Basic Concepts

• Line Configuration
• Topology
• Transmission Mode
• Categories of Networks
• Internetworks

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 Point-to-Point Line Configuration

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 Point-to-Point Line Configuration

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 Point-to-Point Line Configuration

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 Multipoint Line Configuration

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 Mesh Topology

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 Star Topology

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 Tree Topology

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 Bus Topology

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 Ring Topology

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 Hybrid Topology

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 Simplex

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 Half-Duplex

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 Full-Duplex

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 Local Area Network

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 Local Area Network

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 Metropolitan Area Network

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 Wide Area Network

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 Internetwork
(Internet)

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 OSI Model

• The model
• Functions of the layers

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 OSI Model

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 OSI Layers

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 An Exchange Using the OSI Model

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 Physical Layer

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 Data Link Layer

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 Data Link Layer Example

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 Network Layer

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 Transport Layer

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 Session Layer

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 Presentation Layer

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 Application Layer

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 Summary of Layer Functions

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 Signals

• Analog and digital

• Aperiodic and periodic signals
• Analog signals

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 Transformation of Information
to Signals

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 Analog and Digital Clocks

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 Analog and Digital Signals

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 Periodic Signals

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 Aperiodic Signals

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 Sine Wave

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 Phases

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 Amplitude Change

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 Frequency Change

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 Phase Change

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 Time and Frequency Domain

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 Examples

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 Signal with DC Component

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 Digital Signal

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 Amplitude, Period, and Phase
for a Digital Signal

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 Bit Rate and Bit Interval

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 Bandwidth and Data Rate

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 Encoding
• It deals with the basic encoding and
modulation technology used in the Data
communication and Networking

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 Different Conversion Schemes

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 Encoding

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 Types of Digital to Digital
Encoding

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 Unipolar Encoding

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 Types of Polar Encoding

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 NRZ-L and NRZ-I Encoding

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 RZ Encoding
• Return to zero uses three values: positive,
negative, Zero
• In this the bit changes not between bits
but during each bit.
• A bit 1 is actually represented by positive
to zero and bit 0 is represented negative
to zero
• This concept is as explained in following
diagram

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 RZ Encoding

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 Manchester and Differential
Manchester
• Manchester encoding uses the inversion at middle of
each bit interval for both synchronization and bit
representation
• Negative to positive transition represent binary 1 and
positive to negative represent binary 0
• Differential Manchester uses inversion at the middle of
bit for synchronization , but presence or absence of an
additional transition at the beginning of interval is used
to identify the bit
• A transition means binary 0 and no transition mean
binary 1
• It require two signal changes to represent binary 0 but
only one to represent 1
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 Manchester and Diff. Manchester
Encoding

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 Bipolar Encoding
• Bipolar encoding uses three level of voltages: positive
negative and zero
• The zero level in this is used to represent binary 0
• The 1’s are represented by alternating positive and
negative voltage
• If the first bit is represented by positive amplitude the
next is represented by negative amplitude and so on
• The alternation occur even when the 1 bits are not
consecutive

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 Types of Bipolar Encoding

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 Bipolar Alternate Mark Inversion (AMI)
• In this the name alternate mark inversion , the word mark
means 1
• AMI means alternate 1 inversion, zero voltage represent
binary 0
• A variation to these system is termed as pseudoternary,
in which binary zero alternates between positive and
negative voltages
• The advantage are :I) due to inversion DC component is
zero and II) a long sequence of 1’s remain synchronized
• The mechanism is as shown in following diagram

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 Bipolar Alternate Mark Inversion (AMI)

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 Bipolar 8 Zero Substitution (B8ZS)
• It is convention adopted in the North America to provide
Synchronization of long strings of zero
• It is identical to bipolar AMI. Bipolar AMI changes poles
with every 1 its encounter, this change provide
synchronization needed by receiver
• The signal does not change during strings of 0’s, so
synchronization is often lost
• Whenever eight or more 0’s are encountered in the data
stream B8ZS is applied
• The solution is force artificial signal change called
Violation within the Zero Strings

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 Bipolar 8 Zero Substitution (B8ZS)
• Every time consecutive eight 0’s occur in succession, B8ZS
introduce change in pattern based on the polarity of previous
1
• If the previous 1 bit was positive, the eight 0’s are encoded as
zero, zero, zero, positive, negative, zero, negative, positive.
• The receiver is looking for alternating polarity to identify 1’s,
but when it finds two consecutive positive charges surrounded
by three 0’s it recognizes the pattern is deliberately introduce
violations and not an error
• It then looks for second pair of expected violation, on finding
it invert all 8 bit to 0 and revert back to Bipolar AMI
• If the first polarity is negative the pattern of violation is also
inverted
• The pattern is as shown

