Introduction to Analog Communication Systems

Post-Test Quiz — 20 Questions with Answers & Explanations

REMEMBERUNDERSTANDAPPLY

Instructions

This post-test assesses the fundamentals of analog communication systems. All 20 questions are drawn from the first three levels of Bloom's Taxonomy only — no analysis, evaluation, or creation items are included.

LevelSkill TestedQuestionsTypical Cue Words
1 — RememberRecall definitions, facts, and terminologyQ1–Q7define, list, name, state
2 — UnderstandExplain concepts, classify, interpret ideasQ8–Q14explain, distinguish, classify, why
3 — ApplyUse formulas and concepts in numerical/concrete situationsQ15–Q20calculate, determine, find, solve

How to use: Attempt every question by selecting an option, then press "Submit" at the bottom of each question (or "Check All Answers") to see the correct answer and a full explanation. A running score is displayed at the bottom of the page.

Part 1 — Remember (Q1–Q7)

1What are the three essential elements of any communication system?REMEMBER
Answer: b

Every communication system requires a source (origin of the message), a channel (the physical medium the message travels through), and a destination (the intended recipient). The other options list useful building blocks, but only source–channel–destination is universal.

2In the Shannon–Weaver model, which block is responsible for encoding the message into transmittable signals?REMEMBER
Answer: c

In the Shannon–Weaver model the transmitter converts/encodes the message into a signal suitable for the channel (e.g., via a transducer and modulator). The receiver performs the reverse operation at the destination end.

3Which impairment describes random unwanted electrical energy added to a signal in the channel?REMEMBER
Answer: b

Noise is random, unwanted energy (thermal, shot, atmospheric, man-made) added to the signal. Attenuation is loss of signal strength; distortion is alteration of the waveform; multiplexing is a technique for sharing a channel, not an impairment.

4An AM broadcast radio station (listener can only receive) is an example of which communication mode?REMEMBER
Answer: d

Broadcasting is simplex: information flows in one direction only, from the transmitter to many receive-only receivers. Walkie-talkies are half-duplex; telephones are full-duplex.

5What is the approximate frequency range of telephone-quality speech?REMEMBER
Answer: c

Telephone-quality speech occupies roughly 300 Hz – 3.4 kHz (about 3.1 kHz of bandwidth). 20 Hz – 20 kHz is the full high-fidelity audio range, not the telephone band.

6Which frequency band is used for AM medium-wave broadcasting?REMEMBER
Answer: a

AM broadcasting uses the MF band (530–1700 kHz). FM broadcasting uses VHF (88–108 MHz); HF is shortwave; UHF hosts cellular and TV services.

7Which of the following is a guided (wired) communication channel?REMEMBER
Answer: b

Coaxial cable is a guided medium — the signal is physically contained within the cable. Sky wave, satellite links, and line-of-sight microwave are all unguided (wireless) channels.

Part 2 — Understand (Q8–Q14)

8Why must the modulation index μ of an AM signal be kept less than or equal to 1?UNDERSTAND
Answer: c

The envelope of an AM wave is Ac[1 + μ·m(t)]. If μ > 1 the envelope crosses zero and the carrier undergoes phase reversals — the message is no longer contained in the envelope, so an envelope detector recovers a severely distorted signal and splatters interference into adjacent channels.

9In an envelope detector, what causes "diagonal clipping"?UNDERSTAND
Answer: b

The detector requires 1/fc ≪ RC ≪ 1/W. If RC is too large, the capacitor holds its charge and its voltage "cuts diagonally" across the tips of the RF cycles instead of following the envelope downward — producing distortion, especially at high modulation index and high audio frequencies.

10Why does a product (synchronous) detector require carrier synchronization, while an envelope detector does not?UNDERSTAND
Answer: d

The product detector multiplies the incoming wave by a locally generated carrier. The demodulated output scales by cos φ: a phase error attenuates the message (90° error nulls it completely) and a frequency error produces a beat tone. Hence a PLL must lock the local oscillator to the incoming carrier. The envelope detector simply follows the amplitude envelope and needs no reference.

11Why is a compressor/limiter used before AM-modulating a speech signal?UNDERSTAND
Answer: a

Speech spans roughly 30–40 dB of dynamic range, but peaks are infrequent (average μ ≈ 0.3). Without a limiter, loud peaks would over-modulate the carrier (μ > 1), causing distortion and adjacent-channel splatter, while most of the time the signal would be under-modulated with poor SNR. The compressor raises quiet parts and clips peaks, keeping μ near its optimum.

