# DCN-04-Digital-Transmission-Slides

```Data Communication
& Networking CSCI 3342
Dr. Thomas Hicks
Computer Science Department
Trinity University
1
1
Digital To
Digital
Encoding
2
Major 4 Encoding Methods
Digital-ToDigital
Analog-ToDigital
Digital-ToAnalog
Analog-ToAnalog
Binary Data Must Be Encoded/Converted To A
Form That Will Propagate Over A Wire
3
Digital To Digital Encoding
Digital To Digital Encoding – Converting
Binary 0’s and 1’s Into A Sequence Of Voltage
Pulses That Can Propagate Over A Wire.
Transmit Data From Computer To Printer
4
Signal
Encoding
Signal Level
Data Level
5
Signal Level vs. Data Level
6
Pulse Rate
& Bit Rate
7
Pulse Rate = No Pulses Per Second
Bit Rate = No Bits Per Second
If the pulse carries only one bit, the
Bit Rate = Pulse Rate
[Not Always The Case]
8
General Case:
BitRate = PulseRate x log2 L
L = # Data Levels
BitRate = PulseRate x log2
DataLevels
9
Example 1
A signal has two data levels with a pulse
duration of 1 ms. We calculate the pulse rate
and bit rate as follows:
Pulse Rate = 1/ 10-3
= 1000 pulses/s
Bit Rate = Pulse Rate x log2 L
= 1000 x log2 2
= 1000 bps
10
Example 2
A signal has four data levels with a pulse
duration of 1 ms. We calculate the pulse rate
and bit rate as follows:
Pulse Rate = 1/ 10-3
= 1000 pulses/s
Bit Rate = PulseRate x log2 L
= 1000 x log2 4
= 2000 bps
11
DC
Components
12
We Shall Examine Numerous
Coding Schemes
Most Coding Schemes Will Have Values Above & Below The Line Positive & Negative Values.
10110001
13
DC Component
Coding Schemes
Some coding schemes have a residual DC [Direct-Current] that has a
zero frequency. The positive and negative voltages do not cancel each
other.
10110001
This extra energy on the line is useless and will not pass
14
Synchronization
15
Match The Sender's Bit Intervals
If The Signal Is To Be Interpreted
Correctly!
We Must Have Some Way Of
Synchronizing The Signal!
16
Lack Of Synchronization
17
Example 3
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:
1000 bits sent 1001 bits received1 extra bps
At 1 Mbps:
1,000,000 bits sent 1,001,000 bits
18
Many Line
Coding
Schemes
19
Only Some Of Major Encoding Methods!
Unipolar
Polar
Bipolar
AMI
NRZ
NRZ-L
NRZ-I
RZ
Manchester
B8ZS
HDB3
ETC
Differential
Manchester
20
You May Make A 5"x8" Card To Use On
Exam
(May Include Titles & Images)
NRZ-L
21
Unipolar
22
Unipolar Encoding
uses only One Voltage Level.
23
Unipolar Encoding
24
Unipolar Encoding -1
Unipolar uses either a Positive Or Negative
 Unipolar – Very Simple & Very Primitive Encoding
Scheme (almost obsolete)
Unipolar – Only one polarity.
 Sending Voltage Pulses along a medium link (usually a
wire or cable)
Voltage Level = 1’s
Zero Voltage Level = 0’s
Not Essential Assignment,
But Logical!
25
Unipolar Encoding -2
Unipolar Requires DC Component
Average Amplitude Is Non-Zero
Not All Mediums Can Handle A DC Component
Unipolar Requires Synchronization
No Way Receiver Can Determine Beginning Or
End
Problem With A Long, Uninterrupted Series Of 1’s
Problem With A Long, Uninterrupted Series Of 0’s
Solution To Synchronization Problem – Use A
Separate Parallel Line To Carry Clock Pulse
Doubling # Lines Expensive
26
Polar
27
Polar Encoding
uses two voltage levels
(positive and negative?)
28
Types Of Polar Encoding
29
Polar Encoding
Polar Encoding – Uses Two Voltage
Levels –> 0 Positive & 1 Negative – or
Visa Versa
Average Amplitude is 0
DC Component Not Needed
4+ Types Of Polar Encoding
NRZ
RZ
Manchester
Differential Manchester
30
NRZ-L
Encoding
(Polar)
31
Polar NRZ-L Encoding
In Polar NRZ-L the Level of the
Signal is Dependent upon the
State of the Bit.