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 Bipolar 8 Zero Substitution (B8ZS)

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 High DensityBipolar3
• The problem of synchronizing string of consecutive 0’s is
solved differently in the Europe and Japan
• In this the every time consecutive four 0’s are
encountered violation are introduce instead of waiting
for eight 0’s to occur
• In this the violation pattern is based on polarity of
previous bit , but also checks for number of 1’s occurred
since last substitution
• If the number of 1’s since last substitution is odd, it puts
violation in place of fourth consecutive zero
• If polarity of previous bit is positive the violation is
positive ,If it is negative violation is negative

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 High DensityBipolar3
• Whenever the number of 1’s since last substitution is
even, then violation is put in the place of both first and
fourth consecutive 0’s
• If polarity of previous bit was positive , both violation
are negative
• If polarity of previous bit is negative then both violation
are positive
• The violation are used to synchronize the system
• All are four pattern are as shown

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 High DensityBipolar3

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 Problem based on B8ZS and HDB3
• Using B8ZS encode the bit stream
10000000000100.Assume that the polarity
of first 1 is positive

• Using HDB3 encode the bit stream

10000000000100. Assume the number of
1’s so far is odd and first 1 is positive

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 Solution

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 Analog to Digital Encoding
• It is sometime required to digitized the analog signal. To
send human voice over long distance the signal is
digitized since it is less prone to noise
• Analog to digital conversion require a reduction of
potentially infinite number of values in analog message
so that it can be represented as digital stream with
minimum loss of information
• In this method a continuous wave from is represented
as series of digital pulses
• The basic problem is not transmission but how to
translate information from infinite number of values to
discrete number of values

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 Analog to Digital Encoding

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 Pulse Amplitude Modulation (PAM)
• This is the first step of A/D conversion
• This technique takes an analog signal, samples it, and
generate a series of pulses based on result of sampling
• Sampling means measuring the amplitude of signal at
equal interval
• This method is foundation for PCM
• The original signal is sampled at equal interval using the
Sample and Hold technique
• In this at any given moment the signal level is read and
then held briefly
• The entire waveform is as shown in following diagram

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 Pulse Amplitude Modulation (PAM)

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 Pulse code Modulation
• PCM Modifies the pulses created by PAM to create a
completely digital signal

• PCM first quantizes the PAM pulses

• Quantization is method of assigning integral value in

specific range to the sample instance

• The result of quantization is presented in following graph

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 Quantized PAM Signal

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 Pulse Code Modulation
• The next figure shows method of assigning sign and
magnitude value to quantized sample
• Each value is translated into seven bit binary equivalent.
The eight bit represent the sign
• The binary digit are then transformed in to digitial signal
using one of D/D method
• The next diagram shows PCM of original signal encoded
in to unipolar signal
• PCM is made up of four different processes 1) PAM
2) Quantization
3) Binary Encoding
4) Digital to Digital encoding

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 Quantizing Using
Sign and Magnitude

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 Pulse Code Modulation

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 From Analog to PCM

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 From Analog to PCM

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 Analog to PCM

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 Analog to PCM

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 SAMPLING RATE
• The accuracy of any digital reproduction of analog signal
depends upon the number of sample taken. PAM and
PCM can produce the wave form by exactly taking
infinite number of samples.
• But the question is how many samples are sufficient?
• It requires remarkably little information by the for
receiving device to reconstruct analog signal.
• According to Nyquist theorem to ensure the accurate
reproduction of original analog signal using PAM the
Sampling Rate must be twice the highest frequency of
original signal
• So sample telephone voice with max frequency of
4000Hz , need sampling rate of 8000 samples/second

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 Nyquist Theorem

A sampling rate of twice the frequency of x HZ means the

signal must be sampled every ½ x second
That is one sample every 1/8000 second
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 How many Bits per Sample?
• After sample rate is determined we need to determine the
number of bits transmitted for each sample
• This depends on level of precision needed
• The number of bit are chosen such that the original signal
can be reproduce with desired precision in amplitude

• The Next Question is Bit rate?

