--- title: "Pattern lobes and 180° out-of-phase array" subject: "Propagation and Antenna" chapter: 2 key: "semester_6/propagation_and_antenna:lobes-importance-opposite-phase-array" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/lobes-importance-opposite-phase-array site: BCT Engineering --- # Pattern lobes and 180° out-of-phase array Propagation and Antenna · Chapter 2: Antenna Parameters and Arrays - Asked in: 2071 Magh, 3+3 marks - Video: [Two Isotropic Source Array (Basics, Electric Field & Radiation Direction) Explained](https://www.youtube.com/watch?v=GiwqMQikybQ) — 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/lobes-importance-opposite-phase-array - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-2 ## Question Why are antenna radiation pattern lobes important during antenna design? Derive the relation for the field intensity of linear array of 2 isotropic radiators with equal potential and 180° out of phase. ## Short answer Main lobe sets coverage/directivity; side/back lobes waste power, lower gain, pick up interference; nulls suppress unwanted signals. For two isotropic sources equal amplitude 180° out of phase: E = 2E0 sin[(βd/2)cosφ]; d=λ/2 gives end‑fire pattern, max at 0°, 180°, null at 90°. ## Answer #### Importance of radiation pattern lobes in design - The **main lobe** sets the direction and width of coverage (HPBW) and hence directivity and gain. - **Side lobes and back lobe** waste power in unwanted directions, lower gain, and pick up **interference and noise** (e.g. ground noise in satellite receivers). - Low side lobes are needed in radar to avoid false targets and in communication to meet interference limits; a good **front-to-back ratio** is needed in point-to-point links. - **Nulls** can be placed towards interfering sources. - Designers therefore control element spacing, amplitude taper (e.g. binomial, Dolph–Chebyshev) and phase to shape the lobes. #### Field of 2 isotropic radiators, equal amplitude, 180° out of phase Consider two isotropic point sources 1 and 2 spaced a distance d apart on the x-axis, with the reference (origin) at the midpoint. A far point P is in direction φ measured from the array axis. Each source alone gives field E0 at P. ```text to far point P / / φ (1)-----------O-----------(2) |<--- d/2 --->|<--- d/2 --->| path from (2) shorter by (d/2)cosφ path from (1) longer by (d/2)cosφ ``` Path difference between the two waves = d cosφ, so the phase difference due to path is ```text ψ = β d cosφ, β = 2π/λ ``` With respect to the midpoint, wave 2 leads by ψ/2 and wave 1 lags by ψ/2. Currents equal in magnitude but **opposite in phase** (180°). By superposition: ```text E = −E0 e^(−jψ/2) + E0 e^(+jψ/2) = 2jE0 sin(ψ/2) |E| = 2E0 sin[(βd/2) cosφ] ``` For **d = λ/2** (βd/2 = π/2), normalized: E_n = sin[(π/2) cosφ]. - **Maxima:** (π/2)cosφ = ±π/2 → cosφ = ±1 → φ = 0°, 180° (along the axis, end-fire). - **Minima:** (π/2)cosφ = 0 → cosφ = 0 → φ = 90°, 270°. - **Half power:** sin[(π/2)cosφ] = ±1/√2 → cosφ = ±1/2 → φ = 60°, 120°, 240°, 300° → HPBW = 120° per lobe. ```text φ=90° (null) | .--. | .--. 180°( )---(1)O(2)---( ) 0° '--' | '--' | End-fire figure-of-eight ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the main lobe determine? A. Direction and width of coverage (HPBW), hence directivity and gain B. Only the back radiation C. Only the input impedance D. Nothing about gain Answer: A. It is the intended beam. (Key point: Main lobe) 2. Why are side and back lobes undesirable? A. They increase gain B. They waste power and pick up interference and noise C. They narrow the main lobe D. They improve front-to-back ratio Answer: B. Low side lobes avoid false radar targets and interference. (Key point: Side lobes) 3. The field of two equal 180° out-of-phase sources is A. 2E0 cos[(βd/2)cosφ] B. 2E0 sin(βd) C. |E| = 2E0 sin[(βd/2)cosφ] D. E0 Answer: C. The phasors subtract. (Key point: Opposite-phase field) 4. How can nulls help in design? A. They increase side lobe level B. They make the beam wider C. They remove the need for spacing control D. They can be placed toward interfering sources Answer: D. Spacing, amplitude taper and phase shape nulls and lobes. 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