---
title: "Line voltage Fourier series: triplen harmonics absent"
subject: "Power Electronics"
chapter: 5
key: "semester_7/power_electronics:square-wave-inverter-triplen-absent"
source: https://www.bctengineer.com/q/semester_7/power_electronics/square-wave-inverter-triplen-absent
site: BCT Engineering
---

# Line voltage Fourier series: triplen harmonics absent

Power Electronics · Chapter 5: Inverter

- Asked in: 2071 Shrawan, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/square-wave-inverter-triplen-absent
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5

## Question

Obtain the Fourier series of line to line output voltage of square wave inverter and hence show that all triplen harmonics are absent from it.

## Short answer

Line voltage is the difference of pole voltages: v_ab = sum (4Vdc/n pi) sin(n pi/3) cos n(omega t - pi/3). For triplen n=3,9,15, sin(n pi/3)=0, so all triplen harmonics cancel; only 6k±1 harmonics remain. Fundamental peak is 2√3 Vdc/pi.

## Answer

In a three-phase square-wave (180° conduction) inverter, each pole voltage measured from the dc mid-point O is a square wave of $\pm V_{dc}/2$, and the three poles are displaced by 120°. The line voltage is the difference of two pole voltages, and in this difference all triplen harmonics cancel.

#### Pole voltages
$$
\begin{aligned}
v_{aO} &= \sum_{n=1,3,5,\dots}\frac{2V_{dc}}{n\pi}\sin n\omega t \\
v_{bO} &= \sum_{n=1,3,5,\dots}\frac{2V_{dc}}{n\pi}\sin n\Big(\omega t - \frac{2\pi}{3}\Big)
\end{aligned}
$$

(Each is a square wave of amplitude $V_{dc}/2$, so $b_n = \frac{4}{n\pi}\cdot\frac{V_{dc}}{2}$.)

#### Line voltage
$$
\begin{aligned}
v_{ab} &= v_{aO} - v_{bO} = \sum_{n\ \text{odd}}\frac{2V_{dc}}{n\pi}\Big[\sin n\omega t - \sin n\Big(\omega t - \frac{2\pi}{3}\Big)\Big]
\end{aligned}
$$

Using $\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$ with $\frac{A+B}{2} = n\big(\omega t - \frac{\pi}{3}\big)$ and $\frac{A-B}{2} = \frac{n\pi}{3}$:

$$
v_{ab} = \sum_{n=1,3,5,\dots}\frac{4V_{dc}}{n\pi}\sin\frac{n\pi}{3}\,\cos n\Big(\omega t - \frac{\pi}{3}\Big)
$$

This is the Fourier series of the 120° quasi-square line voltage ($+V_{dc}$ for 0–120°, 0 for 60°, $-V_{dc}$ for 120°, 0 for 60°). It is identical, term by term and with the same time origin, to the textbook form $v_{ab} = \sum\frac{4V_{dc}}{n\pi}\cos\frac{n\pi}{6}\sin n(\omega t + \frac{\pi}{6})$.

#### Triplen harmonics are absent
For $n = 3, 9, 15, \dots$ ($n = 3k$):

$$
\sin\frac{n\pi}{3} = \sin k\pi = 0
$$

so every triplen term in $v_{ab}$ is zero. Physically, the triplen components of $v_{aO}$, $v_{bO}$, $v_{cO}$ are equal in magnitude and in phase (a shift of $2\pi/3 \times 3k = 2k\pi$), so they cancel when two pole voltages are subtracted.

#### Remaining harmonics
| $n$ | $\sin(n\pi/3)$ | Peak of $v_{ab}$ harmonic |
|---|---|---|
| 1 | 0.866 | $1.1027\,V_{dc}$ |
| 3 | 0 | 0 |
| 5 | −0.866 | $0.2205\,V_{dc}$ |
| 7 | 0.866 | $0.1575\,V_{dc}$ |
| 9 | 0 | 0 |

Fundamental line voltage: $\hat V_{L1} = \frac{2\sqrt3V_{dc}}{\pi}$, $V_{L1,rms} = \frac{\sqrt6V_{dc}}{\pi} = 0.7797V_{dc}$. Only harmonics of order $6k\pm1$ (5, 7, 11, 13, …) remain, with amplitude $V_{L1}/n$.

## Check yourself

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

1. What is the amplitude of the nth harmonic of the pole voltage v_aO (square wave +/-Vdc/2)?
   A. 4Vdc/(n pi)
   B. 2Vdc/(n pi)
   C. Vdc/(n pi)
   D. 2Vdc/pi for all n
   Answer: B. b_n = (4/(n pi))(Vdc/2) = 2Vdc/(n pi). (Key point: Pole voltage series)

2. How is the line voltage v_ab obtained from the pole voltages?
   A. v_ab = v_aO + v_bO
   B. v_ab = v_aO - v_bO, where v_bO is delayed by 120 degrees
   C. v_ab = v_aO x v_bO
   D. v_ab = 3 v_aO
   Answer: B. The line voltage is the difference of two pole voltages displaced by 2 pi/3. (Key point: Line voltage as difference)

3. Which condition makes every triplen term of v_ab zero?
   A. cos(n pi/3) = 0 for n = 3k
   B. sin(n pi/3) = sin(k pi) = 0 for n = 3k
   C. sin(n pi/3) = 1 for n = 3k
   D. The factor 4Vdc/(n pi) becomes zero
   Answer: B. The amplitude carries sin(n pi/3), which equals sin(k pi) = 0 for n = 3k. (Key point: Why triplens vanish)

4. Physically, why do triplen harmonics cancel in the line voltage?
   A. The load blocks them
   B. The triplen components of the three pole voltages are equal and in phase (shift 2k pi), so they cancel in the difference
   C. They are 90 degrees out of phase
   D. The dc source filters them
   Answer: B. A 120 degree shift at harmonic 3k is 2k pi, i.e. no shift, so the equal components subtract to zero. (Key point: Physical reason)

5. Which harmonics remain in the line voltage and with what amplitude?
   A. Only n = 6k +/- 1, with amplitude V_L1/n
   B. All odd n except 3
   C. Only n = 3, 9, 15
   D. Only n = 6k +/- 1, with amplitude V_L1 x n
   Answer: A. 5, 7, 11, 13... remain, each with amplitude V_L1/n. (Key point: Remaining harmonics)

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