神经网络PID输出的PID参数为何被局限在0-1之间？

小小鲁班

%基于 BP 神经网络的 PID 控制
clear all;
: Q! C# F3 {) S, ?" V' Vx1=0.3;a1=0.07;
3 u5 G/ }; q$ I) ?% XS=1; %信号类型
# F# u! h0 }. z$ s% DIn=4;h=5;Out=3; %神经网络结构
2 {: E$ X4 H3 D7 Lif S==1, %阶跃信号
; s$ l+ h* }5 a  w: U1 w+ `# @wi=[-0.6394 -0.2696 -0.3756 -0.7023;
-0.8603 -0.2013 -0.05024 -0.2596;
* L! e  E2 B: P8 F6 v-1.0749 0.5543 -1.6820 -0.5437;
-0.3625 -0.0724 -0.6463 -0.2859;
5 X" z, N, I/ G7 K7 g7 ~+ @! x4 h0.1425 0.0279 -0.5406 -0.7660];
# G# S  }$ }% h0 j- o%wi=0.50*rands(h,In);
wi1=wi;wi2=wi;wi3=wi;
$ n) k9 f. t, x, `- p# hwo=[0.7576 0.2616 0.5820 -0.1416 -0.1325;
$ p! f1 S  u; m5 P+ W4 v-0.1146 0.2949 0.8352 0.2205 0.4508;
0.7201 0.4566 0.7672 0.4962 0.3632];
. k. d0 Z1 h0 r. S0 ~%wo=0.50*rands(Out,h);
( }! }8 z/ l% e, u; p8 x4 mwo1=wo;wo2=wo;wo3=wo;
end
! X; c) C; L0 t* K. ?/ Sif S==2 %正弦信号
. b5 x/ K7 ]' u$ d+ swi=[-0.2846 0.2193 -0.5097 -1.0668;
-0.7484 -0.1210 -0.4708 0.0988;
3 ~, f/ }/ l5 w& _-0.7176 0.8297 -1.6000 0.2049;
-0.0858 0.1925 -0.6346 0.0347;
6 k' f  x! }8 o: m2 D0.4358 0.2369 -0.4564 -0.1324];
%wi=0.50*rands(h,In);
' u: f+ X1 b7 @. A, p1 uwi1=wi;wi2=wi;wi3=wi;
wo=[1.0438 0.5478 0.8682 0.1446 0.1537;
% G3 J6 ~: c' ?6 h' F0.1716 0.5811 1.1214 0.5067 0.7370;
" C) s. m% F$ h7 j9 B  c6 Z" Q1.0063 0.7428 1.0534 0.7824 0.6494];
%wo=0.50*rands(Out,h);
0 V+ x; U7 v* C! U! I( G& fwo1=wo;wo2=wo;wo3=wo;
6 m- S( B- C0 oend
x=[0,0,0];
/ l6 n( H# D" |$ ?4 |u1=0;u2=0;u3=0;u4=0;u5=0;
y1=0;y2=0;y3=0;
oh=zeros(h,1); %神经网络中间层输出
2 ^0 b5 W  I3 J0 fI=oh;%神经网络中间层输入
" Z/ G( p4 w. J# N( v5 werror1=0;
! a( K3 f  H$ O" u( N1 ^error2=0;
# x: T! _, Q. f" J8 A3 c1 tts=0.003;
for k=1:1:1000,
time(k)=k*ts;
% x& p7 }$ ?4 [5 r4 k& O; eif S==1,
rin(k)=1.0;
& ~0 x$ e6 M' W5 celseif S==2,
4 \# \5 ]8 f* d2 n9 R  ]rin(k)=sin(1*2*pi*k*ts);
3 `! j0 A3 N7 B& h1 hend
5 F* |; m+ \5 {%非线性模型
a(k)=1+0.15*sin(k*pi*ts/25);
y(k)=(a(k)*y1+u1)/(1+y1^2)+u1;
error(k)=rin(k)-y(k);
xi=[rin(k),y(k),error(k),1];
x(1)=error(k)-error1;
; c1 `. \2 B, e9 ?x(2)=error(k);
" f- T. P& _' Dx(3)=error(k)-2*error1+error2;
ed=[x(1);x(2);x(3)];
