Line voltage Fourier series: triplen harmonics absent

In a three-phase square-wave (180° conduction) inverter, each pole voltage measured from the dc mid-point O is a square wave of $\pm V_{dc}/2$, and the three poles are displaced by 120°. The line voltage is the difference of two pole voltages, and in this difference all triplen harmonics cancel.

Pole voltages

$$\begin{aligned} v_{aO} &= \sum_{n=1,3,5,\dots}\frac{2V_{dc}}{n\pi}\sin n\omega t \\ v_{bO} &= \sum_{n=1,3,5,\dots}\frac{2V_{dc}}{n\pi}\sin n\Big(\omega t - \frac{2\pi}{3}\Big) \end{aligned}$$

(Each is a square wave of amplitude $V_{dc}/2$, so $b_n = \frac{4}{n\pi}\cdot\frac{V_{dc}}{2}$.)

Line voltage

$$\begin{aligned} v_{ab} &= v_{aO} - v_{bO} = \sum_{n\ \text{odd}}\frac{2V_{dc}}{n\pi}\Big[\sin n\omega t - \sin n\Big(\omega t - \frac{2\pi}{3}\Big)\Big] \end{aligned}$$

Using $\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$ with $\frac{A+B}{2} = n\big(\omega t - \frac{\pi}{3}\big)$ and $\frac{A-B}{2} = \frac{n\pi}{3}$:

$$v_{ab} = \sum_{n=1,3,5,\dots}\frac{4V_{dc}}{n\pi}\sin\frac{n\pi}{3}\,\cos n\Big(\omega t - \frac{\pi}{3}\Big)$$

This is the Fourier series of the 120° quasi-square line voltage ($+V_{dc}$ for 0–120°, 0 for 60°, $-V_{dc}$ for 120°, 0 for 60°). It is identical, term by term and with the same time origin, to the textbook form $v_{ab} = \sum\frac{4V_{dc}}{n\pi}\cos\frac{n\pi}{6}\sin n(\omega t + \frac{\pi}{6})$.

Triplen harmonics are absent

For $n = 3, 9, 15, \dots$ ($n = 3k$):

$$\sin\frac{n\pi}{3} = \sin k\pi = 0$$

so every triplen term in $v_{ab}$ is zero. Physically, the triplen components of $v_{aO}$, $v_{bO}$, $v_{cO}$ are equal in magnitude and in phase (a shift of $2\pi/3 \times 3k = 2k\pi$), so they cancel when two pole voltages are subtracted.

Remaining harmonics

Fundamental line voltage: $\hat V_{L1} = \frac{2\sqrt3V_{dc}}{\pi}$, $V_{L1,rms} = \frac{\sqrt6V_{dc}}{\pi} = 0.7797V_{dc}$. Only harmonics of order $6k\pm1$ (5, 7, 11, 13, …) remain, with amplitude $V_{L1}/n$.