# Exponential curve fit with MATLAB's fit function does not deliver good fit

I am trying to use MATLAB's fit function to fit a curve through a data set which obviously shows an exponential decay. These are the commands I use:

f = fit(time', intensity', 'exp1');
plot(f, time, intensity);
xlabel('time (min)');
ylabel('intensity (a.u.)');


The result looks like this: Fit with 'exp1'.

It seems to me that this is clearly not the best possible fit. If I use 'exp2' instead of 'exp1', the fit is much better: [Image not linked due to restrictions on number of links.][2]

However, if I'm not wrong, a similarly good fit should be possible with 'exp1' as well. How can I achieve that?

P.S: The respective coefficients are as follows:

 General model Exp1:
f(x) = a*exp(b*x)
Coefficients (with 95% confidence bounds):
a =        1517  (1513, 1521)
b =   -0.003019  (-0.003133, -0.002905)

General model Exp2:
f(x) = a*exp(b*x) + c*exp(d*x)
Coefficients (with 95% confidence bounds):
a =       131.7  (126.9, 136.4)
b =     -0.1347  (-0.1488, -0.1206)
c =        1467  (1462, 1473)
d =    -0.00198  (-0.002086, -0.001875)


EDIT: I plotted the log of the intensity values over time and it looks like this: log(intensity) over time. Not exactly a straight line. So I guess the exp1 model is just not suited in this case, right?

• Before using "exp1" did you plot $\log(y)$ vs $x$ ? If this was a good model, then the scatter plot would show almost a straight line ? Jul 19 '16 at 9:18
• Thanks for the input @ClaudeLeibovici. I edited the question accordingly.
– Dave
Jul 19 '16 at 9:42
• What is the physical background of the problem ? Jul 19 '16 at 9:45
• It's the brightness of an LED. After turning it on, it will get brighter during approximately the first 5 minutes (data not shown). After that, its brightness is reduced as seen in the data.
– Dave
Jul 19 '16 at 9:48
• a polynomial fit with a few coefficients will give you good results. what is exp1 and exp2. For an exponential curve, just take the log and use polynomial fit. it will solve your problem. Jul 19 '16 at 10:01

The fitting to an exponential function is not good because the exponential function isn't a correct model for the physical phenomena of luminescence decay.

You would have a much better result with the function : $$I(t)=I_0\:e^{-\left(\frac{t}{\tau} \right)^\beta}$$

Of course, this requires a non-linear regression with 3 parameters : $I_0$ , $\tau$ , $\beta$

Moreover, one can see on your graph that for the first points ( close $t=0$ ) , during a short time, the function $I(t)$ decreases more slowly than later which is not well consistent with the above function. This would require a more sophisticated model. But, in order to keep a not too sophisticated function, an approximate consists in introducing a delay $t_0$ into the above function, so that : $$I(t)=\begin{cases} I_0 & t<t_0\\ I_0\:e^{-\left(\frac{t-t_0}{\tau} \right)^\beta} & t>t_0 \end{cases}$$

A non-linear regression with 4 parameters $I_0$ , $\tau$ , $\beta$ , $t_0$ will give an excellent fitting, except close to the initial point where a slight discrepancy will remains.

Making this residual discrepancy disappear would require experimental studies and modeling of the phenomena occuring at the very beginning of the luminesce decay.

Example of regression : Below, the approximates for $I_0$ , $\tau$ , $\beta$ are certainly not the best because the coordinates of the points were imported from the Dave's graph with a rough graphical scan.