１５ー１ー９．ステップ関数（RLC回路）
・α > ω0
\(\Large \displaystyle F(s) = \frac{V_0}{L} \frac{1}{s^2 + 2 \ \alpha \ s + \omega_0^2} \)
\(\Large \displaystyle = \frac{V_0}{L} \frac{1}{(s-s_1)(s-s_2)} \)
\(\Large \displaystyle s_{1,2} = - \alpha \pm \sqrt{ \alpha^2 - \omega_0^2} \)
\(\Large \displaystyle s_1 + s_2 = -2 \ \alpha \)
\(\Large \displaystyle s_1 \cdot s_2 = \left(- \alpha + \sqrt{ \alpha^2 - \omega_0^2} \right)\left(- \alpha - \sqrt{ \alpha^2 - \omega_0^2} \right) \)
\(\Large \displaystyle = \alpha^2 - \left( \alpha^2 - \omega_0^2 \right) = \omega_0^2 \)
部分分数分解を考える．
\(\Large \displaystyle \frac{1}{(s-s_1)(s-s_2)}= \frac{A}{s-s_1} + \frac{B}{s-s_2} \)
\(\Large \displaystyle =\frac{A(s-s_2)+B(s-s_1)}{(s-s_1)(s-s_2)} \)
\(\Large \displaystyle 1 = A(s-s_2)+B(s-s_1) \)
どんなｓでも成り立たたないといけないので，s=s<sub>1</sub>，とおくと，
\(\Large \displaystyle s = s_1 \)
\(\Large \displaystyle 1 = A(s_1-s_2) \)
\(\Large \displaystyle A = \frac{1}{s_1-s_2} \)
\(\Large \displaystyle B = -\frac{1}{s_1-s_2} \)
\(\Large \displaystyle F(s) = \frac{V_0}{L} \frac{1}{(s-s_1)(s-s_2)} \)
\(\Large \displaystyle = \frac{V_0}{L}\frac{1}{s_1-s_2} \left( \frac{1}{s-s_1}-\frac{1} {s-s_2}\right) \)
逆ラプラス変換を行うと，
\(\Large \displaystyle F(s) = \frac{V_0}{L}\frac{1}{s_1-s_2} \left( \frac{1}{s-s_1}-\frac{1} {s-s_2}\right) \)
\(\Large \displaystyle I(t) = \frac{V_0}{L}\frac{1}{s_1-s_2} \left( e^{s_1 \ t} - e^{s_2 \ t}\right) \)
\(\Large \displaystyle \sqrt{ \alpha^2 - \omega_0^2} \equiv \omega \)
\(\Large \displaystyle s_{1,2} = - \alpha \pm \sqrt{ \alpha^2 - \omega_0^2} = - \alpha \pm \omega \)
\(\Large \displaystyle s_1 + s_2 = -2 \ \alpha \\ s_1 - s_2 = 2 \sqrt{ \alpha^2 - \omega_0^2} = 2 \omega \)
なので，
\(\Large \displaystyle I(t) = \frac{V_0}{L}\frac{1}{s_1-s_2} \left( e^{s_1 \ t} - e^{s_2 \ t}\right) \)
\(\Large \displaystyle = \frac{V_0}{L}\frac{1}{2 \omega}\left( e^{(- \alpha + \omega ) \ t} - e^{(- \alpha - \omega ) \ t}\right) \)
\(\Large \displaystyle = \frac{V_0}{2 \omega L} \ e^{- \alpha t}\left( e^{\omega \ t} - e^{ -\omega \ t}\right) \)
\(\Large \displaystyle = \frac{V_0}{ \omega L} \ e^{- \alpha t} \cdot sinh \ \omega t \)
となり，通常の解法，と一致します．
次は，α = ω0，です．