Fourier series of odd sawtooth, ±Vo

Reading the figure. v(t) rises linearly from −Vo at ωt = −π, through 0 at ωt = 0, to +Vo at ωt = π, drops to −Vo, and repeats. Period 2π (θ = ωt):

v(θ) = Vo θ/π       −π < θ < π

The wave is odd (v(−θ) = −v(θ)), so a0 = 0 and an = 0; only sine terms exist.

Coefficients

bn = (1/π)∫_{−π}^{π} (Voθ/π) sin nθ dθ = (2/π)∫₀^π (Voθ/π) sin nθ dθ
   = (2Vo/π²)[ −θ cos nθ/n + sin nθ/n² ]₀^π
   = (2Vo/π²)(−π cos nπ/n)
   = 2Vo(−1)^(n+1)/(nπ)

Fourier series

v(t) = (2Vo/π) [ sin ωt − (1/2) sin 2ωt + (1/3) sin 3ωt − (1/4) sin 4ωt + ... ]

Line spectrum (amplitude 2Vo/(nπ), Vo = 1)

  n=1 |##########################| 0.6366
  n=2 |#############             | 0.3183
  n=3 |#########                 | 0.2122
  n=4 |######                    | 0.1592
  n=5 |#####                     | 0.1273

The amplitude falls as 1/n (slow convergence because of the jump at θ = ±π). The phase alternates because bn alternates in sign.

Answer: v(t) = (2Vo/π) Σ (−1)^(n+1) (1/n) sin nωt, n = 1, 2, 3, ...