推导

LADRC同步误差收敛

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1方法一： 李雅普诺夫法证明稳定

原系统：

$$H(x)\ddot x+C(x,\dot x)\dot x=\tau\\ 单G(s)=\frac{K}{s(Ts+1)}\qquad\to\qquad四\begin{cases}H=\begin{bmatrix}T_1/K_1 &&&\\&T_2/K_2&&\\&&T_3/K_3&\\&&&T_4/K_4\end{bmatrix}\\\\\\\\C=\begin{bmatrix}1/K_1&&&\\&1/K_2&&\\&&1/K_3&\\&&&1/K_4\end{bmatrix}\end{cases}$$
状态空间方程：
$$单\begin{cases}\dot x_1=x_2\\\\\dot x_2=-\frac{1}{T}x_2+w+\frac{K}{T}\tau\\\\y=x_1\end{cases}\qquad\to \qquad 单\begin{cases}\dot x_1=x_2\\\dot x_2=x_3+b_0\tau\\\dot x_3=\dot x_3\end{cases}\\ LESO单\begin{cases}e_s=z_1-x_1\\\dot z_1=z_2-\beta_1 e_s\\\dot z_2=z_3-\beta_2 e_s+b_0\tau\\\dot z_3=-\beta_3 e_s\end{cases}\qquad,\qquad b_0=\quad\frac{K}{T}\\\\ \tau=\quad\frac{\tau_0-z_3}{b_0}\quad ,\quad \tau_0=\quad k_p(x^d-z_1)+k_d(0-z_2)$$

假设LESO无差估计

$$则\begin{cases} z_1=x_1\\ z_2=x_2\\ z_3=x_3\\ \end{cases}，\qquad 且原系统变为\ddot x=\tau$$
$\because LESO$无差估计，$\therefore e_s=0$
$$z_{3}=\ddot x-b_0\tau$$
仿照Sun[1]构造位置同步控制器：
$$Sun:\qquad\tau=K_PE+K_D\dot E+(I+\alpha T)^{-1}K_e\dot e\\ We:\qquad \tau=H(K_PE+K_D\dot E-Z_3)$$
$$Z_3=\begin{bmatrix}z_{31}\\z_{32}\\z_{33}\\z_{34}\end{bmatrix},H^{-1}=\begin{bmatrix}b_{01}&&&\\&b_{02}&&\\&&b_{03}&\\&&&b_{04}\end{bmatrix}$$
$$满足条件：\begin{cases}E=(I+\alpha T)e\\e=x^d-x\\\dot e=-\dot x\end{cases}$$

代入上式：

$$\tau=H(K_PE+K_D\dot E-\ddot x)+\tau\\ \ddot x=K_PE+K_D\dot E$$
可以取能量函数为:
$$V=\frac 12 \dot e^T(I+\alpha T)\dot e+\frac 12E^TK_PE$$
易得其肯定$\geq 0$。
$$\dot V=\dot e^T(I+\alpha T)\ddot e+E^TK_P\dot E$$
而
$$\dot E^T\ddot x=\dot E^TK_PE+\dot E^TK_D\dot E\\ -\dot e^T(I+\alpha T)\ddot e=\dot E^TK_PE+\dot E^TK_D\dot E\\ \dot V=-\dot E^TK_D\dot E\leq 0$$
得证。

2 传递函数 - 没写完

$$z_1=\frac{\beta_1 s^2+\beta_2 s+\beta_3}{s^3+\beta_1s^2+\beta_2s+\beta_3}y+\frac{b_0s}{s^3+\beta_1s^2+\beta_2s+\beta_3}u\\ z_2=(s+\beta_1)z_1-\beta_1 y\\z_2=\frac{\beta_2s^2+\beta_3s}{{s^3+\beta_1s^2+\beta_2s+\beta_3}}y+\frac{b_0(s^2+\beta_1s)}{s^3+\beta_1s^2+\beta_2s+\beta_3}u\\ z_3=sz_2+\beta_2(z_1-y)-b_0u\\ z_3=\frac{\beta_3s^2}{s^3+\beta_1s^2+\beta_2s+\beta_3}y-\frac{b_0\beta_3}{{s^3+\beta_1s^2+\beta_2s+\beta_3}}u$$

考虑$\beta_1=3\omega_0,\beta_2=3\omega_0^2,\beta_3=\omega_0^3$。

$$\begin{cases} z_1=\frac{3\omega_0s^2+3\omega_0^2s+\omega_0^3}{(s+\omega_0)^3}y+\frac{b_0s}{(s+\omega_0)^3}u\\ z_2=\frac{3\omega_0^2s^2+\omega_0^3s}{(s+\omega_0)^3}y+\frac{b_0(s^2+3\omega_0s)}{(s+\omega_0)^3}u\\ z_3=\frac{\omega_0^3s^2}{(s+\omega_0)^3}y-\frac{b_0w_0^3}{(s+\omega_0)^3}u \end{cases}$$

原来$u_0=k_p(v-z_1)-k_dz_2,u=(u_0-z_3)/b_0$
现在$U_0=K_P(I+\alpha T)(V-Z_1)+K_D(-Z_2),U=H(U_0-Z_3)$

$$K_P=\begin{bmatrix}\omega_{c1}^2&&&\\&\omega_{c2}^2&&\\&&\omega_{c3}^2&\\&&&\omega_{c4}^2\end{bmatrix}\quad ，\quad I+\alpha T=\begin{bmatrix}1+2\alpha&-\alpha&0&-\alpha&\\-\alpha&1+2\alpha&-\alpha&0\\0&-\alpha&1+2\alpha&-\alpha\\-\alpha&0&-\alpha&1+2\alpha\end{bmatrix}\\ K_P(I+\alpha T)=\begin{bmatrix}(1+2\alpha)\omega_{c1}^2&-\alpha \omega_{c1}^2& 0&-\alpha \omega_{c1}^2\\-\alpha \omega_{c2}^2&(1+2\alpha)\omega_{c2}^2&-\alpha \omega_{c2}^2& 0\\0&-\alpha \omega_{c3}^2&(1+2\alpha)\omega_{c3}^2&-\alpha \omega_{c3}^2\\-\alpha \omega_{c4}^2&0&-\alpha \omega_{c4}^2&(1+2\alpha)\omega_{c4}^2\end{bmatrix}\\ Z_1=\Lambda\left (\begin{bmatrix}\\\frac{3\omega_{01}s^2+3\omega_{01}^2s+\omega_{01}^3}{(s+\omega_{01})^3}&\frac{3\omega_{02}s^2+3\omega_{02}^2s+\omega_{02}^3}{(s+\omega_{02})^3}&\frac{3\omega_{03}s^2+3\omega_{03}^2s+\omega_{03}^3}{(s+\omega_{03})^3}&\frac{3\omega_{04}s^2+3\omega_{04}^2s+\omega_{04}^3}{(s+\omega_{04})^3}\end{bmatrix}\right )$$

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