inverselaplace 1/(s^2-3s+5)

解

逆ラプラス

\frac{1}{s ^{2} - 3 s + 5}

e^{\frac{3 t}{2}} \frac{2}{11} sin (\frac{11 t}{2})

解答ステップ

L^{- 1} {\frac{1}{s ^{2} - 3 s + 5}}

\frac{1}{s ^{2} - 3 s + 5} = \frac{1}{( s - \frac{3}{2} ) ^{2} + \frac{11}{4}}

= L^{- 1} {\frac{1}{( s - \frac{3}{2} ) ^{2} + \frac{11}{4}}}

逆変換ルールを適用:

L^{- 1} {F (s)} = f (t)

ならば

L^{- 1} {F (s - a)} = e^{a t} f (t)

\frac{1}{( s - \frac{3}{2} ) ^{2} + \frac{11}{4}}

向け

: a = \frac{3}{2}, F (s) = \frac{1}{s ^{2} + \frac{11}{4}}

= e^{\frac{3}{2} t} L^{- 1} {\frac{1}{s ^{2} + \frac{11}{4}}}

L^{- 1} {\frac{1}{s ^{2} + \frac{11}{4}}} : \frac{2}{11} sin (\frac{11 t}{2})

= e^{\frac{3 t}{2}} \frac{2}{11} sin (\frac{11 t}{2})