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

Lösung

inverse laplace transformation

\frac{e ^{- 2 s}}{s ( s - 2 )}

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

Schritte zur Lösung

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

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 ( s - 2 )}} (t - 2)

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

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

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

= {0 - \frac{1}{2} + \frac{1}{2} e^{2 (t - 2)} t < 2 t \geq 2

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

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