inverselaplace (5e^{-s})/((s^2-2s+1))

Lösung

inverse laplace transformation

\frac{5 e ^{- s}}{( s ^{2} - 2 s + 1 )}

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

Schritte zur Lösung

L^{- 1} {\frac{5 e ^{- s}}{( s ^{2} - 2 s + 1 )}}

Faktorisiere

s^{2} - 2 s + 1 : (s - 1)^{2}

= L^{- 1} {\frac{5 e ^{- s}}{( s - 1 ) ^{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 - 1) L^{- 1} {\frac{5}{( s - 1 ) ^{2}}} (t - 1)

L^{- 1} {\frac{5}{( s - 1 ) ^{2}}} : e^{t} \cdot 5 t

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