Fundamental and 3rd harmonic of three-phase inverter output

The given phase voltage ($V_{dc}/3$, $2V_{dc}/3$, $V_{dc}/3$ in 60° steps) is the six-step wave of a 180° conduction three-phase inverter. It is odd and has half-wave symmetry, so $a_0 = a_n = 0$ and only odd sine terms are present.

Fourier coefficient

$$\begin{aligned} b_n &= \frac{2}{\pi}\int_0^{\pi}v_{AN}\sin n\theta\,d\theta \\ &= \frac{2}{n\pi}\Big[\tfrac{V_{dc}}{3}\big(1-\cos\tfrac{n\pi}{3}\big) + \tfrac{2V_{dc}}{3}\big(\cos\tfrac{n\pi}{3}-\cos\tfrac{2n\pi}{3}\big) + \tfrac{V_{dc}}{3}\big(\cos\tfrac{2n\pi}{3}-\cos n\pi\big)\Big] \\ &= \frac{2V_{dc}}{3n\pi}\Big[2 + \cos\tfrac{n\pi}{3} - \cos\tfrac{2n\pi}{3}\Big] \quad (n\ \text{odd}) \end{aligned}$$

Fundamental component ($n = 1$)

$$\begin{aligned} b_1 &= \frac{2V_{dc}}{3\pi}\big[2 + 0.5 + 0.5\big] = \frac{2V_{dc}}{\pi} = 0.6366\,V_{dc}\ \text{(peak)} \\ V_{1,rms} &= \frac{2V_{dc}}{\sqrt2\,\pi} = \frac{\sqrt2\,V_{dc}}{\pi} = 0.4502\,V_{dc} \end{aligned}$$ $$v_{AN1} = \frac{2V_{dc}}{\pi}\sin\omega t$$

Third harmonic ($n = 3$)

$\cos\pi = -1$, $\cos 2\pi = +1$:

$$b_3 = \frac{2V_{dc}}{9\pi}\big[2 - 1 - 1\big] = 0$$

So the third harmonic component is zero. All triplen harmonics vanish; the non-zero terms are $n = 6k\pm1$ (5, 7, 11, 13, …), each with peak $\frac{2V_{dc}}{n\pi}$:

$$v_{AN} = \frac{2V_{dc}}{\pi}\Big[\sin\omega t + \frac{1}{5}\sin5\omega t + \frac{1}{7}\sin7\omega t + \frac{1}{11}\sin11\omega t + \dots\Big]$$

Answer: fundamental $= 0.6366V_{dc}$ peak ($0.4502V_{dc}$ rms); 3rd harmonic $= 0$. For example, with $V_{dc} = 600$ V the fundamental is 381.97 V peak (270.09 V rms).