Question

What is the relation between effective length and Effective aperture of antenna?

(a) \(A_e = \frac{dL^2\eta}{4R_{rad}}\)

(b) \(A_e = \frac{dL^2}{4\eta R_{rad}}\)

(c) \(A_e = \frac{dL^2 R_{rad}}{4\eta}\)

(d) \(A_e = \frac{dL^2 \eta ^2}{4R_{rad}}\)

The question was asked by my school principal while I was bunking the class.

I want to ask this question from Effective Aperture topic in division Antenna Parameters of Antennas

Answer 1

Right choice is (a) \(A_e = \frac{dL^2\eta}{4R_{rad}}\)

The best I can explain: Maximum effective aperture is ratio of maximum power received to the average power density. \(A_e=\frac{P_{Rmax}}{P_{avg}}\)

The received power is maximum when load equal to complex conjugate of network resistance.

⇨ \(I_{total}=\frac{V_{oc}}{2R_{rad}}\)

⇨ \(P_{Rmax}=I_{rms}^2 R_{rad}= (\frac{I_{total}}{\sqrt 2})^2 R_{rad}=\frac{V_{oc}^2}{8R_{rad}}=\frac{\mid E_\theta\mid^2 dL^2}{8R_{rad}} and P_{avg}=\frac{\mid E_\theta\mid^2}{2\eta} \)

\(A_e=\frac{P_{Rmax}}{P_{avg}}=\frac{dL^2\eta}{4R_{rad}}\)