# Slide 1 - Home - KSU Faculty Mem

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```					              IS370
Data Communications
and
Computer Networks

Chapter 2 : Physical Layer and Media

Introduction
• This layer is involved in physically carrying information
from one node in the network to the next.
• One of these services is to convert the data link layer stream of bits
into a signal.
– Take care of the transmission medium
• Control the transmission medium.
• Decides on the directions of the data flow.
• Decides on the number of logical channels when data comes from
different sources.
Data & Signals
• A transmission media moves information in the form of electromagnetic
signals.
– Data in the form of 0s and 1s cannot be sent over a network.
– Data must then be transformed into electromagnetic signals to be transmitted.
– Data Transmission : Communication of data by propagation and processing of signals
• Data and the signals that represent them can take one of the following forms:
– Analog form
– Digital form
• Analog and digital data
– Analog data : refers to information that is continuous (i.e. takes on continuous values)
• Ex. Human voice, sound, light, temperature.
– Human voice
o When somebody speaks, a continuous wave is created in the air.
o Could be captured by a microphone and converted to an analog signal.
– Digital data : refers to information that has discrete states (i.e. takes on discrete values).
• Ex.: Text, integers, symbols stored on the computer memory in the form of 1s and 0s.
– It is usually converted to a digital signal or modulated into an analog signal to be transmitted across a
medium.
Analog & Digital
• Analog and digital signals
– Analog signal
• Has infinitely many levels of intensity over a period of time.
• As the signal moves from A to B, it passes and includes an infinity number of values.
– Digital signal
• Has only a limited number of defined values.
• Each value can be any number, as simple as 0 and 1.
–Illustrating signals by plots
• y axis (Strength or value of signal) – x axis (passage of time)
Analog & Digital
• Both analog and digital signals can take on one of the two forms:
– Periodic
– Aperiodic (nonperiodic)
• Periodic signal
–   Completes a pattern within a measurable time frame called period
–   Repeats (continuously) that pattern over identical subsequent periods.
–   The completion of one full pattern is called a cycle.
–   A period, represented by T, is defined as the amount of time (expressed in second)
required to complete one full cycle.
• Aperiodic signals
– Change constantly without exhibiting a pattern or cycle that repeats over time (=> no
repetitive pattern).
• Both analog and digital signals can be periodic and aperiodic
• In data communication, to send data there is a commonly a use of:
– Periodic analog signals
• Need less bandwidth
– Aperiodic digital signals
• Can represent variation in data
Periodic Analog signals
• Periodic analog signal could be classified as:
– Simple periodic analog signal
• i.e. a sine wave -cannot be decomposed into smaller signals-.
– Composite periodic analog signal
• composed of multiple sine waves
• Sine wave
– Sine wave is the most fundamental form of periodic analog signal
– Sine waves can be represented by three parameters
• 1) Peak Amplitude, 2) Frequency, and 3) Phase.
• Mathematically described as: x(t) = Asin(2pft+Ø )
– x(t): signal at time t; A: Peak Amplitude; f: frequency; Ø:the phase

1. Peak Amplitude (A)
• Absolute value of its highest intensity.
• For electric signals, the peak amplitude is measured in volts.
Periodic Analog signals
•Sine wave (cont’d)
2. Period (T) or frequency (f)
•   Period refers to the amount of time, in seconds, a signal needs to complete 1 cycle.
•   Frequency refers to the number of periods in 1s.
•   Frequency = 1/Period; Period = 1/Frequency.
•   Period is expressed in seconds and the frequency in Hertz (Hz) –i.e. cycle per second-.
 ms(milli)=10-3s. µs(micro)=10-6s, ns(nano)=10-9s,ps(pico)=10-12s.
 KHz=103Hz, MHz=106Hz, GHz=109Hz, THz(Tera)=1012Hz.
• If a signal does not change at all (maintains a constant voltage value) => its frequency is zero
 A DC component ??? (Direct Current)???
• If a signal changes instantaneously => its period is zero => its frequency is infinity.
3. Phase (P)
• Describes the position of the waveform relative to time 0.
• In other words, it is the amount of shift (backward or forward) along the time axis.
• Phase is measured in degrees or radians (360 degrees is 2p radians).
• Phase changes often occur on common angles:
• A phase shift of 360 degrees (i.e 2p) corresponds to a shift of a complete period
- A sine wave with a phase of 0 deg is not shifted
• A phase shift of 180 degrees (i.e p) corresponds to a shift of one-half period
- A sine wave with a phase of 180 degrees is shifted to the left by ½ cycle.
• A phase shift of 90 degrees (i.e. p/2) corresponds to a shift of a one-quarter of a period.
- A sine wave with a phase of 90 degrees is shifted to the left by ¼ cycle.
Periodic Analog signals

• Examples
–The home electricity is a sine wave with a maximal amplitude of 155 volts and a frequency of 60Hz. Write the
mathematical equation.
• 2pf = 2*3.14*60 = 377 radians/second
• x(t) = Asin (2pft+Ø ) = 155sin (377t + Ø) ; Ø usually 0.
–Write the mathematical equation of the electricity produced by a constant (I.e. DC) six-volt battery.
• x(t) = Asin(2pft – p/2 ) = Acos(2pft) = Acos(0) = A = 6 volts
–A sine wave has a frequency of 8KHz. What is its period?
• Let T be the period and f the frequency
T = 1/f = 1/8000 = 0.000125 second = 125*10-6 s=125µs
–A sine wave completes one cycle in 4 second. What is its frequency
• Let T be the period and f the frequency
f = 1/T = 1/4 = 0.25 Hz
Periodic Analog signals
• Time and frequency domains
– The previous plots are time-domain plots.
– The time domain plot shows changes in signal amplitude with respect to time
• It is an amplitude versus time plot.
– Phase and frequency are not explicitly measured on a time-domain plot.
– To show the relationship between amplitude and frequency, use what is called a frequency-domain
plot.
• A frequency-domain plot is concerned with only the peak value and the frequency.
– An analog signal is best represented in the frequency domain
• Composite signals
Periodic Analog signals
– A single-frequency sine wave is useless in data communications
• We could only represent alternating 1s and 0s.
• We will hear a buzz if it is used to convey a telephonic conversation.
– To communicate data, we need to send a composite signal.
– A composite signal is made of many simple sine waves.
– According to Fourier analysis, any composite signal, no matter how complex, can be represented
(I.e. decomposed) as a combination of simple sine waves with different frequencies, phases, and
amplitudes
• Fourier Analysis
–Using the Fourier series, a composite periodic signal with period T (frequency f0) can be
decomposed into a series of sine and cosine functions in which each function is an integral harmonic
of the fundamental frequency f0 of the composite signal
• The composite signal is decomposed in a series of signals with discrete frequencies

• f0 is known as the fundamental frequency (or first harmonic).
–T = 1/f0 represents the period of the periodic signal.
• Multiple of f0 are referred to as harmonics.
–f0=f1, 2f0=f2 (second harmonic), 3f0=f3 (third harmonic), and so on.
–Using the Fourier transform, a composite nonperiodic signal can be decomposed into a combination
of sine waves with continuous frequencies (i.e. frequencies with real values).
Periodic Analog signals
• Ex.: Decomposition of a composite periodic signal
– Next two first figures shows the time-domain and the frequency-domain
of the fundamental frequency component and the 3rd and 9th harmonic
components.
– The third figure shows frequency-domain of the composite periodic
signal resulting from the combination of the three sine waves
(fundamental frequency, 3rd and 9th harmonic components).
– The third figure, is something close to a square wave.
• We need to add more harmonics to be close to a square wave (all the odd harmonics up to infinity)

• In general, to approximate a square wave with frequency f and amplitude A,
the terms of the series are as follows:
Frequency      Amplitude
• Frequencies: f, 3f, 5f, 7f, …
f               4A/π
• Amplitude of sine waves with these frequencies                                     3f               4A/3π
5f               4A/5π
7f               4A/7π
Periodic Analog signals
Periodic Analog signals
• Frequency Spectrum
– Frequency spectrum of a composite signal is the collections of all the components
frequencies it contains
– It is shown using a frequency-domain graph.
– The spectrum of a periodic composite signal contains only integer frequencies while
the spectrum of an nonperiodic composite signal contains all the continuous
frequencies??.
• Bandwidth
– The bandwidth of a composite signal is the width of the frequency spectrum.
– In other words, it refers to the range of component frequencies, and frequency
spectrum refers to the element within that range.
– To calculate bandwidth, subtract the lowest frequency from the highest frequency of
the range.
• Example
– The average voice has a frequency range of roughly 300 Hz to 3100 Hz(??).
– The spectrum would thus be 300 - 3100 Hz
– The bandwidth would be 2800
Periodic Analog signals
• Example (cont’d)
– If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700,
and 900 Hz, what is the bandwidth? Draw the frequency spectrum, assuming all components have
a maximum amplitude of 10 V.
• B = fh - fl = 900 - 100 = 800 Hz
The spectrum has only five spikes, at 100, 300, 500, 700, and 900

– A periodic signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest
frequency? Draw the frequency spectrum if the signal contains all integral frequencies of the same
amplitude
B = fh - fl
20 = 60 - fl
fl = 60 - 20 = 40 Hz
– A nonperiodic composite signal has a bandwidth of 200 kHz, with a middle frequency of 140 kHz
and peak amplitude of 20 V. The two extreme frequencies have an amplitude of 0. Draw the
frequency domain of the signal
• Lowest frequency: 40KHz
• Highest frequency: 240KHz.
Digital signals
• In addition to being represented by an analog signal, data can also be represented by a digital
signal.
• For example, a 1 can be encoded as a positive voltage and 0 as zero voltage.
• A digital signal can have more than two levels.
– Can then send more than 1 bit for each level
– In general, if a signal has L levels, each level needs log2L bits.

