Transcript Digital Transmission - Computing Science

Part II. Physical Layer and
Media
Chapter 4. Digital Transmission
COMP 3270 Computer Networks
Computing Science
Thompson Rivers University
Chapter Contents
• Digital-to-digital conversion
• Analog-to-digital conversion
• Transmission mode
Learning Objectives
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Determine the baud rate from a given bit rate and the number of signal
elements for sending one bit, assuming line coding.
Identify the three problems caused by a long string of 0s and1s.
Explain the basic idea to solve the above three problems.
Identify advantages and disadvantages of Manchester and differential
Manchester line coding schemes, comparing to NRZ coding scheme.
Describe the motivation of block coding.
Describe the basic idea of PCM that converts an alalog signat to digital
data.
Use of the Nyquist Theorem to decide the minimum sampling rate.
Distinguish parallel and serial transmission modes.
1. DIGITAL-TO-DIGITAL CONVERSION
In this section, we see how we can represent digital data by
using digital signals. The conversion involves two techniques:
line coding, and block coding. Line coding is always needed;
block coding may or may not be needed.
Topics discussed in this section:
• Line Coding
• Line Coding Schemes
• Block Coding
Line Coding
Line coding and decoding
The process of converting binary data, a sequence of bits,
to a digital signal.
Signal element (or signal level) versus data element
Data rate
(or Bit rate),
Baud rate)
SN
1
r
S  c N 
1
r
(Text book)
Example:
A signal is carrying data in which one bit is encoded as one signal
element (r = 1). If the bit rate is 100 kbps, what is the baud rate?
Solution:
1
1
S  N   100 Kbps 
 100 Kbaud / sec
r
1 bits / baud
Consecutive bit 0s or 1s may cause serious problems in
digital transmission.
 Baseline Wandering
 DC Component
 Lack of Self-Synchronization
We need good encoding schemes.
Baseline Wandering (at the receiver)
• The average of the received signal power – baseline
• Incoming signal power is evaluated against this baseline to
determine the value of the data element. A long string of 0s or 1s
can cause a drift in the baseline.
• A good line coding scheme needs to prevent baseline wandering.
One example:
DC Components
• When the voltage level in a digital signal is constant for a while,
the spectrum creates very low frequencies around zero, called DC
(Direct Current) components.
• Cannot pass the band pass channels, e.g., a telephone line cannot
pass frequencies below 200 Hz.
Self Synchronization
• Correct bit intervals at the sender and the receiver are critical.
• A self-synchronizing digital signal includes timing information in
the data being transmitted, which is transitions of the signal.
Any good idea how to sole these three problems?
Line Coding Schemes
Unipolar NRZ scheme
☺ Are all the three problems solved?
Polar NRZ-L and NRZ-I schemes
Minimum required bandwidth B 
☺ Are all the three problems solved?
1
N
2N
In NRZ-L the level of the voltage
determines the value of the bit.
In NRZ-I the inversion
or the lack of inversion
determines the value of the bit.
NRZ-L and NRZ-I both have the baud
rate of N Bd.
☺ Are all the three problems solved?
NRZ-L and NRZ-I both have a DC
component problem.
☺ Baseline Wandering?
☺ DC Component?
☺ Lack of Self-Synchronization?
Example:
A system is using NRZ-I to transfer 10-Mbps data. What are the baud
rate and minimum bandwidth?
Solution:
The average baud rate is S = N = 10-Mbaud/sec. The minimum
bandwidth for this baud rate is Bmin = S/2 = 5- MHz.
N max
1
1
 2 B log2 L  2 B  B  N  S
2
2
Polar RZ (Return-to-Zero) scheme
2N
Minimum required bandwidth B  N
☺ Baseline Wandering?
☺ DC Component?
☺ Lack of Self-Synchronization?
☺ Bandwidth consumption?
Polar biphase: Manchester and differential Manchester
schemes
Ethernet
Minimum required bandwidth B  N
2N
Token ring
The minimum bandwidth of Manchester
and differential Manchester is 2 times
that of NRZ.
☺ Why is it important?
Bipolar schemes: AMI and pseudoternary
N
Summary of line coding schemes
☺ How to combine?
