inverselaplace 1/(s^2+4s+13)

Lösung

inverse laplace transformation

\frac{1}{s ^{2} + 4 s + 13}

e^{- 2 t} \frac{1}{3} sin (3 t)

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{2} + 4 s + 13}}

\frac{1}{s ^{2} + 4 s + 13} = \frac{1}{( s + 2 ) ^{2} + 9}

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

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

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

= e^{- 2 t} \frac{1}{3} sin (3 t)