| Bloom's Level | What it tests | Questions |
|---|---|---|
| Level 1 — Remember | Recalling multiplex components, frequencies, roles | 1–4 |
| Level 2 — Understand | Explaining DSB-SC choice, pilot role, compatibility, blending | 5–9 |
| Level 3 — Apply | Computing budgets, sideband placement, corner frequencies | 10–12 |
Correct answer: L + R as an ordinary 0–15 kHz baseband signal..
The sum signal L+R occupies the normal audio band, so any mono receiver simply low-pass filters at 15 kHz and reproduces the full programme without modification.
Correct answer: a DSB-SC signal on a 38 kHz subcarrier..
The difference channel modulates a 38 kHz subcarrier in double-sideband suppressed-carrier form, placing sidebands between 23 and 53 kHz. The 19 kHz tone is only the pilot, and 57/67 kHz belong to RDS/SCA services.
Correct answer: regenerate the 38 kHz subcarrier coherently..
The receiver filters the 19 kHz pilot and doubles it (×2) to rebuild the 38 kHz carrier with exactly the right frequency and phase for coherent demodulation of the L−R DSB-SC signal.
Correct answer: 38 kHz..
The L−R difference signal modulates a 38 kHz subcarrier (DSB-SC); 19 kHz is its half-frequency pilot, 57 kHz carries RDS data, and 76 kHz is simply twice the subcarrier frequency.
Correct answer: power/budget savings and pilot-controlled phase..
A full-strength 38 kHz carrier would consume modulation percentage without carrying information, and the receiver would lose phase control. DSB-SC keeps the carrier out of the spectrum so the receiver's pilot-derived reference decides the demodulation phase — a phase error there would leak L into R.
Correct answer: a 38 kHz pilot would sit on top of the DSB-SC spectrum..
The 38 kHz point is the centre of the L−R sideband cluster, so a pilot there would be impossible to filter out. The 19 kHz pilot instead sits in the quiet gap between 15 kHz and 23 kHz, and doubling it in the receiver regenerates 38 kHz exactly.
Correct answer: cos φ..
Multiplying [L−R]cos(ω_c t) by the regenerated cos(ω_c t + φ) and low-pass filtering yields ½[L−R]cos φ. Perfect lock (φ = 0°) gives full L−R; a 90° error kills the difference channel entirely and intermediate values mix L into R.
Correct answer: the full programme as mono (L+R)..
The mono receiver's audio low-pass filter passes only the 0–15 kHz main channel. The 19 kHz pilot and the 23–53 kHz difference channel are rejected, so the listener hears L+R — exactly the compatibility goal of the Zenith–GE system.
Correct answer: the difference channel degrades first..
Under weak-signal conditions the S/N of the recovered L−R collapses faster than L+R (the DSB-SC path is more vulnerable), so the receiver mutes the difference path and outputs L+R to both speakers — trading stereo separation for listenable audio.
Correct answer: 45%..
The modulation budget is shared: 100% − 45% (L+R) − 10% (pilot) = 45% for L−R. This equal 45%/45% split is precisely the standard stereo allocation; exceeding it would push peak deviation past ±75 kHz and cause adjacent-channel splatter.
Correct answer: 34 kHz and 42 kHz..
DSB-SC translates the message to subcarrier ± message frequency: 38 kHz ± 4 kHz = 34 kHz (lower sideband) and 42 kHz (upper sideband). The tone does NOT appear at baseband in the multiplex signal — the L+R band carries the sum instead.
Correct answer: 2.1 kHz..
f_c = 1/(2πτ) = 1/(2π × 75×10⁻⁶) ≈ 2122 Hz ≈ 2.1 kHz. (For the 50 µs standard used in Europe the corner is ≈ 3.2 kHz.) The pre-emphasis at the transmitter uses the same time constant so the net audio response is flat while high-frequency noise is reduced.
| Q | Ans | Bloom level | Q | Ans | Bloom level |
|---|---|---|---|---|---|
| 1 | d | Remember | 7 | b | Understand |
| 2 | b | Remember | 8 | b | Understand |
| 3 | a | Remember | 9 | d | Understand |
| 4 | a | Remember | 10 | c | Apply |
| 5 | d | Understand | 11 | a | Apply |
| 6 | c | Understand | 12 | c | Apply |