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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 3σ) 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).

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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.

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Thank you very much for your clear answer! What kind of further knowledge do we need to say something more? –  Py-ser Apr 2 at 7:58
    
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. –  hotpaw2 Apr 2 at 8:17
    
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. –  Matt L. Apr 2 at 9:52
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