Notes · ElectronicsUK · A-Levels

Digital communications

Digital communication sends information as a stream of discrete symbols, gaining noise immunity and the ability to regenerate a signal perfectly. This chapter develops digital modulation (ASK, FSK and PSK), pulse code modulation, the Nyquist sampling theorem and bit rate, and the regeneration and time-division multiplexing that make digital links robust and efficient.

4 sections~15 min reading time3 competenciesLevel Advanced 4

T·151515 / 16

Exam profile

AO1 · Describe digital modulation, pulse code modulation, signal regeneration and time-division multiplexing.AO2 · Apply the Nyquist criterion and calculate sampling rate, bit rate and multiplexed data rate.AO3 · Evaluate digital versus analogue transmission and justify a sampling and coding scheme.

Operators:describecalculatedetermineapplyanalyseevaluate

basic level

At AS the focus is the idea of digital modulation and the advantage of a two-state signal.

higher level

The full A-Level develops PCM, the Nyquist theorem, bit-rate and multiplexing calculations, and signal regeneration.

Depth

Text

Contents · 4 sections

Digital communications

Digital modulation: ASK, FSK and PSK#

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

Amplitude-shift keying (ASK)

Fig. 1ASK (on-off keying): the carrier is present for a 1 and absent for a 0 — here the pattern

1010

Digital modulation impresses a stream of bits onto a carrier by switching one of its properties between discrete states — the digital counterparts of AM and FM. In amplitude-shift keying (ASK) the carrier is switched on for a 1 and off for a 0 (on-off keying), so the presence or absence of the carrier signals the bit. It is the simplest scheme and easy to generate, but because the information is in the amplitude it shares AM's vulnerability to noise and interference.

In frequency-shift keying (FSK) the carrier is switched between two frequencies, one representing a 1 and the other a 0, while its amplitude stays constant. Like FM, this constant amplitude makes FSK much more robust to noise than ASK, because a receiver can ignore amplitude variations and detect only which of the two frequencies is present. FSK is used where reliability matters more than the last word in efficiency, such as in older modems and many low-data-rate radio links.

In phase-shift keying (PSK) the information is carried in the phase of the carrier — for example, a

180^{\circ}

phase change marks a change of bit. PSK keeps both amplitude and frequency constant and varies only the phase, giving excellent noise immunity and efficient use of bandwidth, which is why it and its multi-level relatives underpin modern high-speed digital communication such as Wi-Fi and mobile data. More phases (or combinations of phase and amplitude) can encode several bits per symbol, raising the data rate.

The common advantage of all digital modulation over analogue is that the receiver only has to decide between a small number of definite states — is the carrier on or off, at which of two frequencies, in which phase — rather than reproduce a continuous value. This decision is far more robust to noise: as long as the noise is not large enough to push one state into another, the bit is recovered perfectly, and any small corruption is simply removed by the decision.

Choosing a scheme trades simplicity, bandwidth efficiency and noise immunity, exactly as AM versus FM did in the analogue case. ASK is simplest but least robust; FSK is robust and moderate; PSK is the most efficient and robust but the most complex to implement. Recognising each scheme from its waveform, and understanding why the constant-amplitude schemes (FSK, PSK) resist noise better than ASK, is the core of digital modulation.

\text{ASK: amplitude}, \quad \text{FSK: frequency}, \quad \text{PSK: phase}

Digital modulation schemes

Each switches one carrier property between discrete states to send bits.

Frequency-shift keying (FSK)

Fig. 2FSK: the carrier switches between two frequencies (high for a 1, low for a 0) at constant amplitude — much more noise-immune than ASK.

Worked example

Choosing a robust scheme

A radio link suffers from strong amplitude noise (interference that adds to the signal's height). Explain, comparing ASK, FSK and PSK, which scheme would give the most reliable reception.

ASK
ASK carries the bits in the carrier's amplitude, exactly where the noise lands, so amplitude noise can turn an 'off' into an apparent 'on' — least reliable here.
FSK and PSK
Both keep the amplitude constant and carry the information in frequency (FSK) or phase (PSK). A receiver can ignore amplitude variations, so amplitude noise barely affects them.
Best choice
FSK or PSK; PSK is the most bandwidth-efficient and robust, but FSK is simpler. Either is far better than ASK on an amplitude-noisy link.

