Question

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 1

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