{9}{5} \cdot \frac{s + \frac{5}{2}}{( s + \frac{5}{2} ) ^{2} + \frac{19}{4}} - \frac{9}{2} \cdot \frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{19}{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{9}{5} L^{- 1} {\frac{s + \frac{5}{2}}{( s + \frac{5}{2} ) ^{2} + \frac{19}{4}}} - \frac{9}{2} L^{- 1} {\frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{19}{4}}}

L^{- 1} {\frac{s + \frac{5}{2}}{( s + \frac{5}{2} ) ^{2} + \frac{19}{4}}} : e^{- \frac{5 t}{2}} cos (\frac{19 t}{2})

L^{- 1} {\frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{19}{4}}} : e^{- \frac{5 t}{2}} \frac{2}{19} sin (\frac{19 t}{2})

= \frac{9}{5} e^{- \frac{5 t}{2}} cos (\frac{19 t}{2}) - \frac{9}{2} e^{- \frac{5 t}{2}} \frac{2}{19} sin (\frac{19 t}{2})

Fasse

\frac{9}{5} e^{- \frac{5 t}{2}} cos (\frac{19 t}{2}) - \frac{9}{2} e^{- \frac{5 t}{2}} \frac{2}{19} sin (\frac{19 t}{2})

zusammen:

\frac{9}{5} e^{- \frac{5 t}{2}} cos (\frac{19 t}{2}) - \frac{9 e ^{- \frac{5 t}{2}} sin ( \frac{19 t}{2} )}{19}

= \frac{9}{5} e^{- \frac{5 t}{2}} cos (\frac{19 t}{2}) - \frac{9 e ^{- \frac{5 t}{2}} sin ( \frac{19 t}{2} )}{19}

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