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

Lösung

inverse laplace transformation

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

\frac{4}{5} H (t) - \frac{4}{5} e^{- \frac{t}{2}} cos (t) - \frac{2}{5} e^{- \frac{t}{2}} sin (t)

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

Schreibe

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

um:

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

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

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

L^{- 1} {\frac{4}{5 s}} : \frac{4}{5} H (t)

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

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

= \frac{4}{5} H (t) - \frac{4}{5} e^{- \frac{t}{2}} cos (t) - \frac{2}{5} e^{- \frac{t}{2}} sin (t)

d er i v a t i v e e^{x - 1}

\frac{\partial}{\partial x} (2 x y + 5 x^{2})

t an g e n t f (x) = \frac{x ^{2}}{x + 9}, at x = 1

x \to 0 lim (\frac{x}{x ^{2} + 5 x})

d er i v a t i v e y = x^{4} f (x)