Instructions to Candidates
This quiz contains 20 questions on vestigial sideband (VSB) amplitude modulation. Questions are mapped to the first four levels of Bloom's taxonomy. Answer all questions, then check your work against the answer key and explanations after each question (click to reveal, or use the quick key at the end). Suggested time: 45 minutes. Total: 40 marks (2 marks per question). Key facts: VSB bandwidth = B + fv; Nyquist slope attenuation at the carrier = 6 dB; complementarity: |H(fc+fv)| + |H(fc−fv)| = constant.
| Level | Cognitive skill | What it tests | Questions |
|---|
| L1 | Remember | Recall of definitions, formulas, and facts | 1–5 |
| L2 | Understand | Explanation of concepts and principles | 6–10 |
| L3 | Apply | Numerical use of formulas in concrete situations | 11–15 |
| L4 | Analyze | Comparison, diagnosis, and analysis of relationships | 16–20 |
Section A — Remember (Questions 1–5)
Q1. A vestigial sideband (VSB) signal transmits:L1 Remember
- A. Both sidebands in full, with the carrier suppressed
- B. One sideband only, with the carrier suppressed
- C. One sideband in full plus a small vestige of the other sideband, with the carrier transmitted
- D. Both sidebands in full together with the carrier
Answer & Explanation
Answer: C
VSB retains one complete sideband, a deliberately limited portion (vestige, typically 0.75–1.25 MHz) of the other sideband, and transmits the carrier so that simple envelope detection remains possible. Option A is DSB-SC, B is SSB, and D is full-carrier DSB-AM.
Q2. The attenuation of the VSB shaping filter at the picture carrier frequency (the Nyquist slope reference point) is conventionally:L1 Remember
- A. 6 dB
- B. 0 dB (full response)
- C. 20 dB
- D. 3 dB at the receiver and 20 dB at the transmitter
Answer & Explanation
Answer: A
The Nyquist slope is odd-symmetric about the carrier, and the symmetry point is set 6 dB down (3 dB per filter in a paired transmitter/receiver design, conventionally quoted as a 6 dB total reference). The response then rolls off linearly to full stop at the lower vestige edge and rises to full pass at the upper edge.
Q3. For a message band-limited to B Hz, the transmission bandwidth of a VSB signal with vestige width fv is:L1 Remember
- A. 2B
- B. B
- C. B − fv
- D. B + fv
Answer & Explanation
Answer: D
VSB occupies the full message band on one side of the carrier plus the retained vestige on the other: BW = B + fv. This places it between SSB (B) and DSB/AM (2B). For TV: 5 MHz video + 1.25 MHz vestige ≈ 6.25 MHz, versus 10 MHz for DSB.
Q4. The classic mass application of VSB amplitude modulation is:L1 Remember
- A. HF amateur radio telephony
- B. Analogue television broadcasting (e.g. NTSC, PAL, SECAM)
- C. FM stereo multiplexing
- D. Satellite digital TV (DVB-S)
Answer & Explanation
Answer: B
Analogue TV adopted VSB worldwide from the 1940s because the video signal (4–6 MHz, DC-coupled) could not be SSB-filtered economically, yet DSB would have wasted 30–50% of scarce VHF/UHF spectrum. Digital descendants include ATSC 8-VSB. Amateur radio and satellite links use SSB and digital QPSK respectively.
Q5. In analogue television VSB practice, the retained vestige of the lower sideband is typically about:L1 Remember
- A. 50 kHz
- B. 4.2 MHz
- C. 0.75–1.25 MHz
- D. 6 MHz
Answer & Explanation
Answer: C
Standard values are 0.75 MHz (NTSC, 6 MHz channels) to 1.25 MHz (System B/G, 7–8 MHz channels). The Nyquist slope then extends from full rejection at −1.25 MHz (or −0.75 MHz), through the −6 dB point at the carrier, to full pass at +0.75 MHz. A vestige of 50 kHz would again demand an impossible filter; 4.2 or 6 MHz is a full video band, not a vestige.
