Senior 5 practice book · Unit 8 of 11

Signals and Communication Systems

50 questions that complete the Senior 5 quiz for this unit: 31 core and 19 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

Digital signals are never affected by noise.Noise is added to every signal, but a digital receiver only has to decide "0 or 1", so moderate noise causes no errors and the pulses can be regenerated cleanly.
Amplifying a weak, noisy signal removes the noise.An amplifier boosts the signal and the noise by the same factor, so the signal-to-noise ratio does not improve; amplify before the noise becomes large.
In AM radio the sound itself travels through the air to the receiver.Only the electromagnetic carrier wave travels; the sound is carried as a variation of its amplitude and is recovered by the demodulator.
A decibel is an amount of power, like a watt.Decibels compare two powers on a logarithmic scale: +10 dB means 10 times the power, −3 dB about half.
All radio waves travel only in straight lines, so no station can be heard behind a hill.Low-frequency waves follow the ground (ground waves) and short waves are reflected by the ionosphere (sky waves); only VHF and higher need a line of sight.

What this unit covers

Topics marked new are not tested much in the quiz, so this book gives them extra questions.

Communication system block diagram: source, transmitter, channel, receiver, destination; noise
Wired and wireless systems; history from telegraph and telephone to modern networks
Analogue and digital signals; noise, signal-to-noise ratio and regeneration
Signal conversion: ADC and DAC and where they are used
Sampling and digitising: sampling rate, bits per sample, number of levels, bit rate, file size
Modulation: carrier wave, AM and FM waveforms, reasons for modulating, aerial length
Bandwidth and AM side frequencies; number of stations in a band
Radio wave propagation: ground waves, sky waves (ionosphere), space waves and line-of-sight range
Attenuation, gain and decibels; amplifiers and repeaters
Optical fibres: total internal reflection, core and cladding, critical angle, advantages
Satellite communication: geostationary satellites, uplink and downlink, delay
Mobile phone networks: cells, base stations, handover, frequency reuse, sharing data rate
Parts of a radio receiver (aerial, tuner, demodulator, amplifier) and transducers

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 19 harder ones are marked ★ Challenge.

1True or false
Noise added in the channel can be completely removed by amplifying the received analogue signal.
Show answer
Answer: False
An amplifier boosts the noise together with the signal, so the noise remains.
Common mistakeThinking "louder means clearer" ignores that the noise is amplified by the same factor as the wanted signal.
2True or false · ★ Challenge
Using more bits per sample makes the digital version of an analogue signal closer to the original, because the steps between levels become smaller.
Show answer
Answer: True
n bits give 2ⁿ levels; more levels means smaller steps and a smaller rounding (quantisation) error.
Common mistakeSome think more bits means more samples per second; the number of bits sets the fineness of each sample, the sampling rate sets how often.
3True or false
Sending speech in digital form usually needs a larger bandwidth than sending the same speech as an analogue signal.
Show answer
Answer: True
Telephone speech uses about 3 kHz as analogue, but 64 kbit/s as digital, which needs a much wider band.
Common mistakeMany learners think digital is better in every way; its main cost is the extra bandwidth.
4True or false
A digital camera needs an ADC to turn the voltages from its light sensor into numbers that can be stored on the memory card.
Show answer
Answer: True
Each pixel gives an analogue voltage depending on the light falling on it; an ADC turns it into a binary number.
Common mistakeSome think a camera sensor is already digital; the light gives a continuously varying charge that must be converted.
5Fill in the blank · ★ Challenge
To record a sound containing frequencies up to 20 kHz without losing information, it must be sampled at least ______ times per second.
Show answer
Answer: 40 000
The sampling rate must be at least twice the highest frequency: 2 × 20 000 = 40 000 per second.
Common mistakeAnswering 20 000 means sampling only once per cycle, which cannot show the ups and downs of the wave.
6True or false
When you travel by bus from Kigali to Huye during a call, the call is passed from one base station to the next (handover) without being cut off.
Show answer
Answer: True
The network measures the signal from nearby base stations and switches the call to the stronger one.
Common mistakeSome think a phone stays connected to one mast for the whole call; the range of each cell is only a few kilometres.
7Fill in the blank · ★ Challenge
An AM station is allowed a bandwidth of 10 kHz. The highest audio frequency it can transmit is ______ kHz.
Show answer
Answer: 5
Bandwidth = 2 × highest audio frequency, so f = 10 ÷ 2 = 5 kHz.
