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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。 - ^1 t, N7 E* w+ U2 h5 U
%BP based PID Control clear all; close all;
# b# g1 A4 j9 \( h3 K u dxite=0.28; alfa=0.001;
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IN=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;
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wo=0.50*rands(Out,H); wo_1=wo;wo_2=wo;wo_3=wo;
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" c) K( x/ Q1 _, R7 @: Q" J' zx=[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;
' s+ \4 g3 }; F8 t2 s$ iOh=zeros(H,1); %Output from NN middle layer I=Oh; %Input to NN middle layer error_2=0; error_1=0; - T. V- l9 w. o! H H l2 i; o
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); + q! n) G1 T0 ]2 M
xi=[rin(k),yout(k),error(k),1]; % j! ^6 V6 M: f T
x(1)=error(k)-error_1; x(2)=error(k); x(3)=error(k)-2*error_1+error_2; & V l+ {% F; {3 ^/ @8 X# X
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)];
. Y1 r- t( ?, \* {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 ( I+ T0 @! I5 ]( B. e7 r) O
dyu(k)=sign((yout(k)-y_1)/(u(k)-u_1+0.0000001));
, l0 ^, H/ c- 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 ; x, i/ x8 I! N- Q, M8 M
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
: p N% ?2 \7 v+ g$ m, u: M) Td_wi=xite*delta2'*xi; wi=wi_1+d_wi+alfa*(wi_1-wi_2);
$ c! Y( U% F* T: d3 k, G7 n%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; 2 l3 e$ C) n. @+ Y3 _/ W/ R
error_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
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