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Operation of infinitesimal dipole
Short answer
An infinitesimal dipole is a very short wire l≪λ carrying uniform current. Its far fields are Eθ≈jηkI0l sinθ e^(−jkr)/(4πr) and Hφ=Eθ/η. Radiated power is η(π/3)(I0l/λ)^2, radiation resistance Rr=80π²(l/λ)², directivity is 1.5, and the pattern is a doughnut shape.
Define an antenna. Explain the operation of infinitesimal dipole with the help of mathematical relations and the field pattern.
An antenna is a transducer that converts guided electromagnetic energy on a transmission line into free-space radiated waves (and vice versa).
Infinitesimal (Hertzian) dipole
An infinitesimal dipole is a very short straight wire of length l ≪ λ (l ≤ λ/50) carrying a uniform current I(z) = I0 along its length, placed at the origin along the z-axis. Real antennas can be built as a sum of such elements.
z
| P(r, θ, φ)
| /
┌┼┐/ r
l │I0│ θ
└┼┘
|
+----------- y
/
x
Vector potential
A = âz μ I0 l e^(−jkr) / (4π r)
Ar = Az cosθ, Aθ = −Az sinθ
Field components (H = ∇×A/μ, E from Maxwell)
Hφ = j k I0 l sinθ e^(−jkr)/(4πr) · [1 + 1/(jkr)]
Er = η I0 l cosθ e^(−jkr)/(2πr²) · [1 + 1/(jkr)]
Eθ = jηkI0 l sinθ e^(−jkr)/(4πr)
· [1 + 1/(jkr) − 1/(kr)²]
Hr = Hθ = Eφ = 0
Field regions:
- kr ≪ 1 (near/reactive field): 1/r² and 1/r³ terms dominate; energy is stored, not radiated.
- kr ≫ 1 (far field): only 1/r terms remain:
Eθ ≈ jη k I0 l sinθ e^(−jkr) / (4πr)
Hφ ≈ j k I0 l sinθ e^(−jkr) / (4πr)
Er ≈ 0, Eθ / Hφ = η = 120π Ω
The far fields are transverse, in phase, and form a TEM wave.
Power, radiation resistance and directivity
P_rad = η (π/3) |I0 l / λ|²
Rr = 2 P_rad / |I0|² = 80π² (l/λ)² Ω
U(θ) ∝ sin²θ → D = 1.5 (1.76 dBi), HPBW = 90°
Example: l = λ/50 → Rr = 80π²/2500 ≈ 0.316 Ω (very small, so efficiency is low).
Field pattern
The far-field pattern is |Eθ| ∝ sinθ:
- E-plane (vertical plane): figure-of-eight, maximum at θ = 90°, nulls along the axis (θ = 0°, 180°).
- H-plane (horizontal plane): circle – omnidirectional.
- 3-D: doughnut shape.
E-plane (sinθ) H-plane
z y
| .---.
.-. | .-. / \
( ) | ( ) | o | x
'-' | '-' \ /
| '---'
null on axis circle
✓ Answer checked by Claude Opus 5.5 (high effort)
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