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Beliebt Rechnen >

inverselaplace ((2s-1))/(s^2(s+1)^3)

Lösung

inverse laplace transformation

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

Lösung

5 H (t) - t - 5 e^{- t} - e^{- t} \cdot 4 t - \frac{3 e ^{- t} t ^{2}}{2}

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{2 s - 1}{s ^{2} ( s + 1 ) ^{3}} : \frac{5}{s} - \frac{1}{s ^{2}} - \frac{5}{s + 1} - \frac{4}{( s + 1 ) ^{2}} - \frac{3}{( s + 1 ) ^{3}}

= L^{- 1} {\frac{5}{s} - \frac{1}{s ^{2}} - \frac{5}{s + 1} - \frac{4}{( s + 1 ) ^{2}} - \frac{3}{( s + 1 ) ^{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{5}{s}} - L^{- 1} {\frac{1}{s ^{2}}} - L^{- 1} {\frac{5}{s + 1}} - L^{- 1} {\frac{4}{( s + 1 ) ^{2}}} - L^{- 1} {\frac{3}{( s + 1 ) ^{3}}}

L^{- 1} {\frac{5}{s}} : 5 H (t)

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

L^{- 1} {\frac{5}{s + 1}} : 5 e^{- t}

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

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

= 5 H (t) - t - 5 e^{- t} - e^{- t} \cdot 4 t - \frac{3 e ^{- t} t ^{2}}{2}

Beliebte Beispiele

integral of (x^5+8x^3-9x^2+7x-5)

\int (x^{5} + 8 x^{3} - 9 x^{2} + 7 x - 5) d x

integral of (e^x+1/(e^x))^2

\int (e^{x} + \frac{1}{e ^{x}})^{2} d x

integral from 0 to 1 of cos(x^2)

\int_{0}^{1} cos (x^{2}) d x

2t*(dy)/(dt)-4y=sqrt(t)

2 t \cdot \frac{d y}{d t} - 4 y = t

integral of ((x^2)/4+4x+19)^2

\int (\frac{x ^{2}}{4} + 4 x + 19)^{2} d x