Fourier series of ramp wave 0 to 20

Reading the figure. v(t) rises linearly from 0 at θ = ωt = 0 to 20 at θ = 2π, drops to 0 and repeats. Period 2π.

v(θ) = 20 θ/(2π) = 10θ/π       0 < θ < 2π

Coefficients

a0 = (1/2π)∫₀^2π (10θ/π) dθ = 10                 (average value)
an = (1/π)∫₀^2π (10θ/π) cos nθ dθ = 0           (∫ θ cos nθ dθ = 0 over a period)
bn = (1/π)∫₀^2π (10θ/π) sin nθ dθ
   = (10/π²)·(−2π/n) = −20/(nπ)

(using ∫₀^2π θ sin nθ dθ = −2π/n).

Fourier series

v(t) = 10 − (20/π) [ sin ωt + (1/2) sin 2ωt + (1/3) sin 3ωt + ... ]
     = 10 − 6.366 sin ωt − 3.183 sin 2ωt − 2.122 sin 3ωt − 1.592 sin 4ωt − ...

Line spectrum

  n=0 |##########################| 10
  n=1 |#################         | 6.366
  n=2 |########                  | 3.183
  n=3 |######                    | 2.122
  n=4 |####                      | 1.592
  n=5 |###                       | 1.273

(−sin nθ = cos(nθ + 90°), so each harmonic has phase +90°.)

Answer: v(t) = 10 − (20/π) Σ (1/n) sin nωt, n = 1, 2, 3, ...