串联RLC电路对阶跃电压的响应



串联RLC电路响应阶跃电压的电流和电压公式

$v_i - v_R - v_L - v_C = 0$       (I)

$v_R = R \; i$

$\displaystyle v_C = \dfrac{1}{C} \; \int i dt$

$\displaystyle v_L = L \; \dfrac{d i}{dt}$

$\displaystyle v_i - R i - L \dfrac{d i}{dt} - \dfrac{1}{C} \int i dt = 0$

$\displaystyle \mathscr{L}\{ v_i - R i - L \dfrac{d i}{dt} - \dfrac{1}{C} \int i dt \} = \mathscr{L}\{ 0 \}$

$\displaystyle \mathscr{L}\{ v_i \} - R \mathscr{L}\{ i \} - L \mathscr{L} \left\{ \dfrac{d i}{dt} \right\} - \dfrac{1}{C} \mathscr{L} \left\{ \int i dt \right\} = 0$

$\mathscr{L} \left\{ \dfrac{d i}{dt} \right\} = s I(s) - i(0) = s I(s)$，因为初始电流为零 $i(0) = 0$
$\displaystyle \mathscr{L} \left\{ \int i dt \right\} = \dfrac{I(s)}{s}$

$\dfrac{V_0}{s} - R \; I(s) - L \; s \; I(s) - \dfrac{I(s)}{C s} = 0$

$V_0 - R \; s \; I(s) - L \; s^2 \; I(s) - \dfrac{I(s)}{C} = 0$

$I(s) (L \; s^2 + R \; s +\dfrac{1}{C}) = V_0$

$I(s) = \dfrac{V_0}{L} \times \dfrac{1}{s^2 + \dfrac{R}{L} s + \dfrac{1}{L C} }$

$I(s) = \dfrac{V_0}{L} \times \dfrac{1}{ \left(s + \dfrac{R}{2 L} \right)^2 + \dfrac{1}{L C} - \left(\dfrac{R}{2 L}\right)^2 }$

$I(s) = \dfrac{V_0}{L} \times \dfrac{1}{ \left(s + \alpha \right)^2 + \dfrac{1}{L C} - \left(\dfrac{R}{2 L}\right)^2 }$

$I(s) = \dfrac{V_0}{\omega L} \times \dfrac{\omega}{ \left(s + \alpha \right)^2 + \omega^2 }$

$i(t) = \dfrac{V_0}{\omega L} \; \sin (\omega t) \; e^{-\alpha t}$
$v_R(t) = R \; i(t) = \dfrac{R V_0}{\omega L} \sin (\omega t) e^{-\alpha t}$
$\quad \quad = V_0 \dfrac{2 \alpha}{\omega } \sin (\omega t) e^{-\alpha t}$
$v_L(t) = L \; \dfrac{d i}{dt} = V_0 \left\{ \cos (\omega t)- \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t}$
$v_C(t) = v_i(t) - v_R(t) - v_L(t)$
$\quad \quad = V_0 - V_0 \dfrac{2 \alpha}{\omega } \sin (\omega t) e^{-\alpha t} - V_0 \left\{ \cos (\omega t)- \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t}$
$\quad \quad = V_0 - V_0 \left\{ \cos (\omega t) + \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t}$

$\dfrac{1}{L C} = 50000$
$\left(\dfrac{R}{2 L}\right)^2 = 156.25$

$\alpha = \dfrac{R}{2L} = \dfrac{10}{2 \times 0.4} = 12.50$
$\omega = \sqrt {\dfrac{1}{L C} - \left(\dfrac{R}{2 L}\right)^2} = \sqrt {\dfrac{1}{0.4 \times 50 \times 10^{-6}} - \left(\dfrac{10}{2 \times 0.4}\right)^2} = 223.26$
$i(t) = \dfrac{1}{223.26 \times 0.4} \; \sin (223.26 t) \; e^{-12.5 t}$

$i(t) = 0.011 \; \sin (223.26 t) \; e^{-12.5 t}$

$v_R(t) = V_0 \dfrac{2 \alpha}{\omega } \sin (\omega t) e^{-\alpha t}$
$\quad \quad = 0.11198 \; \sin (223.26 t) \; e^{-12.5 t}$
$v_L(t) = V_0 \left\{ \cos (\omega t)- \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t}$
$\quad \quad = \left\{ \cos \left(223.26t\right) - 0.0559875 \sin (223.26t ) \right\}e^{-12.5t}$
$v_C(t) = V_0 - V_0 \left\{ \cos (\omega t) + \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t}$
$\quad \quad = 1 - \left\{ \cos (223.26t) + 0.055988 \sin (223.26t) \right\} e^{-12.5t}$

$I(s) = \dfrac{V_0}{\beta L} \times \dfrac{\beta}{ \left(s + \alpha \right)^2 - \beta^2 }$

