Data Transmission and Computer Networks

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Transcript Data Transmission and Computer Networks

Data Transmission and Computer
Networks
The Physical
Layer
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Continuous Signal



No break or discontinuities in a signal.
Example: Speech.
A signal is continuous if:
lim s(t )  s(a)
t a
a
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Discrete Signal


A discrete signal takes on only a finite number
of values.
Example: binary 1s and 0s.
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Definitions
Sine wave: x(t )  A sin( 2πft  f )
where x(t) is the signal at time t,
A is the maximum amplitude of the signal,
f represents the number of cycles per
second, and
f defines the phase of the signal.
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Definitions
Cosine wave:

If the phase shift of a sine wave is –90 degrees (-p/2
radian), the same signal can be expressed as a cosine
wave instead of sine wave.
π
x(t )  A cos( 2πft )  A sin( 2πft  )
2
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Definitions
Amplitude:
–
The amplitude is the instantaneous value of a signal
at any time.
Frequency:
–
The Frequency is the inverse of the period, or the
number of repetitions of periods per second. Its unit
is Hertz (Hz).
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Definitions
Phase:
–
–
–
The phase describes the position of the waveform
relative to time zero. The range of shift is within a
single period of a signal.
The phase is a measure in degree or radian (2p =
360o).
The figure shows two signals that are out of phase by
p/2 radians.
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Definitions
Relationship between different phases:
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Definitions
Example 2.1:
The electricity that comes into a house is a simple sine wave. The
maximum amplitude is approximately 127 volts and the frequency is 60
Hz. Write the mathematical equation.
Solution of Example 2.1:
  2πf
   2
x(t )  Asin(2 πft  φ)
22
 60  377 radians/se cond
7
 x(t )  127sin(377 t )
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Definitions
Example 2.2:
Your voice is a summation of sine waves, each sine wave having its
own frequency, phase, and amplitude. The range of frequencies is
normally between 300 and 3300 Hz. Give a general equation.
Solution of Example 2.2:
x(t )  A1 sin( 2πf 0t  f1 )  A2 sin( 2πf 2t  f2 )    An sin( 2πf nt  fn )
with 300 Hz < fi < 3300 Hz. f0 is called the fundamental frequency,
and f2, f3 … fn are called the harmonics.
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Definitions
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Periodic Signal
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A periodic signal is a signal that repeats itself at
equal time interval.
It is made up of a finite series of sinusoidal
frequency components.
A signal is periodic if and only if:
s(t  T )  s(t )
   t  
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Periodic Signal
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Periodic Signal

The period (T) of the periodic signal
determines the fundamental frequency
component: reciprocal of the period in
seconds yields the frequency in Hz.

The other components have frequencies which
are multiples of the fundamental frequency
component, and known as the harmonics.
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Periodic Signal

Mathematically, we can express any periodic waveform as follows:
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Periodic Signal
Fourier Analysis
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Basic Binary Signals
1- Unipolar Signal:
– The amplitude of a unipolar signal varies between +V and 0 volts.
– It is called Return-to-Zero (RZ) signal.
– A unipolar signal has mean signal level of V/2 volts.
2- Bipolar Signal:
– The amplitude of a bipolar signal varies between +V and -V volts.
– It is called Non-Return-to-Zero (NRZ) signal.
–
A bipolar signal has mean signal level of
zero.
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Basic Binary Signals
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Basic Binary Signals

