inverselaplace ((20)/(10s+1))*1/s

Lösung

inverse laplace transformation

(\frac{20}{10 s + 1}) \cdot \frac{1}{s}

20 H (t) - 20 e^{- \frac{t}{10}}

Schritte zur Lösung

L^{- 1} {(\frac{20}{10 s + 1}) \frac{1}{s}}

Schreibe

\frac{20}{10 s + 1} \frac{1}{s}

um:

\frac{20}{10 s ^{2} + s}

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

Ermittle den Partialbruch von

\frac{20}{10 s ^{2} + s} : \frac{20}{s} - \frac{200}{10 s + 1}

= L^{- 1} {\frac{20}{s} - \frac{200}{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{20}{s}} - L^{- 1} {\frac{200}{10 s + 1}}

L^{- 1} {\frac{20}{s}} : 20 H (t)

L^{- 1} {\frac{200}{10 s + 1}} : 20 e^{- \frac{t}{10}}

= 20 H (t) - 20 e^{- \frac{t}{10}}

t an g e n t f (x) = ln (e^{x^{2}}), at x = 1

\int \frac{2 x + 5}{x ^{2} + 2 x + 5} d x

\int_{1}^{4} (x^{2} - 4 x - 3) d x

\int cos (6 θ) d θ

x \to \frac{1}{5} lim (\frac{x}{( 10 x - 2 )})