Result: FSK or PSK is preferred because their constant amplitude resists amplitude noise, whereas ASK's amplitude signalling is directly corrupted by it.

Typical mistakes

Confusing which property each scheme varies — ASK amplitude, FSK frequency, PSK phase.
Thinking ASK is as robust as FSK; its amplitude-based signalling is more affected by noise.

Active revision

Sketch the ASK and FSK waveforms for the bit pattern $1010$ and state, with a reason, which of ASK, FSK and PSK is least affected by amplitude noise.

Active recall

Recall the key points — then reveal.

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

Pulse code modulation#

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

The PCM chain

Fig. 3The three PCM stages: sample-and-hold, quantiser and encoder, turning the analogue input into a serial binary bitstream.

Pulse code modulation (PCM) is the standard way to turn an analogue signal into a digital bitstream for transmission, and it uses exactly the conversion ideas of the signal-conversion chapter. It has three stages: sample the analogue signal at regular instants (a sample-and-hold captures each value); quantise each sample to the nearest of a finite set of levels (rounding it, as an ADC does); and encode each quantised level as a binary number. The output is a stream of binary codes — the digital representation of the signal.

The first stage, sampling, is governed by the Nyquist criterion (next section): the signal must be sampled at least twice as fast as its highest frequency, or it cannot be reconstructed. The second stage, quantising, introduces the quantisation error of the conversion chapter — the more bits per sample, the finer the levels and the smaller this error. The third stage, encoding, simply writes each level as an

n

-bit binary number, ready to be sent as a serial bitstream.

The bit rate of a PCM system is the product of the sampling rate and the number of bits per sample:

\text{bit rate} = f_s \times n

. Standard telephone-quality PCM samples speech at

8\,\text{kHz}

(twice the

3.4\,\text{kHz}

audio band, with margin) and uses

8

bits per sample, giving the familiar

64\,\text{kbit/s}

of a digital telephone channel. Compact-disc audio samples at

44.1\,\text{kHz}

with

16

bits per channel. These two numbers — sampling rate and bit depth — completely specify a PCM system's rate.

The great benefit of PCM is that the analogue signal, once turned into numbers, inherits all the robustness of digital transmission: it can be regenerated perfectly along a link, stored and copied without loss, error-corrected, and multiplexed with other channels. The cost is a higher bandwidth than the original analogue signal needed and the complexity of the converters, but for quality and reliability this trade is almost always worth it, which is why PCM underlies digital telephony, audio and countless other systems.

PCM ties together the whole digital-communications story: it is the on-ramp from the analogue world to a robust digital bitstream, using sampling (limited by Nyquist), quantising (limited by bit depth) and encoding, and producing a bit rate that the channel must carry. Understanding its three stages and being able to calculate its bit rate from the sampling rate and bit depth is the central quantitative skill of the chapter.

\text{bit rate} = f_s \times n

PCM bit rate

Sampling rate times the number of bits per sample.

Worked example

Bit rate of a telephone PCM channel

A speech channel band-limited to $3.4\,\text{kHz}$ is PCM-encoded at $8\,\text{kHz}$ with $8$ bits per sample. Find the bit rate and confirm the sampling rate satisfies Nyquist.

Nyquist check
The Nyquist rate is $2 \times 3.4 = 6.8\,\text{kHz}$ ; the chosen $8\,\text{kHz}$ exceeds it, with margin for a practical anti-alias filter.
Bit rate
$\text{bit rate} = f_s \times n = 8000 \times 8 = 64000\,\text{bit/s} = 64\,\text{kbit/s}$ .
$\text{bit rate} = 8\,\text{kHz} \times 8 = 64\,\text{kbit/s}$
Stages
The channel samples the speech at $8\,\text{kHz}$ , quantises each sample to one of $256$ levels ( $8$ bits), and encodes it as an $8$ -bit binary word.

Result: The bit rate is $64\,\text{kbit/s}$ — the standard digital telephone channel — and the $8\,\text{kHz}$ sampling satisfies Nyquist.

Typical mistakes

Forgetting the quantising stage, or confusing the number of levels ( $2^n$ ) with the bits per sample ( $n$ ).
Sampling below the Nyquist rate, so the signal cannot be reconstructed.

Active revision

A speech signal band-limited to $3.4\,\text{kHz}$ is encoded by PCM at an $8\,\text{kHz}$ sampling rate with $8$ bits per sample. Calculate the bit rate and state the three PCM stages.

