---
title: "Opposite-phase two-source array maxima and minima"
subject: "Propagation and Antenna"
chapter: 2
key: "semester_6/propagation_and_antenna:two-sources-opposite-maxima-minima"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/two-sources-opposite-maxima-minima
site: BCT Engineering
---

# Opposite-phase two-source array maxima and minima

Propagation and Antenna · Chapter 2: Antenna Parameters and Arrays

- Asked in: 2080 Chaitra, 6+2 marks; 2076 Baisakh, 6+3 marks (asked 2 times)
- Video: [6 Arrays Of 2 Isotropic Point Sources Case 2 Explained | Module 3 7th Sem Microwave & Antenna | ECE](https://www.youtube.com/watch?v=R6BJ97GsEoA) — VTU Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/two-sources-opposite-maxima-minima
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-2

## Question

Derive an expression for the total field in case of two isotropic point sources with same amplitude and out of phase (currents equal in magnitude but opposite in phase). Also show the condition of maxima and minima direction of intensity and draw the radiation pattern.

## Short answer

Total field magnitude: |E| = 2E₀ sin((βd/2) cosφ). Maxima when (βd/2) cosφ = ±(2n+1)π/2, minima when (βd/2) cosφ = ±nπ. For d=λ/2, maxima at0°,180° and nulls at90°,270°; pattern is end-fire figure-of-eight.

## Answer

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.

```text
                    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

```text
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.

#### Total field
Currents equal in magnitude but **opposite in phase** (180°). By superposition:

```text
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.

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

#### General conditions for maxima and minima
- **Maxima** when sin[(βd/2)cosφ] = ±1:

```text
(βd/2) cosφ = ±(2n + 1) π/2
cosφ_max = ±(2n + 1) λ / (2d),   n = 0, 1, 2 ...
```

- **Minima** when sin[(βd/2)cosφ] = 0:

```text
(βd/2) cosφ = ±n π
cosφ_min = ±n λ / d,   n = 0, 1, 2 ...
```

n = 0 always gives a null at φ = 90° (broadside), since the two equal and opposite fields cancel on the perpendicular bisector.

For d = λ/2 this gives maxima at 0° and 180° and nulls at 90° and 270°, as shown in the pattern.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Two isotropic sources are d apart on the array axis. For a far point at angle φ from the axis, what is the phase difference due to path?
   A. βd sinφ
   B. βd cosφ
   C. (βd/2) cosφ only, with no factor for the full spacing
   D. 2βd cosφ
   Answer: B. The path difference between the two waves is d cosφ, so the phase difference is ψ = βd cosφ. (Key point: Phase difference from path)

2. Why does the total field of two equal sources fed 180° out of phase come out as 2E0 sin(ψ/2) rather than 2E0 cos(ψ/2)?
   A. Because the amplitudes of the two sources are different
   B. Because the path difference becomes zero
   C. The 180° reversal gives a minus sign, so the sum is E0(e^(jψ/2) - e^(-jψ/2)) = 2jE0 sin(ψ/2)
   D. Because the sources become non-isotropic
   Answer: C. Opposite currents subtract the two phasors, and the difference of exponentials is a sine. (Key point: Opposite phase gives sine)

3. What is the general condition for maxima of the opposite-phase pair?
   A. (βd/2)cosφ = ±nπ, i.e. cosφ = ±nλ/d
   B. (βd/2)cosφ = 0
   C. (βd/2)cosφ = ±(2n+1)π/2, i.e. cosφ = ±(2n+1)λ/(2d)
   D. cosφ = ±(2n+1)λ/d
   Answer: C. Maxima occur where sin[(βd/2)cosφ] = ±1, which needs an odd multiple of π/2. (Key point: General condition for maxima)

4. What is the general condition for minima of the opposite-phase pair?
   A. sin[(βd/2)cosφ] = 0, i.e. cosφ = ±nλ/d
   B. sin[(βd/2)cosφ] = 1, i.e. cosφ = ±λ/(2d)
   C. cos[(βd/2)cosφ] = 0
   D. cosφ = ±(2n+1)λ/(2d)
   Answer: A. Minima need the sine to be zero, so (βd/2)cosφ = ±nπ. (Key point: General condition for minima)

## Related in Propagation and Antenna

- [Two isotropic sources, equal amplitude, opposite phase](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Atwo-sources-opposite-phase) — Propagation and Antenna, chapter 2, asked 3×
- [Two isotropic sources, same and opposite phase](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Atwo-sources-same-and-opposite-phase) — Propagation and Antenna, chapter 2, asked 3×
- [Pattern lobes and 180° out-of-phase array](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Alobes-importance-opposite-phase-array) — Propagation and Antenna, chapter 2
- [Pattern multiplication, quadrature-phase sources](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Apattern-multiplication-quadrature) — Propagation and Antenna, chapter 2