Telecommunications Demystified

Telecommunications 
Demystified 
A Streamlined Course in Digital Communications
(and some Analog) for EE Students and
Practicing Engineers
by Carl Nassar 
Eagle Rock, Virginia
www.LLH-Publishing.com 
Library of Congress Cataloging-in-Publication Data
Nassar,Carl,1968-
Telecommunicationsdemystified/byCarlNassar.
p.c.m.--(Demystifyingtechnologyseries)
Includes bibliographical references and index. 
ISBN1-878707-55-8(alk.paper) 
1.Telecommunication.I.Title.II.Series.
TK5101.N2572000
621.382--dc21
00-062902
Copyright©2001byLLHTechnologyPublishing.
Allrightsreserved.Nopartofthisbookmaybereproduced,inanyformormeans
whatsoever, without written permission from the publisher. While every precaution
hasbeentakeninthepreparationofthisbook,thepublisherandauthorassumeno
responsibilityforerrorsoromissions.Neitherisanyliabilityassumedfordamages
resultingfromtheuseoftheinformationcontainedherein.
PrintedintheUnitedStatesofAmerica
10987654321
Coverdesign:JimJohnson
Developmental editing: Carol Lewis
Production: Kelly Johnson
Visit us on the web: www.LLH-Publishing.com
For revisions and updates, check www.telecommunicationsdemystified.com
1 Introducing Telecommunications ............................................ 1 
Communication Systems .............................................................................. 1 
Definition ............................................................................................................. 1 
The Parts of a Communication System .............................................................. 2 
An Example of a Communication System .......................................................... 2 
The Transmitter .................................................................................................. 3 
The Channel ....................................................................................................... 3 
The Receiver ...................................................................................................... 3 
Telecommunication Systems........................................................................ 3 
Definition ............................................................................................................. 3 
Four Examples and an Erratic History Lesson ................................................... 4 
Analog and Digital Communication Systems ................................................ 6 
Some Introductory Definitions............................................................................. 6 
Definitions ........................................................................................................... 7 
And Digital Became the Favorite ........................................................................ 8 
Making It Digital .................................................................................................. 9 
Congrats and Conclusion ............................................................................. 10 
Problems....................................................................................................... 11 
2 Telecommunication Networks.................................................. 2 
Telecommunication Network Basics ............................................................. 2 
Connecting People with Telephones .................................................................. 2 
Connecting More People, Farther Apart ............................................................. 14 
Multiplexing�An Alternative to a Lot of Wire ........................................................ 16 
First There Was FDM ......................................................................................... 16 
Along Came TDM ............................................................................................... 18 
POTS: Plain Old Telephone System ............................................................ 19 
Local Calls .......................................................................................................... 19 
Long Distance Calls............................................................................................ 20 
Connecting the Call ............................................................................................ 20 
The Signals Sent from Switching Center to Switching Center ............................ 21 
Class 5 to Class 4 ............................................................................................... 22 
Other Signals between Switching Centers ......................................................... 24 
Communication Channels............................................................................. 24 
Transmission Lines (Wires) ................................................................................ 24 
Terrestrial Microwave ......................................................................................... 26 
Satellite Connections .......................................................................................... 28 
Fiber-optic Links ................................................................................................. 29 
Data Communication Networks .................................................................... 31 
Mobile Communications ............................................................................... 33 
Local Area Networks (LANs) ........................................................................ 35 
Conclusion .................................................................................................... 37 
Problems....................................................................................................... 38 
3 A Review of Some Important Math, Stats, and Systems ....... 3 
Random Variables ........................................................................................ 3 
Definitions ........................................................................................................... 3 
The Distribution Function: One Way to Describe x ............................................. 3 
The Density Function: A Second Way to Describe x .......................................... 40 
The Mean and the Variance ............................................................................... 41 
Example 3.1........................................................................................................ 43 
Multiple Random Variables ................................................................................. 44 
Random Processes ...................................................................................... 45 
A Definition ......................................................................................................... 45 
Expressing Yourself, or a Complete Statistical Description................................ 47 
Expressing Some of Yourself, or a Partial Description ....................................... 47 
And in Telecommunications � .............................................................................. 48 
Example 3.2........................................................................................................ 48 
Signals and Systems: A Quick Peek ............................................................ 50 
A Few Signals ..................................................................................................... 50 
Another Way to Represent a Signal: The Fourier Transform ............................. 51 
Example 3.3........................................................................................................ 52 
Bandwidth ........................................................................................................... 53 
A Linear Time Invariant (LTI) System................................................................. 55 
Some Special Linear Time Invariant (LTI) Systems ........................................... 56 
Onward ......................................................................................................... 58 
Problems....................................................................................................... x 
4 Source Coding and Decoding: ................................................. 4 
Sampling....................................................................................................... 4 
Ideal Sampling .................................................................................................... 4 
The Sampling...................................................................................................... 4 
The Information in the Samples.......................................................................... 62 
Getting Back All the Information from the Samples ............................................ 64 
Some Commonly Used Words ........................................................................... 65 
Example 4.1........................................................................................................ 66 
Zero-order Hold Sampling .................................................................................. 67 
The Information in the Samples.......................................................................... 67 
Example 4.2........................................................................................................ 68 
Natural Sampling ................................................................................................ 69 
The Information in the Samples.......................................................................... 70 
Quantization.................................................................................................. 71 
Meet the Quantizer ............................................................................................. 71 
Example 4.3........................................................................................................ 73 
Who wants it? ..................................................................................................... 73 
Quantizer Terms ................................................................................................. 74 
Types of Quantizers............................................................................................ 75 
Example 4.4........................................................................................................ 76 
The Good Quantizer ........................................................................................... 77 
What Is a Good Quantizer? ................................................................................ 77 
Measures of Performance .................................................................................. 77 
A �Classic� ........................................................................................................... 78 
Creating the Good Quantizer .............................................................................. 81 
Example 4.5........................................................................................................ 86 
The Quantizer and the Telephone ...................................................................... 88 
The Idea.............................................................................................................. 88 
Source Coding: Pulse Code Modulator (PCM) ............................................. 92 
Introducing the PCM ........................................................................................... 92 
PCM Talk ............................................................................................................ 93 
The �Good� PCM ................................................................................................. 94 
Source Decoder: PCM Decoder ......................................................................... 95 
Predictive Coding.......................................................................................... 96 
The Idea Behind Predictive Coding .................................................................... 97 
Why?................................................................................................................... 97 
The Predicted Value and the Predictive Decoder ............................................... 98 
The Delta Modulator (DM) .................................................................................. 99 
How the DM creates an ...................................................................................... 99 
The Block Diagram of the DM............................................................................. 100 
The Sampler and the Quantizer in the DM ......................................................... 100 
The Signals in the DM ........................................................................................ 2 
Example 4.6........................................................................................................ 104 
Overload and Granular Noise ............................................................................. 105 
Differential PCM (DPCM).................................................................................... 107 
The Predicted Value ........................................................................................... 107 
Example 4.7........................................................................................................ 109 
The Block Diagram ............................................................................................. 109 
Congrats and Conclusion ............................................................................. 110 
Problems....................................................................................................... 111 
5 Getting It from Here to There: .................................................. 5.1 
An Introduction.............................................................................................. 5.1 
Modulators .................................................................................................... 116 
Baseband Modulators ......................................................................................... 116 
NRZ Modulators.................................................................................................. 117 
RZ Modulators .................................................................................................... 118 
Phase-encoded Modulators................................................................................ 120 
Which Modulator to Use? ................................................................................... 122 
Example 5.1........................................................................................................ 124 
Bandpass Modulators ......................................................................................... 124 
ASK..................................................................................................................... 125 
PSK..................................................................................................................... 127 
FSK ..................................................................................................................... 129 
QAM.................................................................................................................... 130 
Example 5.2........................................................................................................ 131 
Choosing a Modulation Method .......................................................................... 131 
Just-in- Time Math, or How to Make a Modulator Signal Look Funny .......... 133 
The Idea.............................................................................................................. 134 
Example 5.3........................................................................................................ 136 
Representing Modulated Signals........................................................................ 138 
BPSK .................................................................................................................. 139 
PSK..................................................................................................................... 140 
ASK..................................................................................................................... 143 
QAM.................................................................................................................... 145 
Bring it Home, Baby, or Demodulators ......................................................... 146 
