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Beliebt Rechnen >

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

Lösung

inverse laplace transformation

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

Lösung

- \frac{1}{4} H (t) + \frac{t}{2} + \frac{5}{4} e^{- 2 t}

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{s ^{2} + 1}{s ^{2} ( s + 2 )} : - \frac{1}{4 s} + \frac{1}{2 s ^{2}} + \frac{5}{4 ( s + 2 )}

= L^{- 1} {- \frac{1}{4 s} + \frac{1}{2 s ^{2}} + \frac{5}{4 ( s + 2 )}}

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{1}{4 s}} + L^{- 1} {\frac{1}{2 s ^{2}}} + L^{- 1} {\frac{5}{4 ( s + 2 )}}

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

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

L^{- 1} {\frac{5}{4 ( s + 2 )}} : \frac{5}{4} e^{- 2 t}

= - \frac{1}{4} H (t) + \frac{t}{2} + \frac{5}{4} e^{- 2 t}

Beliebte Beispiele

(\partial)/(\partial y)(e^{-x}sin(yz))

\frac{\partial}{\partial y} (e^{- x} sin (yz))

derivative of (x^2/((x^2-1)))

\frac{d}{d x} (\frac{x ^{2}}{( x ^{2} - 1 )})

tangent f(x)=4-3ln(x),\at x=1

t an g e n t f (x) = 4 - 3 ln (x), at x = 1

derivative y=6+4sqrt(x)csc(x)

d er i v a t i v e y = 6 + 4 x csc (x)

sum from n=1 to infinity of 3/((4n^2-1))

n = 1 \sum \infty \frac{3}{( 4 n ^{2} - 1 )}