--- title: "Single-phase PWM inverter rms and Fourier series" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:single-phase-pwm-inverter-fourier" source: https://www.bctengineer.com/q/semester_7/power_electronics/single-phase-pwm-inverter-fourier site: BCT Engineering --- # Single-phase PWM inverter rms and Fourier series Power Electronics · Chapter 5: Inverter - Asked in: 2074 Asoj, 10 marks - Video: [PULSE WIDTH MODULATION INVERTER](https://www.youtube.com/watch?v=w9Gum2xChMs) — Dr. Shital Sokashe - Ghorpade - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/single-phase-pwm-inverter-fourier - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question Explain the operation of single phase PWM inverter. Derive the expression for rms value of output voltage and also write down the output voltage in the form of Fourier expression. ## Short answer A single-phase PWM inverter chops each half-cycle into p pulses of width δ by comparing a reference with a triangular carrier. The rms output voltage is V_s√(pδ/π) = V_s√M. The Fourier series is v₀(t) = Σ_{n odd} (4V_s/(nπ)) sin(nδ/2) Σ_{m=1}^p sin(nγ_m) sin nωt. ## Answer A single-phase PWM inverter controls its output voltage (and reduces low-order harmonics) by chopping each half cycle of the output into one or more pulses whose widths are varied. The most common form in exam answers is **multiple-pulse (uniform) PWM**, of which single-pulse PWM is the special case $p = 1$; sinusoidal PWM is the same idea with pulse widths varying sinusoidally. #### Circuit ```text +-------+-----------+ | | | | T1 D1 T3 D3 Vs | | | A---[Load]--B | | v0 | | T4 D4 T2 D2 | | | +-------+-----------+ ``` #### Operation (multiple-pulse PWM) 1. A rectangular reference of amplitude $A_r$ (at output frequency $f$) is compared with a triangular carrier of amplitude $A_c$ and frequency $f_c$. 2. Number of pulses per half cycle: $p = \dfrac{f_c}{2f}$. 3. While $A_r >$ carrier, the switch pair is ON: T1, T2 in the positive half ($v_0 = +V_s$), T3, T4 in the negative half ($v_0 = -V_s$). Otherwise $v_0 = 0$ (one switch of the pair is kept ON and current freewheels via a diode). 4. Modulation index $M = A_r/A_c$; each pulse width $\delta = \dfrac{M\pi}{p}$. Changing $M$ changes all pulse widths and so the output voltage. ```text vc /\/\/\/\/\/\/\/\ carrier (p = 2 shown) Ar ----------------- reference v0 +Vs _ _ | | | | 0 --' '--' '--. .--. .--- | | | | -Vs |_| |_| 0 pi 2pi ``` #### RMS value of output voltage Each half cycle contains $p$ pulses of height $V_s$ and width $\delta$: $$ \begin{aligned} V_{0,rms} &= \left[\frac{2p}{2\pi}\int_{\gamma-\delta/2}^{\gamma+\delta/2}V_s^2\,d(\omega t)\right]^{1/2} = \left[\frac{p\,\delta\,V_s^2}{\pi}\right]^{1/2} \\ V_{0,rms} &= V_s\sqrt{\frac{p\,\delta}{\pi}} = V_s\sqrt{M} \end{aligned} $$ For $p = 1$ this gives the single-pulse result $V_s\sqrt{\delta/\pi}$. #### Fourier series Let the $m$th pulse in the positive half cycle be centred at $\gamma_m$ (for uniform PWM, $\gamma_m = \frac{(2m-1)\pi}{2p}$, $m = 1,\dots,p$), with the negative half a mirror image. The wave has half-wave symmetry, so only odd $n$ exist. For one pulse: $$ \begin{aligned} b_{n,m} &= \frac{2}{\pi}\int_{\gamma_m-\delta/2}^{\gamma_m+\delta/2}V_s\sin n\theta\,d\theta = \frac{2V_s}{n\pi}\Big[\cos n\big(\gamma_m - \tfrac{\delta}{2}\big) - \cos n\big(\gamma_m + \tfrac{\delta}{2}\big)\Big] \\ &= \frac{4V_s}{n\pi}\sin n\gamma_m\,\sin\frac{n\delta}{2} \end{aligned} $$ The cosine terms of pulses placed symmetrically about $\pi/2$ cancel ($a_n = 0$). Adding all pulses: $$ v_0(t) = \sum_{n=1,3,5,\dots}\left[\frac{4V_s}{n\pi}\sin\frac{n\delta}{2}\sum_{m=1}^{p}\sin n\gamma_m\right]\sin n\omega t $$ Fundamental: $$ V_{1m} = \frac{4V_s}{\pi}\sin\frac{\delta}{2}\sum_{m=1}^{p}\sin\gamma_m $$ #### Example $V_s = 200$ V, $p = 2$, $\delta = 45^\circ$ (pulses centred at 45° and 135°): $$ \begin{aligned} V_{0,rms} &= 200\sqrt{\frac{2\times45^\circ}{180^\circ}} = 141.42\ \text{V} \\ V_{1m} &= \frac{4\times200}{\pi}\sin22.5^\circ\,(\sin45^\circ + \sin135^\circ) = 254.65\times0.3827\times1.4142 = 137.81\ \text{V} \end{aligned} $$ #### Sinusoidal PWM (note) If a sinusoidal reference is used instead, the pulse widths $\delta_m$ vary sinusoidally; then $V_{0,rms} = V_s\sqrt{\sum_m \delta_m/\pi}$ and $b_n = \sum_m \frac{4V_s}{n\pi}\sin n\gamma_m\sin\frac{n\delta_m}{2}$. Low-order harmonics are almost eliminated; the dominant harmonics move to around $2p$ (the carrier frequency), which are easily filtered. #### Advantages of PWM - Output voltage control inside the inverter, with a fixed dc input. - Lower low-order harmonics, smaller filter. - Drawback: more switchings per cycle, so higher switching loss. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In multiple-pulse PWM with carrier frequency fc and output frequency f, how many pulses appear per half cycle? A. p = fc/f B. p = 2fc/f C. p = fc/(2f) D. p = f/fc Answer: C. p = fc/(2f). (Key point: Pulses per half cycle) 2. What is the effect of the modulation index M = Ar/Ac on the output? A. Each pulse width is delta = M pi/p, so changing M changes all pulse widths and the output voltage B. It changes the output frequency C. It changes the number of pulses only D. It changes the pulse height Vs Answer: A. delta = M pi/p; the output voltage is controlled by M with fixed dc input. (Key point: Modulation index and pulse width) 3. Which expression gives the rms output voltage for p pulses of width delta per half cycle? A. Vs p delta/pi B. Vs sqrt(p delta/pi) = Vs sqrt(M) C. Vs sqrt(delta/(p pi)) D. Vs M Answer: B. V0,rms = [p delta Vs^2/pi]^(1/2) = Vs sqrt(p delta/pi); with delta = M pi/p this is Vs sqrt(M). (Key point: RMS output voltage) 4. For a pulse centred at gamma_m with width delta, the nth harmonic amplitude is: A. (4Vs/(n pi)) cos(n gamma_m) cos(n delta/2) B. (4Vs/(n pi)) sin(n gamma_m) sin(n delta/2) C. (2Vs/(n pi)) sin(n delta) D. (4Vs/pi) sin(n gamma_m) sin(delta/2) Answer: B. b_n,m = (4Vs/(n pi)) sin(n gamma_m) sin(n delta/2); the cosine terms cancel by symmetry. (Key point: Fourier coefficient per pulse) 5. For Vs = 200 V, p = 2, delta = 45 degrees, what is Vo,rms? A. 100 V B. 200 V C. 70.7 V D. 141.42 V Answer: D. 200 sqrt(2 x 45/180) = 200 x 0.7071 = 141.42 V. 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