I have some questions regarding interharmonics. What I'm going to do is first ask just a few, and then as people answer them I would expand this post or create a new question.

Harmonics are sinusoids that have a frequency which is an integer multiple of the fundamental frequency of the original signal \$x(t)\$ they represent. Interharmonics (or inter-harmonics) are defined as sinusoids that have a frequency which isn't an integer multiple of the fundamental frequency of the signal \$x(t)\$. First question: in that definition of interharmonics, is it assumed the signal \$x(t)\$ is periodic, or not?¹

I mean, yes, we use Fourier series usually for periodic signal, but I haven't read a single text book on math, circuit analysis, electronics, or signals and systems where they talk about interharmonics. Fourier's theorem never talks about interharmonics. The only place I've seen a brief discussion on interharmonics is in textbooks about power quality and harmonics. So this makes me wonder whether interharmonics even make sense, or that's just a term invented without any mathematical proof. So, before I ask further questions, I'd like to know the answer to the first question above.

I have another question. As you know, there're various ways to represent a Fourier series. One is the trigonometric form, other is the amplitude-phase form, and the other is the complex exponential form. Writing the amplitudes (i.e. the maximum values or peak values) of the harmonics in terms of the RMS values, the amplitude-phase form is:

\$x(t) = X_0 + \sqrt{2} \displaystyle \sum_{n=1}^\infty X_{\text{rms,} n} \cos{(2 \pi n f_0 t + \theta_n)} \tag*{}\$

My second question is if when a signal has an interharmonic of frequency \$m f_0\$, where \$m\$ is a non-integer positive number, do we sum it to the previous expression as a new sinusoid \$\sqrt{2} X_{\text{rms,} m} \cos{(2 \pi m f_0 + \theta_m)}\$? If not, then how does the interharmonic analitically contribute to the signal \$x(t)\$?

Note ¹: Fourier series can be used to represent a periodic signal with an expression valid for all time \$t\$, or to represent a non-periodic signal in a time interval \$\Delta t\$.

  • 2
    \$\begingroup\$ "... and then as people answer them I would expand this post or create a new question." Create a new question. Please don't change the question after answers have been given as it makes them look like half-answers. Only edit to improve the original question. \$\endgroup\$
    – Transistor
    Commented Sep 25, 2020 at 23:35
  • \$\begingroup\$ @Transistor Okay, thanks for the suggestion! \$\endgroup\$
    – alejnavab
    Commented Sep 28, 2020 at 16:15
  • \$\begingroup\$ In the context of optics: four-wave mixing. \$\endgroup\$ Commented Oct 28, 2020 at 22:30

2 Answers 2


Let’s say you have an interharmonic of 1.5. If you scale your assumed fundamental frequency by 0.5 then you will have two integer harmonic sinusoids contributing to the signal - a 2nd harmonic and a 3rd harmonic. Equivalent, of course, to the original signal - just an integer way to look at it.

First case: Fundamental = f1

$$ x(t) = \cos{(1*2 \pi f_1 t + \theta_n)} + \cos{(1.5*2 \pi f_1t + \theta_m )} $$

Second case: Fundamental = f2 = 0.5f1

$$ x(t) = \cos{(2*2 \pi f_2 t + \theta_n)} + \cos{(3*2 \pi f_2t + \theta_m)} $$

These are equivalent signals (and periodic), I just chose to assume a different fundamental frequency for my analysis so i would have integer harmonics.

  • \$\begingroup\$ Interharmonics are still sinusoids, right? \$\endgroup\$
    – alejnavab
    Commented Sep 24, 2020 at 0:26
  • \$\begingroup\$ Yes sir, that is correct. \$\endgroup\$ Commented Sep 24, 2020 at 0:29
  • \$\begingroup\$ Thanks. Following your answer, if I scale the frequency, wouldn't we get a new, different signal than the previous one? For example, if I scale the frequency of the signal \$\cos{(t)}\$ to \$\cos{(0.2t)}\$, these two signal aren't the same, are they? (This is just a simple example.) \$\endgroup\$
    – alejnavab
    Commented Sep 24, 2020 at 0:35
  • 1
    \$\begingroup\$ Not if you do it correctly, like i show above (edited my answer). I am not changing the frequency of each sinusoid. I'm just playing around and using a different fundamental - so that i end up with only integer harmonics. \$\endgroup\$ Commented Sep 24, 2020 at 0:52
  • 1
    \$\begingroup\$ That trick of scaling frequency was so neat! I checked the two signals in your answer in this GeoGebra app and they are indeed the same. However, this has bursted more questions in my head. For example, in the first expression of \$x(t)\$ in your answer, if we suppose that \$f_1=1 \text{ Hz}\$ and \$\theta_n = \theta_m = 0°\$, you can see in the GeoGebra app (or prove it analitically) that the fundamental frequency of \$x(t)\$ is actually \$1/(2 \text{ s}) = 0.5 \text{ Hz}\$, not \$f_1=1 \text{ Hz}\$. \$\endgroup\$
    – alejnavab
    Commented Sep 24, 2020 at 2:59

The IEEE standard #519, titled IEEE Recommended Practices and Requirements for Harmonic Control in Electrical Power Systems, version of the year 2014 (the latest update), on page 3 defines an interharmonic as follows:

interharmonic (component): A frequency component of a periodic quantity that is not an integer multiple of the frequency at which the supply system is operating (e.g., 50 Hz or 60 Hz).

So, when a signal is said to have interharmonics, the signal must be periodic, at least when using IEEE's definition.

As relayman showed in his answer, the presence of inteharmonics in a periodic signal depends on which period (the fundamental period or a multiple of it) is used to compute the coefficients of the Fourier series. The fundamental period \$T_0\$ of a periodic signal is defined as the smallest positive value of \$T\$ that satisfies \$x(t + T) = x(t)\$ for all \$t\$. When we use the fundamental period to compute the continuous-time Fourier series, no interharmonics are present.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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