22-Elec-B9 Electromagnetic Field, Transmission Lines, Antennas, and Radiation · May 2016

Question 7 of 8: Radar Return from a Flat Plate under the Uniform Main-Beam Model

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2016 — 07-Elec-B9, Electromagnetic Field, Transmission Lines, Antennas, and Radiation. Three hours; closed book; any non-communicating calculator permitted. Eight questions, each of equal value (20 marks); the paper states that five questions constitute a complete exam paper and that the first five appearing in the answer book are the ones marked. The only constants printed on the cover page are the free-space permittivity and permeability, and a Smith chart is bound in as page 4 for use with Question 3. All eight questions are worked here, because this set is a study resource rather than a three-hour sitting.

Reference texts. D. M. Pozar, Microwave Engineering (transmission lines, the Smith chart, waveguides, resonators); M. N. O. Sadiku, Elements of Electromagnetics (Maxwell's equations, plane waves, guided waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics (field theory and wave propagation); C. A. Balanis, Antenna Theory: Analysis and Design (short elements, arrays, radiation resistance); F. T. Ulaby, Fundamentals of Applied Electromagnetics (radar, link budgets, applied propagation).

Units and constants. The paper works in SI throughout. Where a question supplies a rounded propagation velocity (for example 3×10⁸ m/s) that value is used as given; where it does not, the exact free-space constants are used, giving $c = 2.998\times10^{8}\ \text{m/s}$ and $\eta_0 = 376.7\ \Omega$.

Question 7: Radar Return from a Flat Plate under the Uniform Main-Beam Model (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 uniformly illuminated square aperture of side $d = 1\ \text{m}$ radiating $P_t = 1\ \text{MW}$ at $\lambda = 3\ \text{cm}$; a vertical square flat plate of side $L = 10\ \text{m}$ at $R = 10\ \text{km}$, illuminated normally; and the instruction to treat the transmitted power as uniformly spread over the main beam alone.

Find. The power the radar antenna recovers from the reflection.

The two-way geometry. The transmitted power spreads over a 600 m square footprint, of which the 10 m plate captures a small fraction and re-radiates it into its own, ten times narrower, beam.

Approach. Apply the prescribed uniform-beam model twice — once outbound from the radar aperture to the plate, once inbound from the plate acting as a re-radiating aperture — taking the main beam of a uniformly illuminated aperture to run between its first nulls.

Fix the main beam of the transmitting aperture. A uniformly illuminated aperture of width $d$ has its pattern nulls at $\sin\theta = \pm\lambda/d$, so the main beam is $2\lambda/d$ wide between nulls in each principal plane. The aperture is square, so the beam is a square pyramid: $$\theta_a = \frac{2\lambda}{d} = \frac{2(0.03)}{1} = 0.0600\ \text{rad}, \qquad \Omega_a = \theta_a^{2} = 3.60\times10^{-3}\ \text{sr}.$$
Find the power density arriving at the plate. At $R = 10\ \text{km}$ the beam has spread to $$A_{\text{beam}} = \Omega_a R^{2} = (3.60\times10^{-3})(10^{4})^{2} = 3.60\times10^{5}\ \text{m}^{2},$$ a footprint 600 m square. Spreading $P_t$ uniformly over it, $$S_t = \frac{P_t}{\Omega_a R^{2}} = \frac{1\times10^{6}}{3.60\times10^{5}} = 2.778\ \text{W/m}^{2}.$$
Find how much of that the plate captures. The plate is far smaller than the footprint, so it intercepts only the fraction $L^{2}/A_{\text{beam}} = 100/360\,000$ of the transmitted power: $$P_{\text{int}} = S_t L^{2} = (2.778)(100) = 277.8\ \text{W}.$$ This is the step where the target's modest size does most of the work in the final answer.
Treat the plate as a re-radiating aperture. Illuminated normally and uniformly, the flat plate re-radiates the intercepted power as a uniformly illuminated aperture of its own size, so the same main-beam rule applies with $L$ in place of $d$: $$\theta_t = \frac{2\lambda}{L} = \frac{2(0.03)}{10} = 6.00\times10^{-3}\ \text{rad}, \qquad \Omega_t = 3.60\times10^{-5}\ \text{sr}.$$ The plate's beam is ten times narrower than the radar's, which is the physical reason a large flat plate is such a bright radar target: it concentrates its return sharply back along the line of sight.
Find the power density back at the radar. $$S_r = \frac{P_{\text{int}}}{\Omega_t R^{2}} = \frac{277.8}{(3.60\times10^{-5})(10^{4})^{2}} = \frac{277.8}{3600} = 0.07716\ \text{W/m}^{2}.$$
Collect it with the radar aperture. The uniformly illuminated aperture has effective area equal to its physical area, $A_e = d^{2} = 1\ \text{m}^{2}$, so $$P_r = S_r A_e = (0.07716)(1.00)$$ $$\boxed{P_r = 7.72\times10^{-2}\ \text{W} = 77.2\ \text{mW}}$$
Confirm the far-field condition before trusting either aperture calculation. The Rayleigh distance is $2D^{2}/\lambda$: for the radar aperture, $2(1)^{2}/0.03 = 66.7\ \text{m}$, comfortably satisfied. For the plate, $2(10)^{2}/0.03 = 6.67\ \text{km}$, and the range is 10 km — so the radar is in the plate's far field, but only by a factor of 1.5. The re-radiating-aperture model is therefore legitimate here, though it would fail for a closer or larger plate.

