inverselaplace (25)/(s^3(s^2+4s+5))

Lösung

inverse laplace transformation

\frac{25}{s ^{3} ( s ^{2} + 4 s + 5 )}

\frac{11}{5} H (t) - 4 t + \frac{5 t ^{2}}{2} - \frac{11}{5} e^{- 2 t} cos (t) - \frac{2}{5} e^{- 2 t} sin (t)

Schritte zur Lösung

L^{- 1} {\frac{25}{s ^{3} ( s ^{2} + 4 s + 5 )}}

Ermittle den Partialbruch von

\frac{25}{s ^{3} ( s ^{2} + 4 s + 5 )} : \frac{11}{5 s} - \frac{4}{s ^{2}} + \frac{5}{s ^{3}} + \frac{- 11 s - 24}{5 ( s ^{2} + 4 s + 5 )}

= L^{- 1} {\frac{11}{5 s} - \frac{4}{s ^{2}} + \frac{5}{s ^{3}} + \frac{- 11 s - 24}{5 ( s ^{2} + 4 s + 5 )}}

Schreibe

\frac{- 11 s - 24}{5 ( s ^{2} + 4 s + 5 )}

um:

- \frac{11}{5} \cdot \frac{s + 2}{( s + 2 ) ^{2} + 1} - \frac{2}{5} \cdot \frac{1}{( s + 2 ) ^{2} + 1}

= L^{- 1} {\frac{11}{5 s} - \frac{4}{s ^{2}} + \frac{5}{s ^{3}} - \frac{11}{5} \cdot \frac{s + 2}{( s + 2 ) ^{2} + 1} - \frac{2}{5} \cdot \frac{1}{( s + 2 ) ^{2} + 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{11}{5 s}} - L^{- 1} {\frac{4}{s ^{2}}} + L^{- 1} {\frac{5}{s ^{3}}} - \frac{11}{5} L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 1}} - \frac{2}{5} L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 1}}

L^{- 1} {\frac{11}{5 s}} : \frac{11}{5} H (t)

L^{- 1} {\frac{4}{s ^{2}}} : 4 t

L^{- 1} {\frac{5}{s ^{3}}} : \frac{5 t ^{2}}{2}

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

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

= \frac{11}{5} H (t) - 4 t + \frac{5 t ^{2}}{2} - \frac{11}{5} e^{- 2 t} cos (t) - \frac{2}{5} e^{- 2 t} sin (t)

\int 3 x^{2} - x^{3} d x

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\int (x^{6} - 3 x^{2})^{4} (x^{5} - x) d x