Question 7 of 8: Radar Return from a Flat Plate at 10 km

Question 7: Radar Return from a Flat Plate at 10 km (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A monostatic radar illuminating a flat circular plate at normal incidence.

Find. (i) the received echo power, and (ii) whether the approximations used make that figure high or low.

Approach. Use the aid to model the beam as power spread uniformly over a cone of half-angle $\theta_0$; compute the power density at the plate, the power the plate intercepts, the density it sends back through its own diffraction cone, and finally the power the antenna aperture collects. Then compare with the exact radar equation to settle part (ii).

1. Beam spread at the target. $\lambda = c/f = 0.03\ \text{m}$, so $$\theta_0 = \frac{1.2\lambda}{d_a} = \frac{1.2(0.03)}{1.0} = 0.036\ \text{rad} = 2.06^\circ ,$$ and at $R = 10\ \text{km}$ the illuminated spot has radius $\theta_0 R = 360\ \text{m}$ and area $A_{\text{spot}} = \pi(360)^2 = 4.072\times 10^{5}\ \text{m}^2$. The plate is far inside that spot, and well beyond the aperture's far-field distance $2d_a^{2}/\lambda = 66.7\ \text{m}$.
2. Power density on the plate. $$S_1 = \frac{P_t}{A_{\text{spot}}} = \frac{1\times 10^{6}}{4.072\times 10^{5}} = 2.456\ \text{W/m}^2 ,$$ so the plate intercepts $P_{\text{int}} = S_1 A_p = 2.456 \times 0.7854 = 1.929\ \text{W}$.
3. Re-radiation by the plate. A flat conducting disc at normal incidence behaves as a uniformly illuminated aperture of its own, so it returns the intercepted power into a cone of half-angle $1.2\lambda/d_p = 0.036\ \text{rad}$ — the same as the antenna, because the diameters are equal. Back at the radar the return spot again has area $4.072\times 10^{5}\ \text{m}^2$, giving $$S_2 = \frac{1.929}{4.072\times 10^{5}} = 4.738\times 10^{-6}\ \text{W/m}^2 .$$
4. Power collected by the antenna. $$\boxed{\ P_r = S_2 A_a = (4.738\times 10^{-6})(0.7854) = 3.72\ \mu\text{W} = -24.3\ \text{dBm}\ }$$ which is the answer to part (i) on the model the aid invites.
5. Cross-check with the exact radar equation. For a uniformly illuminated aperture $G = 4\pi A_a/\lambda^{2} = 10\,966$ (40.4 dBi), and the normal-incidence radar cross-section of a flat plate is $\sigma = 4\pi A_p^{2}/\lambda^{2} = 8613\ \text{m}^2$. Then $$P_r = \frac{P_t G^{2}\lambda^{2}\sigma}{(4\pi)^{3}R^{4}} = 47.0\ \mu\text{W} = -13.3\ \text{dBm}.$$
6. Answer part (ii). The cone model is smaller — low by a factor of 12.6 (11 dB). Spreading the power uniformly out to the first null implies an effective gain $4/\theta_0^{2} = 3086$ instead of the true $10\,966$, a shortfall of 3.55 on transmit and the same again on the return path.

Part (ii) rewards saying which way each approximation pushes. Uniform-cone spreading dominates and makes the estimate low, because a real aperture concentrates far more power on boresight than a flat-topped cone does. Working the other way, and much smaller, are the idealisations that the plate is a lossless perfect conductor exactly normal to the beam, that polarisation is perfectly matched and that the receiver path is lossless; each of those makes any estimate slightly optimistic. The far-field conditions are comfortably met at both ends, so no correction is needed there. On balance the 3.72 µW figure should be quoted as a lower bound, with 47 µW as the physical-optics value.