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

Lösung

inverse laplace transformation

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

Schreibe

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

um:

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

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

L^{- 1} {\frac{1}{2 s}} : \frac{1}{2} H (t)

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

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

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

y^{^{'}} + y + \frac{1}{x} y = \frac{e ^{- x}}{x}

d er i v a t i v e f (x) = x^{\frac{9}{2}}

3 y^{^{''}} - 4 y^{^{'}} - 15 y = 0

d er i v a t i v e \frac{2}{3} (x^{2} + 1)^{\frac{3}{2}}

\frac{\partial}{\partial x} (arcsin (5 x yz))