inverselaplace (s+5)/(7s^2+26s+25)

Lösung

inverse laplace transformation

\frac{s + 5}{7 s ^{2} + 26 s + 25}

\frac{1}{7} e^{- \frac{13 t}{7}} cos (\frac{6 t}{7}) + \frac{11 2 e ^{- \frac{13 t}{7}} sin ( \frac{6 t}{7} )}{7 3}

Schritte zur Lösung

L^{- 1} {\frac{s + 5}{7 s ^{2} + 26 s + 25}}

Schreibe

\frac{s + 5}{7 s ^{2} + 26 s + 25}

um:

\frac{1}{7} \cdot \frac{s + \frac{13}{7}}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}} + \frac{22}{49} \cdot \frac{1}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}}

= L^{- 1} {\frac{1}{7} \cdot \frac{s + \frac{13}{7}}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}} + \frac{22}{49} \cdot \frac{1}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}}}

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)}

= \frac{1}{7} L^{- 1} {\frac{s + \frac{13}{7}}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}}} + \frac{22}{49} L^{- 1} {\frac{1}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}}}

L^{- 1} {\frac{s + \frac{13}{7}}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}}} : e^{- \frac{13 t}{7}} cos (\frac{6 t}{7})

L^{- 1} {\frac{1}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}}} : e^{- \frac{13 t}{7}} \frac{7}{6} sin (\frac{6 t}{7})

= \frac{1}{7} e^{- \frac{13 t}{7}} cos (\frac{6 t}{7}) + \frac{22}{49} e^{- \frac{13 t}{7}} \frac{7}{6} sin (\frac{6 t}{7})

Fasse

\frac{1}{7} e^{- \frac{13 t}{7}} cos (\frac{6 t}{7}) + \frac{22}{49} e^{- \frac{13 t}{7}} \frac{7}{6} sin (\frac{6 t}{7})

zusammen:

\frac{1}{7} e^{- \frac{13 t}{7}} cos (\frac{6 t}{7}) + \frac{11 2 e ^{- \frac{13 t}{7}} sin ( \frac{6 t}{7} )}{7 3}

= \frac{1}{7} e^{- \frac{13 t}{7}} cos (\frac{6 t}{7}) + \frac{11 2 e ^{- \frac{13 t}{7}} sin ( \frac{6 t}{7} )}{7 3}

\frac{d y}{d t} + 4 y = 0

t an g e n t f (x) = x^{7}, at x = 2

\frac{d v}{d t} = - k t v^{2}, v (0) = 50

l a pl a ce t r an s f or m sin (3 x)

d er i v a t i v e h (θ) = θ cos (θ)