Two-element array and end fire pattern

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.

               P (far)
              /
             / θ
  1 ●-------O-------● 2  -> axis
    path difference = d cosθ

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

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:

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.

 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|.