Single-pulse PWM: modulation index and Fourier series

In single-pulse width modulation (one pulse per half cycle), the inverter output has one pulse of width $\delta$ in each half cycle, centred at $\pi/2$ (and at $3\pi/2$ for the negative half). The pulse width, and so the output voltage, is set by the modulation index.

Generation and effect of modulation index

 Ac      \        /\        /     carrier (2f)
          \      /  \      /
 Ar     ---\----/----\----/---  reference (dc level)
            \  /      \  /
 0           \/        \/
 +Vs        ____
    0   ___|    |____      ___
 -Vs                 |____|
           |<-d>|
            pi/2     3pi/2

• A rectangular reference of amplitude $A_r$ is compared with a triangular carrier of amplitude $A_c$ at twice the output frequency. The switch is ON while the reference exceeds the carrier, so each pulse is centred on a carrier trough ($\pi/2$, $3\pi/2$).
• Modulation index $M = A_r/A_c$ ($0 \le M \le 1$); the pulse width is $\delta = M\pi$.
• RMS output voltage:

$$V_{0,rms} = \left[\frac{2}{2\pi}\int_{(\pi-\delta)/2}^{(\pi+\delta)/2}V_s^2\,d(\omega t)\right]^{1/2} = V_s\sqrt{\frac{\delta}{\pi}} = V_s\sqrt{M}$$

So increasing $M$ widens the pulse and raises the rms output from 0 (at $M = 0$) to $V_s$ (at $M = 1$, square wave). Example: $M = 0.5$ gives $V_{0,rms} = 0.707V_s$.

Fourier series

The wave is odd and has half-wave symmetry, so $a_0 = a_n = 0$ and only odd $n$ exist:

$$\begin{aligned} b_n &= \frac{2}{\pi}\int_{(\pi-\delta)/2}^{(\pi+\delta)/2}V_s\sin n\theta\,d\theta = \frac{2V_s}{n\pi}\Big[\cos n\tfrac{\pi-\delta}{2} - \cos n\tfrac{\pi+\delta}{2}\Big] \\ &= \frac{4V_s}{n\pi}\sin\frac{n\pi}{2}\sin\frac{n\delta}{2} \end{aligned}$$ $$v_0(t) = \sum_{n=1,3,5,\dots}\frac{4V_s}{n\pi}\sin\frac{n\pi}{2}\sin\frac{n\delta}{2}\sin n\omega t$$

(The factor $\sin(n\pi/2) = \pm1$ only fixes the sign; many texts write $v_0 = \sum\frac{4V_s}{n\pi}\sin\frac{n\delta}{2}\sin n\omega t$.)

Fundamental value

$$\begin{aligned} V_{1m} &= \frac{4V_s}{\pi}\sin\frac{\delta}{2}, \qquad V_{1,rms} = \frac{4V_s}{\sqrt2\,\pi}\sin\frac{\delta}{2} = 0.9V_s\sin\frac{\delta}{2} \end{aligned}$$

• At $\delta = \pi$ (square wave), $V_{1,rms} = 0.9V_s$.
• A harmonic $n$ is eliminated when $\sin(n\delta/2) = 0$, i.e. $\delta = 2\pi/n$; e.g. $\delta = 120^\circ$ removes the 3rd harmonic.
• Drawback: at small pulse widths (low output voltage) the harmonic content is high, which led to multiple-pulse and sinusoidal PWM.