ECE 4331, Fall, 2009

1. ECE 4331, Fall, 2009 Zhu Han Department of Electrical and Computer Engineering Class 11 Sep. 29th, 2009

2. scrambleddata   data PN sequence length2m – 1 = 26 – 1 = 63 Scrambling • Make the data more random by removing long strings of 1’s or 0’s. Improve timing • The simplest form of scrambling is to add a long pseudo-noise (PN) sequence to the data sequence and subtract it at the receiver (via modulo 2 addition); a PN sequence is produced by a Linear Shift Feedback Register (LSFR). • In receiver, descrambling using the same PN. • Secure: what is the PN and what is the initial

3. Scrambling • Exercise: 100000000000

4. Scrambling Example • Scrambler • Descrambler

5. Line coding and decoding • 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

6. Signal element versus data element

7. 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

8. 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?

9. Self-synchronization • Receiver Setting the clock matching the sender’s • Effect of lack of synchronization

10. 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.

11. Other properties • DC components • Transmission bandwidth • Power efficiency • Error detection and correction capability • Favorable power spectral density • Adequate timing content • Transparency

12. Line coding schemes

13. Unipolar NRZ scheme

14. 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.

15. 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.

16. RZ scheme • Return to zero • Self clocking

17. 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

18. 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

19. 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

20. 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.

21. 0 0 0 0 0 0 0 1 0 0 1 Amplitude Time Violation Violation 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 general00000000000V(-V)0(-V)V

22. 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

23. Multilevel: 2B1Q scheme • Integrated Services Digital Network ISDN

24. 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

25. 8B6T code table (partial)

26. Multilevel: 4D-PAM5 scheme

27. Multitransition: MLT-3 scheme

28. PSD of various line codes

29. 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.

30. Clock Recovery • Simple Circuit • PLL

31. Summary of line coding schemes Plus HDB3 and B8ZS

32. 0 1 0 1 1 1 1 0 0 Unipolar NRZ Polar NRZ NRZ-inverted (differential encoding) AMI encoding Manchester encoding Differential Manchester encoding Quiz • 101011100 • Unipolar NRZ, polar NRZ, NRZ-I, AMI, Manchester, Def. Manchester