Transcript Class Notes - Electrical and Computer Engineering

Telecommunications
Networking I
Topic 1
Overview of Telecommunications
Networking I-II
Dr. Stewart D. Personick
Drexel University
[email protected]
1
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Telecommunications
Networking
• Sending and receiving messages (couriers,
smoke signals, flashes of light, telegraph,
teletype, facsimile, voice mail, text E-mail,
multimedia E-mail)
• Real-time conversations and collaborative
work (face-to-face, wireline telephone, 2-way
radio-telephone, video teleconferencing,
multimedia teleconferencing)
• Accessing stored or real-time information
(physical file cabinets and folders, FTP,
WWW, remote sensing, imagery)
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• Networked Copyright
information
systems
(mission2001, S.D. Personick. All rights
Telecommunications
Networking
• How to communicate in efficient,
predictable, reliable, and useful ways?
-Representing data or information as a
“signal”
-Defining the quality (or fidelity) of
reception/communication
-Signals in the presence of noise, distortion
and interference
-Point-to-point communication systems
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-Switched networks
(e.g.,
the
Internet,
LANs,
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•
Overview of
Telecommunications
Networking
I
Overview of telecommunications Networking
I-II
• Data, information (voice, audio, images,
video), and signals that represent data and
information
• Signals in noise, measures of quality of
communication
• Wire-pair and coaxial cable, optical fiber, and
wireless transmission systems (link layer)
-radio frequency wireless links
-lasercom (free-space optical) wireless 4
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links
Overview of
Telecommunications
Networking II
• Wireless systems and networks: peer-to-peer
communication using a shared “ether”,
broadcast systems, cellular/PCS, 2-way
satellite systems
• Local area networks (packet switching)
• The Internet
• Battlespace systems
• Next generation systems, networks, issues
and applications
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Telecommunications
Networking I
Topic 2
Data and Information, and Signals that Represent
Data and Information
Dr. Stewart D. Personick
Drexel University
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Data, Information and Signals
• Sound: speech, audio, represented,
modeled and simulated
• Text and Images: character mapped,
scanned, bit mapped, transform
representations
• Video: real and animated, frame-byframe, interframe compression coded
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Capturing Sound
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Capturing Sound
• Sound takes the physical form of an
acoustic wave... i.e., variations in
pressure vs time and space... that
travels through a compressible physical
medium such as air (~1090
feet/second… ~332 meters/second)
• A microphone (transducer) converts
locally received pressure variations into
a varying voltage/current that
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Capturing Sound (cont’d)
• The varying voltage waveform that
represents the captured sound is
communicated to another location using
one of many possible communication
system technologies
• The received varying voltage waveform
is not an exact replica of the transmitted
voltage waveform
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Capturing Sound (cont’d)
• The received varying voltage waveform
is used to “drive” a speaker (transducer)
which produces a new acoustic wave
(sound) that is perceived as an
approximation of the original sound
• Does the reproduced acoustic wave
“sound” like the original acoustic wave?
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Capturing Sound (cont’d)
• The received varying voltage waveform
is used to “drive” a speaker (transducer)
which produces a new acoustic wave
(sound) that is perceived as an
approximation of the original sound
• Does the reproduced acoustic wave
“sound” like the original acoustic wave?
The answer depends upon the
application
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Representing Speech
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Representing Speech (cont’d)
• Speech is one of the most important
analog signals
• Representation qualities include:
-Intelligibility: Can I understand what
you are saying? Can I build a machine
that responds properly to what you are
saying?
-Naturalness: Does it sound like face-toface communication? Can I identify the
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Representing Speech (cont’d)
• Traditional telephone quality speech:
3 kHz high frequency cutoff, small amounts of
noise and
echo
• AM radio quality speech:
5 kHz high frequency cutoff, varying amounts of
noise
and interference
• FM radio, TV, other high quality speech:
10 kHz+ high frequency cutoff, barely
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perceptible noise
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Representing Speech (cont’d)
• Traditional telephone quality speech:
3 kHz high frequency cutoff, small amounts of
noise and
echo [ ~32 – 64 kbps ]
• AM radio quality speech:
5 kHz high frequency cutoff, varying amounts of
noise
and interference [~64 – 128 kbps]
• FM radio, TV, other high quality speech:
10 kHz+ high frequency cutoff, barely
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perceptible noise
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Representing Speech (cont’d)
• Compressed speech
-Uses digital signal processing to
remove redundancies in the original
speech signal.
-This typically impacts on the
naturalness and comfort associated with
the speech signal produced at the
receiving end of a link, but (hopefully)
still provides intelligibility
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Representing Speech (cont’d)
• Compressed speech [~8 – 16 kbps]
-Uses digital signal processing to
remove redundancies in the original
speech signal.
-This typically impacts on the
naturalness and comfort associated with
the speech signal produced at the
receiving end of a link, but (hopefully)
still provides intelligibility
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Representing Speech (cont’d)
• Modeled and simulated speech
-Uses a model of the vocal tract to
generate sounds that are perceived as
intelligible speech
-Real speech is captured by estimating
the vocal tract parameters (that vary
relatively slowly in time v. the sound
speech waveform itself). These
parameters are stored and/or
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transmitted,
and
used
to
recreate
the
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Representing Speech (cont’d)
• Modeled and simulated speech [~1.2
– 4.8 kbps]
-Uses a model of the vocal tract to
generate sounds that are perceived as
intelligible speech
-Real speech is captured by estimating
the vocal tract parameters (that vary
relatively slowly in time v. the sound
speech waveform itself). These
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parametersCopyright
are2001,
stored
and/or
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Representing Audio
• Audio signals, like music typically
demand a high accuracy of
representation to meet users’
expectations
>10 kHz high frequency cutoff
<100 Hz low frequency cutoff
low noise and distortion
• A typical audio system specification
includes a 20-20,000 Hz “frequency
response” Copyright 2001, S.D. Personick. All rights
21
Representing Audio
• Audio signals, like music typically demand a
high accuracy of representation to meet
users’ expectations
>10 kHz high frequency cutoff
<100 Hz low frequency cutoff
low noise and distortion
• A typical audio system specification includes
a 20-20,000 Hz “frequency response” [128
kbps – 1 Mbps]
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Capturing Analog Images
• Use a camera or scanner (transducer)
to produce a signal or a set of data
which represents the image
• Communicate this signal or data to a
receiving location
• Use the received signal or data [which
is not necessarily identical to the
transmitted signal or data] to reconstruct
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a new image
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Representing Images
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Representing Images
• Character-mapped Images
-The image consists of a number of
“characters” or objects selected from a data
base
-To capture the image, one must obtain or
derive its description in the form of: data that
represents each character or object used;
data representing its location on the image;
data describing colors used, fonts, object
sizes, object and character overlaps, etc.
-The set of data is communicated to the
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receiving location
and
used
to
recreate
the
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Character Mapped Images
Figure 1
Figure 1
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Representing Images
• Scanned images
-Scan the image …e.g., left-to-right, and topto-bottom
-Represent the scanned brightness and color
(e.g., red, green, and blue color brightness)
by a set of signals which change in time as
the scanning point moves
-Communicate these signals to the receiving
location; and use them to “paint” a new image
with a complementary scanning process
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Scanned Images
Scan
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Representing Images
• Bit-mapped Images
-Divide the image into an n x m array of
“pixels” (e.g., 800 x 600 = 480,000 pixels)
-Represent the brightness and color of each
pixel (e.g., red, green, and blue color
intensities) by a set of numbers
-Communicate these numbers to the
receiving end, and use them to recreate the
image
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Bit-mapped Images
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Bit Mapped Image Example
An image contains 800 x 600 = 480,000 pixels
3 bytes of information are required to represent
the intensity and color of each pixel
To store this image you would require 480,000 x
3 bytes of memory = 1,440,000 Bytes
To transmit this image in 1 second, you must
transmit at a data rate of 480,000 (pixels) x 3
(bytes per pixel) x 8 (bits per byte) bits per
second = 11,520,000 bits per second
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Representing Images
• Transforms
-Example: Hadamard transform (for a
single color)
What is the average brightness across
the entire image?
What is the difference in the
brightness of the upper left quadrant vs the
upper right quadrant?
Upper left vs lower left? Upper right 32
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vs lower right?
Transform Image
Representation
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Transforms
• Creative use of transform coding can reduce
the amount of information required to
represent an image. Example
-if the intensity and color of an image isn’t
changing across a selected portion of the
image, then a single set of intensity and color
data (e.g., 3-12 bytes) plus some location
information can represent that entire portion
of the image
• Typical compression achieved: 10 – 20 : 1 34
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2001, S.D. Personick.
