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

Lösung

inverse laplace transformation

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

\frac{1}{5} H (t) - \frac{1}{5} e^{- 2 t} cos (t) - \frac{2}{5} e^{- 2 t} sin (t)

Schritte zur Lösung

L^{- 1} ⎩ ⎨ ⎧ \frac{1}{s ( ( s + 2 ) ^{2} + 1 )} ⎭ ⎬ ⎫

Ermittle den Partialbruch von

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

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

Schreibe

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

um:

- \frac{1}{5} \cdot \frac{s + 2}{( s + 2 ) ^{2} + 1} - \frac{2}{5} \cdot \frac{1}{( s + 2 ) ^{2} + 1}

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

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

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

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

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

= \frac{1}{5} H (t) - \frac{1}{5} e^{- 2 t} cos (t) - \frac{2}{5} e^{- 2 t} sin (t)

n = 1 \sum \infty 6 e^{- 8 n}

x \to 0 lim (2 - x)

\int - e^{- s t} d t

\int \frac{2}{e ^{t}} d t

\frac{\partial}{\partial y} (\frac{1}{x ^{2} + y ^{2} - 1})