• Most digital signals are aperiodic and, thus, period or frequency is not appropriate.
• A new term is used to describe digital signals:
– Bit rate (instead of frequency) : the number of bits sent in 1 second.
• Bit rate = 1/(bit duration) ; bit duration = time to send one bit
• Bit rate is expressed in bit per second (bps).
– Bit length
• The distance one bit occupies on the transmission medium
• Bit length = propagation speed * bit duration.
• Example.                               Digital signals
– A digital signal has eight levels. How many bits are needed per level?
• Number of bits per level = log28= 3
–We need to download text documents at the rate of 100 pages per minute. What is the required bit rate
of the channel?
• In average a page contains 24 lines with 80 characters in each line. If one character requires 8 bits the bit rate
is: 100*24*80*8 = 1636000bps = 1.636 Mbps
– HDTV (High-Definition TV), used to broadcast high quality video signals, uses a screen of a ratio
16:9 (i.e. a resolution of 1920*1080 pixels per screen). The screen is renewed 30 times per second. 24
bits will represent one color pixel.
• The bit rate is : 1920*1080*30*24=1492992000 ≈ 1.5Gbps
• A digital signal is a composite analog signal with an infinite bandwidth (frequencies between
0 and infinity).
– Decomposition of a digital signal with Fourier analysis
• If the digital signal is periodic (rare in data transmission) the decomposition leads to an infinite bandwidth and
discrete frequencies.
– If the digital signal is aperiodic the decomposition leads to an infinite bandwidth and continuous
frequencies.
Digital signals
• Two approaches for digital signal transmission: 1) baseband transmission, 2) broadband
transmission.
• Baseband transmission
– Means sending the digital signal over a channel without changing it into an analog signal.

– Baseband transmission of a digital signal that preserves the shape of a digital signal is possible only if
we have a low-pass channel with an infinite or very wide bandwidth.
• A low-pass channel is a channel with a bandwidth that starts from zero.
– This is the case if we have a dedicated medium with a bandwidth constituting only one channel.
• A low-pass channel with infinite bandwidth is ideal and not found in real life.
• Coaxial cable or optical fiber are examples of wide bandwidth medium

– Baseband transmission with a low-pass channel with limited bandwidth is possible by approximating
a digital signal with an analog signal. Approximation depends on available bandwidth.
• The required bandwidth is proportional to the bit rate. Sends bits faster => more bandwidth.
Digital signals
– Means changing the digital signal to an analog signal for transmission when we cannot send the digital signal
directly to the channel.
– Broadband transmission allows the use of a bandpass channel.
• A bandpass channel is a channel with a bandwidth that does not starts from zero.
– More available than low-pass channel.

•   EX.
– An example of a dedicated channel where the entire bandwidth of the medium is used as one single channel is a LAN.
• Almost every wired LAN today uses a dedicated channel for two stations communicating with each other.
• In a bus topology LAN with multipoint connections, only two stations can communicate with each other at each moment in time (timesharing); the
other stations need to refrain from sending data.
• In a star topology LAN, the entire channel between each station and the hub is used for communication between these two entities.
– An example of broadband transmission using modulation is the sending of computer data through a telephone subscriber line, the
line connecting a resident to the central telephone office.
•   These lines are designed to carry voice with a limited bandwidth.
•   The channel is considered a bandpass channel.
•   Convert the digital signal from the computer to an analog signal, and send the analog signal.
•   Can install two converters to change the digital signal to analog and vice versa at the receiving end.
•   The converter, in this case, is called a modem
Transmission impairment
• Transmission media are not perfect.
• These imperfections cause impairments in the signal sent through the medium.
– This means that what is sent is not what is received.
• Three type of impairments usually occur : 1) Attenuation, 2) Distortion, and 3) Noise
• Attenuation
– When a signal, simple or complex, travels through a medium, it loses some of its energy so that it can overcome
the resistance of the medium.
• I.e. Signal strength falls off with distance.
– Amplifiers are used to compensate from this loss.

– Use the concept of decibels (dB) measure to show that a signal has lost or gained strength
• dB measures the relative strength of two signals or a signal at two different points.
– dB = 10log10(P2/P1)
• P1 is the transmitted signal power and P2 is the received signal power.
– dB is negative if a signal is attenuated and positive if the signal is amplified.
– Example
• A signal travels through a transmission medium and its power is reduced to half.
– This means P2=(1/2)P1, the attenuation (loss of power) can be calculated as
10log10(P2/P1)= 10log10(0.5*P1/P1)= 10log10(0.5)= 10*(-0.3)= -3dB
Engineers knows that –3dB, or a loss of 3 dB, is equivalent to losing half the power.
• A signal travels an amplifier and its power is increased 10 times.
– This means that P2=10*P1, the amplification (gain of power) can be calculated as
10log10(P2/P1)= 10log10(10*P1/P1)= 10log10(10)= 10(1)=10dB
Transmission impairment
•Attenuation (cont’d)
– Losses and gains are additive
– => Decibel numbers can be added (or subtracted) when we are talking of several points.
• The signal strength after traveling into three points
– dB = -3 + 7 – 3 = +1 ; which means that the signal has gained power.

– The loss in a cable is usually defined in decibels per Kilometer (db/Km).
– If the signal at the beginning of a cable with -0.3 dB/Km has a power of 2 mW, what is the power of the signal at 5Km.
• The loss in the cable in decibels is (5*(-0.3))=-1.5dB.
• dB = 10 log10(P2/P1) = -1.5 => P2/P1 = 10-0.15 = 0.71 => P2 = 0.71*P1 = 0.7*2 = 1.4 mW.
• Distortion
– Means that the signal changes its form or shape.
– Occurs in composite signal made of different frequencies.
• Each signal component has its own propagation speed through a medium and, so its own delay to arrive at the final destination.
• Signal components at the receiver have different phases from those at the sender.
• Therefore the composite signal shape is not the same.
• Noise                  Transmission Impairment

