Radar PrinciplesRadio Navigation — DGCA CPL practice questions
Question 1 of 14
The factor which determines the MAXIMUM range of a radar is:
All 14 questions — Radar Principles
Radio Navigation · DGCA CPL. The correct option is marked on each.
Q1. The factor which determines the MAXIMUM range of a radar is:
- A.pulse repetition rate✓
- B.pulse width
- C.power
- D.beamwidth
Why: Maximum range is determined by the pulse repetition rate (PRF) . The next pulse cannot be sent until the first returns from maximum range. Lower PRF = greater max range.
Q2. The main advantage of continuous wave (CW) radars is:
- A.no maximum range limitation
- B.better range resolution
- C.no minimum range limitation✓
- D.better bearing accuracy
Why: CW radars have no minimum range limitation (the radio altimeter uses CW). Pulsed radars have a minimum range determined by pulse width.
Q3. If the PRF of a primary radar is 500 pps, the maximum range will be:
- A.324 NM
- B.300 NM
- C.162 NM✓
- D.600 NM
Why: Max range = 81,000 / PRF = 81,000 / 500 = 162 NM .
Q4. To double the range of a primary radar requires power to be increased by a factor of:
- A.2
- B.4
- C.8
- D.16✓
Why: Power ∝ range&sup4;. To double range: power increase = 2&sup4; = 16 .
Q5. Echo time is 1720 µs. Range of target:
- A.139 km
- B.258 km✓
- C.278 km
- D.516 km
Why: Range = 1720 × 300 / 2 = 258,000 m = 258 km .
Q6. Max range required: 100 NM. Maximum PRF?
- A.1620 pps
- B.1234 pps
- C.617 pps
- D.810 pps✓
Why: PRF = 81,000 / 100 = 810 pps .
Q7. PRI = 2100 µs. Maximum radar range:
- A.170 NM✓
- B.315 NM
- C.340 NM
- D.630 NM
Why: PRF = 1,000,000 / 2100 = 476 pps. Max range = 81,000 / 476 ≈ 170 NM .
Q8. To improve resolution of a radar display:
- A.narrow pulse width and narrow beamwidth✓
- B.high frequency and large reflector
- C.wide beamwidth and wide pulse width
- D.low frequency and narrow pulse width
Why: Better resolution = shorter pulse width (radial) + narrower beamwidth (azimuth). Answer: narrow PW + narrow BW .
Q9. Advantage of phased array (slotted antenna):
- A.better resolution
- B.less power required
- C.reduced side lobes and clutter
- D.all of the above✓
Why: Slotted planar array: narrower beam → better resolution, less power needed, reduced side lobes. All of the above.
Q10. Echo received 900 µs after transmission. Range to target:
- A.73 NM✓
- B.270 NM
- C.135 NM
- D.146 NM
Why: Range = 900/12.36 = 72.8 ≈ 73 NM .
Q11. Factor limiting MINIMUM detection range of radar:
- A.pulse repetition interval
- B.transmitter power
- C.pulse width✓
- D.pulse repetition frequency
Why: Minimum range is limited by pulse width . Target must be far enough that its echo returns after the pulse has finished transmitting.
Q12. Doppler MTI generates second trace returns. These are removed by:
- A.using different frequency for Tx/Rx
- B.jittering the PRF✓
- C.regular pulsewidth changes
- D.limiting power output
Why: Second trace returns are removed by jittering the PRF (varying PRI between pulses). This prevents false targets from appearing within range.
Q13. Radar max range: 12 km. Maximum PRF?
- A.25,000 pps
- B.6,700 pps
- C.12,500 pps✓
- D.13,400 pps
Why: PRF = 300,000,000 / (2 × 12,000) = 300,000,000 / 24,000 = 12,500 pps .
Q14. Bearing of primary radar measured by:
- A.phase comparison
- B.searchlight principle✓
- C.lobe comparison
- D.DF techniques
Why: Primary radar bearing uses the searchlight principle — a narrow rotating beam; direction of target = direction of beam when echo is received.