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

Lösung

inverse laplace transformation

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

- H (t) + t + e^{- t}

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

= - H (t) + t + e^{- t}

d er i v a t i v e f (x) = \frac{7}{2} x^{\frac{2}{7}}

x \to 8 lim (\frac{x ^{2} - 64}{x ^{2} - 8 x})

\frac{\partial}{\partial x} (2 x + 1)

\int (\frac{8}{x ^{3}}) d x

t an g e n t f (x) = 2 x^{3} - 3 x^{2} - 16 x + 3, at x = - 4