inverselaplace (10)/((s(s^2+2s+20)))

Lösung

inverse laplace transformation

\frac{10}{( s ( s ^{2} + 2 s + 20 ) )}

\frac{1}{2} H (t) - \frac{1}{2} e^{- t} cos (19 t) - \frac{1}{2 19} e^{- t} sin (19 t)

Schritte zur Lösung

L^{- 1} {\frac{10}{( s ( s ^{2} + 2 s + 20 ) )}}

Ermittle den Partialbruch von

\frac{10}{s ( s ^{2} + 2 s + 20 )} : \frac{1}{2 s} + \frac{- s - 2}{2 ( s ^{2} + 2 s + 20 )}

= L^{- 1} {\frac{1}{2 s} + \frac{- s - 2}{2 ( s ^{2} + 2 s + 20 )}}

Schreibe

\frac{- s - 2}{2 ( s ^{2} + 2 s + 20 )}

um:

- \frac{1}{2} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 19} - \frac{1}{2} \cdot \frac{1}{( s + 1 ) ^{2} + 19}

= L^{- 1} {\frac{1}{2 s} - \frac{1}{2} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 19} - \frac{1}{2} \cdot \frac{1}{( s + 1 ) ^{2} + 19}}

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 + 1}{( s + 1 ) ^{2} + 19}} - \frac{1}{2} L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 19}}

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

L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 19}} : e^{- t} cos (19 t)

L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 19}} : e^{- t} \frac{1}{19} sin (19 t)

= \frac{1}{2} H (t) - \frac{1}{2} e^{- t} cos (19 t) - \frac{1}{2} e^{- t} \frac{1}{19} sin (19 t)

Fasse

\frac{1}{2} H (t) - \frac{1}{2} e^{- t} cos (19 t) - \frac{1}{2} e^{- t} \frac{1}{19} sin (19 t)

zusammen:

\frac{1}{2} H (t) - \frac{1}{2} e^{- t} cos (19 t) - \frac{1}{2 19} e^{- t} sin (19 t)

= \frac{1}{2} H (t) - \frac{1}{2} e^{- t} cos (19 t) - \frac{1}{2 19} e^{- t} sin (19 t)

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

\int (1 + tan (x))^{5} sec^{2} (x) d x

d er i v a t i v e y = 3 cos (2 x)

\int_{1}^{4} (x - 3)^{2} d x

x y^{^{'}} + y = 8 x y^{2}, y (1) = 3