Effacer les filtres
Effacer les filtres

Linear Fitting starting from a value

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Nicola De Noni
Nicola De Noni le 10 Oct 2022
Modifié(e) : Torsten le 10 Oct 2022
Hi everyone!
I need to fit these data [100 90 80 70 50 45 40 43 42 40 35] whom are distanced by 10 seconds.
The problem is that I would fit these data starting from the first value (100) in order to understand the slope of the line.
Is it possible?
Thanks!

Réponse acceptée

Torsten
Torsten le 10 Oct 2022
Modifié(e) : Torsten le 10 Oct 2022
Sure. Use
l(x) = a*x + 100
as model function and fit the parameter a.
xdata = ...; % your xdata as column vector
ydata = ...; % your ydata as column vector
a = xdata\(ydata-100)
But starting from a certain x-value, the slope seems to change. You should separate the fit into two parts:
l1(x) = a*x + 100 0 <= x<= x1
l2(x) = b*x + [(a-b)*x1 + 100] x1 <= x <= x(end)

Plus de réponses (1)

Hiro Yoshino
Hiro Yoshino le 10 Oct 2022
Modifié(e) : Hiro Yoshino le 10 Oct 2022
refline is the easiest way to achieve that.
y = [100 90 80 70 50 45 40 43 42 40 35];
x = 0:10:10*(length(y)-1);
plot(x,y,'o');
refline
if you want to know the coefficients:
p = polyfit(x,y,1)
p = 1×2
-0.6391 89.6818
polyfit for reference.
  1 commentaire
Nicola De Noni
Nicola De Noni le 10 Oct 2022
Thanks @Hiro but i would like that the fitting line passe through the first point 100,0

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