inverselaplace 1/((s-2)^2-4)

Lösung

inverse laplace transformation

\frac{1}{( s - 2 ) ^{2} - 4}

- \frac{1}{4} H (t) + \frac{1}{4} e^{4 t}

Schritte zur Lösung

L^{- 1} {\frac{1}{( s - 2 ) ^{2} - 4}}

Ermittle den Partialbruch von

\frac{1}{( s - 2 ) ^{2} - 4} : - \frac{1}{4 s} + \frac{1}{4 ( s - 4 )}

= L^{- 1} {- \frac{1}{4 s} + \frac{1}{4 ( s - 4 )}}

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}{4 s}} + L^{- 1} {\frac{1}{4 ( s - 4 )}}

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

L^{- 1} {\frac{1}{4 ( s - 4 )}} : \frac{1}{4} e^{4 t}

= - \frac{1}{4} H (t) + \frac{1}{4} e^{4 t}

y^{^{''''}} - 2 y^{^{''}} + y = 0

\frac{d}{d x} (3 cos^{2} (x) - 6 sin (x))

\frac{\partial}{\partial x} (arcsin (3 x))

\frac{d ^{2}}{d x ^{2}} (\frac{x}{x - 1})

x \to 0 + lim (3 x)