Figure 5.

1 Digital-to-analog modulation

„ Modulation of binary data or digital-to-analog

modulation is the process of changing one of the
characteristics of an analog signal based on the
information in a digital signal (0s and 1s)
„ Telephone wires carry analog signals.
„ When we vary any one of these characteristics
[amplitude or frequency or phase], we create a
different version of that wave.
„ Three mechanisms for modulating digital data into an
analog signal: amplitude shift keying (ASK), frequency
shift keying (FSK) , and phase shift keying (PSK).

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 Aspects of Digital-to-Analog Conversion
„ A signal unit is composed of 1 or more bits.
„ Bit Rate and Baud Rate
„ Bit rate is the number of bits per second.
„ Baud rate is the number of signal units per second.
„ Bit rate equals the baud rate times the number of bits represented
by each signal unit.
„ Baud rate equals the bit rate divided by the number of bits
represented by each signal unit.
„ Baud rate is less than or equal to the bit rate.
„ Baud rate determines the bandwidth required to send the
signal.

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 Example 1
An analog signal carries 4 bits in each signal unit. If 1000
signal units are sent per second, find the baud rate and the
bit rate
Solution
Baud rate = 1000 bauds per second (baud/s)
Bit rate = 1000 x 4 = 4000 bps

Example 2
The bit rate of a signal is 3000. If each signal unit carries
6 bits, what is the baud rate?
Solution Baud rate = 3000 / 6 = 500 baud/s
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 Carrier Signal
„ In analog signal, the sending device produces a high-
frequency signal that acts as a basis for the
information signal. This base signal is called the
carrier signal or carrier frequency.
„ Receiving device is tuned to the frequency of the
carrier signal that it expects from the sender.
„ Digital information then modulates the carrier signal
by modifying one or more of its characteristics
(amplitude, frequency, or phase). This kind of
modification is called modulation (or shift keying, and
the information signal is called the modulating signal.

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 Figure 5.3 Amplitude Shift Keying (ASK)
„ The strength of the carrier signal is varied to represent binary 1
or 0.
„ Both frequency and phase remain constant while the amplitude
changes.
„ A bit duration is the period of time that defines 1 bit.
„ Peak amplitude of the signal during each bit duration is constant,
and its value depends on the bit (0 or 1).
„ ASK is susceptible to noise interference. Noise refers to
unintentional voltages introduced onto a line by various
phenomena such as heat or electromagnetic induction created by
other sources.

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 Figure 5.4 Relationship between baud rate and bandwidth in ASK

„ Bandwidth of a signal is the total range of frequencies occupied

by that signal.
„ When we decompose an ASK-modulated signal, we get a

spectrum of many simple frequencies.

„ Most significant ones are those between fc - Nbaud/2 and fc +

Nbaud/2 with the carrier frequency fc at the middle.

„ Bandwidth BW = (1 + d)* Nbaud

„ BW is bandwidth, Nbaud is baud rate, and d is a factor related to

the modulation process (with a minimum value of 0).

„ Although there is only one
carrier frequency, the process of
modulation produces a complex
signal that is a combination of
many simple signals, each with
a different frequency.

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 Example 3
Find the minimum bandwidth for an ASK signal
transmitting at 2000 bps. The 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 2000 Hz.

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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
bandwidths in each direction. Assume there is no gap between the
bands in the two directions.

Solution
For full-duplex ASK, the bandwidth for each direction is
BW = 10000 / 2 = 5000 Hz
The carrier frequencies can be chosen at the middle of each band.
fc (forward) = 1000 + 5000/2 = 3500 Hz
fc (backward) = 11000 – 5000/2 = 8500 Hz

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 Figure 5.6 FSK

„ Frequency of the carrier signal is varied to represent binary 1 or

0.
„ Frequency of the signal during each bit duration is constant, and
its value depends on the bit (0 or 1); both peak amplitude and
phase remain constant.
„ FSK avoids most of the problems from noise.
„ Because the receiving device is looking for specific frequency
changes over a given number of periods, it can ignore voltage
spikes.

