Why should the last point be excluded when performing a least-squares fit of a periodic discrete time signal?

Question

I fitted the function: f(t)=A_o+A_1 cos(wt)+B_1 sin(wt) to the following periodic discrete signal:

t=0:0.15:1.5;

y=[2.200 1.595 1.031 0.722 0.786 1.200 1.805 2.369 2.678 2.614 2.200];

Where w=2*pi / T, and T=1.5 seconds.

It happens that the fitted curve presents a phase shift with respect to the data. I consulted books on this topic and I found that I should exclude the last point of the series in order to get the right answer,...

I suspect that this matter is related with the exclusion of the last point in the calculation of the DFT, but do not find a mathematical argument that can prove this statement.

I would appreciate any mathematical explanation on this matter.

Answer 1

First (wrong) answer (for integrity) The $y$-value of the last point is the same as the first one. As you apparently know the frequency, this point comes in excess of the "fundamental period". It sounds like this additional point comes like an implicit double-weight to the first point of the period.

Second take: I have tried to fit the data, with or without the last point. It seems to fit well, both cases.

%https://dsp.stackexchange.com/questions/71164/why-should-the-last-point-be-excluded-when-performing-a-least-squares-fit-of-a-d
clear;close all
%% Settings
T = 1.5; w=2*pi / T;
t = (0:0.15:1.5)';
y = [2.200 1.595 1.031 0.722 0.786 1.200 1.805 2.369 2.678 2.614 2.200]';

%% Fitting all points
ft = fittype(@(a1,a2,a3,x) a1+a2*cos(w*x)+a3*sin(w*x),'coefficients',{'a1','a2','a3'},'independent', {'x'});
f = fit( t, y, ft );
% Plot fit
subplot(2,1,1)
plot( f, t, y )
axis([t(1) t(end) 0.5 3])
grid on
title('Whole point set')
%% Fitting all points but the last
ft = fittype(@(a1,a2,a3,x) a1+a2*cos(w*x)+a3*sin(w*x),'coefficients',{'a1','a2','a3'},'independent', {'x'});
f = fit( t(1:end-1), y(1:end-1), ft );
% Plot fit
subplot(2,1,2)
plot( f,  t(1:end-1), y(1:end-1) )
axis([t(1) t(end) 0.5 3])
grid on
title('Whole point set minus 1')

Answer 2

As explained in Laurent's answer, including the last point, which equals the first point, just gives twice as much weight to that point compared to all the others. This doesn't explain a phase shift in your approximation. If you do things right you actually get an almost perfect fit, even with the last point included:

t = 0:0.15:1.5;
y = [2.200 1.595 1.031 0.722 0.786 1.200 1.805 2.369 2.678 2.614 2.200];
t = t(:); y = y(:);
L = length(t);
w0 = 2*pi/1.5;
M = [ones(L,1),cos(w0*t),sin(w0*t)];
x = M\y;        % optimal coefficients
f = M*x;        % approximating function
e = f - y;      % approximation error

   f(t)      y

   2.19999   2.20000
   1.59540   1.59500
   1.03076   1.03100
   0.72175   0.72200
   0.78639   0.78600
   1.20001   1.20000
   1.80460   1.80500
   2.36924   2.36900
   2.67825   2.67800
   2.61361   2.61400
   2.19999   2.20000

If you exclude the last point in the optimization the result is virtually identical. The only difference is the approximation error at the first point, which is slightly smaller when the last point (identical to the first point) is included, because in that case that point gets twice the weight compared to when the last point is not included. The approximation error at the first point $f(t_1)-y_1$ is -5.8462e-06 with the last point included, and -7.6001e-06 with the last point excluded.