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MOB sujet 4 Transmission techniques - DS CDMA
Master of Computer Science 1 - MOB Mobile Internet and Surrounding1/10 Subject 4MOB sujet 4Transmission techniques - DS CDMA1. CDMA ExercisesThe principle of the spread spectrum modulation consists in coding the information by the wayof a bandwidth signal larger than the band strictly necessary, with a goal of fighting the interferingsignals and the distortions tied with the propagation.The signal is coded at the beginning: a code is assigned with each user in order to allowdecoding in the arrival. The spread is ensured by a pseudo-random signal called spread code. Atthe reception, the signal is seen as the noise if the receiver does not have the code.The spread spectrum modulation is optimized to fight against the noise, by which it limit betterthe effects.1.0.1 Mathematical reminds- Correlation measure the manner whose two signals vary simultaneously. A strongcorrelation implies a great resemblance between the two compared signals.- Auto-correction quantifies the correlation of a signal with a version of itself shifted intime, according to this temporal shift.- Cross-correlation quantifies the correlation of two shifted signals the one in comparingwith the other in time, according to this temporal shift.- Orthogonality: two periodic signals from period T are known as orthogonal if theircross-correlation is null for a null temporal shift.CDMA modulation (Code Division Multiple Access) is a spread spectrum modulation. Itallows transmitting two signals simultaneously in the same band, their separation being realized atthe reception in getting profit of their orthogonality. Several mobile telephony systems (IS95,UMTS…) have chooses this technique. The CDMA is then also used in order to manage themulti-users aspect. Among the spread spectrum methods are:- Frequency Hopping - CDMA (FH-CDMA), seemed to be the most interesting techniqueat the beginning, but its difficult implementation took the interest back on spread spectrumby directed sequence;- Direct Sequence - CDMA (DS-CDMA), is used in particular in the 3G European UMTSsystem (Mobile Universal Telephone System).The principle of the DSCDMA is as follows. On each transmitting station, is allocated asingle code of m chips called "chip sequence" constituted from "-1" and from "+1". Totransmit a bit to "1", the station sends its sequence chip. To transmit a bit to “0”, the stationsends the opposite of its sequence chip. The duration Tc of transmission of a chip is m timelower than the duration Tb of a bit.Example: let us suppose that the chip sequence of A is "-1-11-1111-1 " (m = 8). To transmit Master of Computer Science 1 - MOB Mobile Internet and Surrounding2/10 Subject 4a binary "1", A sends "-1-11-1111-1"; to transmit a binary "0", A sends "11-11-1-1-11"Power spectral density (PSD) of a signal corresponds to the power distribution of the signalaccording to the frequency. For a code of forming, the PSD is proportional to the square ofthe Fourier transform of this forming. The forming is a form of wave given starting fromwhich one announces a "0" or a "1". If this is about one rectangular function of width T, theFourier transform of such a function is a cardinal sine whose first cancellation is of f = 1/T.One thus finds a PSD which has the form of the square of a cardinal sine:A signal DS-CDMA corresponds to a succession of rectangles of width Tc, the duration ofa chip. Its PSD thus has the form of the square of a cardinal sine in the first cancellation byf = 1/Tc.By opposition, a digital signal is constituted by a succession of rectangles of width Tb, thebit duration, has a PSD of the form of the square of a cardinal sine with the firstcancellation of f = 1/Tb.The chip sequence consists of m chips. Each bit of the digital signal in starting is replacedby the chip code so that Tb = mTc. And