The mathematical expressions for unipolar and
bipolar signals are:
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Time Domain Representation of a Signal
1
sin( 2πf 0t )  sin( 2π(3 f 0 )t )
3
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Frequency Domain Representation of a
Signal
1
s (t )  sin( 2pf 0t )  sin( 2p (3 f 0 )t )
3
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Time Domain and Frequency Domain
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Frequency Spectrum and Bandwidth
The frequency spectrum of a signal is the collection of all
component frequencies it contains and is shown using a
frequency domain graph.
The bandwidth of a signal is the width of the frequency
spectrum. To calculate the bandwidth, subtract the lowest
frequency from the highest frequency of the range.
A digital signal contains an infinite number of frequencies
with different amplitudes. However, if we send only those
components whose amplitudes are significant (above an
acceptable threshold), we can still recreate the digital
signal with reasonable accuracy at the receiver.
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Exact and Significant Spectrums
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Decomposition of a Digital Signal
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Significant Bandwidth
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Significant Bandwidth
Example 2.3:
If a periodic signal is decomposed into five sine waves with
frequencies of 100, 300, 500, 700, and 900 Hz, what is the
bandwidth?
Solution of Example 2.3:
Let fh be the highest frequency, fl be the lowest frequency, and
B be the bandwidth. Then,
B = fh - fl = 900 – 100 = 800 Hz
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Significant Bandwidth
Example 2.4:
A signal has a bandwidth of 20 KHz. The highest frequency is
60 KHz. What is the lowest frequency?
Solution of Example 2.4:
Let fh be the highest frequency, fl be the lowest frequency, and
B be the bandwidth. Then,
B = fh - fl

fl = 60 – 20 = 40 KHz
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Medium Bandwidth and Channel
Capacity
A transmission medium has a limited bandwidth, which
means that it can transfer only some ranges of frequencies.
A transmission medium with particular bandwidth is
capable of transmitting only digital signals, whose
bandwidth is less than the bandwidth of the medium.
If the signal is sent on a transmission medium whose
bandwidth is less than the required significant bandwidth,
the signal may be so distorted that is not recognized at the
receiver.
The channel capacity is the data rate, in bit per second
(bps), at which data can be communicated.
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Medium Bandwidth and Channel
Capacity
Example 2.5:
What bandwidth is required for data being sent at a rate of 10
bps?
Solution of Example 2.5:
In the worst-case scenario, the data consist of alternating 0s
and 1s. This is the situation that will require the largest
bandwidth. Each 1 and 0 bit combination can be considered
one cycle. Therefore, the required bandwidth is 5 Hz.
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Medium Bandwidth and Channel
Capacity
Example 2.6:
We want to transmit 10 pictures per second. Each picture is
made by 5-by-5 pixels (picture elements). What is the
required bandwidth?
Solution of Example 2.6:
Each picture is made of 25 pixels. We assume that we send
one bit per pixel (1 for black, 0 for white). Therefore, we can
send 25 bit per picture and 250 bit per second. Therefore, the
required bandwidth is 125 Hz.
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Medium Bandwidth and Channel
Capacity
Example 2.7:
A television screen composed of a grid of 525 lines by 700
columns (total of 367,500 pixels). A pixel can be black and
white. Thirty complete frames are scanned in one scanned.
What is the bandwidth required?
Solution of Example 2.7:
The number of bits that must be sent per second is 30 ×
367,500 = 11,025,000 bps. If one bit is sent per pixel, then the
required bandwidth is 5,513,500 Hz (6 MHz approximately).
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Attenuation and Distortion

Transmitted electrical signals are attenuated and distorted by the
transmission medium.

The extent of attenuation and distortion is strongly influenced by:
– The type of transmission medium.
– The bit rate of the data being transmitted.
– The distance between the two communicating devices.
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Attenuation and Distortion
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Attenuation And Distortion
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1- Attenuation
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As a signal propagates a long a transmission line, its amplitude
decreases. Therefore, a limit for the cable length must be set.
If the cable is longer, one or more repeaters (amplifiers) must be
inserted.
We measure both attenuation and amplification (gain) in decibels
(dB).
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1- Attenuation

