Transcript No Slide Title - Computer Information Science

Telecommunications System Components
• Computer to process information.
• Terminals or input/output devices (source/destination)
• Communication channels => Communication channels use various
communication media, such as telephone lines, fiber optic cables,
coaxial cables, and wireless transmission.
• Communication processors => modems, controllers, and front-end
processors.
• Communication software to control the function of the network.
Effect of imperfect transmission medium
Transmission medium: Two-wire open lines
Terminating connectors
Single pair
Flat ribbon
Transmission medium: Two-wire open lines
• Simplest transmission medium.Each wire is insulated from
the other and both are open to free space.
• Up to 50 meters of direct connection with 19.2 kbps can be
achieved.
• Two types: single pair and multiple cable/flat ribbon cable.
• Problems:
 Crosstalk- cross-coupling of electrical signals between
adjacent wires in the same cable.
 Noise- The open structure makes it susceptible to the pick up
of spurious noise signals from other electrical signal
sources.
Communication Media: Twisted Wire
We can reduce the effect of cross talk & noise by using twisted wire.
Single pair
Insulating outer cover
Multicore
Communication Media: Twisted Wire
• A transmission medium consisting of pairs of
twisted copper wires.
• We can transmit 1 Mbps over short distances (less
than 100m).
• They are mainly used to transmit analog signals,
but they can be used for digital signals.
• Advantages: inexpensive and already is in use.
• Disadvantages: slow, high-speed transmission
causes interference (crosstalk).
• Limiting factors: skin effect & radiation effect.
Coaxial Cable
In its simplest form, coaxial consists of a core
made of solid copper surrounded by
insulation, a braided metal shielding, and an
outer cover.
Coaxial Cable
It minimizes both effect:
skin effect
radiation effect
Coaxial Cable
• A transmission medium consisting of thickly
insulated copper wire, which can transmit a large
volume of data than twisted wire.
• Advantages: It is often used in place of twisted
wire for important links in a network because it is
a faster, more interference-free transmission
medium (speed: 200 megabits per second).
• Disadvantages: Coaxial cable is thick, is hard to
wire in many buildings. It does not support analog
conversations.
Optical Fiber
Optical
core
Optical
cladding
Plastic coating
Single core
Multicore
Optical Fiber
• Optical fiber consists of a glass core, surrounded
by a glass cladding with slightly lower refractive
index.
• In most networks fiber-optic cable is used as the
high-speed backbone, and twisted wire and
coaxial cable are used to connect the backbone to
individual devices.
• Advantages: faster, lighter, and suitable for
transferring large amount of data.
• Disadvantages: Fiber-optic cable is more difficult
to work with, more expensive, and harder to
install.
The low loss regions of an optical fiber
Loss
db/km
2.0

 50THz
usable bandwidth
200nm
200nm
1.0
800
1000 1200 1400
Wavelength (nm)
1600
The low-loss regions of an optical fiber
1800
Optical fiber
•
•
•
•
•
•
Optical fiber cable differs from both these transmission media in
that it carries the transmitted information in the form of a
fluctuating beam of light in a glass fiber.
Light transmission has much wider bandwidth, thus enabling the
transmission rate of hundreds of megabits per second.
Optical transmission is immune to electromagnetic interference
and crosstalk.
Optical fibers have less loss of signal strength than copper,
after every 30 miles we need to use a repeater, whereas in copper,
we should insert repeaters at an interval of 2.8 miles .
Optical fiber is more secure, no easy tapping on the cable,
like in copper.
Optical fibers are smaller in diameters compared to copper.
Multimode Stepped Fiber
Three types of fiber exist: multimode stepped, multimode graded, and
Single mode fibers.
• In multimode stepped index fiber, the cladding and the core
material each has a different but uniform refractive index.
• All the light emitted by the diode at an angle less than the critical
angle is reflected at the cladding interface and propagates along
the core by means of multiple (internal) reflections.
• The received signal has a wider pulse width than the input signal.
• Therefore, the maximum permissible bit rate is decreased.
Multimode Graded-Index Fiber
•
Dispersion can be reduced by using a core material that has a
variable (rather than constant) refractive index.
• In a multimode graded index fiber, light is refracted by an
increasing amount as it moves away from the core.
• Therefore, the pulse width of the received signal will be
reduced compared with stepped index fiber.
• Therefore, the maximum bit rate will be higher compared to
stepped index fiber.
Single mode Fiber
Single-mode Fiber:
Further improvements can be obtained by reducing the core
diameter to that of a single wavelength (3-10 Mm).
The emitted light propagates along a single (dispersionless) path.
Small core allows propagation in only one mode
The bit rate will be very high.
Wireless Transmission
• Wireless transmission that sends signals through
air or space without any physical wire.
• Common uses of wireless data transmission
include pagers, cellular telephones, microwave
transmissions, communication satellites, mobile
data networks, personal digital assistants,
television remote controls.
Satellites
• Information can also be transmitted using electromagnetic (radio)
waves through free space as in satellite systems.
• Satellites used for communications are generally geostationary.
• Geostationary satellite orbits the earth once every 24 hours in
synchronism with the earth’s rotation .
