# 耦合回路

$k = \frac{\rvert X_{12} \lvert}{\sqrt{X_{11}X_{22}}}$

## 互感耦合回路

$\dot{V}_1 = \dot{I}_1 (Z_1 + j\omega L_1) - \dot{I}_2(j\omega M)=\dot{I}_1 Z_{11} - j\omega MI_2\\ 0=\dot{I}_2 (Z_2 + j\omega L_2) - \dot{I}_1(j\omega M)=\dot{I}_2Z_{22} - j\omega MI_1$

$\dot{I}_1=\frac{\dot{V}_1}{Z_{11}+\frac{(\omega M)^2}{Z_{22}}}\\ \dot{I}_2=\frac{j\omega M \dot{I}_1}{Z_{22}}=\frac{j\omega M \frac{\dot{V}_1}{Z_{11}}}{Z_{22}+\frac{(\omega M)^2}{Z_{11}}}$

$Z_{f1}=\frac{(\omega M)^2}{Z_{22}}=\frac{(\omega M)^2}{R_{22}+jX_{22}}=\frac{(\omega M)^2}{R_{22}^2+X_{22}^2}R_{22} - j\frac{(\omega M)^2}{R_{22}^2+X_{22}^2} X_{22}\\ =R_{f1}+jX_{f1}$

## 耦合振荡回路的频率特性

$\dot{I}_s = \dot{V}_1 G + \frac{\dot{V}_1}{j\omega L}+j\omega(C_1 + C_M)\dot{V}_1-j\omega C_M \dot{V}_2\\ 0=\dot{V}_2G+\frac{\dot{V}_2}{j\omega L}+ j\omega(C_2+C_M)\dot{V}_2 - j\omega C_M \dot{V}_1$

$\dot{I}_s = \dot{V}_1 G(1+j\xi) - j\omega C_M \dot{V}_2\\ 0=\dot{V}_2 G(1+j\xi)-j\omega C_M \dot{V}_1$

$\dot{V_2}=\frac{j\omega C_M \dot{I}_s}{G^2(1+j\xi)^2+\omega^2C_M^2}$ $V_2=\frac{\omega C_MI_s}{G^2 \sqrt{(1-\xi^2+\frac{\omega^2C_M^2}{G^2})^2+4\xi^2}}$

$\eta=\frac{\omega C_M}{G}=\frac{\omega C}{G}\cdot\frac{C_M}{C}=Q\cdot k\\（k是耦合系数）$

$V_2=\frac{\eta I_s}{G \sqrt{(1-\xi^2+\eta^2)^2+4\xi^2}}$

syms xi eta
V = eta./( sqrt((1-xi.^2+eta.^2).^2 + 4*xi.^2) );
ezmesh(V,[-5 5 2*pi 0 5],100)


## η≤1

### 带宽

$\alpha=\frac{V_2}{V_{2\text{max}}}=\frac{2\eta}{\sqrt{(1-\xi^2+\eta^2)^2+4\xi^2}}$

$令 \eta=1, \alpha=\frac{1}{\sqrt{2}} 有\\ \alpha=\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{4+\xi^4}}\\ 解得：\xi=\pm\sqrt{2}\\ \because \xi\approx Q\frac{2\Delta\omega_{0.7}}{\omega_0}\\ \therefore 2\Delta\omega_{0.7}=\sqrt{2}\frac{\omega_0}{Q}\Leftrightarrow \frac{\omega_0}{Q}(串/并联谐振)$

### 峰值

$\alpha = \frac{2\eta}{1+\eta^2} = \frac{2}{\frac{1}{\eta}+\eta} \leq 1$

## η>1

### 谷深

$\delta=\alpha=\frac{2\eta}{1+\eta^2}\\ （\delta越小，谷越深）$

### 峰宽

$\xi(1+\xi^2-\eta^2)=0\\ 解得： \begin{cases} \xi_0=0\\ \xi_1=-\sqrt{\eta^2-1}\\ \xi_2=+\sqrt{\eta^2-1} \end{cases}\\ 当且仅当 \eta\geq1 时有实根。$

### 带宽

$\frac{1}{\sqrt{2}}=\frac{2\eta}{\sqrt{(1-\xi^2+\eta^2)^2+4\xi^2}}\\ \Rightarrow \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{(1+\eta^2)^2+2(1-\eta^2)\xi^2+\xi^4}}\\ 解出： \vert \xi \vert=\sqrt{\eta^2+2\eta-1}\\ 故：2\Delta f_{0.7}=\sqrt{\eta^2+2\eta-1} \cdot \frac{f_0}{Q}$

$2\Delta f_{0.7}=3.1\frac{f_0}{Q}$

$2\Delta f_{0.7}=\frac{f_0}{Q}$ $2\Delta f_{0.7}=\sqrt{2}\frac{f_0}{Q}$ $2\Delta f_{0.7}=3.1\frac{f_0}{Q}$

# 补充

1. 巴特沃思（Butterworth）逼近
2. 切比雪夫（Chebyshev）逼近
3. 贝塞尔（Bessel）逼近
4. 椭圆函数逼近