15 Questions — Bloom's Levels: Remember • Understand • Apply
5th Year BSc. Electrical & Communication Engineering — ECE521E: Radar & Satellite Engineering
Masinde Muliro University of Science and Technology (MMUST)
| Level | Description | Typical Trigger Words | Questions |
|---|---|---|---|
| 1. Remember | Recall facts, definitions, formulae, waveforms | Define, state, list, identify | Q1 – Q5 |
| 2. Understand | Explain concepts, interpret waveforms, compare | Explain, compare, distinguish, interpret | Q6 – Q10 |
| 3. Apply | Use formulae in numerical problems and scenarios | Calculate, determine, apply, solve | Q11 – Q15 |
Question 1 Remember
In FMCW radar, the quantity measured by mixing the received echo with the currently transmitted signal is called the:
Question 2 Remember
The range resolution of an FMCW radar is determined by the formula:
Question 3 Remember
A standard aircraft radar altimeter is an example of FMCW radar using which modulation waveform?
Question 4 Remember
For a stationary target, the beat frequency of a sawtooth FMCW radar with sweep bandwidth B and sweep time Tm is:
Question 5 Remember
Which two target parameters can be measured simultaneously using a triangular FMCW waveform?
Question 6 Understand
Why does a simple (unmodulated) CW radar fail to measure range, and what does frequency modulation provide to solve this?
Question 7 Understand
In a sawtooth FMCW radar observing a moving target, the measured "range" becomes slightly erroneous because:
Question 8 Understand
Why does the triangular FMCW waveform allow range and velocity to be separated, while the sawtooth waveform alone does not?
Question 9 Understand
Sweep non-linearity in an FMCW radar transmitter primarily degrades:
Question 10 Understand
Why is FMCW preferred over a short high-peak-power pulsed radar for a low-altitude aircraft radar altimeter?
Question 11 Apply
An FMCW radar sweeps B = 150 MHz in Tm = 1 ms. The beat frequency from a stationary target is 40 kHz. The target range is:
Question 12 Apply
An X-band (λ = 3 cm) triangular-FMCW radar with B = 300 MHz and Tm = 2 ms per ramp measures fup = 42 kHz and fdown = 58 kHz. The target's radial velocity is:
Question 13 Apply
For the radar in Question 12, the target range is:
Question 14 Apply
An automotive radar requires 15 cm range resolution. The minimum sweep bandwidth required is:
Question 15 Apply
A tank-level FMCW radar operating above a liquid surface measures a beat frequency of 13.3 kHz with B = 200 MHz swept in Tm = 2 ms. The distance to the liquid surface is:
Select an option for each question above by clicking it, then check your score below.
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Q1 — Answer: (b) Beat frequency
Explanation: In FMCW, the echo (a delayed copy of the sweep) is mixed with the instantaneous transmitted frequency; the difference — the beat frequency fb — is proportional to the round-trip delay and hence to range. Doppler is a separate effect that may be superimposed on the beat.
Q2 — Answer: (a) ΔR = c / (2B)
Explanation: Range resolution is set purely by the swept bandwidth B, exactly as in pulsed radar where B ≈ 1/τ. Wider sweeps give finer resolution — e.g., B = 1.5 GHz yields ΔR = 10 cm. The sweep time Tm affects the beat frequency scale, not the resolution.
Q3 — Answer: (b) Sawtooth (linear ramp) frequency sweep
Explanation: Conventional aircraft radar altimeters linearly sweep ~100–400 MHz near 4.3 GHz; the ground echo's delay produces a beat frequency proportional to altitude. Triangular sweeps are used when velocity must also be extracted (e.g., automotive radar), which altimeters generally do not need.
Q4 — Answer: (b) fb = 2RB / (c·Tm)
Explanation: Beat frequency = sweep slope × round-trip delay: fb = (B/Tm)·(2R/c) = 2RB/(c·Tm). This is the fundamental FMCW range relation, inverted as R = c·Tm·fb/(2B). Option (a) is a reciprocal-time expression with no radar meaning.
Q5 — Answer: (b) Range and radial velocity
Explanation: On the up-ramp the Doppler shift subtracts from the range beat; on the down-ramp it adds. Combining the two beat frequencies yields both R (from the sum) and fd (from the difference), hence radial velocity vr = fdλ/2 — the signature capability of the triangular waveform.
