ECE 4371, Fall, 2015
Introduction to Telecommunication
Engineering/Telecommunication Laboratory
Zhu Han
Department of Electrical and Computer Engineering
Class 10
Sep. 24th, 2015
Outline

HW2
– 5.1.1, 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 System

Line Coding
– NRZ and its variance
– AMI and its variance
– Multilevel
– Spectrum
Digital 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 pulses
Line coding and decoding
Signal element versus data element
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
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?
Self-synchronization

Receiver Setting the clock matching the sender’s

Effect of lack of synchronization
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.
Other properties

DC components

Transmission bandwidth

Power efficiency

Error detection and correction capability

Favorable power spectral density

Adequate timing content

Transparency
Line coding schemes
Unipolar NRZ scheme
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.
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.
RZ scheme

Return to zero

Self clocking


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
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
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
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)

Adds synchronization for long strings of 0s
North American system

Same working principle as AMI except for eight consecutive 0s

10000000001  +000+-0-+01
1
0
0
0
in general
0
0
00000000000V(-V)0(-V)V
0
0
0
0
1
Amplitude
Time
Violation

Violation
Evaluation
– Adds synchronization without changing the DC balance
– Error detection possible

Used in T1/DS1
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
Multilevel: 2B1Q scheme

Integrated
Services
Digital
Network
ISDN
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
8B6T code table (partial)
Multilevel: 4D-PAM5 scheme
Multitransition: MLT-3 scheme
PSD of various line codes

Details in next class
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.
Clock Recovery

Simple Circuit

PLL
Summary of line coding schemes
Plus HDB3 and B8ZS
Homework 3

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 10/20/15