## simultaneous fitting of two plots with two functions

### Arun Kumar Bar (view profile)

on 16 Sep 2018
Latest activity Commented on by Arun Kumar Bar

on 25 Sep 2018

### Torsten (view profile)

I have two sets of dependent variable column vectors (y1, y2) for a set of independent variable column vectors (x). I want to fit the (x, y1) plot with the function f1(ca. f1 = (a + (b-a)*(1 + c*x^d))) and the (x, y2) plot with the function f2(ca. f2 = (b-a)*c*x^d)) simultaneously. Could anyone please help?

Kaushik Lakshminarasimhan

### Kaushik Lakshminarasimhan (view profile)

on 16 Sep 2018
What do you mean by simultaneous? Based on your equations, the transformation from x to y_1 is independent of y_2 and likewise x -> y_2 is independent of y_1. So you can just fit the two functions separately.
Arun Kumar Bar

### Arun Kumar Bar (view profile)

on 17 Sep 2018
The parameters a, b, c and d are to be same for both the fits. If I fit separately, it will not guarantee same values of the parameters for both the fits. Anyway, thank you.

R2018a

### Torsten (view profile)

on 17 Sep 2018

Use "lsqcurvefit" with the data set xdata = (x,x),ydata = (y1,y2).
Best wishes
Torsten.

Arun Kumar Bar

### Arun Kumar Bar (view profile)

on 25 Sep 2018
As I mentioned earlier, for the set of independent vectors x1, there are two sets of vectors, y1 and y2. So essentially, I should be able to fit them separately, ca. (x1 vs y1) and (x1 vs y2) plots. When I use APP in MATLAB, it works fine with the following custom equation:
y=f(x1)=m1+((m2-m1)*(1+(((2*pi*x1*m3)^(1-m4))*sin(m4*pi/2)))/(1+(2*((2*pi*x1*m3)^(1-m4)*sin(m4*pi/2)))+((2*pi*x1*m3)^((2-(2*m4))))))
But, it does not work in CommandWindow when I use the same equation in the function:
fun=@(m,x1) (m(1)+((m(2)-m(1)).*(1+(((2.*pi.*x1.*m(3)).^(1-m(4))).*sin(m(4).*pi/2)))/(1+(2.*((2.*pi.*x1.*m(3)).^(1-m(4)).*sin(m(4).*pi/2)))+((2.*pi.*x1.*m(3)).^((2-(2.*m(4))))))));
The error is a bit surprising to me! Both the x1 and y1 are 24x1 matrices (column vectors). While, the fun is returning a 24x24 matrix, with all identical elements.
Suppose, if I use, m0=[1.1; 17; 1.85; 0.54]; % these are the parameters obtained from the best fit using APP with the above custom equation; fun(m0,x1) returns a 24x24 matrix with all identical elements. Could you please suggest where I am going wrong ?
Kind regards, Arun
Torsten

### Torsten (view profile)

on 25 Sep 2018
... .*pi/2)))./(1+(2.* ...
... .*pi/2)))/(1+(2.* ...
Best wishes
Torsten.
Arun Kumar Bar

### Arun Kumar Bar (view profile)

on 25 Sep 2018
That's great! Thanks a lot, Torsten!
Kind regards, Arun

### Torsten (view profile)

on 17 Sep 2018

xdata=[x(:,1); x(:,1)];
ydata=[y1(:,1); y2(:,1)];
fun=@(x,xdata) [f1(xdata(1:numel(xdata)/2),x(1),x(2),x(3),x(4)); f2(xdata(1:numel(xdata)/2),x(1),x(2),x(3),x(4))];
x0=[- - - -];
x=lsqcurvefit(fun,x0,xdata,ydata);
where f1(...),f2(...) must be column vectors each of the same size as x.