--- title: "Yagi vs log-periodic; LP design" subject: "Propagation and Antenna" chapter: 3 key: "semester_6/propagation_and_antenna:yagi-vs-log-periodic-design" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/yagi-vs-log-periodic-design site: BCT Engineering --- # Yagi vs log-periodic; LP design Propagation and Antenna · Chapter 3: Antenna Classification - Asked in: 2071 Bhadra, 2+8 marks - Video: [Log Periodic Antenna (Basics, Log Periodic Dipole Array, Applications & Advantages) Explained](https://www.youtube.com/watch?v=JhcdmMsPhv8) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/yagi-vs-log-periodic-design - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-3 ## Question Compare Yagi antenna with log periodic antenna. Explain the working principle and design of log-periodic antenna. ## Short answer Yagi has one driven element and parasitic directors/reflector, giving higher gain but narrow bandwidth; log-periodic feeds all dipoles and is frequency-independent with very wide bandwidth. In an LPDA, element lengths and spacings grow by ratio τ; only the near-resonant dipoles radiate, so the beam points toward the apex. Design uses τ, σ, α, and N = 1 + ln(Bs)/ln(1/τ). ## Answer #### Yagi vs log-periodic antenna | Point | Yagi–Uda | Log-periodic dipole array (LPDA) | |---|---|---| | Feeding | Only one driven element; rest parasitic | All dipoles fed by a common line | | Element sizes | Nearly equal | Grow by a constant ratio τ | | Bandwidth | Narrow (≈ 2–5%) | Very wide (e.g. 10:1) | | Gain | Higher for same size (≈ 7–15 dBi) | Moderate (≈ 7–10 dBi), constant over band | | Impedance | Changes with frequency | Nearly constant over band | | Design | Empirical, single frequency | Frequency-independent, by τ, σ, α | | Use | TV reception at one channel, VHF links | Wide-band TV, HF, EMC testing | #### Working principle of log-periodic antenna An LPDA is a set of dipoles of lengths L₁ < L₂ < … < Lₙ placed along a boom inside a triangle of half apex angle α and fed by a crossed (transposed) two-wire line from the short end. ```text apex (feed) /|\ α / | \ L1 --+-- shortest (fmax) ---+--- ----+---- active region -----+----- ------+------ longest (fmin) <- beam toward apex (short end) ``` - Dimensions grow geometrically: τ = Lₙ/Lₙ₊₁ = Rₙ/Rₙ₊₁ = dₙ/dₙ₊₁ < 1, where Rₙ is the distance of dipole n from the apex and dₙ the spacing. - At any frequency, only the 3–5 dipoles near half-wave length (the **active region**) carry large current and radiate. Longer dipoles behind act like reflectors; shorter ones in front act like directors, so the beam points toward the apex. - Changing frequency only moves the active region along the array. Because the structure repeats itself when frequency is multiplied by 1/τ, the impedance and pattern repeat periodically with log f, hence the name. If the variation within one period is small, the antenna is effectively frequency-independent. - The feeder is transposed between adjacent dipoles to give the 180° phase shift needed for end-fire toward the short end. #### Design steps (Carrel method) 1. Choose directivity; from Carrel's chart pick the optimum **scale factor τ** (0.8–0.98) and **spacing factor σ** = dₙ/(2Lₙ). 2. Half apex angle: **α = tan⁻¹[(1 − τ)/(4σ)]**. 3. Bandwidth of active region: **Bₐᵣ = 1.1 + 7.7(1 − τ)² cot α**. 4. Bandwidth of structure: **Bₛ = B × Bₐᵣ**, where B = f_max/f_min. 5. Number of elements: **N = 1 + ln(Bₛ)/ln(1/τ)**. 6. Longest element: **L_max = λ_max/2** (at f_min); then Lₙ₊₁ = τLₙ for the rest. 7. Boom length: **L = (λ_max/4)(1 − 1/Bₛ) cot α**. 8. Spacings dₙ = 2σLₙ; choose feeder impedance to match the source (typically 50–75 Ω via a balun). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which comparison is correct? A. Yagi: all elements fed, very wide bandwidth; LPDA: one driven element, narrow bandwidth B. Yagi: one driven element, narrow bandwidth; LPDA: all dipoles fed by a common line, very wide bandwidth C. Both have identical bandwidth D. Yagi has constant impedance; LPDA's changes with frequency Answer: B. A Yagi has one driven element and parasitic ones with narrow bandwidth (2-5%); an LPDA feeds all dipoles and has very wide bandwidth. (Key point: Yagi vs LPDA feeding and bandwidth) 2. How does an LPDA work as frequency changes? A. All dipoles radiate equally at all frequencies B. Only the 3-5 dipoles near half-wave length (active region) radiate, and the active region moves along the array with frequency C. Only the longest dipole ever radiates D. The apex angle changes with frequency Answer: B. At any frequency only dipoles near λ/2 carry large current; longer ones behind act as reflectors and shorter ones in front as directors. (Key point: Active region) 3. Toward which end does the LPDA beam point, and why is the feeder transposed? A. Toward the long end; no transposition is needed B. Toward the apex (short end); transposition gives the 180° phase shift between adjacent dipoles for end-fire C. Broadside; transposition lowers impedance D. Toward the long end; transposition raises gain Answer: B. The crossed feeder gives 180° phase between adjacent dipoles to produce end-fire radiation toward the short end. (Key point: Direction of beam) 4. In Carrel's design method, how are the half apex angle and the number of elements obtained? A. α = tan⁻¹[(1 − τ)/(4σ)]; N = 1 + ln(Bs)/ln(1/τ) B. α = tan⁻¹[τ/σ]; N = ln(Bs) C. α = 90° − τ; N = 1/τ D. α = tan⁻¹(4σ/(1 − τ)); N = Bs x τ Answer: A. The apex angle comes from τ and σ via α = tan⁻¹[(1 − τ)/(4σ)], and N = 1 + ln(Bs)/ln(1/τ), where Bs = B x Bar. 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