Periodic signal whose Fourier series fails

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.

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.