Notes · ElectronicsUK · A-Levels

Communications systems

Every communication system, from a telephone call to a satellite link, follows the same block model and faces the same enemies: limited bandwidth, attenuation and noise. This chapter develops the source-transmitter-channel-receiver-destination model, the need for a carrier and for bandwidth, the decibel treatment of attenuation and the signal-to-noise ratio, and the comparison of wire, radio and optical channels.

4 sections~15 min reading time3 competenciesLevel Standard 2 · Advanced 2

T·131313 / 16

Exam profile

AO1 · Describe the block model of a communication system and the roles of the carrier, bandwidth, attenuation and noise.AO2 · Calculate attenuation and signal-to-noise ratio in decibels and relate bandwidth to information capacity.AO3 · Compare and choose a communication channel for a stated requirement of range, bandwidth and noise.

Operators:describecalculatedeterminecompareanalyseevaluate

basic level

At AS the focus is the communication block model and the ideas of carrier, bandwidth, attenuation and noise.

higher level

The full A-Level develops the decibel treatment of attenuation and signal-to-noise ratio and the quantitative comparison of channels.

Depth

Text

Contents · 4 sections

Communications systems

The communication system block model#

StandardWJEC/Eduqas A Level Electronics — Unit 4 Digital and Communication (systems)

The communication system model

Fig. 1The five-block model. Noise is added in the channel, which also attenuates the signal and limits its bandwidth.

Every communication system, however simple or complex, fits the same five-block model. An information source produces the message; a transmitter converts it into a form suited to the channel (usually by modulating a carrier); the channel is the physical medium that carries the signal (a wire, free space, or an optical fibre); a receiver recovers the message from the received signal (by demodulating); and the destination is the user or device that consumes it. Reasoning with this model lets any system be analysed block by block.

The transmitter's job is to prepare the signal for the channel. A low-frequency information signal — speech, music, data — cannot usually travel efficiently over the channel as it is, so the transmitter impresses it onto a high-frequency carrier by modulation, and often amplifies it. The receiver reverses this at the far end, demodulating to recover the original information and rejecting as much of the interference the channel added as it can. Transmitter and receiver are therefore a matched pair, designed together.

The channel is where the signal is degraded. Whatever the medium, it attenuates the signal (weakens it with distance), adds noise (unwanted random signals), and limits the range of frequencies it will pass (its bandwidth). These three impairments — attenuation, noise and limited bandwidth — are the universal enemies of communication, and the whole art of system design is to deliver an intelligible message to the destination despite them. Every later section quantifies one of these effects.

The model is deliberately general so that it applies to any system: a broadcast radio station (source: studio; transmitter: modulator and aerial; channel: free space; receiver: your radio; destination: your ears), a telephone call, a fibre-optic internet link, or a spacecraft's telemetry. Recognising the five blocks in an unfamiliar system, and identifying what plays each role, is a routine but important exam skill that frames every communication question.

Thinking in blocks also makes fault-finding and improvement systematic: if a link is poor, you ask which block is responsible — is the transmitter power too low, the channel too lossy or noisy, the receiver too insensitive? — and address that block. This is the same systems approach used throughout the course, now applied to getting information from one place to another, and it underlies the quantitative work on bandwidth, attenuation and noise that follows.

\text{source} \to \text{transmitter} \to \text{channel} \to \text{receiver} \to \text{destination}

Communication model

The universal five-block structure of any communication system.

Worked example

Identifying the blocks in a radio link

For a broadcast radio station and a listener's receiver, identify the five blocks and state one impairment the channel introduces.

Source and transmitter
The source is the studio audio; the transmitter modulates a carrier and radiates it from the aerial.
Channel and receiver
The channel is free space (radio propagation); the receiver is the listener's radio, which demodulates the carrier to recover the audio.
Destination and impairment
The destination is the loudspeaker and listener. The channel attenuates the signal with distance and adds noise (static).

Result: Studio (source), transmitter/aerial, free space (channel), radio (receiver), loudspeaker (destination); the channel attenuates and adds noise.

Typical mistakes

Confusing the transmitter (modulates and sends) with the channel (the medium) or the receiver (demodulates).
Forgetting that noise is added in the channel, not by the source.

