inverselaplace (0.5s)/(5s^2+10s+49)

Lösung

inverse laplace transformation

\frac{0.5 s}{5 s ^{2} + 10 s + 49}

0.1 e^{- t} cos (2.96647 \dots t) - 0.03370 \dots e^{- t} sin (2.96647 \dots t)

Schritte zur Lösung

L^{- 1} {\frac{0.5 s}{5 s ^{2} + 10 s + 49}}

Schreibe

\frac{0.5 s}{5 s ^{2} + 10 s + 49}

um:

0.1 \cdot \frac{s + 1}{( s + 1 ) ^{2} + \frac{44}{5}} - 0.1 \cdot \frac{1}{( s + 1 ) ^{2} + \frac{44}{5}}

= L^{- 1} {0.1 \cdot \frac{s + 1}{( s + 1 ) ^{2} + \frac{44}{5}} - 0.1 \cdot \frac{1}{( s + 1 ) ^{2} + \frac{44}{5}}}

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)}

= 0.1 L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + \frac{44}{5}}} - 0.1 L^{- 1} {\frac{1}{( s + 1 ) ^{2} + \frac{44}{5}}}

L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + \frac{44}{5}}} : e^{- t} cos (\frac{2 11 t}{5})

L^{- 1} {\frac{1}{( s + 1 ) ^{2} + \frac{44}{5}}} : e^{- t} \frac{5}{2 11} sin (\frac{2 11 t}{5})

= 0.1 e^{- t} cos (\frac{2 11 t}{5}) - 0.1 e^{- t} \frac{5}{2 11} sin (\frac{2 11 t}{5})

Vereinfache

0.1 e^{- t} cos (\frac{2 11 t}{5}) - 0.1 e^{- t} \frac{5}{2 11} sin (\frac{2 11 t}{5}) : 0.1 e^{- t} cos (2.96647 \dots t) - 0.03370 \dots e^{- t} sin (2.96647 \dots t)

= 0.1 e^{- t} cos (2.96647 \dots t) - 0.03370 \dots e^{- t} sin (2.96647 \dots t)

n = 1 \sum \infty \frac{1}{e ^{n}}

\int \frac{x ^{2} - 8}{8 x} d x

\frac{d}{d x} (ln (arctan (9 x^{4})))

t a y l or f (x) = ln (x)

x \to 0 lim (x ln (3 x))