---
title: "Exponential Fourier series of sawtooth wave"
subject: "Electrical Circuit Theory"
chapter: 6
key: "semester_3/electrical_circuit_theory:fourier-exponential-form-sawtooth"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/fourier-exponential-form-sawtooth
site: BCT Engineering
---

# Exponential Fourier series of sawtooth wave

Electrical Circuit Theory · Chapter 6: Fourier Series and transform

- Asked in: 2067 Asar (old course), 8 marks
- Video: [Complex Exponential Fourier Series of Sawtooth Waveform](https://www.youtube.com/watch?v=-NaCfOWQ0ik) — K MURTHYRAJU
- 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-exponential-form-sawtooth
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-6

## Question

Find the exponential form of Fourier series for the given Saw-tooth wave.
[Figure: f(t) is a sawtooth with f(t) = t for −1/2 < t < 1, rising linearly from −1/2 at t = −1/2 through 0 at t = 0 to 1 at t = 1, then dropping back to −1/2 and repeating (period 3/2)]

## Short answer

The period is T = 3/2 s, angular frequency ω0 = 4π/3 rad/s. The d.c. term is c0 = 1/4, and for n ≠ 0 the complex coefficients are cn = j3 e^(j2πn/3)/(4πn). Hence f(t) = 1/4 + Σ_{n≠0} [j3 e^(j2πn/3)/(4πn)] e^(j4πnt/3).

## Answer

**Reading the figure.** f(t) = t for −1/2 < t < 1, i.e. it rises from −1/2 to 1, then jumps back to −1/2 and repeats. Period T = 3/2 s, so ω0 = 2π/T = 4π/3 rad/s.

The exponential Fourier series is f(t) = Σ cn e^(jnω0t), cn = (1/T)∫ f(t) e^(−jnω0t) dt.

#### d.c. term

```text
c0 = (1/T)∫_{−1/2}^{1} t dt = (2/3)·(1/2)(1 − 1/4) = 1/4
```

#### cn for n ≠ 0

Let a = nω0 = 4πn/3. Use ∫ t e^(−jat) dt = e^(−jat)(jt/a + 1/a²):

```text
cn = (2/3) [ e^(−jat)(jt/a + 1/a²) ] from t = −1/2 to t = 1
   = (2/3) [ e^(−ja)(j/a + 1/a²) − e^(ja/2)(−j/(2a) + 1/a²) ]
```

For a = 4πn/3: e^(−ja) = e^(−j4πn/3) = e^(j2πn/3) = e^(ja/2), so the 1/a² terms cancel:

```text
cn = (2/3) e^(j2πn/3) · (3j/2a) = j e^(j2πn/3)/a = 3j e^(j2πn/3) / (4πn)
|cn| = 3/(4π|n|),    arg cn = 90° + 120°·n   (for n > 0);  c(−n) = cn*
```

#### Series

```text
f(t) = 1/4 + Σ (n ≠ 0) [ 3j e^(j2πn/3) / (4πn) ] e^(j4πnt/3)
```

Real form: f(t) = 1/4 − Σ (n = 1 to ∞) [3/(2πn)] sin(4πnt/3 + 2πn/3).

#### Spectrum (checked numerically in Python)

| n | 0 | ±1 | ±2 | ±3 | ±4 | ±5 |
|---|---|---|---|---|---|---|
| \|cn\| | 0.25 | 0.2387 | 0.1194 | 0.0796 | 0.0597 | 0.0477 |
| arg cn (n > 0) | 0° | −150° | −30° | 90° | −150° | −30° |

```text
  n=0 |##########################| 0.25
 n=±1 |######################### | 0.2387
 n=±2 |############              | 0.1194
 n=±3 |########                  | 0.0796
 n=±4 |######                    | 0.0597
 n=±5 |#####                     | 0.0477
```

**Answer: f(t) = 1/4 + Σ (n ≠ 0) [3j e^(j2πn/3)/(4πn)] e^(j4πnt/3), with ω0 = 4π/3 rad/s.**

## Check yourself

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

1. What is ω0 for period T = 3/2 s?
   A. 4π/3 rad/s
   B. 2π/3 rad/s
   C. 3π/2 rad/s
   D. 2π rad/s
   Answer: A. ω0 = 2π/T = 4π/3. (Key point: Fundamental frequency)

2. How is c0 obtained and what is its value?
   A. The area times T, 3/2
   B. Zero as the function is odd
   C. Peak value, 1
   D. The average over a period, 1/4
   Answer: D. c0 = (2/3)∫t dt from -1/2 to 1 = 1/4. (Key point: DC term)

3. What is |cn|?
   A. 3/(4π n^2)
   B. 1/(2π|n|)
   C. 3/(4π|n|)
   D. 3/(2π)
   Answer: C. cn = j e^(j2πn/3)/a with a = 4πn/3. (Key point: Magnitude)

4. How is c(-n) related to cn?
   A. c(-n) = -cn
   B. c(-n) = cn* (complex conjugate)
   C. c(-n) = cn
   D. c(-n) = 0
   Answer: B. Real f(t) gives conjugate-symmetric coefficients. (Key point: Negative n)

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