---
title: "Fourier series of odd sawtooth, ±Vo"
subject: "Electrical Circuit Theory"
chapter: 6
key: "semester_3/electrical_circuit_theory:fourier-sawtooth-vo-odd"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/fourier-sawtooth-vo-odd
site: BCT Engineering
---

# Fourier series of odd sawtooth, ±Vo

Electrical Circuit Theory · Chapter 6: Fourier Series and transform

- Asked in: 2070 Asar (old course), 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/fourier-sawtooth-vo-odd
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-6

## Question

Find the trigonometric Fourier series for the waveform shown and also sketch the line spectrum.
[Figure: v(t) is a sawtooth that rises linearly from −Vo at ωt = −π through 0 at ωt = 0 to Vo at ωt = π, drops to −Vo, and repeats with period 2π]

## Short answer

The waveform is odd, so a0 = 0 and an = 0. The sine coefficients are bn = 2Vo (−1)^(n+1)/(nπ). The trigonometric Fourier series is v(t) = (2Vo/π) Σ_{n=1}^∞ (−1)^(n+1) (1/n) sin nωt.

## Answer

**Reading the figure.** v(t) rises linearly from −Vo at ωt = −π, through 0 at ωt = 0, to +Vo at ωt = π, drops to −Vo, and repeats. Period 2π (θ = ωt):

```text
v(θ) = Vo θ/π       −π < θ < π
```

The wave is odd (v(−θ) = −v(θ)), so a0 = 0 and an = 0; only sine terms exist.

#### Coefficients

```text
bn = (1/π)∫_{−π}^{π} (Voθ/π) sin nθ dθ = (2/π)∫₀^π (Voθ/π) sin nθ dθ
   = (2Vo/π²)[ −θ cos nθ/n + sin nθ/n² ]₀^π
   = (2Vo/π²)(−π cos nπ/n)
   = 2Vo(−1)^(n+1)/(nπ)
```

#### Fourier series

```text
v(t) = (2Vo/π) [ sin ωt − (1/2) sin 2ωt + (1/3) sin 3ωt − (1/4) sin 4ωt + ... ]
```

#### Line spectrum (amplitude 2Vo/(nπ), Vo = 1)

| n | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| cn | 0.6366 | 0.3183 | 0.2122 | 0.1592 | 0.1273 |
| phase (cos form) | −90° | +90° | −90° | +90° | −90° |

```text
  n=1 |##########################| 0.6366
  n=2 |#############             | 0.3183
  n=3 |#########                 | 0.2122
  n=4 |######                    | 0.1592
  n=5 |#####                     | 0.1273
```

The amplitude falls as 1/n (slow convergence because of the jump at θ = ±π). The phase alternates because bn alternates in sign.

**Answer: v(t) = (2Vo/π) Σ (−1)^(n+1) (1/n) sin nωt, n = 1, 2, 3, ...**

## Check yourself

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

1. Why are only sine terms present?
   A. The wave is odd, v(-θ) = -v(θ)
   B. The wave is even
   C. The wave has half-wave symmetry
   D. The wave has a d.c. term
   Answer: A. Odd functions have a0 = an = 0. (Key point: Symmetry)

2. How can bn be computed?
   A. As zero over symmetric limits
   B. Using only half amplitude
   C. With cos terms
   D. (2/π) times integral from 0 to π, as the integrand is even
   Answer: D. θ sin nθ is even, so double the half-range integral. (Key point: Integral shortcut)

3. What is bn?
   A. -2Vo/(nπ) for all n
   B. 2Vo/(n^2 π^2)
   C. 2Vo(-1)^(n+1)/(nπ)
   D. Vo/(nπ)
   Answer: C. Result alternates in sign. (Key point: Coefficient)

4. Why do amplitudes fall only as 1/n?
   A. The wave is smooth
   B. The jump discontinuity at θ = ±π
   C. It has half-wave symmetry
   D. It has a d.c. term
   Answer: B. Discontinuities give slowly decaying harmonics. (Key point: Convergence)

## Related in Electrical Circuit Theory

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- [Fourier series of half-period ramp pulses](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Afourier-ramp-half-period-vo) — Electrical Circuit Theory, chapter 6
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