analog transmission

11Kyung Hee Universit

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Analog TransmissionAnalog Transmission


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5.1 Digital-to-Analog 5.1 Digital-to-Analog 부호화부호화

ASK(Amplitude Shift Keying)

FSK(Frequency Shift Keying)

PSK(Phase Shift Keying)

QAM(Quadrature Amplitude Modu

lation) : related to Amplitude and

Phase

Shift Keying = modulation


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Digital-to-Analog Digital-to-Analog 부호화부호화 (cont’d)(cont’d)

Type of Digital-to-Analog encoding


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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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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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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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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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ASK encoding


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Relationship between baud rate

and bandwidth in ASK

BW = (1 + d) x N baud

N baud : Baud rated : factor related to the condition of the line (with a minimum value of 0)


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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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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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Example 5

Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz), draw the full-duplex 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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Solution to Example 5


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FSK(Frequency Shift Keying)

the frequency of the signal is varied to represent binary 1 or 0.


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FSK encoding

• Peak amplitude and phase remain constant


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Bandwidth for FSK


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PSK


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PSK(Phase Shift Keying)

the phase is varied to represent binary 1 or 0.

bit phase

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Constellation diagram

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The 4-PSK methodThe 4-PSK method


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The 4-PSK CharacteristicsThe 4-PSK Characteristics


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The 8-PSK CharacteristicsThe 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 PSKRelationship between baud rate and bandwidth in PSK


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Bandwidth for PSKBandwidth 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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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 constellationsThe 4-QAM and 8-QAM constellations


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Time domain for an 8-QAM signal



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16-QAM constellations16-QAM constellations

• The minimum bandwidth for QAM is the same as that required for ASK and PSK


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Bit/Baud ComparisonBit/Baud Comparison


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Bit and Baud rate comparisonBit and Baud rate comparison

ModulationModulation UnitsUnits Bits/BaudBits/Baud Baud rateBaud rate Bit Rate

ASK, FSK, 2-PSKASK, FSK, 2-PSK Bit 1 N N

4-PSK, 4-QAM4-PSK, 4-QAM Dibit 2 N 2N

8-PSK, 8-QAM8-PSK, 8-QAM Tribit 3 N 3N

16-QAM16-QAM Quadbit 4 N 4N

32-QAM32-QAM Pentabit 5 N 5N

64-QAM64-QAM Hexabit 6 N 6N

128-QAM128-QAM Septabit 7 N 7N

256-QAM256-QAM Octabit 8 N 8N


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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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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 Modems5.2 Telephone Modems

A telephone line has a bandwidth of almost 2400 Hz for data

transmission


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Telephone ModemsTelephone 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/demodulationModulation/demodulation


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Modem StandardsModem 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 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 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 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 ModemsTraditional Modems


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56K Modems56K Modems


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56K Modems56K 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 Signals5.3 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 Analog-to-Analog 부호화부호화 (cont’d)(cont’d)

Type of analog-to-analog encoding


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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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Amplitude modulation


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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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AM bandwidth


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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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AM band allocation


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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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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 Analog-to-Analog 부호화부호화 (cont’d)(cont’d)Frequency modulation


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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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FM bandwidth


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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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FM band allocation

Alternate bandwidth allocation


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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

analog transmission

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