Mean of an oscillatory trajectory

Hi,

I have a particle moving in oscillatory path. See the plot of distantce vs time plot. How do I find the mean trajectory ?

Answer 1

[Daniel Bengtson]

p = polyfit(t,X,2); % fit a second order polynomial to the data
newT = linspace(min(t),max(t)); % 100 evenly spaced steps over the range of t
Y = polyval(p,newT); % evaluate the polynomial at these steps

plot(newT,Y);

[Sam Chak]

Hi @Daniel Bengtson

That's a good strategy. However, I wonder if the solution is really the mean trajectory.

Take this for example:

t = linspace(0, 25, 25001);
x = 147/8000*(t + 0.75*sin(4*t)).^1.65;

p    = polyfit(t, x, 2);            % fit a second order polynomial to the data
newT = linspace(min(t), max(t));    % 100 evenly spaced steps over the range of t
Y    = polyval(p, newT);            % evaluate the polynomial at these steps

plot(t, x), grid on
title({'$\frac{147}{8000} (t + 0.75 \sin(4 t))^{1.65}$'}, 'interpreter', 'latex', 'fontsize', 16)
xlabel('t'), ylabel('x'), hold on
plot(newT, Y);
legend('Data', 'Fitted curve', 'location', 'northwest')

[Daniel Bengtson]

Are you thinking something more along the lines of using an fft to isolated the frequency of the oscillation and remove it?

[Sam Chak]

Nope, I didn't think too deep. Your strategy is a good one for this case. I was wondering about the mathematical definition of the mean trajectory, as well as if some data smoothing algorithms may work.

[Daniel Bengtson]

I should have thought on it more before posting. You are right to question my answer. Polyfit is just a least squares regression and given that I did not take any steps to enforce an intercept at zero, and that the error between the theoretical mean trajectory and raw signal would be relatively small near zero compared to near t max, it probably isn't as good of a fit as it could have been.

Answer 2

[Image Analyst]

If you know how many elements are in a period (and you should), then try movmean.