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CS3282 Digital Communications 4.1 25/08/06/BMGC University of Manchester CS3282 : Digital Communications'05-'06 Section 4: Introduction to digital transmission This section is concerned with the transmission of digital signals in the form of suitably shaped pulses over wire-lines or radio channels. Such pulses are often visualised as being rectangular in shape, and this visualisation is not too unrealistic for base-band transmission over short distances, as used for example with a wired Ethernet based wireless LAN. However, a rectangular pulse shape requires infinitely wide frequency bandwidth and is therefore undesirable for transmission over a wire-line or channel where economy of bandwidth utilisation is a consideration. This is usually the case with long distance high speed transmission. A more typical pulse shape is rounded and has ringing before and after the main part of the pulse to reduce its bandwidth. Data (bit-) rate and signalling rate: The 'data-rate' or 'bit-rate' is the number of bits (binary digits) per second. The 'signalling-rate' is the number of 'symbols' per second. Units for the signalling rate are 'bauds'. A symbol is a voltage pulse whose shape and amplitude is chosen from a set of two or more possibilities. With binary signalling, there are two possible symbols, say a rectangular pulse of amplitude +V and a rectangular pulse of amplitude −V. In this case the signalling rate can be equal to the data rate if no redundancy is included for error checking. Often 'ternary' signalling is used with pulses of amplitude +V, 0, and −V. Now the signalling rate can be less than the bit-rate. We could send three-bits using two ternary (+V, 0, -V) symbols because there are 8 possible 3-bit binary numbers and nine different ways of combining two ternary pulses. Hence the bit-rate could be 1.5 times the signalling (baud) rate. With 'quaternary' signalling, there are four possible symbols and therefore the bit-rate can be twice the symbol rate. The symbol period will always be T seconds, therefore the signalling-rate will be 1/T baud. Asynchronous transmission (low data rates): A digital transmitter must apply suitably shaped symbols to the channel at times specified by a timing reference or 'clock'. The clock is a circuit which generates a 'timing waveform' which is usually a regular sequence of rectangular timing pulses. The timing waveform is normally not transmitted. A timing waveform must also be available at the receiver to indicate the time-points at which the channel may be examined to extract a symbol. The receiver's timing waveform must have the same frequency as that used at the transmitter. Also, it must be synchronised with the symbols being received, even though delay will have been introduced by the channel. It is usual for the receiver to extract the exact symbol frequency and symbol synchronisation from the signal received from the channel, even though in many cases this signal will be distorted in various ways. If lengthy transmissions are intended, the receiver clock must be very accurately matched to the clock frequency of the transmitter since any small discrepancy will accumulate over time to produce large timing errors. However for short blocklength transmissions as used for transmitting 8-bit binary numbers between computers and peripherals over short distances, for example, the transmitter and receiver clocks need only be approximately matched and they may resynchronise at the beginning of each short block. This is often referred to as 'asynchronous' transmission and is the basis of the well known RS232 standard.. Data is sent in short words, say 8 bits long, with synchronising start and stop bits. The receiver clock resynchronises itself at each start-bit. Consider the transmission of 8-bit ASCII characters according to the RS232 protocol. When idle, the line remains high at voltage V1. Asynchronous operation usually has a “startbit” to signify the start of a transmission. This bit is always “0”. The eight bits of data are then transmitted using “non-return to zero” (NRZ) pulses and finally a number of “1” stop-bits (in this case two) are transmitted to ensure that the next character is not sent immediately. CS3282 Digital Communications 4.2 25/08/06/BMGC V1 V0 0 1 Start-bit 0 1 1 0 0 Data 0 1 1 1 t Stop-bits The advantage of this form of transmission is the simplicity of the transmitter and especially the receiver. Its disadvantage for some applications would be that it is inefficient in its utilisation of the channel capacity. The receiver waits for a transition from the 'idle state' "1" to "0" indicating a 'start bit'. It delays for half a symbol period according to its own free running clock having approximately the same frequency as that of the transmitter, and then samples