–Several types of noise :
• Thermal noise
– The random motion of electrons in a wire that creates an extra signal not originally sent by the transmitter.
• Induced noise
– Comes from source such as motors.
– These devices act as a sending antenna and the transmission medium acts as the receiving antenna.
• Crosstalk
– The effect of one wire on the other.
– One wire acts a sending antenna and the other as the receiving antenna.
• Impulse noise
– A spike (a signal with high energy in a very short period of time) that comes from power lines, lightning, and so on
– The signal-to-noise (SNR) is the ratio of the signal to the noise.
• SNR = (average signal power/average noise power).
– SNR interpretation
• High SNR => the signal is less corrupted by noise.
• Low SNR => the signal is more corrupted by noise.
– SNR is often described in decibel units, i.e. SNRdB= 10log10SNR
– Example
• The power of a signal is 10 mW and the power of the noise is 1 mW. What are the values of SNR and SNRdB?
– SNR = 10000mW/1mW = 10000
– SNRdB = 10log10(10000) = 10log10(104) = 40.
Data rate limits
• Data rate = How fast can the data be sent, in bits per second, over a channel?
• Data rate depends on three factors: 1) The available bandwidth, 2) The level of the signal, and
3) The quality of the channel (i.e. the level of the noise).
• Noiseless channel : Nyquist bit rate
– The theoretical maximum bit rate in a noiseless channel
• BitRate = 2*Bandwidth*log2L
– Bandwidth : Bandwidth of the channel, L : the number of signal levels used to represent data.
– Given a specific bandwidth, increase the bit rate => increase the signal levels.
• Pb : increasing the signal levels may reduce the system reliability.
– The receiver should be very sophisticated to distinguish between many levels.
– Examples.
• Consider a noiseless channel with a bandwidth of 3000Hz transmitting a signal with two signal levels. What is
the bit rate?
– BitRate = 2*3000*log22 = 6000bps
• Consider the same noiseless channel transmitting a signal with four data levels. How many bits are sent at each
levels? What is the maximum bit rate.
• L = 4 ; number of bits = log2(4) = 2 bits.
• BitRate = 2*3000*log24 = 12000bps
• Noisy channel : Shannon bit rate.
– The reality: a channel is always noisy.
– Theoretical highest data rate for a noisy channel
• Capacity = Bandwidth*log2(1+SNR)
– There is no indication of the signal levels => no matter how many levels we have, we cannot achieve
a data rate higher than the capacity of the channel.
• Formula defines a characteristic of the channel not the method of transmission.
Data rate limits
• Noisy channel : Shannon bit rate (cont’d)
– Example
• Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero (In other
words, the noise is so strong that the signal is faint). What is the capacity of this channel?
– This means that the capacity of this channel is zero regardless of the bandwidth. In other words, we cannot receive
any data through this channel.
– C =B*log2(1 +SNR) =B*log2 (1 +0) =B*log2(1) =B*0 =0
• We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a
bandwidth of 3000. The signal-to-noise ratio is usually 3162. For this channel the capacity is calculated as
– C =B*log2(1 +SNR) =3000*log2(1 +3162) =3000 *log2(3163)
– C =3000*11.62 =34860 bps
– This means that the highest bit rate for a telephone line is 34.860 Kbps. If we want to send data faster than this, we
can either increase the bandwidth of the line or improve the signal-to-noise ratio
• Assume that SNRdB = 36 and the channel bandwidth is 2 MHz. what is the theoretical channel capacity?
– SNRdB = 10log10SNR => SNR = 10SNRdB/10 => SNR = 103.6 = 3981
– C = Bandwidth*log2(1+SNR) = 2*106*log2(3982) = 24 Mbps.
•In practice need to use both methods to find the limits and signal levels.
–Example
• We have a channel with a 1 MHz bandwidth. The SNR for this channel is 63; what is the appropriate bit rate
and signal level?
–First, we use the Shannon formula to find our upper limit.
• C =B*log2(1 +SNR) =106*log2(1 +63)=106*log2(64) =6 Mbps
–Although the Shannon formula gives us 6 Mbps, this is the upper limit. For better performance we choose something
lower, for example 4 Mbps. Then we use the Nyquist formula to find the number of signal levels.
• 4 Mbps =2 × 1 MHz × log2L  L =4
•The Shannon capacity gives us the upper limit; the Nyquist formula tells us how many signal
levels we need.
Performance
• Network Performance: How good is the network?
• Network performances terms: 1) Bandwidth, 2) Throughput, 3) Bandwidth-delay product, and
4) Jitter
• Bandwidth
– This term can be used in two different contexts
• The bandwidth, in hertz, refers to the range of frequencies in a composite signal or the range of frequencies that
a channel can pass.
– Ex. Bandwidth of a subscriber telephone line is 4KHz.
• The bandwidth, in bits per second, refers to the speed of bit transmission in a channel, link, or a network.
– Ex.: Bandwidth of a Fast Ethernet Network is a maximum of 100 Mbps.
• Relationship: an increase in bandwidth in hertz means an increase in bandwidth in bps.
• Throughput
– Measures how fast we can actually send data through a network.
– A network can have a bandwidth of B in bps but can only send T bps throughput -with B>T-
• Ex. A link with a bandwidth of 1Mbps but the connected devices may handle only 200Kbps
– Bandwidth is the potential measurement of a link; throughput is the actual measurement of how fast
we can send data.
– Example
• A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame
carrying an average of 10,000 bits. What is the throughput of this network?
– Throughput = 12000*10000/60 = 2Mbps
– The throughput is almost one-fifth of the bandwidth in this case.
Performance
•Latency (delay)
– Defines how long it takes for an entire message to completely arrives at the destination
from the time the first bit is sent out from the source :
Latency = propagation Time + Transmission Time + Queuing Time + Processing Time
– Latency components:
• Propagation time
– Measures the time required for a bit to travel from the source to the destination :
Propagation Time = Distance/(Propagation Speed)
– Propagation speed of electromagnetic signals depends on the medium and on the signal frequency.
• Light is propagated with a speed of 3*108m/s in a vacuum. It is lower in air; it is much lower in cable.
• Transmission time.
– Necessary time to send a message.
– Depends on the size of the message and the bandwidth of the link :
Transmission Time = (Message size)/Bandwidth
• Queuing time.
– Time needed for each intermediate (ex. Router) or end device to hold the message before it can be
processed. Not a fixed time but changes with the load imposed on the network.
• Processing delay.
– Example
• What is the propagation time if the distance between the two points is 12,000 km? Assume the propagation
speed to be 2.4 × 108 m/s in cable.
• Delay (cont’d)
Performance
–Example (cont’d)
• What are the propagation time and the transmission time for a 2.5-kbyte message (an e-mail) if the bandwidth
of the network is 1 Gbps? Assume that the distance between the sender and the receiver is 12,000 km and that
light travels at 2.4*108 m/s.

• Jitter
–The jitter issue is related to delay.
–Jitter is a problem for time-sensitive applications (ex. Audio and video) if different packets of data
encounter different delays.
• If the delay of the first packet is 20ms, for the second is 45ms, and for the third is 40ms => the real-time
applications that uses the packets endures jitter
• Bandwidth-Delay product
–Defines the number of bits that can fill the link.
–EX.
• Consider a link with 1bps and the link delay is equal to 5s. The product 1*5 is the maximum number of bits that

–This measurement is important if we need to send data in bursts and wait for the acknowledgement of
each burst before sending the next burst.
• To use the maximum capability of the link, need to make the size of a burst = 2*Bandwidth*Delay
conversion
Digital-to-Digitalsignals: 1) Line coding, 2) Block coding,
• Three techniques to convert digital data to digital
and 3) Scrambling.
– Line coding is always needed. The two others may or may not be needed
• Line coding
• At the sender, converts a sequence of bits to a digital signal.
• At the receiver, the digital data are recreated by decoding the digital signal.
– Line coding schemes: 1) Unipolar, 2) Polar, 3) Bipolar, 4) Multilevel, and 5) Multitransition
1) Unipolar scheme
• All the signal levels are on one side of the time axis, either above or below.
1.1) NRZ (Non-Return-to-Zero)
• Positive voltage defines the bit 1; Zero voltage defines the bit 0.
• Normally not used in data communication because it is costly

2) Polar schemes
• The voltage are on both sides of the time axis.
– Ex. The voltage level for 0 can be positive and the voltage level for 1 can be negative.
2.1) Polar Non-Return-to-Zero (polar NRZ)
 In polar NRZ, two levels of voltage amplitude are used.
 Two versions of polar NRZ
2.1.1) NRZ-L (NRZ-Level): the level of the voltage determines the value of the bit.
2.1.2) NRZ-I (NRZ-Inverted): the change or the lack of change in the voltage level determines the value of the bit.
Bit 0 : No change. Bit 1 : there is a change.
Digital-to-Digital conversion: line coding
2) Polar schemes (cont’d)
2.1) polar Non-Return-to-Zero (polar NRZ) (cont’d)

• Uses three voltage values: positive, negative, and zero.
• The signal changes not between bits but during the bit.
- Requires two signal changes to encode a bit => more bandwidth is required.
- It is more complex to create and discern three levels.
- => not used today and has been replaced by Manchester and Differential-Manchester schemes.
Digital-to-Digital conversion: line coding
2.3) Biphase (Manchester and Differential Manchester)
2.3.1) Manchester encoding
• The duration of a bit is divided in two halves.
• The voltage level moves to the other level in the second halve.
2.3.2) Differential Manchester encoding
• There is a transition at the middle of the bit.
• The bit values are determined at the beginning of the bit.
- If next bit is 0, there is then a transition. If the next bit is 1, there is none.
• Drawback
• Require more bandwidth than the NRZ because of the transition at the middle of each bit and perhaps at the
end of each bit.
Digital-to-Digital conversion: line coding
3) Bipolar schemes (also called multilevel binary)
– Three voltage level are used: positive, negative, and zero.
• One data element is at a zero voltage level.
• The other element alternates between positive and negative voltage level.
– AMI and Pseudoternary
3.1) AMI (Alternate Mark Inversion)
• Zero voltage represents the bits 0.
• Bits 1 are represented by alternating positive and negative voltage.
3.2) Pseudoternary
• Bits 1 are represented by zero voltage and bits 0 are represented by altering voltage levels.
• Commonly used for long-distance communication.

