How to do exponential curve fitting like y=a*exp(-b/x)

I can't linearization of this equation.y=a*exp(-b/x). I want to make the eqn like y=mx+c. Then find out the value of a and b by solving two matrix by matlab. Can anyone help me? Thanks in advance.

Answer 1

[Matt J]

If you make the change of variables u=log(y), c=log(a) and v=1/x, then your exponential equation becomes u=c-b*v which is linear in u and v. However, this is usually recommended only as a way of initializing an iterative fit. The latter you can do either with fit(), lsqcurvefit(), or some File Exchange options like fminspleas.

[Matt J]

Atiqur Rahman's comment moved here:

Thank you for replying. I have to use least square method to solve this problem. That's why I have to create 2 matrix. first one is 2*2 and second one is 2*1 matrix. I have given some x and y points. After solving 2 matrix I will get 2 answer. But my problem is I am not allowed to use any buitin function. I have to mannualy find the value of a and b. But stuck on finding b.

[Matt J]

Plugging in data points into the the equation u=c-b*v n will result in multiple linear equations.

[Atiqur Rahman]

Thank you

[Matt J]

You are quite welcome, but please Accept-click the answer if it helped you.

Answer 2

[Image Analyst]

I would recommend using fitnlm() instead. It's pretty much just as easy and it probably gives you a better fit.

% Define the model as Y = a * exp(-b/x) + c
modelfun = @(b,x) b(1) * exp(-b(2) ./ x(:, 1)) + b(3);  

% Now the next line is where the actual model computation is done.
mdl = fitnlm(tbl, modelfun, beta0);
% Now the model creation is done and the coefficients have been determined.

% Extract the coefficient values from the the model object.
% The actual coefficients are in the "Estimate" column of the "Coefficients" table that's part of the mode.
coefficients = mdl.Coefficients{:, 'Estimate'}

Full demo is attached.

[Atiqur Rahman]

Thanks a lot. I solved my problem. Thanks again.