---
title: "Fourier series of falling sawtooth V to 0"
subject: "Electrical Circuit Theory"
chapter: 6
key: "semester_3/electrical_circuit_theory:fourier-falling-sawtooth-v-to-0"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/fourier-falling-sawtooth-v-to-0
site: BCT Engineering
---

# Fourier series of falling sawtooth V to 0

Electrical Circuit Theory · Chapter 6: Fourier Series and transform

- Asked in: 2074 Asoj, 8 marks
- Video: [Electrical Engineering: Ch 18: Fourier Series (16 of 35) Find the Fourier Series: Saw-Tooth Wave***](https://www.youtube.com/watch?v=_s_P5XsaKdY) — Michel van Biezen
- 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-falling-sawtooth-v-to-0
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-6

## Question

Obtain trigonometric Fourier series of the waveform shown in figure below and sketch the line spectra.
[Figure: v(t) falls linearly from V at ωt = 0 to 0 at ωt = 2π, then jumps back to V and falls again to 0 at 4π, then again to 6π; period 2π]

## Short answer

The waveform falls linearly from V to 0 in each period 2π, with average V/2. Its Fourier series contains only sine terms: v(t) = V/2 + (V/π)[sin ωt + (1/2) sin 2ωt + (1/3) sin 3ωt + …].

## Answer

**Reading the figure.** v(t) falls linearly from V at θ = ωt = 0 to 0 at θ = 2π, then jumps back to V. Period 2π.

```text
v(θ) = V (1 − θ/2π)     0 < θ < 2π
```

#### Coefficients

```text
a0 = (1/2π)∫₀^2π V(1 − θ/2π) dθ = V/2          (average value)
an = (1/π)∫₀^2π V(1 − θ/2π) cos nθ dθ = 0      (∫ cos nθ = 0 and ∫ θ cos nθ = 0)
bn = (1/π)∫₀^2π V(1 − θ/2π) sin nθ dθ
   = −(V/2π²)∫₀^2π θ sin nθ dθ
   = −(V/2π²)(−2π/n) = V/(nπ)
```

#### Fourier series

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

(The wave is a d.c. level V/2 plus an odd sawtooth about that level.)

#### Line spectrum (V = 1 shown, amplitude in units of V)

| n | 0 (d.c.) | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| cn | 0.5 | 0.3183 | 0.1592 | 0.1061 | 0.0796 | 0.0637 |
| phase | – | −90° | −90° | −90° | −90° | −90° |

```text
  n=0 |##########################| 0.5
  n=1 |#################         | 0.3183
  n=2 |########                  | 0.1592
  n=3 |######                    | 0.1061
  n=4 |####                      | 0.07958
  n=5 |###                       | 0.06366
```

Phase is shown for the form cn cos(nωt + φn): sin nθ = cos(nθ − 90°).

**Answer: v(t) = V/2 + (V/π) Σ (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. What is a0?
   A. V/2, the average value
   B. 0
   C. V
   D. V/π
   Answer: A. The wave ramps between 0 and V, so the average is V/2. (Key point: DC term)

2. Why is an = 0?
   A. The wave is even
   B. a0 absorbs them
   C. Half-wave symmetry
   D. The cosine integrals of the constant and θ terms vanish over a period
   Answer: D. ∫cos nθ and ∫θ cos nθ are 0 over 0 to 2π. (Key point: Cosine terms)

3. What is bn?
   A. -V/(nπ)
   B. V/(2nπ)
   C. V/(nπ)
   D. 2V/(nπ)
   Answer: C. bn = -(V/2π^2)(-2π/n) = V/(nπ). (Key point: Sine coefficient)

4. Why is the phase -90° in cos(nωt + φ)?
   A. sin nθ = cos(nθ + 90°)
   B. sin nθ = cos(nθ - 90°)
   C. bn is negative
   D. an is negative
   Answer: B. Positive sine coefficient shifts cosine by -90°. (Key point: Phase)

## Related in Electrical Circuit Theory

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