---
title: "Two isotropic sources, equal amplitude, opposite phase"
subject: "Propagation and Antenna"
chapter: 2
key: "semester_6/propagation_and_antenna:two-sources-opposite-phase"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/two-sources-opposite-phase
site: BCT Engineering
---

# Two isotropic sources, equal amplitude, opposite phase

Propagation and Antenna · Chapter 2: Antenna Parameters and Arrays

- Asked in: 2079 Chaitra, 8 marks; 2075 Baisakh, 6 marks (asked 2 times)
- Video: [Case II :Two Isotropic Point Sources of Same Amplitude and  Opposite Phase](https://www.youtube.com/watch?v=mD7PdD7RhJE) — Radiation and Microwave Techniques
- 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-phase
- 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 equal amplitude and opposite phase.

## Short answer

For two isotropic point sources with equal amplitude and opposite phase, total field magnitude is |E| = 2E₀ sin((βd/2) cosφ) where φ is angle from array axis. For d = λ/2, pattern is end-fire figure-of-eight with maxima at 0° and 180°, nulls at 90° and 270°.

## 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.

#### Field for equal amplitude, opposite phase
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 (any d):
- Maxima: (βd/2)cosφ = ±(2n + 1)π/2 → cosφ = ±(2n + 1)λ/(2d)
- Minima: (βd/2)cosφ = ±nπ → cosφ = ±nλ/d

So the opposite-phase pair is a simple **end-fire array**.

## 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 cosφ
   B. βd sinφ
   C. (βd/2) cosφ only, with no factor for the full spacing
   D. 2βd cosφ
   Answer: A. 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. The 180° reversal gives a minus sign, so the sum is E0(e^(jψ/2) - e^(-jψ/2)) = 2jE0 sin(ψ/2)
   C. Because the path difference becomes zero
   D. Because the sources become non-isotropic
   Answer: B. Opposite currents subtract the two phasors, and the difference of exponentials is a sine. (Key point: Opposite phase gives sine)

3. For d = λ/2 and opposite phase, where are the maxima and the nulls of E_n = sin[(π/2)cosφ]?
   A. Maxima at 90° and 270°; nulls at 0° and 180°
   B. Maxima at 60° and 120°; nulls at 0° and 90°
   C. Maxima at φ = 0° and 180° (along the axis); nulls at 90° and 270°
   D. Maxima at 45° and 135°; nulls at 90° only
   Answer: C. sin is 1 when cosφ = ±1 and 0 when cosφ = 0, so the pattern is end-fire with a broadside null. (Key point: Maxima and nulls for d = λ/2)

4. Why is there always a null at φ = 90° for two equal, opposite-phase sources?
   A. The sources do not radiate sideways
   B. The path difference there is λ/2
   C. The fields add to twice the single value
   D. The path lengths are equal there, so the equal and opposite fields cancel
   Answer: D. On the perpendicular bisector the paths are equal, so the 180° current difference makes the fields cancel. (Key point: Broadside null)

## Close variants set in other exams

- [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×

## Related in Propagation and Antenna

- [Opposite-phase two-source array maxima and minima](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Atwo-sources-opposite-maxima-minima) — Propagation and Antenna, chapter 2, asked 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