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TCOM 551 DIGITAL COMMUNICATIONS. SPRING 2005 IN 136 Wednesdays 4:30 – 7:10 p.m. Dr. Jeremy Allnutt jallnutt@gmu.edu. General Information - 1. Contact Information Room: Science & Technology II, Room 269 Telephone (703) 993-3969 Email: jallnutt@gmu.edu Office Manager: TBD

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tcom 551 digital communications

TCOM 551DIGITAL COMMUNICATIONS

SPRING 2005

IN 136 Wednesdays 4:30 – 7:10 p.m.

Dr. Jeremy Allnutt jallnutt@gmu.edu

Lecture number 1

general information 1
General Information - 1
  • Contact Information
    • Room: Science & Technology II, Room 269
    • Telephone (703) 993-3969
    • Email: jallnutt@gmu.edu
    • Office Manager: TBD
  • Office Hours
    • Mondays and Tuesdays 3:00 – 6:00 p.m.Please, by appointment only

Lecture number 1

general information 2

Sorry: the web document is not yet up

General Information - 2
  • Course Outline
    • Go to http://telecom.gmu.eduand click oncourse schedule
    • Scroll down to TCOM 551
  • Bad weather days: call (703) 993-1000
  • You MUST Have The Following
    • Bateman Textbook, preferably also Kolimbiris
    • A Mathematical Calculator – please, simple ones only

Lecture number 1

general information 3
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

Lecture number 1

general information 4

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 give – assumptions used to find the solution

Lecture number 1

general information 5
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)
    • Paper marks depend on delivery date(see slide 9)

Possible Topics?

Lecture number 1

general information 6
General Information - 6
  • Examples of Term Paper Topics
    • TDMA vs. CDMA in various situations
    • LD-CELP: what is it and how does it help?
    • MPEG2: what is it and how does it help?
    • 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
    • Bluetooth vs. IEEE 802.11B

Etc.!!!

Lecture number 1

general information 7
General Information - 7
  • Class Grades
  • Emphasis is on overall effort and results
  • Balance between homework, tests, paper + final exam:
    • Homework - 10%
    • Tests - 25 + 25%
    • Final exam - 30%
    • Term Paper - 10% maximum

Lecture number 1

term paper grade percentage
Term Paper Grade Percentage
  • A mark will be allocated towards the grade as follows:
    • Prior to and on April 6th:  10%
    • Nov. 10th through April 13th:  8%
    • Nov. 17th through April 20th:  6%
    • Nov. 24th through May 27th:  3%
    • After April 27th: 0%

Lecture number 1

tcom 551 course plan

http://ece.gmu.edu/coursepages.htm

TCOM 551 Course Plan

Another alternative

  • Go to http://telecom.gmu.edu,click on course schedule, scroll down to TCOM 551
  • In-Class Tests scheduled for - March 2nd - April 13th
  • In-Class Final exam scheduled for - May 11th

Lecture number 1

tcom 551 lecture 1 outline
TCOM 551 Lecture 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

Lecture number 1

sine wave review 1
Sine Wave Review - 1

We all know that the Sine of an angle is the opposite side divided by the hypotenuse, i.e.

B

Sine (a) = A/B

A

Angle a

But what happens if line B rotates about Point P?

Point P

Lecture number 1

sine wave review 2
Sine Wave Review - 2

Line B now describes a circle about Point P

B

a

What happens if we shine a light from the left and project the shadow of B onto a screen?

Point P

Lecture number 1

sine wave review 3
Sine Wave Review - 3

End of “B” projected onto the screen

B

a

Point P

Light from the left

Screen on the right

Lecture number 1

sine wave review 4
Sine Wave Review - 4

End of “B” projected onto the screen

As line “B” rotates about the center point, P, the projected end of “B” oscillates up and down on the screen. What happens if we move the screen to the right and ‘remember’ where the projected end of “B” was?

