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Beliebt Rechnen >

inverselaplace (16)/(s(s+2)^4)

Lösung

inverse laplace transformation

\frac{16}{s ( s + 2 ) ^{4}}

Lösung

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

Schritte zur Lösung

L^{- 1} {\frac{16}{s ( s + 2 ) ^{4}}}

Ermittle den Partialbruch von

\frac{16}{s ( s + 2 ) ^{4}} : \frac{1}{s} - \frac{1}{s + 2} - \frac{2}{( s + 2 ) ^{2}} - \frac{4}{( s + 2 ) ^{3}} - \frac{8}{( s + 2 ) ^{4}}

= L^{- 1} {\frac{1}{s} - \frac{1}{s + 2} - \frac{2}{( s + 2 ) ^{2}} - \frac{4}{( s + 2 ) ^{3}} - \frac{8}{( s + 2 ) ^{4}}}

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{2}{( s + 2 ) ^{2}}} - L^{- 1} {\frac{4}{( s + 2 ) ^{3}}} - L^{- 1} {\frac{8}{( s + 2 ) ^{4}}}

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

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

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

L^{- 1} {\frac{4}{( s + 2 ) ^{3}}} : e^{- 2 t} \cdot 2 t^{2}

L^{- 1} {\frac{8}{( s + 2 ) ^{4}}} : \frac{4 e ^{- 2 t} t ^{3}}{3}

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

Beliebte Beispiele

limit as x approaches infinity of (e^{-sx})/s

x \to \infty lim (\frac{e ^{- s x}}{s})

(dy)/(dx)y=(8x^2-16+16)e^x

\frac{d y}{d x} y = (8 x^{2} - 16 + 16) e^{x}

derivative (1+2x^2)(x-x^2)

d er i v a t i v e (1 + 2 x^{2}) (x - x^{2})

derivative (x^3+8x)^3

d er i v a t i v e (x^{3} + 8 x)^{3}

integral from 0 to pi/2 of pisin^2(x)

\int_{0}^{\frac{π}{2}} π sin^{2} (x) d x