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chrisUser
Germany
2 Posts |
 Posted - 04/19/2017 : 08:33:38 AM
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Origin Ver. and Service Release: OriginPro9.1.0G (64-bit) Sr3
Operating System: Windows 7
Hey guys,
following situation:
I have a data set which is exponential, so I can fit an exponential decay function, in detail it's the ExpDec1 function that fit's my data best. I now need to define an x range of values in which my fit takes place. Furthermore I need to define the amount of x values that are included in the fit and the function now has to find the best fit (including start value) by itself in the x range that was initially given by me. So I think the x value needs to be a parameter, too, but I don't know how to handle this. Is this possible? I'm not an expert in this programm and in programming itself.
Best regards
Chris
On the picture you can see my situation decribed. The marks symbolize the range where the fit should take place.
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Edited by - chrisUser on 04/19/2017 08:38:28 AM |
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arstern
USA
237 Posts |
Posted - 04/19/2017 : 11:05:37 AM
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Hi Chris,
1) You can specify the x range of values for your fitted curve by selecting the Data Selection under the Settings tab. Under Rows you can select your range by x or by row number.
2) You can specify the amount of x values by selecting Fitted Curve under the Settings tab. Input the number of points under X Data Type that you want to plot.
Thanks,
Aviel
OriginLab |
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YimingChen
1593 Posts |
Posted - 04/19/2017 : 11:57:36 AM
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Hi Chris,
Basically taking the starting x value(x0) as fitting parameter won't work here. Because the fitting algorithm gets best fitting by minimizing the residual sum of square: sum(y_fit - y)^2, which always decreases as the number of data points decreases. You will probably end up with a x0 value so that only few data points are included in the range for fitting. Thanks.
Yiming |
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chrisUser
Germany
2 Posts |
Posted - 04/20/2017 : 04:54:10 AM
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Hey you both,
many thanks for your quick answer!
@Aviel, this is what I already know, but this doen't eliminate the problem of dynamic range for the fit with a varying start value.
@Yiming, not glad to here.. Isn't there a possibility to solve the problem? |
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Sam Fang
291 Posts |
Posted - 04/21/2017 : 04:51:36 AM
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Hi Chris,
Parameter for x start value can't change range of input data in the fitting. You can use loop to fit data by increasing the x start value each time, and find the x start value whose Reduced ChiSqr is the smallest or R-Square value is close to 1.
We once help a customer find the range of linear section.
Sam
OriginLab Technical Services |
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chrisUser
Germany
2 Posts |
Posted - 04/27/2017 : 05:51:44 AM
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| Thank you for your help! |
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ExpDec varying start value