This unit is ≈4% of the A-Level Physics, across 8 lessons. Full syllabus
Lesson 5 of 8 · Options and board-specific extensions
Digital signals, imaging and communication systems
7 min read · about 1 h 5 min with practice3 quick checks<1% of the testCore: Core: tested on most papers
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This topic turns signals and images into numbers and sends them somewhere. It is examined in the AQA Electronics option (Paper 3 Section B) and in OCR B (H557) Papers 1 and 2. The marks come from short calculations (levels, bits, bit rate, image size, bandwidth) and from explaining why digital beats analogue over long distances.
By the end you’ll be able to
Explain sampling, quantisation, the sampling theorem (rate > 2 × highest frequency) and quantisation noise
Calculate bit rate, number of levels N = 2ᵇ, image information content (pixels × bits per pixel) and resolution
Describe how noise, bandwidth and transmission media (wire, fibre, radio) limit communication
Construct truth tables for logic gates and describe counters and latches (AQA)
Compare AM and FM and explain time-division multiplexing
What the exam asks
Sampling and quantisation: the sampling theorem, levels from bits, and quantisation error.
Rates and sizes: bit rate, information in an image, and time to transmit.
Noise: why digital pulses can be regenerated, and (OCR B) how noise limits the useful number of levels.
Images (OCR B): pixels, resolution, median filtering, edge detection and contrast stretching.
Logic and counters (AQA): truth tables, latches, and binary, decade and Johnson counters.
Communication (AQA): AM and FM bandwidths, transmission media and time-division multiplexing.
vii.Check your understanding
3 questions on digital signals, imaging and communication systems. Every option is explained once you answer.
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PromptCard 1 of 3
State the sampling theorem.
b=log2N
not
Core ideas
Analogue to digital
An analogue signal can take any value in a range; a digital signal takes only discrete values, usually 0 and 1. An analogue-to-digital converter (ADC) does two jobs:
Sampling: it measures the signal at regular intervals, fs times per second.
Quantisation: it rounds each sample to the nearest of N levels and codes it in b bits.
N=2bb=log2N
The step size (resolution) of the ADC is its input range divided by the number of levels, about Vrange/2b. Rounding to the nearest level leaves a quantisation error of up to half a step, which is heard or seen as quantisation noise. More bits give smaller steps and less quantisation noise.
Sampling theorem: to reconstruct a signal, you must sample at more than twice its highest frequency, fs>2fmax. Sampling too slowly creates false low-frequency signals (aliasing). This is why CD audio, with fmax≈20 kHz, samples at 44.1 kHz.
bit rate=fs×b×(number of channels)
Noise and why digital wins
Every channel attenuates the signal and adds noise. Amplifying an analogue signal boosts the noise too, so noise accumulates. A digital receiver only decides “high or low”, so a regenerator at each repeater rebuilds clean pulses and noise does not build up, provided it stays below half the gap between levels. The costs are quantisation error and extra bandwidth.
OCR B: noise limits the levels. If the signal varies over Vtotal and the noise is Vnoise, levels closer than the noise cannot be told apart:
Nmax=VnoiseVtotalb=log2(VnoiseVtotal
Round bdown, because you cannot use part of a bit.
Digital images (OCR B)
An image is a grid of pixels, each stored as a number (brightness, or three numbers for colour).
Information content= number of pixels × bits per pixel. 8 bits = 1 byte.
Resolution= the distance on the object covered by one pixel, for example 5.0 m per pixel for a satellite image. A smaller value shows finer detail.
Processing: a median filter replaces each pixel with the median of it and its eight neighbours, which removes isolated noisy pixels without blurring edges. Edge detection subtracts the average of the neighbours from each pixel, so uniform areas go to zero and boundaries stand out. Contrast stretching maps a narrow range of values onto the full range.
Logic gates and counters (AQA)
A
B
AND
OR
NAND
NOR
XOR
0
0
0
0
1
1
0
0
1
0
1
1
0
1
1
0
0
1
1
0
1
1
1
1
1
0
0
0
NOT inverts a single input. Any gate can be built from NAND gates alone. A latch (two cross-coupled NOR or NAND gates) holds its output after the input that set it has gone: it is a one-bit memory.
An astable circuit supplies clock pulses. In a binary counter, each stage changes state once for every two pulses it receives, so each output has half the frequency of the one before. A 4-bit counter counts 0000 to 1111 (16 states). A decade (BCD) counter resets to 0000 on the tenth pulse. A Johnson counter is a shift register with the inverted last output fed back to the start. It has 2n states for n stages, and only one bit changes at each step.
Communication systems (AQA)
A carrier wave is modulated by the information signal of maximum frequency fM.
AM
FM
What changes
carrier amplitude
carrier frequency, by up to ±Δf
Bandwidth
2fM
2(Δf+fM)
Noise
poor: noise adds to the amplitude
better: amplitude noise is clipped off
Range
long (lower frequencies)
shorter, line of sight
Media: copper pairs and coaxial cable are cheap, but their losses rise with frequency. Optical fibre has a huge bandwidth, low attenuation and no electrical interference. Radio: ground waves (below about 3 MHz) follow the Earth’s surface, sky waves (3–30 MHz) reflect off the ionosphere, and space waves (above 30 MHz) travel in lines of sight to masts and satellites.
Time-division multiplexing (TDM) interleaves samples from several channels in successive time slots on one link, so the link’s bit rate is the sum of the channel bit rates.
Worked examples
Exam technique
Write the chain with units: samples s⁻¹ × bits sample⁻¹ × channels = bit s⁻¹. Divide by 8 only if bytes are asked for.
Levels or bits?N is a count of levels; b is a count of bits. Convert with 2b or log2, and round bits up to meet a required number of levels, but down when noise limits them.
“Explain why digital...” answers need the mechanism: a two-level decision, regeneration at repeaters, and noise not accumulating.
Counter questions: list the states in a table, one row per clock pulse. Frequencies halve at each binary stage.
Common mistakes
Quick recap
N=2b; step ≈Vrange/2b; maximum quantisation error = half a step.
Sample at fs>2fmax; bit rate = channels.
Digital signals are regenerated, so noise does not accumulate; the costs are quantisation noise and bandwidth.
OCR B: b=log2(Vtotal/V, rounded down; information pixels bits per pixel; resolution metres per pixel.
AQA: learn the gate truth tables; binary stages halve the frequency; the Johnson counter has 2n states.
AM bandwidth 2fM; FM bandwidth 2(Δf+fM); TDM shares one link in time slots.