inverselaplace (e^{-2s})/(s^2-3s-4)

Lösung

inverse laplace transformation

\frac{e ^{- 2 s}}{s ^{2} - 3 s - 4}

H (t - 2) (- \frac{1}{5} e^{- t + 2} + \frac{1}{5} e^{4 (t - 2)})

Schritte zur Lösung

L^{- 1} {\frac{e ^{- 2 s}}{s ^{2} - 3 s - 4}}

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)

H (t)

Heaviside (Schritt) Funktion ist

= H (t - 2) L^{- 1} {\frac{1}{s ^{2} - 3 s - 4}} (t - 2)

L^{- 1} {\frac{1}{s ^{2} - 3 s - 4}} : - \frac{1}{5} e^{- t} + \frac{1}{5} e^{4 t}

= H (t - 2) (- \frac{1}{5} e^{- (t - 2)} + \frac{1}{5} e^{4 (t - 2)})

= H (t - 2) (- \frac{1}{5} e^{- t + 2} + \frac{1}{5} e^{4 (t - 2)})

x \to 0 lim (1 - \frac{1}{x})

t an g e n t (x - \frac{6}{x}), at m = 0

d er i v a t i v e x^{8} e^{7 x}

t an g e n t f (x) = \frac{x - 1}{x - 4}, at x = 5

\int cos^{2} (x) + cos (x) d x