inverselaplace 1/(s(s^2+100))

Lösung

inverse laplace transformation

\frac{1}{s ( s ^{2} + 100 )}

\frac{1}{100} H (t) - \frac{1}{100} cos (10 t)

Schritte zur Lösung

L^{- 1} {\frac{1}{s ( s ^{2} + 100 )}}

Ermittle den Partialbruch von

\frac{1}{s ( s ^{2} + 100 )} : \frac{1}{100 s} - \frac{s}{100 ( s ^{2} + 100 )}

= L^{- 1} {\frac{1}{100 s} - \frac{s}{100 ( s ^{2} + 100 )}}

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}{100 s}} - L^{- 1} {\frac{s}{100 ( s ^{2} + 100 )}}

L^{- 1} {\frac{1}{100 s}} : \frac{1}{100} H (t)

L^{- 1} {\frac{s}{100 ( s ^{2} + 100 )}} : \frac{1}{100} cos (10 t)

= \frac{1}{100} H (t) - \frac{1}{100} cos (10 t)

\frac{\partial}{\partial x} (x y e^{7 y})

d er i v a t i v e y = ln (x^{3} + 1)

n = 0 \sum \infty \frac{1}{3}

\int_{0}^{1} (8 x^{2} - 2) e^{x} d x

\frac{d}{d t} ((arccos (t))^{2})