Block Coding
Block coding is normally referred to as
mB/nB coding;
it replaces each m-bit group with an
n-bit group,
so that there are not long 0s or 1s.
Block coding concept
Using block coding 4B/5B with NRZ-I line coding scheme
100 Base T: 4B/5B + NRZ-L
FDDI: 4B/5B + NRZ-I
4B/5B mapping codes
The length of the longest 1s is ???.
Example:
We need to send data at a 1-Mbps rate. What is the minimum
required bandwidth, using a combination of 4B/5B and NRZ-I, or
Manchester coding?
Solution:
First 4B/5B block coding increases the bit rate to 1.25 Mbps. The
minimum bandwidth using NRZ-I is N/2 or 625 KHz.
The Manchester scheme needs a minimum bandwidth of 1 MHz.
The first choice needs a lower bandwidth, but has little DC
component problem; the second choice needs a higher bandwidth,
but does not have a DC component problem.
Intermediate Summary
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Bit rate vs. baud rate
Four problems caused by consecutive 0's and 1's
The problem of Manchester scheme
How block coding works
2. ANALOG-TO-DIGITAL CONVERSION
In this section we describe one technique, pulse code modulation.
Topics discussed in this section:
• Pulse Code Modulation (PCM)
Pulse Code Modulation (PCM)
☺ Any good idea to load an analog signal over
a digital signal? (How to send analog
information using a digital signal?)
Components of PCM encoder – conversion of an analog
signal to diginal data
Three different sampling methods for PCM
According to the Nyquist theorem, the
sampling rate must be
at least 2 times the highest frequency
contained in the signal.
Example:
Telephone companies digitize voice by assuming a maximum
frequency of 4000 Hz. The sampling rate therefore is 8000 samples
per second.
If a byte is used to express a sample (i.e., strength), then we need the
data rate:
64 Kbps = 8 bits / sample  8000 samples / sec for all the samples
for a second.
Audio CD: (20Hz – 20KHz) -> 44.1 K sampling rate; 16 bits / sample
-> 705.6 Kbps
Example:
A complex low-pass signal has a bandwidth of 200 KHz. What is the
minimum sampling rate for this signal?
Solution:
The bandwidth of a low-pass signal is between 0 and f, where f is the
maximum frequency in the signal. Therefore, we can sample this
signal at 2 times the highest frequency (200 KHz). The sampling rate
is therefore 400,000 samples per second.
Example:
A complex band-pass signal has a bandwidth of 200 kHz. What is the
minimum sampling rate for this signal?
Solution:
We cannot find the minimum sampling rate in this case because we
do not know where the bandwidth starts or ends. We do not know the
maximum frequency in the signal.
Quantization and encoding of a sampled signal
Quantization Error
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Some sort of noise, called sampling noise, is induced.
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Reproduction of the original signal
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Example:
We want to digitize the human voice. What is the bit rate, assuming 8
bits per sample?
Solution:
The human voice normally contains frequencies from 0 to 4000 Hz.
So the sampling rate and bit rate are calculated as follows:
Components of a PCM decoder
Intermediate Summary
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The idea of PCM
3 components of PCM
Nyquist theorem
Sampling noise
3. TRANSMISSION MODES
The transmission of binary data across a link can be accomplished
in either parallel or serial mode. In parallel mode, multiple bits are
sent with each clock tick. In serial mode, 1 bit is sent with each
clock tick. While there is only one way to send parallel data, there
are three subclasses of serial transmission: asynchronous, and
synchronous.
Topics discussed in this section:
• Parallel Transmission
• Serial Transmission
Data transmission and modes
Parallel Transmission
Serial Transmission
In asynchronous transmission, we send
1 start bit (0) at the beginning and 1 or
more stop bits (1s) at the end of each
byte. There may be a gap between
each byte.
Asynchronous here means
“asynchronous at the byte level,”
but the bits are still synchronized;
their durations are the same.
Asynchronous transmission
In synchronous transmission, we send
bits one after another without start or
stop bits or gaps. It is the responsibility
of the receiver to group the bits.
Synchronous transmission
USB, Ethernet, …
But asynchronous
☺ Advantage?
☺ How to take out frames from the bit stream?