Fundamental and 3rd harmonic of 180° inverter output

The given phase voltage (200 V, 400 V, 200 V steps of 60°) is the six-step wave of a 180° conduction inverter with $V_{dc}/3 = 200$ V, so $V_{dc} = 600$ V. It is odd with half-wave symmetry, so only odd sine terms exist.

Fourier coefficient

$$\begin{aligned} b_n &= \frac{2}{\pi}\int_0^{\pi}v_{an}\sin n\theta\,d\theta \\ &= \frac{2}{n\pi}\Big[200\big(1-\cos\tfrac{n\pi}{3}\big) + 400\big(\cos\tfrac{n\pi}{3}-\cos\tfrac{2n\pi}{3}\big) + 200\big(\cos\tfrac{2n\pi}{3}-\cos n\pi\big)\Big] \\ &= \frac{400}{n\pi}\Big[2 + \cos\tfrac{n\pi}{3} - \cos\tfrac{2n\pi}{3}\Big] \quad (n\ \text{odd}) \end{aligned}$$

Fundamental ($n = 1$)

$$\begin{aligned} b_1 &= \frac{400}{\pi}\big[2 + 0.5 - (-0.5)\big] = \frac{1200}{\pi} = 381.97\ \text{V (peak)} \\ V_{1,rms} &= \frac{381.97}{\sqrt2} = 270.09\ \text{V} \end{aligned}$$

Check with the standard result: $\hat V_{1} = \dfrac{2V_{dc}}{\pi} = \dfrac{2\times600}{\pi} = 381.97$ V.

Third harmonic ($n = 3$)

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

$$b_3 = \frac{400}{3\pi}\big[2 + (-1) - 1\big] = 0, \qquad V_{3,rms} = 0$$

The third harmonic (and every triplen harmonic) is absent from the phase voltage of a 180° inverter with a balanced star load, because triplen components of the three pole voltages are in phase and appear only in the neutral voltage $v_{nO}$.

(For reference: overall rms $= \sqrt{(200^2+400^2+200^2)/3} = 282.84$ V; next harmonic $b_5 = 76.39$ V peak.)

Answer: $V_{1,rms} = 270.09$ V; $V_{3,rms} = 0$.