Introduction to Pulse Radar — 20-Question Quiz | EEEN 567 Satellite Engineering

Instructions

Attempt all 20 questions before checking answers. Each question carries 1 mark. Choose the single best option. Questions are organised in four blocks of five, one block per cognitive level of Bloom's taxonomy:

Level 1 — Remember (Q1–Q5) Level 2 — Understand (Q6–Q10) Level 3 — Apply (Q11–Q15) Level 4 — Analyze (Q16–Q20)

How to use this quiz Work through the blocks in order. On submission you will get an overall score plus a per-level breakdown so you can see whether gaps are in recall, comprehension, computation, or reasoning. Then study the answer key — every explanation teaches the underlying principle, not just the letter. Constants: c = 3×10⁸ m/s, k = 1.38×10⁻²³ J/K.

Your Score

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Block A — Level 1: Remember

Q1.Remember

In radar range measurement, R = c·t/2, the factor 2 is present because:

a) Radio waves travel at half the speed of light in air b) The radar receiver has a two-stage amplifier c) The pulse makes a round trip: transmitter → target → receiver d) Two pulses are required for every measurement

Answer: c. The measured delay t covers the path to the target and back, so R = ct/2. The same factor of 2 reappears in the Doppler shift (two-way Doppler) and in minimum-range/resolution formulas.

Q2.Remember

The duty cycle of a pulse radar is defined as:

a) PRF / τ b) τ × PRF = τ / PRI c) PRI / τ d) P_avg / P_t × PRF

Answer: b. Duty cycle is the fraction of time the transmitter is on: d = τ/PRI = τ·PRF (typically 0.1–10%). It links peak and average power: P_avg = P_t·d.

Q3.Remember

The radar cross-section (RCS) of a target, denoted σ, has units of:

a) m² (equivalent scattering area) b) m²/s (scattering rate) c) W (scattered power) d) dimensionless ratio

Answer: a. σ is an equivalent area: it relates incident power density to the power scattered back toward the radar. It is not a physical area and varies with aspect, frequency and polarisation (a fighter ≈ 1–5 m²; a large airliner 10–100 m²).

Q4.Remember

A matched filter in a radar receiver is designed to:

a) Reject all frequencies except the carrier b) Amplify the echo above the transmitter feedthrough c) Convert the IF signal to baseband d) Maximise the output signal-to-noise ratio for the transmitted waveform

Answer: d. For a known signal in white noise, the filter whose impulse response is the time-reversed, conjugated signal maximises output SNR — the matched-filter theorem from detection theory. In chirp radars it also performs pulse compression, giving processing gain ≈ τB.

Q5.Remember

The A-scope radar display presents:

a) A plan map of target positions in azimuth and range b) Target height versus slant range c) Echo amplitude versus range (a single sweep) d) Doppler frequency versus antenna rotation rate

Answer: c. The A-scope (amplitude vs range) is the simplest display — used for alignment, waveform inspection and precise range read-out. The PPI (Plan Position Indicator) is the familiar rotating-sweep map; B/E scopes show range vs azimuth/elevation.

Block B — Level 2: Understand

Q6.Understand

In the monostatic radar range equation, received echo power falls as 1/R⁴. The best explanation is:

a) Antenna gain automatically decreases with distance b) Transmitted energy spreads over an area ∝ R² on the outbound path, and the scattered energy spreads again ∝ R² on the return path c) The atmosphere absorbs energy proportionally to R⁴ d) The target's RCS decreases as the fourth power of range

Answer: b. Two independent 1/R² spherical-spreading losses multiply to 1/R⁴. A one-way communications link loses only 1/R² — the extra spreading is the "radar penalty" and the reason detection range is so hard-won (16× power for 2× range).

Q7.Understand

Increasing a radar's PRF improves its Doppler (velocity) measurement capability but at the cost of:

a) A shorter maximum unambiguous range — distant echoes fold back into nearer ranges b) A longer minimum range set by the pulse width c) Reduced antenna gain at high elevation angles d) Loss of coherence between transmitted pulses

Answer: a. R_max,unamb = c/(2·PRF) shrinks as PRF rises, while the unambiguous Doppler window widens toward ±PRF/2 (Nyquist on the pulse train). This is the fundamental PRF dilemma of pulse radar; staggered/multiple PRFs resolve both.