4) Multilevel schemes
– Goal: increase the number of bits per baud by encoding a pattern of m data elements into a pattern of n signal
elements.
• A group of m data elements produce a combination of 2m data patterns.
• Different types of signal elements produced by allowing different signal levels.
– With L levels, we can produce Ln combinations of signal patterns.
• Ln=2m => each data pattern encoded into one signal element.
• 2m>Ln => data patterns occupy only a subset of signal patterns.
• 2m<Ln => not possible because some of data patterns cannot be encoded.
Digital-to-Digital conversion: line coding
4) Multilevel schemes (cont’d)
– Code designers have classified these types of coding as : mBnL
• m : length of the binary pattern.
• B : means binary data.
• n: length of the signal pattern.
• L: number of levels in the signaling.
–A letter is often used in place of L: B (binary) for L=2, T (ternary) for L=3, or Q (Quaternary) for L=4
4.1) 2B1Q (two binary, one quaternary)
• Encodes the 2-bit patterns as one signal element belonging to a four-level signal.
–m=2, n=1, and L=4.
• Can send data 2 times faster than using NRZ-L.
• The receiver has to discern four thresholds corresponding to the four signal levels.
• Used in DSL technology to provide high speed connection to the Internet.
Digital-to-Digital conversion: line coding
4) Multilevel schemes (cont’d)
4.2) 8B6T (eight binary, six ternary)
• Encodes a pattern of 8 bits as a pattern of 6 signal elements where the signal has three levels.
– => 28=256 different data patterns and 36=478 different signal patterns.
• 478-256=222 redundant signal elements that provide synchronization and error detection.
• Used with 100BASE-4T

4.3) 4D-PAM5 (four dimensional five-level pulse amplitude modulation)
• 4D means that data is sent over 4 wires at the same time.
• It uses 5 voltage levels as -2, -1, 0, 1, and 2.
• Used by the gigabit LAN.
5) Multitransition: Multiline transmission three level MLT-3
– Uses three levels (+V, 0, and –V) and three transitions to move between the levels
• If the next bit is 0, there is no transition. If the next bit is 1 and the current level is not 0, the next level is 0. if
the next bit is 1 and the current level is 0, the next level is the opposite of the last nonzero level.
Digital-to-Digital conversion: Block Coding
• Gives redundancy to ensure synchronization and inherent error detecting
• self-Synchronization: a characteristic of a digital signal that allow the receiver to correctly interpret the
– Achieved if there are transitions in the signal that alert the receiver to the beginning, middle, or end of the pulse.
– Changes a block of m bits into a block of n bits (n>m).
– Block coding referred as mB/nB encoding technique.
– Involves three steps
• Division: a sequence of bits is divided into groups of m bits.
– Ex. 4B/5B: bit sequence divided into 4-bit groups.
• Substitution: an m-bit group is substituted for an n-bit group.
– Ex. 4B/5B: substitute 4-bit code for 5-bit group.
• Combination: the n-bit groups are combined together to form a stream.
– 4B/5B (four binary/five binary)
• Used in combination with NRZ-I.
• Prior to encoding with NRZ-I, change the bit stream to not contain a long sequence of 0s (i.e. to cancel NRZ-I
synchronization problem).
• The groups not used in the encoding are used for the control and error detection.
– 8B/10B (eight binary/ten binary)
• Similar to 4B/5B except that a group of 8 bits of data is now substituted by a 10-bit code.
Digital-to-Digital conversion: Scrambling
• Encoding of long distance communication
– Biphase codes are not suitable because of their wide bandwidth requirement.
– The combination of block coding and NRZ line coding is not suitable because of the DC
component.
• DC component: occurs when a digital signal is constant for a while which creates very low frequencies.
This presents a problem for the systems that cannot pass low freqeuncies.
– Bipolar AMI, however has a narrow bandwidth and does not create DC-component.
• However, a long sequence of 0s upsets the synchronization.
• Scrambling
– Modify part of AMI to provide synchronization with a long sequence of 0s
– Two techniques : B8ZS and HDB3
• B8ZS (Bipolar with 8-zero substitution)
– Commonly used in north of America.
– Eight consecutive zero-levels are replaced by the sequence 000VB0VB.
• V denotes “violation”: a nonzero voltage that breaks the AMI rule of encoding.
– i.e. same polarity as the polarity of the previous nonzero pulse.
• B denotes “bipolar”: a nonzero voltage in accordance with the AMI rules.
– i.e. an opposite polarity as the polarity of the previous nonzero pulse.
– Such a scrambling does not change the bit rate.
– There is a balance in the positive and negative voltage levels (2 positives and 2 negatives)
• => DC balance is maintained.
Digital-to-Digital conversion: Scrambling

• HDB3 (High-density bipolar 3-zero)
– Commonly used outside the North America.
– Four consecutive zero-level voltage are replaced with a sequence
• 000V : if the number of nonzero pulses after the last substitution is odd.
• B00V: if the number of nonzero pulses after the last substitution is even.
• The first substitution is even (i.e. B00V).
Analog-to-Digital conversion
• A digital signal is superior to an analog signal.
• The tendency today is to change an analog signal to digital data.
• Two techniques can be used
• PCM
• Delta modulation.
• PCM (Pulse Code Modulation)
• Most commonly used technique to change an analog signal to digital data (digitization)
• PCM encoder has three processes
• Sampling
• Quantization
• Encoding

• Sampling
•   Sampling process also called PAM (Pulse Amplitude Modulation)
•   The analog signal is sampled every Ts.
•   Ts : sample interval.
•   1/Ts : sampling rate or sampling frequency.
Analog-to-Digital conversion: PCM
• Sampling
– The sampling method
• Ideal (not easily implemented): pulses from the analog signal are sampled.
• Natural: A high-speed switch is turned on for only the small period of time when sampling occurs.
– The result is a sequence of samples that retains the shape of the analog signal.
• Sample and hold: creates flat-top samples by using a circuit. Most common method.
– The result of the sampling process is still an analog signal
• Analog signal with nonintegral values.

– Sampling rate
• Nyquist theorem: to reproduce the original analog signal the sampling rate should be at least equal to twice the
highest frequency of the original signal.
– Example
• Telephone companies digitize voice assuming a maximum frequency of 4000Hz.
– The sampling rate is therefore 8000 samples per second.
• What sampling rate is needed for a signal with a bandwidth of 10000Hz [1000,11000]Hz?
– The sampling rate must be twice the highest frequency of the signal
– Sampling rate = 2*11000 = 22000 samples/second.
• A complex low-pass signal has a bandwidth of 200KHz. What is the minimum sampling rate for
this signal?
– The bandwidth of a low-pass signal is between 0 and f (f is the maximum frequency).
– So f=200KHz and the sampling is done at : 2*200KHz=400000 samples/s.
Analog-to-Digital conversion: PCM
• Quantization
– Steps of quantization
1. Assume the original analog signal has instantaneous amplitudes between Vmin and Vmax.
2. Divide the range into L zones each of high ∆
V m ax V m in
• Formula :   
L
3. Assign quantized values of 0 to L-1 to the midpoint of each zone.
4. Approximate the value of the sample amplitude to the quantized values.
– Quantization levels
• The choice of L, the number of levels, depend on the range of the amplitudes of the analog signal and the
accuracy to recover the signal.
– If the amplitude fluctuates between two levels => L=2.
– With audio digitizing => L=256.
– Quantization errors
• The input values to the quantizer are the real values.
• The output values are the approximate values and are chosen in the middle of the zone.
• The difference between the real values and the approximate values is the quantization error.
– Example
• Assume a sample signal with the sample amplitudes in [-20,20]V. Nine samples are shown using
ideal sampling.
• Quantization levels with L=8 => ∆=5
Analog-to-Digital conversion: PCM
• Quantization Example (cont’d)
– The value at the top of each sample in the graph is the actual amplitude.
– 1st row : the normalized value for each sample (actual amplitude/∆).
– 2nd row : the normalized quantized value chosen from the middle of each zone.
– 3rd row : the normalized error.
– 4th row: quantization code for each sample based on the quantization levels (left of the graph).
– 5th row: the encoded words which are the final products of the conversion.

•   Encoding
– After each sample has been quantized and the number of bit per samples is decided, each sample can
be changed to an nb-bit code word.
– The number of bits for each sample is determined from the number of levels L: nb = log2L
– Previous example
• nb = log2L = 3bits. A quantization code of 2 is encode as 010.
Analog-to-Digital conversion: PCM
• Encoding (cont’d)
– Bit rate = sampling rate * number of bits per sample => Bit rate = fs* nb
– Example
• What is the bit rate to digitize the human voice assuming 8 bits per sample.
–Frequencies of the human voice [0,4000]Hz. Sampling rate = 4000*2 = 8000 samples/s Bit rate = 8000*8 = 64000 bps = 64
Kbps.
• Signal recovery
– Requires a PCM decoder.
– The decoder converts the code words into a pulse that holds the amplitude until the next pulse.
– The resultant staircase signal is passed through a low-pass filter to smooth it into an analog signal.
• The filter has the same cutoff frequency as the original signal at the sender.
– If the signal has been sampled >= the Nyquist sampling rate and if there is enough quantization levels
=> the original signal will be recreated.
– The minimum and maximum values of the original signal can be achieved by using amplification.
Transmission modes
• Transmission of binary data across a link can be one of the following modes
– Parallel mode
• Data is grouped into groups of n bits.
• n bits are sent with each clock tick.
• Use n wires to send these n bits at a time.
– The n wires are typically bundled into a cable with a connector at each end.
– Can increase the speed by a factor of n over serial transmission
• Drawback : cost
– Requires n communication wires.
– => usually limited to short distance.
– Serial mode
• One bit is sent in each clock tick.
• Needs only one communication channel between the two communicating devices
– Only one communication channel is used.
• Communication within devices is parallel
• => need conversion devices at the interfaces between the sender and the line (parallel-to-serial) and
between the line and the receiver (serial-to-parallel).
• Serial communication occurs in three ways
– Asynchronous
– Synchronous
– Isochronous.
• Parallel transmission
Transmission modes

