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

Lösung

inverse laplace transformation

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

Schreibe um

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

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

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

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

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

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

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

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

Vereinfache

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

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

\frac{d}{d x} (\frac{x cos ( x ) - sin ( x )}{x ^{2}})

x \to 3 lim (10 (x + 3))

d er i v a t i v e f (x) = 6 x^{2} - 5 x

n = 0 \sum \infty \frac{1}{n ^{\frac{1}{n}}}

d er i v a t i v e 9 arctan (x + 1 + x^{2})