控制实验7.18

状态反馈

电机状态空间模型$x=\left[\begin{array}{c}i\\ n\end{array}\right]$：
$$\{\begin{array}{c}\dot{i}=-\frac{R}{L}i-\frac{C_e\varphi }{L}i+\frac{1}{L}u\\ \dot{n}=\frac{30C_T\varphi }{\pi J}i+d(t)\end{array}$$
$$\left[\begin{array}{c}\dot{i}\\ \dot{n}\end{array}\right]=\left[\begin{array}{cc}-\frac{R}{L}& -\frac{C_e\varphi }{L}\\ \frac{30C_T\varphi }{\pi J}& 0\end{array}\right]\left[\begin{array}{c}i\\ n\end{array}\right]+\left[\begin{array}{c}\frac{1}{L}\\ 0\end{array}\right]u+\left[\begin{array}{c}0\\ d(t)\end{array}\right]$$

电机参数	取值
$R/\Omega$	2.785
$L/mH$	0.0085
$J/kg\ast m^2$	0.0044
$C_e$	0.7
$C_T$	6.685
$\Phi$	1

系统矩阵$A=\left[\begin{array}{cc}-338.2352& -82.3529\\ 14508& 0\end{array}\right],B=\left[\begin{array}{c}117.65\\ 0\end{array}\right],Q_c=\left[\begin{array}{cc}B& AB\end{array}\right]=\left[\begin{array}{cc}117.65& 39800\\ 0& 1706900\end{array}\right],rank{Q}_c=2$因此系统完全能控。
系统矩阵A的特征值为$-169.1\pm j1079.9$，位于左半平面，因此系统稳定。
可以通过状态反馈将系统闭环极点配置在-170，-170.
设$u=Kx,K=\left[\begin{array}{cc}{k}_1& k_2\end{array}\right]$,
$$A+BK=\left[\begin{array}{cc}{a}_{11}+b_1{k}_1& a_{12}+b_1{k}_2\\ a_{21}+b_2{k}_1& a_{22}+b_2{k}_2\end{array}\right]=\left[\begin{array}{cc}117.65k_1-338.2353& 117.65k_2-82.3529\\ 14508& 0\end{array}\right]$$
$$|\lambda I-A|=\left|\begin{array}{cc}\lambda +338.2352-117.65k_1& 82.3529-117.65k_2\\ -14508& \lambda \end{array}\right|={\lambda }^2+(338.235-117.65k_1)\lambda +14508(82.35-117.65k_2)={\lambda }^2+340\lambda +170^2$$
$$k_1=-0.015,k_2=0.683$$
$$u=\left[\begin{array}{cc}{k}_1& k_2\end{array}\right]\left[\begin{array}{c}i\\ n\end{array}\right]$$
转速跟踪状态空间模型改写：
$e=n-n_r,\dot{e}=\dot{n}-{\dot{n}}_r$
$$\left[\begin{array}{c}\dot{i}\\ \dot{e}\end{array}\right]=\left[\begin{array}{cc}-\frac{R}{L}& -\frac{C_e\varphi }{L}\\ \frac{30C_T\varphi }{\pi J}& 0\end{array}\right]\left[\begin{array}{c}i\\ e\end{array}\right]+\left[\begin{array}{c}\frac{1}{L}\\ 0\end{array}\right]u+\left[\begin{array}{c}-\frac{C_e\Phi }{L}{n}_r\\ d(t)-{\dot{n}}_r\end{array}\right]$$
令$\dot{i},\dot{e}=0$，系统平衡点是$i=0,e=-n_r$(假设$d(t),{\dot{n}}_r=0$)，为了移动平衡点到$e=0$，令u的加上移动平衡点的项：
$$u=Kx+C_e\Phi n_r$$
Simulink模型图：

