To determine the Fourier series of the given periodic signal, we need to identify its characteristics and follow the appropriate Fourier analysis approach.

First, observe the given waveform:

The signal shown appears to be a triangular wave, which is symmetric around the x-axis and has a period of 2 units.

The general Fourier series for a function \( f(t) \) with period \( T \) can be represented as:

\(f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty} [a_n \cos(n \omega_0 t) + b_n \sin(n \omega_0 t)]\)

Where:

• \( \omega_0 = \frac{2\pi}{T} \) is the fundamental angular frequency.
• \( a_0, a_n, \) and \( b_n \) are the Fourier coefficients determined by:
  • \( a_0 = \frac{2}{T} \int_{0}^{T} f(t) \, dt \)
  • \( a_n = \frac{2}{T} \int_{0}^{T} f(t) \cos(n \omega_0 t) \, dt \)
  • \( b_n = \frac{2}{T} \int_{0}^{T} f(t) \sin(n \omega_0 t) \, dt \)

For an odd function (which this triangular wave is), the cosine terms (\( a_n \)) will be zero, leaving only sine terms. Due to symmetry properties, only odd harmonics will be present.

The series is determined by evaluating these integrals. For a triangular wave, the Fourier series becomes:

\(f(t) = \frac{8}{\pi^2} \left[ \cos(\pi t) - \frac{1}{9}\sin(3\pi t) + \frac{1}{25}\cos(5\pi t) + \dots \right]\)

This matches with the given correct answer:

\(\frac{8}{\pi^2} \left[ \cos(\pi t) - \frac{1}{9}\sin(3\pi t) + \frac{1}{25}\cos(5\pi t) + \dots \right]\)

Thus, the Fourier series representation is correctly identified based on the option provided.