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8/3/2019 MSIS 4523 Ch4.Digital Transmission
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Data Communications SystemsCh 4: Digital Transmission
JinKyu Lee, Ph.D.
Include the course code (MSIS4523) in every email subject!!
Layer1
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A Categorization of Transmission Techniques
Digital
Transmission
Analog
Transmission
Data Transmission
Line
Coding
Digital Source
Conversion
Analog Source
Conversion
Block
CodingPCM DM
ASK
Digital Source
Conversion
Analog Source
Conversion
FSK
AM FM
PSKQAM
PM
Ch.4
Ch.5
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Topics
Digital-to-Digital Conversion Transmission Modes Parallel vs. Serial
Sync vs. Async
Line coding Schemes
Block coding Schemes
Analog-to-Digital Conversion Sampling, Quantization, Encoding
PCM, Delta Modulation
Nyquists theorem
Baud Rate (Signal Rate) vs. Bit Rate
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Data and Signals, Conversion Techniques
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Data and Signal Conversions Examples
Let us transmit the message Sam, what time is themeeting with accounting? Hannah.This message first leaves Hannahs workstation and
travels across a local area network
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Converting Digital Data into Digital SignalsDigital Encoding Compared
NRZ-L
NRZI
Manchester
DifferentialManchester
Bipolar-AMI
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Transmission Modes - Parallel Transmission
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Transmission Modes - Serial Transmission
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Transmission Modes - Synchronous
In synchronous transmission, bits are sent one after
another without start/stop bits or gaps. Doesnt mean
signal never stop. Bits are sent by block (hundreds of
bytes) by block (e.g., Ethernet frames)
Receiver must pickup individual bits from the stream
of bits.
In synchronous transmission, bits are sent one after
another without start/stop bits or gaps. Doesnt mean
signal never stop. Bits are sent by block (hundreds of
bytes) by block (e.g., Ethernet frames)Receiver must pickup individual bits from the stream
of bits.
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Line Coding Schemes
Direct data-signal conversion
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Unipolar Encoding
Unipolar encoding uses only one voltage levelUnipolar encoding uses only one voltage level
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Polar Encoding Schemes
Non return to zero-level (NRZ-L)
Non return to zero inverted (NRZI)
Bipolar -AMI
Pseudoternary
Manchester
Differential Manchester
B8ZS
HDB3 Polar encoding uses twovoltage levels (positive
and negative)
Polar encoding uses twovoltage levels (positive
and negative)
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Return to Zero (RZ) Encoding
A good encoded digital signal must contain a
provision for synchronization
A good encoded digital signal must contain a
provision for synchronization
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NRZ-L and NRZ-I encoding
In NRZ-L the level of the signal is dependent upon the state of
the bit. In NRZ-I the signal is inverted if a 1 is encountered
In NRZ-L the level of the signal is dependent upon the state of
the bit. In NRZ-I the signal is inverted if a 1 is encountered
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The Synchronization Problem
How does the receiver know when the bit period
begins and ends?
Digital communications requires that the receiver
know when to sample the signal
This requires a mechanism to synchronize thetransmitter and receiver
Solutions: Bi-phase signaling
Substitution for line coding
Block coding
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Biphase Alternatives to NRZ
Require at least one transition per bit time, and may even
have two
Advantages Synchronization due to predictable transitions
Error detection based on absence of a transition
However, modulation rate is greater, so bandwidth
requirements are higher
Known as Manchestercoding and Differential
Manchestercoding
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Manchester Encoding
In Manchester encoding, the transition at the middle
of the bit is used for both synchronization and bitrepresentation.
IEEE 802.3 Ethernet uses Manchester encoding
In Manchester encoding, the transition at the middle
of the bit is used for both synchronization and bitrepresentation.
IEEE 802.3 Ethernet uses Manchester encoding
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Differential Manchester Encoding
In differential Manchester encoding, the transition atthe middle of the bit is used only for synchronization.
The bit representation is defined by the inversion or
non-inversion at the beginning of the bit
In differential Manchester encoding, the transition atthe middle of the bit is used only for synchronization.
The bit representation is defined by the inversion or
non-inversion at the beginning of the bit
Differential Encoding Data represented by changes rather than levels Receivers can more reliably detect transitions rather
than specific levels
In complex transmission layouts it is also easy to
lose sense of polarity
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Substitution Techniques
Use substitution to replace sequences that would produce
constant voltage Goal is to provide enough transitions to allow
synchronization Substitution must be recognized by receiver and replaced with
original
Ideally the same length as original
Advantages: No DC component
Eliminates long sequences of zero level line signal
No reduction in data rate Can provide error detection capability
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Bipolar With 8 Zeros Substitution (B8ZS)
Based on bipolar-AMI In Bipolar-AMI, signal patterns of +0+ or -0- should never happen
under normal condition.