Section B — Understand (Questions 6–10)
Q6. Television broadcasting chose VSB rather than SSB principally because:L2 Understand
- A. SSB occupies twice the bandwidth of VSB
- B. An SSB filter must pass the video band and reject an identical band beginning only tens of kilohertz away at RF (fractional bandwidth ~10−3), which is physically unrealizable — while VSB's gradual Nyquist transition is buildable
- C. SSB cannot transmit a carrier at all
- D. VSB gives better video quality than SSB for the same power
Answer & Explanation
Answer: B
The two sidebands of a 5 MHz video signal lie immediately adjacent to the carrier; selecting one and rejecting the other needs a fractional transition of order 10−3 — far beyond any realisable RF filter. VSB deliberately keeps a vestige so the transition becomes a gradual 1–2 MHz slope that LC, SAW or digital filters can realize. SSB is actually narrower (A is false), and quality is not the differentiator (D).
Q7. A simple diode envelope detector can recover the video from a VSB signal because:L2 Understand
- A. The vestige cancels the carrier before detection
- B. VSB signals have no sidebands and therefore no quadrature term
- C. The receiver regenerates the carrier with a Costas loop
- D. The picture carrier is transmitted at high level, making the waveform quasi-AM (self-heterodyne action), while the complementary receiver filter makes the vestige-plus-full-sideband sum flat
Answer & Explanation
Answer: D
Envelope detection needs a strong carrier reference embedded in the signal — which VSB deliberately transmits. The complementary filter restores flat overall response, so the envelope follows the video. This is exactly what DSB-SC lacks (hence synchronous detection there), and no carrier-recovery loop such as a Costas loop is required.
Q8. If the Nyquist slope of the transmitter's VSB filter is not perfectly linear and symmetric about the picture carrier, the visible result in the received picture is:L2 Understand
- A. A fixed tilt of the grey scale (differential gain error), because the vestige and full-sideband paths no longer sum to a constant
- B. Loss of sound carrier lock
- C. Complete loss of the upper sideband
- D. Over-modulation distortion of the sync pulses
Answer & Explanation
Answer: A
Flat video depends on the identity |H(fc+fv)| + |H(fc−fv)| = constant, which holds only for an odd-symmetric (Nyquist) slope. Asymmetry makes the sum frequency-dependent: some video frequencies arrive stronger than others, appearing as a fixed grey-scale tilt — a differential-gain error that the receiver cannot correct, since it is baked into the transmitted spectrum.
Q9. The picture carrier is transmitted at high level in a VSB TV signal mainly so that:L2 Understand
- A. The transmitter power amplifier can operate in class C for high efficiency
- B. The vestige can be filtered out more easily
- C. The millions of domestic receivers can use a simple, cheap diode envelope detector without any carrier-recovery circuitry
- D. Adjacent-channel interference is reduced
Answer & Explanation
Answer: C
The carrier provides the demodulation reference directly inside the received signal (self-heterodyning). This single decision is what kept the consumer TV receiver a tuned diode detector for half a century — an overwhelming economic advantage. The price is power inefficiency (the carrier carries no picture information), which broadcasters accepted because coverage, not transmitter economics per watt, dominates their budgets.
Q10. The complementarity condition in VSB system design states that:L2 Understand
- A. The vestige width must equal the full sideband width
- B. The gains of the shaping filter at mirror frequencies about the carrier sum to a constant: |H(fc+fv)| + |H(fc−fv)| = constant, so the total demodulated video response is flat
- C. The transmitter and receiver must use identical filters
- D. The sound carrier must be exactly 5.5 MHz from the picture carrier
Answer & Explanation
Answer: B
Each video frequency fv reaches the detector via two RF paths — the full sideband at fc+fv and the vestige at fc−fv. The Nyquist slope's odd symmetry guarantees their sum is constant for every fv. Equivalently, in a paired design the transmitter and receiver filters are mirror images of each other (complementary — not identical, so C is wrong).
Section C — Apply (Questions 11–15)
Q11. A TV system uses a video signal band-limited to 5 MHz with a vestige of 1.25 MHz. The occupied video bandwidth of the transmitted VSB signal is:L3 Apply
- A. 6.25 MHz
- B. 5.0 MHz
- C. 10.0 MHz
- D. 3.75 MHz
Answer & Explanation
Answer: A
BW = B + fv = 5 + 1.25 = 6.25 MHz. Compare with DSB: 2B = 10 MHz. The VSB choice saves 3.75 MHz (37.5%) of channel width per programme.
