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Beliebt Rechnen >

inverselaplace 6/(s(s+1)(s+2)+6)

Lösung

inverse laplace transformation

\frac{6}{s ( s + 1 ) ( s + 2 ) + 6}

Lösung

\frac{6}{11} e^{- 3 t} - \frac{6}{11} cos (2 t) + \frac{9 2 sin ( 2 t )}{11}

Schritte zur Lösung

L^{- 1} {\frac{6}{s ( s + 1 ) ( s + 2 ) + 6}}

Ermittle den Partialbruch von

\frac{6}{s ( s + 1 ) ( s + 2 ) + 6} : \frac{6}{11 ( s + 3 )} + \frac{- 6 s + 18}{11 ( s ^{2} + 2 )}

= L^{- 1} {\frac{6}{11 ( s + 3 )} + \frac{- 6 s + 18}{11 ( s ^{2} + 2 )}}

Schreibe um

= L^{- 1} {- \frac{6 s}{11 ( s ^{2} + 2 )} + \frac{6}{11 ( s + 3 )} + \frac{18}{11 ( s ^{2} + 2 )}}

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{6}{11} L^{- 1} {\frac{1}{s + 3}} - \frac{6}{11} L^{- 1} {\frac{s}{s ^{2} + 2}} + \frac{18}{11} L^{- 1} {\frac{1}{s ^{2} + 2}}

L^{- 1} {\frac{1}{s + 3}} : e^{- 3 t}

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

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

= \frac{6}{11} e^{- 3 t} - \frac{6}{11} cos (2 t) + \frac{18}{11} \cdot \frac{1}{2} sin (2 t)

Fasse

\frac{6}{11} e^{- 3 t} - \frac{6}{11} cos (2 t) + \frac{18}{11} \frac{1}{2} sin (2 t)

zusammen:

\frac{6}{11} e^{- 3 t} - \frac{6}{11} cos (2 t) + \frac{9 2 sin ( 2 t )}{11}

= \frac{6}{11} e^{- 3 t} - \frac{6}{11} cos (2 t) + \frac{9 2 sin ( 2 t )}{11}

Beliebte Beispiele

integral of sin(3)x^2

\int sin (3) x^{2} d x

derivative of (x^2/(a^2))

\frac{d}{d x} (\frac{x ^{2}}{a ^{2}})

limit as x approaches infinity of sqrt(x)

x \to \infty lim (x)

limit as x approaches 0 of tan^2(x)

x \to 0 lim (tan^{2} (x))

integral of (e^x+1)(x-4)

\int (e^{x} + 1) (x - 4) d x