How do I apply exponential and logarithmic curve fitting

Hi,
I have some scatterplots and I want to check the relationships between the variables which resemble exponential and logarithmic functions. I have tried to use the functions
nlinfit
fittype
fit
but so far not successfully (poor fitting or code not working at all).
How can I check the curves for the above scatters which seem to have the following functions:
y= a*exp(b*x)+c and y=log_a(x)+b
in matlab?
Thanks
Iro

3 Comments

What do you mean by ‘code not working at all’?
It’s not obvious to me what you mean here: y=log_a(x)+b. (I don’t understand ‘log_a(x)’.)
Iro
Iro on 20 Feb 2014
Edited: Iro on 20 Feb 2014
I mean that I make some kind of mistake in the definition of the function when I use nlinfit so the code does not work.
With log_a(x) I mean the logarithm of x with base a.
Have you tried fitting an exponential of the inverse? If x = f(y), then y = f^-1(x) and you can use 'fit' and the 'exp' for an exponential fit.

Sign in to comment.

 Accepted Answer

The nlinfit function requires that the first argument of the objective function is the parameter vector and the second the vector of independent variables. Using anonymous functions,
y = a*exp(b*x)+c
becomes
y = @(B,x) B(1).*exp(B(2).*x) + B(3); % B(1) = a, B(2) = b, B(3) = c
For the logarithmic fit, all logs to various bases are simply scaled by a constant. Consider:
a^y = b^x
Taking the log to base a (denoted by loga()) of both sides gives:
y = x*loga(b)
so the log to any base will work.
The anonymous function for your logarithmic regression is then:
y = @(B,x) log(x) + B; % B = b
or alternatively,
y = @(B,x) B(1).*log(x) + B(2);
Those should work with nlinfit.

More Answers (1)

You can use cf = fit(x,y,'exp1'); where x and y are your set of points.

1 Comment

Hi, thanks for your answer. I have used this function for the exxponential scatter but it doesn't give me good fitting. Do you know how I can get into account the constant c in the above equation using fit with 'exp1' ?
Thanks

Sign in to comment.

Categories

Asked:

Iro
on 19 Feb 2014

Commented:

on 25 Sep 2016

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!