Screen on the right

Lecture number 1

sine wave review 5
Sine Wave Review - 5

Locus of “B” end-point

We have a Sine Wave!

One oscillation = One wavelength, 

a.k.a. SHM

ScreenPosition 1

ScreenPosition 2

Lecture number 1

sine wave review 51
Sine Wave Review - 5

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

Lecture number 1

sine and cosine waves 1
Sine and Cosine Waves - 1

SineWave

Sine Wave = Cosine Wave shifted by 90o

0o 90o 180o 270o 0 = 360o 90o 180o

CosineWave

Lecture number 1

sine and cosine waves 2
Sine and Cosine Waves - 2

Sine and Cosine waves can therefore be considered to be at right angles, i.e. orthogonal, to each other

“Cosine Wave”

Sine Wave

Lecture number 1

sine and cosine waves 3
Sine and Cosine Waves - 3
  • A Radio Signal consists of an in-phase component and an out-of-phase (orthogonal) component
  • Signal, S, is often written in the generic formS = A cos  + j B sin 

Where j = ( -1 )

In-phase component

Orthogonal component

We will only consider Real signals

Real

Imaginary

Lecture number 1

sine and cosine waves 4
Sine and Cosine Waves - 4
  • 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 signalV(t) = A cos (t +)

Lecture number 1

sine and cosine waves 5
Sine and Cosine Waves - 5

Vary these to Modulate the signal

  • Time varying signalV(t) = A cos (t +)

Phase: PM; PSK

Instantaneous value of the signal

Frequency: FM; FSK

Amplitude: AM; ASK

Note:  = 2  f

Lecture number 1

back to our sine wave 1 defining the wavelength
Back to our Sine Wave – 1Defining the Wavelength

The wavelength is usually defined at the “zero crossings”

Lecture number 1

back to our sine wave 2
Back to our Sine Wave - 2

One revolution = 360oOne 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

Lecture number 1

back to our sine wave 3 two useful equations
Back to our Sine Wave – 3Two useful equations

The time taken to complete one cycle, or wavelength, is the period, T.Frequency is the reciprocal of the period, that isf = 1/T

Phase has changed by The rate-of-change of the phase, d/dt, is the frequency, f.

Lecture number 1

sine wave 4

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:
  • 720o/s = 2 revolutions/s = 2 Hz
  • 18,000o/s = 18,000/360 revs/s = 50 revs/s = 50 Hz

Lecture number 1

d dt digression 1
d/dt Digression - 1

Person walks 16 km in 4 hours.Velocity = (distance)/(time)Therefore, Velocity = 16/4 = 4 km/hVelocity is really the rate-of-change of distance with time.What if the velocity is not constant?

kilometers

1612840

0 1 2 3 4 5 6 7 8 9Time, hours

Lecture number 1

d dt digression 2
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?

1612840

0 1 2 3 4 5 6 7 8 9Time, hours

Answer: you differentiate, which means you find the slope of the line.

Lecture number 1

d dt digression 3
d/dt Digression - 3

kilometers

To differentiate means to find the slope at any instant.The slope of a curve is given by the tangent at that point, i.e., A/BIn this case, A is in km and B is in hours. It could equally well be phase, , and time, t.

1612840

A

B

0 1 2 3 4 5 6 7 8 9Time, hours

Lecture number 1

d dt digression 4
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

Lecture number 1

sine wave continued
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

Lecture number 1

carrier signals 1
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 byc = f 

Wavelength

Velocity of light

Frequency

Lecture number 1

carrier signals 2
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

Lecture number 1

digression units
Digression - UNITS
  • Standard units to use are MKS
    • M = meters written as m
    • K = kilograms written as kgm
    • S = seconds written as s
  • Hence
    • the velocity of light is in m/s
    • The wavelength is in m
    • And the frequency is in Hz = hertz

