TCOM 551
DIGITAL COMMUNICATIONS
FALL 2009
IN 134 Tuesdays 4:30 – 7:10 p.m.
Dr. Jeremy Allnutt
jallnutt@gmu.edu
TCOM 551
Fall 2009
Lecture number 1
1
General Information - 2
• Course Outline
– Go to http://telecom.gmu.edu and click on course
schedule or go to http://ece.gmu.edu/ and go to
“courses” and “Course web pages”; select TCOM 551
• Bad weather days: call (703) 993-1000
• Textbook: no mandatory requirement. The
Kolimbiris book is very useful. The Bateman
book provides additional information
• Mathematical Calculator – simple ones only
TCOM 551
Fall 2009
Lecture number 1
2
General Information - 3
• Homework Assignments
– Feel free to work together on these, BUT
– All submitted work must be your own work
• Web and other sources of information
– You may use any and all resources, BUT
– You must acknowledge all sources
– You must enclose in quotation marks all parts copied
directly – and you must give the full source information
TCOM 551
Fall 2009
Lecture number 1
3
No double jeopardy
General Information - 4
• Exam and Homework Answers
– For problems set, most marks will be given for
the solution procedure used, not the answer
– So: please give as much information as you can
when answering questions: partial credit cannot
be given if there is nothing to go on
– If something appears to be missing from the
question set, make – and spell out clearly –
assumptions used to find the solution
TCOM 551
Fall 2009
Lecture number 1
4
Double-line space, Times New Roman, 12 font size, default margins. References: must be
with source information – listed either as footnotes or tabulated at the end
General Information - 5
• Term Paper
– Any topic in field of Digital Communications
– About 10 pages long + about 4 figures
– Can work alone or in small groups (length of
paper grows with number in group – with
permission only)
– There will be no specific points given for the
paper, but it can help (or ruin) your grade
Possible Topics?
TCOM 551
Fall 2009
Lecture number 1
5
General Information – 6A
• Examples of Term Paper Topics
–
–
–
–
–
–
–
–
TDMA vs. CDMA in various situations
LD-CELP: what is it and how does it help?
What is net-centric communications?
Digital Imaging and its impact on sports casting
DBS: why did digital succeed where analog failed
What is a smart antenna and how will it help?
UWB applications
Bluetooth vs. IEEE 802.11B
And
TCOM 551
Fall 2009
Lecture number 1
6
General Information – 6B
• Examples of Term Paper Topics (Contd.)
–
–
–
–
–
–
MPEG2: what is it and how does it help?
Why has MPEG-4 taken the lead in video streaming?
Where to next with DVD’s?
Consequences of combining RFID with GPS
Is free-space optical communications for real?
What are the comparative merits of different large
screen displays (LCD, DLP, Plasma, etc.)?
– Talking appliances?
Etc.!!!
TCOM 551
Fall 2009
Lecture number 1
7
General Information - 7
• Class Grades
• Emphasis is on overall effort and results
• Balance between homework, tests, final
exam; plus term paper:
–
–
–
–
Homework
Tests
Final exam
Term Paper
TCOM 551
Fall 2009
Lecture number 1
15%
30 + 30%
25%
0%
8
Term Paper Grade Percentage – 1
• Contribution of paper to final grade (a)
– No mark will be allocated towards the paper.
The paper will be graded as quintuple plus (5+),
through dot (·), to quintuple minus (5–). A student
with a final grade close to the borderline between
two grades may be moved up across the borderline
if his/her paper is ≥ +++
A soft copy and a hard copy shall be submitted
TCOM 551
Fall 2009
Lecture number 1
9
Term Paper Grade Percentage – 2
• Contribution of paper to final grade (b)
– (i) A student who does not hand in an adequate
paper by the final exam without prior permission
will have his/her final exam score reduced by half
– (ii) A student who hands in their paper late, even
with permission, will not have their paper
considered for a “grade shift”
TCOM 551
Fall 2009
Lecture number 1
10
Term Paper Grade Percentage – 3
• PLAGIARISM
I plan on using search software on the term
papers, so please:
– No more than 40% by content directly from the
web
– All quoted content should be inside quotation
marks
– Every source should be acknowledged in the
paper at the point of usage.
