Double exponential fits, but single curves away from the data- why?

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IJ 2016-4-27

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评论： dpb 2016-5-2

采纳的回答： J. Webster

dataToFit.txt

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When I try single exponential fit (red line in attached pic) to my data (the part in blue on the attached graph), it seems to curve way from the values, and clearly does not fit. I used:

f=fit(newx(10000:11000), yy2(10000:11000), 'a*exp(-(1/b)*x)');
plot(f, newx(10000:11000), yy2(10000:11000))

I tried to filter and smooth the signal down to a single line, and I also tried cftool. When I use double exponential, it works, but for my calculations, I need a single exponential fit to get the b-coeff. Is there any reason why single exponential just does not work with this trace?

-----edit

I attached the whole trace, and I want to fit to points 10000:10500

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dpb 2016-4-27

Because the data aren't exponential, maybe, would be the firstmost that comes to mind.

Need to attach a dataset if expect somebody to try to do anything other than look at a picture.

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J. Webster 2016-4-27

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编辑：J. Webster 2016-4-28

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very steep power laws can be notoriously difficult to fit, since small changes in x lead to big changes in y, it makes it hard on the fitting algorithms.

A trick you might try is (in your case) take the natural log of the data and then fit it to the natural log of your fit equation...

ln(a e^(-(1/b)*x)) = ln(a) - x/b;

so that now your fit problem has been reduced to a linear fit y = Mx+B, with M=-1/b and B = ln(a).

Also remember in matlab that the function for ln is log() which drives me crazy.

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dpb 2016-4-29

编辑：dpb 2016-4-29

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@J -- I'm referring to the plot of

semilogy(abs(yy))

where yy=y(10000:10500), the range over which the decay is apparent. Restricting it to that range makes the range of the axes more nearly filled with the data of interest and illustrates the extreme nonlinearity.

and the others I posted in earlier remark. semilogy is simply the same thing as your suggested log transform in original coordinates; if that plot isn't linear, the data aren't first-order exponential, same as for the plot of the transformed data.

@IJ - I don't know anything about the subject matter but as is clearly shown by the plot(s) above, the curvature in your data is nearly as great still on the logarithmic scale as it is on linear; it's extremely removed from being simply a first-order exponential.

I can't explain "why"; only can point out that that's the data you have.

As for the fitting problems, log(<0) --> complex value; you can use abs(y) to solve the inverse problem or offset the data or otherwise transform to get it positive but it's still going to be stronger than single exponential.

IJ 2016-4-29

Thanks for pinpointing where the problem was, dpb and J. Webster. Now I can start looking for alternative approaches.

dpb 2016-5-2

Who knows, maybe there's a Nobel prize hidden in the extra curvature waiting to be discovered... :)

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