Mp = 0.4; % Kg
xp0 = 0.001485; % m
Kp = 242650; % N/m
Dp = 135.4; % N/m/s
Ap = 0.006; % m^2
alpha = 4.1054e-6; % m/s
V0 = 7.8e-5; % m^3
Km = 0.00106; % m^2Cm = 42.74; % N
Aorifice = 2.0442e-5; % m^2rho = 880; % kg/m^3
Cd = 0.7;Patm = 1*10^5; % Pa
beta = 17000*10^5; % PaKs = 0.00107; % m^2Cs = 47.52; % NPc = 0; % Centrifugal pressure
delta_T = 0.000375; % Sample time
%% difining displacement velocity and pressure
steps=10;X2=linspace(5,-1,steps);U=linspace(0,2*10^5,20);xfinal=[0.00058 0 187500]';t=0.3;lambda_2=1;lambda_3=1;lambda_4=1;lambda_5=1;for k=1:steps for j=1:steps X1= xfinal(1)-delta_T*X2(j); X3=(1/(Ap-Km*tanh(X2(j)/alpha)))*(xfinal(2)-X2(j))*(Mp/delta_T)-Ap*(Pc-Patm)+Dp*X2(j)+Kp*(X1+xp0)+ (Cm+Km*Pc)*tanh((X2(j)/alpha)); a=[X1 X2(j) X3]'; xfinal=[X1,X2(j),X3]'; W=(xfinal(3)-X3)*((V0+Ap*X1)/(delta_T*beta))+Ap*X2(j); u=sign(W)*(W/(Cd*Aorifice))^.2*(rho/2)+X3; lambda=[lambda_3 0 0;0 lambda_4 0;0 0 lambda_5]; J_temp=lambda_2*(X2(j)-0)^.2+(a-xfinal)'*lambda*(a-xfinal); save Control.mat u endend
MATLAB: How to get sequence of values for ‘u’ in the for loop using the ‘save control.mat u’ syntax
save control.mat u
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