inverselaplace (48)/((s+2)(s^2+16))

Lösung

inverse laplace transformation

\frac{48}{( s + 2 ) ( s ^{2} + 16 )}

\frac{12}{5} e^{- 2 t} - \frac{12}{5} cos (4 t) + \frac{6 sin ( 4 t )}{5}

Schritte zur Lösung

L^{- 1} {\frac{48}{( s + 2 ) ( s ^{2} + 16 )}}

Ermittle den Partialbruch von

\frac{48}{( s + 2 ) ( s ^{2} + 16 )} : \frac{12}{5 ( s + 2 )} + \frac{- 12 s + 24}{5 ( s ^{2} + 16 )}

= L^{- 1} {\frac{12}{5 ( s + 2 )} + \frac{- 12 s + 24}{5 ( s ^{2} + 16 )}}

Schreibe um

= L^{- 1} {- \frac{12 s}{5 ( s ^{2} + 16 )} + \frac{12}{5 ( s + 2 )} + \frac{24}{5 ( s ^{2} + 16 )}}

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{12}{5} L^{- 1} {\frac{1}{s + 2}} - \frac{12}{5} L^{- 1} {\frac{s}{s ^{2} + 16}} + \frac{24}{5} L^{- 1} {\frac{1}{s ^{2} + 16}}

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

L^{- 1} {\frac{s}{s ^{2} + 16}} : cos (4 t)

L^{- 1} {\frac{1}{s ^{2} + 16}} : \frac{1}{4} sin (4 t)

= \frac{12}{5} e^{- 2 t} - \frac{12}{5} cos (4 t) + \frac{24}{5} \cdot \frac{1}{4} sin (4 t)

Fasse

\frac{12}{5} e^{- 2 t} - \frac{12}{5} cos (4 t) + \frac{24}{5} \frac{1}{4} sin (4 t)

zusammen:

\frac{12}{5} e^{- 2 t} - \frac{12}{5} cos (4 t) + \frac{6 sin ( 4 t )}{5}

= \frac{12}{5} e^{- 2 t} - \frac{12}{5} cos (4 t) + \frac{6 sin ( 4 t )}{5}

x \to 1 lim (3 x^{4} - 3 x^{2} + 6)

\int_{0}^{2} \frac{x}{4 - x ^{2}} d x

a re a x = y^{2} - 12, x = y

d er i v a t i v e e^{3 x^{2}}

x \to 1 lim (\frac{( sin ( π x ) )}{sin ( 3 π x )})