inverselaplace 1/(2s^3+10s^2+12s)

Lösung

inverse laplace transformation

\frac{1}{2 s ^{3} + 10 s ^{2} + 12 s}

\frac{1}{12} H (t) - \frac{1}{4} e^{- 2 t} + \frac{1}{6} e^{- 3 t}

Schritte zur Lösung

L^{- 1} {\frac{1}{2 s ^{3} + 10 s ^{2} + 12 s}}

Ermittle den Partialbruch von

\frac{1}{2 s ^{3} + 10 s ^{2} + 12 s} : \frac{1}{12 s} - \frac{1}{4 ( s + 2 )} + \frac{1}{6 ( s + 3 )}

= L^{- 1} {\frac{1}{12 s} - \frac{1}{4 ( s + 2 )} + \frac{1}{6 ( s + 3 )}}

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}{12 s}} - L^{- 1} {\frac{1}{4 ( s + 2 )}} + L^{- 1} {\frac{1}{6 ( s + 3 )}}

L^{- 1} {\frac{1}{12 s}} : \frac{1}{12} H (t)

L^{- 1} {\frac{1}{4 ( s + 2 )}} : \frac{1}{4} e^{- 2 t}

L^{- 1} {\frac{1}{6 ( s + 3 )}} : \frac{1}{6} e^{- 3 t}

= \frac{1}{12} H (t) - \frac{1}{4} e^{- 2 t} + \frac{1}{6} e^{- 3 t}

d er i v a t i v e y = x^{4} cos (4 x^{2})

a re a y = 17 x, y = x^{2}

\int sin^{2} (t) cos (t) d t

\int 0.0025 x - 0.075 d x

x \to 3 π lim (csc (x))