Fourier series of falling sawtooth V to 0

Reading the figure. v(t) falls linearly from V at θ = ωt = 0 to 0 at θ = 2π, then jumps back to V. Period 2π.

v(θ) = V (1 − θ/2π)     0 < θ < 2π

Coefficients

a0 = (1/2π)∫₀^2π V(1 − θ/2π) dθ = V/2          (average value)
an = (1/π)∫₀^2π V(1 − θ/2π) cos nθ dθ = 0      (∫ cos nθ = 0 and ∫ θ cos nθ = 0)
bn = (1/π)∫₀^2π V(1 − θ/2π) sin nθ dθ
   = −(V/2π²)∫₀^2π θ sin nθ dθ
   = −(V/2π²)(−2π/n) = V/(nπ)

Fourier series

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

(The wave is a d.c. level V/2 plus an odd sawtooth about that level.)

Line spectrum (V = 1 shown, amplitude in units of V)

  n=0 |##########################| 0.5
  n=1 |#################         | 0.3183
  n=2 |########                  | 0.1592
  n=3 |######                    | 0.1061
  n=4 |####                      | 0.07958
  n=5 |###                       | 0.06366

Phase is shown for the form cn cos(nωt + φn): sin nθ = cos(nθ − 90°).

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