inverselaplace (0.2)/((10s^3+s^2))

Lösung

inverse laplace transformation

\frac{0.2}{( 10 s ^{3} + s ^{2} )}

- 2 H (t) + 0.2 t + 2 e^{- \frac{t}{10}}

Schritte zur Lösung

L^{- 1} {\frac{0.2}{( 10 s ^{3} + s ^{2} )}}

Ermittle den Partialbruch von

\frac{0.2}{10 s ^{3} + s ^{2}} : - \frac{2}{s} + \frac{0.2}{s ^{2}} + \frac{20}{10 s + 1}

= L^{- 1} {- \frac{2}{s} + \frac{0.2}{s ^{2}} + \frac{20}{10 s + 1}}

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} {\frac{2}{s}} + L^{- 1} {\frac{0.2}{s ^{2}}} + L^{- 1} {\frac{20}{10 s + 1}}

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

L^{- 1} {\frac{0.2}{s ^{2}}} : 0.2 t

L^{- 1} {\frac{20}{10 s + 1}} : 2 e^{- \frac{t}{10}}

= - 2 H (t) + 0.2 t + 2 e^{- \frac{t}{10}}

\int (tan (x) sec^{2} (x)) d x

l a pl a ce t r an s f or m sin (\frac{1}{2} t) \cdot t^{2}

\int - cos (x) 2 x d x

y^{^{'}} = 4 y

\frac{d}{d x} (A cos (2 x))