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lesaout
Switzerland
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 Posted - 05/02/2007 : 03:35:17 AM
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Origin Version (Select Help-->About Origin): OriginPro7.5 SR6
Operating System: XP
Dear All,
I have experimental data D(t) and try to fit them using a numerical series as:
D(t=0) = 0
D (t) = D(t+dt)+ dt*K1 (K1 is the parameter to adjust, the real function used is of course a little bit more complicated).
Do you think it is possible under ORIGIN?
Thanks for your answer,
Gwenn |
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larry_lan
China
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Posted - 05/02/2007 : 07:23:33 AM
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Better to know what your data looks like, and how you define dt. Generally, dt = t[i] - t[i-1]. If so, dt should be equal spaced, or we don't know what's D(t+dt). Another comment is, can you change the equation like:
D(t=0) = 0
D(t[i]) = D(t[i+1])+ dt*K1 --> D(t[i+1]) = D(t[i]) + dt*K1
So could you please provide more information?
Larry
OriginLab Technical Services |
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lesaout
Switzerland
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Posted - 05/02/2007 : 09:41:01 AM
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quote:
Better to know what your data looks like, and how you define dt. Generally, dt = t[i] - t[i-1]. If so, dt should be equal spaced, or we don't know what's D(t+dt). Another comment is, can you change the equation like:
D(t=0) = 0
D(t[i]) = D(t[i+1])+ dt*K1 --> D(t[i+1]) = D(t[i]) + dt*K1
So could you please provide more information?
Larry
OriginLab Technical Services
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lesaout
Switzerland
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Posted - 05/02/2007 : 09:41:48 AM
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Dear Larry,
Experimentally, at early age (from t=0 to 3 days) we have points with a short delta t (delta t=6hours for example) and at later age we have points with a large delta t (delta t=1 month). The experiment time is about 2 years.
For the fitting, dt is defined as the time step. We have to define the fitting function D at early age with a short dt until 3 days then we can increase the step (for ex. dt= 15 days).
dt is defined as: dt = t[i] - t[i-1]
The equation can be changed like: D(t[i+1]) = D(t[i]) + dt*K1
Thanks for all,
Gwenn
Better to know what your data looks like, and how you define dt. Generally, dt = t[i] - t[i-1]. If so, dt should be equal spaced, or we don't know what's D(t+dt). Another comment is, can you change the equation like:
D(t=0) = 0
D(t[i]) = D(t[i+1])+ dt*K1 --> D(t[i+1]) = D(t[i]) + dt*K1
So could you please provide more information?
Larry
OriginLab Technical Services
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larry_lan
China
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Posted - 05/02/2007 : 11:45:33 AM
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Hi Gwenn
There are several methods, such as, you can add one column, say dt, and set dt = x[i] - x[i-1], like the graph below, and define a two Independent Varianble fitting function, like:
Parameter Names: K1
Independent Var: x, dt
Dependent Var: y
Function:
if (x == 0)
{
y = x;
} else
{
y = x + dt*K1;
}
Then you can fit and evaluate K1.
Hope it helps
Larry
OriginLab Technical Services
Edited by - larry_lan on 05/02/2007 12:12:14 PM |
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fitting with numerial series