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

Lösung

inverse laplace transformation

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

e^{- \frac{t}{2}} \frac{2}{3} sin (\frac{3 t}{2})

Schritte zur Lösung

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

\frac{1}{s ^{2} + s + 1} = \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}}

= L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{3}{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{1}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}} : a = - \frac{1}{2}, F (s) = \frac{1}{s ^{2} + \frac{3}{4}}

= e^{- \frac{1}{2} t} L^{- 1} {\frac{1}{s ^{2} + \frac{3}{4}}}

L^{- 1} {\frac{1}{s ^{2} + \frac{3}{4}}} : \frac{2}{3} sin (\frac{3 t}{2})

= e^{- \frac{t}{2}} \frac{2}{3} sin (\frac{3 t}{2})

x \to 5 lim (4 x - 3)

d er i v a t i v e \frac{6}{4 x + 1}

l o n g d i v i s i o n \frac{x ^{2} - 5 x}{x + 1}

l a pl a ce t r an s f or m (t - 2)^{3} (t - 2)

y^{^{'}} = \frac{( x ^{3} - 2 y )}{x}