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Beliebt Rechnen >

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

Lösung

inverse laplace transformation

\frac{1}{s ( s ^{2} + 1 )}

Lösung

H (t) - cos (t)

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{s ( s ^{2} + 1 )} : \frac{1}{s} - \frac{s}{s ^{2} + 1}

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

Wende die lineare Eigenschaftder inversen Laplace-Transformation an:

Für Funktionen

f (s), g (s)

und Konstanten

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {\frac{1}{s}} : H (t)

L^{- 1} {\frac{s}{s ^{2} + 1}} : cos (t)

= H (t) - cos (t)

Beliebte Beispiele

limit as x approaches-2 of 4x^2

x \to - 2 lim (4 x^{2})

integral of (0.7y^3+10+2y^{-3})

\int (0.7 y^{3} + 10 + 2 y^{- 3}) d y

derivative f(x)=x^2-2x-3

d er i v a t i v e f (x) = x^{2} - 2 x - 3

derivative of sin(cot(x))

\frac{d}{d x} (sin (cot (x)))

tangent y=e^xcos(x)

t an g e n t y = e^{x} cos (x)