inverselaplace 1/(s^2-6s+18)

Lösung

inverse laplace transformation

\frac{1}{s ^{2} - 6 s + 18}

e^{3 t} \frac{1}{3} sin (3 t)

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{2} - 6 s + 18}}

\frac{1}{s ^{2} - 6 s + 18} = \frac{1}{( s - 3 ) ^{2} + 9}

= L^{- 1} {\frac{1}{( s - 3 ) ^{2} + 9}}

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 - 3 ) ^{2} + 9} : a = 3, F (s) = \frac{1}{s ^{2} + 9}

= e^{3 t} L^{- 1} {\frac{1}{s ^{2} + 9}}

L^{- 1} {\frac{1}{s ^{2} + 9}} : \frac{1}{3} sin (3 t)

= e^{3 t} \frac{1}{3} sin (3 t)

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