Topic: Electromagnetics and Antennas
Two communication antennas A and B, one operating at 300 MHz and other at 3 GHz respectively and having same gain, are illuminated with identical flux density of $-100~dBW/m^2$. What is the relation between the received powers $(P_A:P_B)$?
- A. 1:10
- B. 10:1
- C. 1:100
- D. 100:1
Answer: D. 100:1
Explanation: The received power $P_r$ by an antenna is the product of incident flux density and its effective aperture area $A_e$. The effective aperture is related to gain by $A_e = \frac{G\lambda^2}{4\pi}$. Since both gain $G$ and incident flux density are identical, the received power is directly proportional to the square of the wavelength, $P_r \propto \lambda^2$. For Antenna A, $\lambda_A = c / (300 \times 10^6) = 1$ meter. For Antenna B, $\lambda_B = c / (3 \times 10^9) = 0.1$ meter. The ratio of received powers is $\frac{P_A}{P_B} = (\frac{\lambda_A}{\lambda_B})^2 = (\frac{1}{0.1})^2 = 100:1$.