Transcript Chapter # 3 Data and Signals

CHAPTER # 3
DATA AND SIGNALS
Introduction
2


One of the major functions of physical layer is to
move data in the form of electromagnetic signals
across a transmission medium.
Thus, the data must be transformed to
electromagnetic signals to be transmitted.
1. Analog and Digital Data
3


Data can be analog or digital.
The term analog refers to information that is
continuous
 e.g.

The term digital data refers to information that has
discrete states.
 e.g.
1.
2.
analog clock hh:mm:ss
digital clock hh:mm
Analog data take on continuous values.
Digital data take on discrete values.
Analog and Digital Signals
4
2. Periodic and Nonperiodic Signals
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


Both analog and digital signals can take one of two
forms: periodic or nonperiodic
A periodic signal
 Completes a pattern within a measureable time
frame.
 Repeats that pattern over subsequent identical period
A nonperiodic signal

Changes without exhibiting a pattern or cycle that repeats
over time.
Periodic and Nonperiodic Signals
6

In data communications, we commonly use:
 Periodic
analog signals ( because they need
less bandwidth).
 and
nonperiodic digital signals ( because they
can represent variation in data)
A. Periodic Analog Signals
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Periodic analog signals can be classified as simple or
composite.


A simple periodic analog signal, a sine wave, cannot
be decomposed into simpler signals.
A composite periodic analog signal is composed of
multiple sine waves.
1) Sine Waves
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

The sine wave is the most fundamental form of a periodic
analog signal.
A sine wave is represented by three parameters: Peak
amplitude, Frequency, and Phase.
1.
Peak amplitude: it is the absolute value of the highest
intensity. It is normally measured in volts.
Sine Waves
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2.
Frequency: it refers to the number of periods in 1 s. It is formally
expressed in Hertz (Hz).
 Period is the amount of time, in seconds, a signal needs to complete one
cycle (the completion of one full pattern).
Therefore , frequency and period are the inverse of each other.
 Note : Frequency is the rate of change with respect to time.
 Change in a short span of time means high frequency.
 Change over a long span of time means low frequency.
 If a signal does not change at all, its frequency is zero

Units of period and frequency
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TABLE 3.1
Example#1
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
The power we use at home has a frequency of 60
Hz. The period of this sine wave can be determined
as follows:
Example# 2
12
Express a period of 100 ms in microseconds.
 Solution
 From Table 3.1 we find the equivalents of 1 ms (1
ms is 10−3 s) and 1 s (1 s is 106 μs). We make the
following substitutions:.
Example# 3
13
The period of a signal is 100 ms. What is its frequency
in kilohertz?
 Solution
 First we change 100 ms to seconds, and then we
calculate the frequency from the period (1 Hz =
10−3 kHz).
Sine Waves
14
3.

Phase:
 It describes the position of the waveform relative to time
0.
 It is measured in degree or radian
To look to the phase is in term of shift or offsit:
1.
2.
3.
A sine wave with a phase 0° is not shifted.
A sine wave with a phase 90° is shifted to the left by
¼ cycle.
A sine wave with a phase 180° is shifted to the left
by ½ cycle.
Three sine waves with the same amplitude and
frequency, but different phases
15
The figure below show Two signals with the same phase
and frequency, but different amplitudes
16
Two signals with the same amplitude and phase, but
different frequencies
17
Wavelength
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

Wavelength binds the period or frequency of the simple sine
wave to the propagation speed of the medium.
Wavelength depends on both the frequency and the medium.
 Wavlength = propgation speed X period = progation
speed/ frequency
Example# 1
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In a vacuum, light is propagated with a speed of 3 X
108 m/s. (that speed is lower in air and cable.) . The
frequency of red light is 4 X 1014
 Wavelength is normally measured in micrometers.
Solution
 Wavelength= c/f = (3 X 108 ) / (4 X 1014)
= 0.75 X 10-6 m= 0.75 μm
Time and Frequency Domain
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
A complete sine wave in the time domain can be represented by one
single spike in the frequency domain.
Example
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This example Shows three sine waves, each with different amplitude
and frequency. All can be represented by three spikes in the frequency
domain.
2) Composite Signals
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
A single-frequency sine wave is not useful in
data communications; we need to send a
composite signal, a signal made of many
simple sine waves.
 e.g.
if we use single sine wave to convey a
conversation over the telephone. It would just
hear a buzz.
2) Composite Signals
23

According to Fourier analysis, any composite signal
is a combination of simple sine waves with different
frequencies, amplitudes, and phases.

