I have readings of the vertical displacement of someone over time, as in the picture below:

vertical displacement over time At some points in time, the person will be exercising or jumping up & down, creating more or less obvious periodic motions (yellow regions of the graph). I am now trying to find a way to automatically spot these yellow areas, by looking for periodicity in the recorded signal.

My initial attempt has been to use a spectrogram, using the specgram built-in function in Octave: segmentLength=150; window=hanning(150); overlap = 80% * segmentLength; specgram(x, segmentLength, 1, window, overlap)

Which results in the following output: Spectrogram output

This seems to be hinting in an ok direction since I see for example that something gets picked up just before frame 2000, however it is not very conclusive. Given my limited understanding of the spectrogram, I am not sure how to proceed from here.

Is the spectrogram the right approach? If yes, how can I improve the output? If no, what approach would you recommend for this task? I am aware that some of the windows that I am trying to identify will be too difficult, but was expecting to at least pick up windows 1, 2 and 5 (starting from the left). Thanks for any pointers!


I have tried to implement a simple ASDF in octave with the following code:

#Parameters for asdf analysis
  N=400; %Window size
  kmin= 12;
  kmax= 40;
  step = 50;

  n0min = floor((N+kmax)/2);
  n0max= size(x,2) - floor((N+kmax)/2) -1;
  Q=zeros(kmax-kmin+1, floor((n0max-n0min)/step)+2);

    for k= kmin : kmax %i

    for n0= n0min : step : n0max %j
      for n= 1 : N-1
        Q(i, j) = Q(i,j) + ( (x(n+n0-n0min) - x(n+n0-n0min+k))^2 * hanning(N)(n+1) );
      Q(i, j) = (2/N) * Q(i,j);

    fprintf('Calculated %d out of %d periods \n', k-kmin+1, kmax-kmin+1);



Tweaking around with the parameters, I ended up with the ones shown in the code, that yield the following result: enter image description here

This is already getting closer to what I expected. Other suggestions welcome :)

  • $\begingroup$ "At some points in time, the person will be exercising or jumping up & down, creating more or less obvious periodic motions (yellow regions of the graph)." Well, there appears to me only two yellow regions (the first and the fifth) that display much periodicity. Normally to measure the degree of periodicity, something like the autocorrelation function is used. $\endgroup$ – robert bristow-johnson Jun 2 '19 at 9:09
  • $\begingroup$ Here's another reference about measuring periodicity. The application area is more about musical pitch detection, but it's about measuring the period and measuring the degree of periodicity. $\endgroup$ – robert bristow-johnson Jun 2 '19 at 9:12
  • $\begingroup$ Could you possibly make a dataset available? $\endgroup$ – Cedron Dawg Jun 2 '19 at 12:39
  • $\begingroup$ I think you are defining the window wrong, in window = 25. The window variable is supposed to be samples from a window such as hanning or blackman. I think that, if you define a proper window, your spectrogram will give you much better information. $\endgroup$ – MBaz Jun 2 '19 at 17:33
  • $\begingroup$ Thank you @robertbristow-johnson for your comments, I will look in more details into AMDF/ASDF. The theory in your linked posts starts to make sense, need to see how I can try this out in practice! $\endgroup$ – Duthopi Jun 3 '19 at 7:07

If you're gonna use ASDF to measure periodicity and you want to window the data with something other than rectangular window, do it after subtracting and squaring:

$$ Q_x[k, n_0] \triangleq \frac{2}{N} \sum\limits_{n=0}^{N-1} \left(x[n+n_0-\left\lfloor \tfrac{N+k}{2}\right\rfloor] \ - \ x[n+n_0-\left\lfloor \tfrac{N+k}{2}\right\rfloor + k] \right)^2 w\left(\tfrac{n}{N}\right) $$


$\left\lfloor \cdot \right\rfloor$ is the floor() function and, if $k$ is even then $ \left\lfloor \frac{k}{2}\right\rfloor = \left\lfloor \frac{k+1}{2}\right\rfloor = \frac{k}{2} $

and $w\left(\tfrac{n}{N}\right)$ is a window function of non-zero width $N$ samples centered at $n=\tfrac{N}{2}$. A Hann window would be

$$ w(u) \triangleq \begin{cases} \tfrac12 - \tfrac12 \cos(2\pi u) \qquad & 0 \le u < 1 \\ \\ 0 & \text{otherwise} \\ \end{cases}$$

To make this ASDF into "autocorrelation" (in the neighborhood of sample $x[n_0]$) defined from the ASDF:

$$ R_x[k,n_0] = R_x[0,n_0] - \tfrac12 Q_x[k, n_0] $$


$$ R_x[0, n_0] \triangleq \frac{2}{N} \sum\limits_{n=0}^{N-1} \Big(x[n+n_0-\left\lfloor \tfrac{N}{2}\right\rfloor]\Big)^2 w\left(\tfrac{n}{N}\right) $$

Since $Q_x[0, n_0] = 0$ and $Q_x[k, n_0] \ge 0$ for all lags $k$, that means that $ R_x[k, n_0] \le R_x[0, n_0] $ for all lags $k$.

Suppose for a minute that $x[n]$ is periodic with period $P$ (and $P$ happens to be an integer), then

$$ x[n+P] = x[n] \quad \forall n $$

and $Q_x[mP, n_0] = 0$ and $R_x[mP, n_0] = R_x[0, n_0] \ge R_x[k, n_0]$ for any integer number of periods ($m$ is an integer). So you get a peak at $k=0$ and at $k$ equal to any other multiple of $P$ if $x[n]$ is periodic. If $x[n]$ is not perfectly periodic, what we might expect is the biggest peak at $k=0$, another peak (but slightly smaller) at $k=P$ (the period we are looking for) and progressively smaller peaks for larger multiples of $P$.

The measure of periodicity in the neighborhood of $n_0$ would be

$$\frac{R_x[P, n_0]}{R_x[0, n_0]}$$

Hmmm, I just thought of something. Because of my audio/music-centric POV, I have been assuming no DC bias (i.e. $x[n]$ should maybe come out of a DC blocking high-pass filter). But I don't think that is the case with your application. So maybe just stick with ASDF, but somewhere you will need a threshold for how low $Q_x[k, n_0]$ can go (for $k \ne 0$) for you to consider that $x[n]$ is periodic in the neighborhood of $n_0$.

  • $\begingroup$ Many thanks for the detailed answer. I edited my post to show my attempt to implement it, and am getting closer to what I need! I was just not sure about what you mean by 'you need a threshold for how low Q can go'? If you have any advice on how to find the optimal parameters for window, step and k-values, I am also interested, as this took me a lot of trying out... $\endgroup$ – Duthopi Jun 4 '19 at 16:38
  • $\begingroup$ $Q_x[k, n_0]$ is always non-negative and $Q_x[P, n_0]$ gets close to zero when there is a match of periodicity. I am not sure how low it must go for you to be able to call it "periodic" in the neighborhood of $n_0$. $\endgroup$ – robert bristow-johnson Jun 4 '19 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.