Fourier series of two-sided sawtooth with V

Reading the figure. In 0 < θ < π (θ = ωt) the wave falls from 0 to −V; at θ = π it jumps to 0 and rises to +V at θ = 2π; then it drops to 0 and repeats. Period 2π:

v(θ) = −V θ/π            0 < θ < π
v(θ) = V (θ − π)/π       π < θ < 2π

Check: v(θ + π) = −v(θ), so the wave has half-wave symmetry; only odd harmonics are present and a0 = 0.

Coefficients (odd n only)

an = (2/π)∫₀^π (−Vθ/π) cos nθ dθ
   = −(2V/π²)·(cos nπ − 1)/n² = 4V/(π² n²)
bn = (2/π)∫₀^π (−Vθ/π) sin nθ dθ
   = −(2V/π²)·(−π cos nπ/n) = (2V/πn) cos nπ = −2V/(nπ)

using ∫₀^π θ cos nθ dθ = (cos nπ − 1)/n² and ∫₀^π θ sin nθ dθ = −π cos nπ/n.

Fourier series

v(t) = Σ (n odd) [ (4V/π²n²) cos nωt − (2V/πn) sin nωt ]
     = V [ 0.405 cos ωt − 0.637 sin ωt + 0.045 cos 3ωt − 0.212 sin 3ωt + ... ]

Combined form: an cos nθ + bn sin nθ = cn cos(nθ + ψ), where cn = √(an² + bn²) = (2V/πn)√(1 + 4/(π²n²)) and ψ = arctan(−bn/an).

Line spectrum (V = 1)

  n=1 |##########################| 0.7547
  n=3 |#######                   | 0.2169
  n=5 |####                      | 0.1284
  n=7 |###                       | 0.0913
  n=9 |##                        | 0.0709

Even harmonics (and the d.c. term) are absent.

Answer: v(t) = Σ (n = 1, 3, 5...) [ (4V/π²n²) cos nωt − (2V/nπ) sin nωt ].