Half-bridge inverter RL load Fourier, THD

With the centre point of the dc source grounded, the single-phase (half-bridge) inverter gives a square wave of $\pm V_s/2$ across the load. Its Fourier series contains only odd harmonics, each of which drives a current through the RL impedance at that frequency.

Given: $V_s = 200$ V (so $\pm100$ V), $R = 5\ \Omega$, $L = 20$ mH, $f = 50$ Hz, $\omega = 314.16$ rad/s.

Fourier series of output voltage

$$v_0 = \sum_{n=1,3,5,\dots}\frac{4(V_s/2)}{n\pi}\sin n\omega t = \sum_{n\ \text{odd}}\frac{2V_s}{n\pi}\sin n\omega t = \sum_{n\ \text{odd}}\frac{127.32}{n}\sin n\omega t$$

Harmonic currents

$$Z_n = \sqrt{R^2 + (n\omega L)^2}, \qquad \theta_n = \tan^{-1}\frac{n\omega L}{R}, \qquad I_{nm} = \frac{V_{nm}}{Z_n}$$

Load current up to the 3rd order:

$$i_0(t) = 15.86\sin(314.16t - 51.49^\circ) + 2.18\sin(942.48t - 75.14^\circ)\ \text{A}$$

Harmonic factor and THD of output voltage

$$\begin{aligned} V_{0,rms} &= \frac{V_s}{2} = 100\ \text{V} \\ V_{1,rms} &= \frac{127.32}{\sqrt2} = 90.03\ \text{V} \\ HF_n &= \frac{V_n}{V_1} = \frac{1}{n} \;\Rightarrow\; HF_3 = \frac{1}{3} = 33.33\%,\ HF_5 = 20\% \end{aligned}$$ $$\begin{aligned} THD &= \frac{\sqrt{V_{0,rms}^2 - V_{1,rms}^2}}{V_{1,rms}} = \frac{\sqrt{100^2 - 90.03^2}}{90.03} \\ &= \frac{43.52}{90.03} = 0.4834 = 48.34\% \end{aligned}$$

Answer: $i_0 = 15.86\sin(\omega t - 51.49^\circ) + 2.18\sin(3\omega t - 75.14^\circ)$ A; $HF_3 = 33.33\%$ (lowest-order harmonic); $THD = 48.34\%$.