What Demodulators Do ...................................................................................... 146 
The Channel and Its Noise ................................................................................. 147 
Building a Demodulator, Part I�the Receiver Front End ...................................... 148 
What it does........................................................................................................ 148 
An orthonormal basis for r(t) ............................................................................... 148 
Representing r(t) as a vector using the orthonormal basis ................................. 149 
Building the Receiver Front End ......................................................................... 151 
Example 5.4........................................................................................................ 152 
The Rest of the Demodulator, Part II�The Decision Makers ............................... 152 
What It Does ....................................................................................................... 152 
How It Works ...................................................................................................... 153 
How to Build It ..................................................................................................... 156 
The Correlator Receiver ..................................................................................... 156 
Example 5.5........................................................................................................ 157 
The Matched Filter Receiver�Version 1 .............................................................. 158 
The Matched Filter Receiver �Version 2 ............................................................. 159 
How Good Is It Anyway (Performance Measures)........................................ 161 
A Performance Measure..................................................................................... 161 
Evaluation of ....................................................................................................... 162 
for Simple Cases ................................................................................................ 162 
The BPSK Modulator Remembered ................................................................... 162 
The BPSK Demodulator: A Summary................................................................. 162 
Evaluating the P( ................................................................................................ 163 
Some Well-known P( .......................................................................................... 166 
s .......................................................................................................................... 166 
What We Just Did ......................................................................................... 166 
Problems....................................................................................................... 167 
6 Channel Coding and Decoding: ............................................... 0 
Simple Block Coding..................................................................................... 172 
The Single Parity Check Bit Coder ..................................................................... 172 
Example 6.1........................................................................................................ 174 
Some Terminology.............................................................................................. 175 
Rectangular Codes ............................................................................................. 175 
Channel Coders for Rectangular Codes ............................................................. 175 
Channel Decoders for Rectangular Codes ......................................................... 176 
Example 6.2........................................................................................................ 176 
Linear block codes ........................................................................................ 177 
Introduction ......................................................................................................... 177 
Example 6.3........................................................................................................ 178 
Understanding Why ............................................................................................ 179 
Systematic Linear Block Codes .......................................................................... 181 
Example 6.4........................................................................................................ 181 
The Decoding ..................................................................................................... 182 
Example 6.5........................................................................................................ 184 
Example 6.6........................................................................................................ 186 
Performance of the Block Coders ................................................................. 188 
Performances of Single Parity Check Bit Coders/Decoders ............................... 188 
The Performance of Rectangular Codes ............................................................ 189 
The Performance of Linear Block Codes............................................................ 189 
Example 6.7........................................................................................................ 190 
Benefits and Costs of Block Coders ............................................................. 192 
Conclusion .................................................................................................... 193 
Problems....................................................................................................... 194 
7 Channel Coding and Decoding: ............................................... 1 
Convolutional Coders ................................................................................... 1 
Our Example ....................................................................................................... 1 
Making Sure We� ve Got It ................................................................................... 199 
Polynomial Representation ................................................................................. 200 
The Trellis Diagram ............................................................................................ 201 
Example 7.1........................................................................................................ 202 
Channel Decoding ........................................................................................ 203 
Using a Trellis Diagram ...................................................................................... 204 
Example 7.2........................................................................................................ 206 
The Viterbi Algorithm .......................................................................................... 206 
Example 7.3........................................................................................................ 212 
Performance of the Convolutional Coder...................................................... 213 
Catastrophic Codes ...................................................................................... 214 
Building Your Own ........................................................................................ 216 
Problems....................................................................................................... 217 
8 Trellis-Coded Modulation (TCM) .............................................. 8 
The Idea........................................................................................................ 222 
Improving on the Idea ................................................................................... 225 
Example 8.1........................................................................................................ 229 
The Receiver End of Things ......................................................................... 230 
The Input ............................................................................................................. 231 
The TCM Decoder Front End ............................................................................. 233 
The Rest of the TCM Decoder ............................................................................ 234 
Example 8.2........................................................................................................ 236 
Searching for the Best Path................................................................................ 237 
Problems....................................................................................................... 242 
9 Channel Filtering and Equalizers............................................. i 
Modulators and Pulse Shaping ..................................................................... i 
Example 9.1........................................................................................................ 248 
The Channel That Thought It Was a Filter .................................................... 249 
Example 9.2........................................................................................................ 250 
Receivers: A First Try ................................................................................... 251 
The Proposed Receiver ...................................................................................... 251 
Making the Receiver a Good One ...................................................................... 254 
The Proposed Receiver: Problems and Usefulness ........................................... 256 
Example 9.3........................................................................................................ 257 
Optimal Receiver Front End ......................................................................... 258 
Optimal Rest-of-the-Receiver ....................................................................... 262 
The Input ............................................................................................................. 262 
A Problem with the Input, and a Solution............................................................ 264 
The Final Part of the Optimal Receiver ............................................................... 265 
Example 9.4........................................................................................................ 270 
An Issue with Using the Whitening Filter and MLSE .......................................... 271 
Linear Equalizers .......................................................................................... 271 
Zero Forcing Linear Equalizer ............................................................................ 272 
MMSE (Minimum Mean Squared Error) Equalizer ............................................. 273 
Example 9.5........................................................................................................ 273 
Other Equalizers: the FSE and the DFE ....................................................... 274 
Conclusion .................................................................................................... 275 
Problems....................................................................................................... 276 
10 Estimation and Synchronization ............................................ 10 
Introduction ................................................................................................... 10 
Estimation: Part 1 ......................................................................................... 280 
Our Goal ............................................................................................................. 280 
What We Need to Get an Estimate of a Given r ................................................. 281 
Estimating a Given r, the First Way .................................................................... 281 
Estimating a Given r, the Second Way ............................................................... 282 
Estimating a Given r, the Third Way ................................................................... 283 
Example 10.1...................................................................................................... 283 
Evaluating Channel Phase: A Practical Example ......................................... 285 
Our Example and Its Theoretically Computed Estimate ..................................... 285 
The Practical Estimator: the PLL ........................................................................ 290 
Updates to the Practical Estimator in MPSK ...................................................... 292 
Conclusion .................................................................................................... 295 
Problems....................................................................................................... 296 
11 Multiple Access Schemes....................................................... 11 
What It Is....................................................................................................... 11 
The Underlying Ideas.................................................................................... 300 
Example 11.1...................................................................................................... 302 
TDMA............................................................................................................ 303 
Example 11.2...................................................................................................... 305 
FDMA............................................................................................................ 305 
CDMA ........................................................................................................... 306 
Introduction ......................................................................................................... 306 
DS-CDMA ........................................................................................................... 310 
FH-CDMA ........................................................................................................... 312 
MC-CDMA .......................................................................................................... 313 
CIMA............................................................................................................. 315 
Conclusion .................................................................................................... 318 
Problems....................................................................................................... 319 
12 Analog Communications ........................................................ 12.1 
Modulation�An Overview ............................................................................... 12.1 
Amplitude Modulation (AM) .......................................................................... 322 
AM Modulators�in Time ....................................................................................... 323 
Example 12.1...................................................................................................... 325 
AM Modulation�in Frequency .............................................................................. 326 
Demodulation of AM Signals�Noise-Free Case .................................................. 328 
An Alternative to AM�DSB-SC ............................................................................ 330 
Example 12.2...................................................................................................... 333 
Frequency Modulation (FM) .......................................................................... 334 
The Modulator in FM........................................................................................... 335 
Example 12.3...................................................................................................... 338 
The Demodulator in FM ...................................................................................... 339 
The Superheterodyne Receiver .................................................................... 339 
Summary ...................................................................................................... 341 
Problems....................................................................................................... 342 
Annotated References and Bibliography ................................... 5. 
