Interpolating through spline and then get corresponding values
2 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
Hello, I have a certain set of points which create a curve. I want to evaluate what is the y value at other x locations, not given in the problem.
Unfortunately, there is no polynomial which can passes though all these points correctly.
I am sure other people might have asked this question, but I could not find anything.
The X values are:
if true
% code
x=[0
0.00420000000000000
0.00800000000000000
0.0125000000000000
0.0177000000000000
0.0236000000000000
0.0305000000000000
0.0385000000000000
0.0476000000000000
0.0581000000000000
0.0701000000000000
0.0838000000000000
0.0993000000000000
0.116700000000000
0.136300000000000
0.158200000000000
0.182500000000000
0.209200000000000
0.238400000000000
0.269800000000000
0.303600000000000
0.339300000000000
0.376800000000000
0.415600000000000
0.455300000000000
0.495400000000000
0.535600000000000
0.575100000000000
0.613700000000000
0.650900000000000
0.686200000000000
0.719500000000000
0.750500000000000
0.779200000000000
0.805400000000000
0.829200000000000
0.850600000000000
0.869800000000000
0.886900000000000
0.902000000000000
0.915300000000000
0.927000000000000
0.937200000000000
0.946100000000000
0.953900000000000
0.960600000000000
0.965400000000000
0.970400000000000
0.975400000000000
0.980200000000000
0.985200000000000
0.990200000000000
0.995100000000000
1];
end
The y values are:
if true
% code
y=[0.127489040000000
0.111766040000000
0.0167797700000000
0.0337462690000000
0.0266396250000000
0.111117580000000
0.130855270000000
0.138929860000000
0.163963910000000
0.134142850000000
0.150978780000000
0.151461170000000
0.142033560000000
0.143054720000000
0.142033560000000
0.139969710000000
0.139452300000000
0.138929860000000
0.137879260000000
0.141520320000000
0.143562680000000
0.146571800000000
0.159900150000000
0.167058910000000
0.178109250000000
0.196599630000000
0.216707350000000
0.225149480000000
0.251330640000000
0.286469960000000
0.304243410000000
0.330819300000000
0.359129760000000
0.389805860000000
0.429068680000000
0.448364490000000
0.485802150000000
0.531624250000000
0.547334130000000
0.565165850000000
0.582247920000000
0.598792800000000
0.619664680000000
0.639333560000000
0.638537430000000
0.668565600000000
0.713292490000000
0.687499690000000
0.660858750000000
0.633757700000000
0.628645890000000
0.672958870000000
0.660858750000000
0.623358350000000];
end
And the values where I want to evaluate are:
if true
% code
xnew=[0.0250000000000000
0.0500000000000000
0.0750000000000000
0.100000000000000
0.150000000000000
0.200000000000000
0.300000000000000
0.400000000000000
0.500000000000000
0.600000000000000
0.700000000000000
0.800000000000000
0.900000000000000
0.950000000000000];
end
The way I would do is by linear interpolation through the different points and then get for each interval the coefficients of the polynomial and input my data and check if they are within certain limits and use poleval. I think this way is a bit long, and it might be possible to fit a spline or other manners.
Thanks a lot for any suggestions.
Antonio
0 Commenti
Risposta accettata
Azzi Abdelmalek
il 12 Set 2012
Modificato: Azzi Abdelmalek
il 12 Set 2012
method='linear'
ynew=interp1(x,y,xnew,method)
%or
method='smoothingspline'
f=fit(x,y,method);
ynew=f(xnew)
Più risposte (0)
Vedere anche
Categorie
Scopri di più su Spline Postprocessing in Help Center e File Exchange
Prodotti
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!