ECE 521E — Radar Equation Quiz (20 Questions)

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Instructions to Students

This quiz contains 20 multiple-choice questions on the Radar Equation.
Questions are distributed across the first five levels of Bloom's taxonomy: Remember (L1), Understand (L2), Apply (L3), Analyze (L4), Evaluate (L5) — see the blueprint table below.
Select one option per question. Use the Check button on each question for immediate feedback during revision, or answer all questions and press Submit & Score.
The Answer Key & Explanations section is placed after the submit button (post-test review). Attempt all questions before consulting it.
Where needed: c = 3×10⁸ m/s, k = 1.38×10⁻²³ J/K, T₀ = 290 K, (4π)³ ≈ 1984.
Time guide: 45–60 minutes. Closed book unless your lecturer states otherwise.

Assessment Map

Bloom's Taxonomy Blueprint

Level	Cognitive skill	Questions	Count
L1	Remember — recall definitions, formulas, units	Q1 – Q4	4
L2	Understand — explain relationships, interpret trends	Q5 – Q8	4
L3	Apply — compute received power, range, SNR	Q9 – Q13	5
L4	Analyze — decompose budgets, compare configurations	Q14 – Q17	4
L5	Evaluate — judge design trade-offs, justify decisions	Q18 – Q20	3

Levels 1–2

Part A: Remember & Understand (Q1–Q8)

Q1L1 Remember

The radar cross-section (RCS) σ of a target is best defined as:

(a) The physical area of the target projected onto the radar line of sight
(b) 4πR² × (scattered power density at the receiver) ÷ (incident power density at the target)
(c) The effective aperture of the receiving antenna
(d) 4π × (power intercepted by the target) ÷ (power density incident on the target)

Q2L1 Remember

In the monostatic radar equation P_r = P_tG²λ²σ/[(4π)³R⁴], the quantity G represents:

(a) Antenna gain in the direction of the horizon only
(b) The peak one-way power gain of the antenna (same antenna used on transmit and receive)
(c) The two-way (transmit × receive) gain product
(d) The aperture efficiency of the antenna

Q3L1 Remember

The thermal noise power at the output of a radar receiver with bandwidth B, noise figure F, and reference temperature T₀ is:

(a) N = k T₀ B / F
(b) N = k T₀ B F
(c) N = k T₀ F / B
(d) N = P_t λ² / (4πR²)

Q4L1 Remember

For a monostatic radar with antenna effective aperture A_e, which expression correctly converts gain to aperture?

(a) G = 4πA_e/λ²
(b) G = A_eλ²/4π
(c) A_e = Gλ/4π
(d) G = 4πλ²/A_e

Q5L2 Understand

Why does received radar power fall as R⁻⁴ rather than R⁻²?

(a) Because antenna gain decreases with the square of range
(b) Because power spreads over 4πR² on the path to the target AND the scattered power spreads over another 4πR² on the return path
(c) Because the target absorbs half the incident energy at long range
(d) Because atmospheric attenuation always varies as R⁻²

Q6L2 Understand

Increasing the radar wavelength λ (all other parameters fixed) has what effect on maximum detection range, and why?

(a) Decreases R_max, because longer wavelengths carry less energy
(b) No effect, because λ cancels between the gain and aperture terms
(c) Increases R_max ∝ λ^1/2 for a fixed-gain antenna, since P_r ∝ λ²
(d) Increases R_max ∝ λ², because the radar equation is linear in λ²

Q7L2 Understand

A Swerling 1 target model assumes the radar cross-section is:

(a) Constant from pulse to pulse and from scan to scan
(b) Exponentially distributed and essentially constant (correlated) during one dwell, but independent from scan to scan
(c) Exponentially distributed and independent from pulse to pulse
(d) Chi-square with 4 degrees of freedom, correlated over a dwell

Q8L2 Understand

Lowering the detection threshold at the radar receiver output will:

(a) Increase P_d but leave P_fa unchanged
(b) Decrease both P_d and P_fa
(c) Increase both P_d and P_fa — the fundamental detection trade-off
(d) Increase P_fa but leave P_d unchanged

Levels 3–5

Part B: Apply, Analyze, Evaluate (Q9–Q20)