Q8.Understand

In a radar receiver chain, the low-noise amplifier (LNA) is placed immediately after the antenna/duplexer because:

a) It compensates for cable loss only at microwave frequencies b) It prevents the transmitter power from reaching the mixer c) It sets the IF frequency of the receiver d) By Friis' formula, the first stage's noise figure dominates the system noise figure, so its gain must be high and its noise figure low

Answer: d. Friis: F_sys = F₁ + (F₂−1)/G₁ + (F₃−1)/(G₁G₂) + … The LNA's noise is added directly; later stages' contributions are divided by its gain. A low-NF, high-gain first stage therefore sets the receiver sensitivity — exactly the HEMT/HBT placement studied in microwave engineering.

Q9.Understand

Pulse compression is used in modern radars primarily because a simple short pulse cannot simultaneously provide:

a) High PRF and low duty cycle b) Good Doppler resolution and low transmitter cost c) Large pulse energy (for detection) and fine range resolution — the long coded pulse is compressed by the matched filter to ~1/B d) Coherence and frequency agility

Answer: c. Detection needs energy (long τ at limited peak power); resolution needs bandwidth (ΔR = c/2B). A chirp or phase-coded long pulse keeps the energy but is compressed in the matched filter to an effective width 1/B, with processing gain ≈ τB (hundreds to thousands). This is the enabling technique behind long-range surveillance and satellite SAR.

Q10.Understand

The Doppler shift measured by a monostatic radar is f_d = 2v_r/λ rather than v_r/λ because:

a) The target moves during the pulse transit time, doubling the effective delay change b) Doppler acts on both the outbound and return paths — the received echo is shifted twice c) The factor 2 converts wavelength to half-wavelength units d) The radar transmits on two polarisations simultaneously

Answer: b. Each path contributes v_r/λ, and the two contributions add on reception — the same round-trip logic as the factor 2 in R = ct/2. Consequence: a 400 MHz beacon on a 7 km/s satellite gives a one-way Doppler of ~9.3 kHz but ~18.7 kHz at a tracking radar.

Block C — Level 3: Apply

Q11.Apply

An echo returns 400 µs after its pulse is transmitted. The target range is:

a) 60 km b) 120 km c) 30 km d) 6 km

Answer: a. R = ct/2 = 3×10⁸ × 400×10⁻⁶ / 2 = 6×10⁴ m = 60 km. Option (b) forgets the round-trip factor — worth 1 µs → 150 m as a mental shortcut.

Q12.Apply

A surveillance radar operates at PRF = 500 Hz. Its maximum unambiguous range is:

a) 600 km b) 60 km c) 300 km d) 150 km

Answer: c. R_max,unamb = c/(2·PRF) = 3×10⁸/(2×500) = 3×10⁵ m = 300 km. Echoes from a target at, say, 350 km arrive 100 µs after the next pulse was sent and are mis-ranged at 50 km.

Q13.Apply

A radar transmits pulses of width τ = 2 µs at PRF = 1 kHz with peak power 200 kW. Its duty cycle and average power are:

a) 0.2 % and 40 W b) 2 % and 4 kW c) 0.2 % and 400 W d) 0.2 % and 400 W — check: d = τ·PRF = 2 µs × 1 kHz = 0.002 = 0.2 %; P_avg = P_t·d = 200 kW × 0.002 = 400 W

Answer: d. Duty cycle d = τ·PRF = 2×10⁻⁶ × 10³ = 2×10⁻³ = 0.2 %. Average power P_avg = P_t·d = 200 000 × 0.002 = 400 W. Note the striking contrast: 200 kW peak, only 400 W average — peak power is set by device/breakdown limits, average power by thermal limits, and it is the average that builds detection capability over time.