All rights
[less for “busy
images”
where
details must be
Video
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Video (continued)
• A video is a sequence of images (called
“frames”), typically presented to a viewer at
24-80 frames per second
• If the images within the video are captured by
scanning, then each image may be scanned
twice, with “interlaced” scans. Thus each
image may be represented by two interlaced
“fields”. (e.g., NTSC video uses 60 interlaced
fields per second = 30 frames per second).
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Interlacing
Field #1
Field #2
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Video (continued)
• Given a fixed number of scan lines per
second, the use of interlacing allows one to
increase the field rate-- to reduce the
perceptual artifact called “flicker”-- while
maintaining a high enough number of scan
lines per frame.
• For some applications (e.g., computer
displays of multimedia information) the use of
“progressive scanning” (no interlacing) is
preferred.
• Experts disagree of the relative merits of 38
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progressiveCopyright
scanning
vs interlaced scanning
Video (continued)
• While a traditional video signal is generated
by a video camera that scans the (moving)
images formed on its focal plane, many
modern “videos” may be created using
computer generated video (i.e., animations,
like the original Walt Disney animations),
where each image in the video sequence is
intrinsically a bit mapped image or some
standard compressed image file
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NTSC Video
Scan lines per frame = 525 (262.5 x 2, interlaced)
Frame rate = 30 frames per second
Field rate = 60 fields per second
Scan rate = 15,750 “lines” per second
Bandwidth of scanned signal (B&W information) =
6 MHz
Data rate required (no compression) ~ 96 Mbps
(B&W) – 288 Mbps (3-component color)
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HDTV Video
1100 lines per frame ( v. 525 lines per frame for
NTSC)
16 : 9 Aspect ratio (v. 4 : 3 for NTSC)
Frame rate = 30 frames per second
Bandwidth of scanned signal ~ 30 MHz per color
(3 colors)
Data rate required (no compression) ~ 1.44 Gbps
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Video Compression Coding
• Remove redundant information within
each frame, as in image compression
coding
• Remove redundancy that exists from
frame-to-frame by
-communicating only the differences
that exist from one frame (or several
prior frames) to the next
-employing motion
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42
Video Compression Coding
• Remove redundant information within each
frame, as in image compression coding
• Remove redundancy that exists from frameto-frame by
-communicating only the differences that exist
from one frame (or several prior frames) to
the next
-employing motion prediction/compensation
for moving objects
• AchievableCopyright
compression:
~100:1
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Telecommunications
Networking I
Topic 3
Quantifying the Performance of
Communication Systems Carrying Analog
Information
Dr. Stewart D. Personick
Drexel University
44
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Signals in Noise
The basic model:
data or information
signal
+
data or information
signal + noise
noise
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Analog Signals in Noise:
Example
noise
Engine
Temperature
Sensor
+
s= ca (volts)
where a=
temperature (C)
r=s+n
=ca+n
c= .01 volt/degree-C
n(av) =0, var(n) = .0001
volt**2
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The distribution of the heights of
Vulcans
children
12”
24
”
adult
males
36”
48
”
adult
females
60”
72”
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Analog Signals in Noise
Example (continued)
a= a number representing information to be
communicated. apriori, a is a Gaussian random variable
with variance A, and zero average value
s = a signal in the form of a voltage proportional to a ….
= ca (volts), where c is a known constant
r = a received signal with additive noise = s + n (volts)
n = a Gaussian random variable with variance N
(volts**2)
How can we estimate
“a”, if we receive “r”?
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Signals in Noise
• What is the criterion for determining
whether we’ve done a good job in
estimating “a” from the received signal
“r”?
• It would have something to do with the
difference between our estimated value
for “a” and the true value for “a”
• Example: minimize E{(a-a)**2}, where a
is the estimated value of “a”, given r
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Signals in Noise
• We will give a proof, on the blackboard, that
a, the estimate of “a” that minimizes the
average value of (a-a)**2 is
a(r) = E (a|r) = the “expected value” of “a”,
given “r”
• The above is true for any probability
distributions of “a” and “n”; and for the
specific cases given,
a(r) = r/c [Ac**2/(Ac**2 +N)]
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50
Signals in Noise
Harder example:
a = a Gaussian random variable with variance A,
representing information
s(t) = a signal of duration T (seconds) where s(t) = a c(t),
and c(t) is a known waveform
r(t) = a received signal = s(t) + n(t), where n(t) is a
“random process” having a set of known statistical
characteristics
How do we estimate a, given r(t)?
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Signals in Noise
• What is the criterion for evaluating how
good an estimate of “a” we have derived
from r(t)?
• How do we describe the noise n(t) in a
way that is useful in determining how to
estimate “a” from r(t)?
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Signals in Noise
• Suppose n(t) = nc(t) + x(t) where:
n is a Gaussian random variable of
variance N; and where, in some loosely
defined sense:
• x(t) is a random process that is
statistically independent of the random
variable “n”, and where …. (continued
on next slide)
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Signals in Noise
• ...x(t) is also “orthogonal” to the known
waveform c(t), then we can construct a
“hand waiving” argument that suggests
that we can ignore x(t), and concentrate
on the noise nc(t), as we attempt to
estimate the underlying information
variable “a”.
• To make this argument more precisely
requires a deep understanding of the
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theory of random
processes
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Signals in Noise
• We will show (using the blackboard)
that we can convert this problem into
the earlier problem of estimating an
information parameter “a” from a
received signal of the form r= ca + n.
• While doing so, we will introduce the
concept of a “matched filter”
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R(t) = a c(t) + n(t)
[0,T]
But : N(t) = n c(t) + x(t)
R(t) = (a + n) c(t) + x(t)
Integral of { r(t) c(t)} = Integral { (a + n) c(t) c(t) + x(t)
c(t) }
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Gaussian Random Processes
• If we look at (“sample”) the random
process n(t) at times: t1,t2,t3,…,tj, then
we get a set of random variables: n(t1),
n(t2), n(t3), …,n(tj).
• If the set of random variables {n(tj)} has
a joint probability density that is
Gaussian, then n(t) is called a Gaussian
random process
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Gaussian Random Processes
(cont’d)
• Any linear combination of samples of a
Gaussian random process is a
Gaussian random variable
• Extending the above, the integral of the
product n(t)c(t) over a time interval T is
also a Gaussian random variable if, n(t)
is a Gaussian random process, and c(t)
is a known function
58
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Gaussian Random Processes
(cont’d)
• Let n(t) be a random process (not
necessarily Gaussian)
• Define “n” as follows:
n= the integral over T of n(t)c(t)/W,
where
W= the integral over T of c(t)c(t)
• Then, we can write n(t) as follows:
n(t)= nc(t) + “the rest of n(t)”
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Gaussian Random Processes
(cont’d)
• If n(t) is a “white, Gaussian random
process”, then:
-n is a Gaussian random variable, and
- “rest of n(t)” is statistically independent
of “n”…I.e., “the rest of n(t)” contains no
information that can help of estimate
either “n” or “a”
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Gaussian Random Processes
(cont’d)
• Furthermore, we can build a correlator
that works as follows. It takes the
received waveform, r(t), multiplies it by
the known waveform c(t), integrates
over the time interval T, and finally
divides by W
z= {the integral over T of r(t)c(t)}/W
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Gaussian Random Processes
(cont’d)
• Going back to the definitions and
equations above, we find that
z= a + n, where “a” is the original
information variable we wish to
estimate, and n is a Gaussian random
variable
• Thus by introducing the correlator, we
convert the new problem to (continued)62
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Gaussian Random Processes
(cont’d)
• …the old problem of estimating a scalar
(“a”) from another scalar (“r”); where r=
a+n, and where n is a Gaussian random
variable
• The correlation operation is also known
as “matched filtering”, because it can be
accomplished by passing r(t) through a
filter whose impulse response is c(-t).