Active recall

Recall the key points — then reveal.

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

The Nyquist theorem and bit rate#

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

Sampling a signal at the Nyquist rate

Fig. 4Sampling a sinusoid: with samples (dots) taken more than twice per cycle, the original wave can be reconstructed exactly (Nyquist satisfied).

The Nyquist sampling theorem states that to capture a signal faithfully it must be sampled at a rate of at least twice its highest frequency:

f_s \geq 2 f_{max}

. Sample fast enough and the original continuous signal can be reconstructed exactly from its samples; sample too slowly and the reconstruction is wrong. This theorem sets the minimum sampling rate for any digital audio, communication or measurement system and is one of the most important results in the whole subject.

If a signal is sampled below the Nyquist rate, a serious error called aliasing occurs: high-frequency components masquerade as lower frequencies in the reconstructed signal, producing spurious tones that were never in the original. Aliasing cannot be removed after sampling, so it is prevented before sampling by an anti-aliasing filter — a low-pass filter that removes any frequencies above half the sampling rate, guaranteeing the Nyquist condition is met. This is a direct, practical use of the low-pass filter from the AC chapter.

The choice of sampling rate is therefore driven by the highest frequency that must be preserved. Telephone speech is band-limited to

3.4\,\text{kHz}

and sampled at

8\,\text{kHz}

; high-fidelity audio extends to

20\,\text{kHz}

and is sampled at

44.1\,\text{kHz}

on a compact disc — comfortably above the

40\,\text{kHz}

Nyquist rate to allow a practical filter. In each case the rate is set by Nyquist plus a margin for the anti-aliasing filter's finite roll-off.

The sampling rate combines with the bit depth to fix the bit rate a digital system must transmit or store,

\text{bit rate} = f_s \times n

, and hence the channel bandwidth it needs. A higher sampling rate or more bits per sample gives better fidelity but demands a higher bit rate and more bandwidth — the same fidelity-versus-resources trade-off met in the conversion chapter, now determining the load a communication channel must carry.

The Nyquist theorem is thus the hinge between the analogue and digital worlds: it fixes how fast a signal must be sampled to be represented digitally without loss, and through the bit-rate relation it links that requirement to the channel capacity needed. Being able to apply the criterion, explain and prevent aliasing, and calculate the resulting bit rate is the quantitative heart of digital communication and a frequent exam requirement.

f_s \geq 2 f_{max}

Nyquist criterion

The sampling rate must be at least twice the highest signal frequency.

Worked example

Sampling rate and bit rate for CD audio

Audio up to $20\,\text{kHz}$ is to be digitised with $16$ bits per sample. State the minimum sampling rate, and find the bit rate for one channel at the CD rate of $44.1\,\text{kHz}$ .

Minimum sampling rate
$f_s \geq 2 f_{max} = 2 \times 20 = 40\,\text{kHz}$ ; the CD rate of $44.1\,\text{kHz}$ exceeds this, leaving room for the anti-aliasing filter.
Bit rate
$\text{bit rate} = f_s \times n = 44100 \times 16 = 705600\,\text{bit/s} \approx 706\,\text{kbit/s}$ per channel.
$\text{bit rate} = 44100 \times 16 \approx 706\,\text{kbit/s}$
Stereo
Two channels double this to about $1.41\,\text{Mbit/s}$ — the raw data rate of CD audio.

Result: The minimum sampling rate is $40\,\text{kHz}$ ; at $44.1\,\text{kHz}$ each channel is about $706\,\text{kbit/s}$ (roughly $1.41\,\text{Mbit/s}$ in stereo).

Typical mistakes

Sampling at the highest frequency rather than at twice it, so the signal cannot be reconstructed.
Thinking aliasing can be filtered out after sampling; it must be prevented before, by an anti-aliasing filter.

Active revision

High-fidelity audio must preserve frequencies up to $20\,\text{kHz}$ and is encoded with $16$ bits per sample. State the minimum sampling rate and calculate the bit rate at the CD rate of $44.1\,\text{kHz}$ .

Active recall

Recall the key points — then reveal.

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

Regeneration and time-division multiplexing#

AdvancedWJEC/Eduqas A Level Electronics — Unit 4 Digital and Communication (regeneration, TDM)

Time-division multiplexing

Fig. 5TDM: a multiplexer gives each channel a time slot on the shared high-speed link, and a synchronised demultiplexer restores the separate channels.