Q9L3 Apply

A monostatic radar has P_t = 100 kW, G = 30 dB, λ = 0.1 m, σ = 5 m², R = 50 km, no losses. The received power P_r is approximately:

(a) −114 dBW
(b) −119 dBW
(c) −124 dBW
(d) −129 dBW

Q10L3 Apply

A radar receiver has B = 2 MHz and noise figure F = 5 dB (F ≈ 3.16). The noise power N is closest to:

(a) −126 dBW
(b) −131 dBW
(c) −136 dBW
(d) −141 dBW

Q11L3 Apply

To double the maximum detection range by increasing transmitter power alone (all else fixed), the required power increase is:

(a) 2× (3 dB)
(b) 4× (6 dB)
(c) 8× (9 dB)
(d) 16× (12 dB)

Q12L3 Apply

An X-band radar (λ = 0.03 m) must detect a target of σ = 0.5 m². All parameters fixed. If the target's RCS drops by 12 dB (stealth coating), R_max becomes:

(a) 1/16 of the original range
(b) 1/8 of the original range
(c) 1/2 of the original range
(d) unchanged, since λ compensates

Q13L3 Apply

A radar needs a minimum SNR of 10 dB. With noise power N = −131 dBW, and 16 pulses coherently integrated (perfect coherent gain = 10·log₁₀16 ≈ 12 dB), the minimum required single-pulse received power is:

(a) −123 dBW
(b) −128 dBW
(c) −133 dBW
(d) −138 dBW

Q14L4 Analyze

A measured radar link shows 3 dB less received power than the free-space radar equation predicts. Which of the following combinations of faults could fully account for the discrepancy?

(a) Feed-line loss of 1 dB on transmit only, plus a 2 dB error in measured antenna gain
(b) Transmit feed-line loss of 1 dB plus receive feed-line loss of 2 dB
(c) One-way atmospheric attenuation of 3 dB
(d) Target RCS overestimated by a factor of 2 (3 dBsm)

Q15L4 Analyze

Two radars operate at the same frequency and must detect the same σ at the same R with the same P_d. Radar X uses an antenna with twice the gain of Radar Y (G_X = 2G_Y, all else equal). The required transmitter power of X relative to Y is:

(a) 4× higher — power must compensate the square of the gain ratio
(b) 2× higher
(c) 4× lower — P_r ∝ G², so doubling G gives 6 dB, i.e. 1/4 the power
(d) 16× lower

Q16L4 Analyze

Comparing a monostatic radar with a bistatic radar of equal transmitter power, antenna gains, and target at the same slant geometry: as the bistatic receiver moves far from the transmitter baseline so that R_rx ≫ R_tx, the received power compared with the monostatic case will:

(a) Increase, because the forward-scatter RCS is always larger
(b) Decrease, because the R_rx² spreading term dominates and R_txR_rx grows beyond R² of the monostatic geometry
(c) Stay equal, since (4π)³ is unchanged
(d) Become zero because the bistatic angle reaches 180°

Q17L4 Analyze

A radar has a system noise temperature T_s = 580 K. Its noise figure F (dB) relative to T₀ = 290 K is:

(a) 0 dB
(b) 2 dB
(c) 3 dB
(d) 6 dB

Q18L5 Evaluate

A design team must extend a radar's range from 100 km to 141 km (a factor of √2) at minimum cost. Three independent options each give exactly the needed factor: (i) +3 dB transmitter power, (ii) +3 dB antenna gain (≈ double effective aperture), (iii) +3 dB via coherent integration of 2× more pulses. Which evaluation is MOST defensible?

(a) Option (i) — power increase is always the most reliable since R ∝ P_t^1/4
(b) Option (iii) — integration is free and adds no hardware
(c) Option (ii) or (iii) — power is the costliest lever (prime power, cooling, supply), whereas aperture raises search time/dwell costs and integration costs only dwell time; the best choice depends on whether dwell time or antenna size is the scarcer resource
(d) All options are exactly equivalent, so choose randomly

Q19L5 Evaluate

A long-range surveillance radar detects a Swerling 1 target with P_d = 0.9 per scan using a fixed-frequency waveform. An engineer proposes adding pulse-to-pulse frequency agility. The BEST critical evaluation is:

(a) Reject — agility widens the spectrum and always reduces SNR at the matched filter
(b) Accept unconditionally — agility converts the target to Swerling 0 and removes fluctuation loss entirely
(c) Accept with caution — decorrelating the echoes (Swerling 1 → 2-like) lowers the required SNR at high P_d by several dB, but only if the frequency steps exceed the target's correlation bandwidth, and wider bandwidth increases processing/Rx complexity
(d) Reject — Swerling models are statistical fictions and cannot be changed by waveform design

Q20L5 Evaluate

For a fixed-volume air-surveillance mission (fixed frame time, coverage solid angle, P_d, and range), two candidate designs meet the power–aperture requirement: Design A — large aperture + low average power; Design B — small aperture + high average power. Which judgement is correct?

(a) Design B is superior because peak power always dominates radar performance
(b) Design A is superior because antenna gain enters the radar equation twice
(c) Both are equivalent in detection performance by the search-radar (power–aperture) equation; the choice should be based on secondary criteria — cost, prime power/cooling, physical size, beam agility, and vulnerability — since P_avgA_e is the invariant figure of merit
(d) Neither works, because the search-radar equation applies only to tracking radars

Marking

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Select an option in each question above, then press the button below. Your score and Bloom's-level breakdown will be shown immediately.

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Post-Test Review

Answer Key & Explanations

Attempt all 20 questions before reading this section. Correct options are marked green in each question when you use "Check".

Q1 — (d) · L1
By definition, σ = 4π × (power scattered back toward the source per unit incident power density), equivalently 4πR² × (S_scattered/S_incident). Option (b) states the same idea but misplaces R² relative to the standard definition; (d) is the canonical form using intercepted power.

Q2 — (b) · L1
In the monostatic equation the same antenna serves Tx and Rx; G is the one-way peak gain, and it appears squared (once for transmit, once for receive). Option (c) would require a single symbol for a product of two different gains, as in the bistatic form.

Q3 — (b) · L1
N = kT₀BF. The noise figure multiplies the available thermal noise: available noise at T₀ is kT₀B, and the receiver degrades it by factor F. Options (a) and (c) invert or misplace F and B.

Q4 — (a) · L1
G = 4πA_e/λ² is the fundamental antenna theorem (with A_e = ηA for physical aperture A). Option (b) inverts it, (c) misses the square on λ, (d) inverts the whole ratio.

Q5 — (b) · L2
Radar is a two-way path: isotropic spreading 4πR² on transmit leg gives S₁ = P_tG/4πR²; the target re-radiates and a second 4πR² spreading applies on return. Hence P_r ∝ 1/R⁴. Option (a) is wrong because antenna gain is range-independent in the far field; (c) and (d) are false physical claims.

Q6 — (c) · L2
P_r ∝ λ², so R_max ∝ λ^1/2 (fourth root of λ²). Option (d) forgets the fourth-root step from setting P_r = MDS. Caveat for design: for a fixed-size (fixed-aperture) antenna rather than fixed gain, G itself falls as 1/λ² and the trend reverses — a favorite exam follow-up.

Q7 — (b) · L2
Swerling 1: slow fluctuation — σ is constant during a dwell but obeys an exponential (many comparable scatterers) distribution from scan to scan. Option (a) is Swerling 0/5 (non-fluctuating); (c) is Swerling 2; (d) is Swerling 3/4.

Q8 — (c) · L2
Lowering the threshold makes crossings more likely for both noise-alone and signal-plus-noise cases, so P_fa and P_d both rise. This is the Neyman–Pearson motivation for fixing P_fa and maximizing P_d through processing gain instead.

Q9 — (c) · L3
G = 10³ (30 dB → 1000), so G² = 10⁶. Numerator: 10⁵ × 10⁶ × (0.1)² × 5 = 5×10⁹. Denominator: (4π)³ × R⁴ = 1984 × (5×10⁴)⁴ = 1984 × 6.25×10¹⁸ ≈ 1.24×10²². P_r = 5×10⁹ / 1.24×10²² ≈ 4.03×10⁻¹³ W, and 10log₁₀(4.03×10⁻¹³) ≈ −124 dBW.

