inverselaplace (50)/(100s+0.01s^2)

Lösung

inverse laplace transformation

\frac{50}{100 s + 0.01 s ^{2}}

\frac{1}{2} H (t) - 0.5 e^{- 10000 t}

Schritte zur Lösung

L^{- 1} {\frac{50}{100 s + 0.01 s ^{2}}}

Ermittle den Partialbruch von

\frac{50}{100 s + 0.01 s ^{2}} : \frac{1}{2 s} - \frac{0.005}{0.01 s + 100}

= L^{- 1} {\frac{1}{2 s} - \frac{0.005}{0.01 s + 100}}

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{1}{2 s}} - L^{- 1} {\frac{0.005}{0.01 s + 100}}

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

L^{- 1} {\frac{0.005}{0.01 s + 100}} : 0.5 e^{- 10000 t}

= \frac{1}{2} H (t) - 0.5 e^{- 10000 t}

im pl i c i t \frac{d y}{d x}, x^{2} = \frac{( 3 x + 5 y )}{3 x - 5 y}

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

\int csc (7 x) d x

\int_{0}^{\frac{π}{2}} 4 cos (\frac{x}{2}) d x

v \to 4 lim (\frac{v ^{2} - 6 v + 8}{2 v ^{2} - 8 v})