Analog+Transmissionch 05

8/6/2019 Analog+Transmissionch 05

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AnalogTransmission


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Modulation of Digital Data

Digital-to-Analog Conversion

Amplitude Shift Keying (ASK)

Frequency Shift Keying (FSK)

Phase Shift Keying (PSK)

Quadrature Amplitude ModulationBit/Baud Comparison


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Converting Binary data or a low-passanalog signal to a band-pass analogsignal is called Modulation

Modulation of Binary data or D-to-Amodulation is the process of changing

one of the characteristics of an analogsignal based on the information in adigital signal(0s & 1s)


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Figure 5.1 Digital-to-analog modulation


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Figure 5.2 Types of digital-to-analog modulation


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Bit rate & Baud rate

In discussing computer efficiency bit rate tellshow long it takes to process each piece ofinformation

In Data Transmission, concern is how efficientlydata can be moved from place to place, whetherin pieces or blocks

Fewer signal units required, more efficient the

system & less bandwidth required to transmitmore bits

Baud rate determines the bandwidth required tosend the signal


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Bit rate is the number of bits per

second. Baud rate is the number ofsignal units per second. Baud rate is

less than or equal to the bit rate.

Note:Note:


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Example 1Example 1

An analog signal carries 4 bits in each signal unit. If 1000signal units are sent per second, find the baud rate and the

bit rate

SolutionSolution

Baud rate = 1000 bauds per second (baud/s)Baud rate = 1000 bauds per second (baud/s)

Bit rate = 1000 x 4 = 4000 bpsBit rate = 1000 x 4 = 4000 bps


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Example 2Example 2

The bit rateof a signal is 3000. Ifeach signal unit carries6 bits, what is the baud rate?

SolutionSolution

Baud rate = 3000 / 6 = 500 baud/sBaud rate = 3000 / 6 = 500 baud/s


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Amplitude Shift Keying (ASK)

Strength of the carrier signal is varied torepresent binary 1 or 0.

Both frequency & phase remain constantwhile the amplitude changes

Highly susceptible to noise interference


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Figure 5.3 ASK


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Figure 5.4 Relationship between baud rate and bandwidth in ASK

BW=(1+d)X Nbaud

Where Nbaud is the baud rate, d is a factor related tothe modulation process


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Example 3Example 3

Find the minimum bandwidth for an ASK signaltransmitting at 2000 bps.Thetransmission mode is half-

duplex.

SolutionSolution

In ASKthe baud rate and bit rate arethe same.The baud

rate is therefore 2000.An ASK signal requires aminimum bandwidth equal to its baud rate.Therefore,

the minimum bandwidth is 2000 Hz.


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Example 5Example 5

iven a bandwidth of 10,000 Hz (1000 to 11,000 Hz),draw the full-duplex ASK diagram ofthe system.Find the

carriers and the bandwidths in each direction.Assume

there is nogap between the bands in thetwo directions.

SolutionSolution

For full-duplex ASK, the bandwidth foreach direction is

BW = 10000 / 2 = 5000 Hz

The carrier frequencies can be chosen atthe middleof

each band (seeFig. 5.5).

fc (forward) = 1000 + 5000/2 = 3500 Hz

fc (backward) = 11000 5000/2 = 8500 Hz


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Figure 5.5 Solution to Example 5


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Frequency Shift Keying (FSK)

Frequency of the carrier signal is varied torepresent binary 1 or 0

Frequency of the signal during each bitduration is constant, and its valuedepends on the bit(0 or 1)

Avoids problems from noise, it can ignore

voltage spikes Limiting factors are the physical

capabilities of the carrier


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Figure 5.6 FSK


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Figure 5.7 Relationship between baud rate and bandwidth in FSK


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Example 6Example 6

Find the minimum bandwidth for an FSK signaltransmitting at 2000 bps.Transmission is in half-duplex

mode, and the carriers are separated by 3000 Hz.

SolutionSolution

ForFSK

BW = baud rate + fBW = baud rate + fc1c1 ffc0c0

BW = bit rate + fc1BW = bit rate + fc1

fc0 = 2000 + 3000 = 5000 Hzfc0 = 2000 + 3000 = 5000 Hz


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Example 7Example 7

Find the maximum bit rates for an FSK signal ifthebandwidth ofthe medium is 12,000 Hz and the difference

between thetwo carriers is 2000 Hz.Transmission is in

full-duplex mode.

SolutionSolution

Becausethetransmission is full duplex, only 6000 Hz is

allocated foreach direction.

BW = baud rate + fc1BW = baud rate + fc1

fc0fc0

Baud rate = BWBaud rate = BW (fc1(fc1 fc0 ) = 6000fc0 ) = 6000 2000 = 40002000 = 4000

But becausethe baud rate is the same as the bit rate, the

bit rate is 4000 bps.


