I just ran the example code, stepped into the tsa function, and saw the line 139:

By looking at the code in the red box, it seems that MATLAB assumes constant rpm in the same rotation and evenly divides the rotation time by number of points, is my guess right? Anyone agrees?

Here is my code:

close all
clear all
clc

fs = 1000;
t = 0:1/fs:5-1/fs;

rpm0 = 2400;

a = 0.1;
f0 = rpm0/60;
T = 0.75;

phi = 2*pi*f0*T*(1-exp(-t/T));

x = a*cos(phi) + randn(size(phi))/200;
y = a*sin(phi) + randn(size(phi))/200;

plot(t,x,t,y)

[rpm,~,tp] = tachorpm(x(t<2.5),fs);
tachorpm(x(t<2.5),fs)

clf
tsa(x,fs,tp,'Method','linear')

I just ran the example code, stepped into the tsa function, and saw the line 139:

By looking at the code in the red box, it seems that MATLAB assumes constant rpm in the same rotation and evenly divides the rotation time by number of points, is my guess right? Is it a valid assumption?

Here is my code:

close all
clear all
clc

fs = 1000;
t = 0:1/fs:5-1/fs;

rpm0 = 2400;

a = 0.1;
f0 = rpm0/60;
T = 0.75;

phi = 2*pi*f0*T*(1-exp(-t/T));

x = a*cos(phi) + randn(size(phi))/200;
y = a*sin(phi) + randn(size(phi))/200;

plot(t,x,t,y)

[rpm,~,tp] = tachorpm(x(t<2.5),fs);
tachorpm(x(t<2.5),fs)

clf
tsa(x,fs,tp,'Method','linear')