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Beliebt Rechnen >

inverselaplace (4s-1)/(s^2(s+1)^3)

Lösung

inverse laplace transformation

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

Lösung

7 H (t) - t - 7 e^{- t} - e^{- t} \cdot 6 t - \frac{5 e ^{- t} t ^{2}}{2}

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{4 s - 1}{s ^{2} ( s + 1 ) ^{3}} : \frac{7}{s} - \frac{1}{s ^{2}} - \frac{7}{s + 1} - \frac{6}{( s + 1 ) ^{2}} - \frac{5}{( s + 1 ) ^{3}}

= L^{- 1} {\frac{7}{s} - \frac{1}{s ^{2}} - \frac{7}{s + 1} - \frac{6}{( s + 1 ) ^{2}} - \frac{5}{( 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{7}{s}} - L^{- 1} {\frac{1}{s ^{2}}} - L^{- 1} {\frac{7}{s + 1}} - L^{- 1} {\frac{6}{( s + 1 ) ^{2}}} - L^{- 1} {\frac{5}{( s + 1 ) ^{3}}}

L^{- 1} {\frac{7}{s}} : 7 H (t)

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

L^{- 1} {\frac{7}{s + 1}} : 7 e^{- t}

L^{- 1} {\frac{6}{( s + 1 ) ^{2}}} : e^{- t} \cdot 6 t

L^{- 1} {\frac{5}{( s + 1 ) ^{3}}} : \frac{5 e ^{- t} t ^{2}}{2}

= 7 H (t) - t - 7 e^{- t} - e^{- t} \cdot 6 t - \frac{5 e ^{- t} t ^{2}}{2}

Beliebte Beispiele

limit as x approaches+5 of (x+1)/(x-5)

x \to + 5 lim (\frac{x + 1}{x - 5})

derivative of (sqrt(x)^{3x})

\frac{d}{d x} ((x)^{3 x})

integral of 1/((4x+1)^3)

\int \frac{1}{( 4 x + 1 ) ^{3}} d x

y*(dy)/(dx)-x=5y^2

y \cdot \frac{d y}{d x} - x = 5 y^{2}

derivative of ln((x^2-1^3))

\frac{d}{d x} (ln ((x^{2} - 1)^{3}))