--- title: "Two-element array and end fire pattern" subject: "Propagation and Antenna" chapter: 2 key: "semester_6/propagation_and_antenna:two-isotropic-endfire-pattern" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/two-isotropic-endfire-pattern site: BCT Engineering --- # Two-element array and end fire pattern Propagation and Antenna · Chapter 2: Antenna Parameters and Arrays - Asked in: 2070 Magh, 4+4 marks - Video: [Derivation of Two point source -End fire array(English &Tamil)](https://www.youtube.com/watch?v=6wg9McntDNM) — ECE VIBE - 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-isotropic-endfire-pattern - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-2 ## Question Derive a mathematical relation to calculate the field intensity of an array for two element isotropic radiators. Draw the resultant pattern for an End Fire Array with dipoles perpendicular to the axis of array with d = λ/2 and α = π, where λ is the wavelength of the operating frequency. ## Short answer Derivation: E = 2E0 cos[(βd cosθ + α)/2]. For end‑fire with d=λ/2, α=π, array factor = |sin(π/2 cosθ)|, giving max at 0°,180°, null at 90°. Dipoles perpendicular to axis are omnidirectional in that plane, so total pattern = figure‑8 along axis with same maxima and nulls. ## Answer For two isotropic sources of equal amplitude, spacing *d* and current phase difference *α*, the far field is **E = 2E₀ cos(ψ/2)** with **ψ = βd cosθ + α**. #### Derivation Sources 1 and 2 lie on the array axis, *d* apart; θ is measured from the axis; the midpoint is the phase reference. ```text P (far) / / θ 1 ●-------O-------● 2 -> axis path difference = d cosθ ``` ```text phase difference ψ = βd cosθ + α, β = 2π/λ E = E₀ e^(-jψ/2) + E₀ e^(jψ/2) = 2E₀ cos(ψ/2) = 2E₀ cos[(βd cosθ + α)/2] ``` Maximum where ψ = 0; broadside for α = 0, end fire for α = ±βd. #### Pattern for d = λ/2, α = π, dipoles ⊥ to array axis Here βd = π, so the **array factor** is ```text AF = cos[(π cosθ + π)/2] = |sin(π/2 · cosθ)| θ = 0° : |sin(π/2)| = 1 (max) θ = 60° : |sin(π/4)| = 0.707 θ = 90° : sin 0 = 0 (null) θ = 180° : 1 (max) ``` By **pattern multiplication**, total pattern = element pattern × AF. Plane perpendicular to the dipoles (contains the array axis): a short dipole is omnidirectional here, so the total pattern = AF, a figure-8 **along the axis**. Plane containing the dipoles and the array axis: the dipole field is proportional to the sine of the angle from the dipole, which is cosθ here. Total: ```text E(θ) = |cosθ| · |sin(π/2 · cosθ)| θ = 0°: 1, 30°: 0.85, 45°: 0.63, 60°: 0.35, 90°: 0 ``` The two lobes along the axis become narrower than the AF alone. ```text Total pattern, plane of dipoles and axis 90° (null) | .--. | .--. 180°( )====1-+-2====( ) 0° max '--' | '--' max | 270° (null) two narrow lobes along the array axis ``` The result is a **bidirectional end-fire pattern**: maxima at θ = 0° and 180° along the array axis, nulls broadside (θ = 90°), half-power points of the AF alone at θ = 60° and 120° (|sin(π/4)| = 0.707); the total pattern is narrower after multiplying by |cosθ|. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For d = λ/2, α = π, the array factor is A. |sin[(π/2)cosθ]| B. |cos[(π/2)cosθ]| C. |sinθ| D. 1 Answer: A. βd = π, so cos[(π cosθ + π)/2] = |sin(π cosθ/2)|. (Key point: Array factor) 2. What is AF at θ = 60°? A. 0 B. 0.707 C. 1 D. 0.5 Answer: B. sin(π/4) = 0.707; this is a half-power point. (Key point: Value at 60°) 3. In the plane perpendicular to the dipoles, the total pattern is A. Zero B. A circle C. Equal to the AF (end-fire figure-8) D. |cosθ| only Answer: C. The short dipole is omnidirectional there. (Key point: Plane perpendicular to dipoles) 4. In the plane containing dipoles and axis, E(θ) is A. |sinθ| only B. |cos[(π/2)cosθ]| C. The same as the AF D. |cosθ|·|sin[(π/2)cosθ]|, narrower lobes than the AF alone Answer: D. The dipole factor is |cosθ|, which narrows the lobes. (Key point: Plane of dipoles) ## Related in Propagation and Antenna - [Two-element array: broadside and end fire](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Atwo-element-array-broadside-endfire) — Propagation and Antenna, chapter 2 - [Pattern multiplication for two short dipoles](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Apattern-multiplication-short-dipoles) — Propagation and Antenna, chapter 2 - [Pattern lobes and 180° out-of-phase array](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Alobes-importance-opposite-phase-array) — Propagation and Antenna, chapter 2 - [Linear array; broadside vs end fire](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Alinear-array-broadside-endfire) — Propagation and Antenna, chapter 2 - [Features of end-fire array](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Aendfire-array-features) — Propagation and Antenna, chapter 2 - [Antenna arrays and two-element expression](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Adefine-array-two-isotropic) — Propagation and Antenna, chapter 2 ## Related in other subjects - [Constitutive relation for 2D isotropic material](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Aconstitutive-2d-isotropic) — Computational Techniques in Civil Engineering, chapter 3