inverselaplace 6/((s^2(s^2+8s+1)))

Lösung

inverse laplace transformation

\frac{6}{( s ^{2} ( s ^{2} + 8 s + 1 ) )}

- 48 H (t) + 6 t + 48 e^{- 4 t} cosh (15 t) + \frac{62 3}{5} e^{- 4 t} sinh (15 t)

Schritte zur Lösung

L^{- 1} {\frac{6}{( s ^{2} ( s ^{2} + 8 s + 1 ) )}}

Ermittle den Partialbruch von

\frac{6}{s ^{2} ( s ^{2} + 8 s + 1 )} : - \frac{48}{s} + \frac{6}{s ^{2}} + \frac{48 s + 378}{s ^{2} + 8 s + 1}

= L^{- 1} {- \frac{48}{s} + \frac{6}{s ^{2}} + \frac{48 s + 378}{s ^{2} + 8 s + 1}}

Schreibe

\frac{48 s + 378}{s ^{2} + 8 s + 1}

um:

48 \cdot \frac{s + 4}{( s + 4 ) ^{2} - 15} + 186 \cdot \frac{1}{( s + 4 ) ^{2} - 15}

= L^{- 1} {- \frac{48}{s} + \frac{6}{s ^{2}} + 48 \cdot \frac{s + 4}{( s + 4 ) ^{2} - 15} + 186 \cdot \frac{1}{( s + 4 ) ^{2} - 15}}

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{48}{s}} + L^{- 1} {\frac{6}{s ^{2}}} + 48 L^{- 1} {\frac{s + 4}{( s + 4 ) ^{2} - 15}} + 186 L^{- 1} {\frac{1}{( s + 4 ) ^{2} - 15}}

L^{- 1} {\frac{48}{s}} : 48 H (t)

L^{- 1} {\frac{6}{s ^{2}}} : 6 t

L^{- 1} {\frac{s + 4}{( s + 4 ) ^{2} - 15}} : e^{- 4 t} cosh (15 t)

L^{- 1} {\frac{1}{( s + 4 ) ^{2} - 15}} : e^{- 4 t} \frac{1}{15} sinh (15 t)

= - 48 H (t) + 6 t + 48 e^{- 4 t} cosh (15 t) + 186 e^{- 4 t} \frac{1}{15} sinh (15 t)

Fasse

- 48 H (t) + 6 t + 48 e^{- 4 t} cosh (15 t) + 186 e^{- 4 t} \frac{1}{15} sinh (15 t)

zusammen:

- 48 H (t) + 6 t + 48 e^{- 4 t} cosh (15 t) + \frac{62 3}{5} e^{- 4 t} sinh (15 t)

= - 48 H (t) + 6 t + 48 e^{- 4 t} cosh (15 t) + \frac{62 3}{5} e^{- 4 t} sinh (15 t)

d er i v a t i v e y = ln ∣ x ∣

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