If we denote transmitted signal power level by
P1 and the received power by P2, then
( P1  P2 )
Attenuation  10 log 10
P1
P2
dB
( P1  P2 )
Amplification  10 log 10
P2
P1
dB
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1- Attenuation
Example 2.8:
A transmission channel between two DTEs is made up of three
section. The first introduces an attenuation of 16 dB, the second an
amplification of 20 dB, and the third an attenuation of 10 dB. Assuming
a mean transmitted power level of 400 mW, determine the mean
output power of the channel.
P1 = 400 mW
16 dB
Attenuation
P2
20 dB
Amplification
P3
10 dB
Attenuation
P4
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1- Attenuation
Solution of example 2.8:
First section:
Attenuation  10 log 10
16  10 log 10
1.6  log 10
10  10
1.6
39.81 
P2 
P1
P2
400
P2
400
P2
log10
400
P2
400
P2
400
 10.0475mW
39.81
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1- Attenuation
Solution of example 2.8 (Continued):
Second section:
Amplification  10 log 10
20  10 log 10
2  log 10
10  10
2
100 
P3
P2
P3
10.0475
P3
10.0475
log10
P3
10.0475
P3
10.0475
P3  10.0475 100  1004.75mW
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1- Attenuation
Solution of example 2.8 (Continued):
Third section:
Attenuation  10 log 10
10  10 log 10
1  log 10
10  10
1
10 
P4 
P3
P4
1004.75
P4
1004.75
P4
log10
1004.75
P4
1004.75
P4
1004.75
 100.475mW
10
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1- Attenuation
Solution of example 2.8 (Continued):
Or, Overall attenuation channel = (16 - 20) + 10 = 6 dB
Attenuation  10 log 10
6  10 log 10
10
0.6
 10
3.981 
P4 
log10
P1
P4
400
P4
400
P4
400
P4
400
 100.475mW
3.981
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2- Limited Bandwidth

Channel Bandwidth specifies the sinusoidal frequency
components from 0 up to some frequency fc that will be
transmitted by the channel undiminished. All frequencies
above this cutoff frequency are strongly attenuated.

In general, channel bandwidth refers to the width of the range
of frequencies that channel can transmit, and not the
frequency themselves.

If the lowest frequency a channel can transmit is f1 and the
highest is f2, the the bandwidth is: f2 – f1.

Because the telephone line can transmit frequencies from
approximately 300 to 3300 Hz, its bandwidth is 3 KHz.
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2- Limited Bandwidth
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LIMITED BANDWITH
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2- Limited Bandwidth


The sequence 101010… generates the highest-frequency
components, while a sequence of all 1s or all 0s is equivalent to a
zero frequency of the appropriate amplitude.
The channel capacity is the data rate, in bit per second (bps), at
which data can be communicated.
In 1928, Nyquest developed the relationship between bandwidth (W)
and the channel capacity (C) in noise-free environment. The Nyquest
relationship is:
C  2W
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2- Limited Bandwidth
Example 2.9:
A binary signal of rate 500 bps is to be transmitted
over a communication channel. Derive the
minimum bandwidth required assuming:
(a)
(b)
(c)
The fundamental frequency only,
The fundamental and third harmonic, and
The fundamental, third, and fifth harmonic of the
worst-case sequence are to be received.
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2- Limited Bandwidth
Solution of Example 2.9:
The worst case sequence 101010… at 500 bps has a
fundamental frequency component of 250 Hz. Hence the
third harmonic is 750 Hz and the fifth harmonic is
1250Hz.
The bandwidth required in each case is as follows:
(a) 0-250 Hz.
(b) 0-750 Hz.
(c) 0-1250 Hz.
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2- Limited Bandwidth
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2- Limited Bandwidth

We can transmit more than one bit with each change in the signal
amplitude, therefore increasing the data bit rate.

With multilevel signaling in noise-free environment, the Nyquest
formulation becomes:
C  2W log 2 M
Where
C is the channel capacity in bps.
W is the bandwidth of the channel in Hz.
M is the number of levels per signaling elements.
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2- Limited Bandwidth
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2- Limited Bandwidth
For Limited-bandwidth channel such as PSTN, we can
often use more than two levels. This means that each
signal element can represent more than a single binary
digit.
In general, if the number of signal levels is M, the number
of bits per signal element m, is given by:

m  log 2 M

The rate of change of signal is known as the signaling
rate (Baud rate) (Rs), and measures in baud.
Rs  2W
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2- Limited Bandwidth

It is related to the data bit rate, R, by the
following expression:
R  Rs m

The signaling element time period, Ts, is given
by:
Ts 

1
Rs
The time duration of each bit, Tb, is:
Tb 
1
R
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2- Limited Bandwidth
M  2  m  1  R  Rs
M  4  m  2  R  2  Rs
M  8  m  3  R  3  Rs
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2- Limited Bandwidth
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2- Limited Bandwidth

