Exponential Fitting function not plotting the same information as the data points?

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).

[David Goodmanson]

Hi Lewis,

could you explain a bit more? Is the second plot a good fit done on another computer?

[Lewis Brammeld]

Hi David,

Thanks for getting back to me. No, the second plot is on the same computer believe it or not, but is just weirdly changing the data? So the idea is to take the experimental data (CH4_exp) and plot it against residence time. Then fit the exp2 to the curve, for which the a-d coefficients are correct to others that have carried out the analysis. From here, the co-efficients are used with the same times, to back calculate the predicted concentrations based off of the fit. But when I do this, the predicted concentrations (CH4_pred) is not correct, and the fit line does not match the points, outputting incorrect concentration values (half of which are suddenly negative- which they shouldn't be). I have had a go with Tolfun etc, but this largely changes my a-d equation co-efficient values, and that is not desireable?

[David Goodmanson]

so are you saying the fit in the second image (fig 5) is not good enough?

[Lewis Brammeld]

Yes, I don't understand how the data points for the predicted can be so accurate, but the fit for the line is so off?

[David Goodmanson]

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]

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]

Attached is the excel file that I am drawing my information off. Apologies for not attaching this earlier, was thinking that it was a simple fix without the requirement of the data.

Answer 1

[John D'Errico]

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.

[Lewis Brammeld]

Hi John, yes the data was given to me, so apologies for that, I understand that it wasn't great! Thanks for looking into this for me, appreciate it!!

[John D'Errico]

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.)