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Beliebt Rechnen >

inverselaplace (2e^{-2s})/(5s^2+45)

Lösung

inverse laplace transformation

\frac{2 e ^{- 2 s}}{5 s ^{2} + 45}

Lösung

H (t - 2) \frac{2}{15} sin (3 (t - 2))

Schritte zur Lösung

L^{- 1} {\frac{2 e ^{- 2 s}}{5 s ^{2} + 45}}

Wende inverse Transformationsregel an: if

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

dann

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

Wo

H (t)

Heaviside (Schritt) Funktion ist

= H (t - 2) L^{- 1} {\frac{2}{5 s ^{2} + 45}} (t - 2)

L^{- 1} {\frac{2}{5 s ^{2} + 45}} : \frac{2}{15} sin (3 t)

= H (t - 2) \frac{2}{15} sin (3 (t - 2))

Beliebte Beispiele

derivative y=e^{((3x^2))/2}

d er i v a t i v e y = e^{\frac{( 3 x ^{2} )}{2}}

derivative (x^{63}-1)/(x-1)

d er i v a t i v e \frac{x ^{63} - 1}{x - 1}

derivative of 3/(3x+1)

\frac{d}{d x} (\frac{3}{3 x + 1})

limit as x approaches infinity of 1/e

x \to \infty lim (\frac{1}{e})

derivative y=e^{x^3}

d er i v a t i v e y = e^{x^{3}}