Exponential Fitting function not plotting the same information as the data points?
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Hi, my brain is pickled with this one. Below are the two graphs I have plotted with exp2 function. The points do not match the curve, and this is ultimately changing my entire answer, as it is giving the wrong values out, and I cannot understand why?
Here is the code I am using, both graphs plot a concentration against time, but yet give different results:
CH4_fit = fit(Res_time, CH4_exp, 'exp2');
CH4_coeff = coeffvalues(CH4_fit); %Co-efficient values for exponential fitting
CH4_pred =(CH4_coeff(1)*exp(CH4_coeff(2)*Res_time)) + ...
(CH4_coeff(3)*exp(CH4_coeff(4)*Res_time));
plot(Res_time,CH4_exp, Res_time, CH4_pred);
Can I just added that the exact same data was run on different computers, and it gave the same equation co-efficients exactly (to 4.dp) and the same times, but yet still outputs different concentrations on my version? I have the R2018b, and I have just used default settings (don't know how to change anything, so I definitely haven't).
7 个评论
David Goodmanson
2019-2-14
编辑：David Goodmanson
2019-2-15
Hi Lewis,
could you explain a bit more? Is the second plot a good fit done on another computer?
Lewis Brammeld
2019-2-14
David Goodmanson
2019-2-14
so are you saying the fit in the second image (fig 5) is not good enough?
Lewis Brammeld
2019-2-14
David Goodmanson
2019-2-14
编辑：David Goodmanson
2019-2-14
I don't know what you mean by 'the fit for the line' (is it in a buried plot?). I think for anyone to weigh in with an opinion you will have to provide more detail on the problem.
John D'Errico
2019-2-14
编辑：John D'Errico
2019-2-14
We don't have your data. That makes it impossible to really help you. Please attach the data points as a .mat file to a comment.
It may be true that a picture is worth a thousand words, but then real data? More like 10 thousand words. ;-)
Lewis Brammeld
2019-2-14
采纳的回答
John D'Errico
2019-2-14
编辑：John D'Errico
2019-2-14
Where should I start? :-)
A terribly important feature of ANY exponential model is you absolutely, positively need good start points.
Asecond important thing is you need good data. Here, you have crappy data. (Sorry, but that is a tough model to fit well, because sums of exponentials are difficult to fit.) So better data would really help.
mdl = fit(x,y,ft,'start',[.05 350 .05 30])
mdl =
General model:
mdl(x) = a*exp(-b*x) + c*exp(-d*x)
Coefficients (with 95% confidence bounds):
a = 0.05937 (0.04962, 0.06911)
b = 368.4 (279.8, 457)
c = 0.01068 (0.0009941, 0.02036)
d = 61.21 (12.76, 109.7)
plot(mdl)
hold on
plot(x,y,'bo')
As exponential fits go, not terrible. I had to spend a few minutes to find good starting values for the parameters. If you did not do that, letting fit choose starting values for you? You would then expect complete crap for a result.
Again, the most important thing is to remember that exponential models NEED good starting values, especially for the rate parameters.
2 个评论
Lewis Brammeld
2019-2-14
John D'Errico
2019-2-14
编辑：John D'Errico
2019-2-14
In my 29 year career of being a consultant on these topics, I may never have had a client who was able to give me data that was as good as I really wanted. (Well, ONCE, a client was willing to get absolutely anything I asked for. They were desperate.)
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