I'm acquiring data which looks like thisLC data:

Theoretically it should be possible to fit it with a Lissajous curve, which is typically defined in the following way:

$x = A \sin(a t + \delta)$, and $y = B \sin(b t)$.

However, neither the cftools or optimization toolbox seem ready to do this or at least I do not know how to parse it in such a way as to get an algorithm to try fitting variables $A, a, t, \delta, B, b$ to my data. If I try joining these two equations together, then something breaks:

Say I solve $x$ for $t$, then $$ t = \frac{\sin^{-1}(\frac{x}{A}) - \delta + 2 \pi n}{a} $$ and then plugging this into $y$ to get: $$y = B \sin(b(\frac{\sin^{-1}(\frac{x}{A}) - \delta + 2 \pi n}{a}))$$, but I had no success using this.


1 Answer 1


It seems that both $x$ and $y$ have an offset, since a Lissajous curve should oscillate around the origin, so: $x=A\sin{\left(at+\delta\right)}+x_0$ and $y=B\sin{\left(bt\right)}+y_0$.

Besides that, your test equations do not seem to be correct. Because the ratio $\frac{a}{b}$ determines the number of "lobes" (points where the curve crosses itself) of the cyclic curve. When assuming that the data contains noise/uncertainty and that at least one complete oscillation has been complete. It seems that there are no lobes, which means that $a=b$ and that there would be only be a phase shift between $x$ and $y$. However this should produce an elliptical curve, which does not seems to be the case. This curve does look similar to a hysteresis curve.

But to answer how you might fit data to a Lissajous curve we first have to look at why you can't calculate the vector $t$. The inverse of the sine function, $\sin^{-1}$, has actually another solution beside the one you mentioned: $\sin{x}=a\rightarrow x=\sin^{-1}{a}+2\pi n\bigvee x=-\sin^{-1}{a}+2\pi+\pi$. These two slopes can be seen by plotting $\sin^{-1}\left(\sin{(x)}\right)$, which gives a triangle wave. However the absolute value of the slopes of both data sets should be the same, so for most data points, using index $i$, the following should be true: $$ \left|\frac{d}{di}t(i)\right|=\left|\frac{d}{di}\frac{1}{a}\left(\sin^{-1}\left(\frac{x(i)-x_0}{A}\right)-\delta\right)\right| $$ Instead of taking the absolute value, you could also square both sides.
However $t$ is still unknown, but this should also hold for $y$, so: $$ \left|\frac{d}{di}\frac{b}{a}\left(\sin^{-1}\left(\frac{x(i)-x_0}{A}\right)-\delta\right)\right|-\left|\frac{d}{di}\sin^{-1}\left(\frac{y(i)-y_0}{B}\right)\right|=0 $$ However I could only think of one solution to be able to use $\frac{d}{di}$ while fitting in MATLAB. And that is taking the equivalent of diff(x), namely splitting $x$ up into two vectors: x1 = x(1:end-1); x2 = x(2:end);. So the custom equation would look like this:
abs(b / a * (asin(x1 / A) - asin(x2 / A))) - abs(asin(y1 / B) - asin(y2 / B)) = 0
This method loses information about $\delta$, but once you know all the other parameters it would be much easier to find $\delta$ with another method. Another downside of this approach is that you will end up with a total of 5 data sets, x1, x2, y1, y2 and a vector of zeros. However MATLAB cftool can only handle 3 data inputs, so I am not sure if you could perform a fit using cftool.

However according to this report it is possible to fit a Lissajous curve using a least squares algorithm.

This report might also be interesting, since it is about least square fitting of ellipses (so $a=b$). I have only taken a quick look at it, so I am not 100% sure how useful it would be to you.

  • $\begingroup$ My problem is how to set up a fit rather than which model to use. $\endgroup$ Jan 9, 2014 at 2:35
  • $\begingroup$ @sciencenewbie I have added an attempt of fitting a Lissajous curve, however I do not know how useful and practical it would be. $\endgroup$
    – fibonatic
    Jan 9, 2014 at 19:17

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