Radio Propagation TheoryRadio Navigation — DGCA CPL practice questions
Question 1 of 8
The process which causes the reduction in signal strength as range from a transmitter increases is known as:
All 8 questions — Radio Propagation Theory
Radio Navigation · DGCA CPL. The correct option is marked on each.
Q1. The process which causes the reduction in signal strength as range from a transmitter increases is known as:
- A.absorption
- B.diffraction
- C.attenuation✓
- D.ionisation
Why: Attenuation is the overall term for loss of signal strength as range increases. It has two components: absorption (energy absorbed by matter) and the inverse square law (energy spread over increasing area). See Section 2.1 . — Classic definition trap. Absorption = one mechanism of attenuation. Attenuation = the overall effect (includes both absorption and inverse square law). The question asks for the overall term.
Q2. Which of the following will give the greatest surface wave range?
- A.243 MHz
- B.500 kHz✓
- C.2182 kHz
- D.15 MHz
Why: Surface wave range decreases as frequency increases . The lowest frequency in the list gives the greatest surface wave range. 500 kHz (MF) < 2182 kHz (MF) < 15 MHz (HF) < 243 MHz (VHF — no surface wave). See Section 4.1 . — Surface wave is maximised at the lowest frequency. Always pick the lowest frequency for maximum surface wave range. The rule: lower frequency = less surface attenuation = more range.
Q3. It is intended to increase the range of a VHF transmitter from 50 NM to 100 NM. This will be achieved by increasing the power output by a factor of:
- A.2
- B.8
- C.16
- D.4✓
Why: By the inverse square law, power ∝ 1/R². To double the range (100/50 = factor of 2), power must increase by 2² = 4 . See Section 2.1 . — Formula to remember: Power factor = (new range / old range)². Here: (100/50)² = 2² = 4. This is a direct application of the inverse square law — perhaps the most common calculation in this chapter.
Q4. The maximum range an aircraft at 2500 ft can communicate with a VHF station at 196 ft is:
- A.79 NM✓
- B.64 NM
- C.52 NM
- D.51 NM
Why: Range = 1.23 × (√h TX + √h RX ) = 1.23 × (√196 + √2500) = 1.23 × (14 + 50) = 1.23 × 64 = 78.72 ≈ 79 NM . See Section 4.2 . — The formula uses heights in feet . Both √196 = 14 and √2500 = 50 are perfect squares — ideal for an exam question. Sum = 64, then × 1.23 = 78.72. Always check: are you using feet, not metres?
Q5. What is the minimum height for an aircraft at a range of 200 NM to be detected by a radar at 1700 ft AMSL?
- A.25,500 ft
- B.15,000 ft✓
- C.40,000 ft
- D.57,500 ft
Why: Rearrange Range = 1.23 × (√h Radar + √h AC ): 200 = 1.23 × (√1700 + √h AC ) 200/1.23 = 162.6 = √1700 + √h AC √1700 ≈ 41.23 √h AC = 162.6 − 41.23 = 121.37 h AC = 121.37² ≈ 14,731 ft ≈ 15,000 ft . See Section 4.2 . — Inverse LOS calculation. The key step is dividing 200 by 1.23 first (≈162.6), then subtracting √(radar height) before squaring. √1700 ≈ 41.2 is not a perfect square — in the exam, approximate to 2 decimal places and square the result.
Q6. Determine which of the following statements concerning atmospheric ionization are correct: 1. The highest levels of ionization will be experienced in low latitudes 2. Ionization levels increase linearly with increasing altitude 3. The lowest levels of ionization occur about midnight 4. The E-layer is higher by night than by day because the ionization levels are lower at night
- A.statements 1, 2 and 3 are correct
- B.statements 1, 3 and 4 are correct
- C.statements 2 and 4 are correct
- D.statements 1 and 4 are correct✓
Why: Statement 1 — TRUE: Ionization increases as latitude decreases (low latitudes = equatorial regions = most solar radiation intensity). Statement 2 — FALSE: Ionization does NOT increase linearly — it forms into discrete layers due to gravitational and magnetic effects at high altitudes. Statement 3 — FALSE: The lowest levels of ionization occur just before sunrise , not midnight. After midnight, ionization continues to decay but reaches minimum just before dawn. Statement 4 — TRUE: The E-layer reduces in altitude at sunrise (more ionization from solar radiation), and increases in altitude aft…
Q7. The average height of the E-layer is …… and the maximum range for sky wave will be ……
- A.60 km, 1350 NM
- B.125 km, 2200 km
- C.225 km, 2200 km
- D.125 km, 1350 NM✓
Why: E-layer average altitude = 125 km . Maximum E-layer sky wave range = 1350 NM . (The F-layer at 225 km gives the 2200 NM range.) See Section 7.5 . — Two pairs to memorise: E-layer = 125 km → 1350 NM ; F-layer = 225 km → 2200 NM . Options (b) and (c) use km instead of NM for range — a unit trap. Check units in every answer option.
Q8. Concerning HF communications, which of the following is correct?
- A.The frequency required in low latitudes is less than the frequency required in high latitudes
- B.At night a higher frequency is required than by day
- C.The frequency required is dependent on time of day but not the season
- D.The frequency required for short ranges will be less than the frequency required for long ranges✓
Why: For HF sky wave comms, short ranges require lower frequencies (smaller skip distances); long ranges require higher frequencies (larger skip distances needed). This is a direct and unambiguous fact. See Section 8 . — Options (a) and (b) are direct reversals of correct facts — classic exam traps. Option (c) contradicts both the source and basic understanding. Option (d) is directly stated in the text: "Short ranges will require lower frequencies and longer ranges will require higher frequency."