inverselaplace {1/(s^3+5s)}

Lösung

inverse laplace transformation

{\frac{1}{s ^{3} + 5 s}}

\frac{1}{5} H (t) - \frac{1}{5} cos (5 t)

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{3} + 5 s}}

Ermittle den Partialbruch von

\frac{1}{s ^{3} + 5 s} : \frac{1}{5 s} - \frac{s}{5 ( s ^{2} + 5 )}

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

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

L^{- 1} {\frac{1}{5 s}} : \frac{1}{5} H (t)

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

= \frac{1}{5} H (t) - \frac{1}{5} cos (5 t)

\int e^{- 3 x} d x

\int \frac{x ^{3}}{3 x ^{2} + 1} d x

n = 1 \sum \infty \frac{1}{4 n + 3}

\frac{\partial}{\partial y} (3 cos (x))

x \to 3 lim (\frac{( x - 3 )}{x - 3})