Q10 — (c) · L3
N = kT₀BF = 1.38×10⁻²³ × 290 × 2×10⁶ × 3.16 ≈ 2.53×10⁻¹⁴ W. In dBW: 10log₁₀(2.53×10⁻¹⁴) ≈ −136 dBW. Quick dB check: kT₀ = −204 dBW/Hz; + 10log₁₀(2×10⁶) = +63 dB; + noise figure 5 dB → −204 + 63 + 5 = −136 dBW ✓.

Q11 — (d) · L3
R_max ∝ P_t^1/4. Doubling R: P_t,new/P_t,old = 2⁴ = 16 (12 dB). This is the famous "fourth-root penalty" that makes brute-force range extension so expensive.

Q12 — (c) · L3
12 dB RCS reduction = factor 1/16 in σ. R ∝ σ^1/4 → factor (1/16)^1/4 = 1/2. The range halves. Option (a) would be the answer if R were linear in σ — a common error.

Q13 — (c) · L3
Coherent integration of n = 16 pulses adds 10log₁₀(16) ≈ 12 dB of SNR gain. The single pulse must therefore supply 10 − 12 = −2 dB of SNR on its own. Hence P_r,min (single pulse) = N + SNR = −131 dBW + (−2 dB) = −133 dBW. A common error is to answer −121 dBW (applying the integration gain in the wrong direction).

Q14 — (b) · L4
Each dB of loss in the transmit chain and each dB in the receive chain enters the link once, so 1 + 2 = 3 dB total shortfall. Option (a): gain error of 2 dB in measured G would appear twice (G²) giving 4 dB + 1 = 5 dB. Option (c): 3 dB one-way becomes 6 dB two-way. Option (d): σ overestimated by 3 dBsm means true σ is lower → predicted P_r too high, which is a 3 dB discrepancy — also plausible in isolation, but the question asks which combination fully accounts for it; only (b) is a consistent loss decomposition. Accept reasoned discussion of (d) with correct dBsm logic in marking.

Q15 — (c) · L4
P_r ∝ G² ⇒ doubling G yields 4× received power (6 dB). To keep P_r constant, P_t can drop by 4×. Option (a) reverses the reasoning; (d) applies the 1/4-root to the gain change, confusing it with the range scaling.

Q16 — (b) · L4
Bistatic received power ∝ 1/(R_tx²R_rx²). With R_rx ≫ R_tx, the product R_txR_rx exceeds the monostatic R² for comparable geometries, so P_r falls. Forward-scatter enhancement (option a) occurs only near the baseline, not asymptotically.

Q17 — (c) · L4
T_s = T₀·F ⇒ F = 580/290 = 2 → F(dB) = 10log₁₀2 ≈ 3 dB. This is the standard conversion between system noise temperature and noise figure, T₀ = 290 K.

Q18 — (c) · L5
Each lever contributes 3 dB = √2 in range via its respective fourth-root scaling: P_t⁺³ᵈᴮ → 2× power; G⁺³ᵈᴮ → √2 × G ≈ 2× aperture; integration⁺³ᵈᴮ → 2× pulses. A defensible evaluation weighs system-level costs: transmitter power is usually most expensive (prime power, cooling, reliability); aperture costs size/weight and slows search; integration costs dwell time and may be limited by target motion. Hence (c) is the most professionally reasoned choice. (d) is wrong because the options are equivalent in detection only, not in cost or secondary effects.

Q19 — (c) · L5
Frequency steps decorrelate target echoes only if they exceed the coherence bandwidth (≈ 1/target extent). When they do, scan-to-scan Swerling 1 behavior becomes pulse-to-pulse (Swerling 2-like), saving several dB at high P_d. But wider bandwidth costs faster digitizers, more processing, and possible range-ambiguity problems — hence "accept with caution". Option (b) is wrong because the target cannot become Swerling 0; option (d) wrongly dismisses waveform effects on fluctuation statistics.

Q20 — (c) · L5
The search-radar equation makes P_avgA_e the invariant: both designs deliver identical search performance. The decision then rests on cost, prime power, cooling, mobility, beam scheduling, and ECCM posture — exactly the systems-engineering judgment in (c). Options (a), (b) privilege one lever using the peak-power intuition of tracking radars, which does not apply to a fixed-volume search mission; (d) misstates the equation's applicability.