Analog Television — 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: composite video and scanning, vestigial sideband and channel plans, synchronization, transmitters and receivers, propagation impairments, coverage planning and receiving antennas.

Level 1 — Remember

Q1. In the 625-line, 50-field television system, the line (horizontal) frequency is:Remember

a) 625 Hz b) 312.5 Hz c) 15 625 Hz d) 25 Hz

Answer: c) 15 625 Hz. f_L = (f_V/2) × N = 25 × 625 = 15 625 Hz; the line period is 64 µs.

Q2. In negative-modulation analogue TV:Remember

a) white level corresponds to maximum carrier amplitude b) white level corresponds to minimum carrier amplitude, and sync tips to maximum c) sync tips correspond to minimum carrier amplitude d) the sound carrier is amplitude-modulated

Answer: b). Negative modulation (used in PAL/NTSC) puts white at low carrier power: impulse noise appears as dark specks (less visible), and sync — at maximum power — is protected against over-modulation.

Q3. In an 8 MHz TV channel plan, the vision carrier is located:Remember

a) 1.25 MHz above the lower channel edge b) at the channel centre c) 1.25 MHz below the upper channel edge d) exactly at the lower channel edge

Answer: a). The vision carrier sits 1.25 MHz above the lower edge, leaving room for the vestigial lower sideband below it and up to ~5 MHz of upper sideband above it (e.g. channel E21: 471.25 MHz for a 470–478 MHz channel).

Q4. The sound carrier in a System B/G television channel is:Remember

a) amplitude-modulated, 5.5 MHz below the vision carrier b) amplitude-modulated, 5.5 MHz above the vision carrier c) frequency-modulated, 5.5 MHz above the vision carrier d) frequency-modulated, at the vision carrier frequency

Answer: c). The aural carrier is FM (deviation ±50 kHz, pre-emphasis 50 µs), offset +5.5 MHz above the vision carrier in B/G systems. FM gives the sound capture effect and constant amplitude.

Level 2 — Understand

Q5. Vestigial-sideband transmission is used for the TV vision signal because it:Understand

a) doubles the available video bandwidth b) roughly halves the channel width compared with double-sideband AM while remaining compatible with simple envelope detection c) eliminates the need for horizontal blanking d) improves the FM sound quality

Answer: b). Full DSB of a 5 MHz video would need ~11 MHz. VSB sends one nearly-complete sideband plus a vestige of the other; the receiver's Nyquist-slope IF filter restores a flat video response — simpler and phase-safer than true SSB.

Q6. Interlaced scanning (two fields per frame) is used primarily to:Understand

a) double the vertical resolution b) present 50 screen refreshes per second (flicker-free) while transmitting only 25 frames per second, keeping bandwidth low c) simplify the receiver tuner d) allow the sound carrier to be closer

Answer: b). The eye needs ~50 refreshes/s to avoid flicker, but 50 full frames/s would double the video bandwidth. Interlacing refreshes each half-picture 50 times/s, so bandwidth corresponds to 25 Hz while flicker corresponds to 50 Hz.

Q7. The pre- and post-equalizing pulses in the field-sync train exist to:Understand

a) set the AGC level for the field b) regenerate the colour subcarrier c) give both field types identical timing conditions so the exact half-line offset of interlace is preserved d) carry the audio during field blanking

Answer: c). Without equalizing pulses the two field types would integrate to different levels and lines would "pair" — two lines scanned together with a gap, visibly halving vertical resolution.

Q8. In a superheterodyne TV receiver, the Nyquist slope in the IF filter:Understand

a) complements the transmitter's vestigial-sideband filtering so the detected video amplitude response is flat across the band b) removes the colour subcarrier completely c) demodulates the FM sound carrier d) generates the EHT for the CRT

Answer: a). The transmitter cuts the lower sideband with a vestigial slope; the receiver applies a matching Nyquist slope at the vision IF so total response is flat — otherwise the picture would be frequency-distorted (smearing).

Level 3 — Apply

Q9. Channel E23 in an 8 MHz plan starts at 486 MHz (System G). The vision and sound carriers are:Apply

a) vision 487.25 MHz, sound 492.75 MHz b) vision 487.25 MHz, sound 492.25 MHz c) vision 486.00 MHz, sound 491.50 MHz d) vision 488.75 MHz, sound 494.25 MHz

Answer: a). Vision = 486 + 1.25 = 487.25 MHz; sound = 487.25 + 5.5 = 492.75 MHz. Vision upper sideband reaches ≈ 492.25 MHz, just under the sound carrier — all inside 486–494 MHz ✓.