12Why does a single BJT with a nonlinear transfer characteristic produce an AM wave?UNDERSTAND
Answer: c

Expanding iC = a₀ + a₁v + a₂v² + … with v = c(t) + m(t), the square-law term yields a₂c²(t) + 2a₂c(t)m(t) + a₂m²(t). The cross-product 2a₂c(t)m(t) is exactly a DSB component at fc ± fm; the tuned tank then selects the carrier-plus-sideband group, completing the AM spectrum.

13Lower-frequency radio bands (e.g., MF/HF) generally achieve longer range than UHF/SHF mainly because…UNDERSTAND
Answer: b

Propagation physics favors low frequencies: ground waves follow the Earth's curvature (LF/MF), and HF sky waves reflect from the ionosphere for intercontinental coverage. UHF/SHF signals travel essentially line-of-sight with little diffraction. The trade-off is that lower frequencies offer far less available bandwidth.

14Two stations use a walkie-talkie (push-to-talk) link. This is an example of:UNDERSTAND
Answer: d

The defining feature of half-duplex is bidirectional capability with alternating, non-simultaneous use of a shared channel. Push-to-talk forces turn-taking: while A transmits, B can only receive, and vice versa. Simplex would mean a station could never transmit (or never receive).

Part 3 — Apply (Q15–Q20)

15An AM signal is modulated by a 4 kHz audio tone. What is its transmission bandwidth?APPLY
Answer: b

AM bandwidth = 2 fm = 2 × 4 kHz = 8 kHz (upper and lower sidebands each extend fm above and below the carrier). This is why AM broadcast stations are spaced 10 kHz apart.

16A carrier of amplitude Ac = 10 V is modulated by a message of peak amplitude Am = 6 V. The modulation index μ is:APPLY
Answer: c

μ = Am/Ac = 6/10 = 0.6. This is under-modulation, so the envelope faithfully carries the message and can be demodulated with an envelope detector.

17An envelope detector must demodulate an AM signal with fc = 1 MHz and a highest message frequency W = 5 kHz. Which RC value satisfies 1/fc ≪ RC ≪ 1/W?APPLY
Answer: a

The bounds are 1/fc = 1 μs and 1/W = 200 μs, so RC should lie comfortably between (a few μs up to ~50–100 μs). 10 μs satisfies this: it smooths the 1 MHz carrier yet discharges fast enough to follow a 5 kHz envelope. RC = 1 ms ≈ 1/W causes diagonal clipping; 0.5–1 μs leaves heavy carrier ripple.

18Using Shannon–Hartley, a channel with B = 4 kHz and S/N = 15 has a capacity of approximately:APPLY
Answer: b

C = B log₂(1 + S/N) = 4000 × log₂(16) = 4000 × 4 = 32 kbps. (log₂16 = 4 since 2⁴ = 16.) This is the theoretical maximum error-free rate — a cornerstone result of information theory.

19A single-sideband (SSB) transmitter carries a 3 kHz speech signal. Approximately what bandwidth does it occupy?APPLY
Answer: c

SSB transmits only one sideband, so its bandwidth equals the message bandwidth: 3 kHz — half that of DSB-AM (2fm = 6 kHz). This spectral efficiency is why SSB dominates amateur radio and analog multiplex telephony.

20An AM transmitter has carrier power Pc = 500 W and μ = 0.8. The total transmitted power is approximately:APPLY
Answer: d

Pt = Pc(1 + μ²/2) = 500 × (1 + 0.64/2) = 500 × 1.32 = 660 W. Of this, 500 W sits in the carrier (which carries no information) and only 160 W in the sidebands — illustrating AM's power inefficiency.

Answer Key (Quick Reference)

QAnswerBloom LevelTopic
1bRememberElements of communication
2cRememberShannon–Weaver model
3bRememberChannel impairments (noise)
4dRememberSimplex mode
5cRememberSpeech bandwidth
6aRememberFrequency bands (MF/AM)
7bRememberGuided channels
8cUnderstandModulation index ≤ 1
9bUnderstandDiagonal clipping
10dUnderstandCarrier synchronization
11aUnderstandSpeech compression/limiting
12cUnderstandSquare-law modulation
13bUnderstandPropagation vs. frequency
14dUnderstandHalf-duplex
15bApplyAM bandwidth B = 2fₘ
16cApplyModulation index μ = Aₘ/A𝒸
17aApplyEnvelope detector RC design
18bApplyShannon–Hartley capacity
19cApplySSB bandwidth
20dApplyAM power Pₜ = P𝒸(1+μ²/2)

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