---
title: "Fourier series of half-wave rectified sine"
subject: "Data Communication"
chapter: 3
key: "semester_5/data_communication:fs-half-wave-rectified"
source: https://www.bctengineer.com/q/semester_5/data_communication/fs-half-wave-rectified
site: BCT Engineering
---

# Fourier series of half-wave rectified sine

Data Communication · Chapter 3: Signal Analysis

- Asked in: 2071 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/fs-half-wave-rectified
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-3

## Question

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]

## 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)].

## Answer

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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The average value a0 is
   A. 2/π
   B. 1/2
   C. 1/π
   D. 0
   Answer: C. (1/2π)×2. (Key point: DC term)

2. The fundamental term is
   A. (1/2) cos t
   B. (2/π) cos t
   C. 0
   D. (1/2) sin t
   Answer: D. b1 = 1/2. (Key point: Fundamental)

3. Odd harmonics (n ≥ 3) are
   A. Zero
   B. Nonzero
   C. Equal to 1/2
   D. Negative
   Answer: A. cos(n±1)π = 1. (Key point: Odd harmonics)

4. an for even n is
   A. 2/(π n)
   B. -2/(π(n^2-1))
   C. -1/(π n)
   D. 0
   Answer: B. From the integral. (Key point: Even harmonics)

## Related in Data Communication

- [Fourier series of full-wave rectified sine](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afs-full-wave-rectified) — Data Communication, chapter 3
- [Fourier series coefficients of a square wave](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afs-coefficients-square-wave) — Data Communication, chapter 3

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