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Pattern multiplication for two short dipoles

Short answer

Pattern multiplication states total field pattern equals element pattern multiplied by array factor. For two short dipoles perpendicular to array axis with d=λ/2: (a) α=0 gives broadside figure-8 in plane perpendicular to dipoles and four-lobed pattern in dipole plane with lobes near 57° and 123°. (b) α=π gives end-fire figure-8 perpendicular to dipoles and sharp two-lobed end-fire pattern along axis in dipole plane.

State the principle of pattern multiplication. Use the principle to obtain a wave pattern for array of two short dipoles for following cases where (d) = dipole separation and (α) = current phase difference. a) Dipoles aligned perpendicular to the array axis with d = λ/2, α = 0 b) Dipoles aligned perpendicular to the array axis with d = λ/2, α = π

Principle of pattern multiplication

The total field pattern of an array of identical, similarly oriented elements equals the element (unit) pattern multiplied by the array factor (the pattern of the same array made of isotropic point sources). The total phase pattern is the sum of the element phase pattern and the array phase pattern.

E_total(θ,φ) = E_element(θ,φ) × AF(θ,φ)

It lets us sketch complex patterns quickly without full integration.

Setup for both cases: two short dipoles, spacing d = λ/2 (βd = π), both perpendicular to the array axis. θ is measured from the array axis. Two planes are considered:

  • Plane ⊥ to dipoles: short dipole pattern is a circle (omnidirectional).
  • Plane containing dipoles and axis: short dipole pattern = |sin(angle from dipole)| = |cosθ|.

Array factor of two isotropic sources: AF = |cos[(βd cosθ + α)/2]|.

a) d = λ/2, α = 0

AF = |cos(π/2 · cosθ)|   max at θ = 90°, nulls at 0°,180°
Plane ⊥ dipoles : E = AF  -> broadside figure-8
Plane of dipoles: E = |cosθ| · |cos(π/2 · cosθ)|
  θ=0°: 0   θ=30°: 0.18   θ=45°: 0.31
  θ=60°: 0.35   θ=75°: 0.24   θ=90°: 0

In the plane of the dipoles, the AF maximum (θ = 90°) falls on the dipole null, so the result is a four-lobed pattern with lobes near θ ≈ 57°, 123°, 237°, 303°.

 element        AF (broadside)      total (dipole plane)
   ( )            ___                  \   /
 --(+)--   x   --1-2--       =         ->1-2<-
   ( )            ‾‾‾                  /   \
 max on axis   max ⊥ axis           4 lobes

In the plane perpendicular to the dipoles the pattern is the broadside figure-8 (max at θ = 90°).

b) d = λ/2, α = π

AF = |sin(π/2 · cosθ)|   max at θ = 0°,180°, null at 90°
Plane ⊥ dipoles : E = AF  -> end-fire figure-8 on axis
Plane of dipoles: E = |cosθ| · |sin(π/2 · cosθ)|
  θ=0°: 1   θ=30°: 0.85   θ=45°: 0.63
  θ=60°: 0.35   θ=75°: 0.10   θ=90°: 0

Here both factors are maximum along the axis, so the total is a two-lobed end-fire pattern, narrower than the AF alone.

 element     AF (end fire)      total
   ( )                         
 --(+)--  x  <=1-2=>    =    <(1-2)>  narrow lobes
   ( )                         along the axis
CasePlane ⊥ dipolesPlane of dipoles
α = 0broadside figure-8four lobes
α = πend-fire figure-8sharp end-fire two lobes

✓ Answer checked by Claude Opus 5.5 (high effort)

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