A composite signals can be periodic or non
periodic.
A
periodic composite signal can be decomposed into a
series of simple sine waves with discrete frequencies (
with integer values [1,2,3 ,….] ).
A
nonperiodic composite signal can be decomposed
into a combination of an infinite number of simple sine
waves with continuous frequencies. ( with real value)
A composite periodic signal
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Explanation
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


The previous figure shows a periodic composite
signal with frequency f. This type of signal is not
typical of those found in data communications.
We can consider it to be three alarm systems, each
with a different frequency.
The analysis of this signal can give us a good
understanding of how to decompose signals.
Decomposition of a composite periodic signal in the
time and frequency domains
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The time and frequency domains of a
nonperiodic signal
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In this case, the composite signal cannot be periodic, because that
implies that we are repeating the same word or words with exactly
the same tone.
This can be the signal created by a microphone or a telephone set
when a word or two is pronounced.
Bandwidth
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
The bandwidth of a composite signal is the
difference between the highest and the lowest
frequencies contained in that signal.
 e.g.
if a composite signal contain frequencies
between 1000 and 5000, its bandwidth is 50001000 = 4000
The bandwidth of periodic and nonperiodic composite signals
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Example#1
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If a periodic signal is decomposed into five sine waves with
frequencies of 100, 300, 500, 700, and 900 Hz, what is its
bandwidth? Draw the spectrum, assuming all components have a
maximum amplitude of 10 V.
Solution
 Let fh be the highest frequency, fl the lowest frequency, and B the
bandwidth. Then

Example#2
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A periodic signal has a bandwidth of 20 Hz. The highest frequency is
60 Hz. What is the lowest frequency? Draw the spectrum if the signal
contains all frequencies of the same amplitude.
Solution
 Let fh be the highest frequency, fl the lowest frequency, and B the
bandwidth. Then