Common mistakeAnswering 10 kHz forgets that AM makes two sidebands, one above and one below the carrier.
8Fill in the blank
In FM, the louder the sound, the ______ the change (deviation) of the carrier frequency.
Show answer
Answer: greater (larger)
The size of the frequency change follows the amplitude (loudness) of the audio; how often it changes follows the pitch.
Common mistakeMany think loudness changes the carrier amplitude in FM; that would be AM.
9Fill in the blank
The outer layer of an optical fibre, which has a lower refractive index than the core, is called the ______.
Show answer
Answer: cladding
The cladding gives a boundary with a smaller refractive index, so total internal reflection can happen at it.
Common mistakeSome call it the "sheath"; the protective plastic sheath is outside; the glass layer next to the core is the cladding.
10Multiple choice · ★ Challenge
Why are FM and TV transmitters in Rwanda put on high hills such as Mount Jali and Mount Karisimbi?
1. AHigh places are nearer the ionosphere, which reflects VHF
2. BGround waves at VHF travel further over rock than soil
3. CVHF space waves need a line of sight, which height extends
4. DThin air at height lets radio waves travel much faster
Show answer
Answer: C. VHF space waves need a line of sight, which height extends
Space waves travel in straight lines; a high mast or hill lets them clear the curve of the Earth and other hills, so they reach further.
Common mistakeChoosing the ionosphere option is wrong: VHF is not reflected by the ionosphere, and a few km of height makes no difference to it.
11Fill in the blank
The ratio of the wanted signal power to the noise power, often given in decibels, is called the signal-to-______ ratio.
Show answer
Answer: noise
Signal-to-noise ratio (S/N): the larger it is, the clearer the signal.
Common mistakeSome write "signal-to-power"; the ratio compares the signal with the noise.
12True or false · ★ Challenge
In AM, the frequency of the audio signal sets how quickly the carrier's amplitude rises and falls, and the loudness sets how much the amplitude changes.
Show answer
Answer: True
The envelope is a copy of the audio wave: its rate is the audio frequency and its depth follows the audio amplitude.
Common mistakeSome think the carrier frequency changes with pitch in AM; the carrier frequency stays fixed.
13Fill in the blank
The layer of charged particles high in the atmosphere that reflects short-wave radio signals back to Earth is called the ______.
Show answer
Answer: ionosphere
Sky waves (about 3–30 MHz) are reflected by the ionosphere, so they can reach places far beyond the horizon.
Common mistakeSome answer "ozone layer"; ozone absorbs UV, while the ionosphere is the charged layer that reflects radio waves.
14Multiple choice
Light stays inside an optical fibre because of
1. Arefraction of light out of the core into the cladding
2. Breflection from a silver coating on the outside surface
3. Cabsorption and re-emission of light by the glass atoms
4. Dtotal internal reflection at the core–cladding boundary
Show answer
Answer: D. total internal reflection at the core–cladding boundary
Light hits the boundary at more than the critical angle and is totally reflected back into the core, again and again.
Common mistakeChoosing a silver coating is wrong: fibres use total internal reflection, which loses almost no light, unlike a mirror.
15True or false · ★ Challenge
A cable loses 3 dB per km. After 10 km a 30 dB amplifier is fitted. The power coming out of the amplifier equals the power put into the cable.
Show answer
Answer: True
Total loss = 3 × 10 = 30 dB; gain = 30 dB; overall change = 0 dB, so the output equals the input.
Common mistakeLearners sometimes multiply the ratios wrongly; in decibels losses and gains simply add.
16Multiple choice
Which conditions are both needed for total internal reflection inside a fibre core?
1. An(core) > n(cladding), and angle > critical angle
2. Bn(core) < n(cladding), and angle > critical angle
3. Cn(core) > n(cladding), and angle < critical angle
4. Dn(core) = n(cladding), and angle equal to 90°
Show answer
Answer: A. n(core) > n(cladding), and angle > critical angle
Light must be going from the denser to the less dense medium and meet the boundary at more than the critical angle.
Common mistakeChoosing a smaller core index reverses the first condition; total internal reflection only happens going into a LESS dense medium.
17Multiple choice
In a radio receiver, which part selects one station from all the signals picked up by the aerial?
1. AThe demodulator (detector)
2. BThe loudspeaker unit
3. CThe audio amplifier
4. DThe tuner (tuned circuit)
Show answer
Answer: D. The tuner (tuned circuit)
The tuner resonates at the chosen carrier frequency, so only that station is passed on.
