Space 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: LOS geometry and Fresnel clearance, multipath and two-ray fading, diffraction loss, path-loss models, link budgets, antennas (Yagi to dish), microwave backhaul and satellite links.

Level 1 — Remember

Q1. The space wave is the dominant propagation mode at:Remember

a) VHF (30 MHz) and above b) MF (300 kHz – 3 MHz) c) VLF (3 – 30 kHz) d) LF only

Answer: a) VHF and above. Below VHF, ground wave (LF/MF) and sky wave (HF) dominate; from 30 MHz upward the direct/space wave with its reflections and diffraction paths rules.

Q2. The free-space path loss formula is:Remember

a) L = 20log f + 10log d b) L(dB) = 32.44 + 20log f(MHz) + 20log d(km) c) L = 32.44 + 10log f + 10log d d) L = f × d / c

Answer: b). 20 dB per decade in both frequency and distance; 32.44 dB is the constant for f in MHz and d in km.

Q3. The standard tropospheric refraction convention for radio-horizon calculations uses:Remember

a) k = 1 earth b) k = 4/3 effective earth radius c) k = 2 earth d) flat earth

Answer: b) k = 4/3. The standard atmosphere bends rays downward, equivalent to a 4/3-radius earth: radio horizon d(km) ≈ 4.12(√h₁ + √h₂).

Q4. A parabolic dish antenna's gain is approximately:Remember

a) G ≈ η(πD/λ)², with η ≈ 0.5–0.7 b) G ≈ η(λ/πD)² c) G ≈ D/λ d) G independent of frequency

Answer: a). Gain grows with the square of aperture-to-wavelength ratio — the reason small dishes do big work at microwave frequencies.

Level 2 — Understand

Q5. Why must at least 60 % of the first Fresnel zone be clear even when the direct line of sight is unobstructed?Understand

a) secondary waves arriving within the Fresnel ellipsoid add in phase with the direct ray; blocking them causes destructive partial-cancellation loss — so "optical LOS" is not enough, the volume must be clear b) to allow for antenna sway only c) it is a bureaucratic requirement d) clearance only matters for radar

Answer: a). By Huygens' principle the received field is the coherent sum over the aperture; the first Fresnel zone contributes the in-phase energy. Obstruction into it costs typically 10–20 dB — the classic "mystery fade" when vegetation grows into the path.

Q6. In the two-ray (flat earth) mobile channel, received power falls as 1/d⁴ beyond the critical distance because:Understand

a) the direct and ground-reflected rays arrive with nearly opposite phase at grazing angles; their sum's magnitude squared then falls with the square of the small path difference, giving a fourth-power distance law — twice the free-space exponent b) the transmitter loses power with distance c) antennas shrink with distance d) the law is only empirical with no physical basis

Answer: a). With Γ ≈ −1 at grazing incidence, E ∝ (1/d)(Δφ) and Δφ ∝ 1/d, so P_r ∝ 1/d⁴. This is why cellular base stations need far more power than free-space estimates suggest.

Q7. Knife-edge diffraction loss at grazing incidence (v = 0) is approximately:Understand

a) 0 dB b) 3 dB c) 6 dB d) 20 dB

Answer: c) ≈ 6 dB. The classic Fresnel knife-edge result: even a grazing edge halves the field (6 dB); v = 1 costs ≈ 16 dB, v = 2 ≈ 22 dB. Geometry, not guessing, gives the number.

Q8. Rain attenuation becomes a first-order budget term roughly above:Understand

a) 100 MHz a) 1 GHz c) 10 GHz d) 100 GHz only

Answer: c) ≈ 10 GHz. Below C-band, rain loss is negligible; from Ku band up (γ_R = kR^α per ITU-R P.838) rain fade dominates link design and drives fade margins or site diversity.

Level 3 — Apply

Q9. The radio horizon between a 30 m tower and a 2 m rooftop (k = 4/3) is approximately:Apply

a) 28.4 km b) 39.0 km c) 19.2 km d) 42.5 km

Answer: a) ≈ 28.4 km. d = 4.12(√30 + √2) = 4.12(5.477 + 1.414) = 28.4 km.

