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Beliebt Rechnen >

inverselaplace 5/(s(s^2+5^2))

Lösung

inverse laplace transformation

\frac{5}{s ( s ^{2} + 5 ^{2} )}

Lösung

\frac{1}{5} H (t) - \frac{1}{5} cos (5 t)

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{5}{s ( s ^{2} + 5 ^{2} )} : \frac{1}{5 s} - \frac{s}{5 ( s ^{2} + 25 )}

= L^{- 1} {\frac{1}{5 s} - \frac{s}{5 ( s ^{2} + 25 )}}

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}{5 s}} - L^{- 1} {\frac{s}{5 ( s ^{2} + 25 )}}

L^{- 1} {\frac{1}{5 s}} : \frac{1}{5} H (t)

L^{- 1} {\frac{s}{5 ( s ^{2} + 25 )}} : \frac{1}{5} cos (5 t)

= \frac{1}{5} H (t) - \frac{1}{5} cos (5 t)

Beliebte Beispiele

laplacetransform f(t)=2(1-e^{-5t})

l a pl a ce t r an s f or m f (t) = 2 (1 - e^{- 5 t})

derivative of ln^2(2*2^x)

\frac{d}{d x} (ln^{2} (2) \cdot 2^{x})

derivative y=ln(x-4)

d er i v a t i v e y = ln (x - 4)

laplacetransform-5e^{2t}

l a pl a ce t r an s f or m - 5 e^{2 t}

sum from n=1 to infinity of n!(x-1)^n

n = 1 \sum \infty n! (x - 1)^{n}