1
$\begingroup$

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:

enter image description here

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:

enter image description here

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

$\endgroup$
  • 1
    $\begingroup$ 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 $T_1$. $\endgroup$ – JJacquelin Sep 29 '16 at 7:38
  • 1
    $\begingroup$ 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. $\endgroup$ – JJacquelin Sep 29 '16 at 7:42
  • $\begingroup$ 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. $\endgroup$ – Merin Sep 29 '16 at 12:02
  • 1
    $\begingroup$ 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 ? $\endgroup$ – JJacquelin Sep 29 '16 at 12:40
2
$\begingroup$

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 ADDITION :

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.

enter image description here

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Merin Sep 30 '16 at 7:30
  • 1
    $\begingroup$ 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) . $\endgroup$ – JJacquelin Sep 30 '16 at 8:09
  • $\begingroup$ 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. $\endgroup$ – Merin Oct 1 '16 at 13:36
  • 1
    $\begingroup$ 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. $\endgroup$ – JJacquelin Oct 1 '16 at 17:39
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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:

enter image description here

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 ?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.