Example#3
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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.
Solution
 The lowest frequency must be at 40 kHz and the highest at 240
kHz. Figure 3.15 shows the frequency domain and the bandwidth.
B- DIGITAL SIGNALS
In addition to being represented by an analog
signal, information can also be represented by a
digital signal. For example, a 1 can be encoded
as a positive voltage and a 0 as zero voltage.
A digital signal can have more than two levels.
In this case, we can send more than 1 bit for each
level.
3.33
Figure 3.16 Two digital signals: one with two signal levels and the
other with four signal levels
3.34
Example # 1
A digital signal has eight levels. How many bits are
needed per level? We calculate the number of bits from
the formula
Each signal level is represented by 3 bits.
3.35
Example #2
A digital signal has nine levels. How many bits are needed
per level?
We calculate the number of bits by using the formula:
Log2 L= number of bits in each level
Log2(9)=3.17bits.
However, this answer is not realistic. The number of bits
sent per level needs to be an integer as well as a power of
2.
For this example, 4bits can represent one level.
3.36
Bit rate and bit interval
Most digital signals are nonperiodic, frequency and period
are not appropriate. Another terms instead of frequency is bit
rate and instead of period: bit interval(bit duration)
Bit rate: number of bits per second bps
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Bit interval=1/bit rate
Bit length =propagation speed x Bit interval
Example
Assume we need to download text documents at the rate
of 100 pages per minute. What is the required bit rate of
the channel?
Solution
A page is an average of 24 lines with 80 characters in
each line. If we assume that one character requires 8bits
The bit rate is:
=100x24x80x8/60
=25.6Kbps
3.38
Digital Signal as a composite Analog Signal
Note
A digital signal is a composite analog signal with an infinite
bandwidth.
Fourier analysis can be used to decompose a digital signal
o If the digital signal is periodic (rare in data
communications), the decomposed signal has a frequency
domain representation with an infinite Bandwidth and
discrete frequencies.
o If it is nonperiodic, the decomposed signal still has infinite
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Bandwidth, but the frequencies are continuous.
Figure 3.17 The time and frequency domains of periodic and
nonperiodic digital signals
3.40
Transmission of Digital Signals
How can we send a digital signal from point A to point B?
We can transmit a digital signal by using one of two
different approach :
1. Baseband transmission
2. Broadband transmission
3.41
1. Baseband transmission
Means sending a digital signal over a channel
without changing the digital signal to an analog
signal
Base band transmission required a low-pass
channel (channel with a B-W that starts from
zero)
3.42
Figure 3.19 Bandwidths of two low-pass channels
3.43
Figure 3.20 Baseband transmission using a dedicated medium
Note
Baseband transmission of a digital signal that preserves
the shape of the digital signal is possible only if we have a
low-pass channel with an infinite or very wide bandwidth.
3.44
2. Broadband Transmission (modulation)
Means changing the digital signal to an analog signal for
transmission. Modulation use a band-pass channel (a
channel with a B-W that doesn't start from Zero). This type
of channel is more available than a low-pass channel.
3.45
Note
If the available channel is a bandpass
channel, we cannot send the digital signal
directly to the channel;
we need to convert the digital signal to an
analog signal before transmission.
3.46
Figure 3.24 Modulation of a digital signal for transmission on a
bandpass channel
3.47
Example
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. We convert the digital signal from the
computer to an analog signal, and send the analog signal.
We 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.
3.48
3-4 TRANSMISSION IMPAIRMENT
Signals travel through transmission media, which are not
perfect. The imperfection causes signal impairment.
• This means that the signal at the beginning of the
medium is not the same as the signal at the end of the
medium. What is sent is not what is received.
Three causes of impairment are attenuation, distortion,
and noise.
3.49
1. Attenuation – a loss of energy
when Signal travels through a medium, it losses some of
its energy in overcoming the resistance of the medium. To
compensate for this loss, amplifiers are used to amplify
the signal.
Decibel: Measure the relative power(attenuation)
dB=10 log10 P2/P1
3.50
Example #1
Suppose a signal travels through a transmission medium
and its power is reduced to one-half. This means that P2 is
(1/2)P1. In this case, the attenuation (loss of power) can be
calculated as
A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.
See the book for more Examples
3.51
2. Distortion
