Fitting experimental data with exponential curve

  • #1

Main Question or Discussion Point

Hello! :-)

I' have a problem...

I fitted experimental data with a second order exponential theoretical curve:

y = A1*exp(-x/t1)+ A2*exp(-x/t2) + y0

and I have these data for 4 repetitions of the same experiment:


first time:

Reduced Chi-Sqr = 6,81066E-4
Adj. R-Square = 0,41005

y0 = 1,60364 (Standard Error = 0,0012)
A1 = 0,03892 (Standard Error = 0,00349)
t1 = 1,80889 (Standard Error = 0,30923)
A2 = 0,29705 (Standard Error = 0,01466)
t2 = 0,05342 (Standard Error = 0,00445)


second time:

Reduced Chi-Sqr = 9,52433E-4
Adj. R-Square = 0,34084

y0 = 1,05106 (Standard Error = 8,31296E-4)
A1 = 0,27818 (Standard Error = 0,02292)
t1 = 0,02764 (Standard Error = 0,00432)
A2 = 0,09086 (Standard Error = 0,01005)
t2 = 0,47223 (Standard Error = 0,06265)


third time:

Reduced Chi-Sqr = 6,13655E-4
Adj. R-Square = 0,45827

y0 = 0,60035 (Standard Error = 0,00425)
A1 = 0,02735 (Standard Error = 0,0032)
t1 = 4,40428 (Standard Error = 1,76829)
A2 = 0,27926 (Standard Error = 0,01113)
t2 = 0,0895 (Standard Error = 0,00593)



forth time:

Reduced Chi-Sqr = 9,87583E-4
Adj. R-Square = 0,36993

y0 = 0,16076 (Standard Error = 8,19131E-4)
A1 = 0,28751 (Standard Error = 0,02368)
t1 = 0,02714 (Standard Error = 0,00429)
A2 = 0,10903 (Standard Error = 0,01115)
t2 = 0,42818 (Standard Error = 0,05112)



I discarded the second because of the big standard error of t1. But, this has the bigger adj. R squared.

I thought that the first and third are the best, because values A1, A2, t1, t2 are near.
Is this correct?
How can I evaluate them by the reduced chi-squared and adj. R squared. R squared is so low because of the noise..
 
Last edited:

Answers and Replies

  • #2
StatGuy2000
Education Advisor
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881
First of all, a reduced chi-squared test only really makes sense if we can assume that the errors in the term follows a normal (Gaussian) distribution. Have you tried fitting a QQ plot of the data?

Assuming that the data is in fact normally distributed (or close enough to it), it may be possible that the exponential theoretical curve you gave may not give you a good fit of the data. In that case, you may have to try doing some exploratory plots of the data and try fitting various generalized linear or additive models and see if those may fit it better.

Without knowing more, that's the best advice I could give.
 
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  • #3
Thank you! :-)

Here is the fitting:

fl1.png


fl2.png


fl3.png


fl4.png
 
  • #4
34,389
10,476
Plots 1 and 2 look different, even by eye. If they are supposed to have the same shape, something went wrong. Looks like there is a group (1 and 3) and one (2 and 4).
The numbering is confusing, the pictures do not match to the order in the first post.

To compare amplitudes, I think it would be better to fit
A*(f1*exp(-x/t1)+ f2*exp(-x/t2) + (1-f1-f2))
where A is the (not relevant?) overall amplitude and f1 and f2 are relative fractions of the signal.
Alternatively, if the constant (background?) is not related to the exponentials, use
A*(f1*exp(-x/t1)+ (1-f1)*exp(-x/t2)) + y0
 
  • #5
Thanks you!

Yes, the numbering was confusing. I corrected it.

Plots 1 and 2 look different, even by eye. If they are supposed to have the same shape, something went wrong. Looks like there is a group (1 and 3) and one (2 and 4).
That's the problem. They supposed to be the same. That's why I'm trying to evaluate the fitting, in order to find the best.. Is it correct? I don't know what else I can do... The first group has higher R^2 than the second. Is it a correct criterion? to said that 1 and 3 are better?

To compare amplitudes, I think it would be better to fit
A*(f1*exp(-x/t1)+ f2*exp(-x/t2) + (1-f1-f2))
where A is the (not relevant?) overall amplitude and f1 and f2 are relative fractions of the signal.
Alternatively, if the constant (background?) is not related to the exponentials, use
A*(f1*exp(-x/t1)+ (1-f1)*exp(-x/t2)) + y0

y0 is not related to the exponentials..
 
  • #6
34,389
10,476
They supposed to be the same.
Then your experiment has a problem somewhere.
The easiest explanation I see are different compositions of the samples, if the two exponentials come from different origins. If 1 and 3 just don't have enough of the longer-living component, that's fine if the composition can vary. You should estimate the longer lifetime based on samples 2 and 4 then. You can then use this as fixed parameter to fit samples 1 and 3.

Alternatively (better, but more complicated), make a fit to all samples at the same time.
 
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  • #7
It is the same sample... :-(

make a fit to all samples at the same time.
I didn't understand.. Could you explain me what you mean?
 
  • #8
34,389
10,476
I didn't understand.. Could you explain me what you mean?
Currently you make a fit to all data points of sample 1, then you make a fit to all data points of sample 2, and so on.

You could make a single fit to all data points at the same time, where all 4 curves would share the same lifetimes, but have different amplitudes.
 

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