TCOM 551
Fall 2009
Lecture number 1
11
Another
alternative
http://ece.gmu.edu/coursepages.htm
TCOM 551 & ECE 463 Course Plan
Go to http://telecom.gmu.edu, click on
course schedule, scroll down to TCOM 551
-
- In-Class Tests scheduled for
- October 6th
- November 17th
- In-Class Final exam scheduled for
- December 15th
TCOM 551
Fall 2009
Lecture number 1
12
TCOM 551 GTA Information
• The TA is TBD
• Email address is: TBD
• Office Hours in room TBD of the new
engineering building are:
– TBD;
TBD p.m.
• Please Email the TA if you would like to
meet with him/her.
TCOM 551
Fall 2009
Lecture number 1
13
TCOM 551 & ECE 463 Lect. 1 Outline
•
•
•
•
Sine Wave Review
Frequency, Phase, & Wavelength
Logarithms and dB (decibel) notation
Core Concepts of Digital Communications
– Source info., Carrier Signal, Modulation
– C/N, S/N, and BER
– Performance & Availability
TCOM 551
Fall 2009
Lecture number 1
14
TCOM 551 & ECE 463 Lect. 1 Outline
•
•
•
•
Sine Wave Review
Frequency, Phase, & Wavelength
Logarithms and dB (decibel) notation
Core Concepts of Digital Communications
– Source info., Carrier Signal, Modulation
– C/N, S/N, and BER
– Performance & Availability
TCOM 551
Fall 2009
Lecture number 1
15
Sine Wave Review – 1A
We all know that the Sine of an angle is the side
opposite to the angle divided by the hypotenuse,
i.e.
A
Sine (a) = A/B
Angle a
Point P
TCOM 551
Fall 2009
Lecture number 1
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Sine Wave Review – 1B
We all know that the Sine of an angle is the side
opposite to the angle divided by the hypotenuse,
i.e.
A
Sine (a) = A/B
Angle a
But what happens
if line B rotates
about Point P?
Point P
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Fall 2009
Lecture number 1
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Sine Wave Review – 2A
The end of Line B
will describe a
circle about Point P
a
Point P
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Fall 2009
Lecture number 1
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Sine Wave Review – 2B
The end of Line B
will describe a
circle about Point P
a
What happens if we now
shine a light from the left
and project the shadow of
the end of line B onto a
screen?
Point P
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Fall 2009
Lecture number 1
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Sine Wave Review - 3
a
End of “B”
projected
onto the
screen
Point P
Light
from
the left
TCOM 551
Fall 2009
Screen on
the right
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Sine Wave Review – 4A
End of “B”
projected
onto the
screen
As line “B” rotates about
the center point, P, the
projected end of line “B”
oscillates up and down on
the screen.
The end of line “B”
moves up and down with
what is called
Simple Harmonic Motion,
Screen on
the right
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Fall 2009
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Sine Wave Review – 4B
End of “B”
projected
onto the
screen
Simple Harmonic Motion
is an oscillation, or a
rotational movement,
about a mean value
(center) that is periodic.
What happens if we move
the screen to the right and
‘remember’ where the
projected end of “B” was?
Screen on
the right
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Fall 2009
Lecture number 1
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Sine Wave Review – 5A
Locus of “B”
end-point
We have a Sine Wave!
One oscillation =
One wavelength, 
a.k.a.
SHM
Screen
Position
TCOM
551 1
Fall 2009
Lecture number 1
Screen
Position 223
Sine Wave Review – 5B
Remember:
Sine 0 = 0; Sine 90 = 1; Sine 180 = 0; Since 270 = -1; Sine 360 = Sine 0 = 0
+1
0
90
180
270
360
Degrees
-1
TCOM 551
Fall 2009
Lecture number 1
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Sine and Cosine Waves – 1
Sine
Wave
0o
TCOM 551
Fall 2009
90o
Sine Wave = Cosine Wave shifted by 90o
180o
Cosine
Wave
270o
0 = 360o
Lecture number 1
90o
180o
25
Sine and Cosine Waves – 2
• There is a useful java applet that will show
you a sine wave derived from circular
motion (simple harmonic motion)
• The applet is found at:
http://home.covad.net/alcoat/sinewav.htm
It is very slow to load: have patience!