Distortion : means that signal changes its form or shape.
Each signal component has its own propagation speed
through the medium and therefore ,its own delay in
arriving final destination
3.52
3. Noises
•Thermal noise: is the random motion of electrons in
a wire which creates an extra signal not originally
sent by the transmitter
•Induced noise: Comes from sources such as motors
and appliances.
•Crosstalk noise: Is the effect of one wire on the
other.
Impulse Noise: is a spike ( a signal with high energy
in a very short time) that comes from power lines,
lighting and so on.
3.53
Figure 3.29 Noise
3.54
Signal-to-Noise Ratio
SNR: ratio between signal power to the noise power
o A high SNR: means the signal is less corrupted by noise
o A low SNR: means the signal is more corrupted by noise.
SNR can be described in db units
SNR db=10 log10 SNR
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Figure 3.30 Two cases of SNR: a high SNR and a low SNR
3.56
Example
The power of a signal is 10 mW and the power of the noise is
1 μW; what are the values of SNR and SNRdB ?
Solution
The values of SNR and SNRdB can be calculated as follows:
SNRdb= 10 log10 10,000 = 40
3.57
3.5 Data Rate Limits
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◻
A very important consideration in data communications is
how fast we can send data, in bits per second, over a
channel. Data rate depends on three factors:
⬜
The bandwidth available
⬜
The level of the signals we use
⬜
The quality of the channel (the level of noise)
Two theoretical formulas were developed to calculate the
data rete :
1. By Nyquist for a noiseless channel
2. By Shannon for a noisy channel
Nyquist Theorem (noiseless channel)
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The relation between bandwidth and data rate in a
noiseless channel (throughput)
⬜ Maximum data rate = 2 x bandwidth x log2 L
where L: No of signal levels used to represent data
Example :
Consider a noiseless channel with a bandwidth of 3 KHz
transmitting a signal with two signal levels.
The maximum bit rate can be calculated as:
◻
Throughput =2*3000 log2 2 = 6000 bps.
Example
We need to send 265 kbps over a noiseless channel
with a bandwidth of 20 kHz. How many signal levels
do we need?
Solution
We can use the Nyquist formula as shown:
Since this result is not a power of 2, we need to either
increase the number of levels or reduce the bit rate. If
we have 128 levels, the bit rate is 280 kbps. If we have
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64 levels, the bit rate is 240 kbps.
Shannon Theorem (Noisy Channel)
The maximum throughput of a noisy channel of
bandwidth B with a signal to noisy ratio of S/N is:
Maximum throughput = B log2(1+S/N) bps.
Shannon Capacity
Capacity = bandwidth x log2(1+SNR)
is the capacity of the channel in bps ( max data
rate)
Example :
Telephone line Bandwidth=3kHz; S/N=30 dB =>
Max throughput = 3000 * log2(1+1000) =~ 30.000 bps = 28.8
kbps
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Example (using both limits)
We have a channel with a 1-MHz bandwidth. The
SNR for this channel is 63. What are the
appropriate bit rate and signal level?
Solution
First, we use the Shannon formula to find the
upper limit.
3.62
Example (continued)
The Shannon formula gives us 6 Mbps, the upper
limit. For better performance we choose something
lower, 4 Mbps, for example. Then we use the
Nyquist formula to find the number of signal
levels.
3.63
Note
The Shannon capacity gives us the upper
limit; the Nyquist formula tells us how many
signal levels we need.
3.64
3-6 PERFORMANCE
One important issue in networking is the
performance of the network—how good is it?
1. Bandwidth
2. Throughput
3. Latency (Delay)
3.65
1. Bandwidth
In networking, we use the term bandwidth in two
contexts.
❏The first, bandwidth in hertz, refers to the range
of frequencies in a composite signal or the range
of frequencies that a channel can pass.
❏The second, bandwidth in bits per second, refers
to the speed of bit transmission in a channel or
link.
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2. Throughput
Is a measure of how fast we can actually send
data through a network.
 The bandwidth is potential measurement of a
link; the throughput is an actual
measurement of how fast we can send data.
 Throughput less than Bandwidth
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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?
Solution
We can calculate the throughput as
The throughput is almost one-fifth of the bandwidth in this
case.
3.68
3. Latency ( Delay)
Latency defines how long it takes for an entire
message to completely arrive at the destination
from the time the first bit is sent out from the
source
Latency (Delay) =
propagation time + transmission time
+queuing time + processing time
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1) Propagation time
• It is the time required for a bit to travel from the
source to the destination
• Propagation speed depend on the medium and on
the frequency of the signal
