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我正在编写代码以在 MATLAB 中使用最小二乘法优化 MIMO 系统的 PID 增益。我正在为其设计 PID 控制器的工厂有 3 个输入和 3 个输出，对于每个输入，我正在设计一个 PID 控制器。整个系统如下所示：

%                   _____________________________________________________________________
%                  |                                                                     | 
%                  |                                                                     |  
%                  |                                                                     |
%                  |                      __________________________________             |  
%                  |    e              u |                                  |   y        |  
% r_phi,pitch,psi--|-->O--> PID array -->|              ssModel             |----------.-|------
%                  |   ^                 |                                  |          | |  
%                  |   | y               |__________________________________|          | |  
%                  |   |                                system                         | |  
%                  |   |                                                               | |   
%                  |   .---------------------------------------------------------------. |                                                      
%                  |_____________________________________________________________________|
%                                                        clSys
%

我的问题是优化逻辑总是返回非常接近初始值的优化增益，显然它总是落在局部最小值。

如何将 9 增益全局最小化？

这是状态空间模型文件：

function [ssModel, Ts] = getSSmodel()
%% Define the parameter of our hardware
Jxx = 0.01; % appr. the inertia of a solid cube with the quad's mass (the best)
Jyy = 0.01;
Jzz = Jxx*2;

l   = 0.225;  % meters, the distance from the center of the propeller to the center of
              % the drone

Ts  = 0.0083; % the sampling time

b1  = l/Jxx;
b2  = l/Jyy;
b3  = 1/Jzz;

%% Define the continuous and discrete SS model
A = [0 1 0 0 0 0;
     0 0 0 0 0 0;
     0 0 0 1 0 0;
     0 0 0 0 0 0;
     0 0 0 0 0 1;
     0 0 0 0 0 0];

 B = [0  0  0;
      b1 0  0;
      0  0  0;
      0  b2 0;
      0  0  0;
      0  0  b3];

  C = [1 0 0 0 0 0;
       0 0 1 0 0 0;
       0 0 0 0 1 0];

  D = zeros(size(C,1),size(B,2));

  continuous_sys = ss(A,B,C,D);

  discrete_sys   = c2d(continuous_sys,Ts);

  ssModel     = discrete_sys;
end

这是我指定要最小化的函数和成本函数的代码：

function costFunction = closedLoopPoles(pidGains, ssModel)

if (nargin < 2) 
    [ssModel, Ts]   =  getSSmodel();
end

%Define the PID controllers array
pidTF_array  = ones(3,3) * tf(0,1,Ts);

for i = 1:size(pidGains,1)
    pidTF_array(i,i)  = pid( pidGains(i,1), pidGains(i,2), pidGains(i,3),'IFormula','BackwardEuler','DFormula','BackwardEuler', 'Ts', Ts); 
end

% Assign the signals names and the systems structure
pidTF_array.u = 'e';          pidTF_array.y = 'u'; 
ssModel.u     = 'u';          ssModel.y     = 'y';
sum1          = sumblk('e = r - y',3);

system      = ssModel;

% Find the closed loop system
clSys               = connect(system,pidTF_array,sum1,'r','y');
clPoles            = pole( clSys );    
maxAbs_clPoles     = max(abs(clPoles));
sumAbs_clPoles     = sum(abs(clPoles));
sumnorm_clPoles    = sum(norm(clPoles));

costFunction       = sumnorm_clPoles;

end

最后是优化函数：

function pidGains_opt = optimizePID_minPole()

clear all
clc

pidGains_ini       = [180 4 20   160 4 20    150 4 10];
func               = @closedLoopPoles;
[pidGains_opt, ~]  = lsqnonlin(func,pidGains_ini)

end

当我运行优化代码时，我在工作区中得到了这个：

 pidGains_ini =

   180     4    20
   160     4    20
   150     4    10

Warning: Trust-region-reflective algorithm requires at least as many equations as variables; using
Levenberg-Marquardt algorithm instead. 
> In lsqncommon at 77
  In lsqnonlin at 235
  In optimizePID_minPole at 11 

Local minimum possible.

lsqnonlin stopped because the final change in the sum of squares relative to 
its initial value is less than the default value of the function tolerance.

<stopping criteria details>


pidGains_opt =

  179.8817    4.0022    3.1099
  159.8839    4.0025    3.0151
  149.9482    4.0022    1.5592
4

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