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Capturing analog
waveforms
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Example
Information
Waveform
Time
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Example
Pulse stream
Time
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Example
Sampling
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Example
Samples
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Example
Pulse Amplitude
Modulation (PAM)
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Example
PAM Stream
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Example
The PAM stream is a representation of the
information signal
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Example
s(t) = transmitted signal
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Example
r(t) = s(t) + n(t) = received signal
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Matched Filtering
y (t)
a x(t)
h(t)= x(-t)
Matched Filter
y(t)= integral of [h(t-u) a x(u)du]
= integral of [ x(u-t) a x(u)du]
y(0)= integral of [a (x(u)x(u)du] = a E
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Matched Filtering
y (t)
a x(t) + n(t)
h(t)= x(-t)
Matched Filter
y(t)= integral of [h(t-u) {a x(u) + n(u)}du]
= integral of [ x(u-t) {a x(u) + n(u)}du]
y(0)= integral of [ (x(u){a x(u) + n(u)}du] = a E + n
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Example
• If each of the random variables “a” is a
Gaussian random variable with variance
A
• and, if n(t) is white Gaussian noise with
“spectral density” N
• Then the optimal estimate of each of the
variables “a” is given by “sampling” the
output of the matched filter and
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multiplyingCopyright
it by:
(1/E)
[AE/(N+AE)]
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Example
• If each of the random variables “a” is a
Gaussian random variable with variance
A
• and, if n(t) is “white” Gaussian noise
with “spectral density” N
• Then the associated mean squared
error of each estimate will be A
[N/(N+AE)]
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Telecommunications
Networking I
Topic 4
Point-to-Point Communication over Metallic
Cables
Dr. Stewart D. Personick
Drexel University
78
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Metallic Cables
Insulator
Insulated Copper
Wires
Shield
Coaxial Cable Center Conductor
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Metallic Cables
• Copper wire (“wire pair”) cable is a very
important medium in the commercial
sector, because
-essentially all homes and small
businesses in the United States, and in
many other parts of the world, currently
access the worldwide
telecommunications infrastructure using
a pair of wires
-wire pairs are very commonly used for
local area networks in offices and
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Metallic Cables (cont’d)
• Coaxial cable is a very important
medium in the commercial sector,
because
-more than 80% of residences in the
U.S., and a large fraction of residences
in many other parts of the world can
access the telecommunications
infrastructure using coaxial cable
-many local area networks utilize
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coaxial cable
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Metallic Cables (cont’d)
• While optical fibers and wireless carry a
growing share of telecommunications
traffic in the commercial sector, metallic
cables will continue to carry the majority
of local area network and access
network traffic for many years to come
• The cost of replacing all of the metallic
access cable in the U.S. with fiber
would be around $1000/home x 100
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million homes
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Metallic Cable System
Informatio
n
cable
s(t)
Transmitter
Signal (e.g., 1
volt peak)
r(t)
Receiver
Information
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What can we say about r(t)?
• r(t) = s(t)*h(t) + n(t) + i(t), where:
s(t) is the signal that enters the cable,
h(t) is the impulse response of the
cable,
n(t) is noise associated with the finite
temperature of the cable,
i(t) is interference from other signals,
and
a*b means:
the convolution of a(t) and
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84
What can we say about h(t)?
• h(t) is the cable’s impulse response,
and is equal to the Fourier transform of
the cable’s frequency response: H(f)
• H(f), the cable’s frequency response,
approximately takes the form:
20 log H(f)= -aL[f**(1/2)], where L is the
cable length (km) and a is a constant
[dB/km-Hz**(1/2)]
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Typical Coaxial Cable Losses
14
12
10
RG-174
RG-58
RG-8
8
dB/100 feet
6
4
2
0
1 MHz
10
MHz
100 1 GHz
MHz
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What can we say about h(t)?
(cont’d)
• The frequency response, H(f), “rolls off”
as 10**[- square root of the frequency:
f], due to the “skin effect” in a metallic
cable (either wire pair cable or coaxial
cable)
• This roll off, by definition, attenuates
higher frequencies more than lower
frequencies; and causes “dispersion” of
the signal: s(t)
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“Dispersion”
Input pulse volts vs.
time (microseconds)
Output pulse
millivolts vs. time
(microseconds)
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What can we say about n(t)?
• Noise, as we define it here, is an
unwanted signal, or the sum of several
unwanted signals….each of which is
caused by a natural phenomenon
• Examples of sources of noise are:
-thermally induced, random fluctuations
of the properties of materials (thermal
noise)
-lightning
89
Copyright 2001, S.D. Personick. All rights
What can we say about n(t)?
• In cables, n(t) is usually well modeled
as white Gaussian noise
• This additive noise is called “thermal
noise”, and results from the combination
of the finite temperature of the cable
(e.g., 293K) and the finite loss of the
cable
90
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What can we say about i(t)?
• i(t) is a man made interference signal
that adds to the desired signal at the
receiver
• Interference can be caused by:
-signals on other pairs of wire in the
same wire pair cable (called “crosstalk”)
-signals generated outside of the cable
that “leak” into the cable (e.g., nearby,
strong radio signals)
91
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Our challenge:
• Figure out how to do the best job we
can of recovering the underlying
information being communicated, given
the received signal r(t)
• Understand what the basic limitations of
cable systems are: e.g.; how far can we
transmit and still recover the underlying
information with adequate fidelity?
92
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Example: Dispersion-Limited
Operation; 6dB maximum roll off
s(t)
r(t)
cable
s(t) is a 1 volt pulse, 100 ns wide;
The cable is RG8X, with a loss of 1.8 dB/100ft @
20MHz;
Suppose that the maximum allowable loss at
1/(100ns) = 10 MHz is 6dB
What is the maximum allowable cable length?
Copyright 2001, S.D. Personick. All rights
93
Example: Dispersion Limited
Operation (cont’d)
• If the cable loss at 20 MHz is 1.8
dB/100ft, then the loss at 10 MHz is 1.8
x [(10/20)**0.5)] =
1.8 [.707] ~ 1.3dB/100ft
• If the maximum allowable roll off, at
10MHz, between the transmitter and the
receiver is 6dB, then the maximum
allowable cable length is (6/1.3) (100ft)~
460 feet
94
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Example: Noise Limited Operation
Cable: H(f)
G(f)
s(t)
Noise=4kTB
Receiver
(includes
equalizer)
Equalizer: G(f)
H(f):
95
Copyright 2001, S.D. Personick. All rights
Noise Limited Operation (cont’d)
• The “equalizer” in the receiver amplifies
higher frequencies more than lower
frequencies, to compensate for the “roll off” in
the cable
• The equalizer also amplifies the higher
frequency noise at its input
• kT= Boltzman’s constant x temperature (K) =
approximately 4 x 10**-21 (J) @ 293K
96
Copyright 2001, S.D. Personick. All rights
Example: Noise Limited Operation
• Suppose the receiver has input noise, within
band B, whose variance is 4kTB (watts), where
B= 1/100ns
• Suppose the signal-to-noise ratio (SNR) at the
receiver input must be > 100:1 = 20dB; where:
SNR=[{s(max)**2}/R] [10**-(aL/10)]/[4kTB]
• Suppose: s(max) = 1 volt; a = 5dB/100ft; T=
293K, and R= 50 (ohms)
• What is the maximum allowable cable length L?
97
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Example: Noise Limited Operation
(cont’d)
• [s(max)**2]/R= [(1)**2]/50 = 0.02 watts
• 4kTB = 4 (4 x10**-21)(10**7)=
1.6x10**-13 watts
• SNR= [.02/(1.6x10**-13)][10**-(aL/10)]
= 1.25x10**11 [10**-(aL/10)]
• If the SNR must be greater than 100
(I.e., 20dB), then [10**(-aL/10)] must be
greater than 1.25x10**-9; I.e., aL<89dB
98
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Example: Noise Limited Operation
(cont’d)
• If aL must be less than 89 dB, and a
equals 5 dB/100ft, then the maximum
cable length is 89/5 (100 ft) ~ 1800 ft
99
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Line Amplifiers
• We can extend the distance of
transmission in a noise-limited situation,
by the use of line amplifiers
• The underlying concept is the equalize
and amplify the signal before it has
become attenuated by the cable to the
point where the signal-to-noise ratio
drops below an acceptable level:
100
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Line Amplifiers (cont’d)
Transmitter
Cable section
#1
Line Amplifier #1
Cable section #2
Line Amplifier
#2
Cable section
#3
101
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Line Amplifiers (cont’d)
• Each line amplifier equalizes and
amplifies its input signal to compensate
the the roll off (frequency dependent
loss) of the previous section of cable
• Each line amplifier also amplifies the
noise arriving at its input, and adds
some additional noise of its own
102
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G(f)
Line Amplifiers (cont’d)
Cable
section: H(f)
Gain=
G(f)~
1/[H(f)]
frequency
103
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Line Amplifiers (cont’d)
• If the gain vs. frequency of the line
amplifier approximately cancels the loss
vs. frequency of the cable section
preceding it…I.e., if G(f)~[1/H(f)] …..
