inverselaplace (s^3+s+6)/(s^4(s+2))

Lösung

inverse laplace transformation

\frac{s ^{3} + s + 6}{s ^{4} ( s + 2 )}

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

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

L^{- 1} {\frac{1}{s ^{3}}} : \frac{t ^{2}}{2}

L^{- 1} {\frac{3}{s ^{4}}} : \frac{t ^{3}}{2}

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

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

\int_{7}^{8} \frac{x}{x ^{2} - 3} d x

x \to 7 + lim (3 x + 5)

d er i v a t i v e x

\frac{d y}{d t} = 2 y - 3 e^{- t}

x \to 0 - lim (- 2 x^{2} + 7 x)