Q14.Apply

A monostatic radar has P_t = 100 kW, G = 30 dB (gain = 1000), λ = 0.1 m, target σ = 10 m², and observes a target at R = 50 km. The received power P_r = P_tG²λ²σ/[(4π)³R⁴] is approximately:

a) 8 × 10⁻⁹ W b) 8 × 10⁻¹³ W c) 8 × 10⁻¹⁷ W d) 8 × 10⁻⁷ W

Answer: b. Numerator: 10⁵ × (10³)² × 0.1² × 10 = 10¹⁰. Denominator: (4π)³ ≈ 1984 and R⁴ = (5×10⁴)⁴ = 6.25×10¹⁸, giving ≈ 1.24×10²². So P_r ≈ 8.1×10⁻¹³ W ≈ −91 dBW — orders of magnitude below a nanowatt, yet comfortably above typical noise floors (kT₀B ≈ 4×10⁻¹⁵ W for B = 1 MHz) with integration. Sanity-check the exponent carefully; (c) and (a) are the classic slip of miscounting the powers of ten.

Q15.Apply

A circular parabolic dish has diameter D = 2 m, aperture efficiency η = 0.6, operating at 3 GHz (λ = 0.1 m). Its gain G = 4πηA/λ² with A = πD²/4 is closest to:

a) 30 dBi b) 27 dBi c) 34 dBi d) 37 dBi

Answer: c. A = π(2)²/4 = π m². G = 4π × 0.6 × π / 0.01 = 2.4π²/0.01 ≈ 2 370, i.e. 10·log₁₀(2370) ≈ 33.7 dBi. (A useful rule: G(dBi) ≈ 20·log₁₀(D/λ) + 10·log₁₀η + 9.9.) The corresponding beamwidth is θ ≈ 1.22λ/D ≈ 3.5°.

Block D — Level 4: Analyze

Q16.Analyze

An air-defence pulse radar is being upgraded to detect small, fast, low-flying targets. The system engineer must increase sensitivity. Which single change yields the greatest increase in detection range?

a) Doubling the transmitter peak power b) Doubling the antenna aperture diameter (gain quadruples, entering the two-way equation squared) c) Doubling the PRF d) Doubling the receiver bandwidth

Answer: b. Work in dB on the R⁴ equation, where range is proportional to the fourth root of each power term. (a) Doubling P_t gives +3 dB → range ×1.19. (b) Doubling the diameter quadruples gain (G ∝ D²); gain appears squared in the two-way equation, i.e. +12 dB, and range ∝ G^1/2 → ×2. (c) PRF does not enter the single-pulse SNR at all — it sets ambiguities, not sensitivity (extra pulses help only if integrated, ~10·log₁₀N dB). (d) Doubling bandwidth raises the noise floor by 3 dB, reducing sensitivity. The aperture is the most powerful single lever because it works twice, on transmit and on receive.

Q17.Analyze

A satellite ground station can track a new LEO spacecraft either by skin echo (scattering from the body) or by a cooperative beacon transponder that replies to an interrogation. The best analysis of the choice is:

a) Beacon tracking is strongly preferred during LEOP: the reply is orders of magnitude stronger and independent of the satellite's RCS, attitude and tumbling state, giving faster acquisition and more accurate range/Doppler b) Skin tracking is preferred because it requires no spacecraft power and cannot fail c) Both are equivalent because the link budget depends only on transmitter power d) Skin tracking is preferred because it avoids the two-way Doppler ambiguity

Answer: a. A transponder amplifies and retransmits, replacing an unpredictable σ (1–100 m², aspect-dependent, tumbling-modulated) with a cooperative, stable link — the same physics as secondary surveillance radar (Mode-S/ADS-B) in aviation. Skin tracking remains essential for non-cooperative space-surveillance of debris and dead satellites, but for your own spacecraft during early orbit phase the beacon wins decisively. Option (d) is wrong: Doppler ambiguities are set by PRF/waveform, not skin vs beacon.

Q18.Analyze

A weather radar wants to map a storm cell 250 km away, but at its current PRF the echoes alias: the cell appears at 50 km. The correct analysis and fix is:

a) The radar is fine — aliasing means the target is genuinely at 50 km b) Reduce the PRF so the unambiguous range becomes 250 km — e.g. from the current 1 kHz toward ~600 Hz c) Either lower the PRF to push R_max,unamb beyond 250 km, or keep the PRF and de-alias using multiple/staggered PRFs (echo moves between PRFs while the alias does not) d) Increase the pulse width so the echo becomes stronger than the aliasing

Answer: c. An apparent range of 50 km for a storm at 250 km means the echo is a second-time-around return: with PRF = 750 Hz the unambiguous range is c/(2·750) = 200 km, and 250 − 200 = 50 km — the classic range-folding signature. Two valid fixes exist: (1) lower the PRF so R_max,unamb exceeds 250 km (e.g. ≤600 Hz), which costs Doppler capability; or (2) keep the PRF and transmit alternating staggered PRFs — the true echo stays at the same delay across PRFs while the alias jumps between apparent ranges, letting the processor de-alias. Weather and ATC radars use option (2) precisely because it preserves Doppler (velocity) measurement.