The bandwidth efficiency of transmission
channel, B, is defined as:
R
B
 2m
W
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2- Limited Bandwidth
Example 2.10:
Data is to be transmitted over the PSTN using a transmission scheme with
eight levels per signaling element. If the bandwidth of the PSTN is 3000 Hz,
determine the Nyquest maximum data transfer rate (C) and the bandwidth
efficiency (B).
Solution of Example 2.10:
C  2W log 2 M
C  2  3000  3  18000
B  2m
B  23  6
bps
bps / Hz
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3- Delay Distortion
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The rate of propagation of a sinusoidal signal along a
transmission line varies with the frequency of the signal.
When we transmit a digital signal with various frequency
components, making up the signal, arrive at the receiver with
varying delays, resulting in delay distortion of the received
signal.
Note that:
v

f
v

f
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3- Delay Distortion
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4- Noise
There are several sources for the channel noise:
I. Crosstalk:
It is caused by unwanted electrical coupling
between adjacent lines. This coupling results in a
signal that is being transmitted in one line being
picked up by adjacent lines as a small noise
signal.
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4- Noise
II. Impulse Noise:
 Impulse noise is a sharp spike of energy for a
small time duration.

Example: A lightning discharge.

If the duration of impulse noise is 0.01 second, it
will destroy 48 bits of data being transmitted at
4800 bps.
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4- Noise
III. Thermal Noise:
 At all temperatures above absolute zero, all transmission media
experience thermal noise, where absolute zero = 0 kelvin (K) = - 273ْ
C.
 The amount of thermal noise to be found in a bandwidth of 1 Hz in any
conductor is:
N o  kT

where No is the noise power density for one Hz (watts/Hz),
k is Boltzmann’s constant (1.3803 x 10-23 joule K-1), and
T is the temperature in Kelvin (K).
The thermal noise in watts present in a bandwidth of W Hz can be
expressed by:
N  W  No
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4- Noise

There are random perturbations on a transmission line
even no signal is being transmitted.

The Signal-to-Noise Ratio (SNR) is expressed in
decibels as:
S
SNR  10 log 10 ( )
N
dB
where S is the average power in a received signal, and
N is noise power.
 High SNR means a high power signal relative to the
prevailing noise level, resulting in a good-quality signal.
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4- Noise

In 1948, Shannon calculated the theoretical maximum bit rate capacity of
a channel of bandwidth W as
S
C  W log 2 (1  )
N
where C is the channel capacity in bps,
W is the bandwidth of the channel in Hz,
S is the average signal power in watts, and
N is the thermal noise power in watts.