Then the net gain through the
combination of the cable section and
the line amplifier is 1 (0 dB) over the
range of frequencies where the above
relationship holds
104
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Line Amplifiers (cont’d)
• If the noise power per unit frequency
(watts per Hz) arriving at the input of the
first line amplifier is defined as Nin1(f),
then the noise power per unit frequency
at the output of that amplifier is: Nout1
(f) = {Nin1(f) [G(f)]**2} +
Ninamp(f)[G(f)**2]};
where
Ninamp(f) is noise added by the
105
amplifier Copyright 2001, S.D. Personick. All rights
Line Amplifiers (cont’d)
• The noise power at the output of the
next cable section, which is also the
noise arriving at the input to the
following amplifier is (approximately):
Nin2(f)=Nout1(f)[H(f)]**2 + Ncable(f)=
Nin1(f) + Ninamp(f) + Ncable(f);
where Ncable(f) is noise added by the
cable section
106
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Line Amplifiers (cont’d)
• Furthermore, if we extend this approach
to a cascade of K amplifiers and K+1
sections of cable, then the noise arriving
at the input to the Kth line amplifier is:
NinK(f) = Nin1(f) + [K-1] [ ninamp(f) +
ncable(f) ];
• Thus in a cascade of K amplifiers, the
noise arriving at the input of the Kth
amplifier is approximately proportional
107
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Line Amplifiers (cont’d)
• If cable noise from the first section of
cable dominates Nin1(f), and if the
receiver adds input noise comparable to
that of a line amplifier….then the total
noise after K sections of cable, including
the noise of the receiver is:
NtotalK(f) = K [ Ncable(f) + Ninamp(f)]
108
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Line Amplifiers (cont’d)
• The result of the above is:
If the signal-to-noise ratio at the input of
the final line amplifier (preceding the
extraction of the arriving information
from the arriving signal) must be greater
than some threshold, SNRmin… then
the allowable loss in each section of
cable decreases as more sections are
added
109
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Line Amplifiers (cont’d)
• Quantitatively…If the allowable loss
between the transmitter and the
receiver in a noise-limited single
segment transmission system is X dB,
then the allowable loss in each
segment of a transmission system with
K sections, and K amplifiers/equalizers
(including the receiver) is X-[10logK] dB
110
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Line Amplifiers (continued)
• In a previous example of a loss limited,
single segment transmission system,
the allowable cable loss (at 10 MHz)
was 89dB If we try to extend the reach
of the system by using 10 cable
segments, then the allowable loss in
each segment is 79dB; and the total
loss of all the segments can be as much
as 790dB (less than 10 x 89dB)
111
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Traditional Cable Television
System
Head
end
Super trunk
Line amplifier
Trunk
Splitter
Feeder
Drop
Home
112
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Frequency Division
Multiplexing (FDM): Typical
Cable System
Video signals (each 6MHz wide) spaced
every 6 MHz*
*except for a 4MHz gap above ch4, and a 32MHz
gap above ch6
Ch 2: 57 MHz +/- 3MHz
Ch 75: 531MHz +/- 3
MHz
500
f (MHz)
88-120
MHz
113
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FDM: Typical Cable System
• If there are 75 television channels
sharing the same composite signal
(stacked up in frequency), and if
distortion considerations limit the output
of the transmitter or a line amplifier to 1
volt (for example)….then each of the
component signals that represent a
single channel must have a peak
amplitude of about 1
114
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volt/[75**(1/2)]~0.1V
Cable TV FDM: Issues
• Distortion: If the transmitter and/or the
line amplifiers are not linear, then we
can end up with square law and higher
order terms that result in interference
among the FDM channels
• Noise; and interference from signals
leaking into the cable system
115
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Impact of Distortion on FDM
• Distortion (simplified example):
s(t) is an FDM composite signal (e.g., 6
MHz TV channels stacked up in
frequency)
r(t) = s(t) + b [s(t)]**2 + c[s(t)]**3
116
Copyright 2001, S.D. Personick. All rights
Impact of Distortion (cont’d)
• If s(t) contains components centered at
around 50MHz (for example) then
b[s(t)**2] will contain frequencies at
around 100MHz…interfering with the
components of s(t) at 100MHz
• If s(t) contains components centered at
100MHz and at 150MHz, then b[s(t)**3]
will contain frequencies at 50 MHz, 200
MHz, 300MHz, 350 MHz, 400 MHz,
and 450 MHz…causing interference at
117
all of thoseCopyright
frequencies
2001, S.D. Personick. All rights
Impact of Noise
• Thermal noise associated with the finite
temperatures of the cable sections, plus
noise associated with the line amplifiers
and the receivers connected to the
system set a lower bound or baseline of
system noise. This noise tends to
appear as “snow” on a television
receiver
• “Ingress noise” (interference) from radio
signals, electric motors, etc, represent
118
the more serious
problem
in
most
cable
Copyright 2001, S.D. Personick. All rights
Radio Frequency Interference
• Another consideration, of importance in
all metallic cable systems, is the
possibility that the cable system will
radiate signals that may interfere with
other communications or information
systems/appliances (e.g., medical
devices in a hospital environment)
• Radiation can occur due to imperfect
shielding on coaxial cables, bad
connectors, unterminated splitter output
119
ports, and Copyright
imbalances
in
wire
pair
cable
2001, S.D. Personick. All rights
“Wiring” Concerns
Unused output port
not terminated
properly
Wrong kind of
“wire”
Poorly
designed
splitter
Bad or loose
connector
120
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Personick’s ad-hoc equalizer
100 MHz = 628 M radians per second
=> 50 C = 1/[6.28 x 10**8] ~ 3.2 pf
50 ohms
C
50
ohms
10 ohms
121
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Telecommunications
Networking I
Topic 5
Quantifying the Performance of
Digital Communication Systems
Dr. Stewart D. Personick
Drexel University
122
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Digital Communication
System
Communication
link
Data input:
1011101
Data output: 1011101
123
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Point-to-Point Digital
Communications: The Concept of
“Errors”
1
0
1
0
1
1
time
Input information: …1 0 1 0 1 1 ...
1
0
1
0
0
1
Output information: …1 0 1 0
01
124
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What Causes Errors?
• Errors are caused by noise, distortion,
and interference in the communications
link, and can also be caused by
“synchronization” errors and
“congestion”
• We can have two types of errors:
- “miss” : we mistake a “1” for a “0”
- “false alarm” : we mistake a “0” for a
“1”
125
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Tradeoffs Between Misses
and False Alarms
• We can reduce the “miss” probability to
zero by declaring that each output bit
value will be a “1”; but then we will get a
“false alarm” every time the true value
of the corresponding input bit is a “0”
• We can reduce the “false alarm”
probability to zero by declaring that
each output bit value will be a “0”; but
then we will get a “miss” every time the126
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Probability of a “False
Alarm”
Trading Off “Miss” and “False Alarm”
Errors
0
1
Receiver Operating
Characteristic
1
Probability of a “Miss”
127
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Probability of a “False
Alarm”
Trading Off “Miss” and “False Alarm”
Errors
0
1
Blended receiver
Operating
Characteristic
1
Probability of a “Miss”
128
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Classical Additive Noise
Detection Problem
+
r(t)
s(t) = 1 volt peak
n(t)= white,
Gaussian noise
129
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Matched Filter
Sample
here
r=s+n
h(t)=a(-t)
a(t)=pulse
shape
Matched Filter
130
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Matched Filter Output: r
r = s + n (volts)
“s” is equal to either: S or 0 (volts);
corresponding to a digital “1” or a digital “0”
n is a Gaussian random variable with variance
N (volts**2)
131
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Matched Filter Output: r
S volts
r= s + n
Decision
Threshold
= D volts
If r > D, then we will
guess that s=S (I.e.,
that we have a “1”.
If r < D, then we will
guess that s=0 (I.e.,
that we have a “0”
0 volts
132
Copyright 2001, S.D. Personick. All rights
Calculating the Error
Probabilities
The probability of a miss is given by:
ERFC* [ (S-D)/(N**0.5)]; where ERFC* [ ]
is called the “error function complement
star”
For example ERFC*[6] = 10**-9
0
D
S
133
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Gaussian Shape
Standard
deviation
134
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Calculating ERFC*(x)
(“earf-see-star”); also known as
the Q function: Q(x)
ERFC* (x) = the integral from x to
infinity of:
{1/[(2 pi)**(0.5)]} {exp -(y**2)/2} dy =
Q(x)
135
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Example
• We have a point-to-point metallic cable
system
• r(t)= s(t) + n(t) (volts) ; where:
s(t) is a sequence of pulses, modulated
on or off, where the pulse shape is a(t)
n(t) is white Gaussian noise with
spectral density, N, equal to 2kTR
(volts**2/Hz)
Copyright 2001, S.D. Personick. All rights
136
Example (cont’d)
• R is the impedance of the metallic cable
over which s(t) is being communicated;
kT = 4 x 10**-21 Joules
• We will assume that the energy in a
single received pulse a(t) = integral of
{[a(t)]**2 dt}/R = E Joules
• If we pass r(t) through a matched filter
with impulse response h(t) = C a(-t),
137
where C is aCopyright
constant,
then
…
2001, S.D. Personick. All rights
Example (cont’d)
… The output of the matched filter will
be
r = s + n; where:
• s= S or 0 volts; n is a Gaussian random
variable with variance N=
2kTR(C**2)ER (volts**2); and S = CER
(volts)
• The ratio of (S/2)/(N**0.5)=
[E/(8kT]**0.5
• If we wish to have an error rate of 10**-138
Copyright 2001, S.D. Personick. All rights
Example (cont’d)
• … we require:
[E/(8kT)] = 36 or more
• Since 8kT ~ 3.2 x 10**-20, we require the
received pulse energy E to equal ~ 10**18 (Joules) or more.
