inverselaplace 3+1/s

Lösung

inverse laplace transformation

3 + \frac{1}{s}

3 δ (t) + H (t)

Schritte zur Lösung

L^{- 1} {3 + \frac{1}{s}}

Wende die lineare Eigenschaftder inversen Laplace-Transformation an:

Für Funktionen

f (s), g (s)

und Konstanten

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {3} + L^{- 1} {\frac{1}{s}}

L^{- 1} {3} : 3 δ (t)

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

= 3 δ (t) + H (t)

\int_{o}^{1} 42 x (1 - x)^{5} d x

\frac{d}{d x} (a^{2} + x^{2})

x \to 1 lim (\frac{1 + x}{1 - x ^{2}})

\int_{0}^{- \frac{π}{6}} x cos (3 x) d x

d er i v a t i v e y = 8 x + 7