- 2072 Magh · 6 marks
Gain, efficiency and beamwidth
Short answer
Gain is G = 4πU/P_in = ηD. Efficiency is η = P_rad/P_in = Rr/(Rr+RL). HPBW is the angle between half-power directions; for a short dipole sin²θ = 1/2 gives θ = 45° and 135°, so HPBW = 90°, while a λ/2 dipole has about 78°.
Describe antenna gain, antenna efficiency and beam width of antenna with mathematical derivations if necessary.
Antenna gain
Gain is the ratio of the radiation intensity in a given direction to the radiation intensity that would be obtained if the input power were radiated isotropically:
G(θ,φ) = 4π U(θ,φ) / P_in
P_rad = η P_in → G = η D
G(dBi) = 10 log10 G
It includes losses, so G ≤ D. Example: λ/2 dipole, D = 1.64; if η = 0.97, G = 1.59 (2.0 dBi). For aperture antennas G = 4πAe/λ².
Antenna efficiency
Radiation efficiency is the ratio of power radiated to power accepted at the terminals:
η = P_rad / P_in = Rr / (Rr + RL)
Rr = radiation resistance, RL = ohmic/dielectric loss resistance. Total efficiency also includes mismatch: e0 = (1 − |Γ|²) · η.
Example: Rr = 73 Ω, RL = 2 Ω → η = 73/75 = 97.3 %.
Beamwidth
HPBW is the angle between the two directions in the main lobe where power falls to half (field to 0.707) of maximum. FNBW is the angle between the first nulls.
Derivation for a short dipole (E ∝ sinθ, power ∝ sin²θ):
sin²θ = 1/2 → sinθ = 1/√2
θ1 = 45°, θ2 = 135°
HPBW = 135° − 45° = 90°
For a λ/2 dipole, E = cos[(π/2)cosθ]/sinθ = 0.707 gives θ ≈ 51° and 129° → HPBW ≈ 78°.
max
.---.
0.707/ | \0.707
/<HPBW>\
/ <-FNBW->\
A narrower beam means higher directivity: D ≈ 41253/(θE° θH°).
✓ Answer checked by Claude Opus 5.5 (high effort)
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