32
Polar Encoding  NRZ-L
Methods Are NRZ-L and NRZ-I
NRZ-L
Usually 0 Positive & 1 Negative {For Us!}
Biggest Problem With Long Stream Of 1’s or
0’s [Clocks Might Not Be Synchronized]
33
NRZ-L Practice
Sketch The NRZ-L Encoding For The Signal
Below.
34
NRZ-I
Encoding
(Polar)
35
In NRZ-I the signal is
Inverted If a 1 is
Encountered.
36
Polar Encoding  NRZ-I
NRZ-I
An Inversion Of The Voltage Represents 1
If Pos  Neg
If Neg  Pos
 No Change Represents 0
Synchronization Occurs With Every 1
Bit
0’s Can Still Cause Problem – More 1’s Than 0’s
0 First  Pos
1 First  Neg
Next Bit Is 1
37
NRZ-I Practice
Sketch The NRZ-I Encoding For The Signal
Below.
38
RZ
Encoding
(Polar)
39
Polar RZ
Every Bit To Assure Synchronization
Several Solutions
RZ
Positive Voltage Means 1
Negative Voltage Means 0
Signal Returns To 0 Voltage Half-Through
40
Polar RZ
RZ – 3 Levels Of Amplitude – 3 Voltage Levels
41
Polar RZ Practice
Sketch The RZ Encoding For The Signal Below.
42
A Good
Encoded Digital Signal
Must Contain a Provision for
Synchronization.
43
Manchester
Encoding
(Polar)
44
In Manchester Encoding, the
Transition at the Middle of the Bit is
used for both Synchronization and Bit
Representation.
45
Polar Manchester
Manchester
Positive To Negative Transition For 0
Negative To Positive Transition For 1
Two Levels Of Amplitude
Inversion At Middle Middle Of Bit Time
46
Polar Manchester (cont)
Manchester
Inversion At Middle Middle Of Bit Time
Synch Signal Change Middle Of Each Bit
Green
Blue
47
Polar Manchester Practice
Sketch The Manchester Encoding For The Signal
Below.
Two Levels Of Amplitude
Same Synchronization As RZ
48
I Would Provide KEY
Manchester
49
Differential
Manchester
Encoding
(Polar)
50
In Differential Manchester
Encoding, the Transition at the
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.
51
Polar Differential Manchester
 Differential Manchester
Synch Signal Change Middle Of Each Bit
Inversion At Beginning Of Bit Time
Transition At Start Of Bit Time = 0
No Transition At Start Of Bit Time = 1
2 Signal Changes For 0, 1 Signal Change for 1
52
Polar Differential Manchester
Practice
Sketch The Differential Manchester Encoding
For The Signal Below.
53
Bipolar
54
In Bipolar
Encoding, we use Three
Levels: positive, zero, and negative.
55
Bipolar
Bipolar – Three Most Most Common
Solutions – AMI, B8ZS, & HDB3
Uses 3 Voltage Levels
Positive, Negative, Zero
Zero Level Is 0
Alternating Positive & Negative Are 1
56
AMI
Encoding
(Bipolar)
57
Bipolar - AMI
AMI – Alternate Mark Inversion
Mark In Telegraphy Means 1
Zero Voltage Represent 0
Alternating Positive & Negative Represent 1
Synchronize Long
Sequence 1’s
No Synchronize
Long Sequence
0’s
DC Component = 0
58
Bipolar AMI Practice
Sketch The AMI Encoding For The Signal Below.
59
Bipolar - Pseudoternary
“A Variation Of Bipolar AMI is called
Pseudoternary, In Which Binary 0’s
Alternate Between Positive & Negative
Voltages.”
60
BZPS
Encoding
(Bipolar)
61
Bipolar – BZPS
B8ZS – Bipolar 8 Zero Substitution
Same As AMI Until 8 Consecutive 0’s
Use Chart Below [Will Be Provided On Exam/
Quiz]
62
Bipolar – B8ZS Practice
Sketch The B8ZS Encoding For The Signal
Below.
