--- title: "Single-pulse PWM: modulation index and Fourier series" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:single-pulse-pwm-fourier" source: https://www.bctengineer.com/q/semester_7/power_electronics/single-pulse-pwm-fourier site: BCT Engineering --- # Single-pulse PWM: modulation index and Fourier series Power Electronics · Chapter 5: Inverter - Asked in: 2079 Baishakh, 8 marks - Video: [Single Pulse Width Modulation - Inverter - Power Electronics](https://www.youtube.com/watch?v=skTumL6LUIo) — Ekeeda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/single-pulse-pwm-fourier - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question How does modulation index change the rms value of output voltage of inverter? Derive the Fourier series of single pulse width modulation and determine the fundamental value. ## Short answer Increasing modulation index M widens the single pulse, so V0,rms=Vs sqrt(delta/pi)=Vs sqrt(M). Fourier series: v0=sum(4Vs/(n pi)) sin(n delta/2) sin(n omega t) for odd n; fundamental rms =0.9Vs sin(delta/2). ## Answer 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 ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In single-pulse modulation, how are M and pulse width related? A. delta = M pi / 2 B. delta = pi / M C. delta = M pi D. delta = M^2 pi Answer: C. M = A_r/A_c and the pulse width is delta = M pi. (Key point: Modulation index) 2. How does the rms output voltage vary with M? A. V0,rms = Vs M B. V0,rms = Vs sqrt(M) C. V0,rms = Vs M^2 D. It is independent of M Answer: B. V0,rms = Vs sqrt(delta/pi) = Vs sqrt(M); M = 0.5 gives 0.707 Vs. (Key point: RMS vs M) 3. What is the peak fundamental of the single pulse wave of width delta? A. (4Vs/pi) sin(delta/2) B. (4Vs/pi) sin(delta) C. (2Vs/pi) sin(delta/2) D. (4Vs/pi) cos(delta/2) Answer: A. b_1 = (4Vs/pi) sin(pi/2) sin(delta/2) = (4Vs/pi) sin(delta/2). (Key point: Fundamental) 4. Which pulse width removes the 3rd harmonic? A. 90° B. 60° C. 180° D. 120° Answer: D. sin(3 delta/2) = 0 when delta = 2pi/3 = 120°. 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