Fourier series of rising sawtooth −V0 to V0

Reading the figure. V(t) rises linearly from −V0 at ωt = 0 to +V0 at ωt = 2π, then jumps back to −V0. The period is 2π (θ = ωt).

V(θ) = −V0 + (V0/π) θ = (V0/π)(θ − π)      0 < θ < 2π

The wave is odd about θ = π (the zero crossing), so the average is zero and the series has only sine terms.

Coefficients

a0 = (1/2π)∫₀^2π V(θ) dθ = 0
an = (1/π)∫₀^2π (V0/π)(θ − π) cos nθ dθ = 0     (∫ θ cos nθ dθ = 0 over a full period)
bn = (1/π)∫₀^2π (V0/π)(θ − π) sin nθ dθ
   = (V0/π²)·∫₀^2π θ sin nθ dθ          (the π·sin term integrates to 0)

Using ∫₀^2π θ sin nθ dθ = [−θ cos nθ/n + sin nθ/n²]₀^2π = −2π/n:

bn = (V0/π²)(−2π/n) = −2V0/(nπ)

Fourier series

V(t) = −(2V0/π) [ sin ωt + (1/2) sin 2ωt + (1/3) sin 3ωt + ... ]

Line spectrum (amplitudes cn = |bn|, V0 = 1 shown)

  n=1 |##########################| 0.6366
  n=2 |#############             | 0.3183
  n=3 |#########                 | 0.2122
  n=4 |######                    | 0.1592
  n=5 |#####                     | 0.1273
  n=6 |####                      | 0.1061

All harmonics are present and the amplitude falls as 1/n. Writing −sin nθ = cos(nθ + 90°), each term has phase +90°.

Answer: V(t) = −(2V0/π) Σ (1/n) sin nωt, n = 1, 2, 3, ...