Fourier series of triangular wave, peak 1

Reading the figure. f(t) is a triangular wave of period 2π with peak +1 at t = 0, zero crossings at ±π/2 and minimum −1 at ±π. It is an even function (f(−t) = f(t)) with half-wave symmetry, so bn = 0, a0 = 0 and only odd cosine harmonics exist.

Equation of the wave

f(t) = 1 − 2t/π     for 0 ≤ t ≤ π    (even, so f(−t) = f(t))

Coefficients

a0 = (1/2π)∫ f dt = 0                       (equal area above and below)
an = (2/π)∫₀^π (1 − 2t/π) cos nt dt
   = −(4/π²)∫₀^π t cos nt dt
   = −(4/π²)·[(cos nπ − 1)/n²]
   = 4[1 − (−1)ⁿ] / (π² n²)
an = 8/(π² n²)  for n odd,   an = 0 for n even
bn = 0

Fourier series

f(t) = (8/π²) [ cos t + (1/9) cos 3t + (1/25) cos 5t + (1/49) cos 7t + ... ]

Line spectrum (amplitude cn = |an|)

  n=1 |##########################| 0.8106
  n=3 |###                       | 0.09006
  n=5 |#                         | 0.03242
  n=7 |#                         | 0.01654
  n=9 |#                         | 0.01001

Lines occur only at ω, 3ω, 5ω ... and fall as 1/n². All harmonics are cosines of phase 0° (the cn are positive here).

Answer: f(t) = (8/π²) Σ (1/n²) cos nt, n = 1, 3, 5, ...