OCR A-Level Physics B H557
Planning
Implementing
Analysis
Evaluation
Practical Skills
Use of Apparatus and Techniques
Fundamental Data Analysis
Imaging and Signalling
Sensing
Mechanical Properties of Materials
Waves and Quantum Behaviour
Space, Time and Motion
Creating Models
Out Into Space
Our Place in the Universe
Matter: Very Simple
Matter: Hot or Cold
Electromagnetism
Charge and Field
Probing Deep Into Matter
Ionising Radiation and Risk
Imaging and Signalling
Noise, resolution and digital transmission
- 3 things to know
- Components 01, 02 and 03
Overview
Recall questions on noise and the limit it sets on useful bits, b = log₂(Vₜₒₜₐₗ/Vₙₒᵢₛₑ) and why it is rounded down, rate of transmission as samples per second × bits per sample, the highest frequency a sampling rate can record, the advantages and disadvantages of digital transmission, and regenerating pulses before noise grows too large.
What you need to know
Answer each one before you reveal it.
Explain why noise sets a limit on the useful number of bits per sample.
State the relationship giving the maximum useful number of bits for a noisy signal.
Questions from this subtopic
Each one with the model answer underneath.
- 1
What is noise on a signal?
Model answer
Random unwanted variation added to the signal by the electronics or the environment, carrying no information.
- 2
Explain why noise sets a limit on the useful number of bits per sample.
Model answer
Levels closer together than the noise cannot be distinguished from one another, so extra bits record only the random variation.
- 3
State the relationship giving the maximum useful number of bits for a noisy signal.
Model answer
b = log₂(Vₜₒₜₐₗ/Vₙₒᵢₛₑ), where Vₜₒₜₐₗ is the full range of the signal and Vₙₒᵢₛₑ the size of the noise.
- 4
A signal has a total range of 4.0 V and a noise level of 0.125 V. Calculate the useful number of bits.
Model answer
4.0/0.125 = 32 and log₂32 = 5, so 5 bits are useful.
- 5
A signal of range 5.0 V carries 20 mV of noise. Calculate the maximum useful number of bits.
Model answer
5.0/0.020 = 250 and log₂250 = 7.97, so only 7 bits are useful.
- 6
Explain why the result of a b = log₂(Vₜₒₜₐₗ/Vₙₒᵢₛₑ) calculation is rounded down rather than to the nearest whole number.
Model answer
Rounding up would create more levels than the noise allows to be told apart, so the extra bit would record noise rather than information.
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