inverselaplace (s-1)/(s(s+2.5))

Lösung

inverse laplace transformation

\frac{s - 1}{s ( s + 2.5 )}

- \frac{2}{5} H (t) + 1.4 e^{- 2.5 t}

Schritte zur Lösung

L^{- 1} {\frac{s - 1}{s ( s + 2.5 )}}

Ermittle den Partialbruch von

\frac{s - 1}{s ( s + 2.5 )} : - \frac{2}{5 s} + \frac{1.4}{s + 2.5}

= L^{- 1} {- \frac{2}{5 s} + \frac{1.4}{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{2}{5 s}} + L^{- 1} {\frac{1.4}{s + 2.5}}

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

L^{- 1} {\frac{1.4}{s + 2.5}} : 1.4 e^{- 2.5 t}

= - \frac{2}{5} H (t) + 1.4 e^{- 2.5 t}

\frac{\partial}{\partial x} (3 x z e^{4 x y})

x \to - 3 lim (3 x^{2} - 4 x + 1)

(x + 1) (\frac{d y}{d x}) + (2 x - 1) y = e^{- 2 x}

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

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