---
title: "Fourier series of triangular wave, peak 20"
subject: "Electrical Circuit Theory"
chapter: 6
key: "semester_3/electrical_circuit_theory:fourier-triangular-wave-peak-20-minus"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/fourier-triangular-wave-peak-20-minus
site: BCT Engineering
---

# Fourier series of triangular wave, peak 20

Electrical Circuit Theory · Chapter 6: Fourier Series and transform

- Asked in: 2073 Shrawan
- Video: [Fourier Series of Triangular Wave | Fourier Series of Different Waveforms | #Ep17](https://www.youtube.com/watch?v=pL82z46B9r0) — School of Physics
- 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-triangular-wave-peak-20-minus
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-6

## Question

For the Waveform shown in figure below. Find the trigonometric form of Fourier Series and plot the line spectrum.
[Figure: triangular wave f(t) with peak value 20 at ωt = 0 and ±2π, minimum at ωt = ±π (about −20), with zero crossings at ωt = ±π/2 (period 2π)]

## Short answer

The triangular wave is even, has zero average, and only odd cosine harmonics. Its series is f(t) = (160/π²)[cos ωt + (1/9) cos 3ωt + (1/25) cos 5ωt + …], with amplitudes decaying as 1/n².

## Answer

**Reading the figure.** f(t) is a triangular wave of period 2π (in ωt) with a peak of +20 at ωt = 0, ±2π, zero crossings at ±π/2 and a minimum of −20 at ±π (assumed; the scan shows the negative peak only roughly). Let θ = ωt.

The wave is even (bn = 0), has zero average (a0 = 0), and has half-wave symmetry f(θ + π) = −f(θ), so only odd cosine harmonics are present.

```text
f(θ) = 20 − 40θ/π     for 0 ≤ θ ≤ π
```

#### Coefficients

```text
an = (2/π)∫₀^π (20 − 40θ/π) cos nθ dθ
   = −(80/π²)∫₀^π θ cos nθ dθ
   = −(80/π²)(cos nπ − 1)/n²
   = 80[1 − (−1)ⁿ] / (π² n²)
an = 160/(π² n²)  for odd n;   an = 0 for even n
a0 = 0,  bn = 0
```

#### Fourier series

```text
f(t) = (160/π²) [ cos ωt + (1/9) cos 3ωt + (1/25) cos 5ωt + (1/49) cos 7ωt + ... ]
     = 16.21 cos ωt + 1.801 cos 3ωt + 0.6485 cos 5ωt + 0.3308 cos 7ωt + ...
```

#### Line spectrum

| n | 1 | 3 | 5 | 7 | 9 |
|---|---|---|---|---|---|
| cn | 16.21 | 1.801 | 0.6485 | 0.3308 | 0.2001 |
| phase | 0° | 0° | 0° | 0° | 0° |

```text
  n=1 |##########################| 16.21
  n=3 |###                       | 1.801
  n=5 |#                         | 0.6485
  n=7 |#                         | 0.3308
  n=9 |#                         | 0.2001
```

**Answer: f(t) = (160/π²) Σ (1/n²) cos nωt, n = 1, 3, 5, ... (amplitudes fall as 1/n²).**

## Check yourself

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

1. Which terms are present for this triangle (peak 20, even, half-wave symmetric)?
   A. Odd cosine harmonics only
   B. Odd sine harmonics only
   C. All cosines and sines
   D. Even cosines plus a d.c. term
   Answer: A. Evenness removes sines; f(θ + π) = -f(θ) removes even harmonics and a0. (Key point: Symmetries)

2. What is an for odd n?
   A. 80/(π n)
   B. 160/(π n)
   C. 40/(π^2 n^2)
   D. 160/(π^2 n^2)
   Answer: D. an = 80[1 - (-1)^n]/(π^2 n^2) = 160/(π^2 n^2). (Key point: Coefficient)

3. What is the fundamental amplitude?
   A. 20
   B. 8.1
   C. 16.21
   D. 32.4
   Answer: C. 160/π^2 = 16.21. (Key point: Fundamental amplitude)

4. How do the harmonics decay?
   A. In proportion to 1/n
   B. In proportion to 1/n^2
   C. Not at all
   D. In proportion to 1/n^3
   Answer: B. The coefficient has n^2 in the denominator. (Key point: Decay)

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