inverselaplace 1/((s^2+4s+5))

Lösung

inverse laplace transformation

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

e^{- 2 t} sin (t)

Schritte zur Lösung

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

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

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

Wende inverse Transformationsregel an: if

L^{- 1} {F (s)} = f (t)

dann

L^{- 1} {F (s - a)} = e^{a t} f (t)

Für

\frac{1}{( s + 2 ) ^{2} + 1} : a = - 2, F (s) = \frac{1}{s ^{2} + 1}

= e^{- 2 t} L^{- 1} {\frac{1}{s ^{2} + 1}}

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

= e^{- 2 t} sin (t)

n = 0 \sum \infty \frac{1}{4 ^{n}}

d er i v a t i v e f (x) = \frac{3 x}{x - 6}

x \to 3 lim (f (x) (x - 1))

\frac{\partial}{\partial y} (x - y)

x \to 0 lim (\frac{( 1 1 ^{x} - 1 )}{x})