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 Figure 5.7 Relationship between baud rate and bandwidth in FSK

„ Although FSK shifts between two carrier frequencies, it is easier

to analyse as two co-existing frequencies.
„ FSK spectrum is a combination of two ASK spectra centered on
fc0 and fc1. The bandwidth required for FSK transmission is equal
to the baud rate of the signal plus the frequency shift (difference
between the two carrier frequencies): BW = fc1 – fc0 + Nbaud
„ Although there are only two carrier frequencies, the process of
modulation produces a composite signal that is a combination of
many simple signals, each with a different frequency.

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 Example 6
Find the minimum bandwidth for an FSK signal
transmitting at 2000 bps. Transmission is in half-duplex
mode, and the carriers are separated by 3000 Hz.

Solution
For FSK
BW = baud rate + fc1 − fc0
BW = bit rate + fc1 − fc0 = 2000 + 3000 = 5000 Hz

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 Example 7
Find the maximum bit rates for an FSK signal if the
bandwidth of the medium is 12,000 Hz and the difference
between the two carriers is 2000 Hz. Transmission is in
full-duplex mode.

Solution
Because the transmission is full duplex, only 6000 Hz is
allocated for each direction.
BW = baud rate + fc1 − fc0
Baud rate = BW − (fc1 − fc0 ) = 6000 − 2000 = 4000
But because the baud rate is the same as the bit rate, the
bit rate is 4000 bps.
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 Figure 5.8 PSK

„ The phase of the carrier is varied to represent binary 1 or 0.

„ Both peak amplitude and frequency remain constant as the phase
changes.
„ If we start with a phase of 00 to represent binary 0, then we can
change the phase to 1800 to send binary 1.
„ The phase of the signal during each bit duration is constant, and
its value depends on the bit (0 or 1).
„ 2-PSK or binary PSK: Two different phases (00 and 1800) are
used. Constellation or phase-state diagram shows the relationship
by illustrating only the phases.

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 Figure 5.10 The 4-PSK method

„ PSK is not susceptible to the noise degradation that affects ASK or

to the bandwidth limitations of FSK.
„ Even smaller variations in the signal can be detected reliably by the
receiver.
„ Instead of utilizing only two variations of a signal, each
representing 1 bit, we can use four variations and let each phase
shift represent 2 bits.
„ A phase of 00 now represents 00; 900 represents 01; 1800 represents
10; and 2700 represents 11. This technique is called 4-PSK or Q-
PSK. The pair of bits represented by each phase is called a dibit.
We can transmit data twice as efficiently using 4-PSK as we can
using 2-PSK.

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 Figure 5.12 The 8-PSK characteristics

„ 8-PSK: Instead of 900, we now vary the signal by shifts of 450.

„ With eight different phases, each shift can represent 3 bits (1 tribit)
at a time.
„ 8-PSK is 3 times as efficient as 2-PSK.
„ Minimum bandwidth required for PSK transmission is the same as
that required for ASK transmission.
„ Maximum bit rate in PSK transmission is much greater than that of
ASK.
„ While the maximum baud rates of ASK and PSK are the same for a
given bandwidth, PSK bit rates using the bandwidth can be 2 or
more times greater.

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 Example 8
Find the bandwidth for a 4-PSK signal transmitting at
2000 bps. Transmission is in half-duplex mode.

Solution
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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 Example 9
Given a bandwidth of 5000 Hz for an 8-PSK signal, what
are the baud rate and bit rate?

Solution
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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 Figure 5.14 The 4-QAM and 8-QAM constellations

„ PSK is limited by the ability of the equipment to distinguish small differences in phase.
„ Why not combine PSK, FSK and ASK?
„ Bandwidth limitations make combinations of FSK with other changes practically
useless.
„ Combining ASK and PSK, we could have x variations in phase and y variations in
amplitude, giving us x times y possible variations and the corresponding number of
bits per variation. Quadrature amplitude modulation does that.
„ 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.
„ In 4-QAM and 8-QAM, number of amplitude shifts is fewer than the number of phase
shifts. Because amplitude changes are susceptible to noise and require greater shift
differences than do phase changes, the number of phase shifts used by a QAM system
is always larger than the number of amplitude shifts.

