---
title: "Fourier series of triangle train, peak 10"
subject: "Electrical Circuit Theory"
chapter: 6
key: "semester_3/electrical_circuit_theory:fourier-train-of-isosceles-triangles-10v"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/fourier-train-of-isosceles-triangles-10v
site: BCT Engineering
---

# Fourier series of triangle train, peak 10

Electrical Circuit Theory · Chapter 6: Fourier Series and transform

- Asked in: 2068 Chaitra, 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-train-of-isosceles-triangles-10v
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-6

## Question

The following figure shows a voltage waveform in the form of a train of isosceles triangles. Determine the Fourier series and plot the line spectrum.
[Figure: voltage rises linearly from 0 at ωt = 0 to 10 at ωt = π, falls to 0 at 2π, rises to 10 at 3π and falls to 0 at 4π (period 2π)]

## Short answer

With peak 10 V and period 2π, the isosceles triangle wave has average 5 V and only odd cosine terms: v(t) = 5 − (40/π²)[cos ωt + (1/9) cos 3ωt + (1/25) cos 5ωt + …] V.

## Answer

**Reading the figure.** The voltage rises linearly from 0 at θ = ωt = 0 to 10 V at θ = π, falls to 0 at θ = 2π, and repeats (period 2π, peak Vo = 10 V; this is the same shape as the previous question with T = 2π/ω).

```text
v(θ) = 10θ/π             0 ≤ θ ≤ π
v(θ) = 10(2π − θ)/π      π ≤ θ ≤ 2π
```

The wave is symmetric about θ = π (v(2π − θ) = v(θ)), so bn = 0, and the average is half the peak value.

#### Coefficients

```text
a0 = 10/2 = 5                                   (average value)
an = (1/π)∫₀^2π v cos nθ dθ = (2/π)∫₀^π (10θ/π) cos nθ dθ
   = (20/π²)·(cos nπ − 1)/n²
   = 20[(−1)ⁿ − 1] / (π² n²)
an = −40/(π² n²)  for odd n;   an = 0 for even n
bn = 0
```

(using ∫₀^π θ cos nθ dθ = (cos nπ − 1)/n²).

#### Fourier series

```text
v(t) = 5 − (40/π²) [ cos ωt + (1/9) cos 3ωt + (1/25) cos 5ωt + ... ]
     = 5 − 4.053 cos ωt − 0.4503 cos 3ωt − 0.1621 cos 5ωt − 0.0827 cos 7ωt − ...
```

#### Line spectrum

| n | 0 (d.c.) | 1 | 3 | 5 | 7 |
|---|---|---|---|---|---|
| cn (V) | 5 | 4.053 | 0.4503 | 0.1621 | 0.0827 |
| phase (cos form) | – | 180° | 180° | 180° | 180° |

```text
  n=0 |##########################| 5
  n=1 |#####################     | 4.053
  n=3 |##                        | 0.4503
  n=5 |#                         | 0.1621
  n=7 |#                         | 0.08271
```

**Answer: v(t) = 5 − (40/π²) Σ (1/n²) cos nωt (V), n = 1, 3, 5, ...**

## Check yourself

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

1. What is a0 for the 10 V triangle train?
   A. 5 V
   B. 10 V
   C. 0
   D. 2.5 V
   Answer: A. Half the peak value. (Key point: Average)

2. Why is bn = 0?
   A. It is odd
   B. It has quarter-wave symmetry
   C. a0 is non-zero
   D. The wave is symmetric about θ = π
   Answer: D. Even symmetry about the centre gives cosines only. (Key point: Symmetry)

3. What is an?
   A. 40/(π^2 n^2)
   B. -20/(π^2 n^2)
   C. -40/(π^2 n^2) for odd n
   D. -40/(π n)
   Answer: C. an = 20[(-1)^n - 1]/(π^2 n^2). (Key point: Coefficient)

4. Why do all harmonics show 180° phase?
   A. The sine coefficients are negative
   B. The cosine coefficients are negative
   C. a0 is positive
   D. The wave starts at 10
   Answer: B. A negative cosine is cos(nθ + 180°). (Key point: Phase)

## Close variants set in other exams

- [Fourier series of triangle train, Vo and T](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Afourier-train-of-isosceles-triangles-vo) — Electrical Circuit Theory, chapter 6, asked 2×

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