TCOM 551
Fall 2009
Lecture number 1
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Sine and Cosine Waves – 3
• Another applet that lets you ‘play’ with two
sine waves to see the combined waveform is:
http://www.udel.edu/idsardi/sinewave/sinewave.html
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Fall 2009
Lecture number 1
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For more details on Sine Waves
Sine and Cosine Waves – 4
Sine
Wave
0o
TCOM 551
Fall 2009
90o
Sine Wave = Cosine Wave shifted by 90o
180o
Cosine
Wave
270o
0 = 360o
Lecture number 1
90o
180o
28
http://en.wikipedia.org/wiki/Image:Sine_Cosine_Graph.png
Sine and Cosine Waves – 5
TCOM 551
Fall 2009
Lecture number 1
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Sine and Cosine Waves – 6
• Any wave that is periodic (i.e. it repeats
itself exactly over succeeding intervals) can
be resolved into a number of simple sine
waves, each with its own frequency
• This analysis of complex waveforms is part
of the Fourier Theorem
• You can build up a complex waveform with
harmonics of the fundamental frequency
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Fall 2009
Lecture number 1
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http://www.sfu.ca/sonic-studio/handbook/Harmonic_Series.html
Harmonics – 1
A harmonic is a multiple of a
fundamental frequency. In the figure
below, a fundamental frequency of
100 Hz is shown with 31 harmonics
(total of 32 “lines”).
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Fall 2009
Lecture number 1
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http://www.sfu.ca/sonic-studio/handbook/Law_of_Superposition.html
Harmonics – 2
In this example, 20
harmonics are mixed
together to form a
saw-tooth waveform
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Fall 2009
Lecture number 1
32
Sine and Cosine Waves - 7
“Cosine
Wave”
Sine Wave
TCOM 551
Fall 2009
Lecture number 1
Sine and Cosine
waves can
therefore be
considered to be
at right angles,
i.e. orthogonal,
to each other
33
Sine and Cosine Waves - 8
• A Radio Signal consists of an in-phase component
and an out-of-phase (orthogonal) component
• Signal, S, is often written in the generic form
S = A cos  + j B sin 
In-phase
component
Orthogonal
component
Real
Imaginary
TCOM 551
Fall 2009
Lecture number 1
Where j = ( -1 )
We will only
consider Real
signals
34
Sine and Cosine Waves - 9
• Two concepts
– The signal may be thought of as a time varying
voltage, V(t)
– The angle, , is made up of a time varying
component,  t, and a supplementary value, ,
which may be fixed or varying
• Thus we have a signal
V(t) = A cos (t + )
TCOM 551
Fall 2009
Lecture number 1
35
Sine and Cosine Waves - 10
• Time varying signal
V(t) = A cos (t + )
Instantaneous
value of the
signal
Vary these
to Modulate
the signal
Phase: PM; PSK
Frequency: FM; FSK
Amplitude: AM; ASK
Note:  = 2  f
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Fall 2009
Lecture number 1
36
Back to our Sine Wave – 1A
Defining the Wavelength


TCOM 551
Fall 2009
Lecture number 1
The wavelength
is calculated
between any
two points on
the wave where
it repeats itself
37
Back to our Sine Wave – 1B
Defining the Wavelength

However:

TCOM 551
Fall 2009
Measuring
between the
peaks or the
“zero crossings”
is often used:
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Back to our Sine Wave – 1C
Defining the Wavelength
The wavelength
is usually
defined at the
“zero crossings”
since these
points are more
precise than
anywhere else

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Fall 2009
Lecture number 1
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Back to our Sine Wave - 2
One revolution = 360o
One revolution also completes
one cycle (or wavelength) of
the wave.
So the “phase” of the wave
has moved from 0o to 360o
(i.e. back to 0o ) in one cycle.
The faster the phase changes,
the shorter the time one cycle
(one wavelength) takes
TCOM 551
Fall 2009
Lecture number 1
40
Back to our Sine Wave – 3
Two useful equations
The time taken to complete one
cycle, or wavelength, is the
period, T.
Frequency is the reciprocal of
the period, that is
f = 1/T

Phase has changed by 
The rate-of-change of the phase,
d/dt, is the frequency, f.
TCOM 551
Fall 2009
Lecture number 1
41
Before we look at d/dt, lets look at rate-of-change of phase
Sine Wave – 4
• What do we mean “Rate-of-change of phase
is frequency”?
