EDA365欢迎您登录!
您需要 登录 才可以下载或查看,没有帐号?注册
x
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。
N1 ?3 W: v1 M4 z- L0 ]4 I%BP based PID Control clear all; close all; & H$ `6 f% V! |3 _
xite=0.28; alfa=0.001; & O3 w y& k; Z
3 |: } E, s' a5 WIN=4;H=5;Out=3; %NN Structure L. V* L1 X6 [( R
: _8 }) o7 p+ z+ y; a: I/ H
wi=0.50*rands(H,IN); wi_1=wi;wi_2=wi;wi_3=wi; # q- e4 y, u; t O: ^" Q
; w, I+ }5 x+ K/ W7 v# U1 B
wo=0.50*rands(Out,H); wo_1=wo;wo_2=wo;wo_3=wo;
+ b/ V' y ]0 g" f2 d% G3 j
2 g- J: N/ f7 A a" o' Fx=[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; K& g! |* M. E
Oh=zeros(H,1); %Output from NN middle layer I=Oh; %Input to NN middle layer error_2=0; error_1=0;
: p/ ^7 ^4 F- Gts=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); ' f5 A0 k7 {# M; m# N4 k' m
xi=[rin(k),yout(k),error(k),1]; : W9 B" I. Y. n- \% c
x(1)=error(k)-error_1; x(2)=error(k); x(3)=error(k)-2*error_1+error_2; 4 o2 ^4 b6 U' G) b( k
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)]; 8 q; [% m" q2 g7 F( O, p: P
du(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 R* F8 p, W( X O- Q0 v
dyu(k)=sign((yout(k)-y_1)/(u(k)-u_1+0.0000001)); ; I% K3 z5 `2 U
%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 * J; D1 u; |9 R$ B9 x; ~3 Z
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
, _& c0 _ @% Rd_wi=xite*delta2'*xi; wi=wi_1+d_wi+alfa*(wi_1-wi_2);
' h9 {, n! l F4 K0 T2 |%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;
9 B5 t4 @1 Q! J( \' l/ cerror_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 ' j0 L$ R# Z7 V. _6 S, V) u3 b( d6 F
|