Fourier series of triangular wave, peak 20

Reading the figure. f(t) is a triangular wave of period 2π (in ωt) with a peak of +20 at ωt = 0, ±2π, zero crossings at ±π/2 and a minimum of −20 at ±π (assumed; the scan shows the negative peak only roughly). Let θ = ωt.

The wave is even (bn = 0), has zero average (a0 = 0), and has half-wave symmetry f(θ + π) = −f(θ), so only odd cosine harmonics are present.

f(θ) = 20 − 40θ/π     for 0 ≤ θ ≤ π

Coefficients

an = (2/π)∫₀^π (20 − 40θ/π) cos nθ dθ
   = −(80/π²)∫₀^π θ cos nθ dθ
   = −(80/π²)(cos nπ − 1)/n²
   = 80[1 − (−1)ⁿ] / (π² n²)
an = 160/(π² n²)  for odd n;   an = 0 for even n
a0 = 0,  bn = 0

Fourier series

f(t) = (160/π²) [ cos ωt + (1/9) cos 3ωt + (1/25) cos 5ωt + (1/49) cos 7ωt + ... ]
     = 16.21 cos ωt + 1.801 cos 3ωt + 0.6485 cos 5ωt + 0.3308 cos 7ωt + ...

Line spectrum

  n=1 |##########################| 16.21
  n=3 |###                       | 1.801
  n=5 |#                         | 0.6485
  n=7 |#                         | 0.3308
  n=9 |#                         | 0.2001

Answer: f(t) = (160/π²) Σ (1/n²) cos nωt, n = 1, 3, 5, ... (amplitudes fall as 1/n²).