---
title: "Square-wave inverter ±48 V: fundamental and third harmonic"
subject: "Power Electronics"
chapter: 5
key: "semester_7/power_electronics:inverter-48v-rl-fundamental-third"
source: https://www.bctengineer.com/q/semester_7/power_electronics/inverter-48v-rl-fundamental-third
site: BCT Engineering
---

# Square-wave inverter ±48 V: fundamental and third harmonic

Power Electronics · Chapter 5: Inverter

- Asked in: 2071 Chaitra, 8 marks
- Video: [Harmonics and Total Harmonic Distortion (THD)](https://www.youtube.com/watch?v=aWGK2jFq1rg) — Engineering with Prof. Kim
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/inverter-48v-rl-fundamental-third
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5

## Question

In figure below shows the schematic diagram of a single phase inverter giving square wave AC output voltage Vo with frequency of 50 Hz. Calculate the magnitude and phase of fundamental and third harmonic component of the output voltage and load current. [Figure: split dc supply +48 V / 0 V / −48 V feeding a single phase inverter; output Vo drives io through R = 10 ohm in series with L = 20 mH]

## Short answer

For the ±48 V half-bridge square wave at 50 Hz with R=10 Ω, L=20 mH, fundamental voltage is 61.12 V peak at 0° and current is 5.175 A peak lagging 32.14°. Third harmonic voltage is 20.37 V peak at 0° and current is 0.955 A peak lagging 62.05°.

## Answer

With a split ±48 V supply, the single-phase (half-bridge) inverter applies $+48$ V and $-48$ V for alternate half cycles, so the output is a square wave of amplitude $V = 48$ V at 50 Hz: $v_0 = \sum_{n\ \text{odd}}\frac{4V}{n\pi}\sin n\omega t$.

**Given:** $V = 48$ V, $f = 50$ Hz ($\omega = 314.16$ rad/s), $R = 10\ \Omega$, $L = 20$ mH. Phase angles are referred to the output voltage (sine reference, 0°).

#### Fundamental ($n = 1$)
$$
\begin{aligned}
V_{1m} &= \frac{4\times48}{\pi} = 61.12\ \text{V}, \qquad V_{1,rms} = 43.22\ \text{V} \\
X_1 &= \omega L = 6.283\ \Omega, \quad Z_1 = \sqrt{10^2 + 6.283^2} = 11.810\ \Omega \\
\theta_1 &= \tan^{-1}\frac{6.283}{10} = 32.14^\circ \\
I_{1m} &= \frac{61.12}{11.810} = 5.175\ \text{A}, \qquad I_{1,rms} = 3.659\ \text{A}
\end{aligned}
$$

#### Third harmonic ($n = 3$)
$$
\begin{aligned}
V_{3m} &= \frac{61.12}{3} = 20.37\ \text{V}, \qquad V_{3,rms} = 14.41\ \text{V} \\
X_3 &= 3\omega L = 18.850\ \Omega, \quad Z_3 = \sqrt{10^2 + 18.850^2} = 21.338\ \Omega \\
\theta_3 &= \tan^{-1}\frac{18.850}{10} = 62.05^\circ \\
I_{3m} &= \frac{20.37}{21.338} = 0.955\ \text{A}, \qquad I_{3,rms} = 0.675\ \text{A}
\end{aligned}
$$

#### Results
| Component | Voltage | Current |
|---|---|---|
| Fundamental | 61.12 V peak ∠0° (43.22 V rms) | 5.175 A peak ∠−32.14° (3.659 A rms) |
| 3rd harmonic | 20.37 V peak ∠0° (14.41 V rms) | 0.955 A peak ∠−62.05° (0.675 A rms) |

$$
\begin{aligned}
v_1 &= 61.12\sin314.16t, & i_1 &= 5.175\sin(314.16t - 32.14^\circ) \\
v_3 &= 20.37\sin942.48t, & i_3 &= 0.955\sin(942.48t - 62.05^\circ)
\end{aligned}
$$

## Check yourself

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

1. With a split +/-48 V supply, what is the amplitude of the output square wave?
   A. 96 V
   B. 24 V
   C. 48 V
   D. 61.12 V
   Answer: C. The load sees +48 V and -48 V alternately, so V = 48 V. (Key point: Output amplitude of half-bridge)

2. What is the peak fundamental output voltage?
   A. 48 V
   B. 61.12 V
   C. 43.22 V
   D. 20.37 V
   Answer: B. V1m = 4 x 48/pi = 61.12 V peak (43.22 V rms). (Key point: Fundamental voltage)

3. For R = 10 ohm, L = 20 mH at 50 Hz, what is the fundamental current?
   A. 5.175 A peak, leading 32.14 degrees
   B. 6.11 A peak in phase
   C. 5.175 A peak, lagging 32.14 degrees
   D. 5.175 A peak, lagging 62.05 degrees
   Answer: C. Z1 = sqrt(100 + 6.283^2) = 11.81 ohm, I1m = 61.12/11.81 = 5.175 A, lag tan^-1(6.283/10) = 32.14 degrees. (Key point: Fundamental current)

4. What are the third harmonic voltage and current peaks?
   A. 20.37 V and 0.955 A lagging 62.05 degrees
   B. 20.37 V and 1.72 A lagging 32.14 degrees
   C. 61.12 V and 0.955 A lagging 62.05 degrees
   D. 20.37 V and 0.955 A leading 62.05 degrees
   Answer: A. V3m = V1m/3 = 20.37 V; X3 = 18.85 ohm, Z3 = 21.34 ohm, I3m = 0.955 A, lag 62.05 degrees. (Key point: Third harmonic)

## Related in Power Electronics

- [Single-phase square-wave inverter rms and fundamental](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asquare-wave-inverter-rms-fundamental) — Power Electronics, chapter 5
- [Half-bridge square-wave inverter at 120 Hz](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asquare-wave-inverter-120hz) — Power Electronics, chapter 5
- [Fundamental and 3rd harmonic of three-phase inverter output](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ainverter-waveform-vdc-fundamental-third) — Power Electronics, chapter 5
- [Single-phase inverter fundamental output voltage](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asingle-phase-inverter-r-fundamental) — Power Electronics, chapter 5
- [Square-wave inverter ±100 V: fundamental load current](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ainverter-100v-rl-fundamental) — Power Electronics, chapter 5, asked 2×
- [Full-bridge square-wave inverter, 400 V, RL load](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-bridge-inverter-400v-rl) — Power Electronics, chapter 5

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