Active revision

For an FM radio broadcast, identify what plays the role of each of the five blocks (source, transmitter, channel, receiver, destination) and state one way the channel degrades the signal.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

Carriers and bandwidth#

StandardWJEC/Eduqas A Level Electronics — Unit 4 Digital and Communication (carriers, bandwidth)

A signal band around a carrier

Fig. 2A modulated signal occupies a band of frequencies (here

10\,\text{MHz}

wide) centred on the carrier (here

50\,\text{MHz}

Information signals — speech, music, data — are low-frequency, and sending them directly over a channel is usually impractical. A carrier solves this: a high-frequency sinusoid onto which the information is impressed by modulation. Using a carrier allows an efficient aerial of reasonable size (an aerial works best when its length is comparable to the wavelength, which is short only at high frequencies), lets many signals share a channel by putting each on a different carrier frequency, and suits the propagation properties of the medium.

The bandwidth of a signal is the range of frequencies it occupies, and it is the single most important measure of how much information the signal carries. A speech signal occupies a few kilohertz; music needs about

20\,\text{kHz}

; a television picture needs several megahertz. When the information is placed on a carrier, it spreads out into a band of frequencies around the carrier, and the width of that band is the bandwidth the transmission requires from the channel.

Bandwidth and information capacity are directly linked: the more information per second a signal must carry, the more bandwidth it needs. A channel of a given bandwidth can therefore carry only so much information per second, which is why bandwidth is a precious, shared resource, carefully allocated by regulators. This is the fundamental reason a high-definition video stream needs far more channel capacity than a phone call — it carries far more information each second.

The channel itself has a bandwidth — the range of frequencies it will pass without excessive attenuation — and the signal's bandwidth must fit inside it. A telephone line passes only about

300\,\text{Hz}

3.4\,\text{kHz}

, so it cannot carry the full audio band; an optical fibre passes an enormous bandwidth and can carry vast amounts of data. Matching the signal's bandwidth to the channel's, and choosing a channel with enough bandwidth for the wanted information rate, is a central design consideration.

Because bandwidth is limited and shared, systems are designed to use it efficiently — restricting a signal to only the bandwidth it truly needs, and packing many signals into a channel by giving each its own carrier frequency (frequency-division multiplexing). The trade-off between the information rate a system delivers and the bandwidth it consumes recurs throughout the communication units, and understanding that more information demands more bandwidth is the key idea that ties them together.

\text{bandwidth} = f_{max} - f_{min}

Bandwidth

The range of frequencies a signal occupies; more information needs more bandwidth.

Worked example

Bandwidth of a speech transmission

A speech signal occupying $300\,\text{Hz}$ to $3.4\,\text{kHz}$ is transmitted on a carrier. State the bandwidth of the speech, and explain why a carrier is needed and why a music signal would need more bandwidth.

Speech bandwidth
$\text{bandwidth} = 3400 - 300 = 3100\,\text{Hz} \approx 3.1\,\text{kHz}$ .
Why a carrier
At $3\,\text{kHz}$ the wavelength is enormous, so an efficient aerial would be kilometres long; placing the speech on a high-frequency carrier allows a practical aerial and lets many signals share the medium on different carriers.
Music
Music extends to about $20\,\text{kHz}$ , so it occupies a wider band (roughly $20\,\text{kHz}$ ) and carries more information, needing more channel bandwidth than speech.

Result: Speech occupies about $3.1\,\text{kHz}$ ; a carrier is used for a practical aerial and sharing, and wider-band music needs more bandwidth.

Typical mistakes

Confusing the carrier frequency (the centre) with the bandwidth (the width of the band around it).
Thinking a channel can carry any signal regardless of bandwidth; the signal's band must fit the channel's.

