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Beliebt Rechnen >

inverselaplace (e^{-2s})/(s^2+16)

Lösung

inverse laplace transformation

\frac{e ^{- 2 s}}{s ^{2} + 16}

Lösung

H (t - 2) \frac{1}{4} sin (4 (t - 2))

Schritte zur Lösung

L^{- 1} {\frac{e ^{- 2 s}}{s ^{2} + 16}}

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{1}{s ^{2} + 16}} (t - 2)

L^{- 1} {\frac{1}{s ^{2} + 16}} : \frac{1}{4} sin (4 t)

= H (t - 2) \frac{1}{4} sin (4 (t - 2))

Beliebte Beispiele

integral from-1 to 1 of 1/2 (9-9x^2)^2

\int_{- 1}^{1} \frac{1}{2} (9 - 9 x^{2})^{2} d x

derivative of 500sqrt(x)-0.5x-500

\frac{d}{d x} (500 x - 0.5 x - 500)

integral of (x+1)(3x-2)

\int (x + 1) (3 x - 2) d x

derivative of (sqrt(x)/(x+1))

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

integral of 8sin^2(x-pi/(12))

\int 8 sin^{2} (x - \frac{π}{12}) d x