1. Least Square Method Curve Fitting

2. Linear Least Square Method

4. ในกรณีที่มีค่าพารามิเตอร์จำนวน ตั ว

6. Example 1 on page 362 % Linear1.m k =[141 138 119 115]'; T = [390 500 1000 1500]'; X = [T,[1 1 1 1]']; y = k; p =inv(X'*X)*X'*y

7. Non Linear System of Equation

8. = Jacobian Matrix

10. Example

12. clear; c1 = 1; c2 = 1; %First Guess NONLINEAR1.m for i=1:10 X = [c1;c2]; f1 = 2*(4/3-2/9*c1-8/27*c2)*(c1+c2)+(1-1/3*c1-2/3*c2)^2; f2 = 2*(5/3-2/9*c1-10/27*c2)*(c1+2*c2)+(1+1/3*c1+1/3*c2)^2; F = [f1;f2]; df1c1 = 2*(4/3-2/9*c1-8/27*c2)- 4/9*(c1+c2)- 2/3*(1-1/3*c1-2/3*c2); df1c2 = 2*(4/3-2/9*c1-8/27*c2)- 16/27*(c1+c2)- 4/3*(1-1/3*c1-2/3*c2); df2c1 = 2*(5/3-2/9*c1-10/27*c2)- 4/9*(c1+2*c2)+ 2/3*(1+1/3*c1+1/3*c2); df2c2 = 4*(5/3-2/9*c1-10/27*c2)- 20/27*(c1+2*c2)+ 2/3*(1+1/3*c1+1/3*c2); Z = [df1c1,df1c2;df2c1,df2c2]; W = -F+Z*X; Xnew = inv(Z)*W; c1 =Xnew(1);c2= Xnew(2); fprintf('%g%g%g\n',i,c1,c2); end

13. Non Linear Least Square Numerically

14. Expand {y} by Taylor

15. Example : Stress Relaxation Time Force.000001, 3.083900.016670, 1.647600.033330, 1.176900.050000,.935450.066670,.784570.083330,.681970.100000,.609550

16. Example : Stress Relaxation

17. Stess.m

18. % Parameters Estimation % Using Stress Relaxation Data % Five parameters clear;clc;clf; Data=... [ 1,.000001, 3.083900 2,.016670, 1.647600 3,.033330, 1.176900 4,.050000,.935450 5,.066670,.784570 6,.083330,.681970 7,.100000,.609550 8,.116670,.555230 9,.133330,.506950 10,.150000,.470740 20,.316670,.289690 30,.483330,.223300 40,.650000,.181050 50,.816670,.156910 60,.983330,.138810 70, 1.150000,.126740 80, 1.316700,.114670 90, 1.483300,.108630 100, 1.650000,.102600 110, 1.816700,.090530 120, 1.983300,.084490 130, 2.150000,.084490 140, 2.316700,.078460 150, 2.483300,.078460 160, 2.650000,.078460 170, 2.816700,.072420 180, 2.983300,.072420 190, 3.150000,.066390 200, 3.316700,.066390 220, 3.650000,.060350 240, 3.983300,.060350 260, 4.316700,.054320 280, 4.650000,.060350 300, 4.983300,.048280 350, 5.816700,.048280 400, 6.650000,.048280 500, 8.316700,.042250 600, 9.983300,.036210];

19. t = Data(:,2); y = Data(:,3)*1000/224.8089; y = y/y(1); plot(t,y,'ro'); legend('Data','Fit'); xlabel('Time'); ylabel('Force'); title ('Non Linear Regression'); axis([min(t) round(max(t)) 0 1]); hold on; Plothandle = plot(t,y,'-','EraseMode','xor');

20. a =[10;10;10;1;1]; % First Guess of three parameters da =1;i=0; while da>1e-20 ym = a(1)*exp(-a(2)*t)+a(3)*exp(-a(4)*t)+a(5); dyda1 = exp(-a(2)*t); dyda2 = -a(1)*t.*exp(-a(2)*t); dyda3 = exp(-a(4)*t); dyda4 = -a(3)*t.*exp(-a(4)*t); dyda5 = 1+zeros(size(t)); Jacob = [dyda1,dyda2,dyda3,dyda4,dyda5]; anew = a +0.1*inv(Jacob'*Jacob)*Jacob'*(y-ym); da = norm(anew-a); a = anew; i=i+1; fprintf( 'Iter = %gPar =%g%g%g%g %g\n',i,a(1),a(2),a(3),a(4),a(5)); set (Plothandle,'ydata',ym); drawnow; hold off; end

21. ybar=mean(y);SSr=sum((ym- ybar).^2); Syy=sum((y-ybar).^2);R2=SSr/Syy; fprintf( ' Parameter =%g%g%g%g %g\n',a(1),a(2),a(3),a(4),a(5)); fprintf( ' R_square =%g \n',R2);