Example :
light propagate by 3x108m/s in vacuum. It is lower
in air ; it is much lower in cable.
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2. Transmission time
• It is the time required for transmission of a
message .
• It depends on the size of the message and the
bandwidth of the channel.
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Example #1 :
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.
Solution
We can calculate the propagation and transmission time
as shown on the next slide:
3.72
Example #1 (continued)
Note that in this case, because the message is short and
the bandwidth is high, the dominant factor is the
propagation time, not the transmission time. The
transmission time can be ignored.
3.73
Example # 2
What are the propagation time and the transmission
time for a 5-Mbyte message (an image) if the bandwidth
of the network is 1 Mbps? Assume that the distance
between the sender and the receiver is 12,000 km and
that light travels at 2.4 × 108 m/s.
Solution
We can calculate the propagation and transmission
times as shown on the next slide.
3.74
Example #2 (continued)
Note that in this case, because the message is very long
and the bandwidth is not very high, the dominant factor
is the transmission time, not the propagation time. The
propagation time can be ignored.
3.75
3. Queuing time
• It is the time needed for each end device to
hold the message before it can be processed.
• It changes with the load imposed on the
network, if there is heavy traffic on the
network , the queuing time increases.
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DIGITAL AND ANALOG
TRANSMISSION
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Digital modulation and multiplexing
78
•
Baseband transmission: the signal occupies
frequencies from zero up to a maximum that depends
on the signaling rate.
•
•
Passband transmission: the signal occupies a band
of frequencies around the frequency of the carrier
signal.
•
•
It is common for wires.
It is common for wireless and optical channels.
Multiplexing: Channels are often shared by multiple
signals.
1. Baseband transmission
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Line codes: (a) Bits, (b) NRZ, (c) NRZI,
(d) Manchester, (e) Bipolar or AMI.
Clock Recovery
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•
•
1.
2.
3.
The receiver must know when one symbol ends and the
next symbol begins to correctly decode the bits.
One strategy is to send a separate clock signal to the
receiver.
NRZ: using a positive voltage to represent 1 and a
negative voltage to represent 0
Manchester: mix the clock signal with the data signal
by XORing them together.
NRZI: coding a 1 as a transition and a 0 as no transition,
or vice versa.
81
4.
5.
6.
4B/5B: Every 4 bits is mapped into a5-bit pattern with a fixed
translation table. The five bit patterns are chosen so that there
will never be a run of more than three consecutive 0s.
Balanced Signals: Signals that have as much positive
voltage as negative voltage even over short periods of time
are called balanced signals.
Bipolar encoding: represent a logical 1, (say +1 V or −1 V)
with 0 V representing a logical zero. To send a 1, the
transmitter alternates between the +1 V and −1 V levels.
2- Passband Transmission
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•
•
1.
2.
3.
We can take a baseband signal that occupies 0 to B Hz and
shift it up to occupy a passband of S to S +B Hz.
Digital modulation is accomplished with passband
transmission by regulating or modulating a carrier signal that
sits in the passband.
ASK (Amplitude Shift Keying), two different amplitudes are used
to represent 0 and 1.
FSK (Frequency Shift Keying), two or more different tones are
used.
PSK (Phase Shift Keying): the carrier wave is systematically
shifted 0 or 180 degrees at each symbol period.
Passband Transmission
(a) A binary signal. (b) Amplitude shift keying.
(c) Frequency shift keying. (d) Phase shift keying.
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Multiplexing
84
1- Frequency Division Multiplexing
FDM: divides the spectrum into frequency bands, with each user
having exclusive possession of some band.
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2- Time Division Multiplexing
• TDM : The users take turns (in a round-robin fashion), each one
periodically getting the entire bandwidth for a little burst of time.
•
Bits from each input stream are taken in a fixed time slot and
output to the aggregate stream.
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3 - Code Division Multiplexing
• CDM : is a form of spread spectrum communication.
(a)
Chip sequences for four stations.
(b)
Signals the sequences represent
CHAPTER # 7
TRANSMISSION MEDIA
McGraw-Hill7.87
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Transmission medium and physical layer
• A transmission media defined as anything that carry
information between a source to a destination
• Located below the physical layer and are directly controlled
