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BP神经网络整定的PID控制算法matlab源程序,系统为二阶闭环系统。 + r- v* T7 `5 f0 M% E8 l
%BP based PID Control clear all; close all; 2 L1 ?. L' K6 F7 y
xite=0.28; alfa=0.001;
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' M8 O0 R: ]$ M% D/ k5 V9 nIN=4;H=5;Out=3; %NN Structure
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wi=0.50*rands(H,IN); wi_1=wi;wi_2=wi;wi_3=wi; 6 h. }1 R4 Z( t, q8 R
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wo=0.50*rands(Out,H); wo_1=wo;wo_2=wo;wo_3=wo; 6 r& H6 J0 ?3 l2 x* W4 h
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x=[0,0,0]; u_1=0;u_2=0;u_3=0;u_4=0;u_5=0; y_1=0;y_2=0;y_3=0; : y# V; E7 i( {: o4 E8 \ O' g* D! _9 I
Oh=zeros(H,1); %Output from NN middle layer I=Oh; %Input to NN middle layer error_2=0; error_1=0; ' X% d. i+ t3 q& ~' p! N1 @
ts=0.01; sys=tf(2.6126,[1,3.201,2.7225]); %建立被控对象传递函数 dsys=c2d(sys,ts,'z'); %把传递函数离散化 [num,den]=tfdata(dsys,'v'); %离散化后提取分子、分母 for k=1:1:2000 time(k)=k*ts; rin(k)=40; yout(k)=-den(2)*y_1-den(3)*y_2+num(2)*u_2+num(3)*u_3;%这一步是怎么推的(问题1) error(k)=rin(k)-yout(k); 7 U9 [, y9 i7 L/ y7 M
xi=[rin(k),yout(k),error(k),1]; - J! f$ v3 W7 ]
x(1)=error(k)-error_1; x(2)=error(k); x(3)=error(k)-2*error_1+error_2;
4 g5 x0 _0 t; p& _2 W5 q+ @% a4 U6 p$ `epid=[x(1);x(2);x(3)]; I=xi*wi'; for j=1:1:H Oh(j)=(exp(I(j))-exp(-I(j)))/(exp(I(j))+exp(-I(j))); %Middle Layer end K=wo*Oh; %Output Layer for l=1:1:Out K(l)=exp(K(l))/(exp(K(l))+exp(-K(l))); %Getting kp,ki,kd end kp(k)=K(1);ki(k)=K(2);kd(k)=K(3); Kpid=[kp(k),ki(k),kd(k)];
7 ]" s a' }: p) q( C8 N" {" _6 \! Pdu(k)=Kpid*epid; u(k)=u_1+du(k); if u(k)>=45 % Restricting the output of controller u(k)=45; end if u(k)<=-45 u(k)=-45; end
' b& V2 c) X$ L. W+ m% k \dyu(k)=sign((yout(k)-y_1)/(u(k)-u_1+0.0000001));
5 B0 o, I# P! p- r. L( w" w$ W%Output layer for j=1:1:Out dK(j)=2/(exp(K(j))+exp(-K(j)))^2; end for l=1:1:Out delta3(l)=error(k)*dyu(k)*epid(l)*dK(l); end * P7 v, e; {# o' i3 r- Q
for l=1:1:Out for i=1:1:H d_wo=xite*delta3(l)*Oh(i)+alfa*(wo_1-wo_2); end end wo=wo_1+d_wo+alfa*(wo_1-wo_2);%这一步似乎有问题(问题2) %Hidden layer for i=1:1:H dO(i)=4/(exp(I(i))+exp(-I(i)))^2; end segma=delta3*wo; for i=1:1:H delta2(i)=dO(i)*segma(i); end
1 B/ C4 ~9 x( q0 t$ d+ g' Ed_wi=xite*delta2'*xi; wi=wi_1+d_wi+alfa*(wi_1-wi_2); 3 Q6 A! I1 m. l0 @, A# y' L
%Parameters Update u_5=u_4;u_4=u_3;u_3=u_2;u_2=u_1;u_1=u(k); y_2=y_1;y_1=yout(k); wo_3=wo_2; wo_2=wo_1; wo_1=wo; wi_3=wi_2; wi_2=wi_1; wi_1=wi;
/ N- c2 ]" H1 e1 ` m1 J! Xerror_2=error_1; error_1=error(k); end figure(1); plot(time,rin,'r',time,yout,'b'); xlabel('time(s)');ylabel('rin,yout'); figure(2); plot(time,error,'r'); xlabel('time(s)');ylabel('error'); figure(3); plot(time,u,'r'); xlabel('time(s)');ylabel('u'); figure(4); subplot(311); plot(time,kp,'r'); xlabel('time(s)');ylabel('kp'); subplot(312); plot(time,ki,'g'); xlabel('time(s)');ylabel('ki'); subplot(313); plot(time,kd,'b'); xlabel('time(s)');ylabel('kd'); 问题(1)和问题(2)都标注出来了。还请各位帮忙看一下,尤其是问题(1),到底如何将已知的传递函数转换成,matlab的仿真模型呢 5 [) O- R5 e1 a& S
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