Q10. The maximum first-Fresnel-zone radius for a 10 km link at 900 MHz is approximately:Apply

a) 2.9 m b) 8.7 m c) 17 m d) 29 m

Answer: d) ≈ 29 m. λ = 0.33 m; r₁max = ½√(λd) = ½√(0.33 × 10 000) ≈ 28.9 m. The 60 % clearance radius is ≈ 17 m — hedges and low buildings genuinely matter at 900 MHz (note option c states the 60 % clearance value, not the full first-zone radius).

Q11. FSPL at 14 GHz over 15 km is approximately:Apply

a) 118.8 dB b) 138.8 dB c) 158.8 dB d) 108.8 dB

Answer: b) ≈ 138.8 dB. 32.44 + 20log(14 000) + 20log(15) = 32.44 + 82.9 + 23.5 = 138.8 dB.

Q12. A 60 cm dish at 12 GHz (η = 0.6) has gain and 3 dB beamwidth of approximately:Apply

a) 18 dBi, 3.5° b) 34.7 dBi, 2.9° c) 28 dBi, 6° d) 40 dBi, 1.4°

Answer: b). G = 0.6(π×0.6/0.025)² ≈ 0.6×(75.4)² ≈ 3410 ≈ 34.7 dBi; HPBW ≈ 70λ/D = 70×0.025/0.6 ≈ 2.9°.

Q13. A 10 W (+40 dBm) transmitter at 14 GHz, 15 km, with 35 dBi dishes each end and 2 dB total feeder loss delivers approximately:Apply

a) −30.8 dBm at the receiver input b) −50.8 dBm c) −10.8 dBm d) +4.2 dBm

Answer: a) ≈ −30.8 dBm. P_r = 40 + 35 + 35 − 138.8 − 2 = −30.8 dBm. With 28 MHz noise bandwidth (N ≈ −94.5 dBm) this gives C/N ≈ 63.7 dB — plenty even after 30 dB of rain fade.

Level 4 — Analyze

Q14. A point-to-point microwave link shows 15 dB deep fades only on hot, calm afternoons. Analysis indicates:Analyze

a) multipath fading from tropospheric stratification: strong temperature inversions create layered refractive-index gradients so the direct ray interferes with a ducted/ground-hugging replica; the classic countermeasure is space diversity (two vertically separated receive antennas) so the two paths never null simultaneously b) sunspots c) rain — impossible in calm weather d) antenna icing in summer

Answer: a). Flat-fading multipath (not rain attenuation — that is frequency-selective in time and coincident with rain) is the signature of super-refractive/ducting conditions. Vertical space diversity with ~10–20λ separation decorrelates the nulls — standard on all long-haul microwave.

Q15. In a mobile receiver, signal strength fluctuates 30–40 dB within half a wavelength while the local mean varies slowly over tens of metres. Analyzing the structure:Analyze

a) the fast component is small-scale Rayleigh fading from many unresolved multipath components adding with random phases; the slow component is large-scale log-normal shadowing from building/terrain obstruction — two distinct statistical processes at two distinct scales, handled by different countermeasures (equalization/diversity vs handoff/margin) b) the fast component is transmitter noise c) both are caused by rain d) shadowing is the fast one

Answer: a). Scale separation is the key diagnostic: Rayleigh = λ-scale, shadowing = 10s of metres scale. Confusing them leads to wrong fixes (you cannot equalize shadowing, and you do not hand off against Rayleigh fades).

Q16. A Ku-band VSAT works perfectly at night but drops every heavy-storm afternoon. Analyzing the budget failure and the correct remedies:Analyze

a) rain attenuation at 12–14 GHz (γ_R = kR^α) can reach 5–10 dB/km in heavy rain — tens of dB over the slant path; the remedies, in order, are higher transmitter EIRP and/or larger receive G/T (dish diameter), adaptive coding and modulation (fall back to lower order), uplink power control, and site diversity for networks b) the satellite moves at night c) sun outage — those occur at midday around equinox, not every storm d) nothing can be done about rain

Answer: a). Rain fade is deterministic enough to engineer: size the clear-weather margin for the rainfall rate at your availability target (ITU-R P.618 methodology), and add adaptive waveforms so the link degrades gracefully instead of dropping.

