Topic: Electromagnetics
A certain antenna with an efficiency of 95% has maximum radiation intensity of 0.5 W/sr. The directivity of the antenna fed by input power of 0.4 W.
- A. 16.53
- B. 12.2
- C. 10.36
- D. 11.31
Answer: A. 16.53
Explanation: Given Efficiency $\eta = 0.95$, Maximum radiation intensity $U_{\text{max}} = 0.5~\text{W/sr}$, Input power $P_{\text{in}} = 0.4~\text{W}$.
$\text{Gain} = \frac{4\pi U_{\text{max}}}{P_{\text{in}}} = \frac{4\pi \times 0.5}{0.4} = \frac{2\pi}{0.4} = 5\pi$.
Since $\text{Gain} = \eta \times \text{Directivity (D)}$,
$D = \frac{\text{Gain}}{\eta} = \frac{5\pi}{0.95} = 16.53$.