| Bloom's Level | What it tests | Questions |
|---|---|---|
| Level 1 — Remember | Recalling definitions, symbols, units, standard values | 1–7 |
| Level 2 — Understand | Explaining concepts, interpreting waveforms and spectra | 8–14 |
| Level 3 — Apply | Calculating modulation index, bandwidth, power, amplitudes | 15–20 |
Correct answer: (b) Amplitude.
By definition, AM varies the instantaneous amplitude of a high-frequency carrier in proportion to the message signal, while the carrier frequency and phase remain constant. Varying frequency gives FM; varying phase gives PM.
Correct answer: (c) A_m / A_c.
The modulation index is the ratio of the message amplitude to the carrier amplitude, m_a = A_m/A_c. It can also be found from the envelope as (A_max − A_min)/(A_max + A_min). Option (a) is its reciprocal, and option (d) gives the carrier amplitude A_c, not the index.
Correct answer: (b) 30 kHz.
The transmission bandwidth of DSB-FC AM is twice the message bandwidth: BW = 2W = 2 × 15 kHz = 30 kHz. This is why adjacent medium-wave stations are assigned 9–10 kHz-spaced channels that must limit audio to about 4.5–5 kHz in many regions.
Correct answer: (c) f_c ± f_m.
Expanding s(t) = A_c[1 + m_a cos(2πf_m t)]cos(2πf_c t) with the product-to-sum identity produces components at f_c (carrier), f_c + f_m (upper sideband) and f_c − f_m (lower sideband).
Correct answer: (b) m_a A_c / 2.
From the expansion of s(t), each sideband term is (m_a A_c/2)cos[2π(f_c ± f_m)t], so the sideband amplitude is m_a A_c/2 — half the product m_a A_c that appears when the envelope swings.
Correct answer: (c) Envelope detector.
Because the message is preserved in the envelope of the AM wave, a simple diode with an RC filter (envelope detector) recovers it. Discriminators and PLLs are FM demodulators, and a mixer is a frequency-translation stage.
Correct answer: (d) 33.3%.
η = m_a²/(2 + m_a²); at m_a = 1 this equals 1/3 ≈ 33.3%. The rest of the power is wasted in the information-free carrier — the fundamental power inefficiency of conventional AM.
Correct answer: (a).
The carrier is a high-frequency "vehicle" that transports the message: it enables practical antenna sizes, propagation over long distances, and frequency multiplexing. It contains no information itself — in fact that is why AM is power-inefficient.
Correct answer: (c) m_a > 1.
A_min = A_c(1 − m_a). When m_a > 1 the minimum envelope value becomes negative, meaning the envelope attempts to cross zero and the carrier suffers 180° phase reversals — the defining symptom of over-modulation. At exactly m_a = 1 the envelope only touches zero.
Correct answer: (b).
Each sideband alone contains the complete message, yet DSB-FC transmits two sidebands and a carrier, occupying BW = 2W. SSB suppresses the carrier and one sideband, halving the bandwidth to W — the same information in half the spectrum.
Correct answer: (b).
The "1" is the DC (carrier) term. Multiplying it by A_c cos(2πf_c t) generates the pure carrier line at f_c. Biasing the message by this DC term is what allows simple envelope detection, and it guarantees A_c + m(t) ≥ 0 when m_a ≤ 1.
Correct answer: (b).
P_T = P_c(1 + m_a²/2), so raising m_a from 0.3 to 1 raises total power and the sideband share (m_a²P_c/2). Bandwidth (2f_m), carrier amplitude, and carrier frequency are unaffected by m_a.
Correct answer: (b).
Ideal AM with m_a ≤ 1 has a strictly limited spectrum of width 2f_m. Over-modulation introduces abrupt phase flips — effectively a multiplication by a non-positive factor — which are rich in harmonics and spread spectral energy beyond the assigned channel, interfering with neighbouring stations ("splatter").
Correct answer: (c).
RC must be much longer than the carrier period (to filter the f_c ripple into a smooth envelope) yet much shorter than the fastest message period 1/W (to track envelope changes). Too large causes diagonal clipping; too small leaves excessive ripple at the output.
Correct answer: (a) 0.6.
m_a = (A_max − A_min)/(A_max + A_min) = (16 − 4)/(16 + 4) = 12/20 = 0.6 (60%). This also gives A_c = (16+4)/2 = 10 V and A_m = m_a A_c = 6 V — a valid, distortion-free modulation.
Correct answer: (b) 40 V.
Sideband amplitude = m_a A_c / 2 = (0.8 × 100)/2 = 40 V. The sidebands appear at 1005 kHz (USB) and 995 kHz (LSB), each 40 V peak.
Correct answer: (b) 9 kW.
P_T = P_c(1 + m_a²/2) = 8000 × (1 + 0.25/2) = 8000 × 1.125 = 9000 W = 9 kW. Sidebands carry the extra 1 kW (500 W each); the carrier remains 8 kW.
Correct answer: (b) 11.1%.
η = (sideband power)/(total power) = 1000 W / 9000 W ≈ 11.1%. Equivalently η = m_a²/(2 + m_a²) = 0.25/2.25 ≈ 11.1% — typical of real AM speech transmission, which is why AM is considered power-hungry.
Correct answer: (b).
BW = 2W = 2 × 4 kHz = 8 kHz. A 2 kHz tone produces the classic triplet: carrier at f_c, USB at f_c + 2 kHz, LSB at f_c − 2 kHz. (With a full 4 kHz message, the sidebands become continuous 4 kHz-wide bands on either side of the carrier.)
Correct answer: (a).
At m_a = 1: P_T = P_c(1 + 1/2) = 1.5 P_c ⇒ P_c = 12/1.5 = 8 kW. Each sideband carries m_a²P_c/4 = 8/4 = 2 kW, and indeed 8 + 2 + 2 = 12 kW. The carrier wastes 2/3 of the power at full modulation.
| Q | Ans | Bloom level | Q | Ans | Bloom level |
|---|---|---|---|---|---|
| 1 | b | Remember | 11 | b | Understand |
| 2 | c | Remember | 12 | b | Understand |
| 3 | b | Remember | 13 | b | Understand |
| 4 | c | Remember | 14 | c | Understand |
| 5 | b | Remember | 15 | a | Apply |
| 6 | c | Remember | 16 | b | Apply |
| 7 | d | Remember | 17 | b | Apply |
| 8 | a | Understand | 18 | b | Apply |
| 9 | c | Understand | 19 | b | Apply |
| 10 | b | Understand | 20 | a | Apply |