Data Communication and OSI Layer

April 8, 2023 Data Communication and Networking

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Basic Concepts

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


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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 HzBut 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-PSKCharacteristics


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8-PSKCharacteristics


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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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PSKBandwidth


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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-QAMConstellations


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8-QAM Signal


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16-QAMConstellation


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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

Interleaving

• 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


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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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Only four line sending data


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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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Switched/56 Service

Digital Service Unit


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DIGITAL DATA SERVICE

Data Communication and OSI Layer

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