inverselaplace (s-1)/((s+1)(s+2)s)

Lösung

inverse laplace transformation

\frac{s - 1}{( s + 1 ) ( s + 2 ) s}

2 e^{- t} - \frac{3}{2} e^{- 2 t} - \frac{1}{2} H (t)

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{s - 1}{( s + 1 ) ( s + 2 ) s} : \frac{2}{s + 1} - \frac{3}{2 ( s + 2 )} - \frac{1}{2 s}

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

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

L^{- 1} {\frac{3}{2 ( s + 2 )}} : \frac{3}{2} e^{- 2 t}

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

= 2 e^{- t} - \frac{3}{2} e^{- 2 t} - \frac{1}{2} H (t)

f (x) = \frac{5}{x ^{5}}

t a y l or f (x) = sin (x), 0

a re a x^{2} + x + 3, [1, 3]

\int \frac{2}{3} cos^{2} (3 x) d x

ma c l a u r in \frac{1}{1 + 4 x ^{2}}