Question

Half-power Beam width in E-plane for a rectangular aperture antenna of a×b  is given by ____

(a) 0.886λ/b

(b) 0.443λ/b

(c) 0.5λ/b

(d) λ/b

Answer 1

Right option is (a) 0.886λ/b

The explanation: By equating the field in E-plane to half power point

\(\frac{sin⁡(0.5kbsin \theta)}{0.5kbsin \theta} = \frac{1}{\sqrt 2}  => \theta = arcsin⁡(\frac{0.443

\lambda}{b})\)

Now HPBW = 2 arcsin⁡\((\frac{0.443\lambda}{b})\)≈0.886λ/b.