ECE 4371, Fall, 2014 Introduction to Telecommunication Engineering/Telecommunication LaboratoryOutlineDigital Communication SystemLine coding and decodingSignal element versus data elementData Rate Vs. Signal RateExampleSelf-synchronizationSlide 9Other propertiesLine coding schemesUnipolar NRZ schemePolar NRZ-L and NRZ-I schemesSlide 14RZ schemePolar biphase: Manchester and differential Manchester schemesBipolar schemes: AMI and pseudoternaryHDB3 (High Density Bipolar of order 3 code)HDB3Bipolar 8-Zero Substitution (B8ZS)Coded Mark Inversion (CMI)Multilevel: 2B1Q schememBnL schemes8B6T code table (partial)Multilevel: 4D-PAM5 schemeMultitransition: MLT-3 schemePSD of various line codesClock RecoverySlide 29Summary of line coding schemesHomework 3ECE 4371, Fall, 2014Introduction to Telecommunication Engineering/Telecommunication Laboratory Zhu HanDepartment of Electrical and Computer EngineeringClass 10Sep. 29th, 2014OutlineOutlineHW2–5.1.1, 5.1.4, 5.2.4, 5.2.7, 5.3.1, 5.4.2, 5.4.3, due 10/8 at class. Will grade before I leaveDigital Communication SystemLine Coding –NRZ and its variance–AMI and its variance–Multilevel–SpectrumDigital Communication SystemDigital Communication SystemSource: sequence of digitsMultiplexer: FDMA, TDMA, CDMA…Line Coder–Code chosen for use within a communications system for transmission purposes. –Baseband transmission–Twisted wire, cable, fiber communicationsRegenerative repeator–Detect incoming signals and regenerate new clean pulsesLine coding and decodingLine coding and decodingSignal element versus data elementData Rate Vs. Signal RateData rate: the number of data elements (bits) sent in 1s (bps). It’s also called the bit rateSignal rate: the number of signal elements sent in 1s (baud). It’s also called the pulse rate, the modulation rate, or the baud rate.We wish to:–increase the data rate (increase the speed of transmission)–decrease the signal rate (decrease the bandwidth requirement) –worst case, best case, and average case of r–N bit rate–c is a constant that depends on different line codes.–S = c * N / r baudExampleExample•A signal is carrying data in which one data element is encoded as one signal element ( r = 1). If the bit rate is 100 kbps, what is the average value of the baud rate if c is between 0 and 1?Solution–We assume that the average value of c is 1/2 . The baud rate is then•Although the actual bandwidth of a digital signal is infinite, the effective bandwidth is finite.•What is the relationship between baud rate, bit rate, and the required bandwidth?Self-synchronizationReceiver Setting the clock matching the sender’sEffect of lack of synchronizationExampleExample•In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 kbps? How many if the data rate is 1 Mbps?Solution–At 1 kbps, the receiver receives 1001 bps instead of 1000 bps.–At 1 Mbps, the receiver receives 1,001,000 bps instead of 1,000,000 bps.Other propertiesDC componentsTransmission bandwidthPower efficiencyError detection and correction capabilityFavorable power spectral densityAdequate timing contentTransparencyLine coding schemesUnipolar NRZ schemePolar NRZ-L and NRZ-I schemes•In NRZ-L, the level of the voltage determines the value of the bit. RS232.•In NRZ-I, the inversion or the lack of inversion determines the value of the bit. USB, Compact CD, and Fast-Ethernet.•NRZ-L and NRZ-I both have an average signal rate of N/2 Bd.NRZ-L and NRZ-I both have a DC component problem.ExampleExampleA system is using NRZ-I to transfer 1-Mbps data. What are the average signal rate and minimum bandwidth?Solution–The average signal rate is S = N/2 = 500 kbaud. The minimum bandwidth for this average baud rate is Bmin = S = 500 kHz.RZ schemeReturn to zero Self clockingPolar biphase: Manchester and differential Manchester schemesIn Manchester and differential Manchester encoding, the transition at the middle of the bit is used for synchronization.The minimum bandwidth of Manchester and differential Manchester is 2 times that of NRZ. 802.3 token bus and 802.4 EthernetBipolar schemes: AMI and pseudoternaryIn bipolar encoding, we use three levels: positive, zero, and negative.Pseudoternary: –1 represented by absence of line signal–0 represented by alternating positive and negativeDS1, E1HDB3 (High Density Bipolar of order 3 code) HDB3 (High Density Bipolar of order 3 code) Replacing series of four bits that are to equal to "0" with a code word "000V" or "B00V", where "V" is a pulse that violates the AMI law of alternate polarity and is rectangular or some other shape. The rules for using "000V" or "B00V" are as follows:–"B00V" is used when up to the previous pulse, the coded signal presents a DC component that is not null (the number of positive pulses is not compensated for by the number of negative pulses). –"000V" is used under the same conditions as above when up to the previous pulse the DC component is null. –The pulse "B" ("B" for balancing), which respects the AMI alternancy rule, has positive or negative polarity, ensuring that two successive V pulses will have different polarity. Used in E1HDB3 HDB3 The timing information is preserved by embedding it in the line signal even when long sequences of zeros are transmitted, which allows the clock to be recovered properly on reception. The DC component of a signal that is coded in HDB3 is null.Bipolar 8-Zero Substitution (B8ZS)Bipolar 8-Zero Substitution (B8ZS)Adds synchronization for long strings of 0sNorth American systemSame working principle as AMI except for eight consecutive 0sEvaluation–Adds synchronization without changing the DC balance–Error detection possibleUsed in T1/DS1AmplitudeTime00 0 0 0 0 0 01 0 1ViolationViolation10000000001  +000+-0-+01 in general 00000000000V(-V)0(-V)VCoded Mark Inversion (CMI)Coded Mark Inversion (CMI)Another modification from AMI: Binary 0 is represented by a half period of negative voltage followed by a half period of positive voltageAdvantages:–good clock recovery and no d.c. offset–simple circuitry for encoder and decoder  compared with HDB3Disadvantages: high bandwidthMultilevel: 2B1Q schemeIntegrated Services Digital Network ISDNmBnL schemes•In mBnL