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Beliebt Rechnen >

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

Lösung

inverse laplace transformation

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

Lösung

e^{- \frac{t}{4}} cos (\frac{7 t}{4}) + \frac{e ^{- \frac{t}{4}} sin ( \frac{7 t}{4} )}{7}

Schritte zur Lösung

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

Schreibe

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

um:

\frac{s + \frac{1}{4}}{( s + \frac{1}{4} ) ^{2} + \frac{7}{16}} + \frac{1}{4} \cdot \frac{1}{( s + \frac{1}{4} ) ^{2} + \frac{7}{16}}

= L^{- 1} {\frac{s + \frac{1}{4}}{( s + \frac{1}{4} ) ^{2} + \frac{7}{16}} + \frac{1}{4} \cdot \frac{1}{( s + \frac{1}{4} ) ^{2} + \frac{7}{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{s + \frac{1}{4}}{( s + \frac{1}{4} ) ^{2} + \frac{7}{16}}} + \frac{1}{4} L^{- 1} {\frac{1}{( s + \frac{1}{4} ) ^{2} + \frac{7}{16}}}

L^{- 1} {\frac{s + \frac{1}{4}}{( s + \frac{1}{4} ) ^{2} + \frac{7}{16}}} : e^{- \frac{t}{4}} cos (\frac{7 t}{4})

L^{- 1} {\frac{1}{( s + \frac{1}{4} ) ^{2} + \frac{7}{16}}} : e^{- \frac{t}{4}} \frac{4}{7} sin (\frac{7 t}{4})

= e^{- \frac{t}{4}} cos (\frac{7 t}{4}) + \frac{1}{4} e^{- \frac{t}{4}} \frac{4}{7} sin (\frac{7 t}{4})

Fasse

e^{- \frac{t}{4}} cos (\frac{7 t}{4}) + \frac{1}{4} e^{- \frac{t}{4}} \frac{4}{7} sin (\frac{7 t}{4})

zusammen:

e^{- \frac{t}{4}} cos (\frac{7 t}{4}) + \frac{e ^{- \frac{t}{4}} sin ( \frac{7 t}{4} )}{7}

= e^{- \frac{t}{4}} cos (\frac{7 t}{4}) + \frac{e ^{- \frac{t}{4}} sin ( \frac{7 t}{4} )}{7}

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