Fit of multiple data sets

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Daniele Sonaglioni 2021-3-30

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评论： Star Strider 2021-3-31

Hello everyone,

I have a problem in making a fit of some set of data. I have 4 arrays of data: the first is my x-axis while the other three are my y-axes. I want to fit the y-arrays with the same function but i want only one set of parameters to fit the three arrays.

My fitting function is the subsequent:

where

. My fitting parameters are

and each y-array is referred to a different value of T.

How can i write the code for this kind of fit?

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Answer 1

Star Strider 2021-3-30

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That is relatively straightforward. If your objective function outputs only one vector (as opposed to three), concatenate the dependent variables in one vector, create a matching independent variable vector, then use whatever nonlinear paramerer estimation routine you want to estimate the parameters.

It would help to have your data and the

and m values.

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Daniele Sonaglioni 2021-3-30

在 MATLAB Online 中打开

I have tried to follow the procedure suggested by @William Rose but i'm not sure that the code works correctly.

In fact i have defined a matrix in which the second colums is an array of the temperature but i am not able to plot the fitting function for a single temperature.

Can you give me some suggestion? Below i report my code and my x,y1,y2,y3 arrays.

x=[0.5215    0.7756    1.2679    1.4701    1.6702    1.8680    2.0633    2.2693    2.4584    2.6442    2.8264    3.0046 3.0890    3.2611    3.4287    3.5917    3.7497    3.9309    4.0774    4.2183    4.3535    4.4827    4.5427    4.6628];
y1=[1.0290    1.0025    1.0158    1.0068    0.9705    0.9646    0.9596    0.9499    0.9811    0.9669    0.9519    0.9573 0.8989    0.9315    0.9618    0.9260    1.0481    0.9245    0.9733    0.8830    1.0203    0.9851    0.9314    0.9204];
y2=[0.9773    1.0156    0.9362    1.0103    0.9414    0.9167    0.9257    0.9225    0.9127    0.9230    0.9380    0.8899 0.8047    0.9339    0.9012    0.9079    0.9497    0.8449    0.8837    0.8250    0.8738    0.8602    0.9106    0.8656];
y3=[1.0433    0.9711    0.9913    0.9902    0.9427    0.9146    0.8849    0.9010    0.8876    0.9175    0.9329    0.8639  0.6970    0.8260    0.8675    0.8568    0.8748    0.8156    0.8041    0.7679    0.8333    0.8443    0.8336    0.8344];
T1=120; %temperature reffered to y1
T2=140; %temperature reffered to y2
T3=160; %temperature reffered to y3
A=[x' y1' y2' y3'];
N=length(x);
B=[A(:,1),ones(N,1)*T1,A(:,2); A(:,1),ones(N,1)*T2,A(:,3); A(:,1),ones(N,1)*T3,A(:,4)];
K=8.26e-23;
m=17246;
x1=B(:,1);
y=B(:,3);
t=B(:,2);
fun=@(r,x1,t) exp(-3*K*x1.*t./m).*(1-2*r(2)*r(3)*(1-sin(x1*r(4)))./(x1*r(4)));
r0=[1,1,1,1];
r=lsqcurvefit(fun,r0,x1,y);

William Rose 2021-3-31

@Daniele Sonaglioni,

My code is very similar to @Star Strider, but mine is less elegant.

I did not square the quantity xu=3x*kB*T/m inside the exp(), since @Daniele Sonaglioni did not, in his code, and because I have seen

in physical models, but I have not seen

. But my experience is limited.

The quantity -x*u, or -(x*u)^2, inside the exp() is the only part of the predicted y that depends on temperature. The magnitude of x*u is on the order of 10^-24, or 10^-48 if you square it, at all three temperatures and for all values of x in this dataset. Therefore exp(-x*u)=1, and exp(-(x*u)^2)=1. Therefore it doesn't matter if you square x*u, and the predictions are independent of temperature. Thus you can replace exp(-x*u), or exp(-(x*u)^2), with 1, and you can discard the temperature information, because it is irrelevant.

I suspect that the temperature information is not actually irrelevant. I suspect that the model equation or the value of one of the constants is incorrect. I would check the value of m.

I think there is another problem. It looks like the predicted values depend only the product of p1 and p2, and are unaffected by replacing (p1,p2) with (p1/c, p2*c), where c is any non-zero constant. Therefore their individual values cannot be estimated with this model. Only their product can be estimated.

Daniele Sonaglioni 2021-3-31

I think that the two methods (mine and your) are equivalent.

Star Strider 2021-3-31

With nlinfit, use nlparci to get the confidence intervals on the parameters and nlpredci to get the confidence intervals on the fit. The fitnlm function calculates the parameter confidence intervals automatically, and the predict function calculates the confidence intervals on the fit.

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Answer 2

William Rose 2021-3-30

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@Daniele Sonaglioni,

@Star Strider is right. Specifically, if your data is in array A, and if A has N rows x 4 columns, and the columns are x, y1, y2, y3, and the 3 temps are T1, T2, T3, then do

B=[A(:,1),ones(N,1)*T1,A(:,2); A(:,1),ones(N,1)*T2,A(:,3); A(:,1),ones(N,1)*T3,A(:,4)]; 

The above command turns your Nx4 array into 3*N by 3 array. Note the use of commas and semicolons in the line above. I inserted a column for temperature. The first N rows have temp T1, the next N rows have temp T2, etc. Then you fit it all at once as @Star Strider said.