15 Questions — Bloom's Levels: Remember • Understand • Apply
ECE 521E — Radar & Satellite Engineering • 5th Year BSc. Electrical & Communication Engineering
Masinde Muliro University of Science and Technology (MMUST)
| Level | Description | Typical Trigger Words | Questions |
|---|---|---|---|
| 1. Remember | Recall facts, definitions, frequencies, formulas | Define, state, list, identify | Q1 – Q5 |
| 2. Understand | Explain concepts, compare, interpret | Explain, compare, distinguish, summarize | Q6 – Q10 |
| 3. Apply | Use formulas and concepts in calculations/scenarios | Calculate, determine, apply, solve | Q11 – Q15 |
Question 1 Remember
The duty cycle of a continuous wave radar transmitter is:
Question 2 Remember
The Doppler frequency shift produced by a target approaching with radial velocity vr is given by:
Question 3 Remember
In a CW Doppler radar receiver, the device that produces the low-frequency beat note between the echo and a sample of the transmitted carrier is the:
Question 4 Remember
In an FMCW radar using a sawtooth frequency sweep, the quantity that is directly proportional to target range is the:
Question 5 Remember
The range resolution of an FMCW radar with sweep bandwidth B is:
Question 6 Understand
Why can a simple unmodulated CW radar not measure the range to a stationary target?
Question 7 Understand
In a CW Doppler radar block diagram, why must the local-oscillator reference fed to the mixer be coherent with the transmitted wave?
Question 8 Understand
The principal reason CW radar systems usually employ separate transmit and receive antennas is:
Question 9 Understand
In a triangular-wave FMCW radar, a moving target produces different beat frequencies on the up-ramp and down-ramp because:
Question 10 Understand
Which statement correctly compares CW and pulsed radar?
Question 11 Apply
A police CW Doppler radar operates at 24.15 GHz. The Doppler shift measured from an approaching vehicle is 3,215 Hz. The vehicle's speed is approximately:
Question 12 Apply
An X-band CW radar (f₀ = 10 GHz) measures a Doppler frequency of 1,667 Hz from a target. The target's radial velocity is approximately:
Question 13 Apply
An FMCW radar altimeter sweeps a bandwidth B = 150 MHz in Tm = 1 ms and measures a beat frequency fb = 50 kHz. The aircraft's altitude is:
Question 14 Apply
A 77 GHz automotive FMCW radar (λ ≈ 3.9 mm) uses a triangular sweep with B = 300 MHz, Tm = 2 ms. It measures fup = 46 kHz and fdown = 54 kHz. The target's range is:
Question 15 Apply
For the same radar in Question 14, the magnitude of the target's Doppler shift is 4 kHz. The target's radial velocity is approximately:
Q1 — Answer: (b) 100%
Explanation: A CW radar transmits without interruption, so the transmitter is on for the whole period — a 100% duty cycle. Consequently peak power equals average power (Pavg = Ppeak), which is the fundamental power advantage of CW systems over low-duty-cycle pulsed radars.
Q2 — Answer: (b) fd = 2vr / λ
Explanation: The factor of 2 arises because the two-way path length changes at rate 2vr (radar-to-target distance closes at vr, and the wave must travel out and back). Equivalently fd = 2vrf0/c. Option (a) inverts the factor of 2; option (d) confuses Doppler with pulse-echo timing (f = c/2R is not a standard relation).
Q3 — Answer: (c) Mixer (phase detector)
Explanation: The echo at f0+fd is combined with the coherent LO sample at f0; the mixer outputs the difference fd (plus the sum f0+fd+f0, removed by filtering). A duplexer belongs to pulsed radar; the LNA only amplifies; a matched filter optimizes pulse/SNR detection, not beat-note generation.
Q4 — Answer: (b) Beat frequency during the sweep
Explanation: With sawtooth modulation, fb = (B/Tm)·(2R/c) — the beat frequency is proportional to range R (slope and c constant). This linear relation, R = c·Tm·fb/(2B), is the core of FMCW range measurement.
Q5 — Answer: (b) ΔR = c / (2B)
Explanation: The resolution of an FMCW radar equals c/(2B), identical in form to pulsed-radar resolution cτ/2 with B ≈ 1/τ. Wider bandwidth gives finer resolution — e.g., B = 1.5 GHz gives ΔR = 10 cm. Options (a), (c), (d) misplace the factor of 2 or invert the ratio.