• If the transmitted pulse is 1 volt peak, and
100ns in duration; and the cable has an
impedance of 50 ohms, then the
transmitted pulse energy is:
.02 x 10**-7 = 2 x 10**-9 (Joules)
139
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Telecommunications
Networking I
Topic 6
Point-To-Point Digital Communications
Dr. Stewart D. Personick
Drexel University
140
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Digital Point to Point
Communications
• In a real digital communication system,
one has to be concerned with noise,
interference and other effects that can
cause errors
• In the previous discussion, we briefly
covered the topic of additive noise, and
its impact on errors (misses and false
alarms)
• In metallic cable systems, “intersymbol
141
interference”
is
a
key
factor
that
we
Copyright 2001, S.D. Personick. All rights
Intersymbol Interference
Transmitted pulse
stream
T
Cable output (dispersion)
+
T
142
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Intersymbol Interference:
Equalization
Cable: H(f)
Equalizer: 1/[H(f)]
143
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Digital Regenerator
(Repeater)
Cable: H(f)
Equalizer &
Matched
Filter
Decision
circuit
Clock signal
Timing
Recovery
144
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Timing recovery
Pulse
stream
Clock
(x)**2
Filter or
PLL
PLL=Phase-locked Loop
145
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Decision Circuit
Pulse
stream
Threshold
Compari
tor
D Flip
Flop
Clock
146
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Eye Diagram
147
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Eye Diagram (cont’d)
148
Copyright 2001, S.D. Personick. All rights
T-Carrier
• First introduced by the former Bell
System in 1962. The first digital
transmission system
• Digital transmission rate is 1.544 Mbps
• Works on 24 gauge wire pair cables
• Repeaters every 6000 ft ~ 2km
• Max cable loss at 772MHz ~35 dB
• The signal is a “DS1” (but everyone
149
calls it a “T1”)
Copyright 2001, S.D. Personick. All rights
T-Carrier (cont’d)
• Maximum cable length between
repeaters limited by crosstalk:
interference from other signals on other
pairs in the same cable
• Transmitted signal format: + or - 3V
equals a logical “1”; 0V equals a logical
“0”, no more than 7 “zeros” in a row are
permitted
• 00000000 is mapped to 00000001
originally; recently mapped to ++0000-150
(“B8ZS”) Copyright 2001, S.D. Personick. All rights
Analog-to-Digital Conversion
(and vice versa)
Sampling Theorem: If we sample an
analog signal at twice its highest
frequency, we can reproduce it
exactly from its samples
151
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A/D Conversion
Example: Voice signals…
-Highest frequency is limited (by a filter) to
4kHz
-We sample this band limited signal at 8000
samples
per second (125 microseconds between
samples)
-We represent each sample with 1 byte
(positive and
152
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negative values
are both captured by 256
D/A Conversion
Samples reconstructed
from the received digital
bit stream
Reconstructed
waveform
filter
153
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Multiplexing
24 inputs, each
at 64 kbps
T1 multiplexer; also
known as a “D-channel
bank”
1 output at
1.544 Mbps
[24 x 64 kbps] + 8kbps = 1.544 Mbps; 8 kbps =
overhead
Output
signal is a “DS1”, but everyone calls it a “T1”
154
Copyright 2001, S.D. Personick. All rights
D-Channel Bank Frame
Format
1st data byte, 2nd data byte, ……, 24th data
byte, F
125 microseconds
F= Framing bit
bits
Frame length is (8 x 24) + 1 = 193
8000 frames per second, corresponding to the rate at
which voice signals are sampled
193 bits per frame x 8000 frames per second = 1.544
155
Mbps
Copyright 2001, S.D. Personick. All rights
Multiplexing Standards
• In the United states, some common multiplex
standards are: (above DS1, they are used with
radio or fiber optic transmission systems)
-DS1 (called T1) 1.544 Mbps
-DS3 (called T3) 44.7 Mbps (28 DS1 +
overhead)
-STS1 (SONET-1) 51.84 Mbps
-STS3 (SONET-3) 155.52 Mbps
-STS 12
622.08 Mbps
-STS 48
2.48832 Gbps
-STS 192
~10 Gbps
156
Copyright 2001, S.D. Personick. All rights
Other Popular Digital Metallic
Cable Transmission Systems
• Ethernet: 10Base-T, 100Base-T; coax
versions
• Telephone modems: up to 56 kbps; limited by
end-to-end switched telephone network
• ADSL: asymmetric digital subscriber line;
e.g., 1.5Mbps downstream, 384 kbps
upstream …using only the existing telephone
loop (not the switched network)
• Cable modems: ~ 20 Mbps downstream,
shared with other users; upstream depends 157
Copyright 2001, S.D. Personick. All rights
Ethernet: 10Base-T
Twisted
Pair
NIC
Hub
Computer
NIC= Network
Interface Card
To other
hubs or
router
158
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Modem
Subscriber loop
A/D
modem
Computer
(Philadelphia)
D/A
Public Switched
Telephone Network
(PSTN)
modem
Computer
(Los Angeles)
159
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ADSL
Loop pair
ADSL
Switched voice
DSLAM
computer
PSTN
Router
Packet data
To the Internet
DSLAM=Digital subscriber line
access multiplexer
160
Copyright 2001, S.D. Personick. All rights
Cable Modem
Splitter
TV
Coaxial Cable
Coaxial
Cable
Twisted pair
Modem
PC
161
Copyright 2001, S.D. Personick. All rights
Telecommunications
Networking I
Topic 7
Fiber Optic Transmission Systems
Dr. Stewart D. Personick
Drexel University
162
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Fiber Optics: Overview
• 1966 C. Kao et. al, propose that
strands of glass can be produced, which
can carry light over long distances (>2
km)
• 1970 First demonstration of a fiber with
less than 20dB/km of loss (Maurer,
et.al., at Corning)
• 1975-77 Experiments and field trials
• 1979 Real systems are placed in
163
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Basic Fiber Optic
Transmission System
Digital
pulses
(On/Off)
Light Source
Detector/Receiver
Glass fiber
Optical Transmitter
Digital pulses
164
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The Radio Spectrum
• AM Radio ~ 1 MHz (300 meter
wavelength)
• Television ~ 50-500 MHz
• Digital cordless phone ~ 900 MHz
• Wireless LAN ~ 2.5 - 5 GHz
• DBS ~ 10 GHz (0.3 meter wavelength)
• Fiber optics ~ 0.9 - 1.55 um (not visible)
• Visible light ~ 4-7.5 x 10**14 Hz (0.8-0.4
um)
165
Copyright 2001, S.D. Personick. All rights
Optical Transmitter: example
Data in
2V Peak
~20 mA peak
Light output
50 ohm resistor
Light Emitting Diode
166
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Optical Transmitter Example
driver
bias
Laser
Light output
Data in
1V peak
current
Light output
current
Copyright 2001, S.D. Personick. All rights
167
Simplified Semiconductor
Injection Laser
Curren
t
168
Copyright 2001, S.D. Personick. All rights
Optical Transmitter: example
• Current = 20mA = .020A
• # electrons per second =
.020 A/[1.6 x 10**-19] Coulombs per
electron = n
• # photons produced/second =n x [Quantum
Efficiency]
• optical power out = n x QE x [~1.5 x 10**19 Joules per photon] ~ .020 QE x
[1.5/1.6] (W) ~20 x [1.5/1.6] x QE (mW)
169
S.D. Personick.
