inverselaplace 1/((s*(s^2+0.4s+4)))

Lösung

inverse laplace transformation

\frac{1}{( s \cdot ( s ^{2} + 0.4 s + 4 ) )}

0.25 H (t) - 0.25 e^{- 0.2 t} cos (1.98997 \dots t) - 0.02512 \dots e^{- 0.2 t} sin (1.98997 \dots t)

Schritte zur Lösung

L^{- 1} {\frac{1}{( s ( s ^{2} + 0.4 s + 4 ) )}}

Ermittle den Partialbruch von

\frac{1}{s ( s ^{2} + 0.4 s + 4 )} : \frac{1}{4 s} + \frac{- 0.25 s - 0.1}{s ^{2} + 0.4 s + 4}

= L^{- 1} {\frac{1}{4 s} + \frac{- 0.25 s - 0.1}{s ^{2} + 0.4 s + 4}}

Schreibe

\frac{- 0.25 s - 0.1}{s ^{2} + 0.4 s + 4}

um:

- 0.25 \cdot \frac{s + 0.2}{( s + 0.2 ) ^{2} + 3.96} - 0.05 \cdot \frac{1}{( s + 0.2 ) ^{2} + 3.96}

= L^{- 1} {\frac{1}{4 s} - 0.25 \cdot \frac{s + 0.2}{( s + 0.2 ) ^{2} + 3.96} - 0.05 \cdot \frac{1}{( s + 0.2 ) ^{2} + 3.96}}

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}{4 s}} - 0.25 L^{- 1} {\frac{s + 0.2}{( s + 0.2 ) ^{2} + 3.96}} - 0.05 L^{- 1} {\frac{1}{( s + 0.2 ) ^{2} + 3.96}}

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

L^{- 1} {\frac{s + 0.2}{( s + 0.2 ) ^{2} + 3.96}} : e^{- 0.2 t} cos (1.98997 \dots t)

L^{- 1} {\frac{1}{( s + 0.2 ) ^{2} + 3.96}} : e^{- 0.2 t} \frac{1}{1.98997 \dots} sin (1.98997 \dots t)

= \frac{1}{4} H (t) - 0.25 e^{- 0.2 t} cos (1.98997 \dots t) - 0.05 e^{- 0.2 t} \frac{1}{1.98997 \dots} sin (1.98997 \dots t)

Vereinfache

\frac{1}{4} H (t) - 0.25 e^{- 0.2 t} cos (1.98997 \dots t) - 0.05 e^{- 0.2 t} \frac{1}{1.98997 \dots} sin (1.98997 \dots t) : 0.25 H (t) - 0.25 e^{- 0.2 t} cos (1.98997 \dots t) - 0.02512 \dots e^{- 0.2 t} sin (1.98997 \dots t)

= 0.25 H (t) - 0.25 e^{- 0.2 t} cos (1.98997 \dots t) - 0.02512 \dots e^{- 0.2 t} sin (1.98997 \dots t)

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