Estimate exponential equation from graph

Question

Output y of a system is a non increasing function of an input x Based on my studies I have found that it can be estimated as below $$ y = be^{ax}+c $$ From measurement I have five values of y for x that are not equally spaced. I.e I have values of y for x= 5,10,20,40,50. Values of y are exponentially decreasing and hence should fit the generic equation above.

The task at hand is to find values of a,b and c from these five measurement data. I understand that I have enough points of data for calculating three unknowns. I want to know a iterative/generic logic to come up with estimations of these unknowns, since I have to do this for hundreds of measurements.

PS: Some points already known to me

• c can be estimated by looking for asymptotes
• a is negative as my function is exponentially decreasing with x

I am new to this community, Please consider that while you answer/comment

Answer 1

If $c$ is known then we have points of the form $$(x_i,be^{ax_i})$$ after shifting the data by $(0,-c)$. This relation can be rewritten as $$y=be^{ax}\implies\ln{(y)}=ax+\ln{(b)}$$ So the variables $\ln{(y)}$ and $x$ are linearly related. The Least Squares Regression line can then be found for $\ln{(y)}$ and $x$ giving a relation of the form $$\ln{(y)}=mx+c=ax+\ln{(b)}$$ Hence the value of $a$ is given by the gradient $m$ of the Least Squares Regression line between $\ln{(y)}$ and $x$. Also, the value of $b$ is given by $e^c$ where $c$ is the $y$-axis intercept of the Least Squares Regression line between $\ln{(y)}$ and $x$.

Answer 2

You must be careful with this type of problems. In the least square sense, what you want is to minimize $$SSQ(a,b,c)=\sum_{i=1}^n(b\,e^{a\,x_i}+c-y_i)^2$$ which is nonlinear because of $a$ and then you need reasonable estimates to start the nonlinear regression.

If $c$ is known, the problem is quite simple because you can, in a first step, linearize the model as $$\log(y-c)=\log(b)+a x=\beta + ax$$ as Peter Foreman did explain. Then, a standard linear regression gives estimates of parameters $(a,b)$. However, you must not stop at this point because what has been measured is $y$ and not any of its possible transforms.

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