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Fourier series of half-wave rectified sine

Short answer

The half‑wave rectified sine wave has Fourier series x(t)=1/π + ½ sin t – (2/π) Σ_{n even} cos(nt)/(n²−1). It contains a DC term 1/π, a fundamental sine amplitude ½, and only even cosine harmonics with amplitudes decreasing as 2/[π(n²−1)].

Find the Fourier series representation of the half-wave rectified sine wave. [Figure: periodic half-wave rectified sine wave, positive half-cycles repeating with period 2π (humps over 0 to π, 2π to 3π, ... and -4π to -3π, ...), zero elsewhere]

x(t) = sin t for 0 < t < π, and 0 for π < t < 2π, with period T₀ = 2π and ω₀ = 1.

Coefficients (trigonometric form)

a₀ = (1/2π) ∫₀^π sin t dt = (1/2π)(2) = 1/π

aₙ = (1/π) ∫₀^π sin t cos nt dt = (1/2π) ∫₀^π [sin(n+1)t - sin(n-1)t] dt = (1/2π) [ (1 - cos(n+1)π)/(n+1) - (1 - cos(n-1)π)/(n-1) ] (n ≠ 1)

  • For odd n ≥ 3, cos(n±1)π = 1, so aₙ = 0.
  • For even n, cos(n±1)π = -1, so aₙ = (1/2π)[2/(n+1) - 2/(n-1)] = -2 / (π(n² - 1)), n = 2, 4, 6, ...

a₁ = (1/π) ∫₀^π sin t cos t dt = 0

b₁ = (1/π) ∫₀^π sin²t dt = (1/π)(π/2) = 1/2

bₙ = 0 for n ≥ 2 (orthogonality of sines).

Result

x(t) = 1/π + (1/2) sin t - (2/π) Σ_{n = 2, 4, 6, ...} cos(nt) / (n² - 1)

= 1/π + ½ sin t - (2/3π) cos 2t - (2/15π) cos 4t - (2/35π) cos 6t - ...

Exponential form

cₙ = 1/(π(1 - n²)) for even n, c₁ = -j/4, c₋₁ = j/4, and cₙ = 0 for the other odd n. Check: c₀ = 1/π = 0.318, c₂ = -1/(3π) = -0.106 (verified numerically).

✓ Answer checked by Claude Opus 5.5 (high effort)

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