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Beliebt Rechnen >

inverselaplace s/(s^2+s+25)

Lösung

inverse laplace transformation

\frac{s}{s ^{2} + s + 25}

Lösung

e^{- \frac{t}{2}} cos (\frac{3 11 t}{2}) - \frac{e ^{- \frac{t}{2}} sin ( \frac{3 11 t}{2} )}{3 11}

Schritte zur Lösung

L^{- 1} {\frac{s}{s ^{2} + s + 25}}

Schreibe

\frac{s}{s ^{2} + s + 25}

um:

\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{99}{4}} - \frac{1}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{99}{4}}

= L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{99}{4}} - \frac{1}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{99}{4}}}

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}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{99}{4}}} - \frac{1}{2} L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{99}{4}}}

L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{99}{4}}} : e^{- \frac{t}{2}} cos (\frac{3 11 t}{2})

L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{99}{4}}} : e^{- \frac{t}{2}} \frac{2}{3 11} sin (\frac{3 11 t}{2})

= e^{- \frac{t}{2}} cos (\frac{3 11 t}{2}) - \frac{1}{2} e^{- \frac{t}{2}} \frac{2}{3 11} sin (\frac{3 11 t}{2})

Fasse

e^{- \frac{t}{2}} cos (\frac{3 11 t}{2}) - \frac{1}{2} e^{- \frac{t}{2}} \frac{2}{3 11} sin (\frac{3 11 t}{2})

zusammen:

e^{- \frac{t}{2}} cos (\frac{3 11 t}{2}) - \frac{e ^{- \frac{t}{2}} sin ( \frac{3 11 t}{2} )}{3 11}

= e^{- \frac{t}{2}} cos (\frac{3 11 t}{2}) - \frac{e ^{- \frac{t}{2}} sin ( \frac{3 11 t}{2} )}{3 11}

Beliebte Beispiele

derivative y=2x-7

d er i v a t i v e y = 2 x - 7

derivative of asin(wx)

\frac{d}{d x} (a sin (w x))

limit as x approaches-infinity of (11x+6)/(4-6x)

x \to - \infty lim (\frac{11 x + 6}{4 - 6 x})

fläche x=e^{2y},y=0,x=e^4

a re a x = e^{2 y}, y = 0, x = e^{4}

fläche y=x^2,y=5

a re a y = x^{2}, y = 5