Common mistakeChoosing the demodulator confuses selecting a station with removing the carrier; demodulation comes after tuning.
18Multiple choice · ★ Challenge
A 3-bit ADC covers 0 to 8 V in steps of 1 V and rounds down. An input of 3.3 V is applied. Which code does it output?
1. A100
2. B110
3. C011
4. D010
Show answer
Answer: C. 011
3.3 V rounds down to level 3, which in 3-bit binary is 011.
Common mistakeChoosing 110 writes the bits of 3 in the wrong order (110 is 6); 100 rounds up to 4.
19Multiple choice
Which is an advantage of a cable link over a wireless link for joining two offices in the same building?
1. AIt lets users move around freely while connected
2. BIt needs no installation work or equipment
3. CIt is more secure and suffers less interference
4. DIt can reach ships at sea and aircraft in flight
Show answer
Answer: C. It is more secure and suffers less interference
Signals in a cable are guided and shielded, so they are harder to intercept and pick up less interference.
Common mistakeChoosing 'move around freely' describes the advantage of wireless, not of cables.
20Multiple choice
Mobile networks divide a country into small cells, each with its own base station. What is the main reason?
1. AEach phone needs its own private nearby base station
2. BThe same frequencies can be reused in cells far apart
3. CSmall cells let radio waves travel much faster
4. DCells prevent phones from using digital signals
Show answer
Answer: B. The same frequencies can be reused in cells far apart
Only a limited band of frequencies is available; low-power base stations let cells far apart use the same frequencies, so millions of calls fit.
Common mistakeChoosing 'each phone needs its own base station' is wrong: one base station serves many phones at once.
21Short answer · ★ Challenge
Long ago, villages on neighbouring hills sent messages by drum beats. Identify the information source, transmitter, channel, receiver and destination in this system, and name one source of noise.
Show answer
Model answer: Source: the person with the message (e.g. the chief). Transmitter: the drummer and drum, which code the message as beats. Channel: the air carrying the sound across the valley. Receiver: the listener on the next hill who decodes the beats. Destination: the people who act on the message. Noise: wind, rain, animals or other sounds.
Common mistakeLearners often call the drum the channel; the drum produces the signal (transmitter), while the air carries it.
22Multiple choice
Long-wave and medium-wave signals that travel close to the surface and follow the curve of the Earth are called
1. Aspace (direct) waves
2. Bsky (reflected) waves
3. Cground (surface) waves
4. Dmicrowave beam links
Show answer
Answer: C. ground (surface) waves
Low-frequency waves diffract around the Earth and follow its surface for hundreds of kilometres.
Common mistakeChoosing sky waves confuses following the ground with bouncing off the ionosphere.
23Multiple choice · ★ Challenge
An aerial amplifier takes in 0.50 mW and gives out 400 mW. What is its gain in decibels?
1. A29 dB
2. B800 dB
3. C2.9 dB
4. D58 dB
Show answer
Answer: A. 29 dB
Gain = 10 log₁₀(400 ÷ 0.50) = 10 log₁₀ 800 = 29 dB.
Common mistakeChoosing 800 dB gives the ratio itself, not its logarithm; 58 dB uses 20 log, which is for voltage ratios.
24Multiple choice
Which type of radio wave travels in a straight line from transmitter to receiver and is used for FM, TV and mobile phones?
1. ASpace wave (line of sight)
2. BGround (surface) wave
3. CSky wave from the ionosphere
4. DSound wave in the air
Show answer
Answer: A. Space wave (line of sight)
VHF and UHF waves pass through the ionosphere and do not follow the ground well, so they travel by line of sight as space waves.
Common mistakeChoosing sky wave is wrong: VHF/UHF waves pass through the ionosphere instead of being reflected.
25Multiple choice
A teacher in Rubavu sends a voice note by phone to a colleague in Huye. In the block diagram of this system, what is the channel?
1. AThe radio links and network between the two phones
2. BThe microphone built into the sending teacher's phone
3. CThe colleague who listens to the voice note
4. DThe loudspeaker of the receiving phone
Show answer
Answer: A. The radio links and network between the two phones
The channel is the medium that carries the signal from transmitter to receiver: here the radio links, cables and network equipment.
Common mistakeChoosing the microphone confuses a transducer in the transmitter with the channel; the channel is what lies BETWEEN transmitter and receiver.
26Short answer · ★ Challenge
Explain, in terms of bandwidth, why music sounds less clear on an AM station than on an FM station.