• Serial transmission

• Asynchronous transmission
– The timing of the signal is unimportant.
– Agreed patterns allow the receiver to retrieve the information without regard to the rhythm in which
it is sent.
• Patterns based on grouping bit stream into bytes (usually 8 bits)
• The sender handles each byte independently.
– It sends the byte into the link whenever ready without regard to a timer
Transmission modes: serial
• Asynchronous transmission (cont’d)
– Start bit – stop bit
• Start bit (usually 0)
– An extra bit added to the beginning of each byte to alert the receiver of the arrival of a new byte.
• Stop bits (usually 1s)
– One or more extra bits added at the end of the byte to let the receiver know the end of the byte.
– The transmission of each byte may then be followed by a gap of varying duration.
• The gap is represented by an idle channel or by a stream of additional stop bits
– Meaning of asynchronous
• At the byte level, the sender and the receiver need not to be synchronized.
• Within each byte, although, the receiver must be synchronized with the incoming stream.
– When the receiver detects the start-bit, it sets a timer and begins counting bits as they come in
– After n bits, the receiver looks for a stop bit.
• Cheaper and effective.
• Used for slow-speed communication (keyboard-computer)
– Drawbacks
• Slow because of the insertion of start-bit, stop-bit, and gaps
Transmission modes: serial
• Synchronous transmission
– The bit stream combined into frames.
• A frame contains multiple bytes.
• No gaps between a byte and the next byte belonging to the same frame.
• Receiver’s responsibility to group the bits into bytes (or in characters) to reconstruct the information.
– There may be uneven gaps between frames filled with idle (special sequence of 0s and 1s).
– Timing is important because to allow the receiver to keep an accurate count of the bits as they come.
• Byte synchronization is accomplished in the data link layer.
• Speed because of no extra bits or gaps.
• Helpful for high-speed transmission (computer-computer)

• Isochronous transmission
– Guarantees that the data arrive at a fixed rate.
• Uneven delays between frames (i.e. synchronous transmission) are not acceptable.
– TV images are broadcast at the rate of 30 images/s to be viewed.
– If each image is sent by using one or more frame, they should be no delays between frames.
• Synchronization between characters is not enough, entire stream of bits must be synchronized.
Analog transmission
• Recall
– Digital transmission needs a low-pass channel.
– Bandpass channel => Analog transmission
• Analog conversions
– Digital-to-analog conversion
• Converting digital data to a bandpass signal.
– Analog-to-analog conversion
• Converting a low-pass analog signal to a bandpass analog signal.
• Digital-to-analog conversion
– Process of changing one of the characteristics of an analog signal based on the information in digital data.

– Change one of the characteristic of a sine wave (amplitude, frequency, and phase) to represent digital data.
• Change of one of these characteristics gives a different version of the wave.
– The techniques to modulate digital data into an analog signal
• FSK (Frequency Shift Keying).
• PSK (Phase Shift Keying).
• QAM (Quadrature Shift Keying) : combines the amplitude and phase. Most efficient technique used today
Analog transmission: digital-to-analog
• Aspect of digital-to-analog transmission
– Carrier signal
• In analog transmission, the sender produces a high-frequency signal that acts as a base for the
information signal called carrier signal (or carrier frequency).
• The receiver is tuned to the frequency of the carrier signal expected from the sender.
• Modulation (shift keying) is the change on the carrier signal by modifying one or more
characteristics (amplitude, frequency, or phase) according to the digital information.
– The amplitude of the carrier signal is varied to create signal elements.
• Both frequency and phase remain constant while the amplitude changes.
– Binary ASK (or on-off keying (OOK))
• Use only two levels (kinds) of signal element each with a different amplitude.
– The peak amplitude of one signal level is 0.
– The other is the same as the amplitude of the carrier frequency.


Analog transmission: digital-to-analog
• More than two amplitude levels are used.
– 4, 8, 16 or more different amplitudes for the signal are used.
– Modulate the data using 2, 3, 4, or more bits at a time.
– Not implemented with pure ASK but implemented with QAM.
• FSK (Frequency Shift Keying)
– The frequency of the carrier is varied to represent data.
• The frequency of the modulated signal is constant during the duration of one signal element but changes for the
next signal element if the data element changes.
• Both amplitude and phase remain constant.
– Binary FSK (BFSK)
• Two carriers frequencies are considered.
– One with f1 (for bit 0) and the other with f2 (for bit 1).

– Multilevel FSK
• Use more than one frequency.
– Ex.: use 4 frequencies (f1, f2, f3, and f4) to send 2 bits at a time.
– Ex.: use 8 frequencies to send 3 bits at a time.
Analog transmission: digital-to-analog
• PSK (Phase Shift Keying)
–The phase of the carrier is varied to represent two or more signal elements.
• Both amplitude and frequency remain constant.
–Binary PSK (BPSK)
• Only two signal elements are used
–One with a phase of 0 degree and the other with a phase of 180 degrees.

• Today PSK is more common than ASK or FSK.
–As simple as ASK but less susceptible to noise.
• Noise can change the amplitude easier than it can change the phase
–PSK is superior to FSK because we do not need two carrier signals.
• 2 bits are encoded in each signal element.
–4 kinds of signal elements in the output signal (L=4).
Analog transmission: digital-to-analog
• PSK (cont’d)
– QPSK (cont’d)
• Constellation diagram
• Help to define the amplitude and the phase of a signal element especially when using two carriers.
• Useful when dealing with multilevel ASK, PSK, and QAM.
• A signal element type is represented by as a dot.
• The bit or combination of bits it can carry is often written next it

• Example
• Show the constellation diagram of an ASK (OOK), BPSK, and QPSK signals.
- ASK: Binary 0 has an amplitude of 0 and binary 1 has an amplitude of 1V for example.
- BPSK: 2 different signal elements are used, one with amplitude 1V and in-phase and the other with an
amplitude 1V and 180 out-of phase.
- QPSK: for example, the point representing 11 is made of two combined signal elements, both with an
amplitude of 1V. One element is represented by an in-phase carrier and the other element by a quadrature carrier.
The amplitude of the final element sent for this 2-bits data element is 21/2, and the phase is 45 degree.
Analog transmission: digital-to-analog
• QPSK (cont’d)
– Example (cont’d)

–PSK is limited by the ability of the equipment to distinguish differences in phases.
–So far, we have been altering only one of the three characteristics of a sine wave. What if we alter two?
–Why not combine ASK and PSK? => QAM
–QAM: use two carriers
• One in-phase
• With different amplitude levels for each carrier.
–Possible variations of QAM are numerous.
Analog transmission: analog-to-analog
– Such modulation is needed if the medium is bandpass or only a bandpass channel is available.
– Governments assign a narrow bandwidth to each radio station.
– The analog signal produced by each station is a low-pass signal, all in the same range.
– The low pass signal of each station need to be shifted to a different range.
– Three possible analog-to-analog conversions
• AM (Amplitude Modulation)
• FM (Frequency Modulation)
• PM (Phase Modulation).
• AM
–The carrier signal is modulated so that its amplitude varies with the changing amplitudes of the
modulating signal.
• Phase and frequency of the carrier remain the same.
Analog transmission: analog-to-analog
• AM (cont’d)
– Standard bandwidth allocation for AM radio
• The bandwidth of an audio signal (speech and music) is usually 5 KHz.
• An AM station bandwidth should be 10KHz.
• Federal Communication Commission (FCC)
– AM stations are allowed frequencies in [530, 1700]KHz
– Each AM station is assigned 10KHz.
– To avoid interference: each station carrier frequency must be separated from those on either sides by at least 10KHz.

• FM
–The frequency of the carrier signal is modulated to follow the voltage level (amplitude) of the
modulated signal.
• The peak amplitude and the phase of the carrier signal remain constant.
Analog transmission: analog-to-analog
• FM (cont’d)
–Standard Bandwidth Allocation for FM Radio
• Bandwidth of an audio signal broadcast in stereo is 15KHz.
• FCC allows 200KHZ for each station.
• FM stations allowed carrier frequencies in [88, 108]MHz.
• Stations separated by 200KHz.
• To avoid more interfering, alternate bandwidth allocation may be used.