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 Figure 5.16 16-QAM constellations

„ The first example, 3 amplitudes and 12 phases, handles noise best because of a
greater ratio of phase shift to amplitude. It is ITU-T recommendation.
„ The second example, four amplitudes and eight phases, is the OSI
recommendation.
„ It is to be noted that every intersection of phase and amplitude is utilized.
„ In fact, 4 times 8 should allow for 32 possible variations. But by using only one-
half of those possibilities, the measurable differences between shifts are increased
and greater signal readability is ensured. In addition, several QAM designs link
specific amplitudes with specific phases. This means that even with the noise
problems associated with amplitude shifting, the meaning of a shift can be
recovered from phase information.
„ QAM has an advantage over ASK as is its less susceptibility to noise.
„ Minimum bandwidth required for QAM transmission is the same as that required
for ASK and PSK transmission. QAM has the same advantages as PSK over ASK.

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 Figure 5.17 Bit and baud

„ Each frequency shift represents a single bit; so it requires 1200 signal units to send
1200 bits. Its baud rate, therefore, is also 1200 bps.
„ Each signal variation in an 8-QAM system, however, represents 3 bits. So a bit rate
of 1200 bps, using 8-QAM, has a baud rate of only 400.
„ A dibit system has a baud rate of one-half the bit rate, a tribit system has a baud
rate of one-third the bit rate, and a quadbit system has a baud rate of one-fourth the
bit rate.

Bits/ Baud Bit

Modulation Units
Baud rate Rate
ASK, FSK, 2-PSK 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

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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

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 Example 12
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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 „ Data rate limits
„ Data rate depends on three factors:
„ The bandwidth available
„ The levels of signals we can use
„ The quality of the signal (the level of the noise)
„ Noiseless channel
„ Nyquist Bit Rate defines the theoretical Maximum bit rate,
„ BitRate = 2 * Bandwidth * log2L
„ Bandwidth is the bandwidth of the channel
„ L is the number of signal levels used to represent data
„ BitRate is the bit rate in bits per second

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 Example D1
Consider a noiseless channel with a bandwidth of 3000
Hz transmitting a signal with two signal levels. The
maximum bit rate can be calculated as

Bit Rate = 2 × 3000 × log2 2 = 6000 bps

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 Example D2
Consider the same noiseless channel, transmitting a signal
with four signal levels (for each level, we send two bits).
The maximum bit rate can be calculated as:

Bit Rate = 2 x 3000 x log2 4 = 12,000 bps

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 „ Noisy Channel: Shannon Capacity
„ To determine the theoretical digital highest data rate for a
noisy channel
„ Capacity = Bandwidth * log2(1+SNR)
„ Bandwidth is the bandwidth of the channel
„ SNR is the signal-to-noise ratio.
„ Capacity is the capacity of the power of the signal to the power
of the noise.
„ Signal-to-noise ratio is the statistical ratio of the power of
the signal to the power of the noise.
„ There is no indication of the signal level, which means that
no matter how many levels we use, we cannot achieve a
data rate higher than the capacity of the channel.
„ The formula defines a characteristic of the channel, not the
method of transmission.

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 Example D3
Consider an extremely noisy channel in which the value
of the signal-to-noise ratio is almost zero. In other words,
the noise is so strong that the signal is faint. For this
channel the capacity is calculated as

C = B log2 (1 + SNR) = B log2 (1 + 0)

= B log2 (1) = B × 0 = 0

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 Example D4
We can calculate the theoretical highest bit rate of a
regular telephone line. A telephone line normally has a
bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-
to-noise ratio is usually 3162. For this channel the
capacity is calculated as

C = B log2 (1 + SNR) = 3000 log2 (1 + 3162)

= 3000 log2 (3163)
C = 3000 × 11.62 = 34,860 bps

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 Example D5
We have a channel with a 1 MHz bandwidth. The SNR
for this channel is 63; what is the appropriate bit rate and
signal level?