• It is calculated by formula given as

Bit rate=Sampling rate X Number of bits Per Sample

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 Problems
• What sampling rate is needed for a signal with a
bandwidth of 10,000 Hz?(1 to 11)
• Sampling rate= 2(11000) =22000 samples/seconds
• A signal is sampled. Each signal requires at least 12 level
of precision (+0 to +5 and -0 to -5).how many bit
should be sent for each sample?
• We need 4 bits. One for sign and three for values
• A three bit value can represent 000 to 111 which is more
than what we need
• A 2 bit value is not enough and 4 bit value is too much
• So the factor is decided depending on original signal

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 Problems
• We want to digitize the human voice. What is the bit
rate assuming eight bits per sample?
• Human voice normally contain frequencies from 0 to
4000Hz. So sapling rate is
• Sampling Rate= 4000 * 2 =8000 sample/second
• Bit rate is calculated as
• Sampling rate * Number of Bits per sample
• 8000 * 8=64000bits/sec=64Kbps

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 Digital to Analog Encoding
• The digital to analog conversion is the process of
changing one of the characteristics of an analog signal
based on the information in the digital signal
• When data is transformed from computer to another
computer across a public access for example telephone
line then data must be converted
• The digital data must be modulated to the analog signal
• The entire process is explained as shown in next figure

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 Digital to Analog Encoding

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 Digital to Analog Encoding
• Sine wave is represented by three characteristic :
Amplitude, Frequency, Phase
• When any one of the characteristic vary we create the
second version of sine wave. Just by changing the one
aspect of electrical signal , we use it to represent the
digital data
• The above mentioned three characteristics are varied and
we have
• Amplitude Shift Keying (ASK)
• Frequency Shift Keying (FSK)
• Phase Shift Keying (PSK)
• Quadrature amplitude Modulation (QAM) include
amplitude and phase Change
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 Types of D/A Methods

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 Bit Rate and Baud Rate
• D/A has two basic issue Bit Rate and Baud Rate
• Bit Rate is number of bit transmitted per second
• Baud rate refer to number of signal units per second that are required
to represent those bits
• Bit rate equals the Baud rate times the number of bits represented by
each signal
• The Baud rate equals the bit rate divided by number of bits
represented by each signal shift
• A Baud is analogous to Car, while bit analogous to passenger
• A Car can carry 1 or more passenger. If 1000 car having 1 passenger
each go from one point to another point then 100 passenger are
transported
• Now if each car has 4 passenger , then 4000 are transported.
• Note number of car and not number of passenger determine traffic.
Same number of Baud determine Bandwidth not the bits

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 Problems
• An analog signal carries 4 bit in each signal elements. If
1000 signal elements are sent per second , find the baud
rate and bit rate
• Baud rate= number of signal elements=1000 bauds/sec
• Bit rate= Baud rate * number of bits per signal elements
• =1000 *4 =4000bps
• The bit rate of signal is 3000. if each signal elements
carries six bits , what is baud rate?
• Baud rate= bit rate/ Number of bits per signal
=3000/6=500 bauds/sec

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 Carrier Signals
• In analog transmission the sending device produce a
high frequency signal that act as basis for information
signal
• The base signal is called as carrier signal
• The receiving device is tuned to the frequency of carrier
signal that it expects from sender
• Digital information is modulated on the carrier
signal/frequency
• This type of modification is termed as modulation and
information is called modulating Signal

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 AMPLITUDE SHIFT KEYING
• The strength of the carrier signal is varied to represent
binary 1’s and 0’s. Both frequency and phase remain
constant
• Which voltage represent 1 and 0 is system designer
decision
• A bit duration is period of time that defines 1 bit. The
peak amplitude of each bit duration is constant and its
value depend upon the bit
• The speed of transmission is limited by physical
characteristic of transmission medium
• Conceptual view of ASK is as follows

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 AMPLITUDE SHIFT KEYING

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 AMPLITUDE SHIFT KEYING
• This method is highly susceptible to noise, which refers to
unintentional voltages introduce in to line by heat or
electromagnetic induction created by other source
• Noise always effect amplitude
• Bandwidth of a signal is the total range of frequencies
occupied by the signal
• When we decompose the ASK Modulated signal we get
spectrum of many simple frequency
• The most significant are between fc – Nbaud /2 and fc +
Nbaud/2
• This as shown in following diagram

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 Bandwidth for ASK

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 AMPLITUDE SHIFT KEYING
• Bandwidth requirement are calculated as follows
• BW=(1+d) * Nbaud
where
BW --- Bandwidth
Nbaud --- Baud Rate
d ---- factor related to condition of line

• Find minimum Bandwidth for an ASK signal transmitting at

the rate 2000bps transmission mode half duplex
• In ASK baud rate and bit rate is same so banwidth
requirement will be 2000 hz

