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Beliebt Rechnen >

inverselaplace 1/(s^2+40s+2000)

Lösung

inverse laplace transformation

\frac{1}{s ^{2} + 40 s + 2000}

Lösung

e^{- 20 t} \frac{1}{40} sin (40 t)

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{2} + 40 s + 2000}}

\frac{1}{s ^{2} + 40 s + 2000} = \frac{1}{( s + 20 ) ^{2} + 1600}

= L^{- 1} {\frac{1}{( s + 20 ) ^{2} + 1600}}

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 + 20 ) ^{2} + 1600} : a = - 20, F (s) = \frac{1}{s ^{2} + 1600}

= e^{- 20 t} L^{- 1} {\frac{1}{s ^{2} + 1600}}

L^{- 1} {\frac{1}{s ^{2} + 1600}} : \frac{1}{40} sin (40 t)

= e^{- 20 t} \frac{1}{40} sin (40 t)

Beliebte Beispiele

8t^2y^{''}-16y=11t^2-1

8 t^{2} y^{^{''}} - 16 y = 11 t^{2} - 1

derivative sqrt(cot(3x))

d er i v a t i v e cot (3 x)

tangent f(x)=2x^3+3x^2-12x+6

t an g e n t f (x) = 2 x^{3} + 3 x^{2} - 12 x + 6

integral of (2x^3)/(\sqrt[3]{x^4-5)}

\int \frac{2 x ^{3}}{3 x ^{4} - 5} d x

maclaurin ln(1+x^3)

ma c l a u r in ln (1 + x^{3})