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

Lösung

inverse laplace transformation

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

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

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

a re a f (x) = x^{2} - 4, - 3, 1

t an g e n t f (x) = (25 - 0.075 x) x, at x = 100

\int \frac{4 x ^{2} - 14 x - 6}{x ^{3} - 2 x ^{2} - 3 x} d x

n = 1 \sum \infty 1 - cos (\frac{1}{n})

\frac{( y ^{^{'}} - e ^{- t} + 6 )}{y} = - 6