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Beliebt Rechnen >

inverselaplace 6/s+(32)/(s^2+16)

Lösung

inverse laplace transformation

\frac{6}{s} + \frac{32}{s ^{2} + 16}

Lösung

6 H (t) + 8 sin (4 t)

Schritte zur Lösung

L^{- 1} {\frac{6}{s} + \frac{32}{s ^{2} + 16}}

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{6}{s}} + L^{- 1} {\frac{32}{s ^{2} + 16}}

L^{- 1} {\frac{6}{s}} : 6 H (t)

L^{- 1} {\frac{32}{s ^{2} + 16}} : 8 sin (4 t)

= 6 H (t) + 8 sin (4 t)

Beliebte Beispiele

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

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