All rights
• If QE~ 20%,Copyright
then2001,power
out
~3.75
Optical Fiber
Optical
Pulses
Optical Pulses
Fiber
Optical output pulses are attenuated and spread in time
compared to optical input pulses
170
Copyright 2001, S.D. Personick. All rights
Causes of Attenuation
• Light is absorbed by fiber impurities and
the principal fiber material itself
• Light is “scattered” out of the fiber
because of the inherently random
density fluctuations of any “glass”
(Rayleigh scattering) as well as by more
macroscopic density fluctuations
• Typical long distance fiber: <0.5 dB per
km
Typical plastic fiber: >100 dB per km 171
Copyright 2001, S.D. Personick. All rights
Causes of Pulse Spreading:
Modal Delay Spread
T (min) = nL/c, where c/n = speed of light in fiber
T (max) = T(min) x [1/cos(max angle that is
captured)]
core
cladding
“Multimode” Fiber
c =300,000,000 m/s, n~1.5... n/c ~ 5ns/m
172
Copyright 2001, S.D. Personick. All rights
Modal Delay Spread
•The rays in the previous slide represent the
solutions of Maxwell’s equations…each of which
is called a “mode”
•If one actually solves Maxwell’s equations, one
finds a discrete set of modes, each
corresponding to a ray at a different angle
 D
relative to the axis
 Dspacing between these allowed rays is
•The
•If
is large enough (e.g., 0.2 radians
corresponding to ~11.4 degrees, then only the
axial ray is below the critical angle.
173
Copyright 2001, S.D. Personick. All rights
Modal Delay Spread
• If only the axial ray is below the critical
angle, then there is only one solution to
Maxwell’s equations (one mode) which
is guided by the fiber.
• Such a fiber is called a single mode
fiber
• With only one ray (mode) there is no
modal pulse spreading!
174
Copyright 2001, S.D. Personick. All rights
Dispersion: ps/km-nm
Causes of Pulse Spreading
Dispersion: a change in the
delay
down the fiber as the
wavelength
Zero dispersion at ~1.3 um
changes- ps/[nm-km]
Wavelength
175
Copyright 2001, S.D. Personick. All rights
Causes of Pulse Spreading
• Pulse spreading can be caused by the
variation of delay vs. angle in multimode
fibers (delay spreading). Typical plastic
multimode fiber: >100 ns/km
• Pulse spreading can also be caused by
the variation of delay with wavelength
(“dispersion”). Typical glass fiber with
900 nm LED source ~5 ns/km; with
1550 nm laser source < 0.1 ns/km
176
Copyright 2001, S.D. Personick. All rights
Pulse Spreading: Examples
• Multimode fiber: Maximum angle
captured in fiber is 0.2 radians (for
example)~11.5 degrees. 1/cos(0.2 rad)
= 1.020. Delay spreading = 5 ns/m x
0.02 = 0.1 ns/m = 100 ns/km
• Single mode fiber +900 nm LED source:
Dispersion at 900 nm wavelength ~100
ps/nm-km. LED spectral width ~50 nm.
Dispersion ~ 100 x 50 = 5000ps/km
177
=5ns/km Copyright 2001, S.D. Personick. All rights
Optical Receiver
Amplifier +
Regenerator
Output pulses
Photodiode
178
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Optical Receiver
• The detector converts photons to
electrons
~ 0.5 mA/mW (output current/input
power)
• The amplifier amplifies the weak current
that is produced by the detector in a
typical optical fiber application
• The regenerator produces a new
179
electrical pulse
stream
(clock
recovery,
Copyright 2001, S.D. Personick. All rights
Causes of Errors in Optical
Fiber Systems
• Noise produced by the amplifier in the
receiver
• “Quantum” noise associated with the
detection process
• Intersymbol interference due to pulse
spreading
• Bottom line: In a typical fiber optic system,
we require ~20,000 received photons per
pulse to produce an error rate of 10**-9;
assuming that we don’t have a significant
180
amount of intersymbol
interference
(pulse
Copyright 2001, S.D. Personick.
All rights
Fiber Optic System: example
Digital pulses
(On/Off)
Light source
Detector/Receiver
Glass fiber
Optical
Transmitter
Assume: Bit rate = 100Mbps; Optical transmitter
output = 1 mW; Coupling loss into fiber = 3dB;
Pulse spreading<0.1 ns/km; Fiber loss = 0.5 db/km;
Required optical energy per received pulse:
20,000 photons x 1.5 x 10**-19 J/photon
Digital pulses
181
Copyright 2001, S.D. Personick. All rights
Optical Fiber System:
example
• Receiver requires 20,000 photons per
received optical pulse = 20,000 x 1.5 x 10**19 J per pulse = 3 x 10**-15 J/pulse
• Bit (pulse) rate is 100Mbps; therefore the
average received power level must be greater
than 0.5 x (3 x 10**-15) x (10**8)= 1.5 x 10**7 watts = 1.5 x 10**-4 mW~
-38.2 dBm
182
Copyright 2001, S.D. Personick. All rights
Optical Fiber System:
example
• Transmitter average power into the fiber is: 1 mW
x 0.5 (coupling loss) x 0.5 (duty cycle) = 0.25 mW
• Allowable loss = 0.25/0.00015 = 1.66 x 10**3 ~
32.2 dB
• At 0.5 dB/km loss, we can allow ~64 km of fiber
[44 km]
• Checking the pulse spreading; we get: 64 km x 0.1
ns/km = 6.4 ns (0.64 x pulse spacing)
• Bottom line, if we allow up to 50 km of fiber, we will
be within the pulse spreading limit, and we will
have about 14 x 0.5 = 7 dB of margin w.r.t. noise
183
limited operation
[pulse
spreading
limit
is
5
km]
Copyright 2001, S.D. Personick. All rights
Telecommunications
Networking I
Topic 8
Wireless Transmission Systems
Dr. Stewart D. Personick
Drexel University
184
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Wireless Point-to-Point Link
Antenna
Feed Line (e.g.,
coaxial cable)
Radio
Transmitter
Radio
Receiver
185
Copyright 2001, S.D. Personick. All rights
The Electromagnetic
Spectrum
• 30-300Hz: SLF
3GHz-30GHz: SHF
(DBS)
• 300Hz - 3kHz ULF
30GHz-300GHz: EHF
• 3kHz - 30kHz: VLF 300,000GHz: 1um light
• 30kHz-300kHz: LF
• 300kHz-3MHz: MF (AM Radio)
• 3MHz-30MHz: HF (Short Wave Radio)
• 30MHz-300MHz: VHF (FM, TV)
• 300MHz-3GHz: UHF (TV, Digital Cordless, 186
Copyright 2001, S.D. Personick. All rights
…)
Wireless Transmitter
Antenna
Informatio
n
Modulator
IF Oscillator
Mixer
RF
Amp
RF Oscillator
187
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Transmitter Subsystems
• 1 milliwatt -100 milliwatts: low r.f.
exposure with hand held appliances,
low battery drain (some hand held
appliances radiate ~5 watts…but I
wouldn’t hold one of these near my
head!)
• 10 watts - 100 watts: okay for consumer
and small business applications, with
115 volt or 12 volt automobile power.
Stay 10-30 feet from the antenna.
188
Copyright 2001, S.D. Personick. All rights
Transmitter Subsystems
R.F. Power = (V**2)/Z
Antenna
V~ 1 volt
Z~50 ohms
FET
Coax
R.F. Amplifier
R.F. Power (launched into
the coaxial cable) = 1/50
Watt = 20 mW
R.F. Power radiated = ?
189
Copyright 2001, S.D. Personick. All rights
The Antenna
Reflected
Power
Radiated
Power
Forward Power
190
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The Antenna
• If the antenna is much smaller in length than
the wavelength of the radiation (c/f), then the
antenna will be a very inefficient radiator
(most of the forward power is reflected)
• If the antenna is 1/4 wavelength or larger in
size, it can be an efficient radiator
• If the antenna is significantly larger in size
than 1 wavelength, it can be an efficient and
directional radiator
191
Copyright 2001, S.D. Personick. All rights
A Directional Radiator
Reverse: fields
cancel
1/4 wavelength
spacing between
two dipoles. 3/4
wavelength delay in
crossover cable
Forward: fields add
It’s actually not quite that simple: dipole
interactions
192
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The Parabolic Dish Antenna
Divergence angle
~Lambda/D
193
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Wireless Receiver
Antenna
Informatio
n
Demodulator
IF Oscillator
Mixer
RF
Amp
RF Oscillator
194
Copyright 2001, S.D. Personick. All rights
The Receiver Subsystem
Signal + Background Noise + Interference
Cable attenuation + thermal
noise
+ Amplifier noise
Low Noise Amplifier
195
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Sources of Thermal Noise
Sun
Background (27K)
Signal
Field of View
Mirror
Hot object (333K)
196
Copyright 2001, S.D. Personick. All rights
The Receiver Subsystem
Example:
Equivalent background temperature: T = 100K
(Kelvins)
Equivalent background noise = kTB (watts)
Coupling loss of antenna into cable ~ 0dB
Cable loss = 3 dB
Cable temperature = 293K
Preamplifier Noise Temperature = 30K
Equivalent amplifier total input noise = ???
