Ground Wave Communication — 20-Question Quiz

Instructions & Bloom's Taxonomy Coverage

Answer all 20 multiple-choice questions. Click "Show Answer & Explanation" after each question to self-check, or use the "Grade My Attempt" button at the end for automatic scoring. Each question is tagged with its Bloom's cognitive level:

L1 — Remember (Q1–Q4) L2 — Understand (Q5–Q8) L3 — Apply (Q9–Q13) L4 — Analyze (Q14–Q17) L5 — Evaluate (Q18–Q20)

Syllabus coverage: surface-wave mechanism and fall-off laws, earth conductivity and skin depth, field-strength prediction, ground-wave antennas and ground systems, day/night behaviour at MF, and link-budget implications.

Level 1 — Remember

Q1. Ground (surface) wave propagation is the dominant mode in the band:Remember

a) VLF, LF and MF (3 kHz – 3 MHz) b) VHF (30 – 300 MHz) c) UHF (300 – 3000 MHz) d) EHF (30 – 300 GHz)

Answer: a). Below ~3 MHz the surface wave hugs the earth; above ~3–5 MHz its loss over land becomes prohibitive and sky-wave/LOS modes take over.

Q2. Beyond the near field, ground-wave field strength over average land falls approximately as:Remember

a) 1/d b) 1/d² c) 1/d³ d) independent of distance

Answer: b) 1/d². Geometrical spreading (1/d) times absorption in the lossy ground (1/d) gives 1/d² — the reason MF broadcasters need tens of kilowatts.

Q3. The conductivity of sea water at MF is approximately:Remember

a) 10⁻⁶ S/m b) 0.001 S/m c) ≈ 5 S/m d) 10 000 S/m

Answer: c) ≈ 5 S/m. Sea water's high salinity makes it an excellent conductor at MF, giving ground-wave ranges of hundreds of kilometres — the maritime advantage.

Q4. The standard reference curves for ground-wave field strength prediction are:Remember

a) ITU-R P.368 (and FCC §73.184 for AM broadcast) b) ITU-R P.533 c) ITU-R P.1546 only d) IEEE 802.11 curves

Answer: a). P.368 covers 10 kHz–30 MHz ground wave; FCC §73.184 is the AM broadcast equivalent. P.533 is HF sky-wave prediction; P.1546 is point-to-area VHF/UHF.

Level 2 — Understand

Q5. Ground-wave antennas must be vertical and mounted low (≤ ~0.1–0.2λ) because:Understand

a) a horizontal antenna's ground image cancels the surface wave at the interface, and raising a vertical too high couples energy into space waves instead of the guided surface wave b) vertical antennas are cheaper c) regulations require it d) horizontal polarization is absorbed by the ionosphere

Answer: a). The image of a horizontal antenna is anti-phase at the surface → cancellation. Height converts the near-grazing launch into elevated lobes; ground-wave stations keep antennas low to feed the surface wave.

Q6. Distant MF AM broadcast stations become receivable after sunset because:Understand

a) ground-wave attenuation is halved at night b) the absorbing D layer disappears at night, so MF sky waves (previously absorbed) return from the E/F layers over 1 000+ km, adding to the local ground-wave signal c) antennas physically lengthen d) receivers become more sensitive at night

Answer: b). Daytime MF sky wave is killed by D-layer absorption; at night it survives and hops long distances — a blessing (distant reception) and a curse (co-channel interference, hence night-time AM power reductions and directional arrays).

Q7. Skin depth in the ground matters to the antenna engineer because it:Understand

a) sets the colour of the insulators b) tells how deep the induced ground currents flow — which determines how much ground the radial system/ground screen must capture, and why sea water (δ ≈ 0.2 m at 1 MHz) behaves so differently from dry soil (δ ≈ 16 m) c) determines the transmitter frequency d) sets the modulation index

Answer: b). δ = √(2/(ωμ₀σ)): currents are shallow in good conductors (sea) and deep in poor ground — radials and ground screens must intercept the shallow return currents to stop them dissipating as heat.

Q8. Loran-C (100 kHz) achieves 1 000+ km ground-wave range mainly because:Understand

a) its 3 km wavelength makes diffraction around the spherical earth almost lossless, so the surface wave persists far beyond the optical horizon b) it uses frequency modulation c) it relies on ionospheric reflection d) it uses parabolic dishes

Answer: a). Diffraction loss ∝ (hills/λ): at 3 km wavelength the earth looks almost flat. Combined with low absorption over sea, this gives Loran its enormous, stable coverage — exactly what navigation needs.

Level 3 — Apply

Q9. A 1 kW MF station's reference field is 100 mV/m at 1 km. Using 1/d² over land, the field at 10 km is approximately:Apply

a) 50 mV/m b) 20 mV/m c) 1 mV/m d) 10 mV/m

Answer: c) 1 mV/m. E = 100/(10)² = 1 mV/m. (Free space 1/d would give 10 mV/m — the 20 dB difference is the ground-wave penalty.)

