Instructions to Candidates
This quiz contains 20 questions on suppressed-carrier amplitude modulation (DSB-SC and SSB). 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 provided after each question (click to reveal, or scroll to the full key at the end). Suggested time: 45 minutes. Total: 40 marks (2 marks per question).
| Level | Cognitive skill | What it tests | Questions |
|---|
| L1 | Remember | Recall of definitions, facts, and terminology | 1–5 |
| L2 | Understand | Explanation of concepts, principles, and why-methods work | 6–10 |
| L3 | Apply | Use of formulas and methods in numerical/concrete situations | 11–15 |
| L4 | Analyze | Comparison, differentiation, and analysis of relationships | 16–20 |
Section A — Remember (Questions 1–5)
Q1. The time-domain expression for a DSB-SC signal with message m(t) and carrier Accos(ωct) is:L1 Remember
- A. Ac[1 + m(t)]cos(ωct)
- B. Acm(t) + cos(ωct)
- C. Accos[ωct + m(t)]
- D. Acm(t)cos(ωct)
Answer & Explanation
Answer: D
DSB-SC is formed by the product of message and carrier: s(t) = Acm(t)cos(ωct). Option A describes full-carrier AM (the "1 +" term restores the carrier); C describes phase/frequency modulation; D is an additive, not multiplicative, combination.
Q2. In a full-carrier AM signal, approximately what fraction of the total transmitted power is contained in the carrier alone at 100% modulation?L1 Remember
- A. Two-thirds
- B. One-third
- C. One-sixth
- D. The whole power
Answer & Explanation
Answer: A
At m = 1, each AM sideband has one-quarter of the carrier power, i.e. Psb = Pc/4, so Pc = 4Psb. AM total power: PAM = Pc + 2Psb = 4Psb + 2Psb = 6Psb. SSB transmits a single sideband: PSSB = Psb. Hence PAM/PSSB = 6 (7.8 dB). This is the source of the widely quoted statement that SSB saves about two-thirds of the power of AM for equal information delivery.
Q3. Which of the following is the standard method of generating SSB in commercial HF transmitters?L1 Remember
- A. Filter method
- B. Phasing method
- C. Envelope detection
- D. Frequency modulation
Answer & Explanation
Answer: A
The filter method (balanced modulator followed by a highly selective crystal/ceramic sideband filter, usually at a low IF) is the dominant practical technique because modern crystal filters provide very high unwanted-sideband suppression at reasonable cost.
Q4. The Costas loop is used in suppressed-carrier receivers primarily to:L1 Remember
- A. Remove one sideband from a DSB-SC signal
- B. Suppress the carrier in the transmitter power amplifier
- C. Increase the modulation index of the received signal
- D. Recover the suppressed carrier for coherent demodulation
Answer & Explanation
Answer: D
Because the carrier is not transmitted, the receiver must regenerate it. The Costas loop generates a local carrier locked in frequency and phase to the original suppressed carrier, enabling synchronous (product) detection.
Q5. The Hilbert transform, used in the phasing method of SSB generation, shifts all frequency components of the message by:L1 Remember
- A. −90° for all frequencies (a −90° phase shift)
- B. +90° for all frequencies
- C. 180° for all frequencies
- D. A frequency-dependent delay of one message period
Answer & Explanation
Answer: A
The Hilbert transform applies a −90° phase shift to every frequency component of the message (and leaves amplitudes unchanged). Combined with the −90° shifted carrier channel, this allows one sideband to be cancelled arithmetically.
Section B — Understand (Questions 6–10)
Q6. Why can a simple diode envelope detector NOT be used to demodulate a DSB-SC signal?L2 Understand
- A. Because its envelope is |m(t)|, which loses the sign (polarity) information of the message, and the carrier phase reversals are unrepresented
- B. Because the DSB-SC signal contains no sidebands
- C. Because the DSB-SC signal is always below the noise floor
- D. Because the diode cannot respond to the high carrier frequency
Answer & Explanation
Answer: A
The envelope of DSB-SC is proportional to |m(t)| and the carrier undergoes 180° phase reversals at every message zero-crossing. An envelope detector outputs |m(t)|, a rectified version of the message — the polarity information is destroyed, so the message cannot be faithfully recovered without a coherent reference.
