inverselaplace (28s)/((s^2+11))

Lösung

inverse laplace transformation

\frac{28 s}{( s ^{2} + 11 )}

28 cos (11 t)

Schritte zur Lösung

L^{- 1} {\frac{28 s}{( s ^{2} + 11 )}}

Wende inverse Transformationsregel an: if

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

dann

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

Für

\frac{28 s}{s ^{2} + 11} : F (s) = \frac{28}{s ^{2} + 11}

= \frac{d}{d t} (L^{- 1} {\frac{28}{s ^{2} + 11}}) + L^{- 1} {\frac{28}{s ^{2} + 11}} (0)

L^{- 1} {\frac{28}{s ^{2} + 11}} : \frac{28}{11} sin (11 t)

\frac{d}{d t} (\frac{28}{11} sin (11 t)) = 28 cos (11 t)

Evaluiere

\frac{28}{11} sin (11 t)

am Punkt

0 = 0

= 28 cos (11 t)

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