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 Problems
• Given a bandwidth of 5000 Hz for an ASK signal what is baud rate
and Bit rate?
• In ask the baud rate is same as that of bandwidth means 5000.But
because the baud rate and bit rate are same the bit rate is 5000
bps.
• Given a bandwidth of 10000 Hz ( 1000 to 11000) draw full duplex
ASK diagram of the system. Find the carrier and Bandwidth in each
Direction. Assume there is no gap between the band in two
direction
• BW=10000/2=5000 Hz
• The Carrier Frequency can be chosen between as middle of each
band
fc(forward)=1000+5000/2=3500 Hz
fc(backward)= 11000-5000/2=8500 Hz
• The wave form is as shown

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 FREQUENCY SHIFT KEYING
• The frequency of carrier signal is varied to represent
binary 0’s and 1’s
• The frequency of signal during each bit is constant and
value depends on the bit, both peak amplitude and phase
remain constant
• Conceptual view is as shown in figure
• It avoid noise problem of ASK, because the receiving
device is looking for specific frequency change over a
given periods, it can ignore voltage spikes
• The limiting factor are physical characteristics of carrier

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 FREQUENCY SHIFT KEYING

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 Bandwidth for FSK
• FSK shift between two carrier frequency, it is easier to
analyze as two coexisting frequency
• FSK spectrum is combination of two ASK spectra centered
around fc0 and fc1
• The bandwidth required for spectra is equal to baud rate
of signal plus the shift in frequency
• This concept is as shown in following figure

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 Bandwidth for FSK

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 Problems
• Find the minimum bandwidth for FSK signal transmitting at 2000
bps. Transmission is half duplex and carrier must be separated by
3000 Hz
• If fc0 and fc1 are carrier frequency then
BW= Baud rate + (fc1 –fc0 )
= 2000 + 3000 =5000 Hz
• Find the maximum bit rate for FSK signal if bandwidth of medium is
12000 Hz and difference between two carrier must be at least 2000
Hz. Transmission is in full duplex
• Because the transmission is full duplex, only 6000 Hz for each
direction
fc1 and fc0 are the carrier frequency
BW= Baud rate + ( fc1 – fc0)
Baud Rate = BW – (fc1 – fc0)= 6000-2000= 4000 Hz
But because the baud rate is same as the bit rate , the bit rate is
4000bps

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 PHASE SHIFT KEYING

• The phase of carrier signal is varied to represent binary 0’s and 1’s
• The phase of signal during each bit is constant and value depends on
the bit, both amplitude and frequency remain constant

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 PHASE SHIFT KEYING
• The above method is always often called 2-PSK or
binary PSK because two different phase are used (0
&180 degree)
• The next diagram termed as constellation or phase
state diagram shows same relationship by
illustrating only phases

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 PHASE SHIFT KEYING
• This method is not susceptible to noise nor has bandwidth
limitation which means smaller variation in signal can be
easily detected by the receiver
• Instead of utilizing two variation of signal, each
representing one bit, we can use four variation and let
each phase shift represent two bit
• The constellation diagram for same is as shown
• A phase of 0 degree represent 00, 90 degree represent
01,180 degree represent 10 and 270 degree represent 11
• The technique is termed as 4-PSK or Q-PSK
• The pair of bit represented by each phase is called as
dibits

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 4-PSK

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 4-PSK
Characteristics

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 8-PSK
Characteristics

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 Bandwidth for PSK

• The minimum bandwidth required for PSK is same as that

of FSK

• The maximum bit rate in PSK transmission is potentially

much greater than that of ASK

• While the maximum baud rate of ASK and FSK are same
of given Bandwidth ,PSK bit rate using same bandwidth
can be two or more times greater

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 PSK
Bandwidth

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 QUADRATURE AMPLITUDE MODULATION
• All the methods seen so far has some problems associated
• Again since analog signal has three properties we had made variation
to one of them at a time
• In this method we will combine ASK and PSK so that we have x
variation in phase and y variation in amplitude giving us x times y
possible variation and corresponding number of bits per variation
• Quadrature modulation does same
• The term quadrature is derived from the restriction required for
minimum performance and is related to trigonometry
• There are many variation possible. The next figure shows two possible
configuration 4-QAM and 8-QAM
• In both cases the amplitude change is fewer as compared to number
of phase shifts
• Since amplitude changes are susceptible to noise and require greater
shift difference than do phase changes ,the number of shift change
used are more
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 4-QAM and 8-QAM
Constellations

