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I have a periodic time domain sensor signal sampled at $500 \text{ Hz}$ whose amplitude spectrum I compute and obtain. From the periodicity of the signal, I was expecting the fundamental frequency to be around $f = 1.2\text{ Hz}$, and this is what I observe with amplitude say $A_1$. But besides the minuscule diminishing harmonics at integer multiples of $f$; what may be the cause of having the harmonic at $f_2$ around $2.4 \text{ Hz}$ that has an amplitude $A_2\approx1.25A_1$ ?

The signal is of $29 \text{ sec.} \Rightarrow N = 14500 \text{ samples}$. Down is a section with a full period of the signal, the blue one.

enter image description here

Below is a section of the amplitude spectrum. NFFT = 2^nextpow2(N); I have rectified the question.

enter image description here

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  • $\begingroup$ Is it OK for you to include any figures? Also please provide what is the length of your signal that you are analysing. It might be a leakage (I hope you are familiar with that phenomena). $\endgroup$
    – jojeck ♦
    Jun 15, 2014 at 19:09
  • $\begingroup$ @jojek The question is edited with correct/extra info. $\endgroup$
    – Gilles
    Jun 15, 2014 at 20:30

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I think this is simply a property of your signal and you indeed have two significant frequency components. I don't really know the nature of your signal, but for me, it makes perfect sense, you can distinguish two periods in your signal, as shown on the plot below:

enter image description here

As you can see there is period $\color{green}{T_1}$ and $\color{brown}{T_2}$ between consecutive "peaks". So by doing very rough approximations from your plot:

  • Corresponding frequency for $\color{green}{T_1}$ can be estimated as: $\dfrac{1}{7.1-6.3}=\dfrac{1}{0.8}=1.25 \mathtt{Hz} $

  • In case of $\color{brown}{T_2}$ you get: $\dfrac{1}{6.7-6.3}=\dfrac{1}{0.4}=2.5 \mathtt{Hz} $

Some quick and dirty plot of such case with two sinusoids of a given frequencies and amplitudes:

enter image description here

So indeed you have two frequencies present in your signal. What's more, component around $2.5 \mathtt{Hz} $ can be stronger. Please consider that your signal also has very prominent DC component (also visible at your spectrum).

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  • $\begingroup$ Why would you decompose these into two distinct frequencies ? The smaller peaks centered at $\approx 6.25 \text{ s}$ and at $\approx 7 \text{ s}$ are repeating themselves at the same rate as the bigger peaks around $\approx 6.65 \text{ s}$ and $\approx 7.4 \text{ s}$. Wouldn't this just result in differences in phase but as far as the amplitude spectrum is concerned they should be hitting same harmonics ? $\endgroup$
    – Gilles
    Jun 15, 2014 at 21:24
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    $\begingroup$ @Gilles: If there is a periodicity in your signal then most likely you can match sinusoid to that. I can see two different sinusoids and their frequencies are related: $f_2=2f_1$. Just take a look at the second figure of two sinusoids - couldn't that be you case? After analysing your signal and frequency spectrum I can say that that would be my result after decomposition into Fourier Series by hand - two frequency components and positive interference. $\endgroup$
    – jojeck ♦
    Jun 15, 2014 at 21:32
  • $\begingroup$ @Gilles: How otherwise you would get $2.5 \mathtt{Hz}$ wave with its amplitude increasing every second peak? Simply add $1.25 \mathtt{Hz}$ sinusoid with correct phase and amplitude and you will get what you want. There is no other way by using this basis. What's more - you can correlate this fluctuations at peaks with your higher harmonics. If you still have some problems with understanding, then please check this applet (or different one) and play with it. $\endgroup$
    – jojeck ♦
    Jun 15, 2014 at 21:41
  • $\begingroup$ Kudos for the "quick and dirty" illustrative figure with the sinusoids. So, the in the amplitude spectrum, what's happening is that the peaks are superimposed every $2k \cdot f_1$ for $k \in N$ ? $\endgroup$
    – Gilles
    Jun 15, 2014 at 21:59
  • $\begingroup$ @Gilles: To be 100% right: peaks are superimposed every $2k\cdot \frac{1}{f_1}$ second. Sometimes it is positive and sometimes negative superposition. Obviously this is definitely not the exact case for your signal (phase, frequency, amplitudes might be slightly different) but you get the big picture. $\endgroup$
    – jojeck ♦
    Jun 15, 2014 at 22:07

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