| T O P I C R E V I E W |
| lsanii |
Posted - 07/27/2004 : 4:30:37 PM
Here is the issue: I have what has been characterized in the literature as a biexponential process. When I try to fit the data to this using Analysis, fit exponential decay, second order..I get the same value for A1 and A2, the same values for T1 and T2. Talking to labmates I was told that this information means I actually have a monoexponential process and I should try fitting it as such. Does anyone else have this issue? Can anyone tell me if this is true?
thanks
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| 3 L A T E S T R E P L I E S (Newest First) |
| Mike Buess |
Posted - 07/28/2004 : 2:52:55 PM
Hi Easwar,
The current initialization procedure might be good enough. At least it gives starting points for adjusting the time constants. You know that one will be longer and the other shorter than the initial value. Initializing from the 1st and 3rd 1/3 of the dataset might work in some cases.
The Fit Exp Decay->2nd Order menu command gives reasonable starting (and sometimes final) values. It would be convenient if after using that command you could open the NLSF tool with the results already plugged in as initial values for further iterations.
Mike Buess
Origin WebRing Member
Edited by - Mike Buess on 07/28/2004 2:53:50 PM
Edited by - Mike Buess on 07/28/2004 2:56:33 PM |
| easwar |
Posted - 07/28/2004 : 12:09:13 PM
Hi Mike,
As you correctly point out, we are simply setting A1=A2 and t1=t2 in our initialization code, which is not very useful - we will try to imporve the initialization code - any suggestions are welcome.
What we do in the initialization is to take the log of the data, perform a linear regresssion on that to extract the A and t values. We do this on the entire dataset. Perhaps with two-term exponential, we should do this separately for the 1st 1/3rd of the data and the last 1/3rd of the data to get possibly different A1/A2 and t1/t2 initial values.
Of course, if the data does not correspond to distinctly different expotents (thus having distinctly different A1/A2 and t1/t2) values, the above will not help either and user intervention/decision is needed as Mike points out.
Easwar
OriginLab
Edited by - easwar on 07/28/2004 12:14:16 PM |
| Mike Buess |
Posted - 07/27/2004 : 5:09:19 PM
Hi,
I deal often with biexp decay and success with Origin's ExpDec2 function (I assume that's what you're using) depends greatly on parameter initialization. (This is common to most fitting functions.) Automatic initialization by the fitting tool sets A1=A2 and T1=T2, which is monoexponential with a decay time somewhere in between the true T1 and T2. After that you need to provide better guesses for the actual time constants and prefactors and try again. If your data are truly biexponential the fitting procedure (with your guidance) will eventually home in on the true parameters.
...After reading back over your post it seems you're using the Analysis->Fit Exponential Decay->Second Order menu command. That does more than initialize the parameters but some follow-up work with the Advanced Fitting Tool is usually required. After you've performed a fit with Fit Exp Decay->Second Order you should do the following...
1. Select Analysis->Non-linear Curve Fit->Advanced Fitting Tool.
2. Select the ExpDec2 function and your dataset.
3. Starting with decay time obtained with the Second Order command, increase t1 and decrease t2 (or vice versa) and select the 100 Iter. button to fit again. You may need to try several times and adjust the other parameters (A1, A2 and y0) as well.
Even if your process is biexponential it's possible that the time constants are nearly the same. I've found that if t1 and t2 differ by less than a factor of 2 or 3 a monoexp fit to the data "looks" just as good as the true biexp fit.
Mike Buess
Origin WebRing Member
Edited by - Mike Buess on 07/27/2004 6:46:28 PM
Edited by - Mike Buess on 07/27/2004 7:02:20 PM |
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