thus,TmTbC1The principal lobe of DS-CDMA signal is thus m time broader than the principal lobe ofthe starting signal. It is said that the signal was spread with a spread factor of m. Twosignals have the same total power, but power of the starting signal is more concentrated inthe band, where the appellation of “signal with narrow-band ".1.0.2 Exercises1. In your opinion, what are the advantages and disadvantages in the use of the CDMA for thecellular networks?Advantages: The spread spectrum modulation is optimized to fight against the noise, bywhich it limit better the effects.Disadvantages: In CDMA, any speaker can talk at any time; however each uses a differentlanguage. Each listener can only understand the language of their partner. As more andmore couples talk, the background noise (representing the noise floor) gets louder, butbecause of the difference in languages, conversations do not mix.2. If one wishes to send information of b bits/s with a key of m chips, what will be the flow inchips per second?The chip sequence is constituted of m chips. Each bit digital signal of departure is replacedby the code chip so that Tb = mTc. So,bbccmfTmTf 1.According to Shannon’s theorem, we deduce: Cc = mCb = mb chips/s.3. Which factor will the frequency band occupied by the signal be increased with?The data bit rate and noise power ratio. Master of Computer Science 1 - MOB Mobile Internet and Surrounding3/10 Subject 4The main lobe of DS-CDMA signal is m times larger than the main lobe of the signal ofdeparture. The frequency band occupied is increased by a factor m. They say that the signalwas unfurled with a spreading factor of m. Both signals have the same overall power, butthe signal of departure is more concentrated in the band, hence the name "narrow bandsignal”.4. Two keys are orthogonal if the standardized internal product of these both keys is null.Remind: internal product of [a, b] [a’, b’] = a a’ + b b’.Show that if two keys S and T are orthogonal, S and__Tare too.Suppose S = [s1, s2, ,sm], T = [t1, t2,…,tm]Thus__T = [t1, t2,…, tm]We have S T = (s1 t1+ s2 t2 + … + sm tm)/m = 0So S __T= (s1 t1 + s2 t2 + … + sm tm)/m = (s1 t1+ s2 t2 + … + sm tm)/m = 0e.g S and__T are orthogonal.5. One considers 4 transmitters having each one a key:A: (-1 -1 -1 +1 +1 -1 +1 +1)B: (-1 -1 +1 -1 +1 +1 +1 -1)C: (-1 +1 -1 +1 +1 +1 -1 -1)D: (-1 +1 -1 -1 -1 -1 +1 -1)a) Show that B, C and D are orthogonal at A.A B = [(-1) (-1) + (-1) (-1) + (-1) (+1) + (+1) (-1) + (+1) (+1) + (-1) (+1) + (+1) (+1) + (-1) (-1)]/8 = (1 + 1 – 1 – 1 + 1 – 1 + 1 – 1)/8 = 0b) A code has a good autocorrelation if its product intern with itself is high and its internalproduct with itself is shifted of 1 or several chips are weak. That means that the receiverwill be able to recognize easily the emitted signal. Show that the key of Barker(+1,-1,+1,+1, -1, +1,+1,+1, -1,-1,-1) is a good autocorrelation.We have: Bk Bk = 1The value is even more important that the key is long, but it takes to get complete theconcept of autocorrelation view the product's internal key with it even when it is offset by 1chip, 2 chips, .Suppose Bk’ is Bk shifted of 1 chip to the left, thus:Bk’ = (-1,+1,+1, -1, +1,+1,+1, -1,-1,-1,+1).We have: Bk Bk’ = [1 x (-1) + (-1) x 1 + 1 x 1 + 1 x (-1) + (-1) x 1 + 1 x 1 + 1 x 1 + 1 x(-1) + (-1) x (-1) + (-1) x (-1) + (-1) x 1]/11 = [(-1) + (-1) + 1 + (-1) + (-1) + 1 + 1 + (-1) + 1+ 1 + 1]/11 = –1/11The product of the key Barker (+1 -1 +1 +1 -1 +1 +1 +1 -1 -1 -1) with the same key butshifted by two chips to the left, Bk” (+1 +1 -1 +1 +1 +1 -1 -1 -1 +1 -1) is:Bk Bk” = (1 – 1 – 1 + 1 – 1 + 1 – 1 – 1 + 1 – 1 + 1)/11 = –1/11We deduce that this key and the same key shifted by one chip or two chips are nearlyorthogonal. The higher the value of domestic product, the lower the receiver can easily Master of Computer Science 1 - MOB Mobile Internet and Surrounding4/10 Subject 4interpret the result (signal codes with keys "pseudo-orthogonal").c) When a station wants to transmit a binary "1", it sends its key and when it wants totransmit a binary "0", it sends the opposite of this key. If only C wishes to send a bit of"1", what the receiver will receive and will make to interpret the answer?If only C wishes to send a bit of "1", C send chip sequence S = (-1 +1 -1 +1 +1 +1 -1 -1).The receiver will calculate, it had: S A = 0 and S B = 0 and S D = 0 and S C = 1, soreceiver know C was sent bit "1".d) If only C wishes to send a bit of "0", what the receiver will receive and will make tointerpret the answer?If C send a binary "0", C send chip sequence S = (+1 -1 +1 -1 -1 -1 +1 +1).The receiver will calculate, it had: S A = 0 and S B = 0 and S D = 0 and S C = –1, soreceiver know C was sent bit "0".e) The signals are added linearly, the transmitters are synchronous and the signals arereceived with the same power by the receiver. If one supposes that the stations A, Band C decide to send at the same time the bit "0", what will be the sequence read byreceiver?The stations A, B and C decide to send at the same time the bit "0", chip sequence S at thereceiver:S = (+3 +1 +1 -1 -3 -1 -1 +1)(Remember we have:A = (+1 +1 +1 -1 -1 +1 -1 -1) andB = (+1 +1 -1 +1 -1 -1 -1 +1) andC = (+1 -1 +1 -1 -1 -1 +1 +1) )f) If the receiver receives the sequence “-1 +1 -3 +1 -1 -3 +1 +1”, what does it deducefrom?Suppose S = (-1 +1 -3 +1 -1 -3 +1 +1)We have:S A = (-1 +1 -3 +1 -1 -3 +1 +1) (-1 -1 -1 +1 +1 -1 +1 +1) = (1 - 1 + 3 + 1 – 1 + 3 + 1 +1)/8 = 1, so A sent bit “1”.S B = (-1 +1 -3 +1 -1 -3 +1 +1) (-1 -1 +1 -1 +1 +1 +1 -1) = (1 – 1 – 3 – 1 – 1 – 3 + 1 –1)/8 = -1, so B sent bit “0”.S C = (-1 +1 -3 +1 -1 -3 +1 +1) (-1 +1 -1 +1 +1 +1 -1 -1) = (1 + 1 + 3 +1 – 1 – 3 – 1 – 1)/8= 0, so C has not sent anything.S D = (-1 +1 -3 +1 -1 -3 +1 +1) (-1 +1 -1 -1 -1 -1 +1 -1) = (1 + 1 + 3 – 1 + 1 + 3 + 1 – 1)/8= 1, so D sent bit “1”.6. Suppose that one wants to transmit a data flow of 56 kbit/s by using a spread spectrumtechnique.a) Find the band-width of the necessary channel when the signal-with-noise report is 0,1;0,01 and 0,001.Case 1: SNR = 0,1C = B log2(1 + SNR) B = C / log2(1 + SNR) = 56 103 / log2(1 + 0,1) = 407.262Hz 407,2KHz Master of Computer Science 1 - MOB Mobile Internet and Surrounding5/10 Subject 4Case 2: SNR = 0,01C = B log2(1 + SNR) B = C / log2(1 + SNR) = 56 103 / log2(1 + 0,01) = 3.901.000Hz 3,9 MHzCase 3: SNR = 0,001C = B log2(1 + SNR) B = C / log2(1 + SNR) = 56 103 / log2(1 + 0,001) = 38.835.647Hz 38,835 MHzb) In a ordinary system (without spread spectrum)In a regular system (without spread-spectrum), so that R = BShannon's formula C = Blog2 (1 + S / N) applied with C = B = R giveslog2 (1 + S / N) = 1. This implies that S / N = 1. Thus, spread-spectrum without a S / Nmuch greater is required.7. Prove that a receiving station can get the data sent by a specific sender if it multiplies theentire data on the channel by the sender’s chip code and then divides it by the number ofstations.Let us prove this for the first station, using our previous four-station example. We can say thatthe data on the channel: D = (d1 c1 + d2 c2 + d3 c3 + d4 c4)The receiver which wants to get the data sent by station 1 multiplies these data by c1. D c1 = (d1 c1 + d2 c2 + d3 c3 + d4 c4) c1= d1 c1 c1 + d2 c2 c1 + d3 c3 c1 + d4 c4 c1= d1 N + d2 0 + d3 0 + d4 0 d1 = (D c1 ) / N2. Practical - Spread spectrum techniquesThe goal of this practical is to understand the impact of spread spectrum techniques on theemitted signal and on the received signal by the recipient. You will be able with this practical:– to visualize and study impact of the code used and of the modulation type on the spectrumof a signal;– to observe the properties of two signals according to the used keys (DS-CDMA);– to visualize quality of reception of a signal 1, while varying the signal with noise report(SNR) as well as