--- title: "Diffraction and Fresnel zone phase difference" subject: "Wireless Communication" chapter: 3 key: "semester_7/wireless_communication:diffraction-fresnel-zone" source: https://www.bctengineer.com/q/semester_7/wireless_communication/diffraction-fresnel-zone site: BCT Engineering --- # Diffraction and Fresnel zone phase difference Wireless Communication · Chapter 3: Radio Wave Propagation in Mobile Networks - Asked in: 2079 Bhadra, 2+6 marks - Video: [Lecture-28|Wireless Communications|Diffraction and Understanding of Fresnel Zones for wireless comm.](https://www.youtube.com/watch?v=mVwyoD1lJCE) — Prof.Ashish Tiwari - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/wireless_communication/diffraction-fresnel-zone - Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3 ## Question What do you mean by diffraction in radio wave propagation? Derive an expression for phase difference in Fresnel zone geometry model of diffraction. ## Short answer Diffraction is bending of radio waves around obstacle edges into shadow regions, explained by Huygens principle. Path difference is approximately h^2(d1+d2)/(2d1d2), giving phase difference phi=2*pi*Delta/lambda=(pi/2)v^2, where v=h*sqrt[2(d1+d2)/(lambda*d1*d2)]. Fresnel zone radii are sqrt[n*lambda*d1*d2/(d1+d2)]. ## Answer **Diffraction** is the bending of radio waves around the edges of obstacles (hills, building edges) so that the signal reaches the shadow region behind them, even without line of sight. It is explained by **Huygens' principle**: every point on a wavefront acts as a source of secondary wavelets, which combine behind the obstacle. Diffraction loss depends on how far the obstacle blocks the **Fresnel zones**. #### Fresnel zone geometry (knife-edge) A transmitter and receiver are separated by an obstacle of effective height *h* above the LOS line, at distances d1 and d2. ```text |\ diffracted path | \ ________ h \________ Tx / α β | γ \ Rx o---------------+----------o LOS line |<---- d1 ----->|<-- d2 -->| ht obstacle hr ``` Assume h ≪ d1, d2 and h ≫ λ. #### Path difference Length of the diffracted path (two straight segments over the edge): ```text √(d1² + h²) + √(d2² + h²) ≈ d1 + h²/(2d1) + d2 + h²/(2d2) (binomial approx.) ``` Path difference over the direct path (d1 + d2): ```text Δ ≈ (h²/2)·(1/d1 + 1/d2) Δ ≈ (h²/2)·(d1 + d2)/(d1·d2) ``` #### Phase difference ```text φ = 2πΔ/λ = (2π/λ)·(h²/2)·(d1 + d2)/(d1·d2) ``` Using small angles, the angle α ≈ h(d1 + d2)/(d1·d2). Defining the **Fresnel–Kirchhoff diffraction parameter**: ```text v = h·√[ 2(d1 + d2) / (λ·d1·d2) ] = α·√[ 2d1d2 / (λ(d1 + d2)) ] ``` the phase difference becomes: ```text φ = (π/2)·v² ``` #### Fresnel zones Fresnel zones are regions where the path difference is nλ/2 (phase difference nπ). The radius of the n-th zone at the obstacle is: ```text rn = √[ nλ·d1·d2 / (d1 + d2) ] ``` - Odd zones add to the direct signal, even zones cancel it. - If the obstacle does not block the first Fresnel zone (about 55% clearance is enough), diffraction loss is very small. As blockage increases (v increases), diffraction loss increases; at v = 0 (edge on the LOS line) the loss is 6 dB. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is diffraction in radio propagation? A. Bending of waves around obstacle edges so signal reaches the shadow region B. Re-radiation of energy by small objects in many directions C. Reflection of waves from a smooth ground surface D. Shift in frequency due to motion Answer: A. Diffraction lets signals reach behind obstacles without line of sight, explained by Huygens' principle. (Key point: Meaning of diffraction) 2. For obstacle height h at distances d1 and d2, what is the path difference (h much smaller than d1, d2)? A. h(d1 + d2) B. (h^2/2)(d1 + d2)/(d1 d2) C. 2h^2/(d1 d2) D. h^2 d1 d2/(d1 + d2) Answer: B. Using the binomial approximation, Δ = (h^2/2)(1/d1 + 1/d2). (Key point: Path difference) 3. In terms of the Fresnel-Kirchhoff parameter v, the phase difference is: A. φ = π v^2 B. φ = (π/2) v C. φ = (π/2) v^2 D. φ = 2π v^2 Answer: C. φ = 2πΔ/λ, which simplifies to (π/2)v^2. (Key point: Phase difference) 4. Which statement about Fresnel zones is correct? A. Even zones add to the direct signal and odd zones cancel it B. At v = 0 the diffraction loss is 0 dB C. The n-th zone has path difference nλ and radius proportional to n D. The n-th zone has path difference nλ/2; odd zones add and even zones cancel Answer: D. Zones are defined by path difference nλ/2 (phase nπ); odd zones reinforce, even zones cancel, and loss is 6 dB at v = 0. 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