电路方程 | 二阶线性时不变电路方程列写及五要素法总结¶
列写方法¶
正规列写（用矩阵参量+端口定义写）；用基尔霍夫定律列写够两个方程
列写二阶微分方程¶
对于一个未知量，可以列写出一个二微分方程如下：
\(\frac{d^{2}}{d t^{2}} x+a \frac{d}{d t} x+b x=s_{x}\)
\(\frac{d^{2}}{d t^{2}} x+2 \xi \omega_{0} \frac{d}{d t} x+\omega_{0}^{2} x=s_{x}\)
其中\(\omega_0\)是自由振荡频率，\(\xi\)是阻尼系数
\(a=2\xi\omega_0,\ b=\omega_0^2;\ \omega_0=\sqrt{b},\ \xi=\frac{a}{2\sqrt{b}}\)
传递函数¶
\(\frac{\dot{X}}{\dot{S}}=\frac{\dot{S}_ {X} / \dot{S}}{s^{2}+2 \xi \omega_{0} s+\omega_{0}^{2}},\ s=j\omega\)
- 典型二阶低通：\(H_{C}(s)=\frac{\dot{V}_ {C}}{\dot{V}_ {S}}=\frac{\omega_{0}^{2}}{s^{2}+2 \xi \omega_{0} s+\omega_{0}^{2}}\)
- 典型二阶带通：\(H_{R}(s)=\frac{\dot{V}_ {R}}{\dot{V}_ {S}}=\frac{2 \xi \omega_{0} s}{s^{2}+2 \xi \omega_{0} s+\omega_{0}^{2}}\)
- 典型二阶高通：\(H_{L}(s)=\frac{\dot{V}_ {L}}{\dot{V}_ {S}}=\frac{s^{2}}{s^{2}+2 \xi \omega_{0} s+\omega_{0}^{2}}\)
解的形态¶
特征方程：\(\lambda^{2}+2 \xi \omega_{0} \lambda+\omega_{0}^{2}=0\)
- \(\xi=0\)无阻尼，理想正弦振荡
- \(0<\xi<1\)欠阻尼，特征方程有两个共轭复根（左半平面）\(\lambda_{1,2}=\left(-\xi \pm j \sqrt{1-\xi^{2}}\right) \omega_{0}\)，幅度指数衰减的正弦振荡
- \(\xi=1\)临界阻尼，特征方程有两个负实重根，
- \(\xi>1\)过阻尼，特征方程有两个负实根\(\lambda_{1,2}=\left(-\xi \pm \sqrt{\xi^{2}-1}\right) \omega_{0}\)，指数衰减
求解方法：待定系数法¶
\(x(t)=x_{\infty}(t)+\left\{\begin{array}{lr}A e^{\lambda_{1} t}+B e^{\lambda_{2} t} & \xi>1 \\ A e^{-\omega_{0} t}+B t e^{-\omega_{0} t} & \xi=1 \\ e^{-\xi \omega_{0} t}\left(A \cos \sqrt{1-\xi^{2}} \omega_{0} t+B \sin \sqrt{1-\xi^{2}} \omega_{0} t\right) & 0<\xi<1\end{array}\right.\)
其中：\(\lambda_{1,2}=\left(-\xi \pm \sqrt{1-\xi^{2}}\right) \omega_{0}\)；A、B为待定系数，由电路初始状态\(x(0^+)\ \frac{d}{dt}x(0^+)\)决定
求解方法：五要素法¶
三要素：稳态响应\(x_\infty(t)\)、阻尼系数\(\xi\)和自由振荡频率\(\omega_0\)、初值和微分初值\(x(0^+)\ \frac{d}{dt}x(0^+)\)
稳态响应¶
使用向量法求解。直流时利用电容电感高频低频特性求解。
阻尼系数和自由振荡频率¶
特殊：RLC串联与GCL并联¶
自由振荡频率均为\(\omega_0=\frac{1}{\sqrt{LC}}\)