Show answer
Model answer: AM stations get channels about 9–10 kHz wide, so they can only carry audio up to about 4.5–5 kHz and the high notes are lost. FM stations have about 200 kHz each, so they carry audio up to 15 kHz, giving fuller sound (and FM is also less affected by noise).
Common mistakeSaying "FM is more powerful" misses the point; the clarity comes from the wider band and the noise resistance, not power.
27True or false
VHF and higher-frequency radio waves pass through the ionosphere instead of being reflected, which is why they are used to communicate with satellites.
Show answer
Answer: True
Above about 30 MHz the ionosphere cannot reflect the waves, so they go straight through to space.
Common mistakeSome think all radio waves bounce off the ionosphere; only lower frequencies (short waves and below) do.
28Multiple choice
A digital thermometer has a temperature sensor whose output voltage changes smoothly with temperature. What must be placed between the sensor and the microprocessor?
1. AA DAC, as the sensor output is digital
2. BA modulator, as the sensor gives a carrier
3. CNothing, as chips accept any voltage
4. DAn ADC, as the sensor output is analogue
Show answer
Answer: D. An ADC, as the sensor output is analogue
The sensor voltage varies continuously (analogue); a microprocessor works with binary numbers, so an analogue-to-digital converter is needed.
Common mistakeChoosing a DAC mixes up the direction: a DAC turns numbers back into a voltage, the opposite of what is needed here.
29Short answer · ★ Challenge
Explain why an optical fibre has a cladding instead of relying on reflection at a glass–air surface.
Show answer
Model answer: At a bare glass surface any scratch, dirt, grease from fingers or contact with another fibre would spoil the total internal reflection and let light leak out or cross into a neighbouring fibre. The cladding protects the reflecting surface, keeps it clean and fixes the critical angle, so the light stays in the core.
Common mistakeThinking the cladding "reflects like a mirror" is wrong; it provides a protected boundary of lower refractive index.
30Fill in the blank
A loss of 10 dB means that the output power is ______ of the input power.
Show answer
Answer: one tenth (0.1)
10 log₁₀(P_in/P_out) = 10, so P_in/P_out = 10 and P_out = 0.1 P_in.
Common mistakeAnswering "10 W less" treats the decibel as an amount of power; it is a ratio.
31Short answer · ★ Challenge
Phone signal in a deep valley in a rural district is often weak. Explain why, and suggest two solutions.
Show answer
Model answer: Mobile phones use UHF/microwaves, which travel by line of sight and are blocked by the hills between the valley and the base station. Solutions (any two): build a mast inside or near the valley; put a repeater on a hilltop that can see both; use a signal booster with an outdoor aerial; use satellite links for the local mast.
Common mistakeSaying "the valley is too far below the satellite" is wrong: mobile phones talk to ground base stations, and the problem is the blocked line of sight.
32Fill in the blank
A 4G cell shares a total data rate of 150 Mbit/s equally among 60 active users. Each user gets ______ Mbit/s.
Show answer
Answer: 2.5
150 ÷ 60 = 2.5 Mbit/s.
Common mistakeMultiplying instead of dividing gives 9000 Mbit/s; sharing a fixed rate among more users gives each user less.
33Short answer
Give three advantages of optical fibres over copper cables for Rwanda's national fibre backbone.
Show answer
Model answer: Any three: much larger bandwidth (more data); much lower loss, so fewer repeaters; not affected by electrical interference or lightning; light and thin; do not corrode; hard to tap, so more secure; no sparks.
Common mistakeSaying "light is faster than electricity" is not the reason; the advantage is mainly bandwidth and low loss.
34Multiple choice · ★ Challenge
A fibre has a glass core of refractive index 1.48 surrounded by a cladding of refractive index 1.40. Above what angle of incidence (measured from the normal) is light trapped in the core?
1. A43°
2. B71°
3. C19°
4. D46°
Show answer
Answer: B. 71°
sin C = n_cladding / n_core = 1.40 ÷ 1.48 = 0.946, so C ≈ 71°.
Common mistakeChoosing 43° uses sin C = 1/1.48, which is for a glass–air boundary; inside a fibre the cladding, not air, is the second medium.
35Multiple choice
Why do satellite TV dishes on houses in Kigali point in one fixed direction and never need to move?
1. AThe satellite is directly overhead above Kigali all the time
2. BThe satellite circles once every 12 hours, always in view
3. CIt is geostationary, fixed above one point on the equator
4. DMicrowaves bend to follow the dish wherever it is pointed
Show answer
Answer: C. It is geostationary, fixed above one point on the equator
A geostationary satellite orbits once every 24 hours above the equator in the same direction as the Earth turns, so it appears fixed in the sky.