• PM
– The phase of the carrier signal is modulated to follow the changing amplitude of the modulated signal.
• The peak amplitude and the frequency of the carrier signal remain constant.
• Proved mathematically: PM is the same as FM.
Bandwidth Utilization
• Bandwidth Utilization is the wise use of the available bandwidth to achieve specific goals.
• Bandwidth utilization categories
– Multiplexing : how to combine several channels into one for an efficiency goal.
– Spreading: how to expand the bandwidth of a channel for privacy and antijamming goals.
• Used with WLANs.
• Multiplexing
– Share a link when its bandwidth is greater than the needs of the devices.
– Multiplexing: set of techniques that allows the simultaneous transmissions of multiple signals across
– Multiplexed system: N lines share the bandwidth of one link
• The multiplexer (MUX) combines the n transmission streams into a single stream (many-to-one).
• The demultiplexer (DEMUX) separates the stream back to its component transmissions and directs them to
their corresponding lines (one-to-many).
• The link refers to the physical path.
• The channel refers to the portion of a link that carriers a transmission between a given pair of lines.
• One link can have many (n) channels.

– Multiplexing techniques
• Frequency-division multiplexing : used for analog signal.
• Wavelength-division multiplexing: used for analog signals.
• Time-division multiplexing: used for digital signals.
Bandwidth Utilization: Multiplexing
• Frequency-Division Multiplexing (FDM)
– Multiplexing process
• Signals generated by each sending device (with a similar frequency range) modulate different carrier
frequencies (f1, f2, and f3).
• These signals are combined into a single composite signal that can be transported by the link having enough
bandwidth to accommodate it.

– Demultiplexing process
• The DMUX uses a series of filter to decompose the multiplexed signal into its component signals.
• Individual signals are passed to a demodulator to extract them from the carrier and pass them to the output
lines.
Bandwidth Utilization: Multiplexing
• Frequency-Division Multiplexing (FDM)
– FDM applications:
cellular telephones.
– EX.
• Assume that a voice channel occupies a bandwidth of 4 kHz. We need to combine three voice channels into a
link with a bandwidth of 12 kHz, from 20 to 32 kHz. Show the configuration, using the frequency domain.
Assume there are no guard bands.
– We shift (modulate) each of the three voice channels to a different bandwidth, as shown in the figure.
– Use the 20- to 24-kHz bandwidth for the first channel, the 24- to 28-kHz bandwidth for the second channel, and the 28- to
32-kHz bandwidth for the third one.
– Then combine them as shown in the Figure.

• Five 100KHz-bandwidth channels are to be multiplexed together. What is the needed minimum bandwidth of
the link if a guard band of 10 KHz is used between the channels.
– Five channels need at least four guards.
– => required bandwidth at least = 5*100 + 4*10 = 540KHz.
Bandwidth Utilization: Multiplexing
•Wavelength-Division Multiplexing (WDM)
–Designed to use the high data-rate capability of optical fiber cable by combining several
lines into one.
–Same idea as FDM
• Combining different signals of different frequencies except that the we are dealing with optical
signals having higher frequencies.
–Wavelength
• Another characteristic of a sine wave (like phase, etc.) which binds the period or the frequency to
the propagation speed.
–Wavelength = propagation speed * period
• Characteristic often used to describe transmission of light in an optical fiber

–WDM applications
• SONET network
Bandwidth Utilization: Multiplexing
• Time-Division Multiplexing (TDM)
– Allows several connections to share the high bandwidth of a link by sharing the time.
• Each connection occupies a portion of time in the link.

– Two techniques: 1) Synchronous TDM, and 2) Statistical TDM.
– Synchronous TDM
• Each input has always a reserved slot in the output frame whether it has data to send or not.
• Time slots and frames
– Data flow of each connection divided into units.
• A unit can be 1 bit, one character, or one block of data.
–   Each input occupies one input time slot.
–   Each input unit becomes one output unit and occupies one output time slot.
–   If an input time slot = Ts => the output time slot = T/n (n is the number of connections)
–   A round of data units from each input connection is collected into a frame.
• If we have n connections => a frame is divided into n time slots with one slot allocated for each unit, one for each input line.
– If an input unit duration = T s => {an slot duration = T/n ; frame duration = T }
– Output link data rate = n * a_connection_data_rate

– Synchronization bits, called framing bits, are added at the beginning of each frame to allow the demultiplexer to synchronize
with the incoming bit stream
Bandwidth Utilization: Multiplexing
• Synchronous TDM (cont’d)
– Example
• In a 3 channels are multiplexed using TDM, the data rate for each input connection is 3Kbps. If 1 bit at a time
is multiplexed (i.e. a unit=1 bit), what is the duration of (a) each input slot, (b) each output slot, and (c) each
frame.
a) Input unit data rate = 1Kbps => input time slot duration = 1/1000 s = 1ms.
b) output time slot duration = 1/3 input time slot duration = 1/3ms.
c) Each frame carries 3 output time slots => frame duration = 3*output time slot = 1ms
• The frame duration = input time slot duration.
• Four channels are multiplexed using TDM. If each channel sends 100bytes/s and we multiplex 1 byte per
channel, find : a) the frame size, b) the frame duration, c) the frame rate, and d) the bit rate for the link.
a) Each frame carries 1 byte from each channel => the frame size = 4 bytes = 32 bits.
b) Frame duration = 1/100s.
c) Frame rate = 100frames/s=400bytes/s = 3200 bps
d) Link date rate = 400bytes/s = 3200 bps.????

– Synchronous TDM drawback
• If a source doesn’t have data to send, the corresponding slot in the frame is empty.
– Three strategies, or a combination of them, can be used to handle the disparity in the input data rates.
• 1) multilevel multiplexing, 2) multiple-slot allocation, and 3) pulse stuffing.
– Applications: 1) Telephone companies (with T lines), 2) 2nd generation cellular telephone companies
Bandwidth Utilization: Multiplexing
• Statistical TDM
– Slots are dynamically allocated to improve the bandwidth efficiency especially when some
input lines do not have data to send.
• A slot is allocated only if the input line has a slot to send.

• A slot needs to carry data as well as the address of the destination.
– Simplest form: n bits to define N different output lines with n=log2N
– Slot size
• The ratio of the data size to address size must be reasonable to make transmission efficient. => a
block of data is usually many bytes while the address is just a few bits.
– Bandwidth
• The capacity of the link is normally less than the sum of the capacities of each channel.
• The capacity of the link is based on the statistics of the load of each channel.
– if on average only x percent of the input slots are filled, the capacity of the link reflects this.
Transmission Media
•Transmission media is below the physical layer and is directly controlled by it.
• Transmission medium is usually free space, metallic cable, or fiber-optic
cable.
• Signals, representing data, are transmitted from one device to another in the
form of electromagnetic energy propagated through the transmission media.
– Electromagnetic energy includes power, radio waves, infrared light, visible light,
ultraviolet light, and X, gamma, and cosmic rays.
• Each of these constitute a portion of the electromagnetic spectrum.
• Transmission media can be divided into two categories:
– Guided media including twisted-pair, coaxial cable, and fiber-optic.
– Unguided media is the free space.
• Guided media
– Guided media provide a conduit from one device to another.
• A signal transmitted along any of these media is directed and contained by the physical medium.
– Guided media include
• Twisted-Pair and coaxial cable : uses metallic (copper) conductors that accept and transport signals
in the form of electric current.
• Optical fiber cable: accepts and transports signals in the form of light.
Transmission Media: Guided Media
• Twisted-Pair (TP) cable
– Consists of two conductors (normally copper) each with its own plastic insulation, twisted
together.

– One of the wires is used to carry signals to the receiver.
– The other wire is used only as a ground reference.
• The receiver uses the difference between them.
– External influences -Interference (noise) and crosstalk- can affect both wires and created
unwanted signals.
– If the two wires were parallel
• The effect of these unwanted signals will not be the same in both wires because of the different
locations relative to the external influences sources.
• This will result in a difference at the receiver.
– If the two wires are twisted
• A balance is maintained
– In one twist, one wire is closet to the noise source and the other is farther.
– In the next twist, the reverse is true.
• Twisting makes it probable that both wires are equally affected by external influences.
• The receiver, which calculate the difference between the two, receives no unwanted signals => unwanted
signals are mostly cancelled out.
• The number of twists per unit of length (i.e. inch) has some effect on the quality of the cable.
Transmission Media: Guided Media
• TP cable (cont’d)
– Most common TP cable used in communication is UTP (Unshielded TP).
• IBM has produced an STP (Shielded TP) cable.
– STP has a metal foil or braided-mesh covering that encases each pair of insulated conductors.
– Such metal casing prevents the penetration of noise and crosstalk but is bulkier and more expensive.

– EIA (Electronic Industries Association) has developed standards to classify UTP cable into seven
categories
• Categories are determined by cable quality (1 lowest and 7 the highest)
• Each category is suitable for a specific use.
Transmission Media: Guided Media
• UTP
– Connectors
• Most common connector is the RJ45 (Registered Jack)

– Performance : compare the attenuation vs. frequency and distance.
• A TP cable can pass a wide range of frequencies.
• However, with frequencies above 100KHz, the attenuation increases.
– The gauge is a measure of the thickness of the wire.

