--- title: "Pattern multiplication for two short dipoles" subject: "Propagation and Antenna" chapter: 2 key: "semester_6/propagation_and_antenna:pattern-multiplication-short-dipoles" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/pattern-multiplication-short-dipoles site: BCT Engineering --- # Pattern multiplication for two short dipoles Propagation and Antenna · Chapter 2: Antenna Parameters and Arrays - Asked in: 2070 Bhadra, 2+3+3 marks - Video: [Pattern Multiplication Method - Applications with Examples](https://www.youtube.com/watch?v=hHt_VdB5ZPg) — Padmasri Naban - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/pattern-multiplication-short-dipoles - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-2 ## Question State the principle of pattern multiplication. Use the principle to obtain a wave pattern for array of two short dipoles for following cases where (d) = dipole separation and (α) = current phase difference. a) Dipoles aligned perpendicular to the array axis with d = λ/2, α = 0 b) Dipoles aligned perpendicular to the array axis with d = λ/2, α = π ## Short answer Pattern multiplication states total field pattern equals element pattern multiplied by array factor. For two short dipoles perpendicular to array axis with d=λ/2: (a) α=0 gives broadside figure-8 in plane perpendicular to dipoles and four-lobed pattern in dipole plane with lobes near 57° and 123°. (b) α=π gives end-fire figure-8 perpendicular to dipoles and sharp two-lobed end-fire pattern along axis in dipole plane. ## Answer #### Principle of pattern multiplication The total field pattern of an array of **identical, similarly oriented** elements equals the **element (unit) pattern multiplied by the array factor** (the pattern of the same array made of isotropic point sources). The total phase pattern is the sum of the element phase pattern and the array phase pattern. ```text E_total(θ,φ) = E_element(θ,φ) × AF(θ,φ) ``` It lets us sketch complex patterns quickly without full integration. Setup for both cases: two short dipoles, spacing d = λ/2 (βd = π), both perpendicular to the array axis. θ is measured from the array axis. Two planes are considered: - Plane ⊥ to dipoles: short dipole pattern is a circle (omnidirectional). - Plane containing dipoles and axis: short dipole pattern = |sin(angle from dipole)| = |cosθ|. Array factor of two isotropic sources: AF = |cos[(βd cosθ + α)/2]|. #### a) d = λ/2, α = 0 ```text AF = |cos(π/2 · cosθ)| max at θ = 90°, nulls at 0°,180° Plane ⊥ dipoles : E = AF -> broadside figure-8 Plane of dipoles: E = |cosθ| · |cos(π/2 · cosθ)| θ=0°: 0 θ=30°: 0.18 θ=45°: 0.31 θ=60°: 0.35 θ=75°: 0.24 θ=90°: 0 ``` In the plane of the dipoles, the AF maximum (θ = 90°) falls on the dipole null, so the result is a **four-lobed** pattern with lobes near θ ≈ 57°, 123°, 237°, 303°. ```text element AF (broadside) total (dipole plane) ( ) ___ \ / --(+)-- x --1-2-- = ->1-2<- ( ) ‾‾‾ / \ max on axis max ⊥ axis 4 lobes ``` In the plane perpendicular to the dipoles the pattern is the broadside figure-8 (max at θ = 90°). #### b) d = λ/2, α = π ```text AF = |sin(π/2 · cosθ)| max at θ = 0°,180°, null at 90° Plane ⊥ dipoles : E = AF -> end-fire figure-8 on axis Plane of dipoles: E = |cosθ| · |sin(π/2 · cosθ)| θ=0°: 1 θ=30°: 0.85 θ=45°: 0.63 θ=60°: 0.35 θ=75°: 0.10 θ=90°: 0 ``` Here both factors are maximum along the axis, so the total is a **two-lobed end-fire pattern**, narrower than the AF alone. ```text element AF (end fire) total ( ) --(+)-- x <=1-2=> = <(1-2)> narrow lobes ( ) along the axis ``` | Case | Plane ⊥ dipoles | Plane of dipoles | |---|---|---| | α = 0 | broadside figure-8 | four lobes | | α = π | end-fire figure-8 | sharp end-fire two lobes | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Pattern multiplication says A. Total pattern = element pattern × array factor, for identical similarly oriented elements B. Total pattern = element pattern + array factor C. Total pattern = AF only D. It works only for isotropic sources Answer: A. It lets complex patterns be sketched without integration. (Key point: Principle) 2. For d = λ/2, α = 0, dipoles perpendicular to axis, the pattern in the plane of the dipoles has A. Two lobes along the axis B. Four lobes (near 57°, 123°, 237°, 303°) C. A single lobe D. A circle Answer: B. The AF maximum at 90° coincides with the dipole null. (Key point: Case a: lobes) 3. For d = λ/2, α = π, the plane of dipoles gives A. Four lobes B. Broadside figure-8 C. A narrow two-lobed end-fire pattern D. Omnidirectional Answer: C. Both |cosθ| and the AF peak along the axis. (Key point: Case b: pattern) 4. In the plane containing the dipoles and axis, the short dipole factor is A. |sinθ| B. 1 C. |cos2θ| D. |cosθ| Answer: D. The angle from the dipole is 90° - θ, so sine of it is cosθ. 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