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Beliebt Rechnen >

inverselaplace 5/(s^2-12)-(4s)/(s^2+8)

Lösung

inverse laplace transformation

\frac{5}{s ^{2} - 12} - \frac{4 s}{s ^{2} + 8}

Lösung

- \frac{5}{4 3} e^{- 23 t} + \frac{5}{4 3} e^{23 t} - 4 cos (22 t)

Schritte zur Lösung

L^{- 1} {\frac{5}{s ^{2} - 12} - \frac{4 s}{s ^{2} + 8}}

Ermittle den Partialbruch von

\frac{5}{s ^{2} - 12} : - \frac{5}{4 3 ( s + 2 3 )} + \frac{5}{4 3 ( s - 2 3 )}

= L^{- 1} {- \frac{5}{4 3 ( s + 2 3 )} + \frac{5}{4 3 ( s - 2 3 )} - \frac{4 s}{s ^{2} + 8}}

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)}

= - L^{- 1} {\frac{5}{4 3 ( s + 2 3 )}} + L^{- 1} {\frac{5}{4 3 ( s - 2 3 )}} - L^{- 1} {\frac{4 s}{s ^{2} + 8}}

L^{- 1} {\frac{5}{4 3 ( s + 2 3 )}} : \frac{5}{4 3} e^{- 23 t}

L^{- 1} {\frac{5}{4 3 ( s - 2 3 )}} : \frac{5}{4 3} e^{23 t}

L^{- 1} {\frac{4 s}{s ^{2} + 8}} : 4 cos (22 t)

= - \frac{5}{4 3} e^{- 23 t} + \frac{5}{4 3} e^{23 t} - 4 cos (22 t)

Beliebte Beispiele

derivative y=3x^2+(8x)/3-2

d er i v a t i v e y = 3 x^{2} + \frac{8 x}{3} - 2

integral of (900)/((2x-5)^2)

\int \frac{900}{( 2 x - 5 ) ^{2}} d x

integral of 1/(x^2sqrt(x^2-6))

\int \frac{1}{x ^{2} x ^{2} - 6} d x

derivative f(x)=(x+2)/(10-3x)

d er i v a t i v e f (x) = \frac{x + 2}{10 - 3 x}

integral from 5 to 7 of x^3

\int_{5}^{7} x^{3} d x