Free-knot spline approximation (BSFK) problem

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@BrunoLuong
My data acquisition system produce periodically 1-D measured noised data with the fixed time window length W. I want to produce smoothed data for each window W separately, with specific constraints on continuous (k=2) or smooth (k = 3 or 4) processed signal connections between consecutive time measurement windows. So, for each window W I get finally separate "pp" structure. How to set proper BSFK options setting to fulfil these constraints?
The second question is: Is there any method how to merge separate "pp" structures to one "pp" structure for several processing windows at one?
Add note: May by some processing windows overlap could be required. Do you have any experiance with using BSFK in streaming regime?
  13 Commenti
Bruno Luong
Bruno Luong il 23 Set 2022
Modificato: Bruno Luong il 23 Set 2022
You could try to recursively enforce the continuity for function/derivative when you call BSFK on the next interval using the pp of the previous interval, just tell BSFK to have function/derivative of the most left knot (current) = previous pp function/derivative at the right knot (previous).
Michal
Michal il 23 Set 2022
Yes, something like this I would like to try, but I don't know how exactly implement this type of fix by "shape" and "pntcon" options. This is the main reason why I need your help.

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Bruno Luong
Bruno Luong il 23 Set 2022
Modificato: Bruno Luong il 12 Gen 2023
Here is the recursive pointwise constraint. You'll see it does the job (zoom in) the transition is not nice
data=load('result_4_8.mat')
data=data.result;
[m,n] = size(data);
x = cellfun(@(data) data.x, data, 'unif', 0);
y = cellfun(@(data) data.y, data, 'unif', 0);
j = 1; % n
close all
figure
hold on
for i = 1:m
% Normalize data so that dy/dx is comparable to y
xij = x{i,j}/10000;
yij = y{i,j}/10000;
options = struct('lambda', 1e-8);
if i >= 2
xleft = pp.breaks(end);
yleft = ppval(pp,xleft);
ydleft = ppval(ppder(pp),xleft);
xij = [xleft; xij];
yij = [yleft; yij];
pntcon = struct('p', {0 1}, 'x', {xleft,xleft}, 'v', {yleft ydleft});
options.pntcon = pntcon;
end
pp =BSFK(xij, yij, 4, [], [], options);
xi = linspace(min(xij),max(xij),1000);
yi = ppval(pp, xi);
plot(xij, yij,'c.');
plot(xi, yi, 'r', 'Linewidth', 2);
drawnow
end
function ppd = ppder(pp)
ppd = pp;
coefs = ppd.coefs;
n = size(coefs,2);
ppd.coefs = coefs(:,1:n-1).*(n-1:-1:1);
ppd.order = ppd.order-1;
end
  5 Commenti
Bruno Luong
Bruno Luong il 29 Set 2022
I haven't not studied the complexity of BSFK.
But it seems to me the linear dependency to number of knots is not quite true, I would say it is more like quadratic.
Michal
Michal il 29 Set 2022
Of course, this is my mistake ... you are right! CPU time dependency to number of knots is ~ quadratic!!!

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