| T O P I C R E V I E W |
| akijitsu |
Posted - 01/13/2012 : 12:34:44 PM
Hello - I hope someone can help with this problem.
I am working with data to follow the equation:
Y=Y0Exp(mX)
where m is gradient
and Y0 the value of Y at X=0 (y axis intercept)
I have plotted some data in log(Y) vs X format. I want to fit this data to find the intercept value Y0.
I am not confident of the answers provided by simply performing a linear fit. |
| 5 L A T E S T R E P L I E S (Newest First) |
| Hideo Fujii |
Posted - 08/13/2012 : 10:43:04 AM
Hi Chris,
I don't see any strange thing here because the intercept is the Y value at X=1 (i.e., 10^0) - far below your current Y scale.
Does it make sense?
--Hideo Fujii
OriginLab |
| cjg |
Posted - 08/12/2012 : 02:34:46 AM
A similar problem on a log/log plot gives a funny answer (see attached).
Certainly not the intercept to the power 10.
Hope I am missing something obvious?
regards
Chris
 |
| Hideo Fujii |
Posted - 01/17/2012 : 11:01:18 AM
Hi akijitsu,
Since it is semi-log scale on Y, i.e., log10(y)=a0+a1*x,
actual intercept should be y(0)=10^a0.
--Hideo Fujii
OriginLab |
| akijitsu |
Posted - 01/17/2012 : 07:14:52 AM
Hi Hideo
Thanks for your reply. I have checked 'apparent fit'. However the intercept value provided by the table is clearly wrong - it gives me a intercept value of -2 While i can clearly see the fitted line is crossing the y axis at about 0.001. Is there another setting which could be causing this error?
|
| Hideo Fujii |
Posted - 01/13/2012 : 2:26:45 PM
Hi akijitsu,
Yes, you can perform the linear fit for the semi-log plot.
Make sure that when you run the linear fit, "Apparent Fit" check box is checked as shown below:
--Hideo Fujii
OriginLab |
