串联RLC二阶电路

零状态RLC串联电路的自然响应

• 拓扑约束
  • KVL: $u_C=u_L+u_R$
  • KCL: $-i_C=i_L=i_R=i$
• 元件约束
  • $i_C=-C\frac{d{u}_C}{dt}$
  • $u_L=L\frac{d{i}_L}{dt}$
  • $R=\frac{u_R}{i_R}$

列写方程：

$$\begin{array}{rl}{u}_L+u_R& =u_C\\ L\frac{d}{dt}{i}_L+R{i}_R& =u_C\\ L\frac{d}{dt}(i_C)+R(i_C)& =u_C\\ L\frac{d}{dt}(-C\frac{d{u}_C}{dt})+R(-C\frac{d{u}_C}{dt})& =u_C\\ -LC\frac{d^2{u}_C}{d^{2}t}-RC\frac{d{u}_C}{dt}& =u_C\\ -LC\frac{d^2{u}_C}{d^{2}t}-RC\frac{d{u}_C}{dt}-u_C& =0\\ \frac{d^2{u}_C}{d^{2}t}+\frac{R}{L}\frac{d{u}_C}{dt}+\frac{1}{LC}{u}_C& =0\end{array}$$

定义：$2\alpha =\frac{R}{L}$，${\omega }_0^2=\frac{1}{LC}$，得：

$$\frac{d^2{u}_C}{d^{2}t}+2\alpha \frac{d{u}_C}{dt}+{\omega }_0^2{u}_C=0$$

特征方程为

$${\lambda }^2+2\alpha \lambda +{\omega }_0^2=0$$

判别式为

$$\begin{array}{rl}\mathrm{\Delta }& =b^2-4ac\\ & =(2\alpha {)}^2-4\times 1\times ({\omega }_0^2)\\ & =4{\alpha }^2-4{\omega }_0^2\\ & =4({\alpha }^2-{\omega }_0^2)\end{array}$$

特征根为

$$\begin{array}{rl}\lambda & =\frac{-b\pm \sqrt{\mathrm{\Delta }}}{2a}\\ & =\frac{-(2\alpha )\pm \sqrt{4({\alpha }^2-{\omega }_0^2)}}{2\times 1}\\ & =\frac{-2\alpha \pm 2\sqrt{{\alpha }^2-{\omega }_0^2}}{2\times 1}\\ & =-\alpha \pm \sqrt{{\alpha }^2-{\omega }_0^2}\end{array}$$

根据判别式的情况，微分方程的解为：

过阻尼情况： $\mathrm{当}{\alpha }^2>{\omega }_0^2$ ( 即 $\mathrm{\Delta }=4({\alpha }^2-{\omega }_0^2)>0$ ) 时

$$\begin{array}{rl}{\lambda }_1,{\lambda }_2& =-\alpha \pm \sqrt{{\alpha }^2-{\omega }_0^2}\\ u_C& =A{e}^{{\lambda }_{1}t}+B{e}^{{\lambda }_{2}t}\end{array}$$

临界阻尼情况： 当${\alpha }^2={\omega }_0^2$ ( 即 $\mathrm{\Delta }=4({\alpha }^2-{\omega }_0^2)=0$ ) 时

$$\begin{array}{rl}\lambda & =-\alpha \pm \sqrt{{\alpha }^2-{\omega }_0^2}\\ & =-\alpha \pm 0\\ & =-\alpha \\ u_C& =A{e}^{\lambda t}+Bt{e}^{\lambda t}\end{array}$$

欠阻尼情况： $\mathrm{当}{\alpha }^2<{\omega }_0^2$ ( 即 $\mathrm{\Delta }=4({\alpha }^2-{\omega }_0^2)<0$ ) 时

$$\begin{array}{rl}{\lambda }_1,{\lambda }_2& =-\alpha \pm \sqrt{{\alpha }^2-{\omega }_0^2}\\ & =-\alpha \pm \sqrt{(-1)\times ({\omega }_0^2-{\alpha }^2)}\\ & =-\alpha \pm \sqrt{-1}\times \sqrt{({\omega }_0^2-{\alpha }^2)}\\ & =-\alpha \pm i\times \sqrt{({\omega }_0^2-{\alpha }^2)}\\ \\ {\omega }_d& =\sqrt{({\omega }_0^2-{\alpha }^2)}\\ u_C& =e^{-\alpha t}(A\cos ({\omega }_{d}t)+B\sin ({\omega }_{d}t))\\ \\ R& =\sqrt{A^2+B^2}\\ \varphi & ={\tan }^{-1}\frac{A}{B}\\ u_C& =R{e}^{-\alpha t}\sin ({\omega }_{d}t+\varphi )\end{array}$$

零阻尼情况： $\mathrm{当}\alpha =0$ ( 即 $\mathrm{\Delta }=4({\alpha }^2-{\omega }_0^2)=-4{\omega }_0^2<0$ ) 时

$$\begin{array}{rl}{\lambda }_1,{\lambda }_2& =-\alpha \pm \sqrt{{\alpha }^2-{\omega }_0^2}\\ & =-0\pm \sqrt{0^2-{\omega }_0^2}\\ & =\pm \sqrt{-1\times {\omega }_0^2}\\ & =\pm \sqrt{-1}\times \sqrt{{\omega }_0^2}\\ & =\pm i\sqrt{{\omega }_0^2}\\ & (\text{特征根为纯虚数})\\ \\ {\omega }_d& =\sqrt{{\omega }_0^2}={\omega }_0\\ u_C& =e^{-\alpha t}(A\cos ({\omega }_{d}t)+B\sin ({\omega }_{d}t))\\ & =e^{-0t}(A\cos ({\omega }_{d}t)+B\sin ({\omega }_{d}t))\\ & =A\cos ({\omega }_{d}t)+B\sin ({\omega }_{d}t)\\ & =A\cos ({\omega }_{0}t)+B\sin ({\omega }_{0}t)\\ \\ R& =\sqrt{A^2+B^2}\\ \theta & ={\tan }^{-1}\frac{A}{B}\\ u_C& =R{e}^{-\alpha t}\sin ({\omega }_{d}t+\theta )\\ & =R{e}^{-0t}\sin ({\omega }_{d}t+\theta )\\ & =R\sin ({\omega }_{d}t+\theta )\\ & =R\sin ({\omega }_{0}t+\theta )\end{array}$$

案例

根据拓扑约束和元件约束写出微分方程；根据特征方程的解写出微分方程的通解形式。

分析计算电路支路量的初值和一阶导，据此求解通解的待定系数。

绿色为电容电压、红色为电感电流

绿色为电容电压、红色为电感电流

绿色为电容电压、红色为电感电流

绿色为电容电压、红色为电感电流

红、绿、蓝、紫分别为过阻尼、临界阻尼、欠阻尼、零阻尼情况下的电感电流。