Masinde Muliro University of Science and Technology

Introduction to Antennas

Post-Test Quiz — Electrical & Communication Engineering (5th Year)

Course Assessment  |  20 Questions  |  Bloom's Taxonomy Levels 1–3  |  Time: 40 Minutes

Instructions to Candidates

Tips: Read each stem carefully. For "Apply" questions, work out the calculation on paper before selecting an option — unit consistency (watts, metres, dB, degrees) is essential.

Bloom's Taxonomy Coverage (Levels 1–3 Only)

LevelSkill TestedQuestion NumbersCount
Level 1 — RememberRecall definitions, symbols, and standard facts about antennasQ1, Q2, Q3, Q4, Q5, Q6, Q77
Level 2 — UnderstandExplain concepts, interpret patterns, and relate parametersQ8, Q9, Q10, Q11, Q12, Q13, Q147
Level 3 — ApplyUse formulas to compute gain, wavelength, beamwidth, power density, and apertureQ15, Q16, Q17, Q18, Q19, Q206

Part A — Quiz (Attempt All Questions)

Part B — Post-Test Answers & Explanations

Each answer below includes a concise explanation and its Bloom's taxonomy level, suitable for post-test review and moderation purposes.

QAnswerExplanationBloom Level
1(c)An antenna is a transducer: it converts electrical energy from a transmission line into radiated electromagnetic waves (transmit mode), and the reverse on reception (reciprocity).Remember
2(b)The far field (Fraunhofer) region is defined by R > 2D²/λ. Inside this region, angular field distribution is independent of distance and the wave approximates a plane wave.Remember
3(d)Isotropic radiator: equal radiation intensity in all directions. It is a theoretical lossless reference source, not physically realisable.Remember
4(c)Directivity is defined as maximum radiation intensity divided by the average radiation intensity: D = Umax/Uavg. It is a dimensionless ratio (expressed in dBi relative to isotropic).Remember
5(b)A half-wave dipole has a feed-point impedance of approximately 73 + j42 Ω; with end effect it is usually quoted near 73 Ω (resonant length slightly less than λ/2).Remember
6(a)Beamwidth (HPBW) is the angular separation between the two directions where the power pattern falls to one-half its maximum value (i.e., −3 dB points).Remember
7(d)Standard relations: HPBW (degrees) ≈ 70 λ/D for large uniform apertures; Aeff = λ²G/(4π); and directivity of a short dipole is 1.5. HPBW ≈ 2π/N is not standard.Remember
8(b)Wire antennas (dipoles, loops) are low-directivity, broad-beam radiators. High directivity is achieved with aperture antennas (horns, dishes), arrays, or Yagi-Uda structures.Understand
9(c)A parabolic reflector collimates rays from a feed at the focus into a narrow beam — high directivity/gain with high aperture efficiency, the standard choice for satellite ground stations.Understand
10(b)Near the horizon (low elevation), sky waves can travel very long distances via ionospheric reflection, and the horizon itself is the limiting geometry for direct waves. Ground wave at VHF is limited; "waves travel slower horizontally" is false.Understand
11(b)Grazing/large separation from a ground plane produces image currents that cancel the radiated field, driving input resistance toward zero — a very poor match to the feed line.Understand
12(c)End-fire means maximum radiation along the array axis (θ = 0° and 180°), as in a Yagi-Uda or properly phased linear array. Broadside is perpendicular to the axis.Understand
13(c)Gain G = ηD. For an efficiency of 55% (0.55) and D = 10, G = 5.5 ≈ 7.4 dB. The distractor 10 dB (10 W/W) ignores the efficiency loss.Understand
14(d)All statements are true: reciprocity holds (same radiation pattern transmit/receive), the same effective aperture formula applies on receive, and polarisation must be matched to avoid polarisation mismatch loss.Understand
15(b)λ = c/f = 3×10⁸ / 3×10⁹ = 0.1 m = 10 cm.Apply
16(b)EIRP = Pin × G(linear) = 50 W × 10 = 500 W. EIRP(dBm) ≈ 57 dBm; 50 W × 100 dB is a common misuse of dB as a linear multiplier.Apply
17(c)HPBW ≈ 70λ/D = 70(0.1)/0.5 = 14°. This illustrates the inverse relation between aperture size and beamwidth.Apply
18(b)S = Prad/(4πR²) = 1000/(4π·10⁶) ≈ 7.96×10⁻⁵ W/m². The 2πR² option is a frequent error (spherical surface area is 4πR²).Apply
19(c)Aeff = λ²G/(4π) = (0.1)²×10/(4π) = 0.1/(4π) ≈ 7.96×10⁻³ m² ≈ 0.008 m² (≈ 80 cm²).Apply
20(c)Prec = S × Aeff. With S from Q18: 7.96×10⁻⁵ × 7.96×10⁻³ ≈ 6.33×10⁻⁷ W ≈ 0.63 μW. Note that idealised Friis with Gt = 10, Gr = 10, R = 1 km, λ = 0.1 m gives Pr = PtGtGr(λ/4πR)² ≈ 6.3×10⁻⁷ W — consistent.Apply