Common mistakeChoosing 'directly above Kigali' is wrong: geostationary satellites stay above the equator (Kigali is about 2° south of it) and most are far to the east or west, so dishes point at the satellite's fixed position, not straight up.
36Multiple choice
Why does a photocopy of a photocopy look worse each time, while a copy of a digital file is identical to the original?
1. ADigital files contain no information that could be lost
2. BA digital copy only needs each bit read as 0 or 1
3. CPhotocopiers use digital signals that collect noise
4. DAnalogue copies are always made at a lower power
Show answer
Answer: B. A digital copy only needs each bit read as 0 or 1
Small errors in an analogue copy add up; in a digital copy each bit is read correctly as 0 or 1 unless the noise is very large, so the copy is perfect.
Common mistakeChoosing 'photocopiers use digital signals' reverses the idea: the gradual loss of quality is typical of analogue copying.
37Multiple choice · ★ Challenge
Music on a CD is sampled 44 100 times per second, with 16 bits per sample, on 2 channels (stereo). What is the bit rate?
1. A7.06 × 10⁵ bit/s
2. B1.41 × 10⁶ bit/s
3. C8.82 × 10⁴ bit/s
4. D2.82 × 10⁶ bit/s
Show answer
Answer: B. 1.41 × 10⁶ bit/s
Bit rate = 44 100 × 16 × 2 = 1 411 200 ≈ 1.41 × 10⁶ bit/s.
Common mistakeChoosing 7.06 × 10⁵ bit/s forgets the second (stereo) channel; 8.82 × 10⁴ forgets the 16 bits per sample.
38Short answer
Name the transducer at each end of (a) a telephone call and (b) a television broadcast, stating the energy change each makes.
Show answer
Model answer: (a) Microphone: sound → electrical; earpiece/loudspeaker: electrical → sound. (b) Camera (and microphone): light (and sound) → electrical; screen (and loudspeaker): electrical → light (and sound).
Common mistakeLearners often name the aerial as a transducer for sound; the aerial changes electrical signals to radio waves, not sound to electricity.
39Short answer · ★ Challenge
An analogue sine-wave signal and a digital pulse signal are each sent along a long cable and pick up noise. Describe how each looks at the far end and explain which one can be recovered exactly.
Show answer
Model answer: The analogue wave arrives with a fuzzy, jagged outline; the noise is now part of its shape and cannot be told apart from the signal. The digital pulses arrive with ragged tops and rounded edges, but each level is still clearly high or low, so a regenerator can rebuild perfect pulses. Only the digital signal can be recovered exactly.
Common mistakeLearners often say digital signals pick up no noise; they do, but the noise can be removed by deciding 0 or 1.
40Short answer
Put these in the order they were invented and give one feature of each: telephone, telegraph, mobile phone, radio broadcasting.
Show answer
Model answer: Telegraph (1830s–40s): coded pulses (Morse) along wires. Telephone (1876): the voice sent as a varying (analogue) electric current along wires. Radio broadcasting (early 1900s): wireless, one transmitter to many receivers. Mobile phone (1980s on): wireless, two-way, now digital, through cells and base stations.
Common mistakeMany learners put the telephone first; the telegraph came before it and only sent coded pulses, not speech.
41Multiple choice · ★ Challenge
An AM wave is shown on an oscilloscope. Its envelope (outline) repeats 2000 times per second, and there are 500 carrier cycles inside each envelope cycle. What are the audio and carrier frequencies?
1. AAudio 2 kHz; carrier 1 MHz
2. BAudio 1 MHz; carrier 2 kHz
3. CAudio 2 kHz; carrier 500 Hz
4. DAudio 500 Hz; carrier 1 MHz
Show answer
Answer: A. Audio 2 kHz; carrier 1 MHz
The envelope follows the audio: 2000 Hz. The carrier: 500 × 2000 = 1 000 000 Hz = 1 MHz.
Common mistakeChoosing a 2 kHz carrier swaps the two: the slow outline is the audio, the fast wave inside it is the carrier.
42Multiple choice
An old telephone line passes frequencies from 300 Hz to 3400 Hz. What is its bandwidth?
1. A3700 Hz
2. B3400 Hz
3. C1850 Hz
4. D3100 Hz
Show answer
Answer: D. 3100 Hz
Bandwidth = highest − lowest = 3400 − 300 = 3100 Hz.