One revolution = 360o = 2 radians
One revolution = 1 cycle
One revolution/s = 1 cycle/s = 1 Hz

Examples:
1.
720o/s = 2 revolutions/s = 2 Hz
2.
18,000o/s = 18,000/360 revs/s
= 50 revs/s = 50 Hz
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Fall 2009
Lecture number 1
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http://www.sfu.ca/sonic-studio/handbook/Simple_Harmonic_Motion.html
Simple Harmonic Motion
“Geometric derivation of simple harmonic motion. A point p moves at constant speed on the
circumference of a circle in counter-clockwise motion. Its projection OC on the vertical axis XOY
is shown at right as a function of the angle . The function described is that of a sine wave.”
From the URL above
TCOM 551
Fall 2009
Lecture number 1
43
d/dt Digression - 1
Person walks 16 km in 4 hours.
kilometers
16
Velocity = (distance)/(time)
Therefore, Velocity = 16/4
= 4 km/h
12
8
Velocity is really the rate-of-change
of distance with time.
4
0
0
TCOM 551
Fall 2009
1
2
3
4
5
Time, hours
What if the velocity is not constant?
6
7
8
9
Lecture number 1
44
d/dt Digression - 2
kilometers
You can compute the
Average Velocity
using distance/time,
(i.e. 16/8 = 2 km/h),
but how do you get
the person’s speed at
any particular point?
16
12
8
4
0
0
TCOM 551
Fall 2009
1
2
3
4
5
6
7
8
9
Time, hours
Answer: you differentiate,
which means you find the
slope of the line.
Lecture number 1
45
d/dt Digression - 3
kilometers
16
12
B
8
4
0
0
TCOM 551
Fall 2009
1
2
3
4
5
6
Time, hours
Lecture number 1
To differentiate
means to find the
slope at any instant.
A
The slope of a curve
is given by the
tangent at that point,
i.e., A/B
In this case, A is in
km and B is in hours.
It could equally well
7
8
9 phase, , and
be
time, t.
46
d/dt Digression - 4
-When we differentiate, we are taking the
smallest increment possible of the parameter
over the smallest interval of (in this case)
time.
- Small increments are written ‘d’(unit)
-Thus: the slope, or rate-of-change, of the
phase, , with time, t, is written as d/dt
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Fall 2009
Lecture number 1
47
Sine Wave Continued
• Can think of a Sine Wave as a Carrier Signal,
i.e. the signal onto which the information is
loaded for sending to the end user
• A Carrier Signal is used as the basis for sending
electromagnetic signals between a transmitter
and a receiver, independently of the frequency
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Fall 2009
Lecture number 1
48
Carrier signals – 1
• A Carrier Signal may be considered to
travel at the speed of light, c, whether it is
in free space or in a metal wire
• Travels more slowly in most substances
• The velocity, frequency, and wavelength of
the carrier signal are uniquely connected by
c=f
Velocity of light
TCOM 551
Fall 2009
Lecture number 1
Wavelength
Frequency
49
Carrier signals – 2
• Example
– WAMU (National Public Radio) transmits at a
carrier frequency of 88.5 MHz
– What is the wavelength of the carrier signal?
• Answer
– c = (3×108) m/s = f × = (88.5  106) × ()
– Which gives  = 3.3898 m = 3.4 m
Remember: Make sure you are using the correct units
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Fall 2009
Lecture number 1
50
Digression – UNITS – 1
• Standard units to use are MKS
– M = meters
– K = kilograms
– S = seconds
written as m
written as kgm
written as s
• Hence
– the velocity of light is in m/s
– The wavelength is in m
– And the frequency is in Hz = hertz
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Fall 2009
Lecture number 1
51
Digression – UNITS – 2
• Standard units to use are MKS
– M = meters
– K = kilograms
– S = seconds
So: do not mix
written as m
feet with meters
written as kgm
and pounds
written as s
with kilograms
• Hence
– the velocity of light is in m/s
– The wavelength is in m
– And the frequency is in Hz = hertz
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Fall 2009
Lecture number 1
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Carrier signals – 3
• A Carrier Signal can
– carry just one channel of information (this is often
called Single Channel Per Carrier = SCPC)
– Or carry many channels of information at the same
time, usually through a Multiplexer
Single
Channel
Tx
Multiplexer
TCOM 551
Fall 2009
Note: The modulator
has been omitted in
these drawings
SCPC
Multi-channel
carrier
Tx
Lecture number 1
Multiplexed
Carrier
53
TCOM 551 & ECE 463 Lect. 1 Outline
•
•
•
•
Sine Wave Review
Frequency, Phase, & Wavelength
Logarithms and dB (decibel) notation
Core Concepts of Digital Communications
– Source info., Carrier Signal, Modulation
– C/N, S/N, and BER
– Performance & Availability
TCOM 551
Fall 2009
Lecture number 1
54
Logarithms – 1
• The use of logarithms came about for two
basic reasons:
– A need to multiply and divide very large numbers
– A need to describe specific processes (e.g. in
Information Theory) that counted in different
bases
• Numbers are to the base 10; i.e. we count in
multiples of tens
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Fall 2009
Lecture number 1
55
Logarithms – 2
• 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
To be easier to see, this should be
written as the series
00, 01, 02, 03, 04, 05, …. 09, 10
•
•
•
•
•
11, 12, 13, 14, 15 …..