I=xi*wi';
for j=1:1:h
  M7 I$ U' @2 j6 hoh(j)=(exp(I(j))-exp(-I(j)))/(exp(I(j))+exp(-I(j)));
end
+ W# z$ ^+ t. x/ d. TK=wo*oh; %输出层
for j=1:1:Out
! X" U- I8 f  d+ W' NK(j)=exp(K(j))/(exp(K(j))+exp(-K(j))); %求 kp,ti,td
end
3 Y/ V% m! r2 S9 D, k9 @kp(k)=K(1);ti(k)=K(2);td(k)=K(3);
Kd=[kp(k),ti(k),td(k)];
  ~) l1 r8 x1 z+ \, V4 g3 x/ tdu(k)=Kd*ed;
u(k)=u1+du(k);
) j/ W2 P: r, A+ Mif u(k)>=10, %限制控制器输出
9 @! a. k$ v6 T6 hu(k)=10;
end
3 T/ ]  g# g, J, |! i- O& U% Iif u(k)<=-10,
# S; o' b' W, B5 Zu(k)=-10;
: l  V7 v3 _3 B4 u' U- o% zend
' y4 `! O1 G  H, N4 ddyu(k)=sign((y(k)-y1)/(u(k)-u1+0.0000001));
for j=1:1:Out,
2 y0 ?1 }8 }8 G3 Z& D* S! d- hdK(j)=2/(exp(K(j))+exp(-K(j)))^2;
end
for i=1:1:Out,
de3(i)=error(k)*dyu(k)*ed(i)*dK(i);
end
+ a+ v- W. x1 Kfor j=1:1:Out,
" J6 i4 u6 k* Z5 ?6 Yfor i=1:1:h,
! O$ f; I$ |$ V1 R2 a5 |; udwo=x1*de3(j)*oh(i)+a1*(wo1-wo2);
end
end
# b' @1 M# W- ^  t' r$ ?0 {* Kwo=wo1+dwo+a1*(wo1-wo2);
%隐含层
for i=1:1:h,
dO(i)=4/(exp(I(i))+exp(-I(i)))^2;
" j+ ^. i' \$ G  |0 Y0 z2 U2 r; s) eend
seg=de3*wo;
for i=1:1:h
de2(i)=dO(i)*seg(i);
* [3 X7 q  ~' o0 ~) s# j  Eend
dwi=x1*de2'*xi;
" t. B: h* h" N+ Hwi=wi1+dwi+a1*(wi1-wi2);
# o  b; h2 E8 x, ^2 Z7 ^%参数更新
0 H, a$ p6 E2 ]4 V% g# [( H% s2 Ju5=u4;u4=u3;u3=u2;u2=u1;u1=u(k);
y2=y1;y1=y(k);
& G  m7 J, ~/ l: h' a& }( Vwo3=wo2;wo2=wo1;wo1=wo;
3 b1 `) ?. c* U2 I) fwi3=wi2;wi2=wi1;wi1=wi;
error2=error1;
error1=error(k);
. Y3 j8 a3 Q7 L4 l& ?end
( ?2 t$ g! u- M5 {7 `; }: K( M6 P5 _7 Dfigure(1);
plot(time,rin,'r',time,y,'c');
) s0 j7 ~1 d. m1 O7 ?  h, rxlabel('时间(秒)');ylabel('输出跟踪输入的响应曲线');
figure(2);
plot(time,u,'r');
xlabel('时间(秒)');ylabel('控制信号 u(k)的变化曲线');
figure(3);
subplot(311);
5 ^$ @' P% b6 O/ m: h( E2 g6 e; W. @plot(time,kp,'r');
xlabel('时间(秒)');ylabel('参数 kp');
7 S: J# j3 z1 l/ ^/ y8 ]. F3 Lsubplot(312);
plot(time,ti,'r');
xlabel('时间(秒)');ylabel('参数 ki');
6 X' T' @! u$ xsubplot(313);
7 P% ^, ]' x- A) g! t# ?& R1 h+ Eplot(time,td,'r');
6 h# Z0 O0 u6 B/ Exlabel('时间(秒)');ylabel('参数 kd');


目标值不能过大怎么解决

pTDbn25

kekek

这个可能跟输入样本关系较大

nkkopd

帮你顶一下