Encoding
Prof. Choong Seon HONG
Kyung Hee
University
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5 장 부호화(Encoding)
5.1 Digital-to-Digital
5.2 Analog-to-Digital
5.3 Digital-to-Analog
5.4 Analog-to-Analog
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부호화(cont’d)
 Information must be encoded into signals before it can be
transported across communication media.
 Different encoding schemes
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5.1 Digital-to-Digital 부호화
~ is the representation of digital information by a digital signal.
 Digital-to-Digital encoding
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Digital-to-Digital 부호화(cont’d)
 Types of digital-to-digital encoding
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Digital-to-Digital 부호화(cont’d)
 Unipolar
uses only one level of value(1: positive value, 0: idle )
Simple and inexpensive
 Unipolar encoding
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Digital-to-Digital 부호화(cont’d)
 Unipolar encoding problems
almost obsolete today
Having DC(Direct Current)Component : 직류성분 (a component
with zero frequency)

cannot travel through media that cannot handle DC
components , such as microwave.
Synchronization(동기)

The receiver has to rely on a timer.
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Digital-to-Digital 부호화(cont’d)
 Polar
~ uses two levels (positive and negative) of amplitude.
 Types of polar encoding
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Digital-to-Digital 부호화(cont’d)
 NRZ(Non-Return to Zero)
NRZ-L : the level of the signal is dependent upon the state of
the bit
NRZ-I : the signal is inverted if a 1 is encountered
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Digital-to-Digital 부호화(cont’d)
 NRZ-L and NRZ-I encoding
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Digital-to-Digital 부호화(cont’d)
RZ(Return to Zero)
using three values (positive, negative, zero)
1 : positive-to-zero
0 : negative-to-zero
The signal changes not between bits but during each bit
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Digital-to-Digital 부호화(cont’d)
RZ encoding
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Digital-to-Digital 부호화(cont’d)
The main disadvantages of RZ encoding
requiring two signal changes to encode one bit and therefore
occupies more bandwidth
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Digital-to-Digital 부호화(cont’d)
Biphase
provides probably the best existing solution to the problem of
synchronization
is implemented in two different ways.

Manchester : used by Ethernet LANs

Differential Manchester : used by Token Ring LANs
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University
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Digital-to-Digital 부호화(cont’d)
 Manchester and Differential Manchester encoding
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Digital-to-Digital 부호화(cont’d)
 In Manchester encoding, the transition at the middle of the
bit is used for both synchronization and bit representation
 In Differential Manchester, the transition at the middle of the
bit is used only for synchronization.
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Digital-to-Digital 부호화(cont’d)
Bipolar
uses three voltage levels(positive, negative,zero)

zero level : binary 0

positive and negative voltage : 1(alternate)
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Digital-to-Digital 부호화(cont’d)
Type of bipolar encoding
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Digital-to-Digital 부호화(cont’d)
Bipolar AMI(Alternate Mark Inversion)
~ is the simplest type of bipolar encoding
Bipolar AMI changes poles with every 1 it encounters

