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Beliebt Rechnen >

inverselaplace (10*e^{-s})/(5s+1)

Lösung

inverse laplace transformation

\frac{10 \cdot e ^{- s}}{5 s + 1}

Lösung

H (t - 1) \cdot 2 e^{- \frac{t - 1}{5}}

Schritte zur Lösung

L^{- 1} {\frac{10 e ^{- s}}{5 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 - 1) L^{- 1} {\frac{10}{5 s + 1}} (t - 1)

L^{- 1} {\frac{10}{5 s + 1}} : 2 e^{- \frac{t}{5}}

= H (t - 1) \cdot 2 e^{- \frac{t - 1}{5}}

Beliebte Beispiele

(\partial)/(\partial x)(x^{2y}+4y)

\frac{\partial}{\partial x} (x^{2 y} + 4 y)

integral of (cot^2(x)+cot^4(x))

\int (cot^{2} (x) + cot^{4} (x)) d x

derivative of ln((x+2/(x-2)))

\frac{d}{d x} (ln (\frac{x + 2}{x - 2}))

tangent f(x)=sqrt(5x),\at x=5

t an g e n t f (x) = 5 x, at x = 5

integral of tan(5x)sec^2(5x)

\int tan (5 x) sec^{2} (5 x) d x