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    發布日期: 2021-06-10
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資源簡介

利用SOA神經網絡實現PID參數的自整定,選擇最優的控制參數

資源截圖

代碼片段和文件信息

function?BsJ=PID_SOA(Kpidi)
ts=0.001;
sys=tf([1.05][6.8e-6?2.47e-3?0.7925]);
dsys=c2d(systs‘z‘);
[numden]=tfdata(dsys‘v‘);
u_1=0.0;u_2=0.0;
y_1=0.0;y_2=0.0;
x=[000]‘;
B=0;
error_1=0;
tu=1;
s=0;
P=100;
for?k=1:1:P
????timef(k)=k*ts;
????r(k)=1;
????u(k)=Kpidi(1)*x(1)+Kpidi(2)*x(3)+Kpidi(3)*x(2);
????if?u(k)>=10
????????u(k)=10;
????end
????if?u(k)<=-10
????????u(k)=-10;
????end
????yout(k)=-den(2)*y_1-den(3)*y_2+num(2)*u_1+num(3)*u_2;
????error(k)=r(k)-yout(k);
????%-------------------Return?of?PID?parameters----------------
????u_2=u_1;u_1=u(k);
????y_2=y_1;y_1=yout(k);
????x(1)=error(k);???????????%?Calculating??P
????x(2)=(error(k)-error_1)/ts;???%??????????D
????x(3)=x(3)+error(k)*ts;???????%???????????I?
????error_2=error_1;
????error_1=error(k);
????if?s==0
????????if?yout(k)>0.95&yout(k)<1.05
????????????tu=timef(k);
????????????s=1;
????????end
????end
end
for?i=1:1:P
????Ji(i)=0.999*abs(error(i))+0.01*u(i)^2*0.1;
????B=B+Ji(i);
????if?i>1
????????erry(i)=yout(i)-yout(i-1);
????????if?erry(i)<0
????????????B=B+100*abs(erry(i));
????????end
????end
end
BsJ=B+0.2*tu*10;

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----

?????文件????????984??2013-09-26?12:49??shuchuquxianSOA.m

?????文件???????3309??2013-09-26?12:49??SOA.m

?????文件???????1148??2013-09-26?12:49??PID_SOA.m

-----------?---------??----------?-----??----

?????????????????5441????????????????????3


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