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criscnunes
8 Posts |
 Posted - 11/03/2010 : 3:39:36 PM
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I want to plot a exponential function:
y=yo.e-at
I can´t find this function in Origin and I don´t know how to type it.
Thank you
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3Ls
USA
2 Posts |
Posted - 11/03/2010 : 4:35:29 PM
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I think this function is installed in the origin.
go to analysis, then select non-linear curve fit,
Once you see the "select function" button, you will see many choices.
Your equation is in the class of expotential.
Good luck
Live, Laugh, and Love |
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criscnunes
8 Posts |
Posted - 11/04/2010 : 07:16:02 AM
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Thank you for your reply, but I couldn´t find my function.
Y=Yo*exp(-a*x)
Im my case, Y ist the concentration and x ist the time. a is the biodegradation coefficient.
I selected non-linear curve fit and didn´t see saw I need.
- ExpAssoc
y = y0 + A1*(1 - exp(-x/t1)) + A2*(1 - exp(-x/t2))
- ExpDec1
y = A1*exp(-x/t1) + y0
- ExpDec2
y = A1*exp(-x/t1) + A2*exp(-x/t2) + y0
- ExpDec3
y = A1*exp(-x/t1) + A2*exp(-x/t2) + A3*exp(-x/t3) + y0
ExpGrow1
y = y0 + A1*exp((x-x0)/t1)
ExpGrow2
y = y0 + A1*exp((x-x0)/t1) + A2*exp((x-x0)/t2)
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criscnunes
8 Posts |
Posted - 11/04/2010 : 07:24:06 AM
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I found another functions that are similar, but not exactly waht I want.
Y=Yo*exp(-a*x)
Exp1p2
y = exp(-A*x)
Exp1P3
y = A*exp(-A*x)
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3Ls
USA
2 Posts |
Posted - 11/04/2010 : 1:46:36 PM
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It is just the matter of name.
For example, you can use this,
- ExpDec1
y = A1*exp(-x/t1) + y0
yo can be fixed at 0. Once you get A1 and t1. You know A1 is your Yo and t1 =1/a. Is that right?
Live, Laugh, and Love |
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criscnunes
8 Posts |
Posted - 11/08/2010 : 08:35:03 AM
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Dear Helper,
I did what you suggested. See the images.
I got a t=0.4995, so my a=2.00
And I want to obtain as the image results excel. I choose a exponential tendence line, and the correct value for the biodegradation coefficent ist approx. -0.67.
How can I solve that?
Thank you for our help.
Cristina |
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criscnunes
8 Posts |
Posted - 11/08/2010 : 08:41:34 AM
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I tried to do as you suggested.
See the images.
My a=2.00 (a=1/t, where t=0.49995) and the correct answer ist approx. 0.67.
 
Thank you for your help.
Cristina |
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easwar
USA
1964 Posts |
Posted - 11/08/2010 : 2:27:37 PM
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Hi Cristina,
In your earlier posts you said you wanted the function
Y=Yo*exp(-a*x)
The closest built-in function is Exp2PMod1 in Exponential category:
y=a*exp(b*x)
so here the "a" would be your "Yo" and the "b" would be your "-a".
When I fit with this function it works fine on your data. Of course
if you expect the a parameter to be of a particular value such as
0.67, you can try fixing that parameter, or setting a bound around
the desired value. If you do not fix or specify bounds, the fitter
will simply try find the best mathematical value for the optimal fit,
and that value may not make physical sense to you, so if you know
what the value should be, you should fix or specify some bounds.
Also note that you can easily define a new fitting function with the
exact parameter names etc, so that is another option. Look at
tutorials on how to create a new fitting function.
In the latest version 8.5, we have added a Fitting Function Builder
tool, which is a wizard that walks you thru the process of creating a
new fitting function. You can download the 8.5 demo and try this
feature.
Easwar
OriginLab
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Topic |
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Exponential Fitting