…..
91, ……, 97, 98, 99, 100
…
991, ….., 997, 998, 999, 1000
TCOM 551
Fall 2009
Lecture number 1
We actually count
from 1 to 10 but
the numbering
goes from 0 to 9,
then we change
the first digit and
go from 0 to 9
again, and so on
56
Logarithms – 3
• Counting to base 10 is the Decimal System
• We could equally well count in a
Duodecimal System, which is a base 12, a
Hexadecimal System, which is a base 16, a
Binary System, which is a base 2, etc.
• Sticking with the Decimal System
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Fall 2009
Lecture number 1
57
Logarithms – 4A
• A Decimal System can be written as a
power of 10, for example
–
–
–
–
–
100 = 1
101 = 10
102 = 100
103 = 1,000
104 = 10,000
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Fall 2009
Lecture number 1
58
Logarithms – 4B
• A Decimal System can be written as a
power of 10, for example
–
–
–
–
–
100 = 1
101 = 10
102 = 100
103 = 1,000
104 = 10,000
TCOM 551
Fall 2009
Do you detect any logic here?
Lecture number 1
59
Logarithms – 4C
• A Decimal System can be written as a
power of 10, for example
–
–
–
–
–
100 = 1
101 = 10
102 = 100
103 = 1,000
104 = 10,000
TCOM 551
Fall 2009
Do you detect any logic here?
The number of zeroes is the
same as the value of the
exponent
Lecture number 1
60
Logarithms – 5
• Let’s look at these again
–
–
–
–
–
The exponent is called the
100 = 1
logarithm of the number
101 = 10
102 = 100
That is:
103 = 1,000 The logarithm of 1 = 0
104 = 10,000 The logarithm of 10 = 1
The logarithm of 100 = 2, etc.
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Fall 2009
Lecture number 1
61
Logarithms – 6
• Question:
– The logarithm of 1 to the base 10 (written as
log101) = 0 and log1010 = 1. What if I want the
logarithm of a number between 1 and 10?
• Answer:
– You know the answer must lie between 0 and 1
– The answer = x, where x is the exponent of 10
– Ummmmmh???? We’ll do an example
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Fall 2009
Lecture number 1
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Logarithms – 7
• Question
– What is the logarithm of 3?
• Answer:
–
–
–
–
–
We want log103
Let log103 = x
Transposing, we have 10x = 3
And 100.4771213 = 3, giving x = 0.4771
Thus log103 = 0.4771
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Fall 2009
Lecture number 1
63
Logarithms – 8
• More Examples
–
–
–
–
–
–
What is log10 4?
What is log10 7?
What is log10 7.654?
What is log10 24?
What is log10 4123.68?
What is log10 0.69?
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Fall 2009
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Logarithms – 9
• More Examples (Answers)
–
–
–
–
–
–
What is log10 4?
What is log10 7?
What is log10 7.654?
What is log10 24?
What is log10 4123.68?
What is log10 0.69?
=
=
=
=
=
=
0.6021
0.8451
0.8839
1.3802
3.6153
-0.1612
0.69 is < 1 so the answer must be below 0
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Fall 2009
Lecture number 1
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Logarithms – 10
• Question
– What if I want to have a logarithm of the value
“x” with a different base?