阻尼系数：RLC串联\(\xi=\frac{R}{2Z_0}=\frac{R}{2\sqrt{\frac{L}{C}}}\)，RCL并联\(\xi=\frac{G}{2Y_0}=\frac{1}{2R\sqrt{\frac{C}{L}}}\)
一般：从传递函数推导¶
\(H(j \omega) \stackrel{j \omega \rightarrow s}{=} \frac{? ? ?}{s^{2}+2 \xi \omega_{0} s+\omega_{0}^{2}}\)
将分母中\(s^{2}\)的系数归一化，然后观察一次项和零次项的系数，得到\(\xi\ \omega_0\)
一般：从二阶微分方程推导¶
\(\frac{d^{2}}{d t^{2}} x+a \frac{d}{d t} x+b x=s_{x}\)
\(\frac{d^{2}}{d t^{2}} x+2 \xi \omega_{0} \frac{d}{d t} x+\omega_{0}^{2} x=s_{x}\)
\(a=2\xi\omega_0,\ b=\omega_0^2\Longrightarrow\omega_0=\sqrt{b},\ \xi=\frac{a}{2\sqrt{b}}\)
写结果¶
\(0<\xi<1\)¶
复杂：\(x(t)=x_{\infty}(t)+\left(X_{0}-X_{\infty 0}\right) e^{-\xi \omega_{0} t} \cos \sqrt{1-\xi^{2}} \omega_{0} t+\left(\frac{\dot{X}_ {0}-\dot{X}_ {\infty 0}}{\xi \omega_{0}}+X_{0}-X_{\infty 0}\right) \frac{\xi}{\sqrt{1-\xi^{2}}} e^{-\xi \omega_{0} t} \sin \sqrt{1-\xi^{2}} \omega_{0} t\)
\(x(t)=x_{\infty}(t)+ e^{-\xi \omega_{0} t} (A\cos \sqrt{1-\xi^{2}} \omega_{0} t+B \sin \sqrt{1-\xi^{2}} \omega_{0} t)\)
\(A=X_{0}-X_{\infty 0}\)
\(B=\left(\frac{\dot{X}_ {0}-\dot{X}_ {\infty 0}}{\xi \omega_{0}}+X_{0}-X_{\infty 0}\right) \frac{\xi}{\sqrt{1-\xi^{2}}}\)
\(\xi=1\)¶
直接带入\(\xi=1\)得到：
复杂：\(x(t)=x_{\infty}(t)+\left(X_{0}-X_{\infty 0}\right) e^{-\omega_{0} t}+\left(\frac{\dot{X}_ {0}-\dot{X}_ {\infty 0}}{\omega_{0}}+X_{0}-X_{\infty 0}\right) \omega_{0} t e^{-\omega_{0} t}\)
\(x(t)=x_{\infty}(t)+e^{-\omega_{0} t} \left [\left(X_{0}-X_{\infty 0}\right) +\left(\frac{\dot{X}_{0}-\dot{X}_{\infty 0}}{\omega_{0}}+X_{0}-X_{\infty 0}\right) \omega_{0} t \right]\)
\(\xi>1\)¶
将欠阻尼\(0<\xi<1\)情况中的\(\sqrt{1-\xi^{2}}\)换成\(\sqrt{\xi^{2}-1}\)，\(\cos\ \sin\)换成\(\cosh\ \sinh\)
\(x(t)=x_{\infty}(t)+ e^{-\xi \omega_{0} t} (A\cosh \sqrt{\xi^{2}-1} \omega_{0} t+B \sinh \sqrt{\xi^{2}-1} \omega_{0} t)\)
\(x(t)=x_{\infty}(t)+\left(X_{0}-X_{\infty 0}\right) e^{-\omega_{0} t}+\left(\frac{\dot{X}_ {0}-\dot{X}_ {\infty 0}}{\omega_{0}}+X_{0}-X_{\infty 0}\right) \omega_{0} t e^{-\omega_{0} t}\)