Solution

First, we use the Shannon formula to find our upper

limit.
C = B log2 (1 + SNR) = 106 log2 (1 + 63) = 106 log2 (64) = 6 Mbps
Then we use the Nyquist formula to find the
number of signal levels.
4 Mbps = 2 × 1 MHz × log2 L Î L = 4
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 Figure 5.18 Telephone line bandwidth

„ Traditional telephone lines can carry frequencies between 300 and 3300
Hz, giving them a bandwidth of 3000 Hz.
„ All this range is used for transmitting voice, where a great deal of
interference and distortion can be accepted without loss of intelligibility.
„ Data signals require a higher degree of accuracy to ensure integrity. For
safety’s sake, therefore, the edges of the bandwidth range are not used for
data communications.
„ We can say that the signal bandwidth must be smaller than the cable
bandwidth. The effective bandwidth of a bandwidth line being used for
data transmission is 2400 Hz, covering the range from 600 to 3000 Hz.
„ Today some telephone lines are capable of handling more bandwidth than
traditional lines.
„ A telephone line has a bandwidth of almost 2400 Hz for data transmission.

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 Figure 5.19 Modulation/demodulation

„ Bandwidth defines a baseband nature, which means we need to modulate

if we want to use this bandwidth for data transmission. Devices that were
traditionally used to do so are called 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.
„ The computer on left sends binary data to the modulator portion of the modem; the
data is sent as an analog signal on the telephones lines. The modem on the right
receives the analog signal, demodulates it through its demodulator, and delivers
data to the computer on the right. The communication can be bidirectional, which
means the computer on the right can also send data to the computer on the left
using the same modulation/demodulation processes.

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 Figure 5.20 The V.32 constellation and bandwidth

„ Most popular modems available are based on the V-series standards

published by ITU-T.
„ V.32
„ Uses a combined modulation and encoding technique called trelliscoded
modulation.
„ Trellis is essentially QAM plus a redundant bit.

„ Data stream is divided into 4-bit sections. Instead of quadbit, however, a

pentabit (5-bit pattern) is transmitted. The value of the extra bit is calculated
from the values of the data bits.
„ In any QAM system, the receiver compares each received signal point to all
valid points in the constellation and selects the closest point as the intended
value. A signal distorted by transmission noise can arrive closer in value to an
adjacent point than to the intended point, resulting in a misidentification of the
point and an error in the received data. The closer the points are in the
constellation, the more likely it is that transmission noise can result in a
signal’s being misidentified. By adding a redundant bit to each quadbit, trellis-
coded modulation increases the amount of information used to identify each
bit pattern and thereby reduces the number of possible matches. For this
reason, a trellis-encoded signal is much less likely than a plain QAM signal to
be misread when distorted by noise.
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 Figure 5.21 The V.32bis constellation and bandwidth

„ The V.32 calls for 32-QAM with a baud rate of 2400. Because only 4 bits
of each pentabit represents data, the resulting speed is 4 * 2400 = 9600bps.
„ V.32bits
„ First introduced by ITU-T to support 14,400-bps transmission.
„ V.32bis uses 128-QAM transmission (7 bits/baud with 1 bit for error control)
at a rate of 2400 baud (2400*6 = 14,400 bps)
„ V.32bis includes an automatic fall-back and fall-forward feature that enables
the modem to adjust its speed upward or downward depending on the quality
of the line or signal.
„ V.34bis: Provides a bit rate of 28,800 with a 960-point constellation to a bit rate of
33,600 with a 1664-point constellation.
„ V.90: Bit of 56,000bps, called 56K modems.
„ These modems may be used only if one party is using digital signature.

„ Supports asymmetric [downloading rate is 56 Kbps; uploading rate is

33.6Kbps].

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 Figure 5.22 Traditional modems

„ After modulation by the modem, an analog signal reaches the telephone

company switching station, where it is sampled and digitized to be passed
through the digital network. The quantization noise introduced into the
signal at the sampling point limits the data rate according to the Shannon
Capacity. This limit is 33.6 Kbps.
„ 56K Modems:
„ Uploading: Analog signal must still be sampled at the switching

station, which means the data in uploading is limited to 33.6 Kbps.

„ Downloading: No sampling involved. Signal is not affected by

quantization noise and not subject to the Shannon capacity limitation.