Q19.Analyze

A spaceborne radar altimeter (e.g., Jason/Sentinel-6 class) must measure sea-surface height to centimetre accuracy from ~1 300 km altitude. Which analysis best explains its waveform design?

a) It uses a very high PRF so the short unambiguous range matches the ocean surface b) It uses a narrow-beam dish so that only the sub-satellite point is illuminated, avoiding any averaging c) It transmits very short, powerful pulses and gates out everything except the first return d) It transmits a long coded (chirp) pulse and compresses it on board — pulse-limited geometry over the ocean footprint means the returned power rise curve maps surface height, and compression provides the fine timing without prohibitive peak power

Answer: d. From 1 300 km, centimetric timing needs ~nanosecond-class resolution with a manageable peak-power budget — exactly what pulse compression delivers (compression ratio τB of hundreds). The altimeter exploits pulse-limited geometry: as the compressed pulse intersects the growing ocean footprint, the return power follows a predictable rise whose midpoint gives sea-surface height; averaging thousands of echoes then reaches cm accuracy. High PRF (a) is irrelevant to altitude precision; (b) is wrong — a wide beam with pulse-limited gating is the classical altimeter geometry; (c) would demand extreme peak power.

Q20.Analyze

A 5 m²-RCS aircraft must be detected at 400 km. Analysis shows the radar is 12 dB short of the required SNR. Considering cost, feasibility and the R⁴ law, the most reasonable recovery plan is:

a) Combine modest levers: +6 dB from pulse integration (≈ four pulses coherently or ~sixteen non-coherently), +3–4 dB from a slightly larger aperture/efficiency, +3 dB from longer coded pulses at the same peak power (pulse compression keeps resolution) — together ≈ +12 dB, doubling range b) Multiply transmitter power by 16, keeping everything else unchanged c) Double the PRF, which doubles sensitivity d) Double the receiver bandwidth to capture more signal energy

Answer: a. +12 dB = ×16 in the R⁴ power product = ×2 range. A 16× power amplifier is usually infeasible (thermal, breakdown, cost). Integration (~10logN per N pulses, saturating with N), aperture growth, and stretching pulse length with compression (average power up, resolution unchanged) are the standard, compounding levers. (c) PRF alone adds nothing to single-pulse SNR; (d) doubling B adds +3 dB of noise, worsening the shortfall by 3 dB.

Post-Quiz Answer Key & Explanations

Complete answer table followed by a short explanation per question. Cross-reference with the Introduction to Pulse Radar study guide sections indicated.

Q	Level	Answer	Core concept
1	Remember	c	Round-trip factor 2 in R = ct/2
2	Remember	b	Duty cycle d = τ·PRF
3	Remember	a	RCS units: equivalent area (m²)
4	Remember	d	Matched filter maximises SNR
5	Remember	c	A-scope: amplitude vs range
6	Understand	b	R⁴ law: two spreading losses
7	Understand	a	High PRF ⇒ short unambiguous range
8	Understand	d	Friis formula; LNA placement
9	Understand	c	Pulse compression: energy + resolution
10	Understand	b	Two-way Doppler factor 2
11	Apply	a	R = ct/2 = 60 km
12	Apply	c	R_unamb = 300 km at 500 Hz
13	Apply	d	d = 0.2 %, P_avg = 400 W
14	Apply	b	P_r ≈ 8×10⁻¹³ W (−91 dBW)
15	Apply	c	G ≈ 34 dBi for 2 m dish @ 3 GHz
16	Analyze	b	Aperture works two-way; strongest lever
17	Analyze	a	Beacon vs skin tracking in LEOP
18	Analyze	c	Range folding & staggered-PRF dealiasing
19	Analyze	d	Altimeter waveform: compression + pulse-limited geometry
20	Analyze	a	Combining dB levers; R⁴ insensitivity