---
title: "Periodic signal whose Fourier series fails"
subject: "Signals and Systems"
chapter: 2
key: "semester_4/signals_and_systems:fourier-series-existence-examples"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/fourier-series-existence-examples
site: BCT Engineering
---

# Periodic signal whose Fourier series fails

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2078 Baisakh, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/fourier-series-existence-examples
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

"Even if a signal is periodic, its Fourier series might not exist." Justify the statement using three different examples.

## Answer

A periodic signal has a Fourier series that converges to it only if it meets the **Dirichlet conditions** over one period: (1) absolutely integrable, (2) finite number of maxima and minima, (3) finite number of finite discontinuities. Periodicity alone is not enough. Three periodic signals (period $T = 1$) that each break one condition:

#### Example 1: not absolutely integrable
$$
x(t) = \frac{1}{t}, \quad 0 < t \le 1, \quad \text{repeated with period } 1
$$

$$
\int_0^1\left|\frac{1}{t}\right|dt = \left[\ln t\right]_0^1 = \infty
$$

The coefficients $a_k = \int_0^1 \frac{1}{t}e^{-j2\pi kt}dt$ are not finite, so the series does not exist.

#### Example 2: infinite number of maxima and minima
$$
x(t) = \sin\left(\frac{2\pi}{t}\right), \quad 0 < t \le 1, \quad \text{period } 1
$$

It is bounded and absolutely integrable ($\int_0^1|x|\,dt < 1$), so its coefficients can be computed. But as $t \to 0$ it oscillates infinitely fast, giving infinitely many maxima and minima in one period, so the series is no longer guaranteed to converge to $x(t)$.

#### Example 3: infinite number of discontinuities
A staircase signal over $0 \le t < 1$ that equals 1 on $[0, 1/2)$, $1/2$ on $[1/2, 3/4)$, $1/4$ on $[3/4, 7/8)$, and so on (each step half the height and half the length of the previous one), repeated every 1 s.

It is bounded and $\int_0^1|x|\,dt < 1$, but it has an infinite number of discontinuities within one period, so it violates the third condition and convergence of its series to $x(t)$ is not guaranteed.

| Example | Condition violated |
|---|---|
| $1/t$ on $(0, 1]$ | absolute integrability |
| $\sin(2\pi/t)$ | finite maxima/minima |
| halving staircase | finite discontinuities |

So even though all three are periodic, a proper Fourier series representation is not guaranteed: for the first signal the coefficients do not even exist, and for the other two the series is not guaranteed to converge to the signal. Such signals are mathematical curiosities; practical signals usually meet the Dirichlet conditions.

## Check yourself

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

1. What must hold for a periodic signal's Fourier series to converge to it?
   A. Dirichlet conditions
   B. It must be even
   C. It must be real and positive
   D. Period must be integer
   Answer: A. Absolute integrability, finite extrema and finite discontinuities per period. (Key point: Why periodic is not enough)

2. Which condition does x(t) = 1/t on (0,1] violate?
   A. Finite maxima
   B. Absolute integrability
   C. Finite discontinuities
   D. Periodicity
   Answer: B. Integral of 1/t from 0 to 1 diverges (ln t). (Key point: 1/t example)

3. Which condition does sin(2pi/t) violate?
   A. Absolute integrability
   B. Finite number of maxima and minima
   C. Finite discontinuities
   D. Boundedness
   Answer: B. It oscillates infinitely as t -> 0. (Key point: sin(2pi/t) example)

4. What is wrong with the halving staircase?
   A. It is unbounded
   B. Not periodic
   C. Infinite discontinuities in one period
   D. Infinite integral
   Answer: C. Steps accumulate near t = 1, giving infinitely many jumps. (Key point: Staircase example)

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