$i(t) = \dfrac{V_0}{\beta L} \; \sinh (\beta t) \; e^{-\alpha t}$
$\quad \quad = \dfrac{V_0}{\beta L} \; \left\{ \dfrac{e^{\beta t} - e^{\beta t}} {2} \right\} \; e^{-\alpha t}$
$\quad \quad = \dfrac{V_0}{2\beta L} \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\}$
$v_R(t) = R \; i(t) = \dfrac{R V_0}{2\beta L} \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\}$
$\quad \quad = V_0 \dfrac{\alpha}{\beta } \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\}$
$v_L(t) = L \; \dfrac{d i}{dt} = \dfrac{V_0}{2\beta} \left\{ (\beta - \alpha) e^{ (\beta - \alpha) t} + (\beta + \alpha) e^{ ( - \beta - \alpha) t} \right\}$
$v_C(t) = v_i(t) - v_R(t) - v_L(t)$
$\quad \quad = V_0 - V_0 \dfrac{\alpha}{\beta } \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\} - \dfrac{V_0}{2\beta} \left\{ (\beta - \alpha) e^{ (\beta - \alpha) t} + (\beta + \alpha) e^{ ( - \beta - \alpha) t} \right\}$
$\quad \quad = V_0 - V_0 \left\{ \dfrac{\beta + \alpha}{2 \beta} e^{(\beta - \alpha) t} + \dfrac{\beta - \alpha}{2 \beta} e^{(-\beta - \alpha) t} \right\}$

$\dfrac{1}{L C} = 50000$
$\left(\dfrac{R}{2 L}\right)^2 = 62500$

$\alpha = \dfrac{R}{2L} = \dfrac{200}{2 \times 0.4} = 250$
$\beta = \sqrt { \left(\dfrac{R}{2 L} - \dfrac{1}{L C}\right)^2} = \sqrt {\dfrac{1}{0.4 \times 50 \times 10^{-6}} - \left(\dfrac{200}{2 \times 0.4}\right)^2} = 111.80339$
$i(t) = \dfrac{V_0}{2\beta L} \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\}$
$\quad \quad = 0.01118 \; \left\{ e^{ -138.2 t} - e^{ -361.8 t} \right\}$
$v_R(t) = V_0 \dfrac{\alpha}{\beta } \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\}$
$\quad \quad = 2.23613 \; \left\{ e^{ -138.2 t} - e^{ -361.8 t} \right\}$

$v_L (t) = \dfrac{V_0}{2\beta} \left\{ (\beta - \alpha) e^{ (\beta - \alpha) t} + (\beta + \alpha) e^{ ( - \beta - \alpha) t} \right\}$
$\quad \quad = -0.617754 e^{ -138.2 t} + 1.617246 e^{ -361.8 t}$

$v_C(t) = V_0 - V_0 \left\{ \dfrac{\beta + \alpha}{2 \beta} e^{(\beta - \alpha) t} + \dfrac{\beta - \alpha}{2 \beta} e^{(-\beta - \alpha) t} \right\}$
$\quad \quad = 1 - 1.61806 e^{ -138.2 t} + 0.61806 e^{-361.8 t}$

$I(s)$ 简化为
$I(s) = \dfrac{1}{ \left(s + \alpha \right)^2} \dfrac{V_0}{L}$

$i(t) = \dfrac{V_0}{ L} \; t \; e^{-\alpha t}$
$v_R(t) = R \; i(t) = \dfrac{R V_0}{ L} \; t \; e^{-\alpha t}$
$\quad \quad = 2 V_0 \alpha \; t \; e^{-\alpha t}$
$v_L(t) = L \; \dfrac{d i}{dt} = V_0 \left( 1 - \alpha t \right) e^{-at}$
$v_C(t) = v_i(t) - v_R(t) - v_L(t)$
$\quad \quad = V_0 - 2 V_0 \alpha \; t \; e^{-\alpha t} - V_0 e^{-at} \left( 1 - \alpha t \right)$
$\quad \quad = V_0 - V_0(1+\alpha t)e^{-\alpha t}$

$\dfrac{1}{L C} = 15625$
$\left(\dfrac{R}{2 L}\right)^2 = 15625$

$\alpha = \dfrac{R}{2L} = \dfrac{100}{2 \times 0.4} = 125$
$i(t) = \dfrac{V_0}{ L} \; t \; e^{-\alpha t}$
$i(t) = 2.5 \; t \; e^{- 125 t}$
$v_R(t) = 2 V_0 \alpha \; t \; e^{-\alpha t}$
$\quad \quad = 250 e^{ - 125 t}$
$v_L(t) = L \; \dfrac{d i}{dt} = V_0 ( 1 - \alpha t ) e^{-at}$
$\quad \quad = (1 - 125) e^{ - 125 t}$
$v_C (t) = V_0 - V_0(1+\alpha t)e^{-\alpha t}$
$\quad \quad = 1 - (1 + 125) e^{ - 125 t}$