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Beliebt Rechnen >

inverselaplace 1/(s^2+0.5s+1)

Lösung

inverse laplace transformation

\frac{1}{s ^{2} + 0.5 s + 1}

Lösung

1.03279 \dots e^{- 0.25 t} sin (0.96824 \dots t)

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{2} + 0.5 s + 1}}

\frac{1}{s ^{2} + 0.5 s + 1} = \frac{1}{( s + 0.25 ) ^{2} + 0.9375}

= L^{- 1} {\frac{1}{( s + 0.25 ) ^{2} + 0.9375}}

Wende inverse Transformationsregel an: if

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

dann

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

Für

\frac{1}{( s + 0.25 ) ^{2} + 0.9375} : a = - 0.25, F (s) = \frac{1}{s ^{2} + 0.9375}

= e^{- 0.25 t} L^{- 1} {\frac{1}{s ^{2} + 0.9375}}

L^{- 1} {\frac{1}{s ^{2} + 0.9375}} : \frac{1}{0.96824 \dots} sin (0.96824 \dots t)

Vereinfache

e^{- 0.25 t} \frac{1}{0.96824 \dots} sin (0.96824 \dots t) : 1.03279 \dots e^{- 0.25 t} sin (0.96824 \dots t)

= 1.03279 \dots e^{- 0.25 t} sin (0.96824 \dots t)

Beliebte Beispiele

limit as x approaches 1 of (x-1)^{ln(x)}

x \to 1 lim ((x - 1)^{l n (x)})

(dy)/(dx)=-10xy

\frac{d y}{d x} = - 10 x y

integral of tsin(t)cos(t)

\int t sin (t) cos (t) d t

limit as x approaches infinity of xy

x \to \infty lim (x y)

(dy)/(dx)+y/x =4x^8y^2

\frac{d y}{d x} + \frac{y}{x} = 4 x^{8} y^{2}