ECE 4371, Fall, 2015

Introduction to Telecommunication Engineering/Telecommunication Laboratory

Zhu Han

Department of Electrical and Computer Engineering

Class 10

Sep. 24th, 2015

ECE 4371 Fall 2008

OutlineHW25.1.1, 5.1.4, 5.2.4, 5.2.7, 5.3.1, 5.4.2, 5.4.3, due 10/13 at class. Digital Communication SystemLine Coding NRZ and its varianceAMI and its varianceMultilevelSpectrum

ECE 4371 Fall 2008

Digital Communication SystemSource: sequence of digitsMultiplexer: FDMA, TDMA, CDMALine CoderCode chosen for use within a communications system for transmission purposes. Baseband transmissionTwisted wire, cable, fiber communicationsRegenerative repeatorDetect incoming signals and regenerate new clean pulses

ECE 4371 Fall 2008

Line coding and decoding

ECE 4371 Fall 2008

Signal element versus data element

ECE 4371 Fall 2008

Data Rate Vs. Signal RateData rate: the number of data elements (bits) sent in 1s (bps). Its also called the bit rateSignal rate: the number of signal elements sent in 1s (baud). Its 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 rN bit ratec is a constant that depends on different line codes.S = c * N / r baud

ECE 4371 Fall 2008

ExampleA 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?SolutionWe 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?

ECE 4371 Fall 2008

Self-synchronizationReceiver Setting the clock matching the sendersEffect of lack of synchronization

ECE 4371 Fall 2008

ExampleIn 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?SolutionAt 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.

ECE 4371 Fall 2008

Other propertiesDC componentsTransmission bandwidthPower efficiencyError detection and correction capabilityFavorable power spectral densityAdequate timing contentTransparency

ECE 4371 Fall 2008

Line coding schemes

ECE 4371 Fall 2008

Unipolar NRZ scheme

ECE 4371 Fall 2008

Polar NRZ-L and NRZ-I schemesIn 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.

ECE 4371 Fall 2008

ExampleA system is using NRZ-I to transfer 1-Mbps data. What are the average signal rate and minimum bandwidth?SolutionThe average signal rate is S = N/2 = 500 kbaud. The minimum bandwidth for this average baud rate is Bmin = S = 500 kHz.

ECE 4371 Fall 2008

RZ schemeReturn to zero Self clocking

ECE 4371 Fall 2008

Polar biphase: Manchester and differential Manchester schemesIn 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 Ethernet

ECE 4371 Fall 2008

Bipolar schemes: AMI and pseudoternaryIn bipolar encoding, we use three levels: positive, zero, and negative.Pseudoternary: 1 represented by absence of line signal0 represented by alternating positive and negativeDS1, E1

ECE 4371 Fall 2008

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 E1

ECE 4371 Fall 2008

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.

ECE 4371 Fall 2008

Bipolar 8-Zero Substitution (B8ZS)Adds synchronization for long strings of 0sNorth American systemSame working principle as AMI except for eight consecutive 0s

EvaluationAdds synchronization without changing the DC balanceError detection possibleUsed in T1/DS110000000001 +000+-0-+01 in general 00000000000V(-V)0(-V)V

ECE 4371 Fall 2008

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 voltageAdvantages:good clock recovery and no d.c. offsetsimple circuitry for encoder and decoder compared with HDB3Disadvantages: high bandwidth

ECE 4371 Fall 2008

Multilevel: 2B1Q schemeIntegrated Services Digital Network ISDN

ECE 4371 Fall 2008

mBnL schemesIn mBnL schemes, a pattern of m data elements is encoded as a pattern of n signal elements in which 2^m L^n.Multilevel: 8B6T scheme, T4

ECE 4371 Fall 2008

8B6T code table (partial)

ECE 4371 Fall 2008

Multilevel: 4D-PAM5 scheme

ECE 4371 Fall 2008

Multitransition: MLT-3 scheme

ECE 4371 Fall 2008

PSD of various line codesDetails in next class

ECE 4371 Fall 2008

Clock RecoveryA timing reference signal can be extracted from the received signal by differentiation and full-wave rectification provided that the signal carries sufficient transitions.This timing reference signal is then used to fine tune the frequency and phase of a local oscillator. The receiver clock is then derived (e.g. add a phase shift) from this local oscillator.

ECE 4371 Fall 2008

Clock RecoverySimple Circuit

PLL

ECE 4371 Fall 2008

Summary of line coding schemesPlus HDB3 and B8ZS

ECE 4371 Fall 2008

Homework 3Draw line codes for 1010 0000 0000 1011 0000 1011 0000NRZNRZ-L, NRZ-IAMI, Pseudoternary, HDB3, B8ZS, CMIManchester and differential Manchester schemes2B1Q, MLT-3If the bit rate is 1Kbps, what are the baud rates for the above line codes.Matlab plot of spectrumDue 10/20/15

ECE 4371 Fall 2008

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