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 Non linear fit: diff. betw. graphs and values
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Bobby64

Spain
5 Posts

Posted - 08/21/2012 :  12:01:58 PM  Show Profile  Edit Topic  Reply with Quote  View user's IP address  Delete Topic
Origin Ver. and Service Release (Select Help-->About Origin): OriginPro 8.5
Operating System: Windows

Hi,

I would need some advice from specialists of non linear fitting with Origin. Here is my problem: I fit a function through experimental data points and I systematically observe a discrepancy between the fitting results (i.e. the numerical values in the table) and the graph.

This graph will show you an example (Bottom: zoom on the low X values)



The fittign function I use is a user defined function (called DSE3), with 5 parameters to fit . The last parameter (called DE) is in theory the value of the function at Y=0. As you can see on the table, this value is of 1967.05. However, when I check on the graph, the real value seems to be between 1500 and 1600... SO my questions are : What is the correct value ? Do you have an idea to explain such a difference between the graph and the table ? Do you think that the values indicated in the table are reliable ?

In fact, when I use a slightly different function (function DSE), I do not have this kind of problem, i.e. the numerical value of the DE parameter and the graph do match (see below)...



So I guess thet the problem may come from the function DSE3 itself, but I do not understand why, since there is no indication of problems with the fitting (chi-2 is reduced, r2 is excellent...)

I would appreciate if some of you could give me some advice to understand this,

Thanks in advance for your help,

Cheers


Bobby64

Sam Fang

293 Posts

Posted - 08/22/2012 :  05:57:13 AM  Show Profile  Edit Reply  Reply with Quote  View user's IP address  Delete Reply
There is no discrepancy for fitting result of DSE3. You can notice that DE is not the value of the function DSE3 at y=0 in theory. In fact DSE3(-DE)=I2*(1-exp(DE/D2)) != 0

And DE is the value of the function DSE at y=0 in theory indeed, DSE(-DE)=0

DSE3 and DSE can be converted into each other. If we assume DSE function is expressed as:
y=I1b*(1-exp(-(D+DEb)/D1))+I2b*(1-exp(-(D+DEb)/D2))

then parameters in DSE corresponding to DSE3 can be calculated:
I1b=22.68773
I2b=39.13803
DEb=1580.787
D1=2388.74029
D2=14732.8549

The value of the function DSE3 at y=0 is -1580.787.

Sam
OriginLab Technical Services
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Bobby64

Spain
5 Posts

Posted - 08/23/2012 :  03:08:03 AM  Show Profile  Edit Reply  Reply with Quote  View user's IP address  Delete Reply
Ok Sam, understood.

Many thanks for this clarification!

Cheers

Bobby64
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