63
Bipolar – HDB3
HDB3 – High Density Bipolar 3
Similar To B8ZS – Except Done In Sets Of 4
Use Chart Below [Will Be Provided On Exam/
Quiz]
64
2B1Q
Encoding
(Bipolar)
65
2B1Q Encoding  2 Binary 1
Quaternary Encoding
2B1Q Encoding  2 Binary 1 Quaternary
Encoding  4 Voltages
66
MLT-3
Encoding
(Bipolar)
67
MLT-3 Encoding
MLT-3 Encoding Similar to NRZ-I
Uses 3 Levels Of Signal +1, 0, -1
The Signal Transitions From One Level To The
Next At The Beginning Of A 1 Bit
There Is No Transition At The Beginning Of A 0
Bit
68
Block
Encoding
69
Figure 4.15
Block coding
Need Some Kind Of Redundancy To Assure
Synchronization!
High Performance!
Need Some Of The Chapter 10 Error Detection
To Assure Delivery!
70
Block Encoding
1. Divide Into
Groups Of M Bits
2. Substitute N-Bit
Code For M-Bit
Group
N>M
3. Use A Line
Encoding Scheme
To Create A Signal
Comes At A Cost - Requires Increase Bandwidth!
71
Figure 4.16
Substitution in block coding
72
4B5B
8B10B
Encoding
73
Table 4.1 4B/5B encoding -- Not All 5 Bit Codes Used!
Data
Code
Data
Code
0000
11110
1000
10010
0001
01001
1001
10011
0010
10100
1010
10110
0011
10101
1011
10111
0100
01010
1100
11010
0101
01011
1101
11011
0110
01110
1110
11100
0111
01111
1111
11101
74
Table 4.1 4B/5B encoding (Continued)
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)
R (Reset)
11001
00111
8B/10B Encoding
Groups of 8 Bits - Substituted Into A 10 Bit Code - More Efficient &
Better Error Detection! Long Table!
75
8B6T
Encoding
76
8B/6T Block Encoding
Take Advantage Of Speed & Error Detection Of Block
Encoding
Requires Much Less Bandwidth
8 Binary Bits Substituted Into A 6 Bit Ternary Table
8 Bits Translated Into 6 Bit of +1, 0, -1 [Table In Appendix D]
28 8-Bit Sequences <== Translated Into ==> 36 Ternary
256
729
77
Analog To
Digital
Encoding
78
Encoding Methods Review
Digital-ToDigital
Analog-ToDigital
Digital-ToAnalog
Analog-ToAnalog
Binary Data Must Be Encoded/Converted To A Form
That Will Propagate Over A Wire
79
Analog To Digital Encoding
Analog To Digital Encoding – Digitizing An
Analog Signal
Reducing The Potentially Infinite Number Of
Values In An Analog Signal So That They Can
Be Represented In A Digital Stream With A
Minimum Loss Of Information.
80
Codec
PAM
PCM
81
Codec – Coder-Decoder
Analog To Digital Converter Called A
Codec
coder –decoder  codec
Conversion Requires Two Steps:
Pulse Amplitude Modulation
(PAM)
Pulse Code Modulation (PCM)
82
Pulse Amplitude Modulation has some
applications, but it is not used by itself in
data communication.
However, it is the first step in another very
popular conversion method called
Pulse Code Modulation.
83
Step 1: Pulse Amplitude
Modulation (PAM)
I. Pulse Amplitude Modulation (PAM)
Sample Analog Signal At Regular Intervals 
Generate Pulses
Accuracy Depends Upon
# Of Samples Selected
84
Step 2: Pulse Code
Modulation (PCM)- 1
II. Pulse Code Modulation (PCM)
3 Steps
Step 1: Quantitize PAM Signals
85
Step 2: Pulse Code
Modulation (PCM) - 2
II. Pulse Code Modulation (PCM)
3 Steps
Step 2: Translate Each Value Into 7 Bit
Binary Equivalent
86
Step 2: Pulse Code
Modulation (PCM) - 3
II. Pulse Code Modulation (PCM)
3 Steps
Step 3: Convert To Digital Using Appropriate
Technique.
Review
Unipolar
Polar
…
Bipolar
…
87
Complete Analog-To-Digital
Conversion Flow Diagram
88
Nyquist
According to the Nyquist Theorem,
the Sampling Rate must be at least
2 times the Highest Frequency.
89
NYQUIST Theorem
Sampling Rate
Remember : Analog-To-Digital Accuracy
Depends Upon # Of Samples Selected
How Many?
Nyquist Theorem : The
Sampling Rate Must Be At
Least Two Times The Highest
Frequency!
