Square-wave inverter ±48 V: fundamental and third harmonic

With a split ±48 V supply, the single-phase (half-bridge) inverter applies $+48$ V and $-48$ V for alternate half cycles, so the output is a square wave of amplitude $V = 48$ V at 50 Hz: $v_0 = \sum_{n\ \text{odd}}\frac{4V}{n\pi}\sin n\omega t$.

Given: $V = 48$ V, $f = 50$ Hz ($\omega = 314.16$ rad/s), $R = 10\ \Omega$, $L = 20$ mH. Phase angles are referred to the output voltage (sine reference, 0°).

Fundamental ($n = 1$)

$$\begin{aligned} V_{1m} &= \frac{4\times48}{\pi} = 61.12\ \text{V}, \qquad V_{1,rms} = 43.22\ \text{V} \\ X_1 &= \omega L = 6.283\ \Omega, \quad Z_1 = \sqrt{10^2 + 6.283^2} = 11.810\ \Omega \\ \theta_1 &= \tan^{-1}\frac{6.283}{10} = 32.14^\circ \\ I_{1m} &= \frac{61.12}{11.810} = 5.175\ \text{A}, \qquad I_{1,rms} = 3.659\ \text{A} \end{aligned}$$

Third harmonic ($n = 3$)

$$\begin{aligned} V_{3m} &= \frac{61.12}{3} = 20.37\ \text{V}, \qquad V_{3,rms} = 14.41\ \text{V} \\ X_3 &= 3\omega L = 18.850\ \Omega, \quad Z_3 = \sqrt{10^2 + 18.850^2} = 21.338\ \Omega \\ \theta_3 &= \tan^{-1}\frac{18.850}{10} = 62.05^\circ \\ I_{3m} &= \frac{20.37}{21.338} = 0.955\ \text{A}, \qquad I_{3,rms} = 0.675\ \text{A} \end{aligned}$$

Results

$$\begin{aligned} v_1 &= 61.12\sin314.16t, & i_1 &= 5.175\sin(314.16t - 32.14^\circ) \\ v_3 &= 20.37\sin942.48t, & i_3 &= 0.955\sin(942.48t - 62.05^\circ) \end{aligned}$$