Q6 — Answer: (b) Because a continuous tone carries no time reference — the echo delay appears only as an unobservable constant phase shift
Explanation: A pure sinusoid is periodic; delaying it by td = 2R/c merely shifts its phase by 2πf0td. There is no pulse edge or modulation mark to "time," so range is fundamentally unobservable — velocity (Doppler) is the only measurement available until modulation (FMCW) is added. Option (a) is a true statement about Doppler (a stationary target produces no Doppler shift) but it does not answer why range cannot be measured. Options (c) and (d) are physically false.
Q7 — Answer: (b) So that the beat note at fd has a stable phase; an incoherent reference makes the Doppler difference phase wander and average to zero
Explanation: The measured beat signal is cos(2πfdt + Δφ) only if the LO is phase-locked to the transmitted carrier (Δφ constant). An independent oscillator's phase drifts randomly, so the low-frequency product averages toward zero and the Doppler tone vanishes. This is why CW radars use stable, low-phase-noise crystal oscillators — oscillator drift is indistinguishable from true Doppler shift.
Q8 — Answer: (b) To provide isolation, since the continuously transmitting signal must not saturate the simultaneously receiving channel
Explanation: CW radar transmits and receives at the same instant; a pulsed radar's duplexer (or simple switch) can protect the receiver between pulses, but CW has no "between." Separate antennas, physical spacing, shielding, and feed-through cancellation circuits are used to keep transmitter leakage (watts) from overwhelming echoes (picowatts). Option (a) is false — two antennas do not double gain; options (c), (d) are unrelated to the TX/RX simultaneity problem.
Q9 — Answer: (b) The Doppler shift adds to the range beat on one ramp and subtracts on the other, allowing range and velocity to be separated
Explanation: With triangular modulation: up-ramp beat fup = fb − fd, down-ramp beat fdown = fb + fd (signs per convention). Adding the beats cancels Doppler and yields range; differencing them cancels the range beat and yields fd, hence velocity. This decoupling is the principal reason triangular sweeps are used when both range and velocity are needed (e.g., automotive radar). The sweep slope does change sign (option a) but the Doppler add/subtract mechanism is what creates the two distinct beats.
Q10 — Answer: (b) CW radar has no minimum-range blind zone, while pulsed radar cannot detect targets closer than about cτ/2
Explanation: In pulsed radar the receiver is blanked during the transmitted pulse, so echoes arriving within the pulse width are lost — the minimum range Rmin ≈ cτ/2. CW transmission is continuous, so arbitrarily close targets (even zero range) produce a beat note. Option (a) is backwards: CW peak power is low (= average). Option (c) reverses the roles — CW measures velocity directly; pulsed radar needs pulse-Doppler/MTI processing. Option (d) contradicts Q1.
Q11 — Answer: (b) 20 m/s (72 km/h)
Explanation: λ = c/f₀ = 3×108/2.415×1010 = 0.01242 m. vr = fdλ/2 = 3215 × 0.01242/2 = 19.97 ≈ 20 m/s ≈ 72 km/h — a typical highway speeding case. This applies the Doppler formula numerically, exactly as a traffic radar's processor does.
Q12 — Answer: (a) 25 m/s
Explanation: λ = c/f₀ = 3×108/1010 = 0.03 m. vr = fdλ/2 = 1667 × 0.03/2 = 25 m/s ≈ 90 km/h. Note how a shift of less than 2 kHz at a 10 GHz carrier corresponds to 90 km/h — illustrating the precision demanded of the oscillator (Q7).
Q13 — Answer: (b) 50 m
Explanation: R = c·Tm·fb/(2B) = (3×108 × 10−3 × 5×104)/(2 × 1.5×108) = (1.5×1010)/(3×108) = 50 m. This is the standard FMCW altimeter calculation — a low-altitude aircraft on final approach.
Q14 — Answer: (b) 50 m
Explanation: For triangular FMCW, R = c·Tm(fup+fdown)/(4B) = 3×108 × 2×10−3 × (46+54)×103 / (4 × 3×108) = 3×108 × 2×10−3 × 105 / 1.2×109 = 50 m. Note the factor 4 (two ramps share the bandwidth) — using 2 instead gives the distractor 100 m.
Q15 — Answer: (b) 7.8 m/s
Explanation: fd = (fdown − fup)/2 = (54 − 46)/2 = 4 kHz ✓ (consistent). vr = fdλ/2 = 4000 × 3.9×10−3/2 = 7.8 m/s ≈ 28 km/h — a plausible closing speed for an ACC scenario. Option (a) omits the factor ½; option (d) doubles the wavelength error. The pair Q14–Q15 demonstrates the complete triangular-FMCW solution: sum the beats → range; difference the beats → velocity.