---
title: "Fourier series of triangular wave, peak 1"
subject: "Electrical Circuit Theory"
chapter: 6
key: "semester_3/electrical_circuit_theory:fourier-triangular-wave-peak-1-period"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/fourier-triangular-wave-peak-1-period
site: BCT Engineering
---

# Fourier series of triangular wave, peak 1

Electrical Circuit Theory · Chapter 6: Fourier Series and transform

- Asked in: 2075 Asoj, 8 marks
- 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-1-period
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-6

## Question

For the given waveform, find the trigonometric form of Fourier series and then plot its line spectrum.
[Figure: triangular wave f(t) with peak value 1 at t = 0 and minimum value −1 at t = ±π, with straight-line segments crossing zero at t = ±π/2 and ±3π/2 (period 2π)]

## Short answer

The trigonometric Fourier series is f(t) = (8/π²)[cos t + (1/9)cos 3t + (1/25)cos 5t + …], with no DC or sine terms. The amplitude line spectrum shows only odd harmonics decaying as 1/n².

## Answer

**Reading the figure.** f(t) is a triangular wave of period 2π with peak +1 at t = 0, zero crossings at ±π/2 and minimum −1 at ±π. It is an even function (f(−t) = f(t)) with half-wave symmetry, so bn = 0, a0 = 0 and only odd cosine harmonics exist.

#### Equation of the wave

```text
f(t) = 1 − 2t/π     for 0 ≤ t ≤ π    (even, so f(−t) = f(t))
```

#### Coefficients

```text
a0 = (1/2π)∫ f dt = 0                       (equal area above and below)
an = (2/π)∫₀^π (1 − 2t/π) cos nt dt
   = −(4/π²)∫₀^π t cos nt dt
   = −(4/π²)·[(cos nπ − 1)/n²]
   = 4[1 − (−1)ⁿ] / (π² n²)
an = 8/(π² n²)  for n odd,   an = 0 for n even
bn = 0
```

#### Fourier series

```text
f(t) = (8/π²) [ cos t + (1/9) cos 3t + (1/25) cos 5t + (1/49) cos 7t + ... ]
```

#### Line spectrum (amplitude cn = |an|)

| n | 0 | 1 | 3 | 5 | 7 | 9 |
|---|---|---|---|---|---|---|
| cn | 0 | 0.8106 | 0.0901 | 0.0324 | 0.0165 | 0.0100 |
| phase | – | 0° | 0° | 0° | 0° | 0° |

```text
  n=1 |##########################| 0.8106
  n=3 |###                       | 0.09006
  n=5 |#                         | 0.03242
  n=7 |#                         | 0.01654
  n=9 |#                         | 0.01001
```

Lines occur only at ω, 3ω, 5ω ... and fall as 1/n². All harmonics are cosines of phase 0° (the cn are positive here).

**Answer: f(t) = (8/π²) Σ (1/n²) cos nt, n = 1, 3, 5, ...**

## Check yourself

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

1. Which symmetry properties does this triangular wave have?
   A. Even and half-wave symmetric, so bn = 0, a0 = 0 and only odd cosines
   B. Odd symmetry, so only sine terms
   C. Quarter-wave only with even harmonics
   D. No symmetry; all terms appear
   Answer: A. Peak at t = 0 gives evenness; f(t + π) = -f(t) kills even harmonics. (Key point: Symmetry)

2. Why is a0 = 0?
   A. The wave is odd
   B. The wave has only odd harmonics so a0 must be 1
   C. Bn = 0
   D. Equal areas above and below the axis
   Answer: D. The average value is zero because positive and negative areas cancel. (Key point: Zero a0)

3. What is an for odd n?
   A. 4/(π n)
   B. 8/(π n)
   C. 8/(π^2 n^2)
   D. 2/(π^2 n)
   Answer: C. an = 4[1 - (-1)^n]/(π^2 n^2) = 8/(π^2 n^2) for odd n. (Key point: Coefficient)

4. How do the line heights vary?
   A. Fall as 1/n
   B. Fall as 1/n^2, with c1 = 0.8106 and c3 = 0.0901
   C. Remain constant
   D. Fall as 1/n^2 with even harmonics present
   Answer: B. c3/c1 = 1/9, and only odd harmonics occur. (Key point: Spectrum)

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