inverselaplace 4/(s(s^2-4s+5))

Lösung

inverse laplace transformation

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

\frac{4}{5} H (t) - \frac{4}{5} e^{2 t} cos (t) + \frac{8}{5} e^{2 t} sin (t)

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{4}{s ( s ^{2} - 4 s + 5 )} : \frac{4}{5 s} + \frac{- 4 s + 16}{5 ( s ^{2} - 4 s + 5 )}

= L^{- 1} {\frac{4}{5 s} + \frac{- 4 s + 16}{5 ( s ^{2} - 4 s + 5 )}}

Schreibe

\frac{- 4 s + 16}{5 ( s ^{2} - 4 s + 5 )}

um:

- \frac{4}{5} \cdot \frac{s - 2}{( s - 2 ) ^{2} + 1} + \frac{8}{5} \cdot \frac{1}{( s - 2 ) ^{2} + 1}

= L^{- 1} {\frac{4}{5 s} - \frac{4}{5} \cdot \frac{s - 2}{( s - 2 ) ^{2} + 1} + \frac{8}{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{4}{5 s}} - \frac{4}{5} L^{- 1} {\frac{s - 2}{( s - 2 ) ^{2} + 1}} + \frac{8}{5} L^{- 1} {\frac{1}{( s - 2 ) ^{2} + 1}}

L^{- 1} {\frac{4}{5 s}} : \frac{4}{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{4}{5} H (t) - \frac{4}{5} e^{2 t} cos (t) + \frac{8}{5} e^{2 t} sin (t)

d er i v a t i v e \frac{- 5 x - 9}{2 x + 8}

ma c l a u r in sin (x), 1.5

d er i v a t i v e (x^{2} + 1)^{5}

\frac{d}{d x} (tan (e^{s i n (x)}))

\int t \cdot e^{4 t} d t