Q6 — Answer: (b) A single tone carries no time reference — the delay is an unobservable constant phase; a frequency sweep gives each instant a distinct frequency, so delay becomes a measurable frequency difference
Explanation: With a pure tone, the echo differs from the transmission only by a fixed phase φ = 2πf0(2R/c), which cannot be resolved without a time marker. Frequency modulation stamps every instant with a unique frequency; after the round-trip delay the echo frequency differs from the current transmit frequency by a measurable amount — converting the time-delay problem into a frequency-offset problem.
Q7 — Answer: (a) The Doppler shift adds to or subtracts from the range-related beat frequency, biasing the range estimate
Explanation: The mixer output contains both the range beat (slope × delay) and the Doppler shift; on a single sawtooth ramp they merge into one measured frequency. For a fast target this contaminates the range estimate — the reason moving-target applications use triangular sweeps (or Doppler-tolerant processing) to separate the two contributions.
Q8 — Answer: (b) The Doppler shift adds to the beat on one ramp and subtracts on the other, so the sum of the two beat frequencies gives range and their difference gives Doppler
Explanation: With fup = fb − fd and fdown = fb + fd, the Doppler cancels in the sum (→ range) and doubles in the difference (→ velocity): R = cTm(fup+fdown)/4B and fd = (fdown−fup)/2. Two equations, two unknowns — one of the most elegant results in radar engineering.
Q9 — Answer: (b) The sharpness of the beat-frequency spectrum, hence range resolution and accuracy
Explanation: If the sweep is not perfectly linear, the beat frequency of a point target wanders over the sweep (residual video phase error), spreading the spectral line and degrading both resolution and peak-detection accuracy. Modern FMCW systems therefore use phase-locked or digitally synthesized sweeps with real-time linearization.
Q10 — Answer: (b) FMCW needs no minimum-range dead zone, avoids high peak power, and achieves fine resolution through sweep bandwidth — well suited to continuous low-altitude operation
Explanation: Pulsed altimeters must blank the receiver during transmission, creating a minimum-range dead zone unacceptable at very low altitude; generating very short high-peak-power pulses is also costly. FMCW transmits continuously at modest power and derives fine resolution (≈1 m) from a few hundred MHz of sweep — ideal for an always-on, low-altitude, power-limited avionics function.
Q11 — Answer: (b) 40 m
Explanation: R = c·Tm·fb/(2B) = (3×108 × 10−3 × 4×104)/(2 × 1.5×108) = 1.2×1010/3×108 = 40 m. This is the standard altimeter-style calculation; option (a) results from forgetting the factor 2 in the denominator.
Q12 — Answer: (c) 120 m/s
Explanation: fd = (fdown − fup)/2 = (58 − 42)/2 = 8 kHz. vr = fdλ/2 = 8000 × 0.03 / 2 = 120 m/s. Option (d) is the trap for using fdown − fup without the factor ½; option (a) misplaces a decimal.
Q13 — Answer: (b) 50 m
Explanation: R = c·Tm(fup+fdown)/(4B) = 3×108 × 2×10−3 × (4.2×104 + 5.8×104) / (4 × 3×108) = 3×108 × 2×10−3 × 105 / 1.2×109 = 50 m. Note the factor 4 (not 2) in the denominator for triangular sweeps — the designed distractor (d) comes from using only one ramp frequency.
Q14 — Answer: (b) 1 GHz
Explanation: B = c/(2ΔR) = 3×108/(2 × 0.15) = 1 GHz. This is why 77 GHz automotive radars sweep bandwidths of 1–4 GHz to achieve 10–15 cm resolution. Option (a) gives 30 cm; (c) gives 10 cm (better than required, but not the minimum).
Q15 — Answer: (b) 20 m
Explanation: R = c·Tm·fb/(2B) = 3×108 × 2×10−3 × 1.33×104 / (2 × 2×108) = 3×108 × 2×10−3 × 1.33×104 / 4×108 ≈ 20 m. Tank-level radars apply exactly this altimeter computation, with the level = empty-tank height − measured distance.