This article will go over all you need to know in the A Level Physics which includes advantages of digital signalling, analogue signalling, converting between digital to analogue and analogue to digital, sampling, polarisation, sensing circuits, resolution, bandwidth and a sampling summary including sampling rate and channel capacity (I think are in chronological order). Your more than welcome to skip to the parts that are of most help to you. As well as that, you may also want to look at other articles I have done for Physics AS revision. If you want a quick look at this article, I would suggest you look at Sampling Summary at the bottom although that is not an excuse to just read that: you will be missing out loads of information and detail from the main article.
There are many advantages to digital signalling. However, before we get onto that, we should know what is an analogue and digital signal.
The parts of a digital signal can have one of two values, 0 or 1.
Examples of digital devices include on/off switch, digital ammeter, digital clock, optic fibre and telephone cable. Time is on the x-axis.
A digital code will look like this:
Because there are only different possibilities.
The parts of an analogue signal can have any value within a fixed range of values . For my example, it’s range is from 0-3 but this can vary from analogue signal to analogue signal. They basically take any values within it’s range.
Examples of analogue devices include dimmer switch, thermometer, analogue ammeter and speedometer. Time is on the x-axis.
Advantages of Digital Signalling
- Digital signals have better quality – this is because noise can be easily removed. More information on noise can be found here.
- Digital signals allow more information to be carried.
- Many signals can be sent down the same optic fibre – this is called multiplexing.
Analogue to Digital Conversion
Analogue signals are turned into digital signals by sampling. The height of the wave is measured at regular time intervals and the height put into binary code
Each sample can only have one of a fixed set of values (in the case 0 to 7 or 8 bit). I will use the above analogue signal and converse it to an 8 bit digital signal.
As you can see, the grey bars are the new conversed digital signal from the analogue signal. You can see that some of the analogue signal in-between the bars have been lost.
As well as that, because it has 8 possible options, it means it is in 3 bit. You can see more on binary codes here
Digital to Analogue Conversion
The analogue signal is the constructed from the digital signal. The reconstructed signal will be in red and follows the line of best fit. After seeing the two lines, can see the differences between the sampled (black) wave and reformed (red) wave?
The signal at the end is very similar to each other. However, at the start, much of the detail in the sampled (original) wave has been lost. Different places throughout the signal have frequencies that have been introduced that were not present in the original sample.
What happens when you view polarised light through a rotating polaroid filter and why?
You will see it go darker at a certain point when rotating the filter. This is because at the point of it darkening, the polaroid filter is stopping the polarised light from going through it causing less light waves to pass through. At the point of darkening, the filter is parallel to the electric filed of the light waves causing the light waves parallel to the filter to go through but all the others waves not to. The amount of light going through is fluctuated between two extremes.
To make more sense of what I’ve just said, I’ll draw a diagram.
If you can’t read the writing, click on the image and a zoomed version of the image will appear.
Here is an example of a sensing circuit:
To work out the sensitivity, you have to do the equation:
Voltage output range / input range
or more general:
output range / input range
The sensitivity is also know as the gradient.
For our example the sensitivity is 2 / 100 and will be measured in Volts per lux.
Therefore, the sensitivity will be 2×10 to the power of -3 Volts/lux.
Resolution = smallest change that can be detected.
To calculate the resolution:
- What is the smallest possible output? In our example, it will be 0.01 Volts as our voltmeter is accurate to two decimal places.
- What input change does this correspond to ?
The sensitivity is 0.002 V/lux. Therefore, resolution = 0.01 / 0.002 which equals 5 lux.
Bandwidth of a signal is the range of frequencies in a signal.
Capacity of a channel is the maximum amount of information it can sends bits per second.
As a rough estimation the bandwidth equal to the maximum number of bits per second.
A square wave contains a fundamental frequency and all its multiples in its signal and therefore has an extremely high bandwidth.
The fundamental frequency is the rate at which each 01 pair is sent (i.e. number of bits per second = 2 x fundamental frequency of sound waves).
Filtering out some of these frequencies changes the shape of the square wave but it is still recognisable and therefore reformable at the other end.
This reduces the bandwidth and means that the signal can be sent down a channel with smaller bandwidth.
It is possible to remove almost all frequencies except the fundamental hence fundamental frequency, new bandwidth and number of bits per second are all roughly equal. (i.e. same order of magnitude).
- The parts of an analogue signal can have any value within a fixed range of values
- The parts of a digital signal can have one of only two values, 0 or 1.
- Streams of 0s and 1s can be used to represent any whole number binary code.
Analogue to Digital Conversion
Analogue signals are turned into digital signals in 3 stages:
- Binary coding
- Further encoding
To sample a wave the height is measured at regular time intervals and put into binary code.
Each sample can only have one of a fixed set of values (0 to 7 in 8 bit binary).
Number of Bits
- Number of possible values = 2 to the power of b where b = number of bits.
- The noise limits the maximum number of bits it is worth sending.
Number of possible values = total signal variation / noise variation.
- The sampling frequency must be at least 2 x the highest frequency component present or else details of the signal may be lost.
- The highest frequency present can be identified in a frequency spectrum.
- This is the rate at which a channel can transmit information, measured in bits per second.
- The amount of information in a signal per second = sampling frequency x bits per sample.
- The range of frequencies used to send information on a particular channel.
- Bandwidth B needed = b bits per second / 2.
Digital Signals have the advantage that:
- They can be regenerated easily, reducing the effects of noise.
- They can be processed and encoded.
- They can represent different kinds of information in the same way.
Signals are coded in such a way that errors can be located and corrected.
E.g. each part of the signal is sent more than once. If the ‘copies’ do not agree there must be an error, and they can be asked again until they agree.
It’s a lot to take in so if you are stuck on any of it, please comment below, subscribe to future comments and I will try my best to answer your questions. Hope this has helped! Please also see other revision material I have done on Physics AS.