Fourier series of triangle train, Vo and T

Reading the figure. v(t) rises linearly from 0 at t = 0 to Vo at t = T/2 and falls to 0 at t = T; period T, fundamental ω0 = 2π/T. Let θ = ω0 t.

The wave is symmetric about θ = π (v(2π − θ) = v(θ)), so it has no sine terms: bn = 0, and the average is Vo/2.

Coefficients (θ = ω0 t over 0 to 2π)

v(θ) = Voθ/π            0 ≤ θ ≤ π
v(θ) = Vo(2π − θ)/π     π ≤ θ ≤ 2π

a0 = (area)/(period) = (Vo·T/2)/T = Vo/2
an = (1/π)∫₀^2π v cos nθ dθ = (2/π)∫₀^π (Voθ/π) cos nθ dθ     (symmetry about π)
   = (2Vo/π²)·(cos nπ − 1)/n²
   = 2Vo[(−1)ⁿ − 1] / (π² n²)
an = −4Vo/(π² n²) for odd n,   an = 0 for even n
bn = 0

Fourier series

v(t) = Vo/2 − (4Vo/π²) [ cos ω0t + (1/9) cos 3ω0t + (1/25) cos 5ω0t + ... ]
     with ω0 = 2π/T
     = 0.5Vo − 0.405Vo cos ω0t − 0.0450Vo cos 3ω0t − 0.0162Vo cos 5ω0t − ...

(Check at t = 0: 0.5 − 0.405 − 0.045 − 0.0162 − ... → 0, as it must since v(0) = 0.)

Line spectrum (amplitudes, in units of Vo)

  n=0 |##########################| 0.5
  n=1 |#####################     | 0.4053
  n=3 |##                        | 0.04503
  n=5 |#                         | 0.01621
  n=7 |#                         | 0.008271

Lines are at 0, ω0, 3ω0, 5ω0, ... = 0, 2π/T, 6π/T, ...

Answer: v(t) = Vo/2 − (4Vo/π²) Σ (1/n²) cos(2πnt/T), n = 1, 3, 5, ...