Analog Transmission
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5.1 Digital-to-Analog 부호화
 ASK(Amplitude Shift Keying)
 FSK(Frequency Shift Keying)
 PSK(Phase Shift Keying)
Shift Keying = modulation
 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 1
An analog signal carries 4 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 2
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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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 3
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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Digital-to-Analog 부호화(cont’d)
 Example 4
Given a bandwidth of 5000 Hz for an ASK signal, what are the baud
rate and bit rate?
 Solution
In ASK the baud rate is the same as the bandwidth, which
means the baud rate is 5000. But because the baud rate and the
bit rate are also the same for ASK, the bit rate is 5000 bps.
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Digital-to-Analog 부호화(cont’d)
 Example 5
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
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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
• Peak amplitude and phase remain constant
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Digital-to-Analog 부호화(cont’d)
 Bandwidth for FSK
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Digital-to-Analog 부호화(cont’d)
 PSK
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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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The 4-PSK method

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The 4-PSK Characteristics
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The 8-PSK Characteristics
 Find the bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission
is in half-duplex mode.
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Relationship between baud rate and bandwidth in PSK

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Bandwidth for PSK
Example 4: Find the bandwidth for a 4-PSK signal transmitting at 2000
bps. Transmission is in half-duplex mode.
Answer) For 4-PSK baud rate is one-half of the bit rate. The baud rate is
therefore 1000. A PSK signal requires a bandwidth equal to its baud rate.
Therefore, the bandwidth is 1000 Hz.
Example 4 : Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the
baud rate and bit rate?
For PSK the baud rate is the same as the bandwidth, which means the
baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the
bit rate is 15,000 bps.
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Digital-to-Analog 부호화(cont’d)
QAM(Quadrature Amplitude Modulation)
Quadrature amplitude modulation is a combination of ASK and
PSK so that a maximum contrast between each signal unit (bit,
dibit, tribit, and so on) is achieved
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The 4-QAM and 8-QAM constellations
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Digital-to-Analog 부호화(cont’d)
 Time domain for an 8-QAM signal
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16-QAM constellations
• The minimum bandwidth for QAM is the
same as that required for ASK and PSK
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Bit/Baud Comparison

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Bit and Baud rate comparison
Modulation
Units
Bits/Baud
Baud rate
Bit Rate
Bit
1
N
N
4-PSK, 4-QAM
Dibit
2
N
2N
8-PSK, 8-QAM
Tribit
3
N
3N
16-QAM
Quadbit
4
N
4N
32-QAM
Pentabit
5
N
5N
64-QAM
Hexabit
6
N
6N
128-QAM
Septabit
7
N
7N
256-QAM
Octabit
8
N
8N
ASK, FSK, 2-PSK
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Bit and Baud rate comparison
 Example 10 : A constellation diagram consists of eight
equally spaced points on a circle. If the bit rate is 4800 bps,
what is the baud rate?
Solution : The constellation indicates 8-PSK with the points 45
degrees apart. Since 23 = 8, 3 bits are transmitted with each signal unit.
Therefore, the baud rate is
4800 / 3 = 1600 baud
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Bit and Baud rate comparison
 Example 11 : Compute the bit rate for a 1000-baud 16-QAM signal.
Solution :
A 16-QAM signal has 4 bits per signal unit since
log216 = 4.
Thus, (1000)(4) = 4000 bps
 Compute the baud rate for a 72,000-bps 64-QAM signal
Solution : A 64-QAM signal has 6 bits per signal unit since
log2 64 = 6.
Thus,
72000 / 6 = 12,000 baud
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5.2 Telephone Modems
 A telephone line has a bandwidth of almost 2400 Hz for data
transmission
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Telephone Modems

Modem stands for modulator/demodulator.
Modulator – creates a band-pass analog signal from
binary data
Demodulator – recovers the binary data from the
modulated signal
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Modulation/demodulation
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Modem Standards
 V.32 – 9600 bps : called Trellis-coded modulation
 V.32bis – 14,400 bps
 V.34bis – 28,800 bps & 33,600 bps
 V.90 – 33,600 upload, 56,000 bps download
 V.92 – 48 Kpbs up, 56 Kpbs down
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Modem Standards -The V.32 constellation and bandwidth
 32-QAM with a baud rate of 2400
 4 data bits x 2400 = 9600 bps
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Modem Standards -The V.32bis constellation and bandwidth
• 128-QAM
• 6 data bits x 2400 baud = 14,400 bps
• fall-back, fall-forward feature enabling modem to adjust speed
depending on line and/or signal quality
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Modem Standards -The V.34bis and V.90
 V.34bis
bit rate of 28,800 with 960-point constellation
bit rate of 33,600 with 1664-point constellation
 V.90
download 56K, upload 33.6K
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Traditional Modems

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56K Modems
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56K Modems
 telephone company samples 8,000 times per second
 8 bits per sample (7 data bits)
 rate = 8,000 x 7 = 56,000 bps
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5.3 Modulation of Analog Signals
 Analog-to-Analog encoding 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.
The total bandwidth required for AM
can be determined from the bandwidth
of the audio signal:
BWtotal = 2 x BWmod.
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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)
 each frequency must be separated by 10 KHz
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Analog-to-Analog 부호화(cont’d)
 AM band allocation
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Analog-to-Analog 부호화(cont’d)
 We have an audio signal with a bandwidth of 4 KHz.
What is the bandwidth needed if we modulate the signal
using AM? Ignore FCC regulations.
 An AM signal requires twice the bandwidth of the
original signal:
BW = 2 x 4 KHz = 8 KHz
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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.
The total bandwidth required for FM
can be determined from the bandwidth
of the audio signal:
BWt = 10 x BWm.
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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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