– Application
• Used in telephone lines to provide voice and data channels.
– The local loop –the line connecting the subscriber to the central telephone office- commonly consists of UTP cables.
• DSL lines are also UTP cables.
• LANs such as, 10Base-T and 100Base-T use UTP cables.
Transmission Media: Guided Media
• Coaxial Cable (or Coax)
– Carries signals of higher frequency ranges than TP cables.
– Coax components
• A central core conductor of solid or stranded wire (usually copper) enclosed in an insulating sheath
• The insulating sheath is encased in an outer conductor or metal foil, braid, or a combination of the two.
– The outer metallic wrapping serves both as shield against noise and crosstalk and as a second conductor.
• The outer conductor is also enclosed in an insulating sheath and the whole cable is protected by a plastic cover.

–Standards
• Coaxial cables are categorized by their RG (Radio Government).
–Each RG number denotes a unique set of physical specifications
• Wire gauge of the inner conductor, thickness and type of the inner insulator, the construction of the shield, and the size and type of
the outer casing.
• Each cable defined by an RG rating is adapted for a specialized function.
Transmission Media: Guided Media
• Coaxial Cable (cont’d)
– Connectors
• Most common type of connector is BNC (Bayone-Neill-Concelman)
– BNC connector : connects the end of the cable to a device, such as a TV set
– BNC T connector: used in Ethernet networks to branch out to a connection, to a computer, or to a device.
– BNC terminator: used at the end of the cable to prevent the reflection of the signal.

– Performance
• Although a coaxial cable has a much higher bandwidth, the signal weakens rapidly and requires the frequent use
of repeaters.

–Application
• Was widely used in analog telephone networks, and later with digital telephone networks.
• Cable TV networks use coaxial cables (RG-59) at the network boundaries.
– Today, fiber-optic cables has replaced the coax cable in telephone and TV networks.
– 10Base-2, or thin Ethernet, uses RG-58 coax cable with BNC connectors.
– 10Base-5, or thick Ethernet, uses RG-11 coax cable with specialized connectors
Transmission Media: Guided Media
• Fiber-Optic Cable
– Optic-cable is made of glass or plastic and transmits signals in the form of light.
– Light characteristic
• Light travels in straight line as long it is moving though a single uniform substance.
• If a ray of light traveling through one substance suddenly enters another substance (of a different density), the
ray changes direction according to the angle of incidence (I).
–   If I < critical angle => the ray refracts and moves closer to the surface.
–   If I = critical angle => the light bends along the interface
–   If I>critical angle => the ray reflects (makes a turn) and travel again in the denser substance.
–   The critical angle is a property of the substance and differs from one substance to another.

– Optical fibers use reflection to guide light through a channel.
• A glass or plastic core is surrounded by a cladding of less dense glass or plastic.
– The difference in density of the two materials must be such that a beam of light moving through the core is reflected off the
Transmission Media: Guided Media
• Fiber-optic cables (cont’d)
– Propagation modes
• Two modes are supported to propagate light along optical channels
– Mulimode : can be implemented in two forms, step-index and graded-index
– Single mode
• Each mode requires fiber with different physical characteristics.
• Multimode
– Multiple beams of light from a light source move through the core in different paths.
– Multimode step-index fiber
• The density of the core remains constant from the center to the edge.
• A beam of light moves through this constant density in a straight line until it reaches the interface of the
• At the interface, there is an abrupt change due to a lower density => alter the angle of beam’s motion.
• There is a distortion of the signal as it passes through the fiber
• The fiber is with a varying density.
• Highest at the center of the core and decrease gradually to its lowest at the edge.
• => Decrease the distortion of the signal through the cable.
• Single-mode
• Uses step-index fiber and a highly focused source of light that limits beam to a small range of
angles, all close to the horizontal.
Transmission Media: Guided Media
• Fiber-optic cables (cont’d)
– Propagation modes (cont’d)
• Single-mode (cont’d)
• Such a fiber is manufactured with a much small diameter and with lower density. The decrease in the density
results in a critical angle of 90 degrees to make the beam propagation almost horizontal
• Propagations of different beams is almost identical and delays are negligible.
• All the beams arrive at the destination together and can be recombined with little distortion of the signal

• Fiber sizes
– Defined by the ratio of the diameter of their core to the diameter of their cladding, both in micrometers
Transmission Media: Guided Media
• Fiber-optic cables (cont’d)
– Cable composition
•   The outer jacket is made of either PVC or Teflon.
•   Inside the jacket are Kevlar (strong material) strands to straighten the cable.
•   Below the Kevlar is another plastic coating to cushion the fiber.
•   The fiber is at the center of the cable and it consists of cladding and core

– Connectors
• Three types of connectors
–SC (Subscriber Channel) connector: used for TV cable.
–ST (Straight-Tip) connector: used for connecting cable to networking devices.
–MT-RJ connector: has the same size as RJ45.
Transmission Media: Guided Media
• Fiber-optic cables (cont’d)
– Performance
• The attenuation (attenuation versus wavelength) is flatter than in the case of twisted-pair cable and
coaxial cable.
• The performance is such that we need fewer repeaters when we use fiber-optic cable.

– Application
• Often used as backbone networks (like SONET) because its wide bandwidth is cost-effective.
– Actually, with WDM we can transfer data at rate of 1600 Gbps.
• Some cable TV companies use a hybrid network: combination of optical-fiber and coaxial cable.
• LAN such as 100Base-FX network (Fast Ethernet) and 1000Base-X.
• 1) Higher Bandwidth, 2) Less signal attenuation, 3) Immunity to electromagnetic interference, 4)
Resistance to corrosive materials, 5) Light weight, and 6) Greater immunity to tapping.
• 1) Installation and maintenance, 2) Unidirectional light propagation, and 3) Coast.
Switching
• With large networks point-to-point or multipoint communications are not practical.
– Point-to-point connection between each pair of devices (i.e. mesh topology): cost, idle links.
– Multipoint connection (i.e. a bus topology) : the distance between the devices and their number increase beyond
the capacity of the media.
• Switching is the solution
– A switched network consists of a series of interlinked nodes, called switches.
• A switch is a device capable of creating temporary connections between two or more devices linked to it.
– Some of these nodes are connected to end systems (computers, telephones, etc.).
– Other are used only for routing.
• End systems (communicating devices) labeled A, B, E, etc.
• Switches labeled I, II, IV, etc. Each switch is connected to multiple links.

• Today network classification
– Circuit-switched networks.
– Packet-switched networks: further divided
• Virtual-circuit networks.
• Datagram networks.
– Message-switched networks
• Phased out but still has networking applications (e-mail)
Switching: Circuit-switched networks
– Each link is divided into n channels using FDM or TDM.
• Each connection uses only one channel on each link.
– Communications phases
• Connection setup phase: End system A needs to request a connection to end system M that must be accepted
by all the switches and by M.
– A circuit (channel) is reserved on each link and the combination of these circuits defines the dedicated path.
• Data transfer phase: data can then be transferred.
• Connection teardown phase: After all data have been transferred, the circuits are torn down.

•

– Circuit switching is done at the physical layer.
– Resources (bandwidth in FDM (channels), switch buffers, switch processing time, etc.) need to be
reserved along the path and during the setup phase.
• These resources remain dedicated for the entire duration of data transfer.
• These resources are released at the teardown phase.
– During data transfer, the data are not delayed at each switch.
– Application
• Traditional telephone networks used the circuit-switching approach.
– Efficiency : such networks are not efficient because of the resource allocation during the entire connection duration.
Switching : Datagram networks
• In a packet-switched network, a message needs to be divided into packets.
– Packets, of fixed or variable size, are then sent from end system to another.
• No resource allocation for a packet transmission.
– Resources are allocated on demand.
• A packet must wait if other packets are being transmitted.
• In a datagram network, each packet is treated independently even if it is a part of multipacket
transmission
– Packets are then refereed to as datagrams.
• Datagram switching done at the network layer.
• The switches are traditionally referred to as routers.
• The following example shows how the datagram approach is used to deliver four packets from A to X.
– The packets may travel through different paths even they belong to the same message.
– The packets may arrive at the destination out of order. They may also be lost or dropped (lack of resources).
• It is the responsibility of the upper layer to reorder datagrams or to ask for the lost datagrams

• Datagram network are sometimes referred to as connectionless networks
– Connectionless = The switch doesn’t keep information about the connection state.
– There are no connection or teardown phases.
Switching : Datagram networks
• Routing table
– To route packets to their destination, each switch has a routing table.
– The routing table records the destination addresses and the corresponding forwarding ports.
– Routing tables are dynamic and are updated periodically.
– Each packet carries a header that contains, among other information, its destination address
• When the switch receives the packet, the packet destination address is examined and the routing table is
consulted to find the corresponding forwarding output port.
• This address remains unchanged during all the entire packet journey (unlike in circuit-switched network).

