inverselaplace (s^2)/(s^4+2s^2+1)

Lösung

inverse laplace transformation

\frac{s ^{2}}{s ^{4} + 2 s ^{2} + 1}

\frac{t cos ( t ) + sin ( t )}{2}

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

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

Vereinfache

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

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

d er i v a t i v e 2 - 7 x

\int \frac{1}{x ^{2}} - \frac{3}{x ^{4}} d x

l a pl a ce t r an s f or m 2 t e^{3 t} + 4 t^{2} e^{- 5 t}

\frac{d}{d x} (\frac{1}{2 x - 3})

d er i v a t i v e 5 - 4 x