RMS and fundamental of stepped inverter waveform

The given wave (steps of 200 V and 400 V, each 60° wide) is the per-phase voltage of a 180°-conduction three-phase inverter with $V_{dc}/3 = 200$ V, i.e. $V_{dc} = 600$ V.

RMS value

Each 60° block occupies $\pi/3$ rad. Using the half cycle (the negative half is a mirror image):

$$\begin{aligned} V_{rms} &= \left[\frac{1}{\pi}\left(200^2\cdot\frac{\pi}{3} + 400^2\cdot\frac{\pi}{3} + 200^2\cdot\frac{\pi}{3}\right)\right]^{1/2} \\ &= \left[\frac{40000 + 160000 + 40000}{3}\right]^{1/2} = \sqrt{80000} \\ &= 282.84\ \text{V} \end{aligned}$$

(Check: $\dfrac{\sqrt2}{3}V_{dc} = 0.4714\times600 = 282.84$ V.)

Peak value of the fundamental

The wave has half-wave and quarter-wave (odd) symmetry, so $a_n = 0$ and

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

For $n = 1$: $\cos 60^\circ = 0.5$, $\cos 120^\circ = -0.5$, $\cos 180^\circ = -1$:

$$\begin{aligned} b_1 &= \frac{2}{\pi}\big[200(0.5) + 400(1) + 200(0.5)\big] = \frac{2}{\pi}\times600 \\ &= 381.97\ \text{V} \end{aligned}$$

(Check: $\dfrac{2V_{dc}}{\pi} = \dfrac{2\times600}{\pi} = 381.97$ V.) RMS of fundamental $= 381.97/\sqrt2 = 270.09$ V.

The same formula gives $b_3 = 0$ (no triplen harmonics), $b_5 = 76.39$ V, $b_7 = 54.57$ V.

Answer: $V_{rms} = 282.84$ V; peak fundamental $V_{1m} = 381.97$ V (270.09 V rms).