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Beliebt Rechnen >

inverselaplace (e^{-4s})/s

Lösung

inverse laplace transformation

\frac{e ^{- 4 s}}{s}

Lösung

H (t - 4)

Schritte zur Lösung

L^{- 1} {\frac{e ^{- 4 s}}{s}}

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 - 4) L^{- 1} {\frac{1}{s}} (t - 4)

L^{- 1} {\frac{1}{s}} : H (t)

= H (t - 4) H (t - 4)

= H (t - 4)

Beliebte Beispiele

integral from 0 to 5/4 of 1/(25+16x^2)

\int_{0}^{\frac{5}{4}} \frac{1}{25 + 16 x ^{2}} d x

derivative of yln(x)

\frac{d}{d x} (y ln (x))

tangent 6cos(t),\at t=-pi/4

t an g e n t 6 cos (t), at t = - \frac{π}{4}

inverselaplace 2s+1

in v erse l a pl a ce 2 s + 1

tangent f(x)=x^4-4x^3+2x^2-6,\at x=1

t an g e n t f (x) = x^{4} - 4 x^{3} + 2 x^{2} - 6, at x = 1