inverselaplace (s+1)/(s(s^2+2s+5))

Lösung

inverse laplace transformation

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

Schreibe

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

um:

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

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

L^{- 1} {\frac{1}{5 s}} : \frac{1}{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{1}{5} H (t) - \frac{1}{5} e^{- t} cos (2 t) + \frac{4}{5} e^{- t} \frac{1}{2} sin (2 t)

Fasse

\frac{1}{5} H (t) - \frac{1}{5} e^{- t} cos (2 t) + \frac{4}{5} e^{- t} \frac{1}{2} sin (2 t)

zusammen:

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

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

t an g e n t \frac{x - 1}{2 - x}

\int_{25}^{x} 1 + 9 t d t

\frac{d}{d x} (6 x^{2} - 6 x)

\int_{1}^{3} 15 r^{3} ln (r) d r

y^{^{''}} + 2 y^{^{'}} + y = sin (x) + 3 cos (2 x)