Topic: Microwave Engineering
A monostatic pulsed radar operating at 30GHz has a transmitter with 2KW O/P power and an antenna with 30dB gain. Minimum detectable signal in the receiver is -100dBm. Determine the maximum range of the radar, if it is required to detect a target having radar cross section of 10sq.m (consider $log_{10}4\pi=1.1$). Assume EM wave propagate under ideal conditions.
- A. 10 km
- B. 21.5 km
- C. 56 km
- D. 100 km
Answer: A. 10 km
Explanation: Operating frequency $f = 30$ GHz gives wavelength $\lambda = c / f = 3 \times 10^{10} / 30 \times 10^9 = 0.01$ m. $P_t = 2000$ W; Gain $G = 30$ dB $= 10^3$; $P_{min} = -100$ dBm $= 10^{-13}$ W; $\sigma = 10 \text{ m}^2$. The maximum radar range is given by $R_{max}^4 = \frac{P_t G^2 \lambda^2 \sigma}{(4\pi)^3 P_{min}}$. Given $\log_{10}(4\pi) = 1.1 \implies 4\pi = 10^{1.1}$, so $(4\pi)^3 = 10^{3.3}$. Substituting values: $R_{max}^4 = \frac{2000 \times (10^3)^2 \times (0.01)^2 \times 10}{10^{3.3} \times 10^{-13}} = \frac{2 \times 10^6}{10^{-9.7}} = 2 \times 10^{15.7} = 2 \times 10^{0.7} \times 10^{15} \approx 2 \times 5.011 \times 10^{15} \approx 10^{16}$. Therefore, $R_{max} = (10^{16})^{1/4} = 10000$ meters = 10 km.