复杂：\(x(t)=x_{\infty}(t)+\left(X_{0}-X_{\infty 0}\right) e^{-\xi \omega_{0} t} \cosh \sqrt{\xi^{2}-1} \omega_{0} t+\left(\frac{\dot{X}_ {0}-\dot{X}_ {\infty 0}}{\xi \omega_{0}}+X_{0}-X_{\infty 0}\right) \frac{\xi}{\sqrt{\xi^{2}-1}} e^{-\xi \omega_{0} t} \sinh \sqrt{\xi^{2}-1} \omega_{0} t\)
过阻尼：最后还需化简为两个指数衰减函数叠加的效果
得到：
\(x(t)=x_{\infty}(t)+\frac{A+B}{2} e^{\lambda_{1} t}+\frac{A-B}{2} e^{\lambda_{2} t}\)
其中：\(\lambda_{1,2}=\left(-\xi \pm \sqrt{1-\xi^{2}}\right) \omega_{0}\)
注¶
- \(\sinh x=\frac{e^{x}-e^{-x}}{2}\)
- \(\cosh x=\frac{e^{x}+e^{-x}}{2}\)
- \(\sin x=\frac{e^{j x}-e^{-j x}}{2 j}\)
- \(\cos x=\frac{e^{j x}+e^{-j x}}{2}\)
总结¶
稳态响应¶
使用向量法求解。直流时利用电容电感高频低频特性求解。
阻尼系数和自由振荡频率¶
特殊：RLC串联与GCL并联¶
自由振荡频率均为\(\omega_0=\frac{1}{\sqrt{LC}}\)
阻尼系数：RLC串联\(\xi=\frac{R}{2Z_0}=\frac{R}{2\sqrt{\frac{L}{C}}}\)，RCL并联\(\xi=\frac{G}{2Y_0}=\frac{1}{2R\sqrt{\frac{C}{L}}}\)
一般：从传递函数推导¶
\(H(j \omega) \stackrel{j \omega \rightarrow s}{=} \frac{? ? ?}{s^{2}+2 \xi \omega_{0} s+\omega_{0}^{2}}\)
将分母中\(s^{2}\)的系数归一化，然后观察一次项和零次项的系数，得到\(\xi\ \omega_0\)
\(0<\xi<1\)¶
\(x(t)=x_{\infty}(t)+ e^{-\xi \omega_{0} t} (A\cos \sqrt{1-\xi^{2}} \omega_{0} t+B \sin \sqrt{1-\xi^{2}} \omega_{0} t)\)
\(x(t)=x_{\infty}(t)+ e^{-\xi \omega_{0} t} \sqrt{A^2+B^2} \sin \left(\sqrt{1-\xi^{2}} \omega_{0} t+\arctan\frac{A}{B}\right)\)
其中：\(A=X_{0}-X_{\infty 0},\ B=\left(\frac{\dot{X}_ {0}-\dot{X}_ {\infty 0}}{\xi \omega_{0}}+X_{0}-X_{\infty 0}\right) \frac{\xi}{\sqrt{1-\xi^{2}}}\)
\(\xi=1\)¶
\(x(t)=x_{\infty}(t)+e^{-\omega_{0} t} \left [\left(X_{0}-X_{\infty 0}\right) +\left(\frac{\dot{X}_ {0}-\dot{X}_ {\infty 0}}{\omega_{0}}+X_{0}-X_{\infty 0}\right) \omega_{0} t \right]\)
\(\xi>1\)¶
\(x(t)=x_{\infty}(t)+\frac{A+B}{2} e^{\lambda_{1} t}+\frac{A-B}{2} e^{\lambda_{2} t}\)
其中：\(\lambda_{1,2}=\left(-\xi \pm \sqrt{\xi^{2}-1}\right) \omega_{0}\)