by the physical layer
7.88
Classes of Transmission Media
7.89
7-1 GUIDED MEDIA
Guided media, which are those that provide a conduit
from one device to another, include twisted-pair cable,
coaxial cable, and fiber-optic cable.
Twisted –pair cables and coaxial cable:
use metallic (copper) conductors that transport signals in
the form of electric current
Optical fiber :
transport signals in the form of the light
7.90
1. Twisted-pair cable
• One of the wire used to carry signal and the other as a
ground. The receiver uses the difference between the two.
• If the two wires are parallel, the effect of interference
noise and crosstalk is big
• Twisting the pair of wire balance the effect of unwanted
signal and reduce it.
7.91
Applications of Twisted pair
Used in
1. Telephone lines to provide voice and data
channels
2. The DSL lines that are used by the telephone
companies to provide high-data-rate connections
3. Local area networks, such as 10-base-Tand
100base-T
7.92
2. Coaxial cable
• Coax cable carries signals of higher frequency ranges
than those in Twisted pair cable because the two media
are constructed quite differently.
• The outer conductor serves both as a shield against noise
and as second conductor, which complete the circuit
7.93
Applications of coaxial cable
1.Analog telephone network where a single cable could
carry 10,000 voice signals. Later it was used in Digital
telephone networks where cable can carry 600Mbps
2.Cable TV network: hybrid network use coaxial cable only
at the network boundaries , near the consumer.
3.Traditional Ethernet LANs.
7.94
3. Fiber Optic Cable
7.95
• Is made of glass or plastic and transmit signals in the
form of light.
• Light travels in a straight line as long as it is moving
through a single uniform substance. If a ray of light
traveling through one substance enters another
substance of different density , the ray change
direction as shown:
Optical fiber
7.96
Fiber Optical : uses reflection to guide light through a
channel. A glass or plastic core is surrounded by a
cladding of less dense glass or plastic
Back to the book for advantages and disadvantages
Applications for Fiber Optic cable
7.97
Used in :
1.Cable TV network: hybrid network use a combination of
optical fiber and coax cable. Optical provides the
backbone while coaxial cable provide the connation to
the user.
2.Local area networks such as ( fast Ethernet)
3.Backbone networks because its wide bandwidth
Propagation modes using fiber optics
98
•
Multimode Fiber: any light ray incident on the
boundary above the critical angle will be reflected
internally, many different rays will be bouncing around
at different angles.
•
Single-mode Fiber: light can propagate only in a
straight line, without bouncing.
Fiber cable composition
99
(a) Side view of a single fiber.
(b) End view of a sheath with three fibers.
Core: 50 microns for multi-mode, 8-10 microns for single mode
Cladding: glass with a lower refraction index, to keep the light
in the core
7- 2 UNGUIDED MEDIA - wireless
• Unguided media transport electromagnetic waves
without using a physical conductor.
• Signals are normally broadcast through free space and
thus are available to anyone who has a device capable
of receiving them
• Unguided signals can travel from the source to
destination in several ways:
1. Ground propagation
2. Sky propagation
3. Line – of – sight propagation
7.100
7.101
The Electromagnetic Spectrum
102
The electromagnetic spectrum and its uses for
communication.
1. Radio Transmission
103
(a) In the VLF, LF, and MF bands, radio waves follow the
curvature of the earth.
(b) In the HF band, they bounce off the ionosphere.
1. Radio Transmission
104
•
•
•
•
•
•
Frequency ranges: 3 KHz to 1 GHz
Omnidirectional
Susceptible to interference by other antennas using same
frequency or band
Ideal for long-distance broadcasting
May penetrate walls
Apps: AM and FM radio, TV, maritime radio, cordless
phones, paging
2. Microwaves
105
•
•
•
•
Frequencies between 1 and 300 GHz
Unidirectional.
Narrow focus requires sending and receiving antennas to
be aligned.
Issues:
•
•
Line-of-sight (curvature of the Earth; obstacles)
Cannot penetrate walls
2. Satellite Microwaves
106




Similar to terrestrial microwave except the signal travels
from a ground station on earth to a satellite and back to
another ground station.
Satellite receives on one frequency, amplifies or repeats
signal and transmits on another frequency
Satellite is relay station
Applications
 Television
 Long distance telephone
 Private business networks
Communication Satellites
107
The principal satellite bands.
3. Infrared
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•
•
•
•
•
•
Frequencies between 300 GHz and 400 THz.
Short-range communication in a closed area.
High frequencies cannot penetrate walls.
Requires line-of-sight propagation.
Advantage: prevents interference between systems in
adjacent rooms.
Disadvantage: cannot use for long-range communication
or outside a building due to sun’s rays.
The End
7.109