Q17. A 5 GHz WiFi bridge across a valley works at 20 MHz channel width but fails at 80 MHz with the same radios, antennas and transmit power. Analysis:Analyze

a) wider bandwidth raises the noise floor by 10log(80/20) = 6 dB while receive power is unchanged, so C/N falls 6 dB; at the marginal link this pushes the higher-order MCS below threshold — the fix is more antenna gain/aperture, better alignment, or accepting lower MCS at the wider channel (or the opposite: use the 20 MHz channel and enjoy 6 dB better sensitivity) b) bandwidth does not affect noise c) the frequency changed from 5 GHz d) antennas absorb wider channels

Answer: a). Thermal noise N = −174 + 10log B: quadrupling B adds 6 dB of noise with zero extra signal. At marginal C/N, wider channels trade sensitivity for throughput — a fundamental budget identity, not an equipment fault.

Level 5 — Evaluate

Q18. An ISP must backhaul 5 Gbit/s over a 45 km rural path in a high-rainfall region. Options: (A) single E-band (80 GHz) link; (B) two parallel 6 GHz links with space diversity; (C) fibre. Evaluating:Evaluate

a) Choose A: E-band offers huge bandwidth so rain does not matter b) Choose B: at 80 GHz rain attenuation is extreme (tens of dB/km in heavy rain) making a 45 km hop nearly impossible in storm season, while 6 GHz suffers little rain loss, licensed channel pairs deliver the rate with lower-order QAM, and space diversity restores the fading margin — fibre (C) beats both on availability if the trenching cost is acceptable, but for rapid deployment B is the engineering-optimal radio solution c) Choose C but only because radio never works d) Choose A with a bigger dish — rain fades vanish with gain

Answer: b). Rain attenuation is an absorption process — antenna gain cannot defeat it (both wanted signal and the noise temperature argument aside, the fade is multiplicative on the path). Lower frequency + diversity + adaptive coding is the standard availability recipe; E-band is reserved for short urban hops.

Q19. A county government must connect 40 dispersed schools. Option A: one 40 m tower with a high-power omni AP; Option B: four 15 m sectorized sites on existing water tanks. Evaluating coverage engineering:Evaluate

a) Choose B: sector antennas concentrate EIRP into 90° beams (≈ 6 dB effective gain vs omni for the same power), shorter paths to each school reduce two-ray fading and Fresnel obstruction risk, water tanks give free height, and overlapping sectors add redundancy; the single tall tower wastes power in all directions at once, maximizes its own Fresnel clearance problems, and is a single point of failure b) Choose A: height always wins c) Both are identical d) Choose A because omnis are simpler to align

Answer: a). Sectorization converts one wasteful omni into several effective beams — the oldest trick in cellular engineering (it is why base stations look like they do). Height helps the horizon (√h law) but cannot substitute for directing energy where the users are.

Q20. An engineer argues a GEO satellite link "cannot work" because FSPL at 36 000 km is ~205 dB. Evaluating the claim:Evaluate

a) The claim confuses path loss with link viability: massive antenna gains on both ends (satellite EIRP ~50 dBW via high-gain antennas, earth stations with 25–35 dBi dishes and G/T ~30 dB/K) recover the budget easily, yielding C/N₀ ≈ 85–90 dBHz — decades of profitable GEO television and data services prove the arithmetic; the real constraints are rain fade at Ku/Ka, latency, and pointing — not the 205 dB itself b) The claim is correct — satellites must use HF c) FSPL does not apply in space d) Satellites work only below 1 GHz

Answer: a). The link budget is a chain: the 205 dB is paid by aperture on both ends (gain ∝ (D/λ)² works for the satellite too). When a single large number in a budget seems fatal, complete the chain before concluding — the discipline this course teaches.

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 — VHF and above	Remember
2	b — 32.44 + 20log f + 20log d	Remember
3	b — k = 4/3	Remember
4	a — η(πD/λ)²	Remember
5	a — in-phase secondary waves	Understand
6	a — direct + reflected cancellation	Understand
7	c — 6 dB	Understand
8	c — ~10 GHz	Understand
9	a — 28.4 km	Apply
10	d — ≈ 29 m	Apply
11	b — 138.8 dB	Apply
12	b — 34.7 dBi, 2.9°	Apply
13	a — −30.8 dBm	Apply
14	a — ducting multipath, space diversity	Analyze
15	a — Rayleigh vs log-normal scales	Analyze
16	a — rain fade remedies	Analyze
17	a — noise floor rises 6 dB	Analyze
18	b — two 6 GHz links w/ diversity	Evaluate
19	a — four sectorized sites	Evaluate
20	a — complete the budget chain	Evaluate

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