Q10. Skin depth in sea water (σ = 5 S/m) at 1 MHz is approximately:Apply

a) 2.2 m b) 0.22 m c) 22 m d) 0.022 m

Answer: b) ≈ 0.22 m. δ = √(2/(2πf μ₀σ)) with f = 10⁶ Hz gives ≈ 0.225 m — currents flow in a shallow sheet, which is why sea water is such a good ground-wave conductor.

Q11. Using E = 300√P·F/d: a 9 kW radiated MF signal with F = 0.5 over average land at 20 km gives approximately:Apply

a) 45 mV/m b) 22.5 mV/m (≈ 87 dBµV/m) c) 7.5 mV/m d) 2.25 mV/m

Answer: b). E = 300 × √9 × 0.5/20 = 300 × 3 × 0.5/20 = 22.5 mV/m — a strong city-grade AM signal.

Q12. A NAVTEX signal gives 25 mV/m at 100 km over sea (1/d²). For a receiver needing 2 mV/m, maximum range is approximately:Apply

a) 125 km b) 250 km c) 354 km d) 1250 km

Answer: c) ≈ 354 km. d = 100 × √(25/2) = 354 km. (A 1/d law would incorrectly give 1250 km.)

Q13. A quarter-wave monopole over perfect ground (efficiency 95%) fed with 5 kW gives approximately:Apply

a) ERP ≈ 8.7 kW, EIRP ≈ 14 kW b) ERP ≈ 5 kW, EIRP ≈ 5 kW c) ERP ≈ 15 kW, EIRP ≈ 8.7 kW d) ERP ≈ 2.7 kW, EIRP ≈ 4.4 kW

Answer: a). P_rad = 4.75 kW; monopole over perfect ground D ≈ 3 → ERP = 4.75 × 3/1.64 ≈ 8.7 kW; EIRP = ERP + 2.15 dB ≈ 14 kW.

Level 4 — Analyze

Q14. A coastal AM station's field at a seaside town 30 km away is 10 dB stronger than predicted from its land curves. Analysis indicates:Analyze

a) the path from the transmitter to that azimuth crosses sea water (σ ≈ 5 S/m): ground conductivity rises sharply along the coast, so the surface-wave attenuation per kilometre is much lower than the "average land" F-factor assumed — a coastal-recovery effect well documented in ground-wave curve sets b) the transmitter power doubled c) the sky wave is absent at that azimuth d) receivers near water are more sensitive

Answer: a). Mixed-path ground-wave prediction (e.g., ITU-R methods for land–sea transitions) shows the coastal boost: the wave launched over land "recovers" once it travels over sea. This is why AM coastal stations get surprisingly long maritime coverage — and why curve-based prediction must use the actual path profile.

Q15. Two identical MF broadcast transmitters 80 km apart must share a frequency to protect each other's ground-wave service area. Analysis of the night-time problem:Analyze

a) by day their 1/d² ground waves already interfere — nothing can be done b) by day the 1/d² ground-wave law protects each city (field at 80 km is ~35 dB down); the real threat is at night, when sky waves skip over and land inside the co-channel service area — the standard remedy is directional antenna arrays (nulls toward the other city) and/or night-time power reduction, not greater separation alone c) sky waves cannot exist at MF d) the stations should move to VHF

Answer: b). The day/night asymmetry is the essence of MF network planning: ground-wave co-channel protection distances are modest (tens of km at useful field levels), but night-time sky-wave protection demands either large separations (hundreds of km), directional arrays, or synchronized power reduction — a classic antenna-engineering solution (pattern nulls).

Q16. An operator replaces a λ/4 monopole with a top-loaded λ/8 vertical on 500 kHz (for a smaller mast) and finds coverage roughly halved. Analysis:Analyze

a) shortening the radiator lowers radiation resistance (R_r falls from ~36 Ω toward a few ohms) while ground-loss and loading-coil resistances stay similar — so efficiency η = R_r/(R_r+R_loss) collapses; top loading helps but cannot fully restore the missing height: the coverage loss is an efficiency loss, fixable with more radials/better ground, more power, or a taller structure b) the wavelength changed c) top loading is illegal d) the pattern became directional

Answer: a). Short verticals live or die by the ratio R_r/R_loss. At λ/8 over average ground, R_r ≈ 9–12 Ω; loss resistance of a modest radial system can be 5–15 Ω — efficiency may fall below 50% (−3 dB, halving range in the 1/d² regime). Ground improvement is the first remedy.

Q17. A receiving site 5 km from an LF transmitter suffers 20 dB fades on an otherwise stable ground-wave signal every few minutes. Neighbouring sites 2 km and 8 km do not. Analysis of the local mechanism:Analyze

a) the site sits in a local interference null where the direct surface wave and a locally reflected/bent component (from a building, hill edge, or buried conductor) cancel; the slow movement of that reflector (traffic, vegetation, tides of the soil moisture) sweeps the null across the site — a coherent multipath cancellation, not a propagation or transmitter fault, mitigated by relocating the antenna a fraction of a wavelength or using a small loop for directivity b) the transmitter is faulty c) sky wave is interfering — impossible at 5 km d) the ground wave cannot fade

Answer: a). At LF the wavelength is kilometres, so a moving reflector (even a steel structure or wet-ground boundary) sweeps interference fringes slowly — deep, slow fades are coherent cancellation in the near field. Half-wavelength relocation or a directive receive antenna solves it; this is antenna-course territory (Fresnel-zone and multipath reasoning).

