inverselaplace 1/((s^2+0.2s+1)s)

Lösung

inverse laplace transformation

\frac{1}{( s ^{2} + 0.2 s + 1 ) s}

- e^{- 0.1 t} cos (0.99498 \dots t) - 0.10050 \dots e^{- 0.1 t} sin (0.99498 \dots t) + H (t)

Schritte zur Lösung

L^{- 1} {\frac{1}{s ( s ^{2} + 0.2 s + 1 )}}

Ermittle den Partialbruch von

\frac{1}{( s ^{2} + 0.2 s + 1 ) s} : \frac{- s - 0.2}{s ^{2} + 0.2 s + 1} + \frac{1}{s}

= L^{- 1} {\frac{- s - 0.2}{s ^{2} + 0.2 s + 1} + \frac{1}{s}}

Schreibe

\frac{- s - 0.2}{s ^{2} + 0.2 s + 1}

um:

- \frac{s + 0.1}{( s + 0.1 ) ^{2} + 0.99} - 0.1 \cdot \frac{1}{( s + 0.1 ) ^{2} + 0.99}

= L^{- 1} {- \frac{s + 0.1}{( s + 0.1 ) ^{2} + 0.99} - 0.1 \cdot \frac{1}{( s + 0.1 ) ^{2} + 0.99} + \frac{1}{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{s + 0.1}{( s + 0.1 ) ^{2} + 0.99}} - 0.1 L^{- 1} {\frac{1}{( s + 0.1 ) ^{2} + 0.99}} + L^{- 1} {\frac{1}{s}}

L^{- 1} {\frac{s + 0.1}{( s + 0.1 ) ^{2} + 0.99}} : e^{- 0.1 t} cos (0.99498 \dots t)

L^{- 1} {\frac{1}{( s + 0.1 ) ^{2} + 0.99}} : e^{- 0.1 t} \frac{1}{0.99498 \dots} sin (0.99498 \dots t)

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

= - e^{- 0.1 t} cos (0.99498 \dots t) - 0.1 e^{- 0.1 t} \frac{1}{0.99498 \dots} sin (0.99498 \dots t) + H (t)

Vereinfache

- e^{- 0.1 t} cos (0.99498 \dots t) - 0.1 e^{- 0.1 t} \frac{1}{0.99498 \dots} sin (0.99498 \dots t) + H (t) : - e^{- 0.1 t} cos (0.99498 \dots t) - 0.10050 \dots e^{- 0.1 t} sin (0.99498 \dots t) + H (t)

= - e^{- 0.1 t} cos (0.99498 \dots t) - 0.10050 \dots e^{- 0.1 t} sin (0.99498 \dots t) + H (t)

x \to - 3 lim (x^{2} + 3 x + 1)

t a y l or cos (π x) \frac{1}{2}

\int x tan^{2} (x) d x

(x + 2 y) d x + (2 x - 3 y) d y = 0

\frac{d ^{4} y}{d x ^{4}} - 3 y = 0