90
Nyquist In The Real World
Sampling Rate Practice
Telephone Voice
Maximum Frequency = 4000 Hz
8,000 Samples/Second
Sampling Rate = __________
A Bandwidth of 11,000 Is Needed To
Transfer A Signal Whose Frequency Range
Is 1,000 Hz to 12,000 Hz.
24,000 Samples/Second
Sampling Rate = __________
91
Bit Rate
Bit Rate = Sampling Rate x Number Bits Per Sample
Bit Rate : Also Called The Data Rate
92
Bit Rate Practice
Want To Digitize Human Voice Using Eight Bit
Samples.
Human Voice Has Frequency Range of 0 to 4 KHz.
Sampling Rate = 2 * Highest Frequency (4000 Hz)
8,000 Samples/Second
Sampling Rate = __________
Bit Rate = Sampling Rate (8,000) * NoBitsPerSample (8)
Bit Rate = 64,000 bps
64
Bit Rate = __________
Kbps
93
Example 4
What sampling rate is needed for a signal with a
bandwidth of 10,000 Hz (1000 to 11,000 Hz)?
Solution
The sampling rate must be twice the highest frequency in
the signal:
Sampling rate = 2 x (11,000) = 22,000 samples/s
94
Example 5
A signal is sampled. Each sample requires at least 12
levels of precision (+0 to +5 and -0 to -5). How many bits
should be sent for each sample?
Solution
We need 4 bits; 1 bit for the sign and 3 bits for the value.
A 3-bit value can represent 23 = 8 levels (000 to 111),
which is more than what we need. A 2-bit value is not
enough since 22 = 4. A 4-bit value is too much because 24
= 16.
95
Example 6
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.
Sampling rate = 4000 x 2 = 8000 samples/s
Bit rate = sampling rate x number of bits per sample
= 8000 x 8 = 64,000 bps = 64 Kbps
96
Transmission
Modes
97
Data Transmission
Parallel
Serial
Synchronous
Asynchronous
98
Parallel Transmission
Parallel Transmission – Eight or More
Lines Are Bundled Together To Send One
Byte At A Time
99
Serial Transmission
Serial Transmission – Requires Only One
Communication Channel
100
Serial or Parallel Transmission
Which Is Faster?
101
Serial or Parallel Transmission
Which Is Least Expensive?
Usually
Limited
To Short
Distances
102
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.
103
Serial Transmission
Asynchronous - 1
Asynchronous – Information Sent &
Timing Is Unimportant!
104
Serial Transmission
Asynchronous - 2
Asynchronous Serial Transmission
8 Bits – 1 Byte – Of Data Transmitted
1-2 Stop Bits [1’s] Is/Are Sent To Let
User Know Finished
A Brief Time Gap Often Follows
Some Type Of Synchronization Must Be
Embedded Within Data
Cheap/Effective Choice For Low Speed
Communication [Great For Terminal –
Computer!]
105
Asynchronous here means
“asynchronous at the byte level,” but
the bits are still synchronized; their
durations are the same.
106
Serial Transmission
Synchronous - 1
Synchronous – Information Combined
Into Frames [Multiple Bytes]; Timing Is
Essential!
107
Serial Transmission
Synchronous - 2
Synchronous Serial Transmission
No Gaps – Unbroken String 1’s & 0’s
Gaps Generally Filled In With Agreed
Upon Sequences Of 1’s & 0’s – Idle
Timing Essential
Much Faster Than Asynchronous
108
In Synchronous
Transmission,
we send bits one after another without
start/stop bits or gaps.
It is the responsibility of the receiver to
group the bits.
109
Good Practice Problem
 Sketch The Encoding Of Signal 11000011 With Each Of
The Following On A New Page. Write NothingElse On
This/These Pages Except Encoding Type & Your Name(s). Each
Person On Team Must Do Their Own Copy Of This Problem!
A. Unipolar
B. NRZ-L
C. NRZ-I
D. RZ
E. Manchester
F. Differential Manchester
G. AMI
H. Pseudoternary
I. B8ZS
J. Quaternary 2B1Q
K. MLT-3
L. 4B5B
110
M. 8B6T
Data Communications & Networking
CSCI 3342
Dr. Thomas E. Hicks
Computer Science Department
Trinity University
Textbook: Computer Networks
By Andrew Tanenbaum
Textbook: Data Communications & Networking
By Behrouz Forouzan
Special Thanks To WCB/McGraw-Hill For Providing Graphics For
Many Text Book Figures For Use In This Presentation.
111
```
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