non linear parameter estimation least squares

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Alex
Alex on 6 Jul 2011
I have the following equation that i want to find the parameters c1, c2 and c3.
E = SUM(yj-(1/(c1*c2))*(c3-asinh(sinh(c3)*exp(c1*j*2*pi))))^2
yj is consecutive timings for a tennis ball, on a string revolving around a pole (the ball cannot change height)
I have found some help with this:
x = [ 1,2,3,4,5 ];
y = [ 2.2, 1.9, 1.7, 1.5, 1.4];
a11 = sum(x.^2); a12 = sum(x); a21 = sum(x); a22 = sum(ones(1,length(x)));
A = [ a11,a12; a21,a22] % the coefficient matrix of the minimization problem
c1 = 1;
c2 = 1;
c3 = 1;
diff_c1 = (c3-asinh(exp(2*pi*x.*c1).*sinh(c3)))/(c1^2*c2)+(2*x.*pi*exp(2*c1*x.*pi).*sinh(c3)*asinh(exp(2*c1*x.*pi).*sinh(c3)))/(c1*c2)
diff_c2 = (c3-asinh(exp(2*c1*x.*pi).*sinh(c3)))/(c1*c2^2)
diff_c3 = (exp(2*c1*x.*pi).*cosh(c2)*asinh(exp(2*c1*x.*pi).*sinh(c3))-1)/(c1*c2)
b1 = sum(diff_c1*y); b2 = sum(diff_c2*y); b3 = sum(diff_c3*y);
b = [ b1; b2; b3 ] % right-hand-side of the minimization problem
c = A \ b % solution of the minimization problem
xApr = 0 : 0.001 : 1; yApr = c(1)*xApr + c(2);
plot(x,y,'*g',xApr,yApr,'b');
Basically diff_c1,c2,c3 are the parts multiplied by the data values yj for least squares minimization, i think this is correct. b1,2,3 are these multiples.
I think the problem is I haven't got my '.''s in the correct place. I know that they are required to say do this for all x and y's but Im not sure exactly where they are required.
I have posted this here because I would also appreciate some help as to whether I am doing this the correct way - because I also require initial values of c1,2,3 which i have stated as 1 but again I am not sure if I have done that correctly
  1 Comment
bym
bym on 6 Jul 2011
is there some reason you are not using function like lsqcurvefit ?

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