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 8-QAM Signal

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 16-QAM
Constellation

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 Bit Rate and Baud Rate

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 BIT and BAUD rate
• Assuming that an FSK signal over voice grad phone line
can send 1200 bps. Each frequency shift represent a
single bit so it require 1200 signal element to send 1200
bits
• So the baud rate is 1200
• Each signal variation in an 8-QAM system represent 3 bits
• So a bit rate of 1200bps using 8-QAM has Baud rate is
400
• The variation for bit, dibit, tribit, and quadbit will have
½,1/3, ¼ baud rate
• The same concept is shown in following figure

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 Bit Rate and Baud Rate

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 BIT and BAUD rate Comparision
Modulation Units Bits /Band Baud Rate Bit Rate
ASK,FSK,2-PSK BITS 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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 Problems
• A constellation diagram consist of eight equally spaced
points on the circle .if the bit rate is 4800 bps what is
baud rate?
• The constellation indicate 8 –PSK with the points 45
degree apart since 23=8 three bits are transmitted with
each signal therefore baud rate is
4800/3=1600 bauds
• Compute the bit rate for a 1000 baud 16-QAM signal.
• A 16-QAM signal mean that there are four bits per signal
elements since 24=16 thus
(1000)(4)=4000 bps
• Compute the baud rate for a 72000 bps 64-QAM
• A 64-QAM signal means that there are six bit per signal
elements 26=64 thus
72000/6= 12000 bauds

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 Analog to Analog Modulation

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 Amplitude Modulation
• In this transmission the carrier signal is modulated so
that it amplitudes varies with changing amplitude of
modulating signal
• Only amplitude changes to follow variation, phase and
frequency remain constant
• The modulating frequency becomes an envelope to
carrier
• The BANDWIDTH of AM signal is equal to twice the
bandwidth of modulating signal and covers a range
centered around the carrier frequency

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 Amplitude Modulation

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 AM Bandwidth

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 Amplitude Modulation
• The bandwidth of audio signal (speech and music) is
usually 5 KHz therefore an AM station needs a minimum
bandwidth of 10 KHz.
• AM station are allowed carrier frequency anywhere
between 530 and 1700 KHz
• However each frequency must be separated from those
on either side by at least 10 KHz to avoid interference
• If one station carriers carrier frequency of 1100 KHz,
then next station carrier’s frequency can not be lower
than 1100 KHz

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 AM Band Allocation

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 Frequency Modulation
• In this the frequency of carrier signal is modulated to
follow changing voltage level of modulating signal
• The peak amplitude and phase of carrier signal remain
constant
• As the amplitude of information signal changes the
frequency of carrier changes accordingly
• The relationship between modulating signal carrier signal
and resultant FM signal is as shown

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 Frequency Modulation

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 Bandwidth and allocation
• The bandwidth of FM signal is equal to 10 times
bandwidth of modulating signal and covers a range
centered around carrier frequency
• The bandwidth of audio signal broadcast in stereo is 15
KHz so it’s need a bandwidth of 150 KHz
• Each FM station needs a minimum bandwidth of 200 KHz
• FM station are allowed carrier frequency anywhere
between 88 and 108 MHz
• Station must be separated by at least 200 KHz

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 FM Bandwidth

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 FM Band Allocation

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 Transmission of
Digital Data
• Once information is encoded the next step is
transmission process
• Information processing equipment generate encoded
signals but require assistance to transmit those signal
over a communication link
• How do we relay encoded data from generating device
to the receiving device?
• We use INTERFACE
• An interface link two devices not necessarily made by
same manufacturer, it’s characteristic must be defined
and standard must be establish

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 Transmission of
Digital Data
• Characteristics of interface include

• Mechanical specification (how many wires are used to

transport signal )

• Electrical specification (frequency amplitude and phase

of expected signal)

• Functional specification (if multiple wire are use what

does each one do)

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 Types of Transmission

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 Parallel Transmission
• Binary data may be organized in to group of n bits
• Computer generates data in group of bits, so by
grouping we can send data n bits at a time instead of
one
• This parallel transmission. The mechanism used for this
is simple use n wires to send n bits of data
• So each bit has its own wire and all n bits of one group
can be transmitted with each clock pulse from one
device to another.
• Next figure depict parallel transmission. The n=8 bits are
represented, all eight wire are bundled in the cable with
connector at each end
• The basic advantage is speed , while disadvantage is
cost that is associated

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 Parallel Transmission

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 Serial Transmission
• In this one bit follows other so we need only one
communication channel to transmit data between two
communication devices
• The advantage is that only one communication channel
is required , which reduces cost of transmission over
parallel roughly by factor n
• Communication within device is parallel, conversion
device are required at the interface between sender and
the line (parallel to serial) and between line and
receiver (serial to parallel)
• This transmission occur in two type
• Synchronous
• Asynchronous
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 Serial Transmission