the ratio (C/I) of the power of the signal has to interpret (Carrier: signal 1)by compared with a second signal (Interferor: signal 2) which will come to disturb thereception of the first;– to trace the curves of Bit Error Rate (BER), according to coding for a SNR or a fixed C/I.For that you will use the simulation tool developed for this effect. Copy the script tme_cdma inyour folder:Carry out the script with the help of Octave, the digital calculation, visualization and programming Master of Computer Science 1 - MOB Mobile Internet and Surrounding6/10 Subject 4software similar to Matlab. Appear the following menu:Choose an option[1] signal key 1 (useful)[2] signal key 2 (interfering)[3] display the spectrum of the useful signal (red) and the interfering signal(green)[4] choose the modulation type[5] trace the autocorrelation of the key 1 and the cross-correlation betweentwo keys[6] trace the constellation diagram in reception of signal 1 (useful) for a SNRand a fixed C/I[7] calculate the bit error rate in reception for a fixed SNR[8] calculate the bit error rate in reception for a fixed C/I[9] quitMenu description[1]: allows visualizing or modifying the spread spectrum key of the "signal 1". The key by defect isthe key of Barker used in Wifi?[2]: allows visualizing or modifying the spread spectrum key of the "signal 2". The key by defectis a sequence of "1".[3]: trace the curves of the spectral density of signals 1 and 2 after modulation with and withoutfiltering. The two traced spectra, in red for signal 1 and in green for signal 2 have both thesame power. This spectrum is obtained for each signal by getting the same continuationrandom of 10 000 bits. Each signal is coded with its own key which can be visualized andmodified while typing [1] or [2] in the menu.[4]: allows choosing the modulation type which will be applied to signal 1. You have the choicebetween: PSK-2 (BPSK), QAM-4 (QPSK), QAM-16 and QAM-64.[5]: allows visualizing the curve of the autocorrelation of the signal 1 key and cross-correlationbetween the keys of signals 1 and 2.[6]: allows visualizing in the form of points in a constellation diagram, the value of the differentsymbols received by the receiver. One can thus see the amplitude and the phase of thesesymbols. The configuration of the receiver is adapted according to the modulation type ofsignal 1, of the CDMA keys of the two signals, and the values of ratios SNR and C/I.[7]: allows tracing curves of BER according to report C/I for a fixed SNR?[8]: allows tracing curves of BER according to report SNR for a fixed C/I?3. Properties and spectral of the emitted signalThe principle of the DS-CDMA is as follows. At each transmitting station is allocated asingle code of m chips called "sequence chip " constituted from "1 " and from "+1". From acode constituted from "0" and from "1" like the code of Barker, it is advisable to replace the"0" by "1" to obtain the sequence chip. To transmit a bit to "1", the station sends theopposite of its sequence chip. To transmit a bit to "0", the station sends the opposite of itssequence chip. The duration Tc of transmission of a chip is m time lower than the durationTb of a bit. Master of Computer Science 1 - MOB Mobile Internet andSurrounding7/10 Subject 41. What is the key value by defect of the signal 2? What does it mean?For the first part of the tests, we keep the key value of the signal 1 on"10110111000" and the key of the signal 2 on "1". There is thus nothing to modify.2. (a) Choose a modulation type by using the menu [4] and display then the spectraldensity of power of signal 1 and signal 2 by using the menu [3]. What do youobserve by comparing the form of the primary and secondary lobes of signals 1 and2 without filtered for the selected modulation? These observations are they stillvalid for the other modulations?(b) What is the power reduction between the primary and secondary lobe for thetwo curves?