Let's suppose I have a signal $F(t)$ that is periodic, with two periodicities $P1$ and $P2$, with $P1>P2$

Suppose that I know the values of the two periodicities. Using the Fast Fourier transform I can show the two values as peaks in a power spectrum.

Now, let's suppose the second periodicity $P2$ (the faster one), has exactly the same value as the first harmonic of the fundamental value, or $P2=2×P1$. This means that I will be not able to distinguish it by using the power spectrum, at least not by looking at the frequency of the peak.

My question is: is there a way to separate the contributions in such a case?

For example, is it possible to predict the power of the first harmonic, so that the difference between the predicted power and the observed power of the harmonic peak gives a result significant enough (i.e., greater than $3\sigma$) to say that the first harmonic also "contains" the contribution from a periodicity?

Please, be plain I am not experienced in this (nonetheless some equations/numbers are ok).


2 Answers 2


If you have two periodic functions $f(t)$ and $g(t)$, where $f(t)$ has fundamental frequency $f_0$ and $g(t)$ has fundamental frequency $2f_0$, then their sum $h(t)=f(t)+g(t)$ is periodic with fundamental frequency $f_0$. Without any further knowledge about $f(t)$ or $g(t)$ there is no way to separate the two, because if the only knowledge you have about $f(t)$ is that it is periodic with fundamental frequency $f_0$, then there is no way to distinguish it from the sum function $h(t)$, which has the same periodicity.

  • $\begingroup$ Thank you very much for your clear answer! What kind of further knowledge do we need to say something more? $\endgroup$
    – Py-ser
    Apr 2, 2014 at 7:58
  • $\begingroup$ If you know what the waveform shape of the lower periodicity alone should look like, the you might be able to predict the sizes of its harmonics. $\endgroup$
    – hotpaw2
    Apr 2, 2014 at 8:17
  • $\begingroup$ You would need some knowledge about the decay of the amplitudes of the harmonics. In the general case I have no solution for this problem. In some trivial cases it might be possible, e.g. if you know that one signal only consists of a few harmonics (because you know it's low-pass filtered with known cut-off frequency). But I must admit that I've never worked on anything like this before. $\endgroup$
    – Matt L.
    Apr 2, 2014 at 9:52
  • $\begingroup$ I presume it is also the case when the fundamental frequency of $g(t)$ is $0.5\cdot f_0$, that separation is not possible? $\endgroup$
    – jomegaA
    Feb 3, 2020 at 20:46
  • 1
    $\begingroup$ @jomegaA: Yes, that's basically the same situation. $\endgroup$
    – Matt L.
    Feb 3, 2020 at 21:21

Assume a model for waveform generation:

$$y(t) = a_0*\sin(2*\pi*f*t) + a_1*\sin(2*\pi*2*f*t)$$

Case 1:

  • $f(t)$ contains $\left[a_0, a_1\right] = \left[1, 0\right]$ and $f = 440Hz$
  • $g(t)$ contains $\left[a_0, a_1\right] = \left[1, 0\right]$ and $f = 880 Hz$

The sum of the sources will be:

$$z_1(t) = f(t)+g(t)=\sin(2*\pi*440*t) + \sin(2*\pi*880*t)$$

Case 2:

  • $f(t)$ contains $\left[a_0, a_1\right] = \left[1, 1\right]$ and $f = 440Hz$

  • $g(t)$ contains $\left[a_0, a_1\right] = \left[0, 0\right]$ and $f = 880 Hz$

The sum of the sources will be:

$$z_2(t) = f(t)+g(t)=\sin(2*\pi*440*t) + \sin(2*\pi*880*t)$$

Observe that $z_1$ and $z_2$ are identical, even though one contains only the source $f$, while the other contains a mix of sources $f$ and $g$.

Thus, armed with only the mixture $z_1$ (or $z_2$) we cannot say if case 1 or case 2 is what really went into the mix.

For a practical problem, you might have two singers mixed to one audio track, wanting to separate them. In that case, the pitch will be variable over time and practically never «perfect». You might also find that pitch and overtone mixture changes smoothly over time and/or in different ways for two different singers. That would be additional information that might make the problem solvable in a specific scenario.


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