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## 3 SIGNALLING Analogue vs. digital signalling

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**3 SIGNALLINGAnalogue 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 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?)**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**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.**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. Example For 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?**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 wave B = 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 question**Quantisation 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 approximately 3.9 mV. Q2. If the ADC records a signal to the nearest digital value what is the quantisation error in mV? Q3. The average signal being recorded is 540 mV, while the average 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 multiplexing**Speech 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.