(c) What is the effect of the filtering on the frequential spectrum?(d) What is the interest of the filtering?4. Properties and correlationThe autocorrelation is studied in considering the profile of the autocorrelation functionobtained by shifting the digital sequence x[k] (of length N completed beforehand by zeros) of agradually increasing value n and by comparing the initial sequence and the shifted sequence.The formula used here to estimate the autocorrelation function is written:The correlation coefficient obtained will be larger as much as the digital sequence and itsshifted version are similar. Conversely, if the sequences are opposed, the correlation will benegative. A null autocorrelation indicates that there is not correlation between the initialsequence and the shifted sequence for the considered shift.The study of the autocorrelation allows us to evaluate the capacity of the receiver has tosynchronize with the flow of chips of the transmitter. In effect, the receiver makes a correlationbetween the received signal and the used key in the emission. A peak of correlation must beobserved when on the duration of a bit, the key used for encoder a bit and the key used by thereceiver are synchronous. In the other hand, the value of γx[n] must be near to zero for a shift notnull representing a desynchronization of the receiver. Thus, a good key must have suchproperties. It will be said that it has good properties of autocorrelation.1. A good autocorrelation must present a maximum for a null shift and must be near to zeroelsewhere. Which interest is presented in the case where the channel is the subject ofmulti-traffic?2. Observe the autocorrelation of the Barker key of the IEEE standard 802.11"10110111000" by using the menu [5]. It appears good to you?3. Which property appears important to you to choose the keys of two distinct transmitters?By basing you on that which was known as the autocorrelation properties, what appears Master of Computer Science 1 - MOB Mobile Internet andSurrounding8/10 Subject 4important to you for the cross-correlation?4. Give the mathematic expression of an estimator of the cross-correlation of twosequences x[k] and y[k].5. Are the keys of signals 1 and 2 orthogonal?6. Modify the keys of two transmitters and observe their autocorrelation and theircross-correlation.5. Reception and interpretation of the signal 1The menu [6] makes it possible to visualize the value of the amplitude and of the phase ofdifferent symbols received by the receiver. The simulator allows observing the impact of themodulation; of interferences linked to the additive noise (SNR) and to the disturbance caused byanother signal (signal 2) synchronous with the first.1. What does it mean SNR or ratio C/I positive, null and negative?2. Reposition the key value of the signal 1 to “10110111000" (Barker key) and that ofsignal 2 to “1”, i.e. to the default values. Using the menus [4] and [6], vary the SNR, theC/I and the modulation type and fill out table 1. What do you observe? what does itoccur when ratio C/I tends towards 0?Tab. 1 – Synthesis of observations on the constellation diagrams.Modulation SNR C/I Quality of the reception of signal(Good/ Medium /Bad)BPSK 100 15QPSK 100 1516QAM 100 1564QAM 100 15BPSK 15 100QPSK 15 10016QAM 15 10064QAM 15 100BPSK 15 0QPSK 15 016QAM 15 064QAM 15 0BPSK 15 3QPSK 15 316QAM 15 364QAM 15 33. Position the key value of the signal 1 to "10110111000" (Barker key) and that of signal 2 Master of Computer Science 1 - MOB Mobile Internet andSurrounding9/10 Subject 4to "01101110001". Signal 1 and signal 2 are then encodes by orthogonal pseudo keys.Tab. 2 – Synthesis of observations on the constellation diagrams.Modulation SNR C/I Quality of the reception of signal(Good/ Medium /Bad)BPSK 15 3QPSK 15 316QAM 15 364QAM 15 3The signal 2 simulates here a signal interfering broad band. Vary the SNR, the C/I andthe