运行结果：

滑模控制

还是采用上面的转速跟踪误差模型
$$\left[\begin{array}{c}\dot{i}\\ \dot{e}\end{array}\right]=\left[\begin{array}{cc}-\frac{R}{L}& -\frac{C_e\varphi }{L}\\ \frac{30C_T\varphi }{\pi J}& 0\end{array}\right]\left[\begin{array}{c}i\\ e\end{array}\right]+\left[\begin{array}{c}\frac{1}{L}\\ 0\end{array}\right]u+\left[\begin{array}{c}-\frac{C_e\Phi }{L}{n}_r\\ d(t)-{\dot{n}}_r\end{array}\right]$$
$s=ci+e=ci+n-n_r$
$\dot{s}=c\dot{i}+\dot{n}-{\dot{n}}_r=c(-\frac{R}{L}i-\frac{C_e\Phi }{L}n+\frac{1}{L}u+\frac{30C_T\Phi }{\pi J}i+d(t)-{\dot{n}}_r)$
定义李雅普诺夫函数为$V(s)=\frac{1}{2}{s}^2$
$$\dot{V}=s\dot{s}$$
令$u=\frac{L}{c}[(\frac{cR}{L}-\frac{30C_T\Phi }{\pi J})i+\frac{c{C}_e\Phi }{L}n+{\dot{n}}_r-Ks-ϵsgn(s)]$


sfun_stablizesys_SMC.m文件内容：

function [sys, x0, str, ts] = sfun_stablizesys_SMC(t, x, u, flag)
switch flag,
    case 0,
        [sys, x0, str, ts] = mdlInitializeSizes;
    case 3,
        sys = mdlOutputs(t, x, u);
    case {2,4,9}
        sys = [];
    otherwise,
        error(['Unhandled flag = ', num2str(flag)]);
end
function [sys, x0, str, ts] = mdlInitializeSizes
sizes = simsizes;
sizes.NumContStates  = 0;   %需要求解的状态变量个数，控制器表达式不需要
sizes.NumDiscStates  = 0;
sizes.NumOutputs     = 2;
sizes.NumInputs      = 4;
sizes.DirFeedthrough = 1;
sizes.NumSampleTimes = 0;
sys = simsizes(sizes);
x0  = [];
str = [];
ts  = [];

function sys=mdlOutputs(t,x,u)  % u = [i,n,nr,dnr]
c = 10;
k = 0.001;
L = 0.0085;
J = 0.0044;
R = 2.875;
CT = 6.685;
Ce = 0.7;
FI = 1;
K = 30;
epsilon = 2.171;

e = u(2)-u(3);
s = c*u(1)+e;
sys(1) = s;
sys(2) = L/c*(u(4)+(c*R/L-30*CT*FI/pi/J)*u(1)+c*Ce*FI/L*u(2)-K*s-epsilon*sign(s));

位置跟踪

假设位置$z={\int }_0^{t}kndt,\dot{z}=kn$，
模型变为：
$$\left[\begin{array}{c}\dot{z}\\ \dot{i}\\ \dot{n}\end{array}\right]=\left[\begin{array}{ccc}0& 0& k\\ 0& -\frac{R}{L}& -\frac{C_e\varphi }{L}\\ 0& \frac{30C_T\varphi }{\pi J}& 0\end{array}\right]\left[\begin{array}{c}z\\ i\\ n\end{array}\right]+\left[\begin{array}{c}0\\ \frac{1}{L}\\ 0\end{array}\right]u+\left[\begin{array}{c}0\\ 0\\ d(t)\end{array}\right]$$
$A=\left[\begin{array}{ccc}0& 0& 0.001\\ 0& -338.2352& -82.3529\\ 0& 14508& 0\end{array}\right],B=\left[\begin{array}{c}0\\ 117.65\\ 0\end{array}\right],Q_c=\left[\begin{array}{ccc}B& AB& A^{2}B\end{array}\right],rank{Q}_c=3$，因此系统完全能控。
A的特征值为0，$-169.1\pm j1079.9$，需要通过状态反馈进行系统镇定。
设$K=[k_1{k}_2{k}_3]$，配置到-170，-170，-170，
$$|\lambda I-(A+BK)|=(\lambda +170{)}^3$$
解得$k_1=-2878.4,k_2=-1.46,k_3=\mathrm{0.6492.}$
$$u=[k_1{k}_2{k}_3]\left[\begin{array}{c}z\\ i\\ n\end{array}\right]$$
镇定后系统状态都能收敛到0：

