Fourier series of triangle train, peak 10

Reading the figure. The voltage rises linearly from 0 at θ = ωt = 0 to 10 V at θ = π, falls to 0 at θ = 2π, and repeats (period 2π, peak Vo = 10 V; this is the same shape as the previous question with T = 2π/ω).

v(θ) = 10θ/π             0 ≤ θ ≤ π
v(θ) = 10(2π − θ)/π      π ≤ θ ≤ 2π

The wave is symmetric about θ = π (v(2π − θ) = v(θ)), so bn = 0, and the average is half the peak value.

Coefficients

a0 = 10/2 = 5                                   (average value)
an = (1/π)∫₀^2π v cos nθ dθ = (2/π)∫₀^π (10θ/π) cos nθ dθ
   = (20/π²)·(cos nπ − 1)/n²
   = 20[(−1)ⁿ − 1] / (π² n²)
an = −40/(π² n²)  for odd n;   an = 0 for even n
bn = 0

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

Fourier series

v(t) = 5 − (40/π²) [ cos ωt + (1/9) cos 3ωt + (1/25) cos 5ωt + ... ]
     = 5 − 4.053 cos ωt − 0.4503 cos 3ωt − 0.1621 cos 5ωt − 0.0827 cos 7ωt − ...

Line spectrum

  n=0 |##########################| 5
  n=1 |#####################     | 4.053
  n=3 |##                        | 0.4503
  n=5 |#                         | 0.1621
  n=7 |#                         | 0.08271

Answer: v(t) = 5 − (40/π²) Σ (1/n²) cos nωt (V), n = 1, 3, 5, ...