一个基于RBF神经网络的PID参数整定的仿真问题
那位兄弟看过《先进PID算法及Matlab仿真》，请教个问题我想按照书上说的方法做一个基于RBF神经网络的PID参数整定的仿真，可是神经网络不是很熟，始终出不来正确结果。那位兄弟能指教下，谢谢
我用的就是书中的程序，程序如下：
#########################################
%Adaptive PID control based on RBF Identification
clear all;
close all;
ts=0.001;
sys=tf(189.32,);
dsys=c2d(sys,ts,'z');
=tfdata(dsys,'v');
xite=0.1;
alfa=0.1;
belte=0.01;
x=';
ci=30000*ones(3,6);
bi=1000*ones(6,1);
w=20000*ones(6,1);
h=';
ci_1=ci;ci_3=ci_1;ci_2=ci_1;
bi_1=bi;bi_2=bi_1;bi_3=bi_2;
w_1=w;w_2=w_1;w_3=w_1;
u_1=0;y_1=0;u_2=0;y_2=0;
xc=';
error_1=0;error_2=0;error=0;
%kp=rand(1);
%ki=rand(1);
%kd=rand(1);
kp0=40;
ki0=5000;
kd0=0.5;
kp_1=kp0;
kd_1=kd0;
ki_1=ki0;
xitekp=400;
xitekd=200;
xiteki=10;
ts=0.001;
for k=1:1:2000
time(k)=k*ts;
rin(k)=1.0;%*sign(sin(2*pi*k*ts));
yout(k)=-den(2)*y_1-den(3)*y_2+num(2)*u_1+num(3)*u_2;%(-0.1*y_1+u_1)/(1+y_1^2); %Nonlinear plant
for j=1:1:6
h(j)=exp(-norm(x-ci(:,j))^2/(2*bi(j)*bi(j)));
end
ymout(k)=w'*h;
d_w=0*w;
for j=1:1:6
d_w(j)=xite*(yout(k)-ymout(k))*h(j);
end
w=w_1+d_w+alfa*(w_1-w_2)+belte*(w_2-w_3);
d_bi=0*bi;
for j=1:1:6
d_bi(j)=xite*(yout(k)-ymout(k))*w(j)*h(j)*(bi(j)^-3)*norm(x-ci(:,j))^2;
end
bi=bi_1+ d_bi+alfa*(bi_1-bi_2)+belte*(bi_2-bi_3);
for j=1:1:6
for i=1:1:3
d_ci(i,j)=xite*(yout(k)-ymout(k))*w(j)*h(j)*(x(i)-ci(i,j))*(bi(j)^-2);
end
end
ci=ci_1+d_ci+alfa*(ci_1-ci_2)+belte*(ci_2-ci_3);
%%%%%%%%%%%%%%%%%%%%%%Jacobian%%%%%%%%%%%%%%%%%%%%%%%
yu=0;
for j=1:1:6
yu=yu+w(j)*h(j)*(-x(1)+ci(1,j))/bi(j)^2;
end
dyout(k)=yu;
%%%%%%%%%%%%%%%%%%%%%%Start of Control system%%%%%%%%%%%%%%%%%%
error(k)=rin(k)-yout(k);
kp(k)=kp_1+xitekp*error(k)*dyout(k)*xc(1);
kd(k)=kd_1+xitekd*error(k)*dyout(k)*xc(2);
ki(k)=ki_1+xiteki*error(k)*dyout(k)*xc(3);
if kp(k)<0
kp(k)=0;
end
if kd(k)<0
kd(k)=0;
end
if ki(k)<0
ki(k)=0;
end
M=1;
switch M
case 0
case 1 %Only PID Control
kp(k)=kp0;
ki(k)=ki0;
kd(k)=kd0;
end
du(k)=kp(k)*xc(1)+kd(k)*xc(2)+ki(k)*xc(3);
u(k)=u_1+du(k);
%Return of parameters
x(1)=du(k);
x(2)=yout(k);
x(3)=y_1;
u_2=u(k);u_1=u(k);
y_2=u(k);y_1=yout(k);
ci_3=ci_2;
ci_2=ci_1;
ci_1=ci;
bi_3=bi_2;
bi_2=bi_1;
bi_1=bi;
w_3=w_2;
w_2=w_1;
w_1=w;
xc(1)=error(k)-error_1; %Calculating P
xc(2)=error(k)-2*error_1+error_2; %Calculating D
xc(3)=error(k); %Calculating I
error_2=error_1;
error_1=error(k);
kp_1=kp(k);
kd_1=kd(k);
ki_1=ki(k);
end
figure(1);
plot(time,rin,'b',time,yout,'r');
xlabel('time(s)');ylabel('rin,yout');
figure(2);
plot(time,yout,'r',time,ymout,'b');
xlabel('time(s)');ylabel('yout,ymout');
figure(3);
plot(time,dyout);
xlabel('time(s)');ylabel('Jacobian value');
figure(4);
subplot(311);
plot(time,kp,'r');
xlabel('time(s)');ylabel('kp');
subplot(312);
plot(time,ki,'r');
xlabel('time(s)');ylabel('ki');
subplot(313);
plot(time,kd,'r');
xlabel('time(s)');ylabel('kd');
#########################################
其中我的传递函数是：
ts=0.001;
sys=tf(189.32,);
dsys=c2d(sys,ts,'z');
=tfdata(dsys,'v');
PID初始参数给定为：
kp0=40;
ki0=5000;
kd0=0.5;
学习率初值给定为：
xite=0.1;
alfa=0.1;
belte=0.01;
x=';
[ 本帖最后由 tpsharq 于 2008-1-29 21:00 编辑 ] 运行了一下，好像不是程序的问题，是发散了，看看是否什么地方搞错了，另外换换初值看看
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