2
\$\begingroup\$

Before explaining my question, I'm going to assume that 1) interharmonics, just like harmonics, are sinusoids; and 2) to analytically represent the interharmonics of a signal, we sum them to the Fourier series. This was discussed in this previous question. If any of these assumptions is wrong, please say it and preferably share a reliable source. If these assumptions are true, then we can express a periodic signal \$x(t)\$ using the amplitude-phase form of Fourier series, with \$k\$ interharmonics, as follows (correct me if I'm wrong):

\$x(t) = \underbrace{X_0 + \sqrt{2} \displaystyle \sum_{n=1}^\infty X_{\text{rms,} n} \cos{(2 \pi n f_0 t + \theta_n)}}_{\text{DC component and infinite harmonics}} + \underbrace{\sqrt{2} \displaystyle \sum_{q=1}^k X_{\text{rms,} m_q} \cos{(2 \pi m_q f_0 t + \theta_{m_{q}})}}_{k \text{ interharmonics}} \tag 1\$

where all the \$m_q\$ are non-integer rational positive numbers. (Short example to clarify the previous notation: a signal has \$k=3\$ interharmonics, where \$m_1 = 1.5\$, \$m_2 = 2.4\$, \$m_3 = 6.3\$.)

However, the sum of sinusoids of different frequency, whose frequencies aren't irrational numbers, results in a periodic signal, which means that in equation (1), \$x(t)\$ is periodic. And since the signal is periodic, we could compute its Fourier series considering the interharmonics, such that we get no interharmonics in the end (this is illustrated in the following example.) In other words, in equation (1), the terms I've labeled as DC component and infinite harmonics, aren't really the Fourier series of \$x(t)\$. Therefore, the fundamental frequency of \$x(t)\$ isn't \$f_0\$ as we'd think from the first summation; for the same reason, the first summation isn't really the harmonics of \$x(t)\$. This is also illustrated below.

An example

Consider a sawtooth signal \$v(t)\$ of period \$T_0 = 1 \text{ s}\$, frequency \$f_0 = 1/T_0 = 1 \text{ Hz}\$, and amplitude \$A = 1 \text{ V}\$, that starts rising at \$ t = 0 \text{ s}\$:

\$ v(t) = \displaystyle \sum_{i=-\infty}^{\infty} v_{1}(t - i T_0) = \displaystyle \sum_{i=-\infty}^{\infty} v_{1}(t - 1i) \tag 2 \$

where

\$ v_1(t) = \left\{ \begin{aligned} \dfrac{A}{T_0} t &, \, 0 < t < T_0 \\ 0 &, \, \text{otherwise} \end{aligned} \right. = \left\{ \begin{aligned} t &, \, 0 < t < 1 \\ 0 &, \, \text{otherwise} \end{aligned} \right. \tag 3 \$

It can be proved its Fourier series is:

\$ v(t) = \dfrac{A}{2} + \dfrac{A}{\pi} \displaystyle \sum_{n=1}^\infty \dfrac{1}{n} \cos{(2 \pi n f_0 t + 90°)} = \dfrac{1}{2} + \dfrac{1}{\pi} \displaystyle \sum_{n=1}^\infty \dfrac{1}{n} \cos{(2 \pi n 1 t + 90°)} \tag 4 \$

Obviously, the fundamental frequency of \$v(t)\$ in equation (4) is 1 Hz. This GeoGebra app shows both the original signal (in green, given by equations (2) and (3)) and its Fourier series approximation (in red, given by equation (4)). Here's a screenshot:

Sawtooth signal

So far so good. Now imagine that according to someone or to a power analyzer, another periodic signal \$v_3(t)\$ has the same harmonics as \$v(t)\$ in equation (4), but also has one interharmonic of frequency 1.6 Hz (thus, \$m_1 = (1.6 \text{ Hz})/(1 \text{ Hz}) = 1.6\$), amplitude 1 V, and 0° phase shift. (I'm aware real-life power analyzers sample a continuous-time signal and only show up to a certain number of harmonics, e.g. 50 or 100, but I think this is irrelevant in this context.) Therefore, according to the second assumption I said in the first paragraph, we can express \$v_3(t)\$ as follows:

\$ \begin{align} v_3(t) &= \dfrac{A}{2} + \dfrac{A}{\pi} \left[ \displaystyle \sum_{n=1}^\infty \dfrac{1}{n} \cos{(2 \pi n f_0 t + 90°)} \right] + \cos{(2 \pi 1.6 f_0 t)} \\ &= \dfrac{1}{2} + \dfrac{1}{\pi} \left[ \displaystyle \sum_{n=1}^\infty \dfrac{1}{n} \cos{(2 \pi n 1 t + 90°)} \right] + \cos{(2 \pi 1.6 t)} \tag 5 \end{align} \$

or in terms of \$v(t)\$,

\$ v_3(t) = v(t) + \cos{(2 \pi 1.6 t)} \tag 6 \$

Now, we would think at a first glance that the (fundamental) frequency of \$v_3(t)\$ was 1 Hz according to the harmonics in equation (5). But this is wrong! In equation (6), since the (fundamental) period of \$v(t)\$ is 1 s, and the period of \$\cos{(2 \pi 1.6 t)}\$ is 1/(1.6 Hz) = 5/8 s, the ratio of these periods is a rational number. Thus, according to this video, we can compute the (fundamental) period \$T_0'\$ of \$v_3(t)\$ as follows:

