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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»

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.

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

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

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


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