GATE 2016 EC – Question 65
The far-zone power density radiated by a helical antenna is approximated as:
$$\vec{W}_{rad} = \vec{W}_{average} \approx \hat{a}_r C_0 \frac{1}{r^2}\cos^4\theta$$
The radiated power density is symmetrical with respect to $\phi$ and exists only in the upper hemisphere: $0 \leq \theta \leq \frac{\pi}{2}$; $0 \leq \phi \leq 2\pi$; $C_0$ is a constant. The power radiated by the antenna (in watts) and the maximum directivity of the antenna, respectively, are
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Correct answer: (B) $1.256C_0$, 10dB
Explanation
The radiated power is the integral of the power density over a sphere: $P_{rad} = \int_0^{2\pi}\int_0^{\pi/2} C_0\cos^4\theta \sin\theta\,d\theta\,d\phi = 2\pi C_0 \times \frac{1}{5} = 1.256\,C_0$. The maximum radiation intensity is $U_{max} = C_0$, so the maximum directivity is $D = \frac{4\pi U_{max}}{P_{rad}} = \frac{4\pi C_0}{2\pi C_0/5} = 10$, which is 10 dB.