inverselaplace 1/(s(12s^2+10s+1))

Lösung

inverse laplace transformation

\frac{1}{s ( 12 s ^{2} + 10 s + 1 )}

H (t) - e^{- \frac{5 t}{12}} cosh (\frac{13 t}{12}) - \frac{5 e ^{- \frac{5 t}{12}} sinh ( \frac{13 t}{12} )}{13}

Schritte zur Lösung

L^{- 1} {\frac{1}{s ( 12 s ^{2} + 10 s + 1 )}}

Ermittle den Partialbruch von

\frac{1}{s ( 12 s ^{2} + 10 s + 1 )} : \frac{1}{s} + \frac{- 12 s - 10}{12 s ^{2} + 10 s + 1}

= L^{- 1} {\frac{1}{s} + \frac{- 12 s - 10}{12 s ^{2} + 10 s + 1}}

Schreibe

\frac{- 12 s - 10}{12 s ^{2} + 10 s + 1}

um:

- \frac{s + \frac{5}{12}}{( s + \frac{5}{12} ) ^{2} - \frac{13}{144}} - \frac{5}{12} \cdot \frac{1}{( s + \frac{5}{12} ) ^{2} - \frac{13}{144}}

= L^{- 1} {\frac{1}{s} - \frac{s + \frac{5}{12}}{( s + \frac{5}{12} ) ^{2} - \frac{13}{144}} - \frac{5}{12} \cdot \frac{1}{( s + \frac{5}{12} ) ^{2} - \frac{13}{144}}}

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{5}{12}}{( s + \frac{5}{12} ) ^{2} - \frac{13}{144}}} - \frac{5}{12} L^{- 1} {\frac{1}{( s + \frac{5}{12} ) ^{2} - \frac{13}{144}}}

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

L^{- 1} {\frac{s + \frac{5}{12}}{( s + \frac{5}{12} ) ^{2} - \frac{13}{144}}} : e^{- \frac{5 t}{12}} cosh (\frac{13 t}{12})

L^{- 1} {\frac{1}{( s + \frac{5}{12} ) ^{2} - \frac{13}{144}}} : e^{- \frac{5 t}{12}} \frac{12}{13} sinh (\frac{13 t}{12})

= H (t) - e^{- \frac{5 t}{12}} cosh (\frac{13 t}{12}) - \frac{5}{12} e^{- \frac{5 t}{12}} \frac{12}{13} sinh (\frac{13 t}{12})

Fasse

H (t) - e^{- \frac{5 t}{12}} cosh (\frac{13 t}{12}) - \frac{5}{12} e^{- \frac{5 t}{12}} \frac{12}{13} sinh (\frac{13 t}{12})

zusammen:

H (t) - e^{- \frac{5 t}{12}} cosh (\frac{13 t}{12}) - \frac{5 e ^{- \frac{5 t}{12}} sinh ( \frac{13 t}{12} )}{13}

= H (t) - e^{- \frac{5 t}{12}} cosh (\frac{13 t}{12}) - \frac{5 e ^{- \frac{5 t}{12}} sinh ( \frac{13 t}{12} )}{13}

\frac{d}{d x} (5 e^{- 5 x})

n = 0 \sum \infty \frac{2}{4 n ^{2} - 1}

d er i v a t i v e y = 1 - x^{2} arcsin (x)

\int \frac{e ^{x} + 1}{2} d x

d er i v a t i v e \frac{- x ^{2} - 8}{( x ^{2} - 8 ) ^{2}}