加入位置跟踪：
$$\left[\begin{array}{c}\dot{e}\\ \dot{i}\\ \dot{n}\end{array}\right]=\left[\begin{array}{ccc}0& 0& k\\ 0& -\frac{R}{L}& -\frac{C_e\varphi }{L}\\ 0& \frac{30C_T\varphi }{\pi J}& 0\end{array}\right]\left[\begin{array}{c}e\\ i\\ n\end{array}\right]+\left[\begin{array}{c}0\\ \frac{1}{L}\\ 0\end{array}\right]u+\left[\begin{array}{c}-{\dot{z}}_r\\ 0\\ d(t)\end{array}\right]$$
$$u=[k_1{k}_2{k}_3]\left[\begin{array}{c}e\\ i\\ n\end{array}\right]=[k_1{k}_2{k}_3]\left[\begin{array}{c}z\\ i\\ n\end{array}\right]-k_1{z}_r$$

改变初值依然是很小的误差

加入负载$T_L=30$，抗干扰效果也不错

滑模轨迹跟踪：

时滞

状态反馈镇定加Smith预估器控制方案：

时滞存在于输出量和反馈量，由于史密斯预估器只适用于单入单出系统，所以将位置作为输出变量，对于位置的反馈单独分离出来。Plant的模型中先不包含位置的反馈量，只镇定$i,n$两个状态变量：
$$\left[\begin{array}{c}\dot{z}\\ \dot{i}\\ \dot{n}\end{array}\right]=\left[\begin{array}{ccc}0& 0& k\\ 0& -\frac{R}{L}& -\frac{C_e\varphi }{L}\\ 0& \frac{30C_T\varphi }{\pi J}& 0\end{array}\right]\left[\begin{array}{c}z\\ i\\ n\end{array}\right]+\left[\begin{array}{c}0\\ \frac{1}{L}\\ 0\end{array}\right]u+\left[\begin{array}{c}0\\ 0\\ d(t)\end{array}\right]$$
$$u_1=[0k_2{k}_3]\left[\begin{array}{c}z\\ i\\ n\end{array}\right]$$
Smith预估器反并联的控制器为：
$$u_2=-k_1(z_r-z)$$
Smith预估器：

滑模控制加Smith预估器

状态反馈时滞控制

定理：

$K=Y{W}^{-1},P=W^{-1}$
代入镇定后的三阶系统，解得K=[4392.9 -26079948404.99 0.5462]
代入未镇定的三阶系统，解得K=[-0.0766 -8293531604.51 -3.1232]
仿真结果都不稳定。

时滞存在于状态变量中，电流传给转速时有时滞
$$\left[\begin{array}{c}\dot{e}\\ \dot{i}\\ \dot{n}\end{array}\right]=\left[\begin{array}{ccc}0& 0& k\\ 0& -\frac{R}{L}& -\frac{C_e\Phi }{L}\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}e\\ i\\ n\end{array}\right]\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& \frac{30C_T\Phi }{\pi J}& 0\end{array}\right]\left[\begin{array}{c}e(t-\tau )\\ i(t-\tau )\\ n(t-\tau )\end{array}\right]+\left[\begin{array}{c}0\\ \frac{1}{L}\\ 0\end{array}\right]u+\left[\begin{array}{c}-{\dot{z}}_r\\ 0\\ d(t)\end{array}\right]$$

状态反馈那个实验必须用三个LMI解出来的K才能镇定系统，单用第一个LMI解出来的K不能使系统稳定，很奇怪

参考资料

1. ​【现代控制理论笔记】——第三章：状态反馈
   ​https://blog.youkuaiyun.com/weixin_52077466/article/details/135450369
2. 【Advanced控制理论】8.5_线性控制器设计_轨迹跟踪(Follow a Desired Path) https://www.bilibili.com/video/BV1HW411s7YC/?spm_id_from=333.999.0.0&vd_source=20d896eca3aa8b68c67b5e833330db7d
3. 滑模变结构控制仿真实验设计——刘金琨著