Q12. Relative to DSB-AM carrying the same 5 MHz video, VSB with a 1.25 MHz vestige saves approximately what percentage of channel bandwidth?L3 Apply
- A. 25%
- B. 12.5%
- C. 50%
- D. 37.5%
Answer & Explanation
Answer: D
DSB needs 2B = 10 MHz; VSB needs B + fv = 6.25 MHz. Saving = (10 − 6.25)/10 = 3.75/10 = 37.5%. In a 200 MHz UHF allocation this raises the channel count from 20 to 32 — the economic justification for VSB.
Q13. In System G, the picture carrier of channel E5 is at 175.25 MHz. The FM sound carrier frequency is:L3 Apply
- A. 174.00 MHz
- B. 180.75 MHz
- C. 176.00 MHz
- D. 170.75 MHz
Answer & Explanation
Answer: B
System G places the sound carrier 5.5 MHz above the picture carrier: 175.25 + 5.5 = 180.75 MHz. (174.00 MHz is the lower vestige edge, fc − 1.25 MHz; 176.00 MHz is the full-response edge, fc + 0.75 MHz.)
Q14. At a certain video frequency, a VSB transmitter filter has gain 0.35 on the vestige path and 0.65 on the full-sideband path. The sum of the two path gains is:L3 Apply
- A. 0.2275, varying with frequency
- B. 1.00, constant as required by the Nyquist slope symmetry
- C. 0.35, the vestige value only
- D. 1.30, which must be corrected by the receiver
Answer & Explanation
Answer: B
0.35 + 0.65 = 1.00. The Nyquist slope is odd-symmetric about the carrier, so mirror-frequency gains always sum to the same constant (here 1.0) at every video frequency — this is exactly the complementarity condition that yields a flat demodulated video response. (Option A, 0.2275, is the product — a common student error.)
Q15. A TV vision transmitter has a picture carrier power of 5 kW and a peak envelope power (sync tips) of 10 kW. The modulation index at sync tips is approximately:L3 Apply
- A. 0.25
- B. 1.00
- C. 0.41
- D. 0.50
Answer & Explanation
Answer: C
PEP = Pc(1 + m)² → 10 = 5(1+m)² → (1+m)² = 2 → m = √2 − 1 ≈ 0.41. TV vision modulation is modest because the carrier is deliberately large relative to the sidebands — the price paid for envelope-compatible receivers.
Section D — Analyze (Questions 16–20)
Q16. For the same message of bandwidth B, which ordering of schemes from narrowest to widest transmission bandwidth is correct?L4 Analyze
- A. VSB < SSB < DSB = AM
- B. SSB = VSB < DSB < AM
- C. SSB < VSB < DSB = AM
- D. AM < DSB < VSB < SSB
Answer & Explanation
Answer: C
SSB occupies B; VSB occupies B + fv (strictly more than SSB, since the vestige is retained); DSB-SC and full-carrier AM both occupy 2B (suppressing the carrier changes power, not bandwidth). Hence SSB < VSB < DSB = AM — VSB's defining position as the intermediate scheme.
Q17. A viewer reports that all pictures from one transmitter show a permanent grey-scale tilt (dark scenes too dark, bright scenes washed out), while a neighbouring channel is perfect. The most probable cause and remedy are:L4 Analyze
- A. Receiver IF filter misalignment; retune the viewer's set
- B. Multipath fading; raise the receiving antenna
- C. Over-deviation of the sound carrier; reduce FM deviation
- D. Transmitter Nyquist-slope asymmetry (slope error); the transmitter VSB filter must be realigned with a sweep-and-marker generator — the receiver cannot correct it
Answer & Explanation
Answer: D
A fault affecting only one channel, appearing as a fixed grey-scale tilt, is a differential-gain error baked into that transmitter's spectrum: the complementarity sum |H(fc+fv)| + |H(fc−fv)| is no longer constant because the Nyquist slope is misaligned or asymmetric. Since the distortion is transmitted, no receiver adjustment can fix it — the transmitter filter must be realigned to a fraction of a dB. Multipath (B) would vary with position; sound deviation (C) affects sound, not grey scale.
