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Beliebt Rechnen >

inverselaplace (s+2)/(s^2+4s+5)

Lösung

inverse laplace transformation

\frac{s + 2}{s ^{2} + 4 s + 5}

Lösung

e^{- 2 t} cos (t)

Schritte zur Lösung

L^{- 1} {\frac{s + 2}{s ^{2} + 4 s + 5}}

\frac{s + 2}{s ^{2} + 4 s + 5} = \frac{s + 2}{( s + 2 ) ^{2} + 1}

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

Wende inverse Transformationsregel an: if

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

dann

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

Für

\frac{s + 2}{( s + 2 ) ^{2} + 1} : a = - 2, F (s) = \frac{s}{s ^{2} + 1}

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

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

= e^{- 2 t} cos (t)

Beliebte Beispiele

y^{''}+25y=tsin(4t)

y^{^{''}} + 25 y = t sin (4 t)

(\partial)/(\partial w)(w/u)

\frac{\partial}{\partial w} (\frac{w}{u})

limit as x approaches 0 of (e^x-x-1)/x

x \to 0 lim (\frac{e ^{x} - x - 1}{x})

derivative of (cot(x)/(1-sin(x)))

\frac{d}{d x} (\frac{cot ( x )}{1 - sin ( x )})

derivative of sin(8xe^{4x})

\frac{d}{d x} (sin (8 x) e^{4 x})