Custom Fit Fit options

I have created a custom fittype and tried let MATLAB itself find the starting point, given lower and upper bounds that I specify. However, those bounds seem to have been ignored when the fit process was performed (see attachments). Does anyone know how I can force MATLAB to search for the coefficients between the bounds I have specified? Thank you.

Answer 1

[Matt J]

The bounds weren't ignored. You defined them in "Options", but just forgot to pass Options to the fitting routine.

[J.S.]

Hello Matt,

Thanks for the suggestion. I tried including Options in my fit call (see attachment), but I'm getting an error. Do you know I can fix my syntax in f_norm? Thanks!

[Matt J]

It would help us both if you would copy/paste your code and errors/output into your posts, rather than communicating them via screenshots. That way, it is easier for people to copy/paste modifications of the code for you.

[J.S.]

Sorry about that.

    x0=0.6; y0=0.6;
    HOF=fittype('a*exp(b*(x-x0))+c*exp(d*(x-x0))+y0','coefficients',{'a','b'...
                        'c','d'},'problem',{'x0','y0'},'independent','x');

   L=[0,(-1/(0.1e-9)),0,(-1/(0.1e-9))];   
   U=[1,(-1/(1e-6)),1,(-1/(1e-6))];
   Op=fitoptions(HOF);
   Op.Lower=L;
   Op.Upper=U;
   f_normTEST=fit(binTimeCol,bindataColNorm,HOF,'problem',{x0,y0},Op);

binTimeCol and bindataColNorm are two column vectors of the same length.

[Matt J]

Well, the problem is probably that you are now constraining b,d to a region where the exp() operations overflow to infinity, as in the following:

>> exp(1000)

ans =

     Inf

You need to double-check the magnitudes of (x-x0) to be sure that range is sensible.

One other remark. I recommend reformulating the model as follows

     U=[1,(-1/(1e-6)),1,-1];

 HOF=fittype('a*exp(b*X)+c*exp((b+d)*X)',...
             'coefficients',{'a','b','c','d'},'independent','X');

 f_normTEST=fit(binTimeCol-x0,bindataColNorm-y0,HOF,Op);

There are 2 main changes. First, the input data is pre-offset by x0,y0 so that the fitting routine doesn't have to repeat it unnecessarily in every iteration. Second, one of the exponential parameters is now represented as b+d. This tells the solver that those parameters are meant to be nonequal, since that would make a,c ambiguous.