HOW TO INTEGRATE THE SLIDING PART OF SLIDING MODE CONTROL SYSTEM

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Aliff Ashraff 2020-6-18

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https://ww2.mathworks.cn/matlabcentral/answers/550812-how-to-integrate-the-sliding-part-of-sliding-mode-control-system

评论： darova 2020-6-22

Hi, i am currently working on sliding mode control for two tank systems. I have a problem to on the integral part. I need to integrate (x-5)dt from 0 to t, where t is the sampling time. The coding that I use is as belows:

x1d=x(2);
x1_des=5;                       
x1_des_d=0;                                  
C = 0.45;     
e1=x(1)-x1_des;                 
e1d=x(2)-x1_des_d;              
syms t;
s=x1d+(2*C*e1)+((C^2)*int(e1,t,0,t)); 

I am using s-function for the coding to be run by the simulink where I think I do not need the loop to integrate the equations. Can someone help me and give me advice to solve this problems? Your help is very appreciated.

Thank you.

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Aliff Ashraff 2020-6-21

编辑：Aliff Ashraff 2020-6-21

在 MATLAB Online 中打开

function sys=mdlDerivatives(t,x,u)
global x1_des s ismc
%%Parameters took from:%%
%%Sliding Mode Control of Coupled Tanks System: Theory and and Application%%
%%Tabrej Alam (2013)%%
g=981;                          %gravity mass (cm/s2)
A=208.2;                        %area of tank 1 and 2 (cm2)
a1=0.58;                        %area of connecting pipe between tank 1 and tank 2 (cm2)
a2=0.24;                        %area of outlet pipe (cm2)
%%Dynamic Model%%
% q1=a1*sqrt(2*g*(h1-h2));      %flow rate from tank 1 to tank 2
% q2=a2*sqrt(2*g*h2);           %outlet flow rate at tank 2
% x1d=(ismc-q1)/A;             %dh1dt=dx1dt
% x2d=(q1-q2)/A;                %fdh2dt=dx2dt
k1=((a1)*(sqrt(2*g))/A);        %gain constant
k2=((a2)*(sqrt(2*g))/A);        %gain constant
h1=((((k1*sqrt(x(1)))+x(2))/k2)^2)+x(1);   %water level tank 1%
h2=x(1);                                   %water level tank 2%
f=((k1*k2)/2)*(((sqrt(h2))/sqrt(h1-h2))-((sqrt(h1-h2))/sqrt(h2)))+((k1^2)/2)-(k2^2);
b=(k2/(2*A))*(1/(sqrt(h1-h2)));
x(1)=h2;                               
x(2)=(-k1*sqrt(h2))+(k2*sqrt(h1-h2));   
%%y=x(1)%% 
%Model derivatives%
x1d=x(2);                       %x(1) dot
x2d=f+(b*ismc);                %x(2) dot
%%ismc design%%
x1_des=5;                       %x2_des = desired x2=h2 (cm)
x1_des_d=0;                     %x2_des_d = x2 desired dot
x1_des_dd=0;                    %x2_des_dd = x2 desired dot dot
C = 0.45; %0.08,0.45,0.05       %positive constant lambda
e1=x(1)-x1_des;                 %e1 = error
e1d=x(2)-x1_des_d;              %e1d = e1 dot
syms t;
s=x1d+(2*C*e1)+((C^2)*int(e1,t,0,t));                   %sliding equation
%%To define ueq%%
%sd=0;                          %sd = s dot
%sd=f+(b*u)+(2*C*e1d)+((c^2)e1);   
%f+(b*u)+(2*C*e1d)+((c^2)e1)=0;
ueq=(-(1/b))*(f+(2*C*e1d)+((C^2)*e1));
%To define udis%%
M=5500; %5500,0.9                %positive constant
udis=(-M)*sign(s); 
%%Conventional smc equation%%
ismc = ueq+udis;
sys = [x1d;x2d];
function sys=mdlOutputs(t,x,u)
global ismc x1_des s 
sys = [x(1); ismc;s;x1_des];

This is the coding that i used. I need to get '[x(1); ismc;s;x1_des]' as my output. The 's=x1d+(2*C*e1)+((C^2)*int(e1,t,0,t));' is the equation that contains the integral. Before this I used the same coding but with different 's' equation and I get the expected result that I want. But when it si changed with integral part it got problem.