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

2. 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

3. 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

4. Line coding and decoding
EE 541/451 Fall 2006

5. Signal element versus data element
EE 541/451 Fall 2006

6. 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

7. 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

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

9. 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

10. 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

11. Line coding schemes
EE 541/451 Fall 2006

12. Unipolar NRZ scheme
EE 541/451 Fall 2006

13. 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

14. 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

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

16. 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 token bus and Ethernet
EE 541/451 Fall 2006

17. 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

18. 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

19. 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

20. 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
 in general 000V(-V)0(-V)V
Amplitude
Time 1
Violation
EE 541/451 Fall 2006

21. 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

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

23. 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

24. 8B6T code table (partial)
EE 541/451 Fall 2006

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

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

27. PSD of various line codes
EE 541/451 Fall 2006

28. 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

29. Clock Recovery
Simple Circuit
PLL
EE 541/451 Fall 2006

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

31. Quiz
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

32. Homework
Draw line codes for
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

33. Questions?
EE 541/451 Fall 2006