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

Lösung

inverse laplace transformation

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

\frac{1}{4} H (t) - \frac{1}{4} e^{- 2 t} - \frac{e ^{- 2 t} t}{2}

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

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

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

= \frac{1}{4} H (t) - \frac{1}{4} e^{- 2 t} - \frac{e ^{- 2 t} t}{2}

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

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} - 3 y = 0

\int \frac{1}{y ^{2} - 2} d y

l a pl a ce t r an s f or m 2 (t - 2)^{2} + 6 (t - 2) + 6

\int \frac{1}{4 x ^{2} - 16} d x