不完全微分PID算法Matlab仿真程序
时间:11-30 14:17 阅读:2847次
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简介:设一被控对象G(s)=50/(0.125s^2+7s),用增量式PID控制算法编写仿真程序(输入分别为单位阶跃、正弦信号,采样时间为1ms,控制器输出限幅:[-5,5],仿真曲线包括系统输出及误差曲线,并加上注释、图例)。
%PID Controler with Partial differential
clear all;
close all;
ts=20;
sys=tf([1],[60,1],'inputdelay',80);
dsys=c2d(sys,ts,'zoh');
[num,den]=tfdata(dsys,'v');
u_1=0;u_2=0;u_3=0;u_4=0;u_5=0;
ud_1=0;
y_1=0;y_2=0;y_3=0;
error_1=0;
ei=0;
for k=1:1:100
time(k)=k*ts;
rin(k)=1.0;
%Linear model
yout(k)=-den(2)*y_1+num(2)*u_5;
error(k)=rin(k)-yout(k);
%PID Controller with partly differential
ei=ei+error(k)*ts;
kc=0.30;
ki=0.0055;
TD=140;
kd=kc*TD/ts;
Tf=180;
Q=tf([1],[Tf,1]); %Low Freq Signal Filter
M=2;
if M==1 %Using PID with Partial differential加在简单PID后的不完全微分
alfa=Tf/(ts+Tf);
u(k)=alfa*u_1+(1-alfa)*(kc*error(k)+kd*(error(k)-error_1)+ki*ei);
u_1=u(k);
elseif M==2 %Using PID with Partial differential只加在微分环节上的不完全微分
alfa=Tf/(ts+Tf);
ud(k)=kd*(1-alfa)*(error(k)-error_1)+alfa*ud_1;
u(k)=kc*error(k)+ud(k)+ki*ei;
ud_1=ud(k);
elseif M==3 %Using Simple PID 简单的PID微分
u(k)=kc*error(k)+kd*(error(k)-error_1)+ki*ei;
end
%Restricting the output of controller
if u(k)>=10
u(k)=10;
end
if u(k)<=-10
u(k)=-10;
end
u_5=u_4;u_4=u_3;u_3=u_2;u_2=u_1;u_1=u(k);
y_3=y_2;y_2=y_1;y_1=yout(k);
error_1=error(k);
end
figure(1);
plot(time,rin,'b',time,yout,'r');
xlabel('time(s)');ylabel('rin,yout');
figure(2);
plot(time,u,'r');
xlabel('time(s)');ylabel('u');
figure(3);
plot(time,rin-yout,'r');
xlabel('time(s)');ylabel('error');
figure(4);
bode(Q,'r');
dcgain(Q);