inverselaplace ((2s+1))/(s(2s+10))

Lösung

inverse laplace transformation

\frac{( 2 s + 1 )}{s ( 2 s + 10 )}

\frac{1}{10} H (t) + \frac{9}{10} e^{- 5 t}

Schritte zur Lösung

L^{- 1} {\frac{( 2 s + 1 )}{s ( 2 s + 10 )}}

Ermittle den Partialbruch von

\frac{2 s + 1}{s ( 2 s + 10 )} : \frac{1}{10 s} + \frac{9}{10 ( s + 5 )}

= L^{- 1} {\frac{1}{10 s} + \frac{9}{10 ( s + 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}{10 s}} + L^{- 1} {\frac{9}{10 ( s + 5 )}}

L^{- 1} {\frac{1}{10 s}} : \frac{1}{10} H (t)

L^{- 1} {\frac{9}{10 ( s + 5 )}} : \frac{9}{10} e^{- 5 t}

= \frac{1}{10} H (t) + \frac{9}{10} e^{- 5 t}

\frac{d}{d x} ((- x^{2} - 4)^{2})

\frac{d}{d x} (3 - 3 x)

\frac{d}{d x} (\frac{1 - x}{x ^{2}})

\int x^{2} - 5 x + 5 d x

\int_{\frac{π}{2}}^{π} 4 cot (\frac{x}{2}) d x