Index .............................................................................................. 175 
Acknowledgments
In this life of mine, I have been blessed with an abundance of won-
derful people. This book would be incomplete without at least a page to
say “thank you,” for these are people alive in me and, therefore, alive in
the pages of this book.
Dr. Reza Soleymani, your careful guidance through the turmoil that
surrounded my Ph.D. days was nothing short of a miracle. You showed
me, through your example, how to handle even the most difficult of
situations with grace and grit, both academically and in all of life.
Dr. Derek Lile, Department Head at CSU—a young faculty could
not ask for better guidance. Your thoughtfulness, caring, and gentle
support have helped nurture the best of who I am. I am grateful.
Steve Shattil, Vice President of Idris Communications, you are indeed
a genius of a man whose ideas have inspired me to walk down new roads
in the wireless world. Arnold Alagar, President of Idris, thank you for
sharing the bigger picture with me, helping guide my research out of
obscure journals and into a world full of opportunity. To both of you, I am
grateful for both our technological partnerships and our friendships.
Bala Natarajan and Zhiqiang Wu, my two long-time Ph.D. students,
your support for my research efforts, through your commitment and
dedication, has not gone unnoticed. Thank you for giving so fully of
yourselves.
Dr. Maier Blostien, who asked me to change my acknowledgments
page in my Ph.D. thesis to something less gushy, let me thank you now
for saving the day when my Ph.D. days looked numbered. I appreciate
your candor and your daring.
Carol Lewis, my publisher at LLH Technology Publishing, thank
you for believing in this project and moving it from manuscript to
“masterpiece.”
Gretchen Brooks Nassar, you hold my hand and invite me to fly off the
cliffs and into Oceans of Wonder. Your support in inviting me to pursue my
dreams is nothing short of a gift straight from the heavens. I love you.
xi 
And to the three of you who have loved me my whole life, and given
me the best of who you are, Mom (Mona), Dad (Rudy), and Christine
(sister)—your love has shaped me and has made this book a possibility.
Wow!
And to all of you I haven’t mentioned, who appeared in my life and
shared your light with me, thank you.
xii 
About the Author
Carl R. Nassar, Ph.D., is an engineering professor
at Colorado State University, teaching telecommu-
nications in his trademark entertaining style. He is
also the director of the RAWCom (Research in
Advanced Wireless Communications) Laboratory,
where he and his graduate students carry out
research to advance the art and science of wireless
telecommunications. In addition, he is the founder
of the Miracle Center, an organization fostering personal growth for
individuals and corporations.
Since Carl’s undergraduate and graduate school days at McGill
University, he has dreamed of creating a plain-English engineering text
with “personality.” This book is that dream realized.
To contact the author, please write or e-mail him at
Carl R. Nassar, Ph.D.
Department of ECE
Colorado State University
Fort Collins, CO 80523-1373
[email protected]
xiii 
Foreword
I first met the author of this book, Professor Carl Nassar, after he
presented a paper at a conference on advanced radio technology. Pro-
fessor Nassar’s presentation that day was particularly informative and
his enthusiasm for the subject matter was evident. He seemed especially
gifted in terms of his ability to explain complex concepts in a clear way
that appealed to my intuition.
Some time later, his editor asked me if I would be interested in
reviewing a few chapters of this book and preparing a short preface. I
agreed to do so because, in part, I was curious whether or not his acces-
sible presentation style carried over into his writing. I was not
disappointed.
As you will soon see as you browse through these pages, Professor
Nassar does have an uncanny ability to demystify the complexities of
telecommunications systems engineering. He does so by first providing
for an intuitive understanding of the subject at hand and then, building
on that sound foundation, delving into the associated mathematical
descriptions.
I am partial to such an approach for at least two reasons. First, it has
been my experience that engineers who combine a strong intuitive under-
standing of the technology with mathematical rigor are among the best in
the field. Second, and more specific to the topic of this book, because of
the increased importance of telecommunications to our economic and
social well-being, we need to encourage students and practicing engineers
to enter and maintain their skills in the field. Making the requisite techni-
cal knowledge accessible is an important step in that direction.
In short, this book is an important and timely contribution to the
telecommunications engineering field.
Dale N. Hatfield
Former Chief, Office of Engineering and Technology
Federal Communications Commission
xv 
What’s on the CD-ROM?
The CD-ROM accompanying this book contains a fully searchable,
electronic version (eBook) of the entire contents of this book, in Adobe
®
pdf format. In addition, it contains interactive MATLAB
®
tutorials that
demonstrate some of the concepts covered in the book. In order to run
these tutorials from the CD-ROM, you must have MATLAB installed on
your computer. MATLAB, published by The MathWorks, Inc., is a
powerful mathematics software package used almost universally by the
engineering departments of colleges and universities, and at many
companies as well. A reasonably priced student version of MATLAB is
available from www.mathworks.com. A link to their web site has been
provided on the CD-ROM.
Using the Tutorials
Each tutorial delves deeper into a particular topic dealt with in the
book, providing more visuals and interaction with the concepts pre-
sented. Note that the explanatory text box that overlays the visuals can
be dragged to the side so that you can view the graphics and other aids
before clicking “OK” to move to the next window. Each tutorial
filename reflects the chapter in the book with which it is associated. I
recommend that you read the chapter first, then run the associated
tutorial(s) to help deepen your understanding. To run a particular tuto-
rial, open MATLAB and choose Run Script from the Command Window
File menu. When prompted, locate the desired tutorial on the CD-ROM
using the Browse feature and click “OK.” The tutorials contain basic
descriptions and text to help you use them. Brief descriptions are also
given in the following pages.