• Answer
– Let’s assume you want to have loga of x, i.e. the
base is “a” and not 10
– Then loga x =(log10 x) / (log10 a)
Example
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Fall 2009
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Logarithms – 11
• Question
– What is log2 10?
(i.e. base “a” = 2 and the number x =10)
• Answer
– Since loga x =(log10 x) / (log10 a)
– Log210 = (log1010) / (log102) = 1/0.301
= 3.3219
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Fall 2009
Lecture number 1
67
Logarithms – 12
• Let’s look at this another way:
– Log2 10 = 3.3219
• Remember, if loga (number) = x, we can
transpose this to ax = (number)
• Thus, another way of looking at
– Log2 10 = 3.3219 is to write
– 23.3219 = 10 But what if the exponent is
always a whole number?
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Fall 2009
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Logarithms – 13
•
•
•
•
•
•
•
20 = 1
21 = 2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
TCOM 551
Fall 2009
log2 1 = 0
This is the
log2 2 = 1
Binary
System
log2 4 = 2
log2 8 = 3
Log2 is
log2 16 = 4 fundamental to
Information
log2 32 = 5
Theory
log2 64 = 6
Lecture number 1
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Logarithms – 14
• Note you can go forwards (logarithm) and
backwards (anti-logarithm), thus
– If log 10 (number) = x
• Then
– The anti-logarithm of a (value = x) is given by
10x
• So the calculator button “log” gives the
logarithm and the calculator button “10x” gives
the anti-logarithm
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Fall 2009
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70
Logarithms – 15
• Standard notations
– A log10 (number) is normally written as
log (number) - i.e. leave off the 10; e.g. log10 = 1
– A logarithm that uses the exponential value, e, as a
base, referred to as a “natural” logarithm, is
written as loge (number), or ln (number)
– All other bases must be included if they are not 10
or e; e.g. log2 (number)
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Fall 2009
Lecture number 1
71
Logarithms – 16
• So how do logarithms help us?
• Answer: by converting to logarithms
– Instead of multiplying you can add
– Instead of dividing you can subtract
– [They are also an intermediate step (see later)]
• How is that possible?
– See example on the next slide
TCOM 551
Fall 2009
Lecture number 1
72
Logarithms – 17
• Example
2+3=5
– 100  1,000 = 102  103 = 105
– 297  4735 = 102.4728  103.6753 = 106.1481
= 1,406,294.998
– 3879  193 = 103.5907  102.2856 = 101.3051
= 20.1917
• Big Deal! My calculator can do that stuff in
zero seconds flat! So: read on!
TCOM 551
Fall 2009
Lecture number 1
73
Logarithms – 18
• What if the numbers are really large or
really small?
• Examples
– (1,387.465  1014)  (893  109)
– (1.38  10-23)  (10, 397)  (283)
• But logarithms are really an intermediate
step to decibels (written as dB)
TCOM 551
Fall 2009
Lecture number 1
74
Decibel (dB) Notation – 1
• Historically the Bel, named after Alexander
Graham Bell, is a unit of sound
• It was developed as a ratio measure: i.e., it
compares the various sound levels
• The Bel was found to be too large a value and so a
tenth of a Bel was used, i.e., the decibel
• A decibel, or 1 dB, was found to be the minimum
change in sound level a human ear could detect
TCOM 551
Fall 2009
Lecture number 1
75
Decibel (dB) Notation – 2
• Question
– How do you get a dB value?
• Answer
– Take the log10 value and multiply it by 10
• Example
– One number is 7 times larger than another. The
dB difference = 10  log107 = 10  0.8451
= 8.5 dB
NOTE: Never quote a dB number to more
than one place of decimals
TCOM 551
Fall 2009
Lecture number 1
76
Decibel (dB) Notation – 3
• Some things to remember
– A dB value is always 10 log10 ; it is never, ever,
20 log10 , however …..
– 10 log10 (x)a = 10  a  log10 (x)
• e.g. 10 log10 (x)2 = 10  2  log10 (x) = 20 log 10 (x)
– The dB ratio may be referenced to a given
level, for example
• 1 W (unit would be dBW)
• 1 mW (unit would be dBm)
TCOM 551
Fall 2009
Lecture number 1
Some examples
77
Decibel (dB) Notation – 4
• Question
– An amplifier increases power by a ratio of 17:1, what is
the dB gain?
