inverselaplace {((s+1)^3)/(s^4)}

Lösung

inverse laplace transformation

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

H (t) + 3 t + \frac{3 t ^{2}}{2} + \frac{t ^{3}}{6}

Schritte zur Lösung

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

Schreibe

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

um:

\frac{1}{s} + \frac{3}{s ^{2}} + \frac{3}{s ^{3}} + \frac{1}{s ^{4}}

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

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

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

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

L^{- 1} {\frac{1}{s ^{4}}} : \frac{t ^{3}}{6}

= H (t) + 3 t + \frac{3 t ^{2}}{2} + \frac{t ^{3}}{6}

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