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Beliebt Rechnen >

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

Lösung

inverse laplace transformation

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

Lösung

\frac{1}{12} H (t) - \frac{1}{8} e^{- 2 t} + \frac{1}{24} e^{- 6 t}

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{s ( s + 2 ) ( s + 6 )} : \frac{1}{12 s} - \frac{1}{8 ( s + 2 )} + \frac{1}{24 ( s + 6 )}

= L^{- 1} {\frac{1}{12 s} - \frac{1}{8 ( s + 2 )} + \frac{1}{24 ( s + 6 )}}

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{1}{12 s}} - L^{- 1} {\frac{1}{8 ( s + 2 )}} + L^{- 1} {\frac{1}{24 ( s + 6 )}}

L^{- 1} {\frac{1}{12 s}} : \frac{1}{12} H (t)

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

L^{- 1} {\frac{1}{24 ( s + 6 )}} : \frac{1}{24} e^{- 6 t}

= \frac{1}{12} H (t) - \frac{1}{8} e^{- 2 t} + \frac{1}{24} e^{- 6 t}

Beliebte Beispiele

integral from-1 to 1 of 1/(x^2-1)

\int_{- 1}^{1} \frac{1}{x ^{2} - 1} d x

tangent f(x)=4xe^x,(0,0)

t an g e n t f (x) = 4 x e^{x}, (0, 0)

integral of x(4+x)^{10}

\int x (4 + x)^{10} d x

derivative (2(1-ln(x)))/(x^2)

d er i v a t i v e \frac{2 ( 1 - ln ( x ) )}{x ^{2}}

limit as x approaches-2 of 3x^2+5x-1

x \to - 2 lim (3 x^{2} + 5 x - 1)