inverselaplace (30)/((10s+1)(s^2+4))

Lösung

inverse laplace transformation

\frac{30}{( 10 s + 1 ) ( s ^{2} + 4 )}

\frac{300 e ^{- \frac{t}{10}}}{401} - \frac{300}{401} cos (2 t) + \frac{15 sin ( 2 t )}{401}

Schritte zur Lösung

L^{- 1} {\frac{30}{( 10 s + 1 ) ( s ^{2} + 4 )}}

Ermittle den Partialbruch von

\frac{30}{( 10 s + 1 ) ( s ^{2} + 4 )} : \frac{3000}{401 ( 10 s + 1 )} + \frac{- 300 s + 30}{401 ( s ^{2} + 4 )}

= L^{- 1} {\frac{3000}{401 ( 10 s + 1 )} + \frac{- 300 s + 30}{401 ( s ^{2} + 4 )}}

Schreibe um

= L^{- 1} {- \frac{300 s}{401 ( s ^{2} + 4 )} + \frac{3000}{401 ( 10 s + 1 )} + \frac{30}{401 ( s ^{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)}

= \frac{3000}{401} L^{- 1} {\frac{1}{10 s + 1}} - \frac{300}{401} L^{- 1} {\frac{s}{s ^{2} + 4}} + \frac{30}{401} L^{- 1} {\frac{1}{s ^{2} + 4}}

L^{- 1} {\frac{1}{10 s + 1}} : \frac{1}{10} e^{- \frac{t}{10}}

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

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

= \frac{3000}{401} \cdot \frac{1}{10} e^{- \frac{t}{10}} - \frac{300}{401} cos (2 t) + \frac{30}{401} \cdot \frac{1}{2} sin (2 t)

Fasse

\frac{3000}{401} \frac{1}{10} e^{- \frac{t}{10}} - \frac{300}{401} cos (2 t) + \frac{30}{401} \frac{1}{2} sin (2 t)

zusammen:

\frac{300 e ^{- \frac{t}{10}}}{401} - \frac{300}{401} cos (2 t) + \frac{15 sin ( 2 t )}{401}

= \frac{300 e ^{- \frac{t}{10}}}{401} - \frac{300}{401} cos (2 t) + \frac{15 sin ( 2 t )}{401}

d er i v a t i v e x^{x + 2}

\frac{d}{d x} (\frac{x ^{2} - 64}{x - 8})

d er i v a t i v e f (x) = 3 - x

x \to - 1 lim (π)

x \to - 9 lim (50)