This method is explained with respect to a single-phase-inverter. Normally, there is one commutation per half-cycle at the end of each half-cycle and this produces a square-wave output. Instead of having commutation at the end of half-cycle,some more commutations can be created in the half cycle and a waveform as shown in figure.1 can be produced.By properly selecting the values of $\alpha_1$ and $\alpha_2$,any two unwanted lower order harmonics can be eliminated from the waveform.Here, waveforms are drawn for a single phase half bridge inverter.This waveform can also be obtained from a single phase full bridge inverter,but then the amplitude of voltage wave would be $E_{dc}$.It employs four extra commutation per cycle instead of one.For the unmodulated square wave,as this voltage waveform quarter wave symmetry, $B_n = 0$

$$A_{n}=\frac{4 E_{\mathrm{dc}}}{\pi 2}\left[\int_{0}^{\alpha_{1}} \sin n \omega t \cdot d(\omega t)-\int_{\alpha_{1}}^{\alpha_{2}} \sin n \omega t \cdot \mathrm{d}(\omega t)+\int_{\alpha_{2}}^{\pi / 2} \sin n \omega t \cdot d(\omega t)\right]$$

$$=\frac{2 E_{\mathrm{dc}}}{\pi}\left[\frac{1-2 \cos n \alpha_{1}+2 \cos n \alpha_{2}}{n}\right]-----(1)$$

If third and fifth harmonics are to be eliminated, then from Eq. $(1)$

$$A_{3}=\frac{2 E_{\mathrm{dc}}}{\pi}\left[\frac{1-2 \cos 3 \alpha_{1}+2 \cos 3 \alpha_{2}}{3}\right]=0$$

and $$\qquad A_{5}=\frac{2 E_{\mathrm{dc}}}{\pi}\left[\frac{1-2 \cos 5 \alpha_{1}+2 \cos 5 \alpha_{2}}{5}\right]=0$$

$$\begin{array}{ll}{\text { Or }} & {1-2 \cos 3 \alpha_{1}+2 \cos 3 \alpha_{2}=0} \\ {\text { and }} & {1-2 \cos 5 \alpha_{1}+2 \cos 5 \alpha_{2}=0}\end{array}$$

The above two simultaneous equations can be solved numerically to calculate $\alpha_1$ and $\alpha_2$ under the conditions that

$$\begin{array}{ll}{0\lt\alpha_{1}\lt90^{\circ}} & {\text { and } \quad \alpha_{1}\lt\alpha_{2}\lt90^{\circ}} \\ {\alpha_{1}=23.62^{\circ}} & {\text { and } \quad \alpha_{2}=33.6^{\circ}}\end{array}$$

Similarly, any two harmonics can be eliminated by calculating the corresponding values of $\alpha_{1}$ and $\alpha_{2} .$ This method produces a fundamental voltage of 83.9$\%$ or 0.839 times the amplitude of fundamental component of unmodulated voltage wave. Thus, with this method of harmonic reduction, inverter is derated by $(100-83.9)$ 16.1$\%$. Another disadvantage of this method is that, there are additional four commutations per cycle and this leads to more switching losses which decreases the efficiency of operation.