Need for arrays and two-element patterns

Why an array instead of a longer single antenna

• Making a single wire longer than about 1.25λ does not keep increasing gain: current phase reversals appear and the beam splits into many lobes (e.g. 3λ/2 dipole has lower directivity than a λ dipole).
• One element gives a fixed, broad pattern; its beam cannot be steered without moving it.
• An array of elements fed with controlled amplitude and phase gives high directivity, narrow beam, low side lobes, nulls towards interference and electronic beam steering, while each element stays a convenient size (e.g. λ/2).
• Power is shared among elements, and failure of one element only slightly degrades performance.

Two isotropic radiators

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.

i) Equal amplitude and phase

Currents equal in magnitude and in phase. By superposition:

E = E0 e^(−jψ/2) + E0 e^(+jψ/2)
  = 2E0 cos(ψ/2)
E = 2E0 cos[(βd/2) cosφ]

For d = λ/2 (βd/2 = π/2), normalized: E_n = cos[(π/2) cosφ].

• Maxima: (π/2)cosφ = 0 → cosφ = 0 → φ = 90°, 270° (broadside).
• Minima: (π/2)cosφ = ±π/2 → cosφ = ±1 → φ = 0°, 180°.
• Half power: cos[(π/2)cosφ] = 1/√2 → cosφ = ±1/2 → φ = 60°, 120° → HPBW = 60°.

        φ=90°
        .-.
       (   )
  (1)---O---(2)   φ=0°  (null on axis)
       (   )
        '-'
       φ=270°
  Broadside figure-of-eight

ii) Equal amplitude and opposite phase

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