Q18. Analysis of the AM family shows VSB passes DC and very-low-frequency video content cleanly, whereas SSB struggles with it. The reason is:L4 Analyze
- A. VSB retains the carrier and uses gradual Nyquist-slope filtering referenced to the carrier, so no broadband 90° phase networks are needed; SSB generation/filtering cannot maintain accurate phase down to DC, corrupting the brightness information near f = 0
- B. SSB transmits too much power for low frequencies
- C. VSB suppresses the lower sideband completely, removing DC
- D. SSB has twice the bandwidth of VSB and therefore cannot reach DC
Answer & Explanation
Answer: A
Picture brightness is DC-coupled information. VSB's shaping is an amplitude slope about a transmitted carrier — phase behavior near DC is benign. SSB, by contrast, requires either a near-ideal steep filter or broadband Hilbert (90°) networks; neither maintains accurate phase and amplitude as f → 0, so the lowest video frequencies are distorted. This single fact — more than any other — ruled SSB out for television.
Q19. A regulator must squeeze more TV programmes into a congested UHF band. Analysis of the options — DSB-AM, SSB, VSB — for wideband video with DC content and cheap domestic receivers leads to the conclusion that:L4 Analyze
- A. DSB-AM is best because its receivers are simplest and no filter precision is required — bandwidth is irrelevant
- B. VSB is the only workable compromise: it recovers 30–40% of the DSB bandwidth, its Nyquist-slope filters are realizable, it passes DC, and its transmitted carrier keeps domestic receivers as simple envelope detectors; SSB fails on filter realizability and DC phase, DSB fails on bandwidth
- C. SSB is best because it is the narrowest, and receiver complexity is unimportant
- D. Any of the three is equally suitable; the choice is arbitrary
Answer & Explanation
Answer: B
The analysis must weigh all three currencies together. DSB doubles the required spectrum (A ignores bandwidth). SSB, though narrowest, demands unrealizable filters for video and corrupts near-DC content, and its suppressed carrier forces complex receivers (C is false). Only VSB simultaneously satisfies: spectrum saving (B + fv ≈ 6.25 MHz vs 10 MHz), buildable Nyquist-slope filters, DC fidelity, and envelope-detector receivers — exactly why every analogue TV standard adopted it.
Q20. Comparing demodulation of VSB and DSB-SC signals at the same received power, an engineer concludes that:L4 Analyze
- A. Both require synchronous detection, since both are sideband schemes
- B. DSB-SC can use envelope detection if a strong carrier is re-inserted, making it identical to VSB in every respect
- C. VSB permits envelope detection because its strong transmitted carrier self-heterodynes the vestige and full sideband into a flat video response, whereas DSB-SC's suppressed carrier destroys envelope-message proportionality, forcing synchronous (coherent) detection with carrier recovery
- D. VSB requires a Costas loop, while DSB-SC works with a diode detector
Answer & Explanation
Answer: C
The decisive difference is the carrier. VSB transmits it, so the sum of vestige and full-sideband paths beats against the carrier in a diode detector to reproduce the video (complementarity keeps the response flat). In DSB-SC the carrier is absent; the envelope is |m(t)|, polarity information is lost, and the receiver must regenerate the carrier (Costas loop, squaring loop, or pilot) for product detection. Option B fails because re-inserting a carrier strong enough for envelope detection wastes the very power DSB-SC was designed to save, and the vestige/full-sideband complementarity is a VSB-specific property.
Quick Answer Key
| Q | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|
| Key | C | A | D | B | C | B | D | A | C | B |
| Level | L1 | L1 | L1 | L1 | L1 | L2 | L2 | L2 | L2 | L2 |
| Q | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
|---|
| Key | A | D | B | B | C | C | D | A | B | C |
| Level | L3 | L3 | L3 | L3 | L3 | L4 | L4 | L4 | L4 | L4 |
Note to instructor: the key is provided for marking convenience. For student use, hide this section before distribution.