inverselaplace ((s+1))/((s*(s^2+s+1)))

Lösung

inverse laplace transformation

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

H (t) - e^{- \frac{t}{2}} cos (\frac{3 t}{2}) - \frac{sin ( \frac{3 t}{2} )}{3}

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

Schreibe

\frac{s}{s ^{2} + s + 1}

um:

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

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

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

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

L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}} - \frac{1}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}}} : e^{- \frac{t}{2}} cos (\frac{3 t}{2}) - \frac{sin ( \frac{3 t}{2} )}{3}

= H (t) - e^{- \frac{t}{2}} cos (\frac{3 t}{2}) - \frac{sin ( \frac{3 t}{2} )}{3}

n = 5 \sum \infty \frac{1}{2 ^{2 n}}

t an g e n t f (x) = 5 - 8 x^{2}

\frac{d y}{d x} = 7 x y

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

d er i v a t i v e f (x) = (7 x - 8)^{2}