# Two-Term Exponential Curve Fitting

I want to fit a pair of experimental data $(M_z(t), t),$ which can theoretically be modelled according to:$$M_z(t)=M_z(0) e^{-t/T_1} + M_0 (1-e^{-t/T_1}) \tag{1}.$$I want to use the equation of the fitting to solve for $T_1.$ The parameters $M_0,$ and $M_z$ are defined in the diagram below: What kind of fitting should be used here?

Attempt:

I have tried a two-term exponential fitting in Matlab, it is a good fit but I can't see how to deduce $T_1$ from the resulting equation:

f(x) = a*exp(b*x) + c*exp(d*x)

a =   2.642e+04  (2.623e+04, 2.662e+04)
b =   1.347e-06  (-2.953e-05, 3.222e-05)
c =   -2.45e+04  (-2.478e+04, -2.423e+04)
d =    -0.07508  (-0.07726, -0.0729)


Which exponent ($d$ or $b$) should be used for finding $T_1$?

This was my experimental data:

x=[2    5   10  20  30  50  100 200 400];
y=[5418.583 9479.431    14828.01    21052.6 23872.96    25784.02    26420.23    26445.85    26433.05];


And the resulting exponential fit: P.S. I can't do a log plot fitting because I don't know the value of $M_0.$

• If you specify the equation : f(x) = a exp(b x) + c exp(d x) Matlab do a regression for 4 parameters a, b, c, d which corresponds to the function $M_z(t)=M_z(0) e^{-t/T_1} - M_0 e^{-t/T_2}$ where $T_1\neq T_2$.This is not your function (1). You have to specify the equation f(x) = a exp(b x) + c . This will be a regression for 3 parameters a, b, c . From the values a, b, c obtained you can compute $M_z(0)$ , $M_0$, and $T_1$. Sep 29 '16 at 7:38
• Note : in your function (1) there is a constant parameter $M_0$. In the equation f(x) = a exp(b x) + c exp(d x) there is no constant parameter. So, this cannot agree. Sep 29 '16 at 7:42
• Thank you for the response. We know what $M_z(0)$ is, it is the value of $M_z(t)$ at what we call t = 0. Do you know how to specify the equation $f(x) = a exp(b x) + c$ in Matlab? Because the f = fit(x,y,'exp1') syntax does not give a constant term. Sep 29 '16 at 12:02
• I think to y=a exp(bt)+ c because you proposed $$M_z(t)=M_z(0) e^{-t/T_1} + M_0 (1-e^{-t/T_1}) =(M_z(0)-M_0) e^{-t/T_1} + M_0$$ So, a$=(M_z(0)-M_0)$ , b$=-1/T_1$ and c=$M_0$. But are you sure that it is the right model for the experimental data ? Sep 29 '16 at 12:40

In the equation (1), there is only one exponential function, not two : $$M_z(t)=M_z(0) e^{-t/T_1} + M_0 (1-e^{-t/T_1}) \tag{1}.$$ $$M_z(t)=(M_z(0)-M_0) e^{-t/T_1} + M_0 .$$

HINT :

Supposing that $y=ae^{bx}+c$ is a convenient model, the only difficulty is to find a good approximate for $b$.

I haven't Matlab at hand, so using another tool, I found : $$b\simeq -0.075$$

The change of variable $X=e^{-0.075 x}$ leads to the linear equation : $$y=aX+c$$ Then, an usual linear regression will give you the approximates of $a$ and $c$.

In order to answer to the questions raised by Merin in the comments section, the procedure of regression (with integral equation) published in https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales is shown below.

Be careful, the notations of the parameters $a,b,c$ are not the same as above. This method isn't very accurate if the number of points is too low, because the numerical integration for $S_k$ cannot be accurate with only few points (this is the case with 9 points). But anyways, the result could be used as an excellent initial value for a further non-linear regression, if necessary.

• Thank you, that makes perfect sense. But how did you calculate $b$? The purpose of this question is to find $T_1.$ Is it possible to find the $b \approx -0.075$ in Matlab? Sep 30 '16 at 7:30
• As I said, I don't have Matlab and I used another software. This is an home made soft based on the method described pp.16-17 in fr.scribd.com/doc/14674814/Regressions-et-equations-integrales . But this is not useful for you since you want to use Matlab. Set f(x) = a exp(b x) + c instead of f(x) = a exp(b x) + c exp(d x) . Sep 30 '16 at 8:09
• Does this software use some form of non-linear least squares fit of the equation to the data? What is the method called? Unfortunately I couldn't completely understand the French text. Oct 1 '16 at 13:36
• If your problem is an academic exercise, better don't use the method of regression with integral equation because this method isn't standard. The standard methods of non-linear regression are implemented in many math softwares. The general principle is explained for example in : mathworld.wolfram.com/NonlinearLeastSquaresFitting.html and involves some iterative process. One the other hand, the method of fitting with integral equation isn't iterative. For example, I will show a procedure in addition to my first answer. Oct 1 '16 at 17:39

You actually know both $M_0$, which is the first point, and $M_\infty$, which is the last point (with a good approximation). Caution, I changed the notation !

Then your model can be written

$$M_t=M_\infty+(M_0- M_\infty)e^{-t/T}$$ or

$$T=-\frac t{\log\left(\dfrac{M_t- M_\infty}{M_0- M_\infty}\right)}$$

and you can just use the average value of $T$. (Preferably, use points not too close from the asymptote as they will be less accurate).

If you consider that $M_0$ or $M_\infty$ is unknown, you can use a two-parameter model,

$$\delta_{Mt}=\delta_{M0}e^{-t/T}$$ which linearizes by taking the logarithm

$$\log(\delta_{Mt})=\log(\delta_{M0})-\frac tT.$$

The three-parameter model requires a numerical solver, but the above simpler models can be used to find good initial values.

What kind of fitting should be used here ?

Try several, e.g. logy, loglog ... one may be useful.

I can't do a log plot fitting because I don't know the value of $M_0$.

But you can estimate; log( 26500 $\, -$ the data ) gives a roughly straight line, if you drop or downweight the last 2 points: This shows up a general point in curve fitting / regression:
where do you want good fit -- in the changing part, or the flat last part ?