So, downloading is limited to 56 Kbps.

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 Figure 5.23 56K modems

„ 56 Kbps because telephone companies sample 8000 times per second

with 8 bits per sample. One of the bits in each sample is used for
control purposes, which means each sample is 7 bits. The rate is
therefore 8000 * 7 or 56000 bps or 56 Kbps
„ V.92
„ Modems can adjust their speed, and if the noise allows, they can

upload data at the rate of 48 Kbps. The downloading rate is still 56

Kbps.
„ Modem can interrupt the Internet connection when there is an

incoming call if the line has call-waiting service.

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 „ Traditional modems impose an upper limit on the data rate available.
„ DSL Technology
„ DSL provide higher-speed access to the Internet over the existing local
loops. Set of technologies: ADSL, VDSL, HDSL and SDSL.
„ ADSL: Asymmetric DSL
„ Provides higher speed (bit rate) in the downstream direction (from the Internet
to the resident) than in the upstream direction (from the resident to the
Internet). So, it is called as asymmetric.
„ Designed for residential users; it is not suitable for businesses.
„ ADSL provides data rate more than traditional modems.
„ The existing local loops can handle bandwidths up to 1.1 MHz but the filter
installed at the end of the line by the telephone company limits the bandwidth
to 4 KHz (sufficient for voice communication). This was done to allow the
multiplexing of a large number of voice channels.
„ Unfortunately, 1.1 MHz is just the theoretical bandwidth of the local loop.
Factors such as the distance between the residence and the switching office,
the size of the cable, the signaling used, and so on affect the bandwidth.
„ Designers are aware of this problem and used an adaptive technology that
tests the condition and bandwidth availability of the line before setting on a
data rate.
„ The data rate of ADSL is not fixed; it changes based on the condition and type
of the local loop cable.

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 Discrete Multitone Technique- DMT
„ Modulation technique of ADSL that combines QAM and FDM.
„ Each system can decide on its bandwidth division.
„ Voice: Channel 0 is reserved for voice communication.
„ Idle: Channel 1-5 are not used to allow a gap between voice and data
communication.
„ Upstream data and control: Channel 6 - 30 (25). One channel for
control and 24 for upstream data, each using 4KHz.
„ Upstream bandwidth = 24*4000*15 = 1.44Mbps

„ Downstream data and control: Channel 31 – 255 (225). One channel

for control and rest are for downstream data.
„ Downstream bandwidth = 224*4000*15 = 13.4 Mbps

„ Actual bit rates are much lower than the above-mentioned rate:
„ Upstream = 64 Kbps to 1 Mbps

„ Downstream = 500 Kbps to 8 Mbps

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 ADSL
„ Customer site: ADSL Modem
„ Local loop connects to the filter which separates voice and data

communication.
„ ADSL modem modulates the data, using DMT, and creates

downstream and upstream channels.

„ Telephone company site: DSLAM
„ Instead of an ADSL modem, a device called a digital subscriber line

access multiplier (DSLAM) is installed that functions similar to ADSL

„ DSLAM packetizes the data to be sent to the Internet (ISP Server)

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 Other DSL Technologies
„ SDSL – Symmetric DSL
„ Suitable for businesses.
„ It divides the available bandwidth equally between down and up streams.
„ HDSL – High-Bit-Rate DSL
„ Alternative to T-1 Lines (1.544 Mbps).
„ T-1lines use AMI (Alternate mark inversion) encoding which is very
susceptible to attenuation at high frequencies. This limits the length of a T-
1 line to 1 km. For long distances, we need repeaters.
„ Uses 2B1Q encoding, which is less susceptible to noise.
„ Provides data rate of almost 2 Mbps and upto 3.6 km without repeater
„ Use two twisted-pair wires for full-duplex transmission
„ VDSL - Very-high-bit-rate DSL
„ Similar to ADSL. Use coaxial, fiber-optic, or twisted-pair cable for short
distance (300 to 1800m).
„ Modulation technique used is DMT with bit rate of 50 to 55 Mbps
downstream and 1.5 to 2.5 Mbps upstream.
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