Active revision

A speech signal from $300\,\text{Hz}$ to $3.4\,\text{kHz}$ is placed on a $1\,\text{MHz}$ carrier. State the bandwidth the speech occupies and explain why a carrier is used at all.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

Attenuation, noise and signal-to-noise ratio#

AdvancedWJEC/Eduqas A Level Electronics — Unit 4 Digital and Communication (attenuation, noise)

Signal attenuation along a channel

Fig. 3The signal level falls linearly (in dB) with distance; where it approaches the noise floor (dashed) it must be regenerated. Marker at

10\,\text{km}

(

-30\,\text{dB}

As a signal travels along a channel it loses power — it is attenuated. Because the loss can be enormous, it is measured logarithmically in decibels: the attenuation is

10\log_{10}(P_{out}/P_{in})

, a negative number for a loss. For a uniform channel such as a cable or fibre the loss is proportional to distance and is quoted as so many decibels per kilometre, so the total loss is simply the loss per kilometre times the length. Adding losses along a path is easy in decibels because they add.

Noise is unwanted random electrical signal added in the channel and in the receiver — thermal noise in resistors, interference from other circuits, atmospheric noise. Unlike attenuation, noise cannot be removed by amplification, because amplifying the received signal amplifies the noise with it. Noise therefore sets the ultimate limit on how weak a signal can be and still be recovered: once the signal has decayed to the level of the noise, no amount of gain will retrieve it cleanly.

The quality of a received signal is captured by the signal-to-noise ratio (SNR): the ratio of the signal power to the noise power, expressed in decibels as

10\log_{10}(P_{signal}/P_{noise})

. A high SNR means the signal stands well above the noise and is easily recovered; a low SNR means the noise obscures the signal. Every communication system is designed to keep the SNR at the receiver above the minimum its modulation scheme needs for reliable reception.

Attenuation and noise together define the reach of a link. As the signal travels it weakens toward the noise floor, and once the SNR falls too low the signal must be regenerated (for a digital signal) or amplified with careful low-noise design (for an analogue one) before it can travel further. The spacing of repeaters or amplifiers along a long link is set precisely by this balance between attenuation per kilometre and the minimum acceptable SNR.

These decibel calculations — total attenuation from a per-kilometre figure, and SNR from signal and noise powers — are the quantitative core of the communication units, and they are exactly why the decibel was introduced in the filter chapter. Being fluent with them, and understanding that noise (not attenuation) sets the fundamental limit because it cannot be amplified away, is essential for both the analogue and the digital communication topics that follow.

\text{attenuation (dB)} = 10\log_{10}\!\left(\dfrac{P_{out}}{P_{in}}\right)

Attenuation

A power ratio in decibels; loss per kilometre times length for a uniform channel.

\text{SNR (dB)} = 10\log_{10}\!\left(\dfrac{P_{signal}}{P_{noise}}\right)

Signal-to-noise ratio

How far the signal stands above the noise; higher is better.

Worked example

SNR and cable loss

A receiver gets a $2\,\text{mW}$ signal with $1\,\mu\text{W}$ of noise. Find the SNR in decibels. A cable loses $3\,\text{dB/km}$ ; find the total loss over $12\,\text{km}$ and the output power if the input was $1\,\text{mW}$ .

Signal-to-noise ratio
$\text{SNR} = 10\log_{10}(P_s/P_n) = 10\log_{10}(2\times 10^{-3}/1\times 10^{-6}) = 10\log_{10}(2000) = 33\,\text{dB}$ .
$\text{SNR} = 10\log_{10}(2000) = 33\,\text{dB}$
Total cable loss
$\text{loss} = 3\,\text{dB/km} \times 12\,\text{km} = 36\,\text{dB}$ .
Output power
A $36\,\text{dB}$ loss is a power ratio of $10^{-36/10} = 10^{-3.6} = 2.5\times 10^{-4}$ , so $P_{out} = 1\,\text{mW} \times 2.5\times 10^{-4} = 0.25\,\mu\text{W}$ .

Result: The SNR is $33\,\text{dB}$ ; the cable loses $36\,\text{dB}$ over $12\,\text{km}$ , cutting $1\,\text{mW}$ to about $0.25\,\mu\text{W}$ — a repeater would be needed.

Typical mistakes

Using the factor 20 (for voltage) instead of 10 (for power) in a power ratio in decibels.
Thinking amplification can remove noise — it amplifies the noise along with the signal.

Active revision

A signal of $2\,\text{mW}$ is received with $1\,\mu\text{W}$ of noise. Find the signal-to-noise ratio in decibels. Separately, a cable loses $3\,\text{dB/km}$ ; find the total loss over $12\,\text{km}$ .

