inverselaplace (5s+5)/(s(s^2+6s+10))

Lösung

inverse laplace transformation

\frac{5 s + 5}{s ( s ^{2} + 6 s + 10 )}

\frac{1}{2} H (t) - \frac{1}{2} e^{- 3 t} cos (t) + \frac{7}{2} e^{- 3 t} sin (t)

Schritte zur Lösung

L^{- 1} {\frac{5 s + 5}{s ( s ^{2} + 6 s + 10 )}}

Ermittle den Partialbruch von

\frac{5 s + 5}{s ( s ^{2} + 6 s + 10 )} : \frac{1}{2 s} + \frac{- s + 4}{2 ( s ^{2} + 6 s + 10 )}

= L^{- 1} {\frac{1}{2 s} + \frac{- s + 4}{2 ( s ^{2} + 6 s + 10 )}}

Schreibe

\frac{- s + 4}{2 ( s ^{2} + 6 s + 10 )}

um:

- \frac{1}{2} \cdot \frac{s + 3}{( s + 3 ) ^{2} + 1} + \frac{7}{2} \cdot \frac{1}{( s + 3 ) ^{2} + 1}

= L^{- 1} {\frac{1}{2 s} - \frac{1}{2} \cdot \frac{s + 3}{( s + 3 ) ^{2} + 1} + \frac{7}{2} \cdot \frac{1}{( s + 3 ) ^{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{1}{2 s}} - \frac{1}{2} L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 1}} + \frac{7}{2} L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 1}}

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

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

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

= \frac{1}{2} H (t) - \frac{1}{2} e^{- 3 t} cos (t) + \frac{7}{2} e^{- 3 t} sin (t)

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

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

in v erse l a pl a ce \frac{2}{s + 1} + \frac{24 s}{s ^{2} + 9}

f^{^{''}} (x) = 4, f^{^{'}} (2) = 10, f (2) = 15

t an g e n t f (x) = 3 x^{3}, at x = 3