MATLAB is a registered trademark of The MathWorks, Inc.
xvii 
ch2.m 
Demonstrates the creation of the DS-1 signal.
ch4_1.m 
Shows the different sampling techniques, and the effects of sampling
at above and below the Nyquist rate.
ch4_2.m 
Demonstrates quantization, and computation of the MSE.
ch4_3.m 
Explains the operation of the DM.
ch5_1.m 
Shows the workings of modulation techniques such as BPSK and
BFSK.
ch5_2.m 
Explains how three signals are represented by two orthonormal basis
functions.
ch5_3.m 
Illustrates the damaging effect of noise and the operation of decision
devices.
ch5_4.m 
Demonstrates the performance curve for BPSK signals.
ch7.m 
Shows how a convolutional coder and convolutional decoder work.
xviii 
ch8.m 
Provides an example of how TCM works at the coder and the
decoder side.
ch9_1.m 
Demonstrates how the sinc and raised cosine pulse shapes avoid ISI.
ch9_2.m 
Shows how the decision device operates in the optimal receiver.
ch11.m 
Provides colorful examples of TDMA, FDMA, MC-CDMA,
DS-CDMA, and CIMA.
ch12.m 
Illustrates the different analog modulation techniques.
Please note that the other files on the CD-ROM are subroutines that
are called by the above-named files. You won’t want to run them on
their own, but you will need them to run these tutorials.
For MATLAB product information, please contact:
The MathWorks, Inc.
3 Apple Hill Drive
Natick, MA, 01760-2098 USA
Tel: 508-647-7000
Fax: 508-647-7101
E-mail: [email protected]
Web: www.mathworks.com
xix 
1 
Chapter 
Introducing 
Telecommunications 
I
canstillrecallsittinginmyfirstclassontelecommunicationsasan
undergrad—theteachergoingoffintoaworldoftechnicaldetailandIinmychair
wondering,“Whatisthisstuffcalledcommunicationsandtelecommunications?”So,
first,somesimpledefinitionsandexamples—thebigpicture.
1.1CommunicationSystems
1.1.1Definition
Acommunicationsystemis,simply,anysysteminwhichinformationistransmitted
fromonephysicallocation—let’scallitA—toasecondphysicallocation,whichwe’ll
callB.I’veshownthisinFigure1.1.Asimpleexampleofacommunicationsystemis
onepersontalkingtoanotherpersonatlunch.Anothersimpleexampleisoneperson
talkingtoasecondpersonoverthetelephone.
Figure1.1Acommunicationsystem
2  ◆  Chapter One 
1.1.2ThePartsofaCommunicationSystem
Anycommunicationsystemismadeupofthreeparts,showninFigure1.2.Firstisthe
transmitter,thepartofthecommunicationsystemthatsitsatpointA.Itincludestwo
items:thesourceoftheinformation,andthetechnologythatsendstheinformationout
overthechannel.Nextisthechannel.Thechannelisthemedium(thestuff)thatthe
informationtravelsthroughingoingfrompointAtopointB.Anexampleofachannel
iscopperwire,ortheatmosphere.Finally,there’sthereceiver,thepartofthecommu-
nicationsystemthatsitsatpointBandgetsalltheinformationthatthetransmitter
sendsoverthechannel.
We’llspendtherestofthisbooktalkingaboutthesethreepartsandhowtheywork.
TRANSMITTER RECEIVER
CHANNEL 
A B
Figure1.2Partsofacommunicationsystem
1.1.3AnExampleofaCommunicationSystem
Now,let’srunthroughasimplebutveryimportantexampleofacommunication
system.We’llconsidertheexampleofGretchentalkingtoCarlaboutwheretogofor
lunch,asshowninFigure1.3.
Figure1.3 
GretchentalkingtoCarlatlunch 
Channel(theair)
Windpipe
Vocalcords
Introducing Telecommunications  ◆  3 
The Transmitter 
Thetransmitter,inthiscase,ismadeupofpartsofGretchen,namelyhervocalcords,
windpipe,andmouth.WhenGretchenwantstotalk,herbraintellshervocalcords
(foundinherwindpipe)tovibrateatbetween100Hzand10,000Hz,dependingonthe
soundshe’stryingtomake.(Isn’titcoolthat,everytimeyoutalk,apartofyouis
shakingatbetween100and10,000timespersecond?)OnceGretchen’svocalcords
begintovibrate,herearethethreethingsthathappennext:
(1) thevibrationsofhervocalcordscausevibrationsintheairinherwindpipe;
(2) thesevibrationsintheairmoveupherwindpipetohermouth;and
(3) asthevibratingairmovesoutthroughGretchen’smouth,theshapeofher
mouthandlips,andthepositionofhertongue,worktogethertocreatethe
intendedsound.
The Channel 
Inourexample,thechannelissimplytheairbetweenGretchenandCarl.Theshaped
vibrationsthatleaveGretchen’smouthcausevibrationsintheair,andthesevibrations
movethroughtheairfromGretchentoCarl.
The Receiver 
ThereceiverinthiscaseisCarl’seardrumandbrain.Thevibrationsintheairhit
Carl’seardrum,causingittovibrateinthesameway.Carl’sshakingeardrumsends
electricalsignalstohisbrain,whichinterpretstheshakingasspokensound.
Thehumaneardrumcanactuallypickupvibrationsbetween50Hzand16,500
Hz,allowingustohearsoundsbeyondtherangeofwhatwecanspeak,includinga
varietyofmusicalsounds.
1.2TelecommunicationSystems
1.2.1Definition
Atelecommunicationsystemistwothings:(1)acommunicationsystem—thatis,a
systeminwhichinformationistransmittedfromonephysicallocation,A,toasecond
physicallocation,B;and(2)asystemwhichallowsthisinformationtobesentbeyond
therangeofusualvocalorvisualcommunications.GretchenandCarl’slunchtimechat
wouldnotqualifyasatelecommunicationsystem,butthetelephonesystemwhich
theyusedlaterforanafternoontalkdoesqualify.
4  ◆  Chapter One 
1.2.2FourExamplesandanErraticHistoryLesson
Herearefourexamplesoftelecommunicationsystems,orderedchronologicallyto
createwhatwe’lloptimisticallycall“abriefhistoryoftelecommunications.”
Smoking Up IntheB.C.’s,smokesignalsweresentoutusingfireandsomesmoke
signalequipment(suchasablanket).Thesmoke,carriedupwardbytheair,wasseen
bypeoplefar(butnottoofar)away,whotheninterpretedthissmoketohavesome
meaning.ItissaidthatafellownamedPolybius(aGreekhistorian)cameupwitha
systemofalphabeticalsmokesignalsinthe100sB.C.,buttherearenoknownre-
cordedcodes.
Wild HorsesUntilthe1850sintheU.S.,thefastestwaytosendamessagefromone’s
hometosomeoneelse’shomewasbyPonyExpress.Here,youwrotewhatyouwanted
tosay(thetransmitter),gavethewritingtoaPonyExpressman,whothenhoppedon
hishorseandrodetothedestination(thechannel),wherethemessagewouldberead
bytheintendedperson(thereceiver).
