In the case of single-pulse width modulation, the width of the pulse is adjusted to reduce the harmonic. In general, the RMS value of the amplitude of harmonic voltage of a single pulse modulated wave is given by.

$$E_{L n}=\frac{4 E_{\mathrm{dc}}}{\sqrt{2} n \pi} \sin \frac{n p}{2}=\frac{2 \sqrt{2} E_{\mathrm{dc}}}{n \pi} \sin \frac{n p}{2}$$

where $P$ is the width of the pulse and $E_{\mathrm{dc}}$ is the supply d.c. voltage. For example, if the third harmonic is to be eliminated, $E_{L_{3}}=0$ $$\therefore \quad E_{L_{3}}=\frac{4 E_{\mathrm{dc}}}{3 \sqrt{2} \pi} \sin \frac{3 p}{2}=0 \quad \therefore P=120^{\circ}$$

Similarly, to eliminate fifth harmonic, $P=72^{\circ}$

By this method, therefore, only one harmonic can be eliminated at a time.