inverselaplace 3/(s(s^2+2s+5))

Lösung

inverse laplace transformation

\frac{3}{s ( s ^{2} + 2 s + 5 )}

\frac{3}{5} H (t) - \frac{3}{5} e^{- t} cos (2 t) - \frac{3}{10} e^{- t} sin (2 t)

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

Schreibe

\frac{- 3 s - 6}{5 ( s ^{2} + 2 s + 5 )}

um:

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

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

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

L^{- 1} {\frac{3}{5 s}} : \frac{3}{5} H (t)

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

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

= \frac{3}{5} H (t) - \frac{3}{5} e^{- t} cos (2 t) - \frac{3}{5} e^{- t} \frac{1}{2} sin (2 t)

Fasse

\frac{3}{5} H (t) - \frac{3}{5} e^{- t} cos (2 t) - \frac{3}{5} e^{- t} \frac{1}{2} sin (2 t)

zusammen:

\frac{3}{5} H (t) - \frac{3}{5} e^{- t} cos (2 t) - \frac{3}{10} e^{- t} sin (2 t)

= \frac{3}{5} H (t) - \frac{3}{5} e^{- t} cos (2 t) - \frac{3}{10} e^{- t} sin (2 t)

d er i v a t i v e 5 (x^{3} - x)^{4}

\int x (2 sec^{2} (x) + tan^{2} (x)) d x

\frac{d}{d y} (4 - y^{2})

\int_{- 7}^{0} (4 + 49 - x^{2}) d x

\int \frac{2 x ^{2} + 5 x + 2}{( x ^{2} + 1 ) ^{2}} d x