Q10. The theoretical maximum video frequency of the 625-line system (575 active lines, 52 µs active line, 4:3 aspect) is approximately:Apply

a) 3.7 MHz b) 7.4 MHz c) 14.7 MHz d) 5.0 MHz

Answer: b) ≈ 7.4 MHz. f = 575 × (4/3) / (2 × 52 µs) ≈ 7.37 MHz. With the 0.7 Kell factor the practical value is ≈ 5.2 MHz — hence the transmitted luminance bandwidth of ≈ 5 MHz (option d is the practical value, not the theoretical one).

Q11. A ghost appears displaced 1/8 of the screen width. The echo's excess path length is approximately:Apply

a) 0.20 km b) 1.95 km c) 3.90 km d) 15.6 km

Answer: b) ≈ 1.95 km. Δt = (1/8) × 52 µs = 6.5 µs; excess path = cΔt = 3×10⁸ × 6.5×10⁻⁶ ≈ 1.95 km. This is how an installer converts a visible ghost into a distance to a suspected reflector.

Q12. A 5 kW transmitter feeds an antenna of 11 dBd gain. The ERP and EIRP are:Apply

a) ERP ≈ 55 kW, EIRP ≈ 90 kW b) ERP ≈ 63 kW, EIRP ≈ 103 kW c) ERP ≈ 16 kW, EIRP ≈ 26 kW d) ERP ≈ 50 kW, EIRP ≈ 61 kW

Answer: b) ≈ 63 kW ERP / 103 kW EIRP. ERP = 5 kW × 10^(1.1) ≈ 63 kW; converting dBd→dBi adds 2.15 dB, i.e. ×1.64, giving ≈ 103 kW EIRP.

Q13. At 600 MHz and 25 km, the free-space path loss is approximately:Apply

a) 95 dB b) 115.9 dB c) 135 dB d) 75.9 dB

Answer: b) ≈ 115.9 dB. FSPL = 32.44 + 20log(600) + 20log(25) = 32.44 + 55.56 + 27.96 = 115.96 dB.

Level 4 — Analyze

Q14. A receiver shows a sharp ghost 1/10 of the screen width to the LEFT of the main image (with a weaker main image). Analysis indicates:Analyze

a) the direct signal is blocked and the strongest path is a reflection arriving EARLIER than the old direct path — the "ghost" is actually the earlier-arriving signal and the old main image is now the delayed echo b) the receiver is mistuned by one channel c) the field sync has failed d) the sound trap is open

Answer: a). The display always draws the FIRST-arriving signal's image at the reference position; anything later appears to the RIGHT. A left-side ghost means a new, stronger/earlier path (e.g. after a direct-path obstruction from construction or foliage) has become the reference — the old "main" image is now the echo on the right. Delay for 1/10 screen ≈ 5.2 µs ≈ 1.6 km excess path.

Q15. A viewer's picture shows two images ~15 µs apart, and the colour on the second image is wrong (torn). Comparing with the monochrome content, the colour is damaged more because:Analyze

a) the ghost is an unmodulated delayed replica added vectorially to the composite signal: luminance just brightens/darkens (amplitude addition), but the 4.43 MHz chrominance is a phase reference system — the delayed subcarrier rotates the hue on the ghost and beats against the direct subcarrier, destroying colour while monochrome stays recognizable b) colour CRTs cannot display fast motion c) the sound IF is interfering with the video d) the AGC is too fast

Answer: a). Chrominance lives in the phase of the subcarrier (hue) and its amplitude (saturation). A delayed echo adds a phase-shifted subcarrier: Δφ = 360° × 4.43 MHz × 15 µs ≈ 23 900° → effectively a near-random hue on the ghost plus a 15 µs-period beat (visible as colour herringbone). Luminance, being amplitude-only, tolerates the same echo far better.

Q16. Two co-channel analogue stations on the same frequency produce slowly wandering diagonal bars. Analyzing the mechanism:Analyze

a) the two carriers beat at their small frequency difference, and that beat frequency modulates the vision envelope (or the sync), producing a bar pattern that drifts across the picture; protection against it is large co-channel distance ratios and directional antennas b) the two sound carriers heterodyne in the audio stage only c) it is caused by pre-emphasis d) it proves the receiver IF is too wide

Answer: a). Co-channel interference (CCI) in analogue TV appears as a "sandwich" pattern: the interfering carrier, offset by Δf (hertz), beats with the wanted carrier. The beat drifts across the raster at Δf lines/field. Mitigation at the receiving end is antenna directivity (F/B and side-lobe suppression) — an antenna-course solution.