The decisive advantage of digital transmission is that a digital signal can be regenerated. Along a link the signal is attenuated and picks up noise, distorting its clean edges; but because it can only be a 1 or a 0, a regenerator can decide which each bit is and rebuild a perfect, noise-free copy to send onward. An analogue signal, by contrast, can only be amplified, which magnifies its accumulated noise along with the signal — so noise builds up on an analogue link but is wiped out at every regenerator on a digital one.

This is why long digital links can be essentially perfect while long analogue links degrade. As long as the noise at a regenerator is not large enough to make a 1 look like a 0, every bit is recovered exactly and the signal is restored to full quality, so errors do not accumulate stage after stage. This noise immunity, combined with error-detecting and error-correcting codes that catch the rare bit that does slip through, gives digital communication its remarkable reliability over any distance.

Digital signals also multiplex neatly in time. Time-division multiplexing (TDM) sends several digital channels down one link by giving each channel a brief time slot in turn, interleaving their bits (or bytes) so that the single high-speed link carries them all. A multiplexer at the sending end cycles through the channels, filling successive slots, and a synchronised demultiplexer at the far end sorts the slots back into the separate channels. The whole frame repeats fast enough that each channel appears continuous.

The aggregate bit rate of a TDM link is simply the sum of the channel rates (plus a little for synchronisation). Combining, say, thirty

64\,\text{kbit/s}

telephone channels needs a link carrying about

30 \times 64 = 1920\,\text{kbit/s}

, which with framing and signalling becomes the standard

2.048\,\text{Mbit/s}

primary rate. TDM thus lets one expensive high-capacity link — a fibre or a microwave hop — carry many low-rate channels efficiently, sharing the cost across all of them.

Regeneration and multiplexing together explain why the world's communication has gone digital: signals travel any distance without accumulating noise, and one link carries many channels by sharing it in time. These advantages — perfect regeneration, error correction, and efficient multiplexing — outweigh the extra bandwidth and converter complexity that digitisation costs, which is the evaluate-and-justify conclusion the digital-communications topic is building toward.

\text{aggregate rate} = N \times (\text{channel rate}) \ (+ \text{framing})

TDM data rate

The multiplexed link carries the sum of the channel rates plus synchronisation overhead.

Worked example

Aggregate rate of a multiplexed link

Thirty telephone channels, each $64\,\text{kbit/s}$ , are time-division multiplexed. Find the aggregate data rate (ignoring framing), and explain the regeneration advantage over an analogue link.

Aggregate rate
$\text{rate} = N \times \text{channel rate} = 30 \times 64 = 1920\,\text{kbit/s} = 1.92\,\text{Mbit/s}$ (with framing and signalling this becomes the standard $2.048\,\text{Mbit/s}$ ).
$30 \times 64\,\text{kbit/s} = 1.92\,\text{Mbit/s}$
Regeneration
The digital signal has only two states, so each regenerator decides 1 or 0 and rebuilds perfect edges, discarding the noise — errors do not accumulate.
Analogue contrast
An analogue link can only amplify, which magnifies the accumulated noise, so quality falls steadily with distance.

Result: The link carries $1.92\,\text{Mbit/s}$ (plus framing); the digital signal is regenerated noise-free at each repeater, whereas an analogue link accumulates noise.

Typical mistakes

Thinking analogue amplification removes noise as digital regeneration does — amplification magnifies the noise too.
Forgetting to add the individual channel rates (and framing overhead) when finding the TDM aggregate rate.

Active revision

Thirty telephone channels, each $64\,\text{kbit/s}$ , are combined by time-division multiplexing. Calculate the aggregate data rate and explain why the digital link can be regenerated but an analogue one cannot.

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

Digital communications

4 sections~15 min reading time3 competenciesLevel Advanced 4

Digital communications

Key points

Choosing a robust scheme

Key points

Bit rate of a telephone PCM channel

Key points

Sampling rate and bit rate for CD audio

Key points

Aggregate rate of a multiplexed link

Sources

Digital communications

Key points

Choosing a robust scheme

Key points

Bit rate of a telephone PCM channel

Key points

Sampling rate and bit rate for CD audio

Key points

Aggregate rate of a multiplexed link

Sources