• Answer
– 10 log10 17 = 12.3 dB
• Question
– The amplifier is fed with 1W, how many watts are
output?
• Answer
– 17 Watts which is equivalent to 12.3 dBW
TCOM 551
Fall 2009
Lecture number 1
78
Decibel (dB) Notation – 5
• NOTE:
– Whenever you have just “dB” after a number,
then it is merely a ratio. EG: 3dB bigger just
means twice as big. It gives you no measure of
the absolute amount.
– Whenever you have additional letters after “dB”,
this will tell you the absolute value. EG: 3dBW
means 3dB bigger than a watt = 2 watts.
TCOM 551
Fall 2009
Lecture number 1
79
Decibel (dB) Notation – 6
• Examples of dB notations of power, etc.
–
–
–
–
–
–
425 W  26.3 dBW
425 W = 425,000 mW  56.3 dBm
0.3 W  -5.2 dBW
0.3W = 300 mW  24.8 dBm
24,500 K  43.9 dBK
-273 K  Error – you cannot take a logarithm
of a negative number
TCOM 551
Fall 2009
Lecture number 1
80
TCOM 551 & ECE 463 Lect. 1 Outline
•
•
•
•
Sine Wave Review
Frequency, Phase, & Wavelength
Logarithms and dB (decibel) notation
Core Concepts of Digital Communications
– Source info., Carrier Signal, Modulation
– C/N, S/N, and BER
– Performance & Availability
TCOM 551
Fall 2009
Lecture number 1
81
Core Concepts of Digital
Communications – 1
Frequency
Amplification
and transmission
RF
to
IF
Modulation
Channel coding
Multiplexing
Source encoding
Source;
Frequency
Transmission medium
Reception and
amplification
RF
to
IF
Demodulation
Channel decoding
Demultiplexing
Sink;
Information user
Distance
TCOM 551
Fall 2009
Lecture number 1
82
Core Concepts of Digital
Communications – 2
Frequency
Amplification
and transmission
RF
to
IF
Modulation
Channel coding
Multiplexing
Source encoding
Source;
Frequency
Transmission medium
Lectures 2, 6, 7, 11, 12, &14
Lectures 3, 4, & 8 Lectures
9 & 10
Lecture 13
Lecture 4
Lectures 3 & 5
Reception and
amplification
RF
to
IF
Demodulation
Channel decoding
Demultiplexing
Sink;
Information user
Distance
TCOM 551
Fall 2009
Lecture number 1
83
Key Design Issues – 1
• S/N
– Signal-to-Noise Ratio (Analog)
• Need to be above user’s threshold for Required QoS
• C/N
– Carrier-to-Noise Ratio (Analog and Digital)
• Need to be above demodulation threshold
for useful transfer of information
• BER
We will
look at
each of
these
– Bit Error Rate (Sometimes Bit Error Ratio)  S/N
• Need to satisfy the Performance and Availability
Specifications
TCOM 551
Fall 2009
Lecture number 1
84
Signal-to-Noise Ratio – 1
• Signal-to-Noise, written as S/N, is mainly
used for Analog Systems
• S/N is specified at the
Baseband of the Information Channel
Baseband is a
range of
frequencies
close to zero
TCOM 551
Fall 2009
Information is what is sent
to the user and the channel
over which it is sent is the
Information Channel
Lecture number 1
85
Signal-to-Noise Ratio – 2
• What S/N value gives a good reception?
– Telephone and TV channels require a minimum
of 50 dB
50 dB  ratio of 100,000
IE:the Signal power is 100,000 > the Noise power
• Analog signals have “graceful degradation”
characteristics
TCOM 551
Fall 2009
Lecture number 1
86
Signal-to-Noise Ratio – 3A
Digital signal
Analog signal
S/N
Good
Level
Marginal
or
Bad
Eb/N
o100
TCOM 551
Fall 2009
80
60
40
20
Percentage Time above Threshold
Lecture number 1
0
87
Many times you will find performance and
availability curves with this perspective
Signal-to-Noise Ratio – 3B
Digital signal
S/N
Analog signal
Good
Level
Marginal
or
Bad
Eb/N
o0
TCOM 551
Fall 2009
20
40
60
80
Percentage Time above Threshold
Lecture number 1
100
88
Signal-to-Noise Ratio – 4
• The S/N is what the user perceives, but it is
usually measured at the demodulator output
Received
signal
Demodulator
Output
S/N
User’s
Application
Device
• The C/N at the demodulator input will
determine the output S/N
TCOM 551
Fall 2009
Lecture number 1
89
Carrier-to-Noise Ratio – 1
• Carrier-to-Noise, written as C/N, is used for both
Analog and Digital Systems
• The Carrier signal has information from the
sender impressed upon it, through modulation.
