InterviewSolution
Saved Bookmarks
| 1. |
The first null beam width in the E-plane of a rectangular aperture of a×b is given by _______________(a) 2sin^-1(λ/b)(b) sin^-1(λ/a)(c) 2sec^-1(λ/b)(d) 2cos^-1(λ/a)The question was posed to me in an online quiz.My question is taken from Aperture Antenna in chapter Aperture Antenna of Antennas |
|
Answer» CORRECT answer is (a) 2sin^-1(λ/b) To explain I would say: The area the power is radiated is GIVEN by Beam-width. The beam-width between the FIRST nulls is the FNBW. \(\frac{KB}{2}\) sinθ=nπ θ= sin^-1\((\frac{n\LAMBDA}{b})\) Therefore, the FNBW in E-plane is given by FNBW=2 θ = sin^-1\((\frac{\lambda}{b})\). |
|