• Therefore the satellites appear stationary from the ground.
• Geosynchronous satellite rotate around the earth a 6900 miles/hour
and remained positioned over the same point at 22300 miles above
the equator.
• Worldwide coverage can be achieved with three geosynchronous
satellite spaced at 120 degrees interval from one another
Terrestrial Microwave
• Terrestrial microwave links are widely used to provide
communication links when it is impractical or too expensive to
install physical transmission media ( e.g. across a river).
• As the collimated microwave beam travels through the earth’s
atmosphere, it can be affected by weather conditions.
• However, with a satellite link the beam travels mainly through
free space, therefore less prone to such effects (weather
conditions).
Radio Waves
• Radio waves links are widely used to provide communication links
when it is too expensive to install fixed-wire cables.
•
Radio waves are used to connect a large number of data gathering
computers distributed throughout a rural area to a remote data
logging/monitoring computer.
Sources of attenuation and distortion
Signal Attenuation
• Any signal carried on a transmission medium is affected by
attenuation, limited bandwidth, delay distortion, and noise.
• When a signal propagates along a transmission medium its
amplitude decreases. This is known as signal attenuation.
• If the cable is longer, a number of repeaters (amplifiers) are
inserted at some intervals so that the receiver can detect it.
• We measure both attenuation and amplification in decibels (db).
Attenuation = 10 log 10 (P1/P2) db
Amplification = 10 log 10 (P1/P2) db
Where P1 => transmitted signal power level
P2 => received signal power level
Relationship between bandwidth and the
transmission capacity of a channel.
• Bandwidth => The bandwidth of a channel is the
range of frequencies (difference between the
highest and the lowest frequencies) that can be
transmitted by that channel.
• The grater the range of frequencies, the greater the
channel’s transmission capacity.
• Baud => A change in signal from positive to
negative or vice versa that is used as a measure of
transmission speed.
Limited Bandwidth
• Since a communications channel has a limited bandwidth, when a
signal is transmitted over a channel, only those frequency
components that are within the channel bandwidth will be received.
• The larger the channel bandwidth, the more higher-frequency
components are received and hence the closer is the received signal
to the original (transmitted) signal.
A formula derived by Nyquist can be used to find the capacity of the
channel (cable) as a function of the bandwidth is:
C = 2B log2 M
where C = maximum capacity in bits per second
B = bandwidth of the cable
M = signaling level ( 8 - bit byte)
Bandwidth
Data is to be transmitted over the PSTN using a transmission scheme
with eight levels per signaling element ( 8-bit byte). If the bandwidth
of the cable is 2600 Hz, deduce the Nyquist maximum data transfer
rate.
C = 2 x 2600 x log2 8
= 2 x 2600 x 3
= 15 600 pbs
This formula is for noiseless cable, in practice, there are noises on the
cable and we should use Shannon's formula.
C = B log2(1+S/N)
Delay Distortion
• A Digital signal consists of components with various frequencies.
• The rate of propagation of a sinusoidal signal along a transmission
line varies with the frequency of the signal.
• Therefore, when we transmit any signal through a transmission
line, all its components reach at the destination with varying delays.
• This results in delay distortion.
Noise
In the absence of a signal, a transmission line ideally has zero
electrical signal present.
In practice, however, there are random perturbations on the
line even when no signal is being transmitted.
This is called line noise level.
The ration of the average power in a received signal S, to the
power in noise level, N is called signal-to-noise ratio (SNR).
SNR = 10 log 10 ( S / N) dB
High SNR ratio indicates good-quality signal
Low SNR ratio indicates low-quality signal.
Problem
In practice, there are noises on the cable and we should use Shannon's
formula to calculate the theoretical maximum information rate (C) of a
transmission channel.
C = B log2(1+S/N)
where C = maximum capacity in bps
B = bandwidth of the channel in Hz
S/N = ratio of signal power (S) to Noise power (N)
expressed in decibels or dB.
Let us suppose that a phone line has a signal-to-noise power ratio of
20 dB.
SNR = 10 log10 (S/N)
20 = 10 log10 (S/N)
20/10 = log10 S/N
2 = log10 S/N
10 2 = S/N
Therefore S/N = 100.
Problem
If this line has a bandwidth of 2600Hz. Find C
(maximum theoretical information rate that can be
achieved).
C = 2600 x log2 (1 + 100)
C = 2600 x log2 (101)
= 2600 x log10 101 / log10 2
= 2600 x 2 / 0.3
= 2600 x 6.643
= 17, 270 bps
Signal Propagation Delay
• There is always a finite (short) time delay for a signal to
propagate from one end of the transmission line to the other.
• This is called transmission propagation delay, Tp.
• Data is generally transmitted in blocks (frames) of bits.
• When a block of data is reached to its destination, an
acknowledgement is sent to the sender.
• Round-trip delay is the time delay between the first bit of a
block being transmitted by the sender and the last bit of its
associated acknowledgement being received.
a = Tp / Tx
where Tp = Propagation delay = D/V
where D = distance and V = velocity of electrical signal
inside the medium
Tx = Transmission delay = N/R
where N = number of bits to transfer
R = bit rate of the transmission.