%PID?Controler?with?intergration?sturation
clear?all;
close?all;

ts=0.001;
sys=tf（5.235e005[187.351.047e0040]）;
dsys=c2d（systs‘z‘）;
[numden]=tfdata（dsys‘v‘）;

u_1=0.0;u_2=0.0;u_3=0.0;
y_1=0;y_2=0;y_3=0;

x=[000]‘;

error_1=0;

um=6;
kp=0.85;ki=9.0;kd=0.0;???
rin=30;???????????%Step?Signal

for?k=1:1:800
time（k）=k*ts;
???
u（k）=kp*x（1）+kd*x（2）+ki*x（3）;???%?PID?Controller

if?u（k）>=um
???u（k）=um;
end
if?u（k）<=-um
???u（k）=-um;
end

%Linear?model
yout（k）=-den（2）*y_1-den（3）*y_2-den（4）*y_3+num（2）*u_1+num（3）*u_2+num（4）*u_3;

error（k）=rin-yout（k）;

M=2;
if?M==1??%Using?intergration?sturation
if?u（k）>=um
???if?error（k）>0
??????alpha=0;
???else?
??????alpha=1;
???end
elseif?u（k）<=-um
???if?error（k）>0
??????alpha=1;
???else?
??????alpha=0;
???end
else
???alpha=1;
end

elseif?M==2??%Not?using?intergration?sturation
	alpha=1;???
end

%Return?of?PID?parameters
u_3=u_2;u_2=u_1;u_1=u（k）;???
y_3=y_2;y_2=y_1;y_1=yout（k）;
error_1=error（k）;
???
x（1）=error（k）;????????????????%?Calculating?P
x（2）=（error（k）-error_1）/ts;???%?Calculating?D
x（3）=x（3）+alpha*error（k）*ts;??%?Calculating?I

xi（k）=x（3）;
end
figure（1）;
subplot（311）;
plot（timerin‘b‘timeyout‘r‘）;
xlabel（‘time（s）‘）;ylabel（‘Position?tracking‘）;
subplot（312）;
plot（timeu‘r‘）;
xlabel（‘time（s）‘）;ylabel（‘Controller?output‘）;
subplot（313）;
plot（timexi‘r‘）;
xlabel（‘time（s）‘）;ylabel（‘Integration‘）;

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----

?????文件??????48915??2018-01-21?16:47??抗積分飽和PID控制算法及仿真.pdf

?????文件???????1441??2004-06-13?12:22??chap1_15.m

-----------?---------??----------?-----??----

????????????????50356????????????????????2