Fitting curve : difficultu to do the modelized

3 visualizzazioni (ultimi 30 giorni)
Hello
I use for the first time the trial version of Maltlab to test it before buy it. I thought that it will be more simple for modelize.
I collected data with some pressure recorders and I want to know the equation about the all data.
At this time , for each data recorder, I used Excel to put my data and I drew. After, I used the modelisation (poly, linear, exp, ...) and I asked the equation => It's very simple.
Because I have a result with my data recorder than I cannot have the equation with Excel, I decide to test Matlab.
I sucessed to download my data (collected directly by Excel), I successed to plot but how can I have the equation ?. I Itried to test fitting curv but impossible to understand.
Thanks
Regards,
RM
  4 Commenti
Star Strider
Star Strider il 30 Ago 2019
Your curves look like the step response of a control system. Your best option is likely to identify it as such with the System Identification Toolbox.
Rémi MOREAU
Rémi MOREAU il 30 Ago 2019
Thanks a lot Star Strider.
I will try to test with System Identification Toolbox. .
Regards, RM

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darova
darova il 30 Ago 2019
I changed your time span
% A = xlsread('data.xls','B2:E162');
% A = flipud(A); % flip up-down matrix
rec1 = A(:,1);
rec2 = A(:,2);
rec3 = A(:,3);
rec4 = A(:,4);
t = (1:length(rec1))/15 + 0.5; % time span
y = rec4;
ix = ~isnan(y); % extract only numbers
f1 = @(a,b,x) a+b./exp(x); % function to fit
F1 = fit(t(ix)',y(ix),f1)
plot(t,y,t,F1(t),'r')
legend('record 4')
Here is what this script produces
rec.png rec2.png rec3.png rec4.png
Also look in help for fit, you can boundaries of coefficients and start points
1Capture.PNG
  1 Commento
Rémi MOREAU
Rémi MOREAU il 30 Ago 2019
THANKS A LOT :).
I will try to understand (my Matlab experience is very poor).
I will contact another time on monday if I not success in what your write.
Regards and good WE.
RM

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