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 Asynchronous Transmission
• It is named so because timing of signal is unimportant
• Information is received and translated by agreed upon
patterns
• As long as the pattern is followed the receiver can
retrieve information without regard to rhythm in which
they are sent
• Patterns are based on grouping the bit stream in to
bytes
• Each group usually eight bits are sent along the link as
unit
• The sending system handles each group independently
without regard of timer

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 Asynchronous Transmission
• Without synchronizing pulse the receiver cannot use
timing to predict when the next group will arrive
• In order to alert receiver that a new group has arrived
an extra bit is added at start of byte generally 0 termed
as “start bit”
• Also to let receiver know that the byte has ended an
additional bit usually 1 is added at the end termed as
“stop bit”
• By these method each byte is increased to 10 bits of
which 8 bit is information
• In addition the transmission of byte can be followed by
gap of varying duration

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 Asynchronous Transmission
• The start and stop bit along with the gap allows each
byte to synchronize with the data stream
• The mechanism is termed as asynchronous because at
byte level sender and receiver are not synchronized
• Within each byte there is synchronization
• When the receiving detect a start bit it sets timer and
begins counting bits as they come
• After n bits receiver looks at the stop bits
• As soon as it detects stop bit it ignores any received
pulse until it detect start bit

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 Asynchronous Transmission

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 Synchronous Transmission
• The bit streams are combined in to longer frames which
may contain multiple bytes

• Each byte is introduce in to transmission line without

gap between it and next one

• It is left to the receiver to separate bit stream in to bytes

for the purpose of decoding

• The timing is most important in this transmission

• Advantage is speed and more useful for high

transmission medium requirement
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 Synchronous Transmission

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 DTE-DCE interface
• This is important part of interface which is termed as
DATA TERMINAL EQUIPMENT and DATA CIRCUIT
TERMINATING EQUIPMENT
• There are usually four basic functional unit involved in
the communication of data
• A DTE and DCE at one end and DCE and DTE at other
end
• This concept is as shown in the figure
• The DTE generate data and passes it to DCE along with
some control Characters
• DCE convert signal to a format appropriate to the
transmission medium and introduce it on to network link

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 DTEs and DCEs

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 DTE-DCE interface

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 DATA TERMINAL EQUIPMENT (DTE)
• It includes any unit that function as source or
destination of binary digital data
• At physical layer it can be terminal computer , printer,
Fax Machine
• They do not communicate directly with Each other
• They generate and communicate information
• Its like the brain that work we can not transmit what we
think directly to our friend mind
• It need conversion from one from to another form

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DATA CIRCUIT TERMINATING EQUIPMENT (DCE)
• It includes any functional unit that transmit or receives
data in form of analog or digital signal through a
network
• At physical layer DCE take data generated by DTE
convert them in to appropriate signal and introduce it on
telecommunication link
• In any network DTE generates digital data, passes them
to DCE
• DCE convert the data to the form acceptable by
transmission medium and sends converted signal to
another DCE on network

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 EIA-232 (RS-232) Standard
• This is an important standard defined by the Electronics
Industries Association (EIA)
• It defines mechanical, electrical and functional
characteristic of interface between DTE and DCE
• MECHANICAL SPECIFICATION
• It defines interface as a 25 wire cable with male and
female pin connecter attached to both ends
• The length does not exceed 15 meters
• The EIA 232 standards call for 25 wire cable terminated at
one end by male connecter while at other end by female
connector
• Male connector means each wire in cable connecting pin
• Female connector means receptacle with each wire
connecting to a metal tube
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 EIA-232 (RS-232) Standard
• ELECTRICAL SPECIFICATION
• IT DEFINES VOLTAGE LEVELS AND TYPE OF DATA TO BE
TRANSMITTED IN BOTH DIRECTION
• Sending data : All data must be transmitted as 0 and 1
(called space and mark) using NRZ-L encoding with 0 as
+ve and 1 as –ve voltage
• The amplitude of data must fall within the specific range
that is +3 to +15 and -3 to -15 so that it is recognized at
other end
• It also avoid noise problem
• The concept is as shown

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 Sending Data

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 EIA-232 (RS-232) Standard
• CONTROL AND TIMING

• Only 4 wire out of 25 are use for data function

• The remaining 21 are reserved for the various other

function

• Any of other function is considered ON of it transmit a

voltage of at least +3 and OFF if the value is less

• A positive voltage means ON while negative voltage means

OFF

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 Control

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 EIA-232

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 Data Pins

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 Control Pins

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 Timing Pins

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 Other Pins

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 Synchronous Full-Duplex
Transmission