Telegraph In1844,afellownamedSamuelMorsebuiltadevicehecalledthetele-
graph,thebeginningoftheendofthePonyExpress.Thetransmitterconsistedofa
personandasendingkey,whichwhenpressedbytheperson,createdaflowofelec-
tricity.Thiskeyhadthreestates:“Off”whichmeantthekeywasnotpressed;“Dot,”
whichmeantthekeywaspressedforashorttimeandthenreleased;and“Dash,”
whichmeantthekeywaspressedforalongertimeandthenreleased.Eachletterof
thealphabetwasrepresentedbyaparticularsequenceofdotsanddashes.Tokeepthe
timetosendamessageshort,themostcommonlyusedlettersinthealphabetwere
representedbythefewestpossibledotsordashes;forexample,thecommonlyused“t”
wasrepresentedbyasingledash,andthemuch-loved“e”wasrepresentedbyasingle
dot.Thissystemofrepresentinglettersisthewell-knownMorsecode.Thechannel 
wasanironwire.Theelectricitycreatedbythepersonandthesendingkey(the
transmitter)wassentalongthiswiretothereceiver,whichconsistedofanaudio-
speakerandaperson.Whentheelectricityenteredtheaudio-speakerfromtheiron
wire,itmadeabeepingsound.A“Dot”soundedlikeashortbeep,anda“Dash”
soundedlikealongerbeep.Theperson,uponhearingthesebeeps,wouldthendecode
thelettersthathadbeensent.Theoverallsystemcouldsendabouttwolettersa
second,or120lettersaminute.Thefirstwordssentoverthetelegraph,byinventor
Morsehimself,were“WhathasGodwrought!”(IhavesincewonderedwhatMorse,
whobasicallyinventedasimpledot-dashsendingsystem,wouldhavesaidabout,oh,
say,anuclearbomb.)
The Telephone Thetelephonewasinventedin1876byAlexanderGrahamBell,
whosefirstwordsonthephonewere,“Mr.Watson,comeatonce,Ineedyou.”Alex
hadjustspilledbatteryaciddownhispantsand,asyoucanimagine,wasinquite
urgentneedofhisassistant’shelp.Figure1.4showsanillustrationoftwopeople,who
Introducing Telecommunications  ◆  5 
we’llcallCarlandMonica,usingthetelephone.Whatfollowsisawordydescriptionof
howthetelephoneworks.RefertoFigure1.4tohelpyouwiththeterms.
ThetransmitterconsistsofMonica(whoistalking)andthetransmitting(bottom)
endofthetelephone.Monicaspeaks,andhervocalcordsvibrate.Thiscausesvibra-
tionsintheair,whichtravelthroughandouthermouth,andthentravelto the
bottomendofthetelephone.Insidethebottomendofthetelephoneisadiaphragm.
Whenthevibrationsoftheairarriveatthisdiaphragm,it,likeaneardrum,beginsto
vibrate.Directlybehindthediaphragmareabunchofcarbongranules.Thesegran-
ulesarepartofanelectricalcircuit,whichconsistsofa4-Vsource,copperwire,and
thecarbongranules.Thecarbongranulesactasaresistor(withvariableresistance)in
thecircuit.Whenthediaphragmispushedbackbythevibratingair,itcausesthe
carbongranules(rightbehindit)tomushtogether.Inthiscase,thegranules
actlikealow-resistanceresistorinthecircuit.Hence,thecurrentflowingthoughthe
electriccircuitishigh(usingthewell-knownV R ⋅ I rule).Whenthediaphragmis
poppedoutbythevibratingair,itcausesthecarbongranules(rightbehindit)to
separateout.Inthiscase,thosecarbongranulesareactinglikeahigh-resistance
resistorintheelectricalcircuit.Hence,thecurrentflowingthoughthecircuitislow.
Overall,vibrationsinthediaphragm(its“pushingback”and“poppingout”)causethe
samevibrations(frequencies)toappearinthecurrentoftheelectricalcircuit(via
thosecarbongranules).
Thechannelisacopperwire.Thevibratingcurrentgeneratedbythetransmitter
iscarriedalongthiswiretothereceiver.
Windpipe
Carl
Channel(copperwire)
Vocal
cords
electromagnet
eardrum
4vpowersupply
carbongranules
diaphragm
Monica
TRANSMITTER RECEIVER
Figure1.4
MonicaandCarltalkingonatelephone
6  ◆  Chapter One 
Thereceiverconsistsoftwoparts:thereceiving(top)partofthetelephone,and
Carl’sear.Thecurrent,sentalongthecopperwire,arrivesatthetopendofthetele-
phone.Insidethistopendisadevicecalledanelectromagnetandrightnexttothatis
adiaphragm.Thecurrent,containingallofMonica’stalkingfrequencies,entersinto
theelectromagnet.Thiselectromagnetcausesthediaphragmtovibratewithallof
Monica’stalkingfrequencies.Thevibratingdiaphragmcausesvibrationsintheair,and
thesevibrationstraveltoCarl’sear.Hiseardrumvibrates,andthesevibrationscause
electricalsignalstobesenttohisbrain,whichinterpretsthisasMonica’ssound.
1.3AnalogandDigitalCommunicationSystems
Thelastpartofthischapterisdedicatedtoexplainingwhatismeantbyanalog and
digitalcommunicationsystems,andthenexplainingwhydigitalcommunication
systemsarethewayofthefuture.
1.3.1SomeIntroductoryDefinitions
Ananalog signalisasignalthatcantakeonanyamplitudeandiswell-definedatevery
time.Figure1.5(a)showsanexampleofthis.Adiscrete-time signalisasignalthatcan
takeonanyamplitudebutisdefinedonlyatasetofdiscretetimes.Figure1.5(b)
showsanexample.Finally,adigital signalisasignalwhoseamplitudecantakeononly
afinitesetofvalues,normallytwo,andisdefinedonlyatadiscretesetoftimes.To
helpclarify,anexampleisshowninFigure1.5(c).
x(t) x(t) x(t)
t t
0
1
t
T 2T 3T 4T ... T 2T 3T 4T ...
(a) (b) (c)
Figure1.5(a)Ananalogsignal;(b)adiscretetimesignal;and(c)adigitalsignal
Introducing Telecommunications  ◆  7 
1.3.2Definitions
Ananalog communication systemisacommunicationsystemwheretheinformation
signalsentfrompointAtopointBcanonlybedescribedasananalogsignal.An
exampleofthisisMonicaspeakingtoCarloverthetelephone,asdescribedinSection
1.2.2.
Adigital communication systemisacommunicationsystemwheretheinformation
signalsentfromAtoBcanbefullydescribedasadigitalsignal.Forexample,con-
siderFigure1.6.Here,dataissentfromonecomputertoanotheroverawire.The
computeratpointAissending0sor1stothecomputeratpointB;a0isbeingrepre-
sentedby–5VforadurationoftimeTanda1isbeingrepresentedbya+5Vforthe
samedurationT.AsIshowinthatfigure,thatsentsignalcanbefullydescribedusing
adigitalsignal.
