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Pattern lobes and 180° out-of-phase array

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°.

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.

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.

                    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

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:

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.
            φ=90° (null)
                |
     .--.       |       .--.
 180°(    )---(1)O(2)---(    ) 0°
     '--'       |       '--'
                |
  End-fire figure-of-eight

✓ Answer checked by Claude Opus 5.5 (high effort)

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