Note that:
ln x
log 2 x 
ln 2
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4- Noise
Example 2.11:
Assuming that a PSTN has a bandwidth of 3000 Hz and a signal-to-noise ratio
of 20 dB, determine the maximum theoretical data rate that can be achieved.
Solution of Example 2.11:
S
SNR  10 log 10 ( )
N
C  W log 2 (1 
S
20  10 log 10 ( )
N
S
)
N
C  3000 log 2 (1  100)
C  3000
S
 10 2  100
N
ln 101
 19963bps
ln 2
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Transmission Media
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The transmission medium is the physical path between transmitter
and receiver in data transmission system.
The transmission media may be classified as guided or unguided
media.
With guided media, the waves are guided along the physical path;
example of guided media are twisted pair, coaxial cable, and optical
fiber.
Unguided media provide a means for transmitting electromagnetic
waves, but do not guide them; example propagation through air,
vacuum and seawater.
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Transmission Media
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For unguided media, at lower frequencies, signals are
omnidirectional; that is, the signal propagates in all directions
from the antenna. At higher frequencies, it is possible to focus the
signal into a directional beam.
Three most important unguided transmission techniques:
terrestrial an satellite microwave, and radio.
Microwave frequencies cover a range of about 2 to 40 GHz. At
these frequencies, higher directional beam are possible.
Signals in range 30 MHz to1 GHz are radio waves.
Omnidirectional transmission is used and signals at these
frequencies are suitable for broadcast operations.
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Transmission Media
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Transmission Media
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Transmission Media
1. Guided Media
I. Two-Wire Open Lines:
 It is adequate for connecting equipment that is up to 50 meters apart.
 Bit rate is less than 19.2 kbps.
 Problems of Two-wire open Lines:
– Crosstalk: It is caused by cross-coupling of electrical signals
between adjacent wires in the same cable.
– It can pick-up noise signals from other electrical sources caused by
electromagnetic radiation.
 These problems contribute to the limited length of line and bit rates.
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Transmission Media
I. Two-Wire Open Lines (Continued):
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Transmission Media
II. Twisted Pair:
 The most common transmission media for both analog and digital
data is twisted pair.
 A twisted pair consists of two insulated copper wires arranged in a
regular spiral pattern.
 The proximity the signal and ground reference wires means that
any interference signal is picked up by both wires reducing its
effect on the difference signal.
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Unshielded Twisted Pair (UTP):
The
UTP has several categories. These categories define the quality level of the
cable. The categories are:
–
Category 1: This is a common telephone cable.
–
Category 2: CAT2 cable was first networking UTP but is now considered
obsolete. It supports data rate up to 1Mbps.
Category 3: CAT3 is seldom used outside an IBM environment and has been
–
replaced by CAT5. CAT3 is used in 10Mbps Ethernet LANs and 4Mbps Token
Ring LANs.
- Category 4: CAT4 has been specified for up to 20Mbps and is specific
for 16Mbps Token Ring LANs.
–
–
Category 5: CAT5 has been specified for data rates up to 100Mbps. This is the
preferred cabling installation for Ethernet and Token Ring.
–
Category 6: CAT6 has been specified for data rates up to 1000Mbps.
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Transmission Media
II. Twisted Pair (Continued):
 EIA/TIA-568A/B compliant refers to which of the four pairs in the UTP
cable are designated as transmit, and which are designated as receive.
Use the following as a guide:
– EIA/TAI-569A: Devices transmit over pair 3, and receive over pair
2.
– EIA/TAI-569B: Devices transmit over pair 2, and receive over pair
3.
 It is important to terminate all cables at a location to the same standard,
but not both at the same facility.
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Transmission Media
III. Coaxial Cable:
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Transmission Media
III. Coaxial Cable (Continued):
 Networking coaxial cable has a maximum data rate of 10Mbps,
and fair noise immunity. However, it is more expensive to install
than twisted wire.
 Cable TV networks can utilize coaxial cable with a 400MHz (or
higher) bandwidth. Therefore, 52 TV channels can be carried on a
single cable.
 The disadvantages to the coaxial cable is the cost and lack of
compatibility with twisted wires.
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Transmission Media
III. Coaxial Cable (Continued):
 The usable bandwidth of a coaxial cable can be as much as 350
MHz (or higher). We can utilize the high bandwidth in one of the
two ways:
– Baseband mode, in which all the available bandwidth is
used to derive a single high bit rate transmission channel.
– Broadband mode, in which the available bandwidth of the
cable is divided into a number of smaller subfrequency bands
or channels.
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Transmission Media
III. Coaxial Cable (Continued):
Types of the coaxial cables:
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Transmission Media
III. Coaxial Cable (Continued):
a. Thicknet (10BASE5):
 The diameter of Thicknet is 0.5 inch with a maximum segment
length 500 meter.
 It supports 100 transceivers on each segment. The number of
connections is limited to prevent signal attenuation.
 The transceiver is a device to send and receive data to and from
the cable.
 The minimum spacing of transceiver taps is 2.5 meters.
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Transmission Media
III. Coaxial Cable (Continued):
b. Thinnet (10BASE2):
 The diameter of Thinnet is 0.25 inch with a maximum
segment length 185 meter.
 It supports 30 transceivers on each segment.
 The minimum spacing of transceiver taps is no
closer than 0.5 meters.
 To connect a transceiver, the cable is cut and the
ends prepared for BNC connectors.
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Transmission Media
IV. Optical Fiber




An optical fiber is a thin (2 to 125 mm), flexible medium
capable of conducting an optical ray.
Various glasses and plastics can be used to make
optical fibers.
An optical fiber cable has a cylindrical shape and
consists of three concentric sections: the core, the
cladding, and the jacket.
The core is the innermost section and consists of one
or more very thin fibers made of glass or plastic.
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Transmission Media
IV. Optical Fiber (Continued):


