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Beliebt Rechnen >

inverselaplace (0.1s)/(s^2+10s+9.8)

Lösung

inverse laplace transformation

\frac{0.1 s}{s ^{2} + 10 s + 9.8}

Lösung

0.1 e^{- 5 t} cosh (3.89871 \dots t) - 0.12824 \dots e^{- 5 t} sinh (3.89871 \dots t)

Schritte zur Lösung

L^{- 1} {\frac{0.1 s}{s ^{2} + 10 s + 9.8}}

Schreibe

\frac{0.1 s}{s ^{2} + 10 s + 9.8}

um:

0.1 \cdot \frac{s + 5}{( s + 5 ) ^{2} - 15.2} - 0.5 \cdot \frac{1}{( s + 5 ) ^{2} - 15.2}

= L^{- 1} {0.1 \cdot \frac{s + 5}{( s + 5 ) ^{2} - 15.2} - 0.5 \cdot \frac{1}{( s + 5 ) ^{2} - 15.2}}

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 + 5}{( s + 5 ) ^{2} - 15.2}} - 0.5 L^{- 1} {\frac{1}{( s + 5 ) ^{2} - 15.2}}

L^{- 1} {\frac{s + 5}{( s + 5 ) ^{2} - 15.2}} : e^{- 5 t} cosh (3.89871 \dots t)

L^{- 1} {\frac{1}{( s + 5 ) ^{2} - 15.2}} : e^{- 5 t} \frac{1}{3.89871 \dots} sinh (3.89871 \dots t)

= 0.1 e^{- 5 t} cosh (3.89871 \dots t) - 0.5 e^{- 5 t} \frac{1}{3.89871 \dots} sinh (3.89871 \dots t)

Vereinfache

0.1 e^{- 5 t} cosh (3.89871 \dots t) - 0.5 e^{- 5 t} \frac{1}{3.89871 \dots} sinh (3.89871 \dots t) : 0.1 e^{- 5 t} cosh (3.89871 \dots t) - 0.12824 \dots e^{- 5 t} sinh (3.89871 \dots t)

= 0.1 e^{- 5 t} cosh (3.89871 \dots t) - 0.12824 \dots e^{- 5 t} sinh (3.89871 \dots t)

Beliebte Beispiele

integral of 5xsqrt(x)+2

\int 5 x x + 2 d x

derivative of 2(e^xcos(x-e^xsin(x)))

\frac{d}{d x} (2 (e^{x} cos (x) - e^{x} sin (x)))

derivative f(x)= 1/2 x-1/10

d er i v a t i v e f (x) = \frac{1}{2} x - \frac{1}{10}

derivative f(x)= 4/3 pix^3

d er i v a t i v e f (x) = \frac{4}{3} π x^{3}

sum from k=5 to infinity of (k+7)/(k+6)

k = 5 \sum \infty \frac{k + 7}{k + 6}