Level 5 — Evaluate

Q18. A broadcaster must cover a 120 km diameter rural region. Options: (A) one 20 kW MF ground-wave station with a λ/4 monopole; (B) three 2 kW FM (VHF) stations on hills. Evaluating for coverage and cost:Evaluate

a) Choose A — ground wave always covers any distance if power is high enough b) Choose B: 120 km exceeds comfortable MF ground-wave range over average land (50–100 km at useful field strengths, and the 1/d² law makes the marginal kilometres exorbitantly expensive in power), whereas three low-power VHF hilltop sites give uniform high-quality coverage with cheap receivers; A would also need a costly high-power transmitter, large ground system, and suffer night-time sky-wave interference — the classic engineering reason most national broadcasters migrated to FM c) Both cost and cover identically d) Choose A but with a horizontal antenna for more gain

Answer: b). The 1/d² law is merciless: extending an MF ground-wave contour from 60 to 120 km costs ≈ 12 dB (×16 power). Three FM sites of 2 kW each outperform one 320 kW MF monster for a fraction of the cost — and option (d) is doubly wrong (horizontal antennas cancel the surface wave).

Q19. A maritime authority must specify a transmit antenna for a new NAVTEX station (518 kHz) to serve a 400 km coastal sector. Evaluating the options:Evaluate

a) A tall tower-mounted directional Yagi for maximum gain b) A ground-mounted (or low) λ/4 or top-loaded vertical monopole with an extensive radial ground system, omnidirectional in azimuth: ground wave needs low take-off and low loss, the service is broadcast to all bearings, and elevating/directing the antenna would waste energy into space waves and sky wave while the radial system (not the mast height) is what delivers the 13 dB sea-vs-land advantage over the sector c) A horizontally polarized dipole at λ/2 height d) A parabolic dish

Answer: b). NAVTEX is an omnidirectional maritime broadcast: the correct spec is a low vertical with the best possible ground system (≥ 0.25λ radials, sea-adjacent site if possible). Height and directivity are actively harmful to surface-wave launch; horizontal polarization cancels at the interface.

Q20. An engineer evaluates eLoran (100 kHz) as a GPS backup for East African ports. Evaluating the propagation argument:Evaluate

a) The case is strong: ground wave at 100 kHz is stable, predictable to tens of nanoseconds, penetrates indoors and underwater (skin depth metres), and is independent of GPS satellites — its weaknesses are large antennas (λ/4 ≈ 750 m, so top-loaded ~200–400 m masts), atmospheric noise at night, and multi-path from mountains, which are manageable with modern receivers and differential monitoring stations at the ports b) eLoran cannot work because LF signals never travel far c) eLoran is identical to GPS in every way d) LF is absorbed by sea water so ports cannot use it

Answer: a). Loran's engineering magic is precisely its ground wave: stable phase (timing to ~20 ns), no ionospheric dependence, sea-water paths are excellent, and differential eLoran removes residual path errors — the reason several nations reinstated it as a GNSS backup. The antenna size is the price of the long wavelength; everything else in the list (b–d) is factually wrong.

Grade Your Attempt

Click below to score all 20 questions and see your performance by Bloom's level.

Level 1 — Remember:

Level 2 — Understand:

Level 3 — Apply:

Level 4 — Analyze:

Level 5 — Evaluate:

Post-Test Answer Key (Quick Reference)

Q	Answer	Bloom Level
1	a — VLF/LF/MF	Remember
2	b — 1/d²	Remember
3	c — ≈ 5 S/m	Remember
4	a — ITU-R P.368 / FCC 73.184	Remember
5	a — image cancellation / low launch	Understand
6	b — night sky wave returns	Understand
7	b — depth of induced currents	Understand
8	a — λ = 3 km diffraction	Understand
9	c — 1 mV/m	Apply
10	b — 0.22 m	Apply
11	b — 22.5 mV/m	Apply
12	c — 354 km	Apply
13	a — 8.7 kW / 14 kW	Apply
14	a — coastal conductivity recovery	Analyze
15	b — night sky-wave interference	Analyze
16	a — efficiency collapse of short radiator	Analyze
17	a — coherent local multipath null	Analyze
18	b — three FM sites beat one MF brute	Evaluate
19	b — low vertical + radials	Evaluate
20	a — eLoran as GNSS backup	Evaluate

Prepared for ECE 523E — Antenna & Radio Wave Propagation, Department of Electrical & Communication Engineering, Masinde Muliro University of Science and Technology.