Common mistakeChoosing 3700 Hz adds the two limits; bandwidth is the width of the range, so subtract.
43Multiple choice
An AM signal contains the frequencies 1197 kHz, 1200 kHz and 1203 kHz. What are the carrier frequency and the audio frequency?
1. ACarrier 1203 kHz; audio 6 kHz
2. BCarrier 1200 kHz; audio 3 kHz
3. CCarrier 1197 kHz; audio 3 kHz
4. DCarrier 1200 kHz; audio 6 kHz
Show answer
Answer: B. Carrier 1200 kHz; audio 3 kHz
The carrier is in the middle (1200 kHz); the side frequencies are carrier ± audio, so audio = 1203 − 1200 = 3 kHz.
Common mistakeChoosing 6 kHz takes the difference between the two side frequencies, which is the bandwidth (2 × audio), not the audio frequency.
44Short answer · ★ Challenge
Satellite links use different frequencies for the uplink (e.g. 6 GHz) and the downlink (e.g. 4 GHz), and use microwaves rather than medium waves. Explain both choices.
Show answer
Model answer: Different frequencies stop the strong signal sent down by the satellite from swamping the very weak signal it receives from the ground (and the same at the ground station). Microwaves pass straight through the ionosphere, can be sent in narrow beams by small dishes, and have a large bandwidth for many channels; medium waves would be reflected by the ionosphere.
Common mistakeSome say the uplink must be faster; both travel at c, the frequencies differ to avoid interference.
45Short answer
Describe how an FM wave looks when the audio signal is at its positive peak, at zero, and at its negative peak.
Show answer
Model answer: The amplitude is the same everywhere. At the positive peak the waves are squeezed closest together (highest frequency); at zero they have the normal carrier spacing; at the negative peak they are most spread out (lowest frequency).
Common mistakeDrawing the amplitude changing is the AM picture; in FM only the spacing of the waves (the frequency) changes.
46Short answer
Give three reasons why a radio station in Kigali stores its music in digital form rather than on magnetic tapes.
Show answer
Model answer: Any three: digital copies are perfect and do not wear out; files can be compressed to save space; music can be edited and found quickly by computer; it can be sent over the internet or to other studios without loss; storage is cheap and compact.
Common mistakeAnswering "digital sounds louder" is not a reason; loudness is set by the amplifier, not the storage method.
47Short answer · ★ Challenge
The line-of-sight range of a transmitting aerial of height h is about d = √(2Rh), where R = 6.4 × 10⁶ m is the radius of the Earth. Find the range of a TV mast 200 m high, and say how the range changes if the height is made four times larger.
Show answer
Model answer: d = √(2 × 6.4 × 10⁶ × 200) = √(2.56 × 10⁹) ≈ 5.1 × 10⁴ m ≈ 51 km. Since d ∝ √h, four times the height doubles the range (about 100 km).
Common mistakeThinking four times the height gives four times the range ignores the square root.
48Short answer
Explain why long cables need amplifiers or repeaters at intervals, and why they must be placed before the signal becomes very weak.
Show answer
Model answer: The signal loses power along the cable (attenuation) while noise is picked up all the way. If the signal is boosted while it is still much stronger than the noise, the signal-to-noise ratio stays good. If it is left until it is as weak as the noise, amplifying it boosts the noise too and the information cannot be recovered.
Common mistakePlacing one huge amplifier at the end does not work, because by then the noise is as big as the signal.
49Multiple choice
A sensor's ADC gives 4-bit binary numbers. Counting 0000 as one value, how many different values can it output?
1. A4
2. B8
3. C15
4. D16
Show answer
Answer: D. 16
Number of levels = 2ⁿ = 2⁴ = 16 (from 0000 to 1111).
Common mistakeChoosing 15 is the largest value (1111 = 15) but forgets that 0000 is also a level; choosing 8 uses 2 × 4.
50Short answer · ★ Challenge
A school radio club records a 2-minute interview at 22 050 samples per second, 8 bits per sample, on one channel. Calculate the size of the file in megabytes (1 byte = 8 bits).
Show answer
Model answer: Bits = 22 050 × 8 × 120 = 2.12 × 10⁷ bits. Bytes = 2.12 × 10⁷ ÷ 8 = 2.65 × 10⁶ bytes ≈ 2.6 MB.
Common mistakeForgetting to change 2 minutes to 120 s, or forgetting to divide by 8 to change bits into bytes, are the usual slips.