If octet of all zeros and last voltage pulse preceding was
positive (+ 00000000 ): encode as + 000+-0-+
If octet of all zeros and last voltage pulse preceding wasnegative (- 00000000 ): encode as - 000-+0+-
Causes two violations of AMI code
Unlikely to occur as a result of noise
Receiver detects and interprets as octet of all zeros
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Indirect data-signal conversion
Block Coding Schemes
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Substitution in Block Coding
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Converts four bits of data into five-bit quantitiesNo five-bit code has more than 2 consecutive zeroes
Five-bit code is then transmitted using an NRZ-I encoded signal
Converting Digital Data into Digital Signals -4B/5B Digital Encoding Scheme
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4B/5B Encoding Table
Data Code
Q (Quiet) 00000
I (Idle) 11111
H (Halt) 00100
J (start delimiter) 11000
K (start delimiter) 10001
T (end delimiter) 01101
S (Set) 11001
R (Reset) 00111
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Analog-to-Digital Conversion
Encoding digital data is easysimply represent 0s and 1s
with different voltage levels on the carrier wave
Not so easy with continuous data like music or voice
Have to convert the continues data (analog) to discrete
data (digital)
Do this with Pulse Code Modulation (PCM)
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Sampling
Analog voice data must be translated into a series ofbinary digits before they can be transmitted across digitaltransmission facilities
To convert analog data into a digital signal, there are twobasic techniques: Pulse code modulation (PCM) (used by telephone systems)
With PCM, the amplitude of the sound wave is sampled atregular intervals and translated into a binary number
Delta modulation
The difference between the original analog signal and the
translated digital signal is called quantizing error
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Quantization Process
Quantization process assigns a bit value
corresponding to the amplitude of each sampled
sound wave.
Similar concept to pixelization 8-bit range allows for 256 possible sample levels
If 128 levels, then each sample is 7 bits (2 ^ 7 =
128)
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Pulse Code Modulation (PCM)
64K PCM uses a sampling rate of 8000 samples
per second
Each sample is an 8 bit sample resulting in a
digital rate of 64,000 bps (8 x 8000)
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Analog waveform is sampled at specific intervalsSnapshots are converted to binary values
Pulse Code Modulation (PCM) - Sampling
Pulse
Amplitude
Modulation
(PAM) value
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Pulse Code Modulation (PCM) - Pulse Amplitude Modulation (PAM)
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Because the human voice has a fairly narrow
bandwidth
Telephone systems digitize voice into either 128 levels or
256 levels
Called quantization levels
More bits means greater detail, fewer bits means lessdetail
Pulse Code Modulation (PCM) - Quantization
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+1 +1 +1
+0 +0+0
+0
+0+0
+0+0
+0
-0
-1
-0
Pulse Code Modulation (PCM) - Quantizing of PAM Signals
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Converting Analog data to PCM Digital Signal The whole Picture!
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An analog waveform is tracked using a binary 1 torepresent a rise in voltage and a 0 to represent a drop
Delta Modulation
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The more snapshots taken in the same amount of time,or the more quantization levels, the better theresolution
Sampling Frequency, Quantization Level,
and the Quality of Reconstructed Data
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Binary values are later converted to an analog signalWaveform similar to original results
Sampling Frequency, Quantization Level,
and the Quality of Reconstructed Data (cont.)
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Nyquists Theorem Sampling Rate
According to Nyquists theorem, the sampling rate must be at
least 2 times the highest frequency
According to Nyquists theorem, the sampling rate must be at
least 2 times the highest frequency
=
==
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What sampling rate is needed for a signal with a
bandwidth of 10,000 Hz (1000 to 11,000 Hz)?
The sampling rate must be twice the highest frequency in
the signal:
Sampling rate = 2 x (11,000) = 22,000 samples/s
SolutionSolution
Example 15
Nyquists theorem: If a signal is sampled at regular intervalsof time and at a rate higher than twice the significant signal
frequency, the samples contain all the information of the
original signal
Nyquists theorem: If a signal is sampled at regular intervalsof time and at a rate higher than twice the significant signal
frequency, the samples contain all the information of the
original signal
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Baud Rate: The number of times a signal changes
value per second = Signal Rate.Manchester schemes: Baud rate is twice the bps
Two signal changes per bit
Baud Rate (baud/s) vs. Bit Rate (bps)
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An analog signal carries 4 bits in each signal unit. If 1000
signal units are sent per second, find the baud rate and
the bit rate.
Baud rate = 1000 bauds per second (baud/s)
Bit rate = 1000 x 4 = 4000 bps
Baud Rate vs. Bit Rate Example1
SolutionSolution
Bit rate is the number of bits per second. Baud
rate is the number of signal units per second.
Bit rate is the number of bits per second. Baud
rate is the number of signal units per second.
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The bit rate of a signal is 3000. If each signal unit carries
6 bits, what is the baud rate?
Baud rate = 3000 / 6 = 500 baud/s
Baud Rate vs. Bit Rate Example 2
SolutionSolution