Exponential fitting of data not working

I have a vector of 1666 values with an exponential curve. Since this vector is the result of a experimental measurement I don't have the precise law that describes it but I need its horizontal asymptotic value. To get it I thought about an exponential fitting following the model a+b*exp(-c*t) but all the methods I tried result in either an error or a straight line. What am I missing?

I've attached a .mat file containing the T_A vector (temperature measurements). t_A vector is time and can be assumed as:

t_A=linspace(0,83.3,1666);

Edit: I included another vector T_B, with which fit or other methods don't work. You can assume t_B is defined the same way as t_A.

Things I've tried:

g = fittype('a-b*exp(-c*x)'); % a is asymptotic value 
fA = fit(t_A,T_A,g,'StartPoint',[[ones(size(t_A)), -exp(-t_A)]\T_A; 1]);

Gives back "Inf computed by model function, fitting cannot continue.". I also tried to change the upper and lower bounds like it suggests but no luck. Changed starting points as well, no luck.

f = @(A,t_A) A(1)+A(2).*exp(-t_A.*(A(3)));
beta = lsqcurvefit(f,[[ones(size(t_A)), -exp(-t_A)]\T_A; 1],t_A,T_A);

Gives back "Local minimum possible" but the plot looks like a spike in 0 and then goes back to 0 and stays there

f = @(A,t_A) A(1)+A(2).*exp(-t_A.*(A(3)));
A_s = fminsearch(@(A) norm(T_A - f(A,t)), [-exp(-t_A); 1; ones(size(t_A)); 1]);

Runs infinitely, doesn't stop.

I also tried curveFitter but gives out the same results as fit (or an ugly exponential curve when it works), but it's not an optimal solution since I would have to do this numerous timesn and I'd like an automated approach.

I've never done curve fitting and I'm trying now since I need it. Would love a solution but also some insight as to why it doesn't work or how I can improve. Thank you very much to all of you.

Answer 1

[Cris LaPierre]

For attempts 2 and 3, what is A?

First one seems to work.

load T_A.mat
t_A=linspace(0,83.3,1666)';

g = fittype('a-b*exp(-c*x)'); % a is asymptotic value 
fA = fit(t_A,T_A,g,'StartPoint',[[ones(size(t_A)), -exp(-t_A)]\T_A; 1])

fA = 
     General model:
     fA(x) = a-b*exp(-c*x)
     Coefficients (with 95% confidence bounds):
       a =       62.01  (61.98, 62.04)
       b =       35.24  (35.21, 35.27)
       c =     0.03484  (0.03475, 0.03493)

plot(fA,t_A,T_A)

[Henry]

Thank you for your response. For 2 and 3, A's are the coefficients of the exponential function, the ones the fitting needs to find.

I can't believe that it works for you. I can't believe that it works for me now too. I swear I've been trying all morning. Would you be willing to try with another vector? This one doesn't seem to work.

[Cris LaPierre]

What happens when you try it? What does 'doesn't work' mean?

load T_B.mat
t_A=linspace(0,83.3,1666)';

g = fittype('a-b*exp(-c*x)'); % a is asymptotic value 
fA = fit(t_A,T_B,g,'StartPoint',[[ones(size(t_A)), -exp(-t_A)]\T_B; 1])

fA = 
     General model:
     fA(x) = a-b*exp(-c*x)
     Coefficients (with 95% confidence bounds):
       a =       62.04  (62.01, 62.08)
       b =          35  (34.97, 35.03)
       c =     0.03482  (0.03473, 0.03492)

plot(fA,t_A,T_B)

Answer 2

[Torsten]

A = load("T_A.mat");
A = A.T_A;
A = [linspace(0,83.3,1666).',A];
plot(A(:,1),A(:,2),'o')
hold on
p0 = [25 1 1];
f = @(p)p(1)+p(2)*exp(-p(3)*A(:,1));
F = @(p)f(p)-A(:,2);
p = lsqnonlin(F,p0,[],[],optimset('MaxIter',1000000,'MaxFunEvals',1000000))

Local minimum possible.

lsqnonlin stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.

p = 1×3

62.0057 -35.2389 0.0348

norm(F(p))

ans = 6.4488

plot(A(:,1),f(p))
plot([0 100],[p(1),p(1)],"k")
hold off

[Henry]

Hello, thank you for your response. Can you explain how your code works? Especially why you put the T_A vector inside another vector with the t_A vector. Couldn't I use them separately?

Also, I'm assuming that the asymptotic value is p(1), correct? How can I plot the obtained function to visualize the asymptote?

[Torsten]

Can you explain how your code works? Especially why you put the T_A vector inside another vector with the t_A vector. Couldn't I use them separately?

Yes, you could.

Also, I'm assuming that the asymptotic value is p(1), correct? How can I plot the obtained function to visualize the asymptote?

Done.