0
+5v
-5v
s(t)
t
B A
Signalsentis:
0
1
t
Canberepresentedby:
1 0 1 0
Figure1.6Acomputersendinginformationtoanothercomputer
8  ◆  Chapter One 
1.3.3AndDigitalBecametheFavorite
Digitalcommunicationsystemsarebecoming,andinmanywayshavealreadybe-
come,thecommunicationsystemofchoiceamongustelecommunicationfolks.
Certainly,oneofthereasonsforthisistherapidavailabilityandlowcostofdigital
components.Butthisreasonisfarfromthefullstory.Toexplainthefullbenefitsofa
digitalcommunicationsystem,we’lluseFigures1.7and1.8tohelp.
Let’sfirstconsiderananalogcommunicationsystem,usingFigure1.7.Let’s
pretendthetransmittersendsouttheanalogsignalofFigure1.7(a)frompointAto
pointB.Thissignaltravelsacrossthechannel,whichaddssomenoise(anunwanted
signal).ThesignalthatarrivesatthereceivernowlookslikeFigure1.7(b).Let’snow
consideradigitalcommunicationsystemwiththehelpofFigure1.8.Let’simaginethat
thetransmittersendsoutthesignalofFigure1.8(a).Thissignaltravelsacrossthe
channel,whichaddsanoise.ThesignalthatarrivesatthereceiverisfoundinFigure
1.8(b).
s(t) r(t)
Noise
t t
(a)
(b)
Figure1.7(a)Transmittedanalogsignal;(b)Receivedanalogsignal
0
+5v
-5v
s(t)
t
0
+5v
-5v
s(t)
t
1 1 0
(a)
(b)
Noise
Figure1.8(a)Transmitteddigitalsignal;(b)Receiveddigitalsignal
Introducing Telecommunications  ◆  9 
Here’sthekeyidea.Inthedigitalcommunicationsystem,evenafternoiseis
added,a1(sentas+5V)stilllookslikea1(+5V),anda0(–5V)stilllookslikea0(–5
V).So,thereceivercandeterminethattheinformationtransmittedwasa101.Sinceit
candecidethis,it’sasifthechanneladdednonoise.Intheanalogcommunication
system,thereceiverisstuckwiththenoisysignalandthereisnowayitcanrecover
exactlywhatwassent.(Ifyoucanthinkofaway,pleasedoletmeknow.)So,ina
digitalcommunicationsystem,theeffectsofchannelnoisecanbemuch,muchless
thaninananalogcommunicationsystem.
1.3.4MakingItDigital
Anumberofnaturallyoccurringsignals,suchasMonica’sspeechsignal,areanalog
signals.Wewanttosendthesesignalsfromonepoint,A,toanotherpoint,B.Because
digitalcommunicationsystemsaresomuchbetterthananalogones,wewanttousea
digitalsystem.Todothis,theanalogsignalmustbeturnedintoadigitalsignal.The
deviceswhichturnanalogsignalsintodigitalonesarecalledsource coders,andwe’ll
spendallofChapter4exploringthem.Inthissection,we’lljusttakeabriefpeekata
simplesourcecoder,onethatwillturnMonica’sspeechsignal(andanyoneelse’sfor
thatmatter)intoadigitalsignal.ThesourcecoderisshowninFigure1.9.
ItallbeginswhenMonicatalksintothetelephone,andhervibrationsareturned
intoanelectricalsignalbythebottomendofthetelephonetalkedaboutearlier.This
electricalsignalistheinputsignalinFigure1.9.Wewillassume,asthetelephone
companydoes,thatallofMonica’sspeechliesinthefrequencyrangeof100Hzto
4000Hz.
TheelectricalversionofMonica’sspeechsignalentersadevicecalledasampler.
Thesampleris,inessence,aswitchwhichclosesforabriefperiodoftimeandthen
opens,closingandopeningmanytimesasecond.Whentheswitchisclosed,the
electricalspeechsignalpassesthrough;whentheswitchisopen,nothinggets
through.Hence,theoutputofthesamplerconsistsofsamples(pieces)oftheelectrical
input.
Quantizer
Symbol-to-bit
Mapper
Sampler
Monica'sspeechsignal
Figure1.9Asimplesourcecoder
10  ◆  Chapter One 
Assomeofyoumayknow(andifyoudon’t,we’llreviewitinChapter4,sohave
noworries),wewanttheswitchtoopenandcloseatarateofatleasttwotimesthe
maximumfrequencyoftheinputsignal.Inthecaseathand,thismeansthatwewant
theswitchtoopenandclose2 × 4000=8000timesasecond;infancywords,wewanta
samplingrateof8000Hz.
Aftertheswitch,thesignalgoesthroughadevicecalledaquantizer.Thequan-
tizerdoesasimplething.Itmakestheamplitudeofeachsamplegotooneof256
possiblelevels.Forexample,thequantizermayberoundingeachsampleofthe
incomingsignaltothenearestvalueintheset{0,0.01,0.02,...,2.54,2.55}.
Now,here’ssomethinginteresting.Thereisalossofinformationatthequantizer.
Forexample,inroundingasampleofamplitude2.123totheamplitude2.12,informa-
tionislost.Thatinformationisgoneforever.Whywouldweputinadevicethat
intentionallylostinformation?Easy.Becausethat’stheonlywayweknowtoturnan
analogsignalintoadigitalone(andhencegainthebenefitsofadigitalcommunication
system).Thegoodnewshereisengineers(likeyouandme)buildthequantizer,and
wecanbuilditinawaythatminimizesthelossintroducedbythequantizer.(We’lltalk
atlengthaboutthatinChapter4.)
Afterthequantizer,thesignalentersintoasymbol-to-bitmapper.Thisdevice
mapseachsample,whoseamplitudetakesononeof256levels,intoasequenceof8
bits.Forexample,0.0mayberepresentedby00000000,and2.55by11111111.We’ve
nowcreatedadigitalsignalfromourstartinganalogsignal.
1.4CongratsandConclusion
Congratulations—youmadeitthroughthefirstchapter.Justtorecap(andI’llbebrief),
inthischapterwedefinedthewordscommunicationandtelecommunication system.
Next,Ipresentedawholegangofexamples,togiveyouafeelforafewkeycommuni-
cationandtelecommunicationsystems.Finally,wetalkedaboutanaloganddigital
communications,discoveringthatmosttelecommunicationengineersdreamindigital.
MeetyouinChapter2!
Introducing Telecommunications  ◆  11 
Problems
1. Brieflydescribethefollowing:
(a) telecommunicationsystem
(b)communicationsystem
(c) thedifferencebetweenacommunicationsystemanda
telecommunicationsystem.
(d)digitalcommunications
(e) analogcommunications
(f) themainreasonwhydigitalcommunicationsispreferred
(toanalogcommunications).
2. Describethefunctionofthefollowing:
(a) sourcecoder
(b)quantizer
(c) sampler
(d)symbol-to-bitmapper.
2 
Chapter 
Telecommunication 
Networks 
F
irst,andalwaysfirst,adefinition.Atelecommunication networkisatelecommuni-
cationsystemthatallowsmanyuserstoshareinformation.
