--- title: "Two-element array: broadside and end fire" subject: "Propagation and Antenna" chapter: 2 key: "semester_6/propagation_and_antenna:two-element-array-broadside-endfire" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/two-element-array-broadside-endfire site: BCT Engineering --- # Two-element array: broadside and end fire Propagation and Antenna · Chapter 2: Antenna Parameters and Arrays - Asked in: 2071 Bhadra, 5+6 marks - Video: [End Fire Array and Broadside Array | Radiation Pattern | Conditions for Radiation | Antenna Array](https://www.youtube.com/watch?v=Et1cX1yaCbc) — 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/two-element-array-broadside-endfire - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-2 ## Question Derive a relation for the field intensity for the array of two elements isotropic radiators in various conditions. Show the condition for broad side and end fire array with necessary diagrams. ## Short answer General field: E = 2E0 cos[(βd cosθ + α)/2] with ψ = βd cosθ + α, α is current phase lead of element 2. Broadside: α=0, max at θ=90° (cosθ=0). End‑fire: α = −βd, max at θ=0° (cosθ=1). For d=λ/2, α=π yields end‑fire figure‑8 along axis. ## Answer An **antenna array** is a group of similar antennas whose fields add at a distant point. For two isotropic point sources, the total field is the element field multiplied by an **array factor** that depends on the spacing *d* and the current phase difference *α*. #### Derivation of field intensity Assumptions: two isotropic sources 1 and 2 on the x-axis, spacing *d*, equal amplitude *E₀*, and the current in source 2 leads that in source 1 by *α*. A far-field point P is at angle θ from the array axis. Take the midpoint O as the phase reference. ```text to far point P / / θ 1 ●--------O--------● 2 -> array axis <-- d/2 -><- d/2 --> path difference between 1 and 2 = d cosθ ``` Phase difference of the two waves at P: ```text ψ = βd cosθ + α β = 2π/λ ``` Relative to O, the field of source 1 lags by ψ/2 and the field of source 2 leads by ψ/2: ```text E = E₀ e^(-jψ/2) + E₀ e^(+jψ/2) = 2E₀ cos(ψ/2) E = 2E₀ cos[(βd cosθ + α)/2] ``` Normalised array factor: **AF(θ) = |cos[(βd cosθ + α)/2]|**. #### Case 1: equal amplitude and phase (α = 0) — broadside ψ = βd cosθ, so E = 2E₀ cos[(πd/λ) cosθ]. - Maximum when cosθ = 0, i.e. θ = 90° and 270°: **perpendicular to the array axis**. - For d = λ/2: E = 2E₀ cos(π/2 cosθ); nulls at θ = 0° and 180°. ```text d = λ/2, α = 0 (figure-8, broadside) 90° .-"""-. \ / 180° ----1--O--2---- 0° (nulls on axis) / \ '-...-' 270° ``` #### Case 2: equal amplitude, opposite phase (α = π) E = 2E₀ cos[(βd cosθ + π)/2] = 2E₀ |sin(πd/λ · cosθ)|. For d = λ/2 the maximum is along the axis (θ = 0°, 180°) with nulls at θ = 90°: a bidirectional end-fire pattern. #### Case 3: equal amplitude, phase α = −βd — end fire ψ = βd(cosθ − 1). The maximum (ψ = 0) is at θ = 0°, so the beam is **along the array axis** in one direction. For d = λ/4, α = −π/2 the pattern is a cardioid: ```text d = λ/4, α = -90° (cardioid, end fire) .--. ____/ \___ | 1 2 > max at θ = 0° ‾‾‾‾\ /‾‾‾ '--' null at θ = 180° ``` #### Conditions summarised | Array | Phase condition | Direction of max | |---|---|---| | Broadside | α = 0 (ψ = 0 at θ = 90°) | θ = 90°, normal to axis | | End fire | α = −βd (ψ = 0 at θ = 0°) | θ = 0°, along axis | | General | βd cosθₘ + α = 0 | cosθₘ = −α/(βd) | In general, the beam points where the path difference is exactly cancelled by the current phase, so changing α steers the beam (the basis of phased arrays). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With current phase difference α, the total phase difference is A. ψ = βd sinθ + α B. ψ = βd cosθ + α C. ψ = βd cosθ - α only D. ψ = α/βd Answer: B. The path difference d cosθ adds to the current phase. (Key point: Phase difference) 2. The broadside condition is A. α = 0, giving the maximum at θ = 90° B. α = -βd, maximum at θ = 90° C. α = π, maximum at 90° D. α = 0, maximum at θ = 0° Answer: A. ψ = βd cosθ = 0 at θ = 90°. (Key point: Broadside condition) 3. The end-fire condition is A. α = 0 B. α = +βd C. α = -βd, giving ψ = 0 at θ = 0° D. α = π/2 for any d Answer: C. The current lag cancels the path phase along the axis. (Key point: End-fire condition) 4. For d = λ/4 and α = -90°, the pattern is A. A figure-8 broadside B. A circle C. A cardioid with maximum at 180° D. A cardioid with maximum at 0° and null at 180° Answer: D. ψ = (π/2)(cosθ - 1), zero at 0° and -π at 180°. 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