• Efficiency: better than in a circuit-switched network.
– The resources may be reallocated to packets from another sources between two packet transmissions
from the same source.
• Delay : greater than in a circuit-switched network.
• Switching in the Internet is done by using the datagram approach to packet switching at the
network layer
– It uses the universal addresses defined in the network layer to route the packets from the source to the
destination
Switching: Virtual-Circuit networks
• A cross between circuit-switched and datagram networks.
• Characteristics
– There are setup, data transfer, and teardown phases.
– Resources can be allocated during the setup phase or on demand.
– Data are packetized.
– The address in each packet header contains the next switch and the channel on which the packet is
being carried.
– All packets follow the same path during the connection.
– Implemented in the data link layer.
• Global addressing: each source or destination has a global address which is unique.
– Used only to create a VCI.
• Virtual-circuit identifier (VCI): identifier used for data transfer and has only a switch scope.
– When a frame arrives, it has a VCI. When it leaves, it has a different VCI.
• Each switch has a table with an entry for each active virtual circuit.

• Used with switched WANs such as Frame Relay and ATM networks.
Switching
• Packet switch structure
– Packet switch components
•   Input ports: performs the physical and data link functions of the switch.
•   Output ports: same as above but in the reverse order.
•   Routing table: performs the function of the network layer.
•   Switching fabric: responsible to move packets from the input queue to the output queue.

• Circuit switch structures
– Two possible technologies
• Space division switch: the paths are separated from one another spatially.
• Time division switch: uses TDM inside the switch.

Space division : crossbar switch `
Using telephone network for data transmission
• Telephone networks have been originally created for voice communication.
• To communicate digital data over the telephone networks:
– Dial-up modems.
– Digital Subscriber Line (DSL) for high speed Internet access.
• Telephone network
–Telephone networks, or POTS (Plain Old Telephone System), uses circuit switching.
– It was originally an analog system using analog signals to transmit voice.
– In 1980, telephone network began to carry data in addition to voice.
– The network is now digital and analog.
– Telephone network major components
• 1) Local loops, 2) Trunks, and 3) Switching offices: with several levels : end offices, tandem offices, and
regional offices.

• Local loops
– A twisted-pair cable that connects the subscriber to the nearest end office or local central office.
– When used for voice, it has a bandwidth of 4KHz.
– The first three digits of a local phone number define the office, and the next four digits define the local loop number.
• Trunks
– Transmission media that handle the communication between offices usually optical fibers or satellite links.
– A trunk normally handles hundreds of thousands of communication through multiplexing.
Using telephone network for data transmission
• Telephone Network major components (cont’d)
–Switching offices
• Switches are located in a switching offices to avoid having a permanent physical link between any two subscribers.
• A switching office connects several local loops or trunks and allow a connection between different subscribers.
• LATAs (Local-Access Transport Area)
–   Ex.: US is divided into 200 LATAs.
–   The services inside a LATA are called intra-LATA services.
–   The carrier that handles these services is called a local exchange carrier (LEC).
–   The services between LATAs are handled by interexchange carriers (IXCs).
• A telephone call going through an IXC is normally digitized, with the carrier using several types of networks to provide the
service.
• Signaling : required to establish a call
–The phase of setup and teardown needed in a circuit-switched network.
–In-band signaling
• Used in the past and done by human operators.
• The same circuit is used for both voice communication and signaling.
–Out-band signaling
• Signaling becomes automatic.
• The switches use, the digital signal defining the caller telephone number, to create a connection with the called parties.
• A portion of the voice bandwidth is used for signaling. The voice bandwidth and the signaling bandwidth were separate.
• Provide dial-tone (ring tone, busy tone), keeping billing information, provide caller ID, voice mail, etc.
–Actually, the telephone network could be thought as two networks
• A signaling network and a data transfer network. The two networks use different channels of the same links.
–The data transfer network is, for the most part, a circuit-switched network although it can also be a packet-switch network.
–The signaling network is a packet-switched network.
• The protocol used by the signaling system is called SS7 (Signaling System Seven) and is similar to the OSI model but with
different layer names.
Using telephone network for data transmission
• Services provided by Telephone networks
– Analog services : can be categorized as
• Analog switched services
– The familiar dial-up service encountered when a home phone is used.
– The signal on the local loop is analog and the bandwidth in [0,4]KHz
• Analog leased service
– Offers the customer to lease a line (dedicated line) that is permanently connected to another customer.
– Digital services: less sensitive to noise and interference. The two most common services
are
• Switched/56 service: is the digital version of an analog switched line. Data rates up to 56Kbps.
• Digital data service: is the digital version of an analog leased line. Data rates up to 64Kbps.
• Dial-up modems
– Traditional telephone lines carry frequencies in [300,3300]Hz
• This bandwidth = 3000Hz is used for voice transmission.
– To ensure data integrity, only the range [600,3000](B=2400Hz) Hz is used for data
communication.
Using telephone network for data transmission
•Dial-up modems (cont’d)
–Modem stands for modulator/demodulator.
• A modulator creates a bandpass analog signal from binary data.
• A demodulator recovers the binary data from the modulated signal.

–Modem standards: most of the popular modems are based on the V-series standards
• V.32 and V.32bis
– V.32 uses a combined modulation and encoding technique called Trellis-coded modulation.
• Essentially QAM plus a redundant bit (=> 5 bits) for error detection.
• => Data rate = 4*2400=9600bps.
– V.32bis uses 128-QAM transmission (7 bit/baud with one bit for error control)
• => Data rate = 6*2400=14400bps.
• Inclusion of automatic fall-back and fall-forward feature to adjust the speed depending on the quality of the line.
Using telephone network for data transmission
• Dial-up modems (cont’d)
– Modem standards
• V.34bis
– Provides a bit rate of 28800 with a 960-point constellation and a bit rate of 336000bps with a 1664-point constellation.
• V.90
–   Traditional modems have a data rate limit of 33,6Kbps (Shannon capacity).
–   V.90 modems (called 56K modems) have a rate of 56000bps.
–   These modems may be used only if one party is using digital signaling (ex. Internet provider).
• In uploading: the analog signal must still be sampled. Quantization noise reduces the SNR which limits the rate to
33.6Kbps
• In the downloading: there is no sampling. There is no quantization noise and the signal is not subject to Shannon formula.
This allows a 56 Kbps.
- The Telephone company samples 8000 times/s with 8-bit/sample (one bit for control) => 8000*7=56Kbps

• V.92
Using telephone network for data transmission
• Digital Subscriber Line (DSL)
– Technology used to provide higher-speed access to the Internet over the local loop
• Such a technology appeared after the traditional modems reached their peak data rate.
– A set of technologies, referred as xDSL, each differing in the first letter: ADSL, VDSL, etc
• Asymmetric: Higher bit rate in the downstream direction than in the upstream direction.
• The available bandwidth of the local loop is divided unevenly for the residential customer
• Not suitable for business customers.
• The filter at the end office (where the local loop terminates) is removed
• => This makes the entire 1.1MHz of a twisted-pair available for data and voice communications.
• Factors like the distance, the cable size, etc. affect this bandwidth.
• The ADSL data rate is not fixed (adaptive) and changes according to conditions and type of local loop.
• Modulation technique used is DMT (Discrete Multitone Technique) which combines QAM and
FDM.
• Typically, the available bandwidth of 1.104MHz is divided into 256 channels each uses a bandwidth of 4.312KHZ
• Voice (channel 0) ; idle (channels 1 to 5) provide a gap between voice and data communication.
• Upstream data and control (channels 6 to 30) ; Downstream data and control (channels 31 to 255) with
Using telephone network for data transmission
• DSL (cont’d)
– Installed at the customer’s site.
– The local loop connects to a splitter which separates voice and data communications.

• DSLAM (Digital Subscriber Line Access Multiplexer)
– Installed at the telephone company site.
– It packetizes the data to be sent to the ISP.

• The ADSL installation is expensive.
• ADSL Lite technology allows an ADSL Lite modem to be plugged directly into a telephone jacket
and connected to the computer.
• The splitting is done at the telephone company.
• Uses 256 DMT carriers with 8-bit modulation.
Using telephone network for data transmission
• DSL (cont’d)
– HDSL (High-bit-rate DSL)
•   An alternative to T-1 line.
•   Uses 2B1Q encoding which is less susceptible to attenuation.
•   A data rate of 1.544 (sometimes 2) Mbps can be achieved without repeaters.
•   Uses two twisted-pairs (one for each direction) to achieve full-duplex transmission.
– SDSL (Symmetric DSL)
• One twisted-pair version HDSL.
• Provide full-duplex symmetric communication in each direction.
– Supports up to 768 Kbps in each direction.
• Suitable for business that need to send and to receive data in large volume in both directions.
– VDSL (Very high-data-rate DSL)
• Alternative to ADSL, uses coaxial, fiber optic, or twisted-pair for short distances.
• Modulation technique is DMT.
• Provides a range of bit rates (25 to 55 Mbps) for upstream and 3.2Mbps for downstream.

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