the channel eleven times at intervals of T seconds. The samples will hopefully lie in or close to the centre of each symbol, but the timing will drift over the eleven samples. The drift is acceptable because of the frequent resynchronisation. Synchronous transmission: Synchronous techniques are used for the efficient transmission of continuous data for long periods of time, often at data rates close to the maximum possible over a channel of specified bandwidth. A synchronising code (say 10101010) is sent at start of transmission, and thereafter, the receiver clock must be kept synchronised in frequency and symbol-timing from the transmission itself. To achieve the required bandwidth efficiency, pulses must be appropriately shaped, usually by a filter whose impulse response is the required shape. More about this later. To detect the presence or absence of a pulse, the receiver samples the received waveform at the correct symbol timing point. Base-band synchronous transmission over wire-lines: When considering how to synchronously transmit digital information over wires, two factors must be borne in mind:(i) We would like to keep the average voltage level as close as possible to zero since any voltage offset carries no data and just wastes power. In many cases, the DC component of a signal is lost over wire lines sometimes because of AC coupling, the use of transformers and/or because the line is used to carry power as well as the data. This is certainly the case with telephone lines. (ii) For synchronous transmission, we need to ensure that the signal always has a frequency component at the signalling rate (or an exact multiple or sub-multiple of the signalling rate) to allow a timing waveform to be extracted at the receiver for synchronising the detection process. For binary transmission, we could try to achieve a zero average voltage by making the amplitudes of the two pulses +V and -V, hoping that, on average the same number of ones and zeros will occur. However, a long sequence of consecutive '0 0 0 0 ... 0' or '1 1 1 1 1 ... 1' would clearly cause problems with this scheme. One solution to this problem is to use ternary coding with alternate mark inversion (AMI) i.e. to transmit 0 volts for logic "0" and ±V volts, used alternately, for logic "1". For example, to transmit: ' 1, 0, 1, 1, 0 , 1, 1 ' , we send pulses whose amplitudes are: V, 0, -V, V, 0, -V, V The average voltage (the "dc level") is now guaranteed to be zero. Using AMI, as described above, timing waveform extraction and synchronisation is straightforward when, say, '1 1 1 1 1 1 1 ...' is transmitted. If 'non-return to zero' (NRZ) rectangular pulses of width T are used, the signal will be a rectangular wave with period 2T and hence of frequency half the CS3282 Digital Communications 4.3 25/08/06/BMGC signalling rate. If 'return to zero' (RZ) pulses are used, synchronisation is even easier as there will be a strong harmonic at the signalling rate. However when a significant number of consecutive zeros: '0 0 0 0 0 0 .... 0' is transmitted, the receiver can lose synchronisation as the received signal will be zero. A commonly used solution is known as HDB3 coding. +V +V T t T t −V −V '..1111111..' by NRZ AMI ' ..1111111' by RZ AMI HDB3 coding: (high density bipolar, order 3): This scheme uses ternary coding to send binary coded data, as described above for AMI, but places an incorrectly signed pulse in place of any 4th consecutive. zero. E.g. for '1 1 1 ' +V -V 0 +V 0 0 0 0 0 0 0 0 0 0 0 1 0 1 … ' we send: +V 0 0 0 +V 0 -V 0 +V … ' The incorrect "+V" pulse is included only for clock synchronisation. It is taken to be a “0” at the receiver. The average voltage is no longer zero in the short block above, but over a longer time-span the average will still remain zero since incorrect +V pulses and -V pulses will occur equally often. Other base-band signalling waveforms: NRZ-AMI, RZ-AMI and their variants NRZ-HDB3 and RZ-HDB3 are commonly used. However there many other schemes as illustrated in any textbook. These schemes are often referred to as baseband "line-codes" or PCM waveforms. Schemes known as NRZ-L, NRZ-M & NRZ-S are binary signalling methods in that there is only +V and –V. NRZ-L (level) is the most straightforward with +V representing “1” and “-V” representing “0”. NRZ-M (mark) has logic “1” represented by a change from +V to -V and “0” represented by no change. NRZ-S (space) represents “0” by a change, and “1” by no change. Uni-polar RZ has binary 'return to zero' pulses (0 and +V). Bi-polar RZ has +V & -V 'return to zero' pulses. RZ-AMI has been discussed. Another group known as bi-phase-L, bi-phase-M & bi-phase-S are used in magnetic recording systems, optical communications, satellite links, and many other applications including Ethernet. Bi-phase-L is better known as “Manchester