Copyright 2001, S.D. Personick. All rights
197
The Receiver Subsystem
Example (continued):
Equivalent input noise = kB [100/2 + 293/2 +
30]=226.5kB
I.e., half of the background noise + half of the cable
noise + the preamplifier noise
198
Copyright 2001, S.D. Personick. All rights
The End-To-End System
Example (continued from prior example):
Assume that the transmitter power amplifier produces 20
milliwatts. The coupling loss from the transmitter into its
coaxial antenna feed cable is 0 dB. The transmitter
antenna feed cable has 1.5 dB of loss. The antenna
radiates 90% of the power that arrives from the
transmitter antenna feed cable, and reflects the rest
back into the cable.
How much power is radiated? Assume that the
bandwidth,
B = 6 MHz. How much total propagation loss can we
199
allow if the required
signal-to-noise
ratio
is
40
dB?
Copyright 2001, S.D. Personick. All rights
The End-to-End System
Example: (continued)
The transmitter produces 20 mW = +13 dBm
The loss of the cable, plus the impact of a 10%
reflection (90% radiation) is:
1.5 dB (cable loss) - 10 log (0.9) = 1.95 dB
The total radiated power = +11.05 dBm = 12.7 mW
The required power at the preamplifier input =
226.5 x k x 6 x 10**6 x 10**4 (watts)... I.e., 40 dB
larger than the noise
200
Copyright 2001, S.D. Personick. All rights
The End-to-End System
• The radiated power = 12.7 mW
• The required power at the receiver
preamplifier input = 1.85 x 10**-7 mW
• The required power at the receiver
antenna is 2 x 1.85 x 10**-7 mW = 3.7 x
10**-7 mW
• The allowed propagation loss is the ratio
of these ~3.4 x 10**7 ~ 75.3 dB
201
Copyright 2001, S.D. Personick. All rights
Calculating the Propagation
Loss
Antenna equivalent
area = A
Area of
surface
= 4 r**2
Copyright 2001, S.D. Personick. All rights
202
Calculating the Propagation
Loss
Suppose the frequency is 100 MHz, and the
wavelength = 3 meters. Assume that the
equivalent area of the antenna is 2.25
square meters.
If the allowable loss is 75.3 dB, then the
distance from the transmitter to the receiver,
r , can be derived from:
2.25/[4 r**2) > 1/[3.4 x 10**7];
r < 2.5 km (Line-Of-Sight)
203
Copyright 2001, S.D. Personick. All rights
Why Do Broadcast
Transmitters Have Such High
Power?
In the previous example, we could cover 2.5
km with 20 mW. If we want to cover a 100 km
radius, we would need 40**2 or 1600 times
the power.
That would correspond to 32 watts << 10 kW
But, what about: high antenna-feeder-cable
losses, splitters, noisy preamplifiers,
inadequate antennas (low effective area, and
poor matching to the feeder cable),
204
attenuation through
buildings,
and
fading
Copyright 2001, S.D. Personick. All rights
Adaptive Antennas
Combiner
Multiple input antennas
Jamming signal
Desired input signal
Coherently combined output
signal
Copyright 2001, S.D. Personick. All rights
205
Adaptive Antennas
• An adaptive antenna can combine the
signals received by a multiplicity of
component antenna (in various
amplitude and phase relationships) to
null out one or more jamming signals
206
Copyright 2001, S.D. Personick. All rights
207
Copyright 2001, S.D. Personick. All rights
208
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Lasercom
209
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Telescopes
Receiver
Transmitter
210
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Telescope
Receiver
Transmitter
211
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T
R
212
Copyright 2001, S.D. Personick. All rights
R
T
Pointing errors and beam
wander due to clear air
turbulence
213
Copyright 2001, S.D. Personick. All rights
Bio-Complexity Network
Van Rensselear
Hall (33rd St)
Gbps
Ethernet
[fiber]
Light
beams
Telescope/Optical
Transceiver
Remote observation and
control
(Dr. Banu Onaral et. al).
Commonwealth Hall
(31st St)
Gbps
Ethernet
[fiber]
Cellular Observatory
Microscope System
(Dr. J.Yasha Kresh et. al.)
MCP Hahnemann
(15th St)
214
Copyright 2001, S.D. Personick. All rights
cey
215
Copyright 2001, S.D. Personick. All rights
216
Copyright 2001, S.D. Personick. All rights
Telecommunications
Networking I
Topic 9
Switching Fundamentals
Dr. Stewart D. Personick
Drexel University
217
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Switching Fundamentals
The Classical Switching Opportunity
# of links: N(N-1)/2
End system
End system
e.g., Telephone
218
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The Classical Switching
Opportunity
# of links: N
Switch
219
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What’s a Switch?
Crossbar switch
Signal path
closed crosspoint
wires
wires
# of crosspoints:
N
2
220
Copyright 2001, S.D. Personick. All rights
Mechanical Crosspoint: Reed
Relay
Apply a current to the coil,
and the resulting magnetic
field forces the reeds to
touch
Coil
Reeds
Glass tube
Control
Signal
221
Copyright 2001, S.D. Personick. All rights
Multi-stage Switching Fabric
# of cross points: 2N
~ 3/ 2
1
1
9
3x3
#1
3x3
#1
3x3
#2
3x3
#2
3x3
#3
3x3
#3
9
9x9 crosspoints vs. 9x6 cross points (with blocking)
222
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Blocking
Blocking occurs when a input cannot reach an
idle
output
Previous example: Blocking on group 1 occurs if:
2 or 3 inputs try to reach the same output group
Probability of blocking = [1/3 + 1/3] - 1/9
=5/9
Switching theory: how to design switches
with an acceptable # of cross points, and
an acceptable level of blocking
223
Copyright 2001, S.D. Personick. All rights
Banyan Switching Fabric
# of cross points =2 N log 2
1
N
1
2x2
8
8
224
Copyright 2001, S.D. Personick. All rights
Time Division Switching
Channel bank
Time-Space-Time Digital Switch
7
TSI
Space
Switch
TSI
19
TSI= Time Slot Interchanger
Input and output streams are time-division multiplexed
225
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There’s More to Switching
than Switches
• For example:
-On-hook/Off-hook detection
-Providing dial tone
-Signaling (e.g., tones or dial pulses)
-Call processing
-Advanced services (e.g., caller ID)
-Routing (finding a path to the
destination)
-Automatic Message Accounting (AMA)226
Copyright 2001, S.D. Personick. All rights
On-Hook/Off-Hook Detection
Line Card in Central
Office
Telephone
Line
Interface
Circuit
inductor
Hook
switch
Specification: “Off hook” = 20-120 ma of “loop” current
227
Copyright 2001, S.D. Personick. All rights
Signaling
a
b
d
1
2
e
4
5
7
8
*
0
c
3
Send two tones…
Choose: 1 of {a,b,c} and
1 of {d,e,f,g}
Total of 12 combinations
6
f
DTMF*=Dual tone multifrequency
9
g
#
*Also known as:
TouchTone(R)
228
Copyright 2001, S.D. Personick. All rights
Signaling
STP
SS7
Switch
Switch
Trunk Group
STP=Signaling Transfer Point
SS7= Signaling System Number 7
229
Copyright 2001, S.D. Personick. All rights
Finding a route to the
Destination
• Telephone switches are uploaded with
“translation tables” which map dialed
telephone numbers to predetermined
output ports on the switch. Thus, when I
dial 1-215-895-6208 from home in NJ,
my local switch “translates” the first six
numbers (plus the “1”), and knows that
this is a call that must be passed to an
“access tandem” switch…for routing on230
Copyright 2001, S.D. Personick. All rights
to my chosen
long distance carrier.
Automatic Message
Accounting
• Telephone switches collect “automatic
message accounting” (AMA) information:
called number, calling number, start time, end
time, special features used; and store it on a
disk. This information is periodically
downloaded by a billing center for offline
processing into telephone bills. A telephone
company’s billing system is a strategic
competitive asset.