modulation type and fill out table 2. What do you observe?4. In basing itself on the data of the exercise 5 of part 1.3.2 of the subject of precedepractical, one associates now the key "00011011" to signal 1 and the key B "00101110"to signal 2. Fill out table 3. What does one observe then? Explain.Tab. 3 – Synthesis of observations on the constellation diagrams5. What does it occur if one associates the Barker key to the two signals?a) Fill out the table 4.Tab. 4 – Synthesis of observations on the constellation diagrams for codes of Barker.Modulation SNR C/I Quality of the reception of signal(Good/ Medium /Bad)BPSK 15 3BPSK 15 0BPSK 15 3b) What does one observe then? Explain.c) Why would the results on the constellation diagrams have the same oneswithout use of CDMA codes (key of the signals is "1")?d) What would have been different without use of CDMA codes (key of signals is"1")?e) If one preserved the Barker key for the two signals, one could have had a resultdifferent from which is showed here by the simulator? Explain.Modulation SNR C/I Quality of the reception of signal(Good/ Medium /Bad)QPSK 15 016QAM 15 064QAM 15 0QPSK 15 316QAM 15 364QAM 15 3 Master of Computer Science 1 - MOB Mobile Internet andSurrounding10/10 Subject 4f) How can you highlight with help of the simulator?6. Curve of BERTrace a curve of BER in defining the type of modulation, the fixed size (SNR or C/I), theintervals of variations interesting for the variable ratio. The calculation of the curve takingseveral minutes, it is advised to limit the number of points to be displayed . Does one obtain thesame thing if the modulation is changed, the fixed size? [...]... c3 + d4 c4) The receiver which wants to get the data sent by station 1 multiplies these data by c1. D c1 = (d1 c1 + d2 c2 + d3 c3 + d4 c4) c1 = d1 c1 c1 + d2 c2 c1 + d3 c3 c1 + d4 c4 c1 = d1 N + d2 0 + d3 0 + d4 0 d1 = (D c1 ) / N 2. Practical - Spread spectrum techniques The goal of this practical is to understand the impact of spread spectrum techniques. .. Science 1 - MOB Mobile Internet and Surrounding 5/10 Subject 4 Case 2: SNR = 0,01 C = B log 2 (1 + SNR) B = C / log 2 (1 + SNR) = 56 10 3 / log 2 (1 + 0,01) = 3.901.000Hz 3,9 MHz Case 3: SNR = 0,001 C = B log 2 (1 + SNR) B = C / log 2 (1 + SNR) = 56 10 3 / log 2 (1 + 0,001) = 38.835. 647 Hz 38,835 MHz b) In a ordinary system (without spread spectrum) In a regular system (without spread-spectrum),... able with this practical: – to visualize and study impact of the code used and of the modulation type on the spectrum of a signal; – to observe the properties of two signals according to the used keys (DS- CDMA) ; – to visualize quality of reception of a signal 1, while varying the signal with noise report (SNR) as well as the ratio (C/I) of the power of the signal has to interpret (Carrier: signal 1) by... 1. This implies that S / N = 1. Thus, spread-spectrum without a S / N much greater is required. 7. Prove that a receiving station can get the data sent by a specific sender if it multiplies the entire data on the channel by the sender’s chip code and then divides it by the number of stations. Let us prove this for the first station, using our previous four-station example. We can say that the data on... the reception of the first; – to trace the curves of Bit Error Rate (BER), according to coding for a SNR or a fixed C/I. For that you will use the simulation tool developed for this effect. Copy the script tme _cdma in your folder: Carry out the script with the help of Octave, the digital calculation, visualization and programming . considers 4 transmitters having each one a key:A: (-1 -1 -1 +1 +1 -1 +1 +1)B: (-1 -1 +1 -1 +1 +1 +1 -1 )C: (-1 +1 -1 +1 +1 +1 -1 -1 )D: (-1 +1 -1 -1 -1 -1 +1 -1 )a). thereceiver:S = (+3 +1 +1 -1 -3 -1 -1 +1)(Remember we have:A = (+1 +1 +1 -1 -1 +1 -1 -1 ) andB = (+1 +1 -1 +1 -1 -1 -1 +1) andC = (+1 -1 +1 -1 -1 -1 +1 +1) )f) If