expert_pid_controller.m

一个利用专家PID算法进行控制的matlab仿真程序

%Expert PID Controller
clear all;
close all;
ts=0.001;
sys=tf(5.235e005,[1,87.35,1.047e004,0]); %Plant
dsys=c2d(sys,ts,'z');
[num,den]=tfdata(dsys,'v');
u_1=0;u_2=0;u_3=0;
y_1=0;y_2=0;y_3=3;
x=[0,0,0]';
x2_1=0;
kp=0.6;
ki=0.03;
kd=0.01;
error_1=0;
for k=1:1:500
    time(k)=k*ts;
    r(k)=1.0; %Tracing Step Signal
    u(k)=kp*x(1)+kd*x(2)+ki*x(3)%PID Controller
    %Expert control rule
    if abs(x(1))>0.8 %Rule1:Unclosed control rule
        u(k)=0.45;
    elseif abs(x(1))>0.40
        u(k)=0.40;
    elseif abs(x(1))>0.20
        u(k)=0.12;
    elseif abs(x(1))>0.01
        u(k)=0.10;
    end
    if x(1)*x(2)>0|(x(2)==0) %Rule2
        if abs(x(1))>=0.05
            u(k)=u_1+2*kp*x(1);
        else
            u(k)=u_1+0.4*kp*x(1);
        end
    end
    if(x(1)*x(2)<0&x(2)*x2_1>0)|(x(1)==0) %Rule3
        u(k)=u(k);
    end
    if x(1)*x(2)<0&x(2)*x2_1<0 %Rule4
        if abs(x(1))>=0.05
            u(k)=u_1+2*kp*error_1;
        else
            u(k)=u_1+0.6*kp*error_1;
        end
    end
    if abs(x(1))<=0.001 %Rule5:Integration separation PI control
        u(k)=0.5*x(1)+0.010*x(3);
    end
    %Restricting the output of controller
    if u(k)>=10
        u(k)=10;
    end
    if u(k)<=-10
        u(k)=-10;
    end
    %Linear model
    y(k)=-den(2)*y_1-den(3)*y_2-den(4)*y_3+num(1)*u(k)+num(2)*u_1+num(3)*u_2+num(4)*u_3;
    error(k)=r(k)-y(k)
    %Return of parameters
    u_3=u_2;u_2=u_1;u_1=u(k);
    y_3=y_2;y_2=y_1;y_1=y(k);
    x(1)=error(k); %Calculating P
    x(2)=(error(k)-error_1)/ts;%Calculating D
    x2_1=x(2)
    x(3)=x(3)+error(k)*ts; %Calculating I
    error_1=error(k);
end
figure(1);
plot(time,r,'b',time,y,'r');
xlabel('time(s)');ylabel('r,y');
figure(2);
plot(time,r-y,'r');
xlabel('time(s)');ylabel('error');