Effacer les filtres
Effacer les filtres

Minimum and Maximum value of fitted curve

80 vues (au cours des 30 derniers jours)
Zack Trahem
Zack Trahem le 20 Mar 2022
Réponse apportée : Nipun le 20 Déc 2022
I am trying to find the minimum and maximum value of fitted curve. I will use the min and max point in another calculation. I used 'poly4' fit and its gives 4th order polynomial function. I tried polyder but couldnt go further. How can find the maximum and minimum point of fitted curve?

Réponse acceptée

Image Analyst
Image Analyst le 20 Mar 2022
Modifié(e) : Image Analyst le 20 Mar 2022
So I assume you have the red, fitted curve since you plotted it (if not see first 2 lines below), so why can't you just simply do
coefficients = polyfit(x, y, 4); % fit to a 4th order polynomial.
yfit = polyval(coefficients, x); % Evaluate the fit at the same x values.
minFit = min(yFit); % Find min of fitted curve in the range we have fit it over.
maxFit = max(yFit); % Find max of fitted curve in the range we have fit it over.
If you want, you can use an x in polyval() that has finer resolution than the original x. Something like
numSamplePoints = 2000; % However many you need to get the resolution you want.
xFit = linspace(min(x), max(x), numSamplePoints);
yfit = polyval(coefficients, xFit); % Evaluate the fit at the same x values.
Of course you could get the exact value of the apex just by computing the derivative from the coefficients vector.
  2 commentaires
Zack Trahem
Zack Trahem le 20 Mar 2022
Thank you so much thats exactly what i looking for:) .just one more thing, that is coresponded x values of for min and max data.
scatter(Bw(:,1),Bw(:,2))
f=fit(length,Bw(:,2),'poly4')
plot(f,length,w0)
c=coeffvalues(f);
yfit=polyval(c,length);
minfit=min(yfit)
maxfit=max(yfit)
for now i made this way and gives the values Thank you again.
Zack Trahem
Zack Trahem le 20 Mar 2022
I solved it thanks a lot!
minindex=find(yfit==minfit);
maxindex=find(yfit==maxfit);
maxPoint=xFit(minindex);
minPoint=xFit(maxindex);

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Plus de réponses (2)

Torsten
Torsten le 20 Mar 2022
Modifié(e) : Torsten le 20 Mar 2022
p = polyfit(x,y,4);
dp=polyder(p);
ddp=polyder(dp);
r=roots(dp);
r=r(abs(imag(r) ) < 1e-6);
maxima=r(polyval(ddp,r))<=1e-6 )
minima=r(polyval(ddp,r))>=1e-6 )
Extrema at the boundaries are not accounted for.

Nipun
Nipun le 20 Déc 2022
Use the curve of best fit to find the maximum and minimum flow rates separately during the inflow and the outflow

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