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The physical aperture of an isotropic radiator is _______(a) \(\frac{4\pi \eta}{\lambda^2}\)(b) \(\frac{4\pi}{\lambda^2 \eta}\)(c) \(\frac{\lambda^2}{4\pi \eta}\)(d) \(\frac{\lambda^2\eta}{4\pi}\)The question was asked in my homework.I would like to ask this question from Effective Aperture topic in division Antenna Parameters of Antennas

Answer»

Right choice is (c) \(\frac{\lambda^2}{4\pi \eta}\)

To explain: For ISOTROPIC radiator, DIRECTIVITY is 1. So the effective aperture is GIVEN by \(A_{EM}=D \frac{\lambda^2}{4\pi} = \frac{\lambda^2}{4\pi}\)

Then physical aperture =\(\frac{Effective\, aperture}{Aperture\, EFFICIENCY}=\frac{\lambda^2}{4\pi \eta}\)


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