So: do not mix feet with meters and pounds with kilograms

Lecture number 1

carrier signals 3
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

Note: The modulator has been omitted in these drawings

Tx

SCPC

Multiplexer

Multiplexed Carrier

Multi-channel carrier

Tx

Lecture number 1

logarithms 1
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

Lecture number 1

logarithms 2
Logarithms - 2
  • 1, 2, 3, 4, 5, 6, 7, 8, 9, 10To be easier to see, this should be written as the series00, 01, 02, 03, 04, 05, …. 09, 10
  • 11, 12, 13, 14, 15 …..
  • …..
  • 91, ……, 97, 98, 99, 100
  • 991, ….., 997, 998, 999, 1000

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

Lecture number 1

logarithms 3
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

Lecture number 1

logarithms 4a
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

Lecture number 1

logarithms 4b
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

Do you detect any logic here?

Lecture number 1

logarithms 4c
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

Do you detect any logic here?

The number of zeroes is the same as the value of the exponent

Lecture number 1

logarithms 5
Logarithms - 5
  • Let’s look at these again
    • 100 = 1
    • 101 = 10
    • 102 = 100
    • 103 = 1,000
    • 104 = 10,000

The exponent is called the logarithm of the number

That is:The logarithm of 1 = 0The logarithm of 10 = 1The logarithm of 100 = 2, etc.

Lecture number 1

logarithms 6
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

Lecture number 1

logarithms 7
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

Lecture number 1

logarithms 8
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?

Lecture number 1

logarithms 9
Logarithms - 9
  • More Examples (Answers)
    • What is log10 4? = 0.6021
    • What is log10 7? = 0.8451
    • What is log10 7.654? = 0.8839
    • What is log10 24? = 1.3802
    • What is log10 4123.68? = 3.6153
    • What is log10 0.69? = -0.1612

0.69 is < 1 so the answer must be below 0

Lecture number 1

logarithms 10
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 logax =(log10 x) / (log10 a)

Example

Lecture number 1

logarithms 11
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

Lecture number 1

logarithms 12
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?

Lecture number 1

logarithms 13
Logarithms - 13
  • 20 = 1 log2 1 = 0
  • 21 = 2 log2 2 = 1
  • 22 = 4 log2 4 = 2
  • 23 = 8 log2 8 = 3
  • 24 = 16 log2 16 = 4
  • 25 = 32 log2 32 = 5
  • 26 = 64 log2 64 = 6

This is the Binary System

Log2 is fundamental to Information Theory

Lecture number 1

logarithms 14
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 by10x
  • So the calculator button “log” gives the logarithm and the calculator button “10x” gives the anti-logarithm

Lecture number 1

logarithms 15
Logarithms - 15
  • Standard notations
    • A log10 (number) is normally written as log (number) - i.e. leave off the 10; e.g. log 10 = 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)

Lecture number 1

logarithms 16
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

Lecture number 1

logarithms 17
Logarithms - 17

2 + 3 = 5

  • Example
    • 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!

Lecture number 1

logarithms 18
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)

Lecture number 1

decibel db notation 1
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

Lecture number 1

decibel db notation 11
Decibel (dB) Notation - 1
  • 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

Lecture number 1

decibel db notation 2
Decibel (dB) Notation - 2
  • 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)

Some examples

Lecture number 1

decibel db notation 3
Decibel (dB) Notation - 3
  • 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

Lecture number 1

decibel db notation 4
Decibel (dB) Notation - 4
  • 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

Lecture number 1

core concepts of digital communications 1
Core Concepts of Digital Communications - 1

Frequency

Frequency

Amplification and transmission

Reception and amplification

Transmission medium

RFtoIF

RFtoIF

Modulation

Demodulation

Channel coding

Channel decoding

Multiplexing

Demultiplexing

Source encodingSource;

Sink;Information user

Distance

Lecture number 1

core concepts of digital communications 2
Core Concepts of Digital Communications - 2

