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3 SIGNALLING Analogue vs. digital signalling. Recap advantages and disadvantages of analogue and digital signalling Calculate signal transmission rates for various types of communication. Starter: A colour photograph is taken with a 6 megapixel digital - PowerPoint PPT Presentation
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3 SIGNALLINGAnalogue vs. digital signalling
o Recap advantages and disadvantages of analogue and digital signalling
o Calculate signal transmission rates for various types of communication
Starter: A colour photograph is taken with a 6 megapixel digital camera. The camera stores 3 colours (R,G,B) for each pixel, each colour being stored as an 8 bit number. How much information (in Mbits and Mbytes) is in the image? How long would it take to transmit the image at a rate of 2 Mbit s-1?)
Signals and noise
Analogue signals are spoilt by noise Digital signals resist effects of noise
analogue signal without noise
analogue signal plus noise
signal recovered from noise loses detail
digital signal without noise
digital signal plus noise
signal accurately regenerated from noise
Sending a photo
Time to send a photo:
1 photo = 5 million pixels1 pixel = 24 bit
broadband capacity = 10 million bits per second
A digital photo is made of millions of pixels, each coded for colourand intensity by a number
01111 000 01111 0
5 million pixels × 24 bits per pixel
Photo made of about5 million pixels.
Convert pixels to streamof bits.
time to send a photo =
Run length compression:
Recode signal as runs of 1 or 0
11 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 11 1 1 11 0 0
2of 1s
13of 0s
8of 1s
2of 0s
Send these numbers instead.Typical compression 9 : 1
10 million bits per second
= 12 s
Sending an e-mail
An e-mail is a set of numbers coding for characters, and sentas 1s and 0s
Time to send one page:
1 page = 500 words = 3000 characters approx.
Thank you84 104 97 110 107 32 121 114 117
117 = 01110101
Computer and e-mailpackage encode letters asnumbers using ASCII code.
Code numbers stored asbinary digits. One byte (8bits) per character.
More coding:
E-mail also sends data which check and correct errors.
Messages are often divided into small packets, each sentby the best route available at the moment.
Packets have to be re-assembled into messages at thereceiving end.
= 3000 bytes= 24000 bits
digital line capacity = 64000 bits per second
Time to send one page = 24000 bits per page64000 bits per second
= 0.4 s approx.
D e a r s t u d e n t s ,
I h o p e t h a t y o u f i n d t h i s t e x t b o o ka s i n t e r e s t i n g t o r e a d a s w e h a v ef o u n d i l l u s t r a t i n g i t . A l o t o ft h o u g h t a n d d e d i c a t i o n h a s g o n ei n t o p r o d u c i n g t h i s , s o w e a l lh o p e a t I O P t h a t e a c h a n de v e r y o n e o f y o u p a s s y o u r A l e v e l s .G o o d l u c k t o y o u a l l .
T h a n k y o u
Sampling a signal
• Explain how to determine the rate at which a signal must be sampled to retain all detail
• Explain problems associated with sampling too slowly
Sampling a signal
original signal
sampling pulses0 or 1 from clock
sampled waveform
original waveformreconstructedfrom samples
The original signal can be exactly reconstructed if the samplingis frequent enough
Problems with sampling
Sampling too slowly misses high frequency detail in the original signal
Original signal
Samples taken fromsignal
Samples alone
Signal ‘reconstructed’from samples
Samples takenfrom signal
Sampling too slowly creates spurious low frequencies (aliases)
Original signal
Samples alone
Signal ‘reconstructed’from samples
Problems with sampling
Digitising (encoding) a signal• Explain how an analogue to digital
converter (ADC) works• Explain the terms quantisation error
and voltage resolution, and calculate them
• Explain the term signal to noise ratio and use it to determine the number of ADC bits to use
Starter: When recording digitally, music is normally sampled at a rate of 44000 samples per second. Explain why.
Sampling speech for digital telephone signals is normally done at a rate of 8000 samples per second. Explain why a lower sampling rate is used compared to music.
Digitising a signal
The greater the number of bits the better the resolution
7=111
6=110
5=101
4=100
3=011
2=010
1=001
0=000
CD uses16-bitcoding
3-bit coding: an example
quantisationerror
sample
originalsignal
nearest digitalvalue chosen
001 100 101 110 111 110 100 011 100
binary values
digital stream of bits
number of bits N
number of levels
2N
23=8 28=256 216=65536
3 8 16
8 256 65536
Voltage Resolution is the smallest change in signal voltage that can be detected with an ADC:Voltage resolution = Voltage range of ADC / 2N
where N = Number of ADC bits and 2N is the number of signal levels
Example: For an 8-bit ADC that can measure voltages in the range 0 to 5 V, the number of signal levels is 28 = 256, so the voltage resolution is 5 V / 256 = 0.02 V.
Quantisation Error is a measure of the difference between the digitally encoded value and the true value of a signal, expressed as a percentage:Worst case quantisation error = 1 / 2N
Example: For an 8-bit ADC, the QE is (1/28) x 100 = 0.4 %What is the quantisation error for a 16-bit ADC?
Signal to Noise Ratio (S:N ratio)
This is the ratio of the variation in a signal to the variation in the noise in the signal.
