TCOM 551/ECE 463 Spring 2005 Lecture number 11 TCOM 551 DIGITAL COMMUNICATIONS SPRING 2005 IN 136...

TCOM 551/ECE 463 Spring 2005

Lecture number 1 1

TCOM 551DIGITAL

COMMUNICATIONS

SPRING 2005

IN 136 Wednesdays 4:30 – 7:10 p.m.

Dr. Jeremy Allnutt [email protected]


Lecture number 1 2

General Information - 1

• Contact Information– Room: Science & Technology II, Room 269– Telephone (703) 993-3969– Email: [email protected]– Office Manager: TBD

• Office Hours– Mondays and Tuesdays 3:00 – 6:00 p.m.

Please, by appointment only


Lecture number 1 3

General Information - 2• Course Outline

– Go to http://telecom.gmu.edu and click on course 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

Sorry: the web document is not yet up


Lecture number 1 4


• 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 5


• 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

No double jeopardy


Lecture number 1 6

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 7


• 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 8


• 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 9

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 10

TCOM 551 Course Plan

- 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

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

alternative


Lecture number 1 11

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 12

Sine Wave Review - 1

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

AB Sine (a) = A/B

Angle a

Point P

But what happens if line B rotates about Point P?


Lecture number 1 13


Ba

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

onto a screen?

Point P

Line B now describes a circle

about Point P


Lecture number 1 14


Light from the left

Screen on the right

B

End of “B” projected onto the

screena

Point P


Lecture number 1 15


End of “B” projected onto the

screen

Screen on the right

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?


Lecture number 1 16


ScreenPosition 1

ScreenPosition 2

Locus of “B” end-point

One oscillation =

One wavelength,

We have a Sine Wave!

a.k.a. SHM


Lecture number 1 17


0 90 180 270 360 Degrees

+1

-1

Remember:Sine 0 = 0; Sine 90 = 1; Sine 180 = 0; Since 270 = -1; Sine 360 = Sine 0 = 0


Lecture number 1 18

Sine and Cosine Waves - 1Sine

Wave

CosineWave

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

Sine Wave = Cosine Wave shifted by 90o


Lecture number 1 19

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 20


• 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

Where j = ( -1 )

Real Imaginary

We will only consider Real

signals


Lecture number 1 21


• 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 22


• Time varying signal

V(t) = A cos (t + )

Instantaneous value of the

signal

Phase: PM; PSK

Frequency: FM; FSK

Amplitude: AM; ASK

Vary these to Modulate

the signal

Note: = 2 f


Lecture number 1 23

Back to our Sine Wave – 1Defining the Wavelength

The wavelength is usually

defined at the “zero crossings”


Lecture number 1 24

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


Lecture number 1 25

Back to our Sine Wave – 3Two useful equations

Phase has changed by

The rate-of-change of the phase, d/dt, is the frequency, f.

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


Lecture number 1 26

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

Before we look at d/dt, lets look at rate-of-change of phase


Lecture number 1 27

d/dt Digression - 1kilometers

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

16

12

8

4

0

Person walks 16 km in 4 hours.

Velocity = (distance)/(time)Therefore, Velocity = 16/4

= 4 km/h

Velocity is really the rate-of-change of distance with time.

What if the velocity is not constant?


Lecture number 1 28


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

16

12

8

4

0

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?

Answer: you differentiate, which means you find the

slope of the line.


Lecture number 1 29


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

16

12

8

4

0

A

B

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.


Lecture number 1 30

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 31

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 32

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 Wavelength

FrequencyVelocity of light


Lecture number 1 33

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 mRemember: Make sure you are using the correct units


Lecture number 1 34

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 35

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

TxSingle Channel SCPC

Multiplexer TxMulti-channel

carrierMultiplexed

Carrier

Note: The modulator has been omitted in

these drawings


Lecture number 1 36

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 37

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 38

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 39

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 40

Logarithms – 4B


Do you detect any logic here?


Lecture number 1 41

Logarithms – 4C


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

exponent

Do you detect any logic here?


Lecture number 1 42

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 43

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 44

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 45

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 46

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? = -0.16120.69 is < 1 so the answer must be below 0


Lecture number 1 47

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


Lecture number 1 48

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 49

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 50

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 51

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


Lecture number 1 52

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 53

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 54

Logarithms - 17

• 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!

2 + 3 = 5

So: read on!


Lecture number 1 55

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 56

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 57

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 58


• 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 59


• 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 60


• 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 61

Core Concepts of Digital Communications - 1

Distance

Frequency

Source encodingSource;

Modulation

Multiplexing

RFtoIF

Amplification and transmission

Reception and amplification

Channel coding

RFtoIF

Demodulation

Channel decoding

Demultiplexing

Sink;Information user

Transmission medium

Frequency


Lecture number 1 62

Core Concepts of Digital Communications - 2

Distance

Frequency

Source encodingSource;

Modulation

Multiplexing

RFtoIF

Amplification and transmission

Reception and amplification

Channel coding

RFtoIF

Demodulation

Channel decoding

Demultiplexing

Sink;Information user

Transmission medium

Frequency

Lectures 2, 6, 7, 11, 12, &14Lectures 3, 4, & 8 Lectures

9 & 10Lecture 13Lecture 4

Lectures 3 & 5


Lecture number 1 63

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 64

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 65


• 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:the Signal power is 100,000 > the Noise power


Lecture number 1 66


Analog ReceptionS/NLevel

Good

Marginal

Bad

100 80 60 40 20 0Percentage Time above Threshold

AB


Lecture number 1 67


• 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

DemodulatorReceived

signal

User’s Application

Device

Output S/N


Lecture number 1 68

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 69


Information to be sent Modulator

Mixer Mixer

Transmitter Receiver

Information receivedDemodulator

= Wideband (passband) signal with modulation= Baseband signal with raw information

IF IF

RF RF

The C/N at the input to the

demodulator is the key design point in any

communications system


Lecture number 1 70


DemodulatorInput C/N

Useful output?

C/N121086420

Conservative design Level (10 dB) with no coding

Can use these C/N levels with

Coding, etc.


Lecture number 1 71


• Useful design reference for uncoded QPSKBER = 10-6 at 10.6 dB input C/N to Demodulator

BER10-3

10-4

10-5

10-6

10-7

10-8

0 10 20 30 C/N

10.6 dB

BER Voice Maximum

BER Data Maximum

Goal is 10-10

BER?


Lecture number 1 72

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 73

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 74

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 75

BER - 4

Fig. 8.4 in Pratt et al., Satellite Communications


Lecture number 1 76

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 77

BER - 6

100 10 1 0.1 0.01 0.001 Percentage of the Time

20

16

12

8

4

0

Attenuation, dB

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

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

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

3 dB

6 dB

19 dB


Lecture number 1 78

BER – 7Performance & Availability


10-10

10-8

10-6

10-4

10-2

BER

Exceeds Availability Spec.

Exceeds Performance Spec.

Does not meetPerformance or

Availability Specs.


Lecture number 1 79

BER – 8Performance & Availability


10-10

10-8

10-6

10-4

10-2

BER

Without Coding

With Coding

TCOM 551/ECE 463 Spring 2005 Lecture number 11 TCOM 551 DIGITAL COMMUNICATIONS SPRING 2005 IN 136...

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