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

Lösung

inverse laplace transformation

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

e^{- t} \cdot 2 cos (2 t)

Schritte zur Lösung

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

Schreibe

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

um:

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

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

Wende inverse Transformationsregel an: if

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

dann

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

Für

\frac{2 ( s + 1 )}{( s + 1 ) ^{2} + 4} : a = - 1, F (s) = \frac{2 s}{s ^{2} + 4}

= e^{- t} L^{- 1} {\frac{2 s}{s ^{2} + 4}}

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

= e^{- t} \cdot 2 cos (2 t)

\int e^{- x^{2}} \cdot x d x

k = 1 \sum \infty \frac{k ^{2}}{k !}

\frac{d}{d x} (\frac{x ^{3} + 2 x ^{2} - 1}{x + 5})

\int \frac{1}{2} x^{2} \cdot \frac{1}{x} d x

x \to \infty lim ((\frac{x ^{p}}{e ^{x}})^{\frac{1}{n}})