Fourier series of half-period ramp pulses

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

v(θ) = Vo θ/π     0 < θ < π
v(θ) = 0          π < θ < 2π

Coefficients

a0 = (1/2π)∫₀^π (Voθ/π) dθ = (1/2π)(Vo π/2) = Vo/4
an = (1/π)∫₀^π (Voθ/π) cos nθ dθ = (Vo/π²)[ (cos nπ − 1)/n² ]
   = Vo[(−1)ⁿ − 1]/(π² n²)
   → an = −2Vo/(π² n²) for odd n, an = 0 for even n
bn = (1/π)∫₀^π (Voθ/π) sin nθ dθ = (Vo/π²)(−π cos nπ/n)
   = Vo(−1)^(n+1)/(nπ)

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

Fourier series

v(t) = Vo/4 − (2Vo/π²)[cos ωt + (1/9) cos 3ωt + ...]
       + (Vo/π)[ sin ωt − (1/2) sin 2ωt + (1/3) sin 3ωt − ... ]

Line spectrum (cn = √(an² + bn²), Vo = 1)

  n=0 |#################         | 0.25
  n=1 |##########################| 0.3773
  n=2 |###########               | 0.1592
  n=3 |#######                   | 0.1085
  n=4 |#####                     | 0.0796
  n=5 |####                      | 0.0642

The wave has no even, odd or half-wave symmetry, so the d.c. term, cosine terms and sine terms are all present, and lines appear at every harmonic. The phase of each line is ψn = arctan(−bn/an), taken in the correct quadrant, e.g. ψ1 = −122.5° (in the form cn cos(nθ + ψn)).

Answer: v(t) = Vo/4 + Σ [ an cos nωt + bn sin nωt ], with an = [(−1)ⁿ − 1]Vo/(π²n²) and bn = (−1)^(n+1) Vo/(nπ).