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Beliebt Rechnen >

inverselaplace (3(s+1)*e^{-2s})/(s(s+1)(2s+1))

Lösung

inverse laplace transformation

\frac{3 ( s + 1 ) \cdot e ^{- 2 s}}{s ( s + 1 ) ( 2 s + 1 )}

Lösung

H (t - 2) (3 - 3 e^{- \frac{t - 2}{2}})

Schritte zur Lösung

L^{- 1} {\frac{3 e ^{- 2 s} ( s + 1 )}{s ( s + 1 ) ( 2 s + 1 )}}

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{3 ( s + 1 )}{s ( s + 1 ) ( 2 s + 1 )}} (t - 2)

L^{- 1} {\frac{3 ( s + 1 )}{s ( s + 1 ) ( 2 s + 1 )}} : 3 H (t) - 3 e^{- \frac{t}{2}}

= H (t - 2) (3 H (t - 2) - 3 e^{- \frac{t - 2}{2}})

H (t - 2) = {01 t < 2 t \geq 2

= {0 3 - 3 e^{- \frac{t - 2}{2}} t < 2 t \geq 2

= H (t - 2) (3 - 3 e^{- \frac{t - 2}{2}})

Beliebte Beispiele

derivative (2x+1)/(2x-1)

d er i v a t i v e \frac{2 x + 1}{2 x - 1}

derivative f(x)=-x^3

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

integral from-3 to 5 of 2y-(y^2-15)

\int_{- 3}^{5} 2 y - (y^{2} - 15) d y

(\partial)/(\partial x)(3xarctan(x/y))

\frac{\partial}{\partial x} (3 x arctan (\frac{x}{y}))

derivative 4e^x+x

d er i v a t i v e 4 e^{x} + x