inverselaplace (10)/(s(s^2+8s+25))

Lösung

inverse laplace transformation

\frac{10}{s ( s ^{2} + 8 s + 25 )}

\frac{2}{5} H (t) - \frac{2}{5} e^{- 4 t} cos (3 t) - \frac{8}{15} e^{- 4 t} sin (3 t)

Schritte zur Lösung

L^{- 1} {\frac{10}{s ( s ^{2} + 8 s + 25 )}}

Ermittle den Partialbruch von

\frac{10}{s ( s ^{2} + 8 s + 25 )} : \frac{2}{5 s} + \frac{- 2 s - 16}{5 ( s ^{2} + 8 s + 25 )}

= L^{- 1} {\frac{2}{5 s} + \frac{- 2 s - 16}{5 ( s ^{2} + 8 s + 25 )}}

Schreibe

\frac{- 2 s - 16}{5 ( s ^{2} + 8 s + 25 )}

um:

- \frac{2}{5} \cdot \frac{s + 4}{( s + 4 ) ^{2} + 9} - \frac{8}{5} \cdot \frac{1}{( s + 4 ) ^{2} + 9}

= L^{- 1} {\frac{2}{5 s} - \frac{2}{5} \cdot \frac{s + 4}{( s + 4 ) ^{2} + 9} - \frac{8}{5} \cdot \frac{1}{( s + 4 ) ^{2} + 9}}

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{2}{5 s}} - \frac{2}{5} L^{- 1} {\frac{s + 4}{( s + 4 ) ^{2} + 9}} - \frac{8}{5} L^{- 1} {\frac{1}{( s + 4 ) ^{2} + 9}}

L^{- 1} {\frac{2}{5 s}} : \frac{2}{5} H (t)

L^{- 1} {\frac{s + 4}{( s + 4 ) ^{2} + 9}} : e^{- 4 t} cos (3 t)

L^{- 1} {\frac{1}{( s + 4 ) ^{2} + 9}} : e^{- 4 t} \frac{1}{3} sin (3 t)

= \frac{2}{5} H (t) - \frac{2}{5} e^{- 4 t} cos (3 t) - \frac{8}{5} e^{- 4 t} \frac{1}{3} sin (3 t)

Fasse

\frac{2}{5} H (t) - \frac{2}{5} e^{- 4 t} cos (3 t) - \frac{8}{5} e^{- 4 t} \frac{1}{3} sin (3 t)

zusammen:

\frac{2}{5} H (t) - \frac{2}{5} e^{- 4 t} cos (3 t) - \frac{8}{15} e^{- 4 t} sin (3 t)

= \frac{2}{5} H (t) - \frac{2}{5} e^{- 4 t} cos (3 t) - \frac{8}{15} e^{- 4 t} sin (3 t)

\frac{\partial}{\partial x} (x arctan (yz))

x \to 0 lim (\frac{sin ( \frac{x}{2} )}{x})

x \to \infty lim ((1 + x)^{\frac{2}{x}})

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 2, y^{^{'}} (0) = 3

\int_{0}^{2} 5 d x