Frequency

Frequency

Amplification and transmission

Reception and amplification

Transmission medium

RFtoIF

RFtoIF

Lectures 2, 6, 7, 11, 12, &14Lectures 3, 4, & 8 Lectures 9 & 10Lecture 13Lecture 4Lectures 3 & 5

Modulation

Demodulation

Channel coding

Channel decoding

Multiplexing

Demultiplexing

Source encodingSource;

Sink;Information user

Distance

Lecture number 1

key design issues 1
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 thresholdfor useful transfer of information
  • BER
    • Bit Error Rate (Sometimes Bit Error Ratio)  S/N
      • Need to satisfy the Performance and Availability Specifications

We will look at each of these

Lecture number 1

signal to noise ratio 1
Signal-to-Noise Ratio - 1
  • Signal-to-Noise, written as S/N, is mainly used for Analog Systems
  • S/N is specified at theBaseband of the Information Channel

Baseband is a range of frequencies close to zero

Information is what is sent to the user and the channel over which it is sent is the Information Channel

Lecture number 1

signal to noise ratio 2
Signal-to-Noise Ratio - 2
  • What S/N value gives a good reception?
    • Telephone and TV channels require a minimum of 50 dB
  • Analog signals have “graceful degradation” characteristics

50 dB  ratio of 100,000IE:theSignal power is 100,000 > the Noise power

Lecture number 1

signal to noise ratio 3
Signal-to-Noise Ratio - 3

Analog Reception

S/NLevel

A

B

Good Marginal

Bad

100 80 60 40 20 0Percentage Time above Threshold

Lecture number 1

signal to noise ratio 4
Signal-to-Noise Ratio - 4
  • The S/N is what the user perceives, but it is usually measured at the demodulator output
  • The C/N at the demodulator input will determine the output S/N

User’s Application Device

Received signal

Output S/N

Demodulator

Lecture number 1

carrier to noise ratio 1
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

???

Lecture number 1

carrier to noise ratio 2
Carrier-to-Noise Ratio - 2

= Wideband (passband) signal with modulation

= Baseband signal with raw information

Transmitter

Receiver

The C/N at the input to the demodulator is the key design point in any communications system

RF

RF

Mixer

Mixer

IF

IF

Information to be sent

Information received

Demodulator

Modulator

Lecture number 1

carrier to noise ratio 3
Carrier-to-Noise Ratio - 3

Input C/N

Useful output?

Demodulator

C/N121086420

Conservative design Level (10 dB) with no coding

Can use these C/N levels with Coding, etc.

Lecture number 1

carrier to noise ratio 4
Carrier-to-Noise Ratio - 4
  • Useful design reference for uncoded QPSKBER = 10-6 at 10.6 dB input C/N to Demodulator

BER?

10.6 dB

BER10-310-410-510-610-710-8

BER Voice Maximum

BER Data Maximum

Goal is 10-10

0 10 20 30 C/N

Lecture number 1

ber 1
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

Lecture number 1

ber 2
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

Lecture number 1

ber 3
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

Lecture number 1

ber 5
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)
      • Interference
      • Enhanced noise input

We will look at this one

Lecture number 1

ber 6
BER - 6

Attenuation, dB

99.999% = 0.001% outage is a typical single-hop specification

201612840

 19 dB

99.99% = 0.01% outage is a typical high availability spec.

99.7% = 0.03% outage is a typical VSAT spec.

 6 dB

3 dB

100 10 1 0.1 0.01 0.001 Percentage of the Time

Lecture number 1

ber 7 performance availability
BER – 7Performance & Availability

BER

Exceeds Performance Spec.

10-10

Exceeds Availability Spec.

10-8

10-6

Does not meetPerformance orAvailability Specs.

10-4

10-2

100 10 1 0.1 0.01 0.001 Percentage of the Time

Lecture number 1

ber 8 performance availability
BER – 8Performance & Availability

BER

10-10

With Coding

10-8

10-6

Without Coding

10-4

10-2

100 10 1 0.1 0.01 0.001 Percentage of the Time

Lecture number 1