cscvn of a function, coefficients and breaks

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David
David 2014 年 7 月 9 日
編集済み: Bruno Luong 2023 年 11 月 24 日
Hey all together,
I have another question concerning the cscvn function. When I create a spline for a function with cscvn
yy = cscvn(xges)
and take a look at yy.coefs and yy.breaks
form: 'pp'
breaks: [1x25 double]
coefs: [48x4 double]
pieces: 24
order: 4
dim: 2
then I have nearly twice as much coefs as breaks. How is this possible and what does that mean? How can I create the piecewise polynomial functions out of my coefs? At the end I want to calculate the curvature of the function, but first I have to understand what this coefs mean, and how I can create the piecewise polynoms.
Thank you in advance,
David
  2 件のコメント
Lucie
Lucie 2023 年 11 月 24 日
Hello, I am facing the same issue. Did you receive an answer regarding the meaning of the coefficients?
Mathieu NOE
Mathieu NOE 2023 年 11 月 24 日

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回答 (1 件)

Bruno Luong
Bruno Luong 2023 年 11 月 24 日
編集済み: Bruno Luong 2023 年 11 月 24 日
This is how to use pp
m = 2;
n = 5; % 25 in the original question
points = rand(m,n);
pp = cscvn(points)
pp = struct with fields:
form: 'pp' breaks: [0 0.8735 1.5529 2.4191 3.2527] coefs: [8×4 double] pieces: 4 order: 4 dim: 2
m = pp.dim;
p = pp.pieces;
% https://pages.mtu.edu/~shene/COURSES/cs3621/NOTES/INT-APP/PARA-centripetal.html
%D = diff(points,1,2);
%d = sqrt(sqrt(sum(D.^2,1)));
%t = cumsum([0 d]); % Eugene Lee's, equal to pp.breaks, see https://www.mathworks.com/help/curvefit/cscvn.html
t = pp.breaks;
ni = 200;
ti = linspace(t(1),t(end), ni);
loc = discretize(ti, t);
x = zeros(length(ti), m);
coefs = reshape(pp.coefs, m, p, []);
for k=1:p
ink = loc == k;
tik = ti(ink);
dti = tik - t(k);
for d=1:m
P = coefs(d, k, :);
x(ink,d) = polyval(P(:), dti);
end
end
close all
hold on
if m == 2
plot(points(1,:),points(2,:),'or'),
plot(x(:,1), x(:,2),'b.')
elseif m == 3
plot3(points(1,:),points(2,:),points(3,:),'or'),
plot3(x(:,1), x(:,2), x(:,3), 'b.')
else
warning('plotting not supported')
end

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