inverselaplace s/(s^2+64)

Lösung

inverse laplace transformation

\frac{s}{s ^{2} + 64}

cos (8 t)

Schritte zur Lösung

L^{- 1} {\frac{s}{s ^{2} + 64}}

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{s}{s ^{2} + 64} : F (s) = \frac{1}{s ^{2} + 64}

= \frac{d}{d t} (L^{- 1} {\frac{1}{s ^{2} + 64}}) + L^{- 1} {\frac{1}{s ^{2} + 64}} (0)

L^{- 1} {\frac{1}{s ^{2} + 64}} : \frac{1}{8} sin (8 t)

\frac{d}{d t} (\frac{1}{8} sin (8 t)) = cos (8 t)

Evaluiere

\frac{1}{8} sin (8 t)

am Punkt

0 = 0

= cos (8 t)

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