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

Lösung

inverse laplace transformation

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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}{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}} : H (t)

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)

= 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)

Vereinfache

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) - 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)

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