Each fiber is surrounded by its own cladding, a glass or plastic
coating that has optical properties different from those of the core.
The outermost layer, surrounding one or a bundle of cladded
fibers, is the jacket. The jacket is composed of plastic and other
materials to protect against moisture, abrasion, and crushing
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Transmission Media
IV. Optical Fiber (Continued):
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Transmission Media
IV. Optical Fiber (Continued):
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Transmission Media
IV. Optical Fiber (Continued):
Light:
 Light is a form of electromagnetic energy.
 The speed of light depends on the density of the medium which it is
travelling (the higher the density, the slower the speed).
 It travels at it fastest in a vacuum: 3×108 m/s. The speed decreases
as the medium becomes denser.
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Transmission Media
IV. Optical Fiber (Continued):
Advantages of Optical Fibers:





Great bandwidth.
Smaller size and lighter weight.
Lower attenuation.
Electromagnetic isolation:
– Fiber-optic uses light for transmission rather that electricity.
– External light is the only possible interference and can be blocked by
the outer jacket.
– The system is not vulnerable to interference, impulse noise, or
crosstalk.
Greater repeater spacing.
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Transmission Media
IV. Optical Fiber (Continued):
Disadvantages of Optical Fibers:



Cost.
Installation/maintenance:
– All connections must be perfectly aligned and matched the core size.
– The connection must be complete and not overly tight. A gap between
two cores results in a dissipated signal, and overly tight connection can
compress the two core and alter the angle of refraction.
Fragility: Glass fiber is more easily broken than metallic wire.
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Transmission Media
2. Unguided Media:

Unguided media transport electromagnetic waves
without using a physical conductor.

Signals are broadcast through air.
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Transmission Media
2. Unguided Media:
Radio Frequency Allocation


The electromagnetic spectrum is divided into eight
ranges, called bands, each regulated by governmental
authorities.
These bands are rated from very low frequency (VLF)
to extremely high frequency (EHF).
Radio Communication
Radio, microwave, satellite
Sami Al-Wakeel
Transmission Media
2. Unguided Media:
Radio Frequency Allocation:
3 KHz
300 GHz
Radio Communication
Radio, microwave, satellite
3 KHz
30 KHz
VLF
Surface
300 KHz
LF
3 MHz
MF
Troposphere
30 MHz
HF
300 MHz
VHF
Ionosphere
3 GHz
UHF
SHF
Space and
line-of-sight
300 GHz
30 GHz
EHF
Space
Sami Al-Wakeel
Transmission Media
2. Unguided Media:
Radio Frequency Allocation:
VLF
LF
MF
Very Low Frequency
Low Frequency
Middle Frequency
HF
VHF
UHF
High Frequency
Very High Frequency
Ultra High Frequency
SHF
EHF
Super High Frequency
Extremely High Frequency
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Transmission Media
2. Unguided Media:
Propagation of Specific Signals:
I. Very Low Frequency (VLF):
 VLF waves are propagated as surface waves.
 Subject to high levels of of atmospheric noise (heat and electricity).
 Used
for long-range radio navigation and for submarine
communications.
II. Low Frequency (LF):
 LF waves are also propagated as surface waves.
 Used for long-range radio navigation and for radio bacons or
navigational locators.
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Transmission Media
2. Unguided Media:
Propagation of Specific Signals:
III. Middle Frequency (MF):
 MF signals are propagated in the troposphere.
 These frequencies are absorbed by the ionosphere.
 The distance they can cover limited by the angle needed to reflect
the signal within the troposphere without entering the ionosphere.
 Absorption increases during the daytime.
 Used for AM radio and emergency frequencies.
300 KHz
3 MHz
AM Radio
535 KHz
1.602 MHz
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Transmission Media
2. Unguided Media:
Propagation of Specific Signals:
IV. High Frequency (HF):
 HF signals use ionospheric propagation.
 These frequencies moves into the ionosphere, which reflects them
back to the earth.
 Used for international broadcasting, military communication,
telephone, facsimile.
3 MHz
30 MHz
Frequency range for HF
Sami Al-Wakeel
Transmission Media
2. Unguided Media:
Propagation of Specific Signals:
V. Very High Frequency (VHF):
 VHF signals use line-of-sight propagation.