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 Synchronous Full-Duplex
Transmission

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 Modem

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 Null Modem

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 Transmission Media
• The signals are transmitted from one device to another
device in the form of electromagnetic energy

• Electromagnetic signal can travel through vacuum, air,

or other transmission media

• Electromagnetic energy is combination of electrical and

magnetic field vibrating in relation to each other include
power voice radio waves infrared light ultraviolet light
cosmic rays

• It is as shown in next figure

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 Electromagnetic Spectrum

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 Types of Media

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 Guided Media

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 Guided Media
• This are those that provide a conduit from one device to
another device
• The types are as mentioned twisted pair, coaxial cable
and fiber optics
• A signal traveling along these media is directed and
contained by physical limits of the medium
• Twisted pair and coaxial cable use metallic conductor
that accept and transport signal in the form of electrical
current
• Fiber optics accept and transport signal in the form of
light

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 Twisted-Pair Cable
• Twisted pair comes in two different type : Unshielded
and shielded
• Unshielded is the most common used in the telephone
system since it carries the data and voice due the large
frequency range it has
• The twisted pair consist of two conductor, each with its
own color plastic insulator
• The plastic insulation is color banded for identification
• Colors are used both for identification as well as to
indicate which wire belong in pairs and how they relate
to other pairs in bundle

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 Twisted-Pair Cable

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 Effect of Noise on Parallel Lines

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 Noise on Twisted-Pair Lines

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 Unshielded Twisted-Pair Cable

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Advantages and Category of UTP
• It is cheap and easy to use
• Cheap flexible easy to install and used in Ethernet and token
ring
• Depending on cable quality categories are defined as follows
• Category 1: used in telephone system level of quality is fine
for voice but inadequate for low speed data communication
• Category 2: suitable for voice and data transmission up to 4
Mbps
• Category 3: requires at least 3 twist per foot and used for
data transmission of up to 10 Mbps
• Category 4: require at least 3 twist per foot and transmission
up to 16 Mbps
• Category 5 : used for Data transmission up to 100 Mbps

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 UTP Connectors
• Avoids cross talk due to pin penetration and avoid noise level

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 Shielded Twisted-Pair Cable
• It has a metal foil or braided mesh covering that encases each pair
of insulated conductor
• The metal casing prevent the penetration of electromagnetic noise
and eliminates phenomenon called cross talk

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 Coaxial Cable

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 Critical Angle

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 Reflection

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 Propagation Modes

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 Propagation Modes
• Current technology require two mode for propagating
light along optical channel each requiring fiber with
different physical characteristics
• Multimode is so named because multiple beams from
light source move through the core in different paths
• How the move within cable depends upon the structure
of cable
• In multimode step index fiber the density of core remain
constant from centre to the edges
• A beam of light moves through this constant density in
the straight line until it reaches interface of core and the
cladding
• At interface there is abrupt change to lower density that
alter the angle of beam motion

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 Multimode Step-Index

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 Multimode Graded-Index

High density at centre

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 Single Mode

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 Unguided Media
• Radio Communication Band

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 Propagation of Radio Waves
• It utilizes 5 type of propagation
• Surface , topospheric, ionospheric , line of sight and space
• In surface the radio waves travel through the lowest portion
of atmosphere
• Signal emanates in all direction from transmitting antenna and
follow curvature of planet
• Distance depend upon the power of signal

• In Tropospheric a signal can be directed in straight line from

antenna to antenna or it can be broadcast into the upper layer
of troposphere where it is reflected back to earth atmosphere
• First method require setting of receiver and sender whereas
second method allow greater distance to be covered

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 Propagation of Radio waves

• Ionospheric propagation allow higher frequency radio wave

radiate upward in ionosphere where they are reflected back to
earth
• The density difference between troposphere and ionosphere
cause the radio wave to speed up and change direction
backing to earth
• LINE of sight require both sender and receiver to be in one
line without any hurdle in between
• Space propagation uses utilizes satellite relays in place of
atmospheric reflection

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 Propagation Types

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 Cellular System

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 Transmission Impairment
• The imperfection cause impairment in the signal sent through
the transmission medium
• Three types of impairment exists
• Attenuation Distortion Noise
• Attenuation means loss of energy
• When a signal passes through the medium it loses energy to
overcome the resistance of the medium
• That is the reason why wire carrying signal gets warm
• To compensate for loss amplifiers are used
• To show whether a signal has loss or gained strength concept
of decibel is used
• Decibel measure signal at two different points
• dB=10 log10(P2/P1) P1 and P2 power of signal at 1 and 2