The carrier, plus the modulated information, will
pass through the wideband portion of transmitter
and receiver, and also over the transmission path
???
TCOM 551
Fall 2009
Lecture number 1
90
Carrier-to-Noise Ratio – 2
= Wideband (passband) signal with modulation
= Baseband signal with raw information
Transmitter
Receiver
RF
Information
to be sent
TCOM 551
Fall 2009
RF
Mixer
Mixer
IF
IF
Modulator
Demodulator
Lecture number 1
The C/N at the
input to the
demodulator is
the key design
point in any
communications
system
Information
received
91
Carrier-to-Noise Ratio – 3
Input
C/N
C/N
12
10
8
6
4
2
0
TCOM 551
Fall 2009
Demodulator
Useful output?
Conservative design Level
(10 dB) with no coding
Can use these
C/N levels with
Coding, etc.
Lecture number 1
92
Carrier-to-Noise Ratio – 4
• Useful design reference for uncoded QPSK
BER = 10-6 at 10.6 dB input C/N to Demodulator
10.6 dB
BER
10-3
10-4
10-5
10-6
10-7
10-8
BER Voice Maximum
BER Data Maximum
0
TCOM 551
Fall 2009
BER?
10
20
Lecture number 1
30
C/N
Goal is
≤ 10-10
93
BER – 1
• BER means Bit Error Rate, however some
people refer to it as the Bit Error Ratio (i.e.
the ratio of bad to good bits)
• Strictly speaking, it is the Probability that a
single Bit Error will occur
• BER is usually given as a power exponent,
e.g. 10-6, which means one error in 106 bits
TCOM 551
Fall 2009
Lecture number 1
94
BER – 2
• A BER of 10-6 means on the order of one
error in a page of a FAX message
• To improve BER, channel coding is used
– FEC codes
– Interleaved codes
• Communications systems are specified in
many ways, but the two most common are
performance and availability
TCOM 551
Fall 2009
Lecture number 1
95
BER – 3
• Performance
– Generally specified as a BER to be maintained
for a very high percentage of the time (usually
set between 98% and 99% of the time)
• Availability
– Generally specified as a minimum BER below
which no information can be transmitted
successfully - i.e. an outage occurs
TCOM 551
Fall 2009
Lecture number 1
96
Fig. 8.4 in Pratt et al.,
Satellite Communications
BER – 4
TCOM 551
Fall 2009
Lecture number 1
97
BER – 5
• What causes the change in BER?
• Since BER is determined by C/N, change in
BER is caused either by
– Changes in C (i.e. carrier power level)
• Antenna loses track
• Attenuation of signal
– Changes in N (i.e. noise power level)
We will
look at
this one
• Interference
• Enhanced noise input
TCOM 551
Fall 2009
Lecture number 1
98
BER – 6
Attenuation,
dB
20
16
99.999% = 0.001% outage is a
typical single-hop specification
 19 dB
99.99% = 0.01% outage is a
typical high availability spec.
12
8
99.7% = 0.03% outage
is a typical VSAT spec.
 6 dB
4
0
100
TCOM 551
Fall 2009
3 dB
10
1
0.1
0.01
Percentage of the Time
Lecture number 1
0.001
99
BER – 7
Performance & Availability
BER
10-10
Exceeds Performance Spec.
10-8
10-6
10-4
10-2
100
TCOM 551
Fall 2009
Exceeds Availability Spec.
Does not meet
Performance or
Availability Specs.
10
1
0.1
0.01
Percentage of the Time
Lecture number 1
0.001
100
BER – 8
Performance & Availability
BER
10-10
With Coding
10-8
10-6
Without Coding
10-4
10-2
100
TCOM 551
Fall 2009
10
1
0.1
0.01
Percentage of the Time
Lecture number 1
0.001
101