The broadside condition is

α = 0, giving the maximum at θ = 90°

α = -βd, maximum at θ = 90°

The end-fire condition is

α = -βd, giving ψ = 0 at θ = 0°

For d = λ/4 and α = -90°, the pattern is

A cardioid with maximum at 0° and null at 180°

A cardioid with maximum at 180°

2071 Bhadra · 5+6 marks

Two-element array: broadside and end fire

Short answer

General field: E = 2E0 cos[(βd cosθ + α)/2] with ψ = βd cosθ + α, α is current phase lead of element 2. Broadside: α=0, max at θ=90° (cosθ=0). End‑fire: α = −βd, max at θ=0° (cosθ=1). For d=λ/2, α=π yields end‑fire figure‑8 along axis.

Derive a relation for the field intensity for the array of two elements isotropic radiators in various conditions. Show the condition for broad side and end fire array with necessary diagrams.

An antenna array is a group of similar antennas whose fields add at a distant point. For two isotropic point sources, the total field is the element field multiplied by an array factor that depends on the spacing d and the current phase difference α.

Derivation of field intensity

Assumptions: two isotropic sources 1 and 2 on the x-axis, spacing d, equal amplitude E₀, and the current in source 2 leads that in source 1 by α. A far-field point P is at angle θ from the array axis. Take the midpoint O as the phase reference.

                 to far point P
                /
               /  θ
   1 ●--------O--------● 2   -> array axis
     <-- d/2 -><- d/2 -->
 path difference between 1 and 2 = d cosθ

Phase difference of the two waves at P:

ψ = βd cosθ + α        β = 2π/λ

Relative to O, the field of source 1 lags by ψ/2 and the field of source 2 leads by ψ/2:

E = E₀ e^(-jψ/2) + E₀ e^(+jψ/2)
  = 2E₀ cos(ψ/2)
E = 2E₀ cos[(βd cosθ + α)/2]

Normalised array factor: AF(θ) = |cos[(βd cosθ + α)/2]|.

Case 1: equal amplitude and phase (α = 0) — broadside

ψ = βd cosθ, so E = 2E₀ cos[(πd/λ) cosθ].

Maximum when cosθ = 0, i.e. θ = 90° and 270°: perpendicular to the array axis.
For d = λ/2: E = 2E₀ cos(π/2 cosθ); nulls at θ = 0° and 180°.

 d = λ/2, α = 0   (figure-8, broadside)
            90°
          .-"""-.
          \     /
 180° ----1--O--2---- 0°  (nulls on axis)
          /     \
          '-...-'
            270°

Case 2: equal amplitude, opposite phase (α = π)

E = 2E₀ cos[(βd cosθ + π)/2] = 2E₀ |sin(πd/λ · cosθ)|. For d = λ/2 the maximum is along the axis (θ = 0°, 180°) with nulls at θ = 90°: a bidirectional end-fire pattern.

Case 3: equal amplitude, phase α = −βd — end fire

ψ = βd(cosθ − 1). The maximum (ψ = 0) is at θ = 0°, so the beam is along the array axis in one direction. For d = λ/4, α = −π/2 the pattern is a cardioid:

 d = λ/4, α = -90°  (cardioid, end fire)
        .--.
   ____/    \___
  |   1  2       >  max at θ = 0°
   ‾‾‾‾\    /‾‾‾
        '--'      null at θ = 180°

Conditions summarised

Array	Phase condition	Direction of max
Broadside	α = 0 (ψ = 0 at θ = 90°)	θ = 90°, normal to axis
End fire	α = −βd (ψ = 0 at θ = 0°)	θ = 0°, along axis
General	βd cosθₘ + α = 0	cosθₘ = −α/(βd)

In general, the beam points where the path difference is exactly cancelled by the current phase, so changing α steers the beam (the basis of phased arrays).

✓ Answer checked by Claude Opus 5.5 (high effort)

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