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Beliebt Rechnen >

laplacetransform t-2+e^{-5t}-sin(5t)+1.5

Lösung

laplace transformation

t - 2 + e^{- 5 t} - sin (5 t) + 1.5

Lösung

\frac{1}{s ^{2}} - \frac{0.5}{s} + \frac{1}{s + 5} - \frac{5}{s ^{2} + 25}

Schritte zur Lösung

L {t - 2 + e^{- 5 t} - sin (5 t) + 1.5}

Verwende die lineare Eigenschaft der Laplace-Transformation:

Für Funktionen

f (t), g (t)

und Konstanten

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {t} - L {2} + L {e^{- 5 t}} - L {sin (5 t)} + L {1.5}

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

L {2} : \frac{2}{s}

L {e^{- 5 t}} : \frac{1}{s + 5}

L {sin (5 t)} : \frac{5}{s ^{2} + 25}

L {1.5} : \frac{1.5}{s}

= \frac{1}{s ^{2}} - \frac{2}{s} + \frac{1}{s + 5} - \frac{5}{s ^{2} + 25} + \frac{1.5}{s}

Fasse

\frac{1}{s ^{2}} - \frac{2}{s} + \frac{1}{s + 5} - \frac{5}{s ^{2} + 25} + \frac{1.5}{s}

zusammen:

\frac{1}{s ^{2}} - \frac{0.5}{s} + \frac{1}{s + 5} - \frac{5}{s ^{2} + 25}

= \frac{1}{s ^{2}} - \frac{0.5}{s} + \frac{1}{s + 5} - \frac{5}{s ^{2} + 25}

Beliebte Beispiele

taylor xsin(2x),0

t a y l or x sin (2 x), 0

tangent f(x)=5-2x^2,(-2,-3)

t an g e n t f (x) = 5 - 2 x^{2}, (- 2, - 3)

limit as x approaches infinity of-2x+1/x

x \to \infty lim (- 2 x + \frac{1}{x})

laplacetransform-3cos(5t)

l a pl a ce t r an s f or m - 3 cos (5 t)

derivative (x-5)/(x-2)

d er i v a t i v e \frac{x - 5}{x - 2}