---
title: "Fundamental and 3rd harmonic of three-phase inverter output"
subject: "Power Electronics"
chapter: 5
key: "semester_7/power_electronics:inverter-waveform-vdc-fundamental-third"
source: https://www.bctengineer.com/q/semester_7/power_electronics/inverter-waveform-vdc-fundamental-third
site: BCT Engineering
---

# Fundamental and 3rd harmonic of three-phase inverter output

Power Electronics · Chapter 5: Inverter

- Asked in: 2074 Chaitra, 8 marks
- Video: [Lecture 3 | Harmonics in inverter | Harmonics Power Electronics  | Harmonics in Electrical systems |](https://www.youtube.com/watch?v=JfYDkr441is) — ELECTRIC VIBE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/inverter-waveform-vdc-fundamental-third
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5

## Question

The given waveform in figure below is the output voltage of three phase inverter. Calculate the fundamental component and 3rd harmonic component of the output voltage. [Figure: phase voltage VAN — +⅓Vdc for the first 60°, +⅔Vdc for the next 60°, +⅓Vdc for the next 60°, then −⅓Vdc, −⅔Vdc, −⅓Vdc over the negative half cycle, repeating]

## Short answer

The six-step phase voltage has fundamental peak 2Vdc over pi, equal to 0.6366Vdc, and rms 0.4502Vdc. The third harmonic is zero because triplen harmonics vanish in 180 degree conduction; only harmonics of order 6k plus or minus 1 remain.

## Answer

The given phase voltage ($V_{dc}/3$, $2V_{dc}/3$, $V_{dc}/3$ in 60° steps) is the six-step wave of a 180° conduction three-phase inverter. It is odd and has half-wave symmetry, so $a_0 = a_n = 0$ and only odd sine terms are present.

#### Fourier coefficient
$$
\begin{aligned}
b_n &= \frac{2}{\pi}\int_0^{\pi}v_{AN}\sin n\theta\,d\theta \\
&= \frac{2}{n\pi}\Big[\tfrac{V_{dc}}{3}\big(1-\cos\tfrac{n\pi}{3}\big) + \tfrac{2V_{dc}}{3}\big(\cos\tfrac{n\pi}{3}-\cos\tfrac{2n\pi}{3}\big) + \tfrac{V_{dc}}{3}\big(\cos\tfrac{2n\pi}{3}-\cos n\pi\big)\Big] \\
&= \frac{2V_{dc}}{3n\pi}\Big[2 + \cos\tfrac{n\pi}{3} - \cos\tfrac{2n\pi}{3}\Big] \quad (n\ \text{odd})
\end{aligned}
$$

#### Fundamental component ($n = 1$)
$$
\begin{aligned}
b_1 &= \frac{2V_{dc}}{3\pi}\big[2 + 0.5 + 0.5\big] = \frac{2V_{dc}}{\pi} = 0.6366\,V_{dc}\ \text{(peak)} \\
V_{1,rms} &= \frac{2V_{dc}}{\sqrt2\,\pi} = \frac{\sqrt2\,V_{dc}}{\pi} = 0.4502\,V_{dc}
\end{aligned}
$$

$$
v_{AN1} = \frac{2V_{dc}}{\pi}\sin\omega t
$$

#### Third harmonic ($n = 3$)
$\cos\pi = -1$, $\cos 2\pi = +1$:

$$
b_3 = \frac{2V_{dc}}{9\pi}\big[2 - 1 - 1\big] = 0
$$

So the **third harmonic component is zero**. All triplen harmonics vanish; the non-zero terms are $n = 6k\pm1$ (5, 7, 11, 13, …), each with peak $\frac{2V_{dc}}{n\pi}$:

$$
v_{AN} = \frac{2V_{dc}}{\pi}\Big[\sin\omega t + \frac{1}{5}\sin5\omega t + \frac{1}{7}\sin7\omega t + \frac{1}{11}\sin11\omega t + \dots\Big]
$$

**Answer:** fundamental $= 0.6366V_{dc}$ peak ($0.4502V_{dc}$ rms); 3rd harmonic $= 0$. For example, with $V_{dc} = 600$ V the fundamental is 381.97 V peak (270.09 V rms).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why does the Fourier series of this six-step phase voltage contain only odd sine terms?
   A. It is even and has quarter-wave symmetry
   B. It has only one level, so all harmonics cancel
   C. It is odd and has half-wave symmetry, so a0 = an = 0 and even harmonics vanish
   D. It is a dc-offset wave so only odd cosines remain
   Answer: C. Odd symmetry removes cosine terms and a0; half-wave symmetry removes even harmonics. (Key point: Symmetry simplifies Fourier series)

2. What is the peak of the fundamental component of this phase voltage?
   A. 0.6366 Vdc (2Vdc/pi)
   B. 1.273 Vdc (4Vdc/pi)
   C. 0.4502 Vdc
   D. 1.1027 Vdc
   Answer: A. b1 = (2Vdc/3pi)(2+0.5+0.5) = 2Vdc/pi = 0.6366 Vdc peak; 0.4502 Vdc is the rms. (Key point: Fundamental peak value)

3. Why is the third harmonic component zero?
   A. Because the wave has no half-wave symmetry
   B. Because for n = 3 the bracket 2 + cos(pi) - cos(2pi) = 2 - 1 - 1 = 0
   C. Because the third harmonic is filtered by the load
   D. Because the bracket becomes 2 + 1 + 1 = 4
   Answer: B. With cos(pi) = -1 and cos(2pi) = +1 the bracket is 0, so b3 = 0; all triplens vanish. (Key point: Third harmonic is zero)

4. Which harmonics are present in the phase voltage, and with what peak?
   A. All odd harmonics, each 2Vdc/(n pi)
   B. Only n = 6k +/- 1, each 2Vdc/(n pi)
   C. Only n = 6k +/- 1, each 4Vdc/(n pi)
   D. Only triplen harmonics, each 2Vdc/(n pi)
   Answer: B. Triplens are zero, leaving 5, 7, 11, 13... with peak 2Vdc/(n pi). (Key point: Surviving harmonics)

## Related in Power Electronics

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