Check: the prescribed model is deliberately conservative — by a fixed factor. The question's uniform-cone assumption implies an effective gain of $4\pi/\Omega = \pi d^{2}/\lambda^{2}$ on each leg, whereas the true gain of a uniformly illuminated aperture is $G = 4\pi A/\lambda^{2} = 4\pi d^{2}/\lambda^{2}$ — a factor of 4 higher, on each of the two legs. Working the same problem with the exact radar equation, $P_r = P_t G \sigma A_e/[(4\pi)^{2}R^{4}]$ with $G = 1.396\times10^{4}$ (41.4 dBi) and a flat-plate radar cross-section $\sigma = 4\pi L^{4}/\lambda^{2} = 1.396\times10^{8}\ \text{m}^{2}$, returns $P_r = 1.23\ \text{W}$. The ratio is exactly 16, or 12.0 dB, and it is independent of range, power, wavelength and target size — so the two answers never diverge further than this. The 77.2 mW figure is the answer to the question as set; the 1.23 W figure is quoted only to show the size and the sign of the simplification, which is that the uniform-beam model always under-estimates the return. The paper prints no beamwidth aid, so the factor rests on reading “main beam” as the null-to-null width $2\lambda/d$; a candidate who instead takes the beam as $\lambda/d$ wide (close to the half-power width $0.886\lambda/d$) obtains an implied gain of exactly $4\pi d^{2}/\lambda^{2}$ on each leg and so recovers the 1.23 W figure. State the beamwidth convention used.

Question 7 — results
Quantity	Symbol	Value
Radar main-beam width / solid angle	$\theta_a$, $\Omega_a$	$0.0600\ \text{rad}$, $3.60\times10^{-3}\ \text{sr}$
Beam footprint at 10 km	$\Omega_a R^{2}$	$3.60\times10^{5}\ \text{m}^{2}$ (600 m square)
Power density at the target	$S_t$	$2.778\ \text{W/m}^{2}$
Power intercepted by the plate	$P_{\text{int}}$	$277.8\ \text{W}$
Plate main-beam width / solid angle	$\theta_t$, $\Omega_t$	$6.00\times10^{-3}\ \text{rad}$, $3.60\times10^{-5}\ \text{sr}$
Power density back at the radar	$S_r$	$0.07716\ \text{W/m}^{2}$
Power intercepted by the radar antenna	$P_r$	$\mathbf{77.2\ \text{mW}}$
Exact radar-equation cross-check	—	$1.23\ \text{W}$ (16×, i.e. 12.0 dB higher)

Topic. Two-way radar link budget built from aperture geometry rather than from the radar equation.

Key relations. Uniform aperture: nulls at $\sin\theta = \lambda/d$, so the main beam is $2\lambda/d$ wide and $\Omega \approx (2\lambda/d)^{2}$ for a square aperture; $S = P/(\Omega R^{2})$; $A_e = A$ for uniform illumination; exact forms $G = 4\pi A/\lambda^{2}$ and flat-plate $\sigma = 4\pi A_{\text{plate}}^{2}/\lambda^{2}$; radar equation $P_r = P_tG\sigma A_e/[(4\pi)^{2}R^{4}]$.

Why this works. The uniform-beam model replaces the true pattern by a flat-topped beam of the same null-to-null extent, which is a fair caricature: it keeps the essential $1/R^{2}$ spreading on each leg and the essential dependence on aperture size, and it makes the two-way budget a chain of four elementary steps that can be done without the radar equation at all. What it loses is the concentration of real aperture patterns, where the peak is well above the null-to-null average — hence the fixed 12 dB shortfall. Recognising that the discrepancy is a constant, not a function of the scenario, is what makes the simplification safe to use and easy to correct.

Common pitfall. Applying the spreading loss only once and forgetting that the return leg spreads again, which overstates the answer enormously. Two others: taking the plate to scatter isotropically (which understates the return by the plate's own directivity, about 3.5×10⁵ in this model and 1.4×10⁶ for the true aperture gain), and mixing beamwidth conventions between the two legs (null-to-null $2\lambda/d$ on one, half-power $0.886\lambda/d$ on the other), which shifts the answer by a factor of about five without any stated reason.

Source. Ulaby, Fundamentals of Applied Electromagnetics, Ch. 9 (antenna aperture, gain and the radar equation); Balanis, Antenna Theory, §12.5 (aperture antennas and radar cross-section).

← Question 6 Next: Question 8 →