231
Copyright 2001, S.D. Personick. All rights
Telecommunications
Networking I
Topic 10
Telecommunications Network Management
Dr. Stewart D. Personick
Drexel University
232
Copyright 2001, S.D. Personick. All rights
Telecommunications Network
Management
• Some questions to ask if you are the
owner/operator of a large
telecommunications network:
-How do I characterize the quality of
service I will provide in a way that can
be measured and engineered into my
networks; and in a way that my
customers will find useful
Copyright 2001, S.D. Personick. All rights
233
Telecommunications Network
Management
• Some questions to ask if you are the
owner/operator of a large
telecommunications network:
-How will I ensure that I have enough
network capacity to provide my
customers with a specified, quantifiable
quality of service…but not more network
capacity than what is required to do so 234
Copyright 2001, S.D. Personick. All rights
Telecommunications Network
Management
• Some questions to ask if you are the
owner/operator of a large
telecommunications network:
-How do I activate services for
customers without having to send
technicians out to the customers’
physical location (service activation)
-How do I monitor each customer’s
Copyright 2001, S.D. Personick. All rights
quality of service
(service assurance)
235
Telecommunications Network
Management
• Some questions to ask if you are the
owner/operator of a large
telecommunications network:
-How to I monitor the current state
(configuration, alarm conditions, and
usage measurement data) of my
network equipment; and how can my
network automatically recover from
Copyright 2001, S.D. Personick. All rights
various fault
conditions
236
Operations Support Systems
(OSS’s)
• Operations Support Systems (OSS’s), not to
be confused with operating systems, are
complex computer applications which are
used to automate many of the tasks that were
done manually, in telecommunications
systems, a few decades ago.
• The objectives are to increase
responsiveness to customers needs, and to
reduce cost
• Similar to networked information systems in 237
Copyright 2001, S.D. Personick. All rights
other industries
Telecommunications
Management Networks (TMN)
Security
Enterprise Management Systems
Billing
Service Management Systems Traffic Mgmt/QOS
Connectivity
Network Management Systems Redundancy
Routing
Element Management Systems ...
Network Elements
238
Copyright 2001, S.D. Personick. All rights
A Classic Traffic Engineering
Problem
Concentrator
1
1
Lines
N
Trunks
M<N
Erlang’s formulas (queuing theory
239
Copyright 2001, S.D. Personick. All rights
The Poisson (Random)
Process
Events
x
x
Time
x x
x
x
The probability of an “event” in any small
 is dt,
 where “ ” is called
interval of length dt (seconds)
the intensity of the Poisson random process. The occurre
of “events” in any interval is statistically independent
of the occurrence of “events” in any other disjoint
interval
240
Copyright 2001, S.D. Personick. All rights
The Poisson (Random)
Process
Events
x
x
x x
x
x
The probability of N “events” in an interval of length
T seconds is:
 e
N
Where
is

/ N!
T
241
Copyright 2001, S.D. Personick. All rights
Example
Suppose there are 100 telephone lines connected to
a
concentrator with M outgoing trunks. Each telephone
line generates calls (events) approximately as a
Poisson process, at an intensity of 3 calls per hour;
and each call lasts exactly 3 minutes. (Clearly a
simplification)
How many outgoing trunks, N, do we need to ensure
the probability of “all trunks busy” at any given time
is less than .01?
“All trunks busy” would occur if we had M or more
“events” in the Copyright
last 3 2001,
minutes
S.D. Personick. All rights
242
Example (Cont’d)
Probability of “all trunks busy” =


M
Where 
N 
 e
/ N!
< .01
=300 x 3/60 = 15
Answer M = 26
243
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Intelligent
Networks
244
Copyright 2001, S.D. Personick. All rights
Conventional Telephone Network with SS7
STP
Janet
Switch
STP
Switch
Bo
b
STP= Signaling Transfer Point
245
Copyright 2001, S.D. Personick. All rights
Call Processing
Bob’s
Call State:
Network
Janet’s
Call State:
Time
Idle (on hook)
Active (off hook)
Dial tone
Dialing (digit collection)
Post Dial (waiting)
Active/Busy (talking)
Idle (on hook)
Idle (on hook)
Ringing
Active/Busy
Idle (on hook)
246
Copyright 2001, S.D. Personick. All rights
Intelligent network-based 800 Service
800 SMS
800 SCP
STP
Janet
Switch
STP
Switch
Bo
b
STP= Signaling Transfer Point
SCP = Service Control Point
SMS=Service Management System
247
Copyright 2001, S.D. Personick. All rights
Call Processing
Bob’s
Call State:
Network
Janet’s
Call State:
Time
Idle (on hook)
Active (off hook)
Dial tone
Dialing (digit collection)
Play announcement
Digit Collection
Post Dial (waiting)
Active/Busy (talking)
Idle (on hook)
Idle (on hook)
Ringing
Active/Busy
Idle (on hook)
248
Copyright 2001, S.D. Personick. All rights
Service Creation Environment
Non-real-time download
SCE
STP
ISCP (data + call
processing)
Switch
Intelligent Peripheral (IP)
249
Copyright 2001, S.D. Personick. All rights
Advanced Services: Example
“Calling Party Name”
LIDB
Query/response
SW
SW
Bob
Incoming call
Janet
Bob dials Janet. Bob’s switch queries the line information data
base (LIDB) for the name associated with Bob’s line. This
information is passed to Janet’s switch using SS7. Janet’s
switch delivers this information between the first and second
ring of Janet’s phone, using a modem-like signal.
250
Copyright 2001, S.D. Personick. All rights
“Calling Party Name”
Bob dials Janet. Bob’s switch queries the line
information data base (LIDB) for the name associated
with Bob’s line. This information is passed to Janet’s
switch using SS7. Janet’s switch delivers this
information between the first and second ring of
Janet’s phone, using a modem-like signal.
It is also possible to implement this functionality by
having Janet’s switch send a query to Bob’s LIDB only
if Janet subscribes to the calling name delivery service
(rather than sending the information for every call).
251
Copyright 2001, S.D. Personick. All rights
Service Creation Example
• Customer
IP
off-hook
Switch
SCP
Prov inst
252
Copyright 2001, S.D. Personick. All rights
Service Creation Example
• Customer
IP
Switch
off-hook
SCP
Prov inst
Play announcement #1 &
Collect 5 digits
253
Copyright 2001, S.D. Personick. All rights
Service Creation Example
• Customer
IP
Switch
off-hook
SCP
Prov inst
Play announcement #1 &
Collect 5 digits
“Please enter you access
pin number”
254
Copyright 2001, S.D. Personick. All rights
Service Creation Example
• Customer
IP
Switch
off-hook
SCP
Prov inst
Play announcement #1 &
Collect 5 digits
“Please enter you access
pin number”
54321#
Authenticate
Authenticate
255
Copyright 2001, S.D. Personick. All rights
Service Creation Example
• Customer
IP
Switch
off-hook
SCP
Prov inst
Play announcement #1 &
Collect 5 digits
“Please enter you access
pin number”
54321#
Authenticate
Authenticate
OK
Play announcement # 2256&
Process call
Copyright 2001, S.D. Personick. All rights
Service Creation Example
• Customer
IP
Switch
off-hook
SCP
Prov inst
Play announcement #1 &
Collect 5 digits
“Please enter you access
pin number”
54321#
Authenticate
Authenticate
OK
Play announcement # 2257&
Process call
Copyright 2001, S.D. Personick. All rights
Advanced Intelligent Network –based Mobility Management
Janet’s
HLR
Janet’s current
VLR
Radio port controller
Switch
Switch
Radio port
Janet
Bo
b
HLR=Home Location Register
VLR=Visited Location Register
258
Copyright 2001, S.D. Personick. All rights
Advanced Intelligent Network –Domino’s Pizza
Time? Date?
Bob’s number?
Bob’s zip code?
Load management
Domino’s
973 829 1703
SCP
Switch 2
Switch 1
Bo
b
1. Bob (in area code 908) dials: 366-4667 (Dominos)
2. Switch 1 recognizes this as a number requiring “translation”
3. Switch 1 queries SCP.
4. SCP determines and returns the number of nearest open
Domino’s:
(973) 829-1703
Copyright 2001, S.D. Personick. All rights
259
Advanced Intelligent Network –Domino’s Pizza
Call for fire support
Time? Date?
Bob’s ID?
Bob’s location
Load management
Fire Support
(nearest)
Switch 2
SCP
Switch 1
1. Bob (in location XYZ) originates a request to “fire support”
2. Switch 1 recognizes this as origination as requiring
“translation”
3. Switch 1 queries SCP.
4. SCP determines and returns the network address of the
nearest
S.D. Personick. All rights
available sourceCopyright
of fire 2001,
support
Bo
b
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