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

Lösung

inverse laplace transformation

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

e^{\frac{t}{2}} \frac{2}{3} sin (\frac{3 t}{2})

Schritte zur Lösung

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

\frac{1}{s ^{2} - s + 1} = \frac{1}{( s - \frac{1}{2} ) ^{2} + \frac{3}{4}}

= L^{- 1} {\frac{1}{( s - \frac{1}{2} ) ^{2} + \frac{3}{4}}}

Wende inverse Transformationsregel an: if

L^{- 1} {F (s)} = f (t)

dann

L^{- 1} {F (s - a)} = e^{a t} f (t)

Für

\frac{1}{( s - \frac{1}{2} ) ^{2} + \frac{3}{4}} : a = \frac{1}{2}, F (s) = \frac{1}{s ^{2} + \frac{3}{4}}

= e^{\frac{1}{2} t} L^{- 1} {\frac{1}{s ^{2} + \frac{3}{4}}}

L^{- 1} {\frac{1}{s ^{2} + \frac{3}{4}}} : \frac{2}{3} sin (\frac{3 t}{2})

= e^{\frac{t}{2}} \frac{2}{3} sin (\frac{3 t}{2})

n = 0 \sum \infty 7 (\frac{2}{3})^{n}

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