These changes provide the synchronization needed by the
receiver
But, there is no mechanism to ensure the
synchronization of a long string of 0s
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Digital-to-Digital 부호화(cont’d)
Bipolar AMI encoding
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Digital-to-Digital 부호화(cont’d)
B8ZS(Bipolar 8-Zero Substitution)
is the convention adopted in North America to
provide synchronization of long strings of 0s.
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Digital-to-Digital 부호화(cont’d)
 B8ZS encoding
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Digital-to-Digital 부호화(cont’d)
HDB3(High-Density Bipolar 3)
is the convention adopted in Europe and Japan.
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Digital-to-Digital 부호화(cont’d)
 HDB3 encoding
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Digital-to-Digital 부호화(cont’d)
Example 5.1
Using B8ZS, encode the bit stream 10000000000100. Assume
that the polarity of the first 1 is positive.
Solution
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Digital-to-Digital 부호화(cont’d)
Example 5.2
using HDB3, encoded the bit stream 10000000000100. Assume
that the number of 1s so far is odd and the first 1 is positive.
Solution
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Digital-to-Digital 부호화(cont’d)
 Exercise
Amplitude
1
1
0
0
0
0
0
0
0
0
0
1
0
1
0
Time
AMI
Time
B8ZS
Time
HDB3
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5.2 Analog-to-Digital 부호화
is the representation of analog information by a digital
signal. (recording singer’s voice onto a compact disc)
Analog-to-Digital Encoding
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Analog-to-Digital 부호화(cont’d)
PAM(Pulse Amplitude Modulation)
This technique takes analog information, samples it, and
generates a series of pulses based on the results of the
sampling.
* Term sampling means measuring the amplitude of the signal at
equal intervals.
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Analog-to-Digital 부호화(cont’d)
PAM
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Analog-to-Digital 부호화(cont’d)
PCM(Pulse Code Modulation)
Quantization is a method of assigning integral values in a
specific range to sampled instances.
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Analog-to-Digital 부호화(cont’d)
Quantized PAM signal
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Analog-to-Digital 부호화(cont’d)
Quantized sample  Assign sign and magnitude value
(seven-bit binary equivalent)
Quantizing using sign and magnitude
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Analog-to-Digital 부호화(cont’d)
 The binary digits are then transformed into a digital signal
using one of the digital-to-digital encoding techniques(예 :
unipolar)
 PCM
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Analog-to-Digital 부호화(cont’d)
From analog signal to PCM digital code
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Analog-to-Digital 부호화(cont’d)
Sampling Rate
How many samples are sufficient ?
According to the Nyquist theorem

The sampling rate must be at least two times the highest
frequency
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Analog-to-Digital 부호화(cont’d)
 Example 5.3
What sampling rate is needed for a signal with a bandwidth of
10,000Hz (1000 to 11,000Hz) ? If the quantization is eight bits
per sample, what is the bit rate ?
Solution
The sampling rate must be twice the highest frequency in the
signal :

Sampling rate = 2 (11,000) = 22,000 samples/s
Data rate = (22,000 samples/s) (8 bits/sample) = 176 Kbps
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University
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Analog-to-Digital 부호화(cont’d)
 Example 5.4
A signal is sampled. Each sample requires at least 12 levels of
operation (+0 to +5 and –0 to –5). How many bits should be sent for
each sample ?
 Solution
need for 4 bits : one bit for the sign and three bits for the value
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Analog-to-Digital 부호화(cont’d)
 Example 5.5
We want to digitize the human voice. What is the bit rate assuming
eight bits per sample ?
 Solution
Human voice frequencies : 0 to 40000 Hz
Sampling rate : 4000 x 2 = 8,000 samples/second
Bit Rate = sampling rate x Number of bits per sample = 8000 x 8 =
64,000 bits/s = 64 Kbps
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University
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5.3 Digital-to-Analog 부호화
 ASK(Amplitude Shift Keying)
 FSK(Frequency Shift Keying)
Shift Keying = modulation
 PSK(Phase Shift Keying)
 QAM(Quadrature Amplitude Modulation) : related to Amplitude and
Phase
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Digital-to-Analog 부호화(cont’d)
 Type of Digital-to-Analog encoding
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Digital-to-Analog 부호화(cont’d)
 Bit rate : the number of bits per second.
 Baud rate : the number of signal units per second.
Baud rate is less than or equal to the bit rate.
Bit rate equals the baud rate x the number of bits represented by each
signal unit
 반송신호 또는 주파수 (Carrier Signal or Carrier Frequency)
base signal for the information signal
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Digital-to-Analog 부호화(cont’d)
 Example 5.6
An analog signal carries four bits in each signal element. If 1000
signal elements are sent per second, find the baud rate and the bit
rate.
 Solution
Baud rate = Number of signal elements = 1000 bauds per second
Bit rate = Baud rate x Number of bits per signal element = 1000 x 4 =
4000 bps
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Digital-to-Analog 부호화(cont’d)
 Example 5.7
The bit rate of a signal : 3000
If each signal element carries six bits, what is the baud rate ?
 Solution
Baud rate = Bit rate/ number of bits per signal element = 3000/6 = 500
baud per second
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University
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Digital-to-Analog 부호화(cont’d)
ASK(Amplitude Shift Keying)
Both frequency and phase remain constant while the amplitude
changes.
Highly susceptible to noise interference

