pid 控制 很好的例子

%线性模型仿真 只对下水箱液位进行PID控制 clc clear all close all u=zeros(2,1); u(1)=0.0;%稳态初值为0 u(2)=0; Kp=0.1; Ti=10; Td=0; e=[0 0 0]; h2_set1=2.6; %下水箱液位设定值 h2_set=h2_set1-1.4; h2=1.4; h=zeros(1,2); h(1,1)=0; h(1,2)=0; hStep = 10; Hlevel = h; nCounter = 250; for t=0:hStep:(nCounter-1)*hStep e(1)=h2_set-h2; u(1)=u(1)+Kp*(e(1)-e(2))+(Kp*hStep)/Ti*e(1)+(Kp*Td/hStep)*(e(1)-2*e(2)+e(3));%增量PID h = Linear_RK4(hStep,t,h,u); h2=h(1,2); Hlevel=[Hlevel;h]; e(3)=e(2); e(2)=e(1); end Hlevel(:,1) = Hlevel(:,1)+1.5; Hlevel(:,2) = Hlevel(:,2)+1.4; figure(3) plot([0:hStep:nCounter*hStep]',h2_set1,'r');hold on; plot([0:hStep:nCounter*hStep]',Hlevel(:,2)) Grid