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Beliebt Rechnen >

inverselaplace (10)/(s^2+s+10)

Lösung

inverse laplace transformation

\frac{10}{s ^{2} + s + 10}

Lösung

\frac{20}{39} e^{- \frac{t}{2}} sin (\frac{39 t}{2})

Schritte zur Lösung

L^{- 1} {\frac{10}{s ^{2} + s + 10}}

Schreibe

\frac{10}{s ^{2} + s + 10}

um:

10 \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{39}{4}}

= L^{- 1} {10 \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{39}{4}}}

Wende die konstante Multiplikationseigenschaft der inversen Laplace-Transformation an:

Für Funktionen

f (t)

und Konstante

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= 10 L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{39}{4}}}

L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{39}{4}}} : e^{- \frac{t}{2}} \frac{2}{39} sin (\frac{39 t}{2})

= 10 e^{- \frac{t}{2}} \frac{2}{39} sin (\frac{39 t}{2})

Fasse

10 e^{- \frac{t}{2}} \frac{2}{39} sin (\frac{39 t}{2})

zusammen:

\frac{20}{39} e^{- \frac{t}{2}} sin (\frac{39 t}{2})

= \frac{20}{39} e^{- \frac{t}{2}} sin (\frac{39 t}{2})

Beliebte Beispiele

derivative of sqrt(e^{2x)+1}

\frac{d}{d x} (e^{2 x} + 1)

tangent f(x)= 3/(x+1),\at x=7

t an g e n t f (x) = \frac{3}{x + 1}, at x = 7

2y^{''}+8y^'+80y=60cos(8t)

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 60 cos (8 t)

(\partial)/(\partial x)(e^{x+y})

\frac{\partial}{\partial x} (e^{x + y})

sum from n=0 to infinity of (x)^n

n = 0 \sum \infty (x)^{n}