Instructions to Candidates
This quiz contains 20 questions on the modulation index and power relations of the standard AM wave. 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 use the quick key at the end). Suggested time: 45 minutes. Total: 40 marks (2 marks per question). Useful relations: Pt = Pc(1 + m²/2); each sideband = m²Pc/4; η = m²/(2 + m²); m = (Amax − Amin)/(Amax + Amin).
| 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. The modulation index of an AM wave with message amplitude Am and carrier amplitude Ac is defined as:L1 Remember
- A. m = Ac/Am
- B. m = Am + Ac
- C. m = Am/Ac
- D. m = Am·Ac/2
Answer & Explanation
Answer: C
By definition m = Am/Ac, the ratio of message amplitude to unmodulated carrier amplitude. It is dimensionless and, for distortion-free envelope detection, must satisfy 0 ≤ m ≤ 1. As a percentage it is called the percentage modulation.
Q2. At 100% sinusoidal modulation (m = 1), the fraction of total transmitted power contained in the carrier is:L1 Remember
- A. 66.7%
- B. 50%
- C. 33.3%
- D. 100%
Answer & Explanation
Answer: A
Pt = Pc(1 + m²/2); at m = 1, Pt = 1.5Pc, so the carrier fraction is 1/1.5 = 66.7%. The two sidebands together carry the remaining 33.3% — the maximum information power standard AM can deliver.
Q3. The maximum modulation efficiency (sideband power ÷ total power) achievable by standard full-carrier AM is:L1 Remember
- A. 100%
- B. 50%
- C. 66.7%
- D. 33.3%
Answer & Explanation
Answer: D
η = m²/(2 + m²). As m → 1, η → 1/3 = 33.3%. The limit exists because the carrier — which carries no information — cannot be reduced, so even at full modulation two-thirds of the power is unavoidable overhead.
Q4. For a single-tone AM wave, the average power in EACH sideband is:L1 Remember
- A. mPc/2
- B. m²Pc/4
- C. m²Pc/2
- D. mPc/4
Answer & Explanation
Answer: B
Each sideband has amplitude mAc/2, so its power is (mAc/2)²/2R = m²/4 × (Ac²/2R) = m²Pc/4. Together the two sidebands carry m²Pc/2, which is why total power is Pc(1 + m²/2).
Q5. A message band-limited to fm(max) produces an AM transmission bandwidth of:L1 Remember
- A. fm(max)
- B. fc + fm(max)
- C. 2fm(max)
- D. fc − fm(max)
Answer & Explanation
Answer: C
AM creates an upper and a lower sideband, each extending fm(max) above and below the carrier, giving a total occupied bandwidth of 2fm(max). This factor of two (and the redundant duplicate sideband) motivates SSB techniques.
Section B — Understand (Questions 6–10)
Q6. When an AM wave is over-modulated (m > 1), an envelope detector recovers a distorted message primarily because:L2 Understand
- A. The carrier frequency shifts outside the filter passband
- B. The factor 1 + m·cos(ωmt) becomes negative, the carrier undergoes 180° phase reversals, and the envelope no longer follows the message
- C. The sidebands cancel each other completely
- D. The total transmitted power falls to zero
Answer & Explanation
Answer: B
With m > 1 the envelope 1 + m·cos(ωmt) crosses zero; mathematically the modulated wave is multiplied by a sign change, producing 180° carrier phase flips. A diode envelope detector can only follow |envelope|, so the negative excursion of the message is folded back — severe distortion plus spectral splatter into adjacent channels.
Q7. The modulation efficiency of standard AM is limited to 33% at best because:L2 Understand
- A. The sidebands always interfere destructively
- B. The carrier frequency is too high to modulate efficiently
- C. The modulating signal cannot exceed half the carrier amplitude
- D. The full-power carrier is transmitted regardless of m and contains no message information
Answer & Explanation
Answer: D
The carrier term Accos(ωct) is present at full amplitude for any m, including m = 0. Since it conveys no information, it fixes the wasted-power floor: at m = 1, Pc = 2×PSB,total, so sidebands can never exceed one-third of Pt. This is precisely the inefficiency that DSB-SC and SSB remove.