Used for VHF television, FM radio, and aircraft AM radio.
Channels 2-6
30 MHz
FM
TV
54
Channels 7-13
88
Aircraft
108
300 MHz
TV
174
216
Frequency range for VHF
Sami Al-Wakeel
Transmission Media
2. Unguided Media:
Propagation of Specific Signals:
VI. Ultra High Frequency (UHF):
 VHF signals always use line-of-sight propagation.
 Used for UHF television, mobile telephone, and microwave.
 Note that microwave communication begins at 1 GHz in the UHF
band.
Mobile
telephone Channels 14-69
300 MHz
UHF TV
470 MHz
806 MHz
Microwave
3 GHz
1 GHz
Frequency range for UHF
Sami Al-Wakeel
Transmission Media
2. Unguided Media:
Propagation of Specific Signals:
VII. Super High Frequency (SHF):
 VHF signals use line-of-sight and space propagation.
 Used for terrestrial and satellite microwave, and radar
communications.
3 GHz
Microwave
30 GHz
Frequency range for SHF
Sami Al-Wakeel
Transmission Media
2. Unguided Media:
Propagation of Specific Signals:
VII. Extremely High Frequency (EHF):
 EHF signals use space propagation.
 Used for radar, satellite and experimental communications.
30 GHz
Microwave
300 GHz
Frequency range for EHF
Sami Al-Wakeel
Transmission Media
2. Unguided Media:
I. Satellites:




Data can be transmitted using electromagnetic waves through the
free space.
A satellite receives and retransmits (relays) the data to the
predetermined destinations.
A typical satellite channel has high bandwidth (500 MHz) and can
provide many hundreds of high bit rate data links using multiplexing
technique.
Multiplexing: Total channel capacity is divided into a number of
subchannels, each can support high bit rate link.
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Transmission Media
I. Satellites (Continued):

Satellites are geostationary, which means that the satellite
orbits the earth once every 24 hours in synchronism with
earth’s rotation.
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Transmission Media
I. Satellites (Continued):
Satellite Transmission Modes:
1. Point-to-Point Link:
Long distance
telephone transmission.
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Transmission Media
I. Satellites (Continued):
2. Broadcast Link:
Television distribution.
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Transmission Media
I. Satellites (Continued):
3. Multipoints using Very Small Aperture Terminals (VSAT):
–
–
VSATs is used for private business networks.
The concept of VSAT is:
 Typically, a computer is connected to each VSAT and can
communicate with the central computer connected to the hub.
 Normally, the central site broadcasts to all VSATs on a single
frequency, while in the reverse each VSAT transmit at a
different frequency.
Sami Al-Wakeel
Transmission Media
I. Satellites (Continued):



To communicate with a particular VSAT, the central
site broadcasts the message with the identity of the
intended VSAT at the head of the message.
For VSAT-to-VSAT communication, all messages are
first sent to the central site – via the satellite- which
then broadcasts them to the intended recipients.
Note that direct VSAT-to-VSAT is possible without
passing through a central site.
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Transmission Media
I. Satellites:
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Transmission Media
I. Satellites:
Frequency Bands for Satellite Communications:
Each satellite sends and receives over two different bands.
 Transmission from the earth to the satellite is called uplink.
 Transmission from the satellite to the earth is called downlink.
Band
Downlink
Uplink
C
3.7 - 4.2 GHz
5.925 – 6.425
GHz
Ku
11.7 - 12.2
GHz
14 – 14.5 GHz
Ka
17.7 – 21.0
GHz
27.5 – 31 GHz
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Transmission Media
II. Terrestrial Microwave:



Terrestrial microwave links are widely used to provide
communication links when it is impractical or too expensive to
install physical transmission media, for example across a river
or desert.
The distance coverable by a line-of-sight signal depends on the
height if the antenna: the taller the antenna, the longer sight
distance.
Height allows the signal to travel further without being stopped
by the curvature of the planet.
Sami Al-Wakeel
Transmission Media
II. Terrestrial Microwave:





Microwave signal propagate in one direction at a time, which
means that two frequencies are necessary for two-way
communication such as telephone conversation.
Transceiver is an single piece equipment, which allows a single
antenna to transmit and receive frequencies.
The common type of microwave antenna is the “dish”.
A typical size is about 10 ft in diameter.
The antenna focuses a narrow beam to achieve line-of-sight
transmission to the receiving antenna.
Sami Al-Wakeel
Transmission Media
II. Terrestrial Microwave (Continued):

Microwave antenna are usually located at substantial heights above
ground to extend the range between antennas and to be able to
transmit over intervening obstacles.

The maximum distance between antennas:
d  7.14 Kh
where d is the distance between antennas in kilometers,
h is the antenna height in meters, and
K is the adjustment factor.
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Transmission Media
II. Terrestrial Microwave:



To achieve long-distance transmission, a series of relay
microwave towers (repeaters) is used.
Microwave
communication
through
the
earth’s
atmosphere can be used reliably over distances in excess
of 50 km.
Microwave beam travels through the earth’s atmosphere,
therefore, it can be disturbed by the weather conditions.
Sami Al-Wakeel
Transmission Media
III. Cellular Telephony:



Cellular telephony is designed to provide connections
between twp moving devices or between one mobile unit
and one land unit.
A service provider must locate and track the caller, assign
channel to the call, and transfer the signal from channel to
channel as the caller moves out of the range of one
channel and into the range of another.
Each cellular service area is divided into small regions
called cells. Each cell contains an antenna and is
controlled by a switching office called a mobile telephone
switching office (MTSO).
Sami Al-Wakeel
Transmission Media
III. Cellular Telephony:


The MTSO coordinates communications between all the cell
offices and the telephone central office. It is computerized and
responsible for connecting cells, recording call information, and
billing.
The typical radius of a cell is 1 to 12 miles.
Sami Al-Wakeel
Transmission Media
III. Cellular Telephony:
Sami Al-Wakeel
Transmission Media
III. Cellular Telephony:
Cellular bands:





It is assigned two bands for cellular use.
The band between 824 and 849 MHz carries
communications that initiate from mobile phones.
The band between 869 and 894 MHz carries
communications that initiate from land phones.
Carriers are spaced every 30 KHz, allowing each band to
support up to 833 carriers.
For full-duplex communication, the required width for each
channel is 60 KHz and leaves only 416 channels per band.
Sami Al-Wakeel
Transmission Media
III. Cellular Telephony:
Cellular bands:
416 channels
824 MHz
849 MHz
416 channels
869 MHz
894 MHz
Sami Al-Wakeel
Transmission Media
III. Cellular Telephony:
Transmitting:
To place a call from a mobile phone, the caller enters a code of 9
digits and pressed a send buttons.
 The mobile phone scans the band, seeking a setup channel.
 It sends the phone number to the closest cell office using that
channel.
 The cell office relays the data to the MTSO.
 The MTSO sends the data to the telephone central office. IF the
called party is available, a connection is made and the result is
relayed to the MTSO.
 The MTSO assigns an unused voice channel to the call and a
connection is established.
 The mobile phone automatically adjust its tuning to the new
channel an voice communication can begin.
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Transmission Media
III. Cellular Telephony:
Receiving:
When a land phone places a call to a mobile phone, the
telephone office sends the number to the MTSO.
 The MTSO searches for the location of the mobile phone by
sending query signals to each cell in a process called
paging.
 One the mobile phone is found, the MTSO transmits a
ringing signal.
 When the mobile phone is answered, assigns a voice
channel to the call, allowing voice communication to begin.
Sami Al-Wakeel
Transmission Media
III. Cellular Telephony:
Handoff:




During the conversation, the mobile phone may move from one
cell to another. When it does, the signal may become weak.
To solve this problem, the MTSO, monitors the level of the signal
every few seconds.
If the strength of the signal is diminished, the MTSO seeks a new
cell that can accommodate the communication better.
The MTSO changes the channel carrying the call (hands the
signal off from the old channel to the new one).
Sami Al-Wakeel