inverselaplace 3/((s+2)*(4s^2+4s+5))

Lösung

inverse laplace transformation

\frac{3}{( s + 2 ) \cdot ( 4 s ^{2} + 4 s + 5 )}

\frac{3}{13} e^{- 2 t} - \frac{3}{13} e^{- \frac{t}{2}} cos (t) + \frac{9}{26} e^{- \frac{t}{2}} sin (t)

Schritte zur Lösung

L^{- 1} {\frac{3}{( s + 2 ) ( 4 s ^{2} + 4 s + 5 )}}

Ermittle den Partialbruch von

\frac{3}{( s + 2 ) ( 4 s ^{2} + 4 s + 5 )} : \frac{3}{13 ( s + 2 )} + \frac{- 12 s + 12}{13 ( 4 s ^{2} + 4 s + 5 )}

= L^{- 1} {\frac{3}{13 ( s + 2 )} + \frac{- 12 s + 12}{13 ( 4 s ^{2} + 4 s + 5 )}}

Schreibe

\frac{- 12 s + 12}{13 ( 4 s ^{2} + 4 s + 5 )}

um:

- \frac{3}{13} \cdot \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + 1} + \frac{9}{26} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + 1}

= L^{- 1} {\frac{3}{13 ( s + 2 )} - \frac{3}{13} \cdot \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + 1} + \frac{9}{26} \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{3}{13 ( s + 2 )}} - \frac{3}{13} L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + 1}} + \frac{9}{26} L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + 1}}

L^{- 1} {\frac{3}{13 ( s + 2 )}} : \frac{3}{13} e^{- 2 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{3}{13} e^{- 2 t} - \frac{3}{13} e^{- \frac{t}{2}} cos (t) + \frac{9}{26} e^{- \frac{t}{2}} sin (t)

\frac{d}{d x} (x^{4} - 3 x^{3} + 2 x^{2})

x \to \infty lim (\frac{3 x + 5}{6 x - 8})

\frac{d}{d x} (sin^{\frac{2}{3}} (x))

\frac{d}{d x} (4 sin (x) + 4 x cos (x))

\frac{\partial}{\partial x} (ln (x + 5 y + 8 z))