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Exponential rise fit 1

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Brad1979

Electrical
Jul 16, 2010
126
An exponential decay ( y = C*e^(B*x) ) is straightforward to fit by taking the LN(y) and using LINEST in Excel. But does anybody have a good way to fit the equation y = A - C*e^(B*x), where x and y are data points and A, B, and C are all unknown?

 
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For any function you want to fit to data use the Solver tool to minimise the sum of the square of the differences between the actual and predicted Y value.

Have a column of data with initial estimates of A,B, and C and, using the function you want to fit, have a column of the square of the difference between the estimated function and the Y data values, with the total of these squared differences in another cell. Use the Solver to minimise that cell of the sum using the cells of A,B,C as the parameters to vary. You might have to play around with the initial estimates of A,B and C to get the best fit.

Tara
 
Yi = A - C*e^(B*Xi)

You have an estimate of A, it is simply the average of the measurements Yi.

You can build a new random variable Y1i as follows:
Y1i = -1 * (Yi – Average{Yi} )

Then you can solve C and B using your old method based on:
Y1i = C*e*(B*Xi).

BUT, that does impose a constraint that the actual mean is exactly equal to your estimated mean, when in fact there is variability that must be apportioned to all the unknowns model parameters, including the mean. So, you will get a lower residual sum of squares (better fit) if you use the method suggested by Corus.


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(2B)+(2B)' ?
 
Sorry, I was way off base. The method I proposed assumed the mean of the 2nd term was zero, when there was no basis for that.

Never mind...

=====================================
(2B)+(2B)' ?
 
Thanks corus, I think your solution should work for me.
 
why not convert your "y" to "y-A" ?

instead of ln(y) linear fit to ln(y-A), choose A to maximise the r^2 term ??
 
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