coding”, and represents a “one” by a pulse of width T/2 positioned during the first half bit-interval. A zero has a pulse of width T/2 in the second half interval. +V +V 'one' −V T t 'zero' −V T Manchester coding CS3282 Digital Communications 4.4 25/08/06/BMGC Question: What are the advantages and disadvantages of Manchester coding as compared with NRZHDB3? Before completing this survey of base-band signalling waveforms, one more must be mentioned, i.e. 4B3T. 4B3T coding: (4-bits re-coded as 3 ternary digits): In four bits we can have 16 possible numbers. In 3 ternary digits we can have 27 possible numbers. We can represent each 4-bit number by a 3 ternary digit number, and have some 3-ternary digit numbers left over. Therefore we allocate alternative codes to some of the binary numbers and use these (i) to keep the average signal level zero, and (ii) to ensure significant carrier content for receiver synchronisation. The ternary codes are: BINARY (a) TERNARY (b) 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 −−− − −0 −0− 0 −− −−+ −+− +−− −00 0−0 0 0 -− +++ ++0 +0+ 0++ ++− +−+ −++ +00 0+0 00+ 0+− 0−+ +0− −0+ +−0 −+0 Represent “+V” by "+", “−V” by "−", & “0 volts” by “0”. Choose either column (a) or column (b) where there is a choice. When decoded they give the same sequence of 4-bits. If the "accumulated disparity" is "+", i.e. if we have previously sent more "+" pulses than "−" pulses, choose column (a) to redress the balance. Otherwise choose column (b). Given the same pulse shaping, “4B3T” would require less bandwidth than AMI as it makes better use of the 3 levels –V, 0 and +V For example, if each pulse of width T seconds were shaped so that its bandwidth were 1/T Hz, AMI would have a 'bandwidth efficiency' of 1/T bits/second in 1/T Hz , i.e. 1 bit/second per Hz. 4B3T would have 4/(3T) bits/second in 1/T Hz, i.e. 4/3 bits/second per Hz. Example: Encode in 4B3T: 0010, 0000, 1111, 0001, 0001, 0001, .... Answer: Assuming "accumulated disparity" to be 0 at start, + 0 +, - - -, - + 0, + + 0, - - 0, + + 0, ... Estimation of 'bit-error probability' using the 'complementary error function Q(z)': The channel and the receiver may be assumed to add white Gaussian noise of zero mean and fixed variance σ2 to the transmitted signal. The receiver of a binary coded signal, receiving +V volt rectangular pulses for '1' and zero volts for '0' may set a threshold at +V/2 and decide whether this is exceeded at sampling points taken in the centre of each rectangular pulse. If the amplitude of the noise exceeds +V/2 at a sampling point, an error may occur. CS3282 Digital Communications 4.5 25/08/06/BMGC +V +V/2 t The probability of a white Gaussian noise sample with zero mean and variance σ2=1 being greater than some voltage z is ∞ Q( z ) = ∫ p (t )dt z (normalised error probability) Q(z) is the "normalised error function", where p(t) = (1/(√2π))exp(-t2/2) is the probability density function of unit variance (σ2 = 1) Gaussian noise as plotted below. Q(z) = probability of signal exceeding z. Q(z) is plotted as a graph against z in the figure attached. P(t) z t For white Gaussian noise of variance σ2, the probability of a given sample exceeding z is Q(z/σ), therefore the probability of a given sample exceeding A/2 is Q(A/(2σ)). Q(z) may be obtained from the attached graph of Q(z) against z, tabulations of erfc, using the MATLAB 'erfc' function, or the following approximation which is valid for z > 3: Q(z) = 0.5 erfc( z / √2 ) ≈ (0.4 / z) exp (-z2 / 2) when z > 3. Example 4.1: A receiver receives +1 volt and 0 volt pulses with Gaussian noise of variance σ2 = 0.01 Estimate the error rate assuming a 0.5 volt threshold and an equal number of 1s and 0s. Solution: Probability of noise exceeding 0.5 v when "0" transmitted, times 0.5, plus probability of noise being less than -0.5 when "1" transmitted, times 0.5 is simply Q(0.5/0.1) = 3 x 10-7. The error rate = 1 in 0.33 x 107. Example 4.2: A receiver receives +1 volt and 0 volt pulses and has an error probability of 10-3. What is the variance of the received noise? Solution: Q(0.5/σ) = 10-3. From graph, 0.5/σ ≈ 3.2 Therefore, σ = 0.16, σ2=0.026. There is often little one can do to reduce the noise level, and hence, if we need to decrease the error probability, our only option is to increase the signalling pulse amplitude as it reaches the receiver. This can be done by increasing the gains of the regenerative repeaters, or, if this is not possible because it would increase the level of interference (cross-talk), the repeater spacing may be reduced. CS3282 Digital Communications 4.6 25/08/06/BMGC Example 4.3: A synchronous transmission system using cable with 30 dB attenuation per km and regenerative repeaters every 5 km is affected by additive Gaussian noise. A binary bipolar linecode is used with equally spaced detection thresholds at