2.1TelecommunicationNetworkBasics
2.1.1ConnectingPeoplewithTelephones
Let’ssayIhavesixpeoplewithtelephones.Iwanttoconnectthemtogethersothey
canspeaktooneanother.Oneeasywaytoconnecteveryoneistoputcopperwires
everywhere.Bythat,Imeanuseacopperwiretoconnectperson1toperson2,awire
toconnectperson1toperson3,...,awiretoconnectperson1toperson6,...,anda
wiretoconnectperson5toperson6.I’veshownthissolutioninFigure2.1.But,ugh,
allthesewires!Ingeneral,IneedN(N–1)/2wires,whereNisthenumberofpeople.
SowithonlysixpeopleIneed15wires,with100
peopleIneed49,950wires,andwithamillionpeople
1
2 3
4
5 6
Copperwire
Ineed499,999,500,000wires.Toomanywires.
Let’sconsideranotherwaytoconnectusers:
putaswitchingcenterbetweenthepeople,as
showninFigure2.2.Theearlyswitchboards
workedlikethis:Gretchenpicksupherphone
tocallCarl.Aconnectionisimmediatelymade
toMavis,asweetelderlyoperatorseatedatthe
switchingcenter.MavisasksGretchenwho
shewantstotalkto,andGretchensays“Carl.”
Mavisthenphysicallymoveswiresattheswitch-
ingcenterinsuchawaythatthewirefrom
Gretchen’sphoneisdirectlyconnectedtoCarl’s
wire,andGretchenandCarlarenowreadytobegin
theirconversation.
Figure2.1
Asinglewirebetweeneachphone
1
14  ◆  Chapter Two 
2 3
4
5 6
SwitchingCenter
Themanualswitchingcenter,oper-
atedbypeoplelikeMavis,wastheonlyway
togountilalittleafter1889whenAlmonB.
Strowger,amorticianbytrade,invented
thefirstautomaticswitchingcenter.It
seemsthatAlmonsuddenlyfoundthathe
wasnolongergettinganybusiness.Sur-
prised,heinvestigatedanddiscoveredthat
thenewtelephoneoperatorwasalso
marriedtothecompetition...andshewas
switchingallfuneralhomecallstoher
husband.Determinedtokeephismortician
businessalive(nopunintended),Almon
createdthefirstautomaticswitchingcenter.
Figure2.2
Theseswitchingcentersdonotrequire
Phonesconnectedbyaswitchingcenter
anyone(andhencenocompetitor’swife)to
transfercalls.Enteringa7-digitphonenumberautomaticallysetsuptheconnectionat
theswitchingcenter.Today,allswitchingcentersareautomatic.
Justbriefly,howmanywiresareneededbyanetworkusingaswitchingcenter?
OnlyN,whereNisthenumberofusers.That’sfarfewerthanthemany-wiresystem
introducedfirst.
2.1.2ConnectingMorePeople,FartherApart
Let’stakethisswitchingcenterideaabitfurther.ConsiderFigure2.3.Abunchof
peopleareconnectedtogetherinFortCollins,Colorado,byaswitchingcenterin
downtownFortCollins.Then,innearbyBoulder,anothergroupofpeoplearecon-
nectedtogetherbyasecondswitchingcenter.HowdoessomeoneinBouldertalktoa
friendinFortCollins?Oneeasywayistosimplyconnecttheswitchingcenters,as
showninFigure2.3.IfweputseveralwiresbetweentheFortCollinsandBoulder
switchingcenters,thenseveralpeopleinFortCollinscantalktopeopleinBoulderat
thesametime.
1 1
2 2 3 3
4 4
5 5 6 6
SwitchingCenter SwitchingCenter
Figure2.3
Connections between
FortCollins,CO switchingcenters Boulder,CO
Telecommunication Networks  ◆  15 
Considernowanumberofothernearbycities,sayLongmontandDenver.The
folksinthesetownswanttotalktotheirfriendsinBoulderandFortCollins.Wecould
connectalltheswitchingcenterstogether,asshowninFigure2.4.Alternatively,we
couldhavea“super”switchingcenter,whichwouldbeaswitchingcenterforthe
switchingcenters,asshowninFigure2.5.
Switching
Center
Switching
Center
Switching
Center
Switching
Center
Longmont
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
5
5
5
5
6
6
6
6 Denver
Boulder FortCollins
Figure2.4 Connectingalltheswitchingcenterstogether
Switching
Center
Switching
Center
Switching
Center
Switching
Center
Longmont
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
5
5
5
5
6
6
6
6
Switching
Center
Denver
Boulder FortCollins
"Super"
Figure2.5 Connectingswitchingcentersusinga“super”switchingcenter
16  ◆  Chapter Two 
2.1.3Multiplexing—AnAlternativetoaLotofWire
Let’sgobacktotheFortCollinsandBoulderpeopleinFigure2.3.Let’ssaywe’ve
connectedtheirswitchingcenterssotheycantalktoeachother.Asmorepeoplein
FortCollinswanttotalktomorepeopleinBoulder,moreandmorewiresneedtobe
addedbetweentheirswitchingcenters.Itcouldgettothepointwheretherearefartoo
manywiresrunningbetweentheswitchingcenters.TheskiesaroundFortCollins
couldgrowdarkunderthecoverofallthesewires.Thisprobablywouldn’tbetrueofa
smallertownlikeFortCollins,butitwastrueofbigcitieslikeNewYork.
Finally,someonesaid,“Somethingmustbedone!’’andmultiplexingwasinvented.
Multiplexingreferstoanyschemethatallowsmanypeople’scallstoshareasingle
wire.(We’lltalkmoreaboutthisinChapter11,butabriefintroductionnowisuseful.)
First There Was FDM 
FDM,shortforfrequencydivisionmultiplexing,wasthefirstschemecreatedtoallow
people’scallstoshareawire.Let’ssayCarl,Gretchen,andMonicaallwanttomakea
callfromFortCollinstoBoulder.Weonlywanttouseonewiretoconnectcallsbe-
tweenthetwotowns.ThisisshowninFigure2.6.Carl’sspeech,turnedintoacurrent
onawire,containsthefrequencies100to4000Hz.Hisspeechisleftasis.Gretchen’s
speech,turnedintoanelectricalsignalonawire,alsocontainsthefrequencies100to
4000Hz.Asimpledevicecalledamixer(operatingat8000Hz)isappliedtoherspeech
signal.Thisdevicemovesthefrequencycontentofherspeechsignal,andthefrequen-
ciesfoundat100Hzto4,000Hzaremovedtothefrequencies8,100Hzto12,000Hz,
asshowninFigure2.7.
Switching
Center
Carl,Gretchen
andMonica'svoices
Switching
Center
FortCollins
allononewire
Boulder
Figure2.6
People’sspeechononewire
Telecommunication Networks  ◆  17 
BecauseCarlandGretchen’scallsarenowmadeupofdifferentfrequencycom-
ponents,theycanbesentonasinglewirewithoutinterferingwithoneanother.This
tooisshowninFigure2.7.WenowwanttoaddMonica’sspeechsignal,sinceshetoo
ismakingacallfromFortCollinstoBoulder.Amixer,thisoneoperatingat16,000Hz,
isappliedtoherspeechsignal.ThismovesthefrequencycontentofMonica’sspeech
to16,100Hzto20,000Hz,asshowninFigure2.7.BecauseMonica’sspeechsignalis
nowatdifferentfrequenciesthanCarlandGretchen’sspeechsignals,hersignalcan
beaddedontothesamewireasCarlandGretchen’s,withoutinterfering.(Again,take
apeekatFigure2.7.)
