Question

An antenna has directivity of 100 and operates at 150 MHz. The maximum effective aperture is

• A. \(31.8 \, \text{m}^2\)
• B. \(62.4 \, \text{m}^2\)
• C. \(26.4 \, \text{m}^2\)
• D. \(13.2 \, \text{m}^2\)

Hint:

Use \( A_e = \frac{D\lambda^2}{4\pi} \) for max aperture—always compute \(\lambda\) from frequency first.

Answer 1

The Correct Option is A

Effective aperture \( A_e \) is related to directivity \( D \) and wavelength \( \lambda \) by: \[ A_e = \frac{D \cdot \lambda^2}{4\pi} \] Given: \[ D = 100, f = 150\, \text{MHz} \Rightarrow \lambda = \frac{c}{f} = \frac{3 \times 10^8}{150 \times 10^6} = 2\, \text{m} \] Now, \[ A_e = \frac{100 \cdot (2)^2}{4\pi} = \frac{400}{4\pi} = \frac{100}{\pi} \approx 31.8 \, \text{m}^2 \]