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 Transmission Impairment
• Distortion means the signal changes it shape or form
• It occur in composite signal which is made up of different
frequency
• Each signal component has it’s own propagation speed
through the medium and therefore its own delay in arriving at
final destination
• Noise are of several type like thermal, induce, crosstalk,
impulse which will corrupt signal
• Signal to noise ratio is given by Shannon as
• C= Blog2(1+ S/N)
• B is Bandwidth of Channel
• S/N is signal to noise ratio
• C is capacity of channel

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 Multiplexing
• In multiplex system n device share capacity of one link
• The next figure depicts the basic concept of multiplexing
• The four device left there transmission to multiplexer
which combine them in to single stream
• At the receiving end the stream is fed in to demultiplexer
which separates the stream back in to components
transmission and direct them to intended receiving
devices path refers to physical link
• While channel refer to portion of path that carries
transmission between given pair of devices

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 Multiplexing vs. No Multiplexing

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 Type of Multiplexing

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 Frequency Domain Multiplexing
• It is analog technique that can be applied when
bandwidth of link is greater than combined bandwidth of
the signal to be transmitted
• The signal generated by each sending device modulate
different carrier frequency
• These modulated frequencies are then combined in to
single composite signal that can be transported by a link
• Carrier frequencies are separated by enough bandwidth
to accommodate modulated signal which pass through
channel
• Channels must be separated by strips of unused
bandwidth termed as guard bands

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 FDM

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 FDM, Time Domain
• The diagram depicts conceptual time domain illustration of MUX process
• FDM is analog process so telephone is used as input and output
• Each telephone generate a signal of equal frequency. Inside multiplexer these similar
signal are modulated on to different frequencies
• The resulting modulated signal are combined in to single composite signal that is sent
out over media link that enough bandwidth to accommodate it

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 Multiplexing, Frequency Domain
• The diagram depicts conceptual frequency domain illustration of MUX process
• All the three frequencies exists at same time within bandwidth
• Signals are modulated on to separate carrier frequencies using either AM or FM
frequency

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 Demultiplexing, Time Domain
• The DEMUX uses a series of filter to decompose the multiplexed
signal in to constituent component signal
• The individual signal are then passed to a demodulator that
separates them form their carriers and passes them to waiting
receivers

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 Demultiplexing, Frequency
Domain

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 TDM

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 Time Domain Multiplexing
• It is digital process that can be applied when data rate
capacity of transmission medium is greater than rate
required by the sending and receiving device
• Multiple transmission can occupy single link by
subdividing them and interleaving
• TDM can be implemented in two ways
• Synchronous
• Asynchronous

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 Synchronous TDM
• Here synchronous means that the multiplexer allocates exactly the same time
slot to each device at all time irrespective of device being transmitting or not
• Each time its allocated slot come up, device has opportunity to send its data
• If a device is unable to transmit or does not have data to send ,it’s slot remain
empty
• Times slot are grouped into frame. A frame consists of time slots including one
or more slot dedicated to each sending device
• With n input line each frame has at least n slot with each slot allocated to carry
data from specific input line

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 TDM, Multiplexing
• Synchronous TDM can be compared as fast rotating switch, as the switch
opens in front of device , the device has opportunity to transmit specified
amount of data
• The switch moves from device to device at constant rate and fixed order this
process is termed as interleaving
• Interleaving can be done byte by byte or by bit. It actually means multiplexer
will take one byte from each device , then another byte from each device
• At receiver it is demultiplexed in opposite way of multiplexing

Interleaving

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 TDM, Demultiplexing

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 Framing Bits
• The time slot order does not vary in the from frame to frame, moreover
demultiplexer knows where to direct each slot. Because of these reason no
addressing is required
• But various other factor can cause inconsistencies so one or more synchronization
bits are usually added at the beginning of each bit
• This bit are termed as framing bit and they follow a particular pattern that allows
demultiplexer to synchronize
• Most of the time it is alternating 1’s and 0’s

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 Asynchronous TDM
• Here it is designed to avoid the waste of capacity line
• It means flexible, not fixed. The total speed of input line can be greater than
capacity of the path
• If we have n input line then the frame contain no more than m slots with m < n
• The concept is as shown in following figure

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 Frames and Addresses

Only three line sending data

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 Frames and Addresses

Only four line sending data

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 Frames and Addresses

All 5 line sending data

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 Multiplexing and Inverse
Multiplexing

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 Telephone Network

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 Analog Switched Service

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 Analog Leased Service

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11/2/21 Data Communication and 227
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 Switched/56 Service

Digital
Service Unit

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 DIGITAL DATA SERVICE

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