MATLAB curve-fitting, exponential vs linear

Question

I have an array of data which, when plotted, looks like this.

I need to use the polyfit command to determine the best fitting exponential for the time roughly between 1.7 and 2.3. I must also compare this exponential fit to a simple linear fit.

I'm given the equation Temp(t) = Temp0 * exp(-(t-t0)/tau), where t0 is the time corresponding to temperature Temp0 (I can select where to begin my curve-fitting, but it must be confined to the area roughly between 1.7 and 2.3). Here is my attempt.

% Arbitrarily defined starting point
t0 = 1.71;

%Exponential fit
p = polyfit(time, log(Temp), 1)
tau = -1./p(1)
Temp0 = exp(p(2))

tm = 1.8:0.01:2.3;
Temp_t = Temp0*exp(-(tm)/tau);
plot(time, Temp, tm, Temp_t)

figure(2)

%Linear fit
p2 = polyfit(time, Temp, 1);
Temp_p = p2(1)*tm + p2(2);
plot(time, Temp, tm, Temp_p)

My exponential fit ends up looking like . My linear fit looks like . (virtually identical). What am I doing incorrectly? Should the two fits be so similar? I am told that circshift may help, but I couldn't grasp the applicability of the command after reading the help file.

Answer 1

Things are behaving just as you are expecting. The problem is that the function you are trying to fit is not a very good approximation of the data. Eyeballing the curve, it seems that the exponential part of the curve tends asymptotically to a value around 16; but the function you are using will eventually tend to a temperature of 0. Thus, fitting to a part that's going from 22 to 16 will give you an almost linear relationship. To illustrate this I wrote a few lines of code that approximately match the data points you have - and that show how different functions (one that tends to 0, and another that tends to 16) will give you a very different shape of the curve. The first (your original function) is almost linear between T values of 22 and 16 - so it will look like the linear fit.

I suggest that you think about the "right" shape of the function to fit - what is the underlying physics that makes you choose a particular form? Getting that right is essential...

Here is the code:

time = linspace(1.5, 2.5, 200);
t0 = 1.7;
t1 = 2.3;
tau = 2.0;

% define three sections of the function:
s1 = find(time < t0);
s2 = find(time >= t0 & time < t1);
s3 = find(time > 2.3);

% compute a shape for the function in each section:
tData(s1) = 28 - 50*(time(s1)-1.5).^2;
tData(s2) = 22*exp(-(time(s2)-t0)/tau);
tData(s3) = tData(s2(end)) + (s3 - s3(1))*12 / numel(s3);

figure
plot(time, tData)

% modify the equation slightly: assume equilibrium temperature is 16
% with a bit of effort one could fit for this as a second parameter
Teq = 16;
tData2 = tData;
tau2 = tau / 8; % decay more strongly to get down to approx the same value by t1
tData2(s2) = (22 - Teq) * exp( - (time(s2) - t0) / tau2) + Teq;
tData2(s3) = tData2(s2(end)) + (s3 - s3(1))*12 / numel(s3);

hold on;
plot(time, tData2, 'r')

This results in the following plot:

I conclude from this that the main reason your plots look so similar is that the function you are trying to fit is almost linear over the domain you are choosing - a different choice of function will be a better match.

Answer 2

As I mentioned in the comments, there is a difference between fitting a linear model in log-space vs. fitting a non-linear model (both in the least-squares sense).

There a nice demo in the Statistics toolbox that explains the situation. I'm adapting the code below:

%# sample data
x = [5.72 4.22 5.72 3.59 5.04 2.66 5.02 3.11 0.13 2.26 ...
     5.39 2.57 1.20 1.82 3.23 5.46 3.15 1.84 0.21 4.29 ...
     4.61 0.36 3.76 1.59 1.87 3.14 2.45 5.36 3.44 3.41]';
y = [2.66 2.91 0.94 4.28 1.76 4.08 1.11 4.33 8.94 5.25 ...
     0.02 3.88 6.43 4.08 4.90 1.33 3.63 5.49 7.23 0.88 ...
     3.08 8.12 1.22 4.24 6.21 5.48 4.89 2.30 4.13 2.17]';

xx = linspace(min(x), max(x), 100);

%# linear regression in log-space
%#           y = p2 * exp(p1*x)
%#   => log(y) = log(p2) + p1*x
p_exp = polyfit(x, log(y), 1);
yy1 = exp(p_exp(2)) .* exp(xx .* p_exp(1));

%# linear regression
p_lin = polyfit(x, y, 1);
yy2 = polyval(p_lin, xx);

%# non-linear regression (using previous result as initial coeff)
f = @(p,x) p(2)*exp(p(1)*x);
p_nonlin = nlinfit(x, y, f, [p_exp(1) exp(p_exp(2))]);
yy3 = f(p_nonlin, xx);

plot(x,y,'o', xx,yy1,'-', xx,yy2,'-', xx,yy3,'-')
legend({'data points','linear in log-space','linear','non-linear'})