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

Lösung

inverse laplace transformation

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

\frac{3}{5} H (t) - \frac{3}{5} e^{- \frac{t}{2}} cos (\frac{19 t}{2}) - \frac{3 e ^{- \frac{t}{2}} sin ( \frac{19 t}{2} )}{5 19}

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

Schreibe

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

um:

- \frac{3}{5} \cdot \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{19}{4}} - \frac{3}{10} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{19}{4}}

= L^{- 1} {\frac{3}{5 s} - \frac{3}{5} \cdot \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{19}{4}} - \frac{3}{10} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{19}{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 + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{19}{4}}} - \frac{3}{10} L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{19}{4}}}

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

L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{19}{4}}} : e^{- \frac{t}{2}} cos (\frac{19 t}{2})

L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{19}{4}}} : e^{- \frac{t}{2}} \frac{2}{19} sin (\frac{19 t}{2})

= \frac{3}{5} H (t) - \frac{3}{5} e^{- \frac{t}{2}} cos (\frac{19 t}{2}) - \frac{3}{10} e^{- \frac{t}{2}} \frac{2}{19} sin (\frac{19 t}{2})

Fasse

\frac{3}{5} H (t) - \frac{3}{5} e^{- \frac{t}{2}} cos (\frac{19 t}{2}) - \frac{3}{10} e^{- \frac{t}{2}} \frac{2}{19} sin (\frac{19 t}{2})

zusammen:

\frac{3}{5} H (t) - \frac{3}{5} e^{- \frac{t}{2}} cos (\frac{19 t}{2}) - \frac{3 e ^{- \frac{t}{2}} sin ( \frac{19 t}{2} )}{5 19}

= \frac{3}{5} H (t) - \frac{3}{5} e^{- \frac{t}{2}} cos (\frac{19 t}{2}) - \frac{3 e ^{- \frac{t}{2}} sin ( \frac{19 t}{2} )}{5 19}

d er i v a t i v e e^{t} sin (2 t)

\int_{1}^{2} x 2^{(x^{2})} d x

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

\int_{0}^{\infty} x^{- 3} d x

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} - 40 y = 1 - 4 t