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Phase Shift Keying (PSK)

Phase of the carrier is varied

Peak amplitude and frequency remain

constant as the phase changes Not susceptible to the noise degradation

Smaller variations in the signal can be

detected reliably by the receiver


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Figure 5.8 PSK


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Figure 5.9 PSK constellation


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Figure 5.10 The 4-PSK method

Instead of utilizing only two variations of a signal,each representing 1 bit, we can use 4 variationsand each phase shift represents 2 bits


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Figure 5.11 The 4-PSK characteristics

Pair of Bits represented by each phase is called dibit

Data can be transmitted twice as efficiently using 4-PSK as using 2-PSK


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Figure 5.12 The 8-PSK characteristics


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Figure 5.13 Relationship between baud rate and bandwidth in PSK

Max. Bit rate is potentially much greater than ASK

So Max. baud rates of ASK & PSK are the same for a given bandwidth,PSK bit rates using the same bandwidth can be 2 or more times greater


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Example 8Example 8

Find the bandwidth for a 4-PSK signal transmitting at2000 bps.Transmission is in half-duplex mode.

SolutionSolution

For PSKthe baud rate is the same as the bandwidth,

which means the baud rate is 5000.But in 8-PSKthe bit

rate is 3 times the baud rate, sothe bit rate is 15,000 bps.


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Example 9Example 9

Given a bandwidth of 5000 Hz for an 8-PSK signal, whatarethe baud rate and bit rate?

SolutionSolution

For PSKthe baud rate is the same as the bandwidth,

which means the baud rate is 5000.But in 8-PSKthe bit

rate is 3 times the baud rate, sothe bit rate is 15,000 bps.


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Quadrature Amplitude Modulation

(QAM)

Number of amplitudes shifts is fewer than

the number of phase shifts

Amplitude changes are susceptible tonoise and require greater shift differences

than do phase changes


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Quadrature amplitude modulation is a

combination of ASK and PSK so that amaximum contrast between each

signal unit (bit, dibit, tribit, and so on)

is achieved.

Note:Note:


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Figure 5.14 The 4-QAM and 8-QAM constellations


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Figure 5.15 Time domain for an 8-QAM signal


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Figure 5.16 16-QAM constellations

ITU-T OSI


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Figure 5.17 Bit and baud


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Table 5.1 Bit and baud rate comparison

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

ASK, FSK, 2ASK, FSK, 2--PSKPSK Bit 1 N N

44--PSK, 4PSK, 4--QAMQAM Dibit 2 N 2N

88--PSK, 8PSK, 8--QAMQAM Tribit 3 N 3N

1616--QAMQAM Quadbit 4 N 4N

3232--QAMQAM Pentabit 5 N 5N

6464--QAMQAM Hexabit 6 N 6N

128128--QAMQAM Septabit 7 N 7N

256256--QAMQAM Octabit 8 N 8N


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Example 10Example 10

A constellation diagram consists ofeightequally spacedpoints on a circle. Ifthe bit rate is 4800 bps, what is the

baud rate?

SolutionSolution

The constellation indicates 8-PSK with the points 45

degrees apart. Since 23 = 8, 3 bits aretransmitted with

each signal unit.Therefore, the baud rate is4800 / 3 = 1600 baud


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Computethe bit rate for a 1000-baud 16-QAM signal.

SolutionSolution

A 16-QAM signal has 4 bits per signal unit sincelog216 = 4.

Thus,

(1000)(4) = 4000 bps


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Computethe baud rate for a 72,000-bps 64-QAM signal.

SolutionSolution

A 64-QAM signal has 6 bits per signal unit sincelog2 64 = 6.

Thus,

72000 / 6 = 12,000 baud


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5.2 Telephone Modems

Modem Standards


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A telephone line has a bandwidth of

almost 2400 Hz for data transmission.

Note:Note:


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Figure 5.18 Telephone line bandwidth


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Modem stands for

modulator/demodulator.

Note:Note:


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Figure 5.19 Modulation/demodulation


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Figure 5.20 The V.32 constellation and bandwidth


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Figure 5.21 The V.32bis constellation and bandwidth


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Figure 5.22 Traditional modems


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Figure 5.23 56K modems


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5.3 Modulation of Analog Signals5.3 Modulation of Analog Signals

Amplitude Modulation (AM)

Frequency Modulation (FM)

Phase Modulation (PM)


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Figure 5.24 Analog-to-analog modulation


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Figure 5.25 Types of analog-to-analog modulation


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i 26 A li d d l i


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Figure 5.26 Amplitude modulation

Fi 5 27 AM b d id h


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Figure 5.27 AM bandwidth

Fi 5 28 AM b d ll i


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Figure 5.28 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 modulatethe signal

usingAM? IgnoreFCC regulations.

SolutionSolution

An AM signal requires twicethe bandwidth ofthe

original signal:

BW = 2 x 4 KHz = 8 KHz


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The total bandwidth required for FM

can be determined from the bandwidthof the audio signal:

BWt = 10 xBWm.

Note:Note:

Fig re 5 29 F d l ti


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Figure 5.29 Frequency modulation

Figure 5 30 FM bandwidth


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Figure 5.30 FM bandwidth


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The bandwidth of a stereo audio signal

is usually 15 KHz. Therefore, an FM

station needs at least a bandwidth of150 KHz. The FCC requires the

minimum bandwidth to be at least 200

KHz (0.2 MHz).

Note:Note:

Figure 5 31 FM band allocation


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Figure 5.31 FM band allocation


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We have an audio signal with a bandwidth of 4 MHz.What is the bandwidth needed if we modulatethe signal

usingFM? IgnoreFCC regulations.

SolutionSolution

An FM signal requires 10 times the bandwidth ofthe

original signal:

BW = 10 x 4 MHz = 40 MHz

Analog+Transmissionch 05

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