inverselaplace (3s+1)/(s^2(s^2+4))

Lösung

inverse laplace transformation

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

Schreibe um

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

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)}

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

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

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

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

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

= \frac{3}{4} H (t) + \frac{1}{4} t - \frac{3}{4} cos (2 t) - \frac{1}{4} \cdot \frac{1}{2} sin (2 t)

Fasse

\frac{3}{4} H (t) + \frac{1}{4} t - \frac{3}{4} cos (2 t) - \frac{1}{4} \frac{1}{2} sin (2 t)

zusammen:

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

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

x y^{^{'}} - y = \frac{x ^{2}}{y ^{2}}

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

l a pl a ce t r an s f or m 25

a re a 4 - x^{2}, 3 x

\int (3 x^{2} - 2 x + 1) d x