| T O P I C R E V I E W |
| Bobby64 |
Posted - 08/21/2012 : 12:01:58 PM
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 |
| 2 L A T E S T R E P L I E S (Newest First) |
| Bobby64 |
Posted - 08/23/2012 : 03:08:03 AM
Ok Sam, understood.
Many thanks for this clarification!
Cheers
Bobby64 |
| Sam Fang |
Posted - 08/22/2012 : 05:57:13 AM
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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