\$T_0' = \text{LCM} (1, \frac{5}{8}) = \dfrac{\text{LCM} (1, 5)}{\text{GCD} (1, 8)} = \dfrac{5}{1} = 5 \text{ s} \tag*{} \$

The following screenshot proves this, taken from this GeoGebra app, where \$v(t)\$ is shown in green (given by equations (2) and (3)), \$\cos{(2 \pi 1.6 t)}\$ in orange, and \$v_3(t)\$ in purple (given by equation (6).)

Sawtooth wave summed with a sinusoid

Therefore, the harmonics of \$v_3(t)\$ aren't actually those given in equation (5), because the fundamental frequency of \$v_3(t)\$ isn't 1 Hz, but instead \$f_0' = 1/T_0' = 1/(5 \text{ s}) = 0.2 \text{ Hz}\$. To find the actual harmonics of \$v_3(t)\$, we can use equation (6) and substitute \$v(t)\$ by equations (2) and (3). Now we find the Fourier coefficients of that expression. After some math, the result would be

\$ v_3(t) = \dfrac{1}{2} + \cos{\left(2 \pi 1.6 t \right)} - \dfrac{1}{\pi} \displaystyle \sum_{n=1}^\infty \dfrac{1 + \cos{(\frac{2 \pi n}{5})} + \cos{(\frac{4 \pi n}{5})} + \cos{(\frac{6 \pi n}{5})} + \cos{(\frac{8 \pi n}{5})}}{n} \sin{\left(\dfrac{2 \pi n}{5} t \right)} \tag 7 \$

The following image proves the previous expression, where the original signal \$v_3(t)\$ is shown in blue (given by equations (6), (2) and (3)) and its Fourier series approximation in yellow (given by equation (7)):

Sawtooth wave summed with a sinusoid, original and Fourier series

While both equations (5) and (7) correctly represent \$ v_3(t)\$, the former is misleading for the reasons I explained in the previous paragraph. Also, notice initially we thought \$v_3(t)\$ had an interharmonic of 1.6 Hz according to equation (5), however, in equation (7) there're no interharmonics. So, if the assumptions said in the first paragraph are true, then interharmonics are misleading (if you think otherwise, please explain why.) And this makes me wonder why are they even defined by IEEE and IEC.

\$\endgroup\$
4
  • \$\begingroup\$ Interharmoics are to measure frequencies to the side of the fundamental (by IEC/IEEE), the usually the loads that are switching (from periodic noise from a power supply) or sometimes at random (arcing load). The goal of regulatory boards is to place some kind of limit on the noise, so that it can be measured by equipment. The definitions are enough to place a boundary on interharmonic noise, test with equipment and limit devices to limit conducted noise in power systems. We don't usually use a meter stick to measure km's, and exactness is not needed for testing/limiting noise or set a boundary \$\endgroup\$ – Voltage Spike Oct 2 '20 at 20:29
  • \$\begingroup\$ The textbook Power System Harmonics (2nd edition, 2003) by the authors Jos Arrillaga and Neville Watson, says on section 2.11.5 (page 45), titled Application to Inter-Harmonic Analysis, the following: "If frequencies not harmonically related to the sampling period are present or the waveform is not periodic over the sampling interval, errors are encountered due to spectral leakage." Basically, it looks like interharmonics can exist simply because the physical measuring devices are not perfect. \$\endgroup\$ – Alejandro Nava Jan 25 at 21:46
  • \$\begingroup\$ @VoltageSpike "sometimes at random (arcing load)." If the signal is random, it's not periodic, so why would we use Fourier series to represent it? Shouldn't we use Fourier transform? I know a Fourier series can be computed for a non-periodic signal, but its FS will represent such signal only in the time interval where the Fourier coefficients were computed. \$\endgroup\$ – Alejandro Nava Apr 6 at 5:44
  • \$\begingroup\$ you can still represent noise with the forier spectrum, the average of white noise is equal amplitude across all frequencies. \$\endgroup\$ – Voltage Spike Apr 6 at 16:02
0
\$\begingroup\$

Because we have the flexibility to declare what we want to be the "fundamental" frequency. If we have a sinusoid composed of a 60Hz component + a 120Hz component + a 90Hz component we can refer to them as a 1st harmonic, a 2nd harmonic, and a 1.5 harmonic (the interharmonic one).

Viewing them this way is equally valid to the approach you elucidate above (with LCM and GCD to find the true fundamental/period).

It is just a semantic difference. Saying my fundamental is 60Hz and my signal has 1, 2, and 1.5 components is equivalent to saying my fundamental is 30Hz and my signal has 2, 4, and 3rd components.

\$\endgroup\$

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.