inverselaplace 1/(s(s^2-1))

Lösung

inverse laplace transformation

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

- H (t) + \frac{1}{2} e^{- t} + \frac{1}{2} e^{t}

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

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

= - H (t) + \frac{1}{2} e^{- t} + \frac{1}{2} e^{t}

t an g e n t f (x) = \frac{1}{x + 2}, at x = 7

d er i v a t i v e f (x) = \frac{x ^{2}}{x + 5}

\int_{4}^{9} \frac{1}{9 - x} d x

\frac{\partial}{\partial x} (3 x + 8 y + 2)

\frac{\partial}{\partial x} (3 x^{2} - 4 y^{3})