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

Comparing channels#

AdvancedWJEC/Eduqas A Level Electronics — Unit 4 Digital and Communication (channels)

Comparison of communication channels

Fig. 4The channels compared: fibre leads on bandwidth and low loss, radio on mobility, copper on cost.

The three main channels — copper wire, radio (free space) and optical fibre — differ enormously in bandwidth, attenuation, immunity to interference and cost, and choosing between them is a real engineering decision. Copper wire (twisted pair or coaxial cable) is cheap and simple and needs no line of sight, but has limited bandwidth, significant attenuation that rises with frequency, and picks up electromagnetic interference — so it suits short, low-to-moderate-rate links.

Radio uses free space as the channel and needs no physical connection at all, which makes it the only choice for mobile and broadcast communication. But the radio spectrum is a shared, finite and regulated resource, so bandwidth is limited and allocated; radio is also prone to interference, fading and eavesdropping, and its range and behaviour depend strongly on frequency. Its unique advantage — reaching mobile and remote receivers without wires — is decisive where a physical link is impossible.

Optical fibre carries the signal as light and offers by far the greatest bandwidth and the lowest attenuation, so it can carry vast data rates over long distances with widely spaced repeaters. Because the signal is light in a glass fibre, it is immune to electromagnetic interference and very secure (hard to tap without detection), and the fibre is thin and light. Its drawbacks are higher cost, the need for precise connectors and conversion between electrical and optical signals, and that it cannot be bent sharply.

The choice follows the requirement. For a long-distance, high-capacity backbone — the internet's trunk routes — optical fibre is the clear winner on bandwidth and low loss. For mobile phones and broadcasting, radio is essential because no wire can reach a moving receiver. For a short, cheap connection between nearby equipment, copper wire is perfectly adequate and least expensive. A good answer weighs bandwidth, distance, interference, security and cost against the specific need.

In practice modern networks combine all three: a mobile call travels by radio to a mast, then over optical fibre across the country, and perhaps over copper for the last short hop into a building. Recognising that each channel has a domain where it wins, and being able to justify a channel for a stated requirement of range, data rate, mobility and cost, is the evaluative skill that completes the communication-systems chapter and frames the detailed wireless, digital and optical topics that follow.

\text{bandwidth} \ \uparrow, \ \text{attenuation} \ \downarrow, \ \text{interference} \ \downarrow \ \Rightarrow \ \text{fibre}

Channel trade-offs

Fibre leads on bandwidth and low loss; radio on mobility; copper on cost.

Worked example

Choosing a channel

Recommend a channel for (a) a transcontinental high-capacity internet backbone and (b) a link to a moving vehicle, giving reasons and one drawback of each.

Backbone
Optical fibre: it offers the highest bandwidth and lowest attenuation, so it carries huge data rates over long distances with widely spaced repeaters. Drawback: high installation cost and the need for electrical-optical conversion.
Moving vehicle
Radio: only a wireless link can reach a moving receiver. Drawback: limited, shared bandwidth and susceptibility to interference and fading.
Justification
The requirements decide it: capacity and distance favour fibre; mobility forces radio.

Result: Fibre for the fixed high-capacity backbone (cost is the drawback); radio for the mobile link (limited bandwidth and interference are the drawbacks).

Typical mistakes

Claiming one channel is always best; each wins in a different domain of range, rate and mobility.
Overlooking that radio is essential for mobile links, or that fibre needs electrical-optical conversion.

Active revision

Recommend, with reasons, a channel for (a) a transcontinental high-capacity internet link and (b) a link to a moving vehicle, and state one drawback of each choice.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

References & sources

Sources

WJEC / Eduqas

WJEC/Eduqas GCE Electronics specification

Communications systems

4 sections~15 min reading time3 competenciesLevel Standard 2 · Advanced 2

Communications systems

Key points

Identifying the blocks in a radio link

Key points

Bandwidth of a speech transmission

Key points

SNR and cable loss

Key points

Choosing a channel

Sources

Communications systems

Key points

Identifying the blocks in a radio link

Key points

Bandwidth of a speech transmission

Key points

SNR and cable loss

Key points

Choosing a channel

Sources