Q7. Why is a crystal (or ceramic) filter needed in the filter method of SSB generation rather than a simple LC bandpass filter?L2 Understand
- A. LC filters cannot operate at radio frequencies
- B. LC filters introduce too much 90° phase shift
- C. Crystal filters amplify the carrier to improve power efficiency
- D. The two sidebands lie very close together (separated by twice the lowest message frequency), requiring extremely sharp selectivity that only high-Q crystal filters provide
Answer & Explanation
Answer: D
For voice (300–3400 Hz), the inner edges of the two sidebands are separated by only 600 Hz at the carrier. An LC filter at RF with such a narrow transition is physically unrealizable; quartz crystal filters offer Q values of 10,000+ and the required shape factor, especially when the SSB is first formed at a low IF.
Q8. In coherent detection of SSB, a small frequency error Δf in the local oscillator causes all demodulated audio frequencies to be shifted by Δf. For voice communication this is serious mainly because:L2 Understand
- A. The whole spectrum is transposed, distorting the harmonic relationships of speech so that intelligibility and speaker recognition are destroyed
- B. The transmitted power drops to zero
- C. The sideband filter overheats
- D. The carrier is no longer suppressed
Answer & Explanation
Answer: A
Human hearing tolerates modest level (amplitude/phase) errors, but a uniform frequency shift destroys the fixed ratios between harmonics of the voice, producing the characteristic "Donald Duck" effect. Voice tolerates only about ±20–50 Hz of LO error, which is why SSB receivers need fine, stable frequency control.
Q9. In a balanced (ring) modulator, the carrier is suppressed at the output because:L2 Understand
- A. The diodes absorb the carrier energy
- B. A low-pass filter removes the carrier after rectification
- C. The symmetrical circuit topology causes equal and opposite carrier currents that cancel at the output, while the message-dependent products add
- D. The carrier frequency is converted to DC by the transformer
Answer & Explanation
Answer: C
Balance means geometric/electrical symmetry: both halves of the circuit carry identical carrier currents in opposite phase, so the carrier component cancels at the centre tap of the output transformer, whereas the product terms (sidebands) from the two halves reinforce. Residual carrier results from any circuit imbalance, which is why a balance trimmer is fitted.
Q10. Why does SSB require substantially less transmitted power than full-carrier AM for the same received signal quality?L2 Understand
- A. Because the SSB transmitter uses a more efficient power amplifier
- B. Because the message is amplified before modulation
- C. Because SSB occupies twice the bandwidth of AM
- D. Because no power is spent in the carrier and only one sideband is transmitted, so all radiated power carries information
Answer & Explanation
Answer: D
At 100% modulation full-carrier AM uses 2/3 of its power in the carrier and splits the remaining 1/3 between two sidebands. SSB removes the carrier entirely and transmits only one sideband, concentrating all power in information and halving the bandwidth — a combined saving of up to about 9 dB.
Section C — Apply (Questions 11–15)
Q11. A DSB-SC signal is generated from a 1 kHz tone message and a 100 kHz carrier. Which pair of frequencies appears in the transmitted spectrum?L3 Apply
- A. 99 kHz and 101 kHz
- B. 1 kHz and 100 kHz
- C. 100 kHz and 101 kHz
- D. 98 kHz and 102 kHz
Answer & Explanation
Answer: A
From s(t) = AcAmcos(2πfmt)cos(2πfct) = (AcAm/2)[cos2π(fc+fm)t + cos2π(fc−fm)t], the components lie at fc ± fm = 100 ± 1 kHz = 99 kHz and 101 kHz. The 100 kHz carrier itself is absent.
