Topic: Antennas and Wave Propagation
Find the maximum directivity of an antenna whose power density is given by $W_{rad} = \hat{a}_r \frac{A_0 \sin \theta}{r²}\text{ W/m}^2$ where $A_0$ is peak value of power density. $\theta$ is the usual spherical coordinate and $\hat{a}_r$ is the radial unit vector.
- A. $4/\pi$
- B. $2/\pi$
- C. $3/\pi$
- D. $5/\pi$
Answer: A. $4/\pi$
Explanation: The radiation intensity is $U = r² W_{rad} = A_0 \sin \theta$. The total radiated power is $P_{rad} = \iint U d\Omega = \int_0^{2\pi} \int_0^\pi A_0 \sin² \theta d\theta d\phi = 2\pi A_0 (\pi/2) = \pi² A_0$. The maximum directivity is $D_{max} = \frac{4\pi U_{max}}{P_{rad}} = \frac{4\pi A_0}{\pi² A_0} = \frac{4}{\pi}$.