the sampling points. If the repeater gains cannot be increased because of the interference this would cause to other lines, how can the error probability be reduced from its current value of 10-5 to 10-7. (Use the graph of Q(z) against z). Solution: Each regenerative repeater has within it a receiver and a re-transmitter. At each sampling point, the receiver receives a +V or a –V corrupted by noise and possibly other distortion. It must decide whether a “0” or a “1” was intended and then the re-transmitter reconstructs a perfectly shaped symbol of appropriate amplitude and sends it on to the next repeater. Assume that, at a receiver, the symbols are expected to be +V and –V at the sampling points to signal “1” and 0 respectively. We could take the threshold to be 0 volts, and decide a +V symbol was intended if the voltage is greater than 0 and a –V symbol was intended if the received voltage is negative at the sampling point. To produce an error, the noise must exceed +V when –V (logic “0”) is sent, or be less than –V when +V (logic “1”) is sent. Hence the probability of an error is: (Prob of a “0”)x (Prob of noise sample being greater than +V) + (Prob of “1”) x (Prob of noise sample being less than –V) ) We assume an equal probability of “1”s and “0”s. Therefore probability of an error is: 0.5 x Q(V/σ) + 0.5 x Q(V/σ) where σ is the standard deviation (σ2 = variance) of the noise. The probability of a noise sample being less than –V is same as the probability of a noise sample being greater than V. Q(z) as plotted in these notes is the probability of a noise sample being greater than z for Gaussian noise of zero mean and standard deviation equal to one. Therefore Q(z/σ) is the probability of a noise sample being greater than z/σ for Gaussian noise of zero mean and standard deviation equal to one. But, most importantly, Q(z/σ) is also the probability of a Gaussian noise sample being greater than z when the noise has zero mean and standard deviation equal to σ. It is also the probability of a Gaussian noise sample being less than –z when the noise has zero mean and standard deviation equal to σ. If Q(V/σ) = 10-5, from the graph we find that (V/σ) = 4.25 Therefore σ = V/4.25. This is the standard deviation of the zero mean Gaussian noise that is causing the errors. With the same noise level, to produce an error probability of 10-7 rather than 10-5, we need the regenerative repeaters to receive higher voltage levels for the symbols. Assume they are raised from ±V to amplitude ±U at the sampling points. Then Q(U/σ) = 10-7 and from the graph, we find that U/σ = 5.2. This means that U = 5.2σ = (5.2/4.25)V = 1.22 V. 20log10(1.22) = 1.75dB. Therefore we need to arrange that the voltages as received at the receiver are raised by 1.75 dB. If we cannot raise the voltages transmitted by the regenerative repeaters because of cross-talk, we can only reduce the distance between the repeaters so that less attenuation occurs between one regenerative repeater and the next. The attenuation must be reduced from 5x30 = 150 dB to 148.25 dB. Distance between repeaters must be reduced to 148.25/30 = 4.94 km instead of 5 km. Wow! Problems 4.1. How would the bit-error rate be affected by sampling at the end or at the beginning of each rectangular pulse rather than in the centre. CS3282 Digital Communications 4.7 25/08/06/BMGC 4.2. Considering again Example 4.1, how would the bit-error rate be affected by sampling each rectangular pulse three times, rather than just once in the centre, and averaging over the three measurements? 4.3: A mobile phone receives a signal over "line of sight" radio (no reflections) from a base station and is affected by additive white Gaussian noise mainly introduced by the mobile phone itself. The received power from the base-station decreases with increasing distance according to an "inverse square power law", i.e. P(d) is proportional to 1/d2 where P(d) is the received power at distance d. Binary bipolar signalling is used with optimal detection threshold at the sampling points. If the distance from the base-station is currently 800 metres, how much nearer must we move towards it if the error probability is be reduced from its current value of 10-5 to 10-7. (Use the graph of Q(z) against z). 4.4. A signalling system has eight symbols which are rectangular pulses of amplitude -4, -3, -2, -1, 0, 1, 2, 3 volts. If the signalling rate is 10 kbaud, what is the maximum achievable bit-rate. 4.5. A system has a bit-rate of 32 kbits/second. How could you achieve this with a signalling rate of 1 kbaud. 4.6. What is the bandwidth efficiency of (i) AMI and (ii) 4B3T coding if binary or ternary pulses of duration T seconds require a bandwidth of 3T/4 Hz. CS3282 Digital Communications 4.8 Graph of Complementary Error function, Q ( z ) = 25/08/06/BMGC ∫ ∞ z t2 exp − 2π 2 1 dt