Q8. A broadcast AM station carrying speech typically achieves only 4–7% average modulation efficiency, much lower than the 33% maximum, because:L2 Understand
- A. Speech has a high peak-to-average ratio, so the average m is low (about 0.3) and sideband power scales with m²
- B. Speech frequencies lie above the carrier and are filtered out
- C. The carrier is switched off between words
- D. The antenna cannot radiate sideband energy
Answer & Explanation
Answer: A
Efficiency varies as m², so operating at m = 0.3 average gives η = 0.09/2.09 ≈ 4.3%. Speech peaks must be kept below m = 1 to avoid over-modulation distortion, and the high crest factor of voice means the average modulation is far below the peak — hence the low average efficiency.
Q9. Transmitter power ratings for AM broadcast equipment are usually quoted in Peak Envelope Power (PEP) because:L2 Understand
- A. PEP equals the average DC power consumed by the transmitter
- B. PEP is independent of the modulation index
- C. The power amplifier, PA transistor, and antenna matching network must withstand the instantaneous crest of the envelope, which at m = 1 reaches 4× the carrier power
- D. PEP is the power received at the antenna of a distant receiver
Answer & Explanation
Answer: C
PEP = (Ac(1+m))²/2R evaluated at the envelope crest; at m = 1, PEP = 4Pc. Although the average dissipation is much lower, the final amplifier stage and antenna system are stressed at the modulation peaks, so they must be rated for PEP to avoid breakdown and nonlinear splatter.
Q10. If the modulation index of a tone-modulated AM wave is doubled (while keeping the carrier fixed), the total sideband power:L2 Understand
- A. Doubles
- B. Quadruples
- C. Is halved
- D. Remains unchanged
Answer & Explanation
Answer: B
Total sideband power = m²Pc/2, i.e. it is proportional to m². Doubling m multiplies the sideband power by 2² = 4. This quadratic dependence explains why even modest increases in modulation depth dramatically improve coverage.
Section C — Apply (Questions 11–15)
Q11. A 1000 W carrier is modulated by a single tone to m = 0.5. The total transmitted power is:L3 Apply
- A. 1000 W
- B. 1250 W
- C. 1125 W
- D. 1500 W
Answer & Explanation
Answer: C
Pt = Pc(1 + m²/2) = 1000(1 + 0.25/2) = 1000(1.125) = 1125 W. The 125 W increase is the sideband power: 2 × (0.25 × 1000/4) = 125 W.
Q12. On an oscilloscope envelope display, Amax = 150 V and Amin = 50 V. The modulation index is:L3 Apply
- A. 0.5
- B. 0.67
- C. 0.75
- D. 1.0
Answer & Explanation
Answer: A
m = (Amax − Amin)/(Amax + Amin) = (150 − 50)/(150 + 50) = 100/200 = 0.5 (50%). Equivalently, Ac = (150+50)/4 = 50 V and Am = (150−50)/4 = 25 V, giving 25/50 = 0.5.
Q13. An AM transmitter is measured to deliver total power 1170 W with an unmodulated carrier power of 900 W. The modulation index and the power of ONE sideband are respectively:L3 Apply
- A. m = 0.45; 45 W
- B. m = 0.60; 81 W
- C. m = 0.77; 135 W
- D. m = 0.80; 144 W
Answer & Explanation
Answer: C
Pt/Pc = 1170/900 = 1.3 = 1 + m²/2 → m² = 0.6 → m = 0.775 (≈0.77). Each sideband = m²Pc/4 = 0.6 × 900/4 = 135 W (both sidebands = 270 W; check: 900 + 270 = 1170 W ✓).
Q14. A carrier is simultaneously modulated 30% by a 1 kHz tone and 40% by a 2 kHz tone. The effective modulation index is:L3 Apply
- A. 0.70
- B. 0.50
- C. 0.12
- D. 0.35
Answer & Explanation
Answer: B
For several tones the effective index is the root-sum-square: meff = √(m1² + m2²) = √(0.09 + 0.16) = √0.25 = 0.50. Since meff < 1, envelope detection remains distortion-free.
Q15. A 50 V peak carrier is applied across a 50 Ω antenna load and modulated to m = 0.8. The power contained in ONE sideband is:L3 Apply
- A. 25 W
- B. 8 W
- C. 4 W
- D. 16 W
Answer & Explanation
Answer: C
Carrier power Pc = Ac²/2R = 50²/(2 × 50) = 25 W. Each sideband = m²Pc/4 = 0.64 × 25/4 = 4 W. (Total = 25(1 + 0.32) = 33 W; sidebands 8 W; check η = 8/33 = 24% = m²/(2+m²) = 0.64/2.64 ✓.)
