EE 551/451, Fall, 2006 Communication Systems                                                             Zhu Han Department of Electrical and Computer Engineering Class 17 Oct. 19th, 2006

Bandpass sampling theory 2B sampling rate for signal from 2B to 3B B 2B 3B -3B -2B -B X(f) X(f-fs) 4B 5B X(f+fs) -5B -4B Xs EE 541/451 Fall 2006

Digital Communication System 7.1 and 7.2 (part) 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 pulses EE 541/451 Fall 2006

Line coding and decoding EE 541/451 Fall 2006

Signal element versus data element EE 541/451 Fall 2006

Data 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 baud EE 541/451 Fall 2006

Example 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? EE 541/451 Fall 2006

Self-synchronization Receiver Setting the clock matching the sender’s Effect of lack of synchronization EE 541/451 Fall 2006

Example 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. EE 541/451 Fall 2006

Other properties DC components Transmission bandwidth Power efficiency Error detection and correction capability Favorable power spectral density Adequate timing content Transparency EE 541/451 Fall 2006

Line coding schemes EE 541/451 Fall 2006

Unipolar NRZ scheme EE 541/451 Fall 2006

Polar 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. EE 541/451 Fall 2006

Example 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. EE 541/451 Fall 2006

RZ scheme Return to zero Self clocking EE 541/451 Fall 2006

Polar 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 Ethernet EE 541/451 Fall 2006

Bipolar 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, E1 EE 541/451 Fall 2006

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 EE 541/451 Fall 2006

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. EE 541/451 Fall 2006

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/DS1 10000000001  +000+-0-+01 in general 00000000000V(-V)0(-V)V Amplitude Time 1 Violation EE 541/451 Fall 2006

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 bandwidth EE 541/451 Fall 2006

Multilevel: 2B1Q scheme Integrated Services Digital Network ISDN EE 541/451 Fall 2006

mBnL schemes In 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 EE 541/451 Fall 2006

8B6T code table (partial) EE 541/451 Fall 2006

Multilevel: 4D-PAM5 scheme EE 541/451 Fall 2006

Multitransition: MLT-3 scheme EE 541/451 Fall 2006

PSD of various line codes EE 541/451 Fall 2006

Clock Recovery A 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. EE 541/451 Fall 2006

Clock Recovery Simple Circuit PLL EE 541/451 Fall 2006

Summary of line coding schemes Plus HDB3 and B8ZS EE 541/451 Fall 2006

Quiz 101011100 Unipolar NRZ, polar NRZ, NRZ-I, AMI, Manchester, Def. Manchester NRZ-inverted (differential encoding) 1 Unipolar NRZ AMI encoding Manchester Differential Polar NRZ EE 541/451 Fall 2006

Homework Draw line codes for 1010 0000 0000 1011 0000 1011 0000 NRZ NRZ-L, NRZ-I AMI, Pseudoternary, HDB3, B8ZS, CMI Manchester and differential Manchester schemes 2B1Q, MLT-3 If the bit rate is 1Kbps, what are the baud rates for the above line codes. Matlab plot of spectrum Due 10/26/06 EE 541/451 Fall 2006

Questions? EE 541/451 Fall 2006