S:N ratio = Vsignal / Vnoise
Estimate the S:N ratio for the noisy signal shown on the right.
The S:N ratio is a very important property of a signal, as it lets us determine the minimum number of ADC bits we can get away with using when digitising the signal. We don’t want to use more ADC bits than we need, as we want to minimise the amount of data that we store.
ExampleFor example, in a particular noisy signal, the S:N ratio is measured to be about 20:1. How many ADC bits can we get away with using when digitising this signal?
The number of ADC bits to use, b, is given by the expression
2b = S:N ratio Can you explain the origin of this equation?
In this case, as 24 = 16 and 25 = 32, we would use 5 bit encoding. It would be pointless to use 6-bit or more as we would just end up digitising noise.
Polarisation• Explain the meaning of the term polarisation• Observe and explain polarisation effects involving
microwaves and visible light
Starter:A song is recorded by sampling in stereo at 44100 Hz using 16 bit encoding. What is the average bit transmission rate (in bits s-1) during recording? How much data (in bits and bytes) would a recording of the song contain if it is 5 minutes long? How could the amount of data be reduced without loss of quality?
How long would it take to download the song over a fast broadband link operating at 12 Mbits s-1?
When the permitted direction of vibration or polarisation of the filter is parallelto the direction of the polarisation of the wave, it is transmitted by the filter.
When the permitted direction of vibration or polarisation of the filter is perpendicularto the direction of the polarisation of the wave, it is absorbed or reflected butnot transmitted.
Representing signals• Learn how to represent
signals as waveforms, frequency spectra and frequency-time traces
• Generate and analyse signals using data processing software
Starter: Which organ, the eye or the ear, is capable of distinguishing between several frequencies arriving at the same part of the organ at the same time? Explain your answer.
Bandwidth
• Explain why modulating a carrier wave broadens its frequency spectrum
• Estimate the bandwidth required to transmit data at a specified rate
• Explain how and why data compression techniques are used to reduce transmission rates and required bandwidth
Starter: What does “broadband” actually mean? What is the “band” and why is it important that it is as “broad” as possible?
Review prior knowledge
Q1. Put all the wave types in the electromagnetic spectrum in order of frequency, starting with the lowest. Draw a circle around the wave with the longest wavelength.
Q2. How many MHz are in 1 GHz?
Q3. How many kHz are in 1 MHz?
Q4. Express 120 MHz in GHz
MODULATION
cos A cos B = ½ [cos(A+B) + cos(A-B)]
A = carrier waveB = modulating wave
Modulating (switching on and off) the carrier wave A using the modulating wave B broadens the frequency spectrum of the carrier.The higher the modulation frequency, the more the carrier gets broadened.
Why is this a problem when transmitting data rapidly?
Warm up questionQuantisation Error is the difference between the digitised value of a signal and its actual value.
Q1. An 8-bit ADC is used to digitise a signal. The signal range is 0 to 1000 mV. Show that the voltage resolution is approximately3.9 mV.
Q2. If the ADC records a signal to the nearest digital value what is thequantisation error in mV?
Q3. The average signal being recorded is 540 mV, while theaverage noise level is 5 mV. What is the signal : noise ratio? What is the maximum number of bits it is worth using to digitise such a signal?
Time division multiplexingSpeech on a telephone is sampled at 8000 Hz. Each sample is encoded as an 8 bit block of duration 1 ns.
Q1. Show that the time interval between samples is 125 µs.
Q2. Explain why there is a lot of “dead time” when a single telephone call is being sent across a link. Draw a diagram to support your answer.
Q3. Time division multiplexing takes multiple calls and interleaves them so that the dead-time “gaps” are filled with other calls. Estimate how many calls could be carried on the same link for the situation described above.
8 bit sample 8 bit sample 8 bit sample 8 bit sample 8 bit sample 8 bit sample
Sending many telephone calls down the same pipe: time-division multiplexingOne callerSpeech sampled 8000 times per second: approximately at 100 s intervals
Many callersSlot 8-bit samples from other callers into the 100 s gap between samples from one caller
caller 1
caller 2
100 s
caller 1 caller 2 caller 3
other callers
100 s
Advantage of digital signalling: digits can easily be switched
100 s100 s 100 s 100 s
one sample
one sample
next sample
next sample
Microwave link sending digital pulses at 10 GHzTime to send 8-bit sample approximately 1 ns. Number of bits able to be sent in 100 s = 1 million
Signal, noise and bandwidth
b log2(Vtotal )Vnoise
2b Vtotal
Vnoise
Maximum signalfrequency W2 W samples per secondneeded
Smallest voltagevariation worth detecting= noise voltage variation
Divide total voltagevariation into slices equalto noise voltage variation
= 2b slicesb = number of bits
Rate of transmission of informationbits per second = samples per second × bits per sample
bits per second = 2 Wb
Bandwidth B needed to transmit n bits per secondB = n /2
Bandwidth to transmit 2 Wb bits per secondB = Wb
Rate of transmission of information increases with bandwidthNoise decreases the possible rate of transmission of information
2b s lices
1
0 0noise
signal
noisevoltage
variation
1
time time
signal + noise
