1.

A rectangular aperture a ×b is placed in xy-plane, The HPBW in H-plane is given by _____(a) 0.886λ/a(b) 0.443λ/a(c) 0.5λ/b(d) λ/bThe question was posed to me in semester exam.This intriguing question originated from Radiation from Rectangular Aperture in division Aperture Antenna of Antennas

Answer»

Correct option is (a) 0.886λ/a

Explanation: By equating the field in H-plane to half POWER point

\(\frac{sin⁡(0.5kasin\THETA)}{0.5kasin\theta} = \frac{1}{\sqrt{2}} => \theta =arcsin⁡(\frac{0.443\lambda}{a})\)

Now HPBW = \(2 arcsin⁡(\frac{0.443\lambda}{a}) \approx 0.886\lambda/a.\)



Discussion

No Comment Found

Related InterviewSolutions