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

Lösung

inverse laplace transformation

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

- 3 H (t) - 2 t + 3 e^{t} - e^{t} t

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

L^{- 1} {\frac{3}{s}} : 3 H (t)

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

L^{- 1} {\frac{3}{s - 1}} : 3 e^{t}

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

= - 3 H (t) - 2 t + 3 e^{t} - e^{t} t

sin (t) cos (x) d t - sin (x) cos (t) d x = 0

a re a y^{2} = x, x = 2 y

n = 1 \sum \infty \frac{n}{n + n ^{2}}

l a pl a ce t r an s f or m 2 e^{t}

a re a y = 4 x, y = 2 x