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

# Two isotropic sources, same and opposite phase

Propagation and Antenna · Chapter 2: Antenna Parameters and Arrays

- Asked in: 2081 Chaitra, 4+4 marks
- Video: [Two Isotropic Source Array (Basics, Electric Field & Radiation Direction) Explained](https://www.youtube.com/watch?v=GiwqMQikybQ) — Engineering Funda
- 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-same-and-opposite-phase
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-2

## Question

Derive the expressions for the total field in case of two isotropic point sources with (i) an equal amplitude and opposite phase (ii) an equal amplitude and same phase.

## Short answer

For equal amplitude same phase: total field E = 2E0 cos[(βd/2)cosφ]; d=λ/2 gives broadside pattern (max at 90°, HPBW 60°). For opposite phase: E = 2E0 sin[(βd/2)cosφ]; d=λ/2 gives end‑fire pattern (max at 0°, 180°, HPBW 120° per lobe).

## 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.

#### (i) 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
```

#### (ii) Equal amplitude, same phase
Currents equal in magnitude and **in phase**. By superposition:

```text
E = E0 e^(−jψ/2) + E0 e^(+jψ/2)
  = 2E0 cos(ψ/2)
E = 2E0 cos[(βd/2) cosφ]
```

For **d = λ/2** (βd/2 = π/2), normalized: E_n = cos[(π/2) cosφ].
- **Maxima:** (π/2)cosφ = 0 → cosφ = 0 → φ = 90°, 270° (broadside).
- **Minima:** (π/2)cosφ = ±π/2 → cosφ = ±1 → φ = 0°, 180°.
- **Half power:** cos[(π/2)cosφ] = 1/√2 → cosφ = ±1/2 → φ = 60°, 120° → HPBW = 60°.

```text
        φ=90°
        .-.
       (   )
  (1)---O---(2)   φ=0°  (null on axis)
       (   )
        '-'
       φ=270°
  Broadside figure-of-eight
```

| Case | Field | Max (d = λ/2) | Type |
|---|---|---|---|
| Same phase | 2E0 cos[(βd/2)cosφ] | 90°, 270° | Broadside |
| Opposite phase | 2E0 sin[(βd/2)cosφ] | 0°, 180° | End-fire |

## Check yourself

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

1. What is the total field of two equal in-phase sources?
   A. 2E0 sin[(βd/2)cosφ]
   B. 2E0 cos[(βd/2)cosφ]
   C. E0 cos[βd cosφ]
   D. 2E0 cos(βd)
   Answer: B. Adding e^(-jψ/2) and e^(jψ/2) gives 2cos(ψ/2). (Key point: Same phase field)

2. What is the field of two equal sources fed 180° out of phase?
   A. 2E0 cos[(βd/2)cosφ]
   B. E0 sin[βd]
   C. |E| = 2E0 sin[(βd/2)cosφ]
   D. 0 in all directions
   Answer: C. The phasors subtract, giving 2jE0 sin(ψ/2). (Key point: Opposite phase field)

3. For d = λ/2 in-phase, where are maxima and half-power angles?
   A. Maxima at 90°, 270° (broadside); half-power at 60° and 120°, HPBW = 60°
   B. Maxima at 0°, 180°; HPBW = 120°
   C. Maxima at 90°; HPBW = 120°
   D. Maxima at 60°; HPBW = 30°
   Answer: A. cos[(π/2)cosφ] = 0.707 at cosφ = ±1/2. (Key point: In-phase pattern d = λ/2)

4. Which pairing is correct?
   A. Same phase: end-fire; opposite phase: broadside
   B. Both broadside
   C. Both end-fire
   D. Same phase: broadside; opposite phase: end-fire
   Answer: D. Maxima are at 90° for in-phase and at 0°/180° for out-of-phase. (Key point: Array type)

## Close variants set in other exams

- [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×

## 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×