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For a circular aperture the FNBW is ______(a) 140λ/D(b) 70λ/D(c) 140D/λ(d) 70D/λThis question was posed to me by my college professor while I was bunking the class.My doubt is from Aperture Antenna topic in division Aperture Antenna of Antennas

Answer»

Right answer is (a) 140λ/D

The EXPLANATION is: The AREA the POWER is radiated is given by BEAM-width. The beam-width between the FIRST nulls is the FNBW. For circular aperture the FNBW is given by 140λ/D. The half-power beam width is given by 70λ/D.



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