Q12. A DSB-SC transmitter uses Ac = 10 V and a tone message with Am = 0.8 V (into 1 Ω). The total average power is:L3 Apply
- A. 100 W
- B. 16 W
- C. 64 W
- D. 8 W
Answer & Explanation
Answer: B
For DSB-SC with a tone, each sideband has peak AcAm/2 = 4 V, so each carries 4²/2 = 8 W; total = 16 W. Equivalently Pt = (AcAm)²/4 = 64/4 = 16 W. Note no carrier power term appears.
Q13. A voice signal band-limited to 300 Hz–3.4 kHz modulates a carrier in SSB (upper sideband). If the carrier frequency is 2 MHz, the transmitted signal occupies the band:L3 Apply
- A. 2,000,300 Hz to 2,003,400 Hz
- B. 1,996,600 Hz to 2,003,400 Hz
- C. 1,996,600 Hz to 2,003,400 Hz including 2 MHz
- D. 2,003,400 Hz to 2,006,800 Hz
Answer & Explanation
Answer: A
USB occupies fc + fm for each message frequency: from 2,000,000 + 300 = 2,000,300 Hz to 2,000,000 + 3,400 = 2,003,400 Hz. The carrier at 2 MHz is suppressed and the lower sideband is rejected. Option A describes the LSB band; C wrongly includes the carrier.
Q14. In an SSB receiver the local oscillator is set 40 Hz above the suppressed carrier frequency. The received SSB signal contains a 1000 Hz audio tone. After product detection and low-pass filtering, the tone appears at:L3 Apply
- A. 1040 Hz
- B. 1000 Hz
- C. 960 Hz
- D. 2000 Hz
Answer & Explanation
Answer: A
The SSB tone appears in the RF domain at fc + 1000 Hz (USB). Mixing with fc + 40 Hz and low-pass filtering gives (fc + 1000) − (fc + 40) = 1040 Hz. The 40 Hz LO error shifts every audio component upward by 40 Hz — audible and degrading to voice quality.
Q15. A full-carrier AM transmitter and an SSB transmitter deliver the same sideband (information) power into the same load, with the AM tone at m = 1. The ratio of AM total power to SSB total power is:L3 Apply
- A. 2:1
- B. 1:1
- C. 6:1
- D. 8:1
Answer & Explanation
Answer: C
Let the carrier amplitude be Ac; the envelope peaks at 2Ac (m = 1). AM: total power PAM = Ac²/2 + 2×(mAc/2)²/2 = Ac²/2 + Ac²/4 = 3Ac²/4. To reach the same peak envelope, SSB must have peak amplitude 2Ac, i.e. its single-sideband amplitude is 2Ac/√2, giving PSSB = (2Ac)²/(2×2) = Ac²/2. Therefore PAM/PSSB = (3Ac²/4)/(Ac²/2) = 3/2… this equal-envelope comparison shows the carrier absorbs two-thirds of AM power. However, the standard exam comparison holds the information (sideband) power equal: AM then needs Pc = 4Psb extra, so PAM = 6Psb vs PSSB = Psb → 6:1, i.e. the well-known 9 dB (≈ factor 8) SSB saving quoted at 100% modulation.
Section D — Analyze (Questions 16–20)
Q16. Consider three schemes for the same 3 kHz voice channel: (i) full-carrier AM, (ii) DSB-SC, (iii) SSB. Which ordering correctly ranks them from narrowest to widest transmission bandwidth?L4 Analyze
- A. SSB < DSB-SC = AM
- B. AM < DSB-SC < SSB
- C. DSB-SC < SSB < AM
- D. SSB = DSB-SC < AM
Answer & Explanation
Answer: A
AM and DSB-SC both transmit two sidebands (2B = 6 kHz); suppressing the carrier changes power, not bandwidth. SSB transmits one sideband only (B = 3 kHz). Therefore SSB is narrowest, and DSB-SC and AM are equal.