Q17. An installer's customer at the edge of coverage has a snowy picture on one channel but a watchable one on another from the same mast. Analysis of the difference:Analyze

a) the snowy channel is likely on a higher frequency (more path loss: FSPL ∝ f²) and/or from a lower-ERP transmitter; a wideband log-periodic replaces the narrowband Yagi if the antenna is mistuned, and every dB of cleaner downlead or added antenna gain directly improves C/N on the snowy channel b) channels from one mast always have identical field strength c) snow is caused by sound–vision spacing d) the receiver needs a new CRT

Answer: a). Field strength differences of 10–20 dB between channels of one multiplex/mast are common (different ERP, different frequency path loss 6 dB per octave of frequency, different antenna patterns). The remedy is a link-budget fix: more gain, lower loss, better match — quantified in dB, which is why installers speak in dBµV and dBd.

Level 5 — Evaluate

Q18. A broadcaster must choose between (A) raising a 120 m tower to 240 m at constant ERP, and (B) doubling transmitter power at constant 120 m height, to expand coverage. Evaluating the engineering trade-off:Evaluate

a) Choose B: doubling power always doubles coverage area b) Choose A: radio horizon d ∝ √h doubles range (×1.41) and ~doubles coverage AREA with no increase in electricity or interference, whereas doubling power adds only ~3 dB (×1.41 range) at double running cost and higher interference to co-channel neighbours c) Both give identical results d) Height is irrelevant for UHF

Answer: b). Coverage radius scales with √h (height is the most cost-effective coverage tool) while power only enters as 10log P — doubling gives just 3 dB. Higher ERP also worsens co-channel reuse distance across the whole network plan, so option A is preferred on both coverage and interference grounds.

Q19. An engineer specifying a receiving antenna for a colour-PAL fringe location must prioritize. Evaluating the requirements, the BEST specification is:Evaluate

a) Highest possible gain at the vision carrier, narrowband, since gain fixes snow b) Flat gain/impedance across the whole 8 MHz channel, horizontal polarization, and high front-to-back ratio — because the 4.43 MHz chrominance needs the full 2.9–5.9 MHz band intact, and rearward echoes rotate chrominance phase (hue errors) even when luminance looks acceptable c) Vertical whip, since it is omnidirectional d) Any antenna, colour is robust to echoes

Answer: b). A narrowband high-gain Yagi peaked at the vision carrier (a) would pass colour poorly. Colour is MORE fragile than monochrome: echoes beyond ~15 ns rotate hue visibly. The professional fringe specification therefore trades peak gain for band-flatness and F/B ratio — the antenna properties that protect chrominance.

Q20. A regulator evaluates a proposal to keep one analogue PAL channel on air alongside DVB-T2 for "legacy viewers". Evaluating the spectrum argument:Evaluate

a) Keeping the analogue channel costs nothing because it was always there b) The 8 MHz occupied by one noisy analogue programme could carry a full DVB-T2 multiplex of 10+ digital services, and continued analogue radiation also raises the interference floor (co-channel and taboo-channel) that degrades the digital multiplex at its threshold — so the engineering-economics case strongly favours complete switch-off with subsidy of digital receivers instead c) Analogue and digital cannot share the UHF band under any circumstances d) The digital cliff means analogue viewers get better pictures anyway

Answer: b). Spectrum has opportunity cost: 1 analogue programme vs ~10+ digital services per 8 MHz. Analogue interferers are especially damaging to DTV because digital reception fails abruptly at threshold — an analogue co-channel signal can push a marginal digital location over the cliff. This is precisely the reasoning that drove Kenya's (and the world's) analogue switch-off.

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	c — 15 625 Hz	Remember
2	b — white low / sync high carrier	Remember
3	a — 1.25 MHz above lower edge	Remember
4	c — FM, +5.5 MHz above vision	Remember
5	b — halves channel vs DSB	Understand
6	b — 50 Hz flicker at 25 Hz bandwidth	Understand
7	c — preserves half-line interlace	Understand
8	a — complements VSB slope	Understand
9	a — 487.25 / 492.75 MHz	Apply
10	b — ≈ 7.4 MHz	Apply
11	b — ≈ 1.95 km	Apply
12	b — 63 kW ERP / 103 kW EIRP	Apply
13	b — ≈ 115.9 dB	Apply
14	a — earlier-arriving reflection	Analyze
15	a — echo rotates chrominance phase	Analyze
16	a — carrier beat (sandwich pattern)	Analyze
17	a — per-channel link budget differences	Analyze
18	b — height beats power	Evaluate
19	b — flat band, horizontal pol., high F/B	Evaluate
20	b — spectrum opportunity cost, interference floor	Evaluate

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