ezfit using a function

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aneps 2022-7-8

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编辑： Walter Roberson 2022-7-13

I use ezfit to fit a gaussian in my data. It is straight forward as it is in the ezfit examples:

f1 = ezfit(x,counts,'gauss');

showfit(f1)

I want to fit my own function. So before doing that, I attempted using the equation of gaussian, exaclty the same equation in the ezfit documentation to fit the data ( http://www.fast.u-psud.fr/ezyfit/html/ezfit.html ). I am doing this to check if I am doing any mistakes. I did as follows

f1 = ezfit(x,counts,'a*exp(-(x-x0)^2/(2*s^2))');

showfit(f1)

This time it doesn't fit! Could you please tell me what is wrong in it?

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Natalia Demidova 2022-7-12

Hi,

Notice in the ezfit documentation the way functions are specified are with syntax of "y=" as in a usual function. Thus it should look like:

f1 = ezfit(x,counts,'y = a*exp(-(x-x0)^2/(2*s^2))');

Further for your own function you can specify specific guesses for parameters (such as a and s here). The first guesses for the function (ex. a=1 and s=2) to try would be after semicolon in the same string and look like:

f1 = ezfit(x,counts,'y = a*exp(-(x-x0)^2/(2*s^2)); a=1; s=2');

Have fun :)

aneps 2022-7-13

Data.txt

Still it doesn't fit! I used f1 = ezfit(x,counts,'y = a*exp(-(x-x0)^2/(2*s^2))');, it doesn;t work. But if I use f1 = ezfit(x,counts,'gauss'); it do the job! Don't understand what cause this difference. Data attached.. Here, first I plot the histogram of column 9. Then I do the curve fit on the histogram.

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