Q17. An engineer finds that the output of her balanced modulator contains a strong carrier component. The most probable cause is:L4 Analyze
- A. Circuit imbalance (unequal diode characteristics, transformer asymmetry, or unequal carrier drive) that prevents complete cancellation of the carrier
- B. The message frequency is too high
- C. The output bandpass filter is tuned to the wrong centre frequency
- D. Excessive modulation index
Answer & Explanation
Answer: A
Carrier suppression relies on exact symmetry so that the two carrier currents cancel at the output transformer. Any component mismatch, unequal drive amplitude/phase, or transformer asymmetry unbalances the bridge and leaks carrier through. This is diagnosed and corrected with the balance (null) adjustment, not by changing message frequency or filter tuning.
Q18. A receiver must demodulate a DSB-SC signal with no transmitted pilot tone. The designer is choosing between (i) a Costas loop and (ii) a squaring loop. Which analysis is correct?L4 Analyze
- A. The squaring loop is preferred because it has no phase ambiguity
- B. Neither works without a pilot tone
- C. The Costas loop is preferred because it tracks the carrier continuously and provides the message directly from its in-phase branch; the squaring loop's divide-by-two introduces a 180° phase ambiguity
- D. Both require an envelope detector as a first stage
Answer & Explanation
Answer: C
Both loops regenerate the carrier without a pilot, so C is false. The squaring loop's divide-by-two can settle in either of two phases (180° ambiguity), which for analog voice only flips the recovered polarity (inaudible) but must be resolved by differential encoding in data systems. The Costas loop avoids the divider and yields the message at the I-branch output directly, with the Q-branch supplying the phase-error control voltage.
Q19. A crowded HF band must carry the maximum number of 3 kHz voice channels. Comparing SSB (3 kHz spacing) with full-carrier AM (6 kHz spacing), and assuming equal transmitter peak power, which conclusion follows from analysis of bandwidth and power?L4 Analyze
- A. AM carries more channels because its carrier aids reception
- B. SSB carries more channels but with lower received SNR than AM
- C. The two systems carry equal channel counts
- D. SSB doubles the channel count and, for equal peak power, delivers up to ~6 dB more information power per channel (both carrier and one sideband removed)
Answer & Explanation
Answer: D
Halving bandwidth doubles the number of channels in a fixed allocation. Power analysis shows SSB radiates only information power: at equal peak envelope the ratio of AM total power to SSB power is 6:1 (≈7.8 dB), and even comparing information power alone the SSB advantage is ~3 dB; additionally the SSB receiver admits half the noise bandwidth, further improving SNR. Thus SSB wins on both counts — D is wrong.
Q20. In the phasing method of SSB generation, the unwanted sideband is only 30 dB below the wanted sideband. The best explanation and remedy is:L4 Analyze
- A. The carrier frequency is unstable; replace the oscillator
- B. The Hilbert transform introduces distortion; remove it
- C. The two balanced-modulator outputs are not in exact amplitude balance and 90° phase quadrature across the message band, so sideband cancellation is incomplete; remedy: tighten amplitude matching and phase accuracy, or add a cleanup sideband filter (hybrid approach)
- D. The message bandwidth is too narrow; widen it
Answer & Explanation
Answer: C
Sideband cancellation in the phasing method is arithmetic: the unwanted sideband is formed by vector residue of the two channels. Finite tolerance in the 90° networks (typically ±2°) and amplitude mismatch leaves a residual. Since cancellation improves by ~1 dB per 0.1° of phase accuracy near 90°, practical phasing systems achieve only 30–40 dB; a following filter (hybrid/phasing-filter technique) raises suppression to 60 dB+.
Quick Answer Key
| Q | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|
| Key | D | A | A | D | A | A | D | A | C | D |
| Level | L1 | L1 | L1 | L1 | L1 | L2 | L2 | L2 | L2 | L2 |
| Q | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
|---|
| Key | A | B | A | A | C | A | A | C | D | C |
| Level | L3 | L3 | L3 | L3 | L3 | L4 | L4 | L4 | L4 | L4 |
Note to instructor: answer keys are shown for marking convenience. For student use, hide the key section before distribution.