Noise usually affects the amplitude.
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Digital-to-Analog 부호화(cont’d)
 ASK encoding
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Digital-to-Analog 부호화(cont’d)
 Relationship between baud rate and bandwidth in ASK
BW = (1 + d) x N baud
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N baud : Baud rate
d : factor related to the
condition of the line (with a
minimum value of 0)
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Digital-to-Analog 부호화(cont’d)
 Example 5.8
Find the minimum bandwidth for an ASK signal transmitting at 2000
bps. Transmission mode is half-duplex
 Solution
In ASK the baud rate and bit rate are the same. The baud rate is
therefore 2000. An ASK signal requires a minimum bandwidth equal to
its baud rate. Therefore, the minimum bandwidth is 2000Hz
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University
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Digital-to-Analog 부호화(cont’d)
 Example 5.10
Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz), draw the fullduplex ASK diagram of the system. Find the carriers and the
bandwidth in each direction. Assume there is no gap between the
bands in two directions.
 Solution
Bandwidth for each direction : 10000/2 = 5000 Hz
Carrier frequencies : fc (forward) = 1000 + 5000/2 = 3500 Hz
fc (backward) = 11000 – 5000/2 = 8500 Hz
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Digital-to-Analog 부호화(cont’d)
Solution to Example 5.10
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Digital-to-Analog 부호화(cont’d)
FSK(Frequency Shift Keying)
the frequency of the signal is varied to represent binary 1 or 0.
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Digital-to-Analog 부호화(cont’d)
 FSK encoding
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Digital-to-Analog 부호화(cont’d)
 Bandwidth for FSK
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Digital-to-Analog 부호화(cont’d)
PSK(Phase Shift Keying)
the phase is varied to represent binary 1 or 0.
bit
phase
0
1
0
180
1
0
Constellation diagram
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Digital-to-Analog 부호화(cont’d)
 PSK
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Digital-to-Analog 부호화(cont’d)
QAM(Quadrature Amplitude Modulation)
means combining ASK and PSK in such a way that we have
maximum contrast between each bit, dibit, tribit, quadbit, and
so on.
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Digital-to-Analog 부호화(cont’d)
 Time domain for an 8-QAM signal
011
010
101 100
000 001
110
111
2 amplitudes, 4 phases
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5.4 Analog-to-Analog 부호화
 is the representation of analog information by an analog
signal.
 Analog-to-Analog encoding
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Analog-to-Analog 부호화(cont’d)
 Type of analog-to-analog encoding
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Analog-to-Analog 부호화(cont’d)
AM(Amplitude Modulation)
~ The frequency and phase of the carrier remain the same; only
the amplitude changes to follow variations in the information.
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Analog-to-Analog 부호화(cont’d)
Amplitude modulation
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Analog-to-Analog 부호화(cont’d)
AM bandwidth
The total bandwidth required for AM can be determined from
the bandwidth of the audio signal.
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Analog-to-Analog 부호화(cont’d)
AM bandwidth
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Analog-to-Analog 부호화(cont’d)
 Audio signal(음성과 음악) bandwidth : 5 KHz
 Minimum bandwidth : 10 KHz (bandwidth for AM radio station)
 AM stations are allowed carrier frequencies anywhere between 530
and 1700 KHz(1.7 MHz)
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Analog-to-Analog 부호화(cont’d)
 AM band allocation
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Analog-to-Analog 부호화(cont’d)
FM(Frequency Modulation)
as the amplitude of the information signal changes, the
frequency of the carrier changes proportionately.
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Analog-to-Analog 부호화(cont’d)
Frequency modulation
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Analog-to-Analog 부호화(cont’d)
FM Bandwidth
The bandwidth of an FM signal is equal to 10 times the
bandwidth of the modulating signal.
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Analog-to-Analog 부호화(cont’d)
 FM bandwidth
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Analog-to-Analog 부호화(cont’d)
 Bandwidth of an audio signal(음성과 음악) broadcast in stereo : 15
KHz
 minimum bandwidth : 150 KHz
 allows generally 200 KHz(0.2 MHz) for each station
 FM station are allowed carrier frequencies anywhere 88 and 108
MHz(each 200 KHz)
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Analog-to-Analog 부호화(cont’d)
 FM band allocation
Alternate bandwidth allocation
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Analog-to-Analog 부호화(cont’d)
PM(Phase Modulation)
~ is used in some systems as an alternative to frequency
modulation.
The phase of the carrier signal is modulated to follow the changing
voltage (amplitude) of the modulating signal
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