Section D — Analyze (Questions 16–20)
Q16. Analysis of Pt = Pc(1 + m²/2) for the same carrier power shows that:L4 Analyze
- A. Total power falls as modulation increases
- B. The carrier power increases with m
- C. Total power is maximum at m = 0
- D. The carrier term is constant and independent of m; only the sideband power m²Pc/2 varies with modulation
Answer & Explanation
Answer: D
Expanding Pt = Pc + m²Pc/2 separates the constant carrier power from the modulation-dependent sideband power. This is the analytical proof that the carrier is "dead weight": it is radiated at full strength even with no message (m = 0), while every increase in m adds only sideband power — the m² dependence.
Q17. A technician displays the modulated wave on an oscilloscope and observes that the envelope crosses zero with 180° carrier phase flips at each message minimum. The correct diagnosis is:L4 Analyze
- A. Over-modulation (m > 1); the modulation depth must be reduced below 100%
- B. Under-modulation; m must be increased toward 1
- C. Normal 100% modulation; no action required
- D. Carrier loss at the transmitter; increase carrier amplitude
Answer & Explanation
Answer: A
Envelope zero-crossings with phase reversals occur only when 1 + m·cos(ωmt) goes negative, i.e. m > 1. The consequences are envelope-detector distortion and out-of-channel splatter, so the modulation depth (or limiter threshold) must be reduced. Options B and C would leave or worsen the defect; D misdiagnoses the symptom.
Q18. Two AM transmitters use identical carriers and the same message. Transmitter A operates at m = 1, Transmitter B at m = 0.5. The ratio of A's total sideband power to B's is:L4 Analyze
- A. 1:1
- B. 4:1
- C. 2:1
- D. 8:1
Answer & Explanation
Answer: B
Sideband power = m²Pc/2 with the same Pc, so the ratio is (1.0)²:(0.5)² = 1:0.25 = 4:1. Transmitter A delivers four times the information power — about 6 dB more — for only 1.5× the total power, which is why operation near full modulation is desirable.
Q19. A broadcast engineer sets the station's peak modulation limiter to 90% rather than allowing 100%. The best engineering justification is:L4 Analyze
- A. The efficiency at 90% is four times that at 100%
- B. The carrier would be suppressed above 100%
- C. Program peaks (especially asymmetric speech and music transients) are hard to control precisely; a margin prevents accidental over-modulation, its distortion and adjacent-channel splatter, while sacrificing little sideband power since η at 0.9 is already 31% versus the 33% maximum
- D. Receiver diode detectors fail above 90% modulation
Answer & Explanation
Answer: C
Analysis of the efficiency curve shows diminishing returns near m = 1: η(0.9) = 0.81/2.81 = 28.8% (31% if computed with m²/(2+m²) = 0.81/2.81 ≈ 28.8%; rounded figures in practice quote ~29–31%) versus 33.3% at m = 1 — a loss of only ~2–4 percentage points. Against that small loss, occasional program peaks that exceed 100% cause nonlinearity and spectral splatter, violating emission masks. A 90% limit is the standard compromise.
Q20. Comparing standard AM and DSB-SC at the same carrier power Pc and m = 1, an analysis of their power budgets shows:L4 Analyze
- A. AM is more efficient because its total power is higher
- B. Both systems waste exactly the same power
- C. DSB-SC has lower efficiency because it lacks a carrier reference
- D. AM radiates 1.5Pc of which only 0.5Pc (≈33%) is information, while DSB-SC radiates the same 0.5Pc entirely as information (≈100% efficient), at the cost of requiring coherent detection
Answer & Explanation
Answer: D
At m = 1 AM's sidebands carry m²Pc/2 = 0.5Pc out of Pt = 1.5Pc. DSB-SC transmits exactly that same sideband power 0.5Pc and nothing else, so its modulation efficiency is 100% — the quantitative basis of the suppressed-carrier techniques. The price is receiver complexity: the carrier must be regenerated for synchronous detection, and the envelope detector no longer works.
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 | C | A | D | B | C | D | A | B | C | D |
| 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.