OverinBoulder,we’vegotawirewithCarl,Gretchen,andMonica’sspeechonit,
andweneedtoseparatethisintothreesignals.First,togetCarl’ssignal,weusea
devicecalledalow-passfilter(LPF).Thefilterweuseonlyallowsthefrequenciesin0
to4000Hztopassthrough;allotherfrequencycomponentsareremoved.So,inour
example,thisfilterpassesCarl’sspeech,butstopsGretchen’sandMonica’sspeech
cold.ThisisshowninFigure2.8.
Carl'sspeechsignal
100-4000Hz
100-4000Hz
100-4000Hz
8000Hz
16,000Hz
+ +
Gretchen'sspeechsignal
Monica'sspeechsignal
onewire
Mixer
Mixer
Figure2.7
PuttingthreesignalsononewireusingFDM
18  ◆  Chapter Two 
Next,wewanttorecoverGretchen’sspeechsignal.Thisisatwo-partjob.First,a
bandpassfilter(BPF)isapplied,withstartfrequency8,000Hzandstopfrequency
12,000Hz.Thisfilterallowsonlythefrequenciesbetween8,000Hzand12,000Hzto
passthrough,cuttingouteveryotherfrequency.Inthecaseathand,thismeansthat
onlyGretchen’sspeechsignalgetsthroughthefilter.ThisisshowninFigure2.8.We
stillhaveonetaskleft.Gretchen’sspeechsignalhasbeenmovedfrom100–4,000Hzto
8,100–12,000Hz.Wewanttobringitbacktotheoriginal100–4000Hz.Thisisdoneby
applyingamixer(operatingat8,000Hz),whichreturnsGretchen’svoicesignaltoits
originalfrequencycomponents.
Monica’ssignalisrecoveredonasinglewireinmuchthesamewayas
Gretchen’s,and,ratherthanusemanywords,I’llsimplyreferyoutoFigure2.8.
Figure2.8 Gettingthreespeech
signalsbackfromonewireinFDM
LPF
0-4000Hz
BPF
8000-12000Hz
BPF
16000-20000Hz
8000Hz
16000Hz
Carl's
speechsignal
speechsignal
speechsignal
Carl,Gretchen
andMonica'sspeech
ononewire
Mixer
Mixer
Gretchen's
Monica's
Along Came TDM 
TDM,shortfortime-divisionmultiplexing,isthesecondcommonlyusedtechniqueto
letseveralpeople’sspeechshareasinglewire.TDMworkslikethis.Let’ssaywe’ve
againgotCarl,Gretchen,andMonica,whoallwanttomaketheircallsfromFort
CollinstoBoulder.Carl,Gretchen,andMonica’sspeechsoundsarefirstturnedintoan
electricalsignalonawirebytheirphones,asexplainedinChapter1.Then,their
electricalspeechsignalsareturnedintodigitalsignals,againasexplainedinChapter
1.Thedigitized,electricizedversionsofthespeechsignalaretheincomingsignals
thatwillshareawire.Figure2.9showstheseincomingsignals.
Thesesignals,comingalongthewire,thenmeet“thebigswitch,”asshownin
Figure2.9.Thebigswitchmakescontactwitheachofthethreeincomingsignals,
touchingeachsignalforT/3secondsineveryT-secondinterval.Theoutputofthis
switch,againshowninFigure2.9,consistsofonepieceofCarl’sspeech,thenone
pieceofGretchen’sspeech,thenonepieceofMonica’sspeechineveryT-second
interval.Inthisway,apartofeverybody’sspeechsamplegetsontoonewire.These
speechsamplesarenowsharingtimeonthewire,andhencethenametime-division
multiplexing.
Telecommunication Networks  ◆  19 
T
t
t
T
Carl'sdigitalspeech
...
...
digitalspeech
Gretchen'sdigitalspeech
Carl,Gretchen&Monica's
"TheBigSwitch"
Gretchen
t
...
Carl
Monica
...
t
T/3
Monica'sdigitalspeech 
T 
T
Figure2.9 HowthreesignalsshareonewireinTDM
2.2 POTS:PlainOldTelephoneSystem
Enoughofthebasics.Letmenowintroduceyoutoatelecommunicationnetwork
currentlyinuse.Infact,it’sthemostfrequentlyusedtelecommunicationnetworkin
theworld.It’scalledPOTS,shortforPlainOldTelephoneSystem.We’llbeconsider-
ingthephoneconnectionsexclusivelyinCanadaandtheUnitedStates,butkeepin
mindthatsimilarsystemsexistworldwide.
2.2.1LocalCalls
Let’ssayGretchen,athomeinFortCollins,decidestocall
Carl,whoishardatworkatCSUwritingthisbook.Here’s
howthecallgetsfromGretchentoCarl.First,
Gretchen’sphoneturnshersoundsintoananalog
Class5
SwitchingCenter
(endoffice)
Gretchenathome
electricalsignal,asexplainedinSection1.2.2.
Thisanalogelectricalsignalissentalonga
copperwire(calledatwisted-paircable)tothe
switchingcentercalledtheClass 5switching
center,orend of fice(Figure2.10).
Carlattheoffice
Figure2.10Connectingalocalcall:thelocalloop
20  ◆  Chapter Two 
TheswitchingcenterknowsGretchen’scallisalocalcalltoCarl’soffice(based
ona7-digitnumbersheinitiallydialed),anditsendsGretchen’sspeechsignaldown
thecopperwirethatconnectstoCarl’soffice.ThissignalthenentersCarl’sphone,
whichturnstheelectricalsignalbackintoGretchen’sspeechsounds,andtherewe
haveit.Thispartofthephonesystemiscalledthelocal loop.Thereareabout20,000
endofficesinCanadaandtheUnitedStates.
2.2.2 LongDistanceCalls
Connecting the Call 
Let’ssaythatCarl’smom,Mona,wholivesinthecoldofMontreal,Canada,wantsto
callCarlinColoradoandseeifhe’seatingwell(yes,Mom,I’meatingwell).How
wouldthetelephonesystemconnectthiscall?We’lluseFigure2.11asourguide.First,
Mona’sAre you eating well?soundsareturnedintoananalogelectricalsignalbyher
telephone.Thiselectricalsignalissentalongcopperwiretotheendoffice(orclass5
switchingcenter).Theendoffice,realizingthatthisisn’talocalcall,doestwothings:
(1)itmixesMona’ssignalwithotherpeople’svoicesignals(thatarealsonotlocal
calls),usingmultiplexing;then,(2)ittakesthemixofMona’sspeechsignalandthe
otherpeople’sspeechsignalsandsendsthismixtoabiggerswitchingcenter,calleda
Class 4 switching center,ortoll center,showninFigure2.11.Thetollcenteriscon-
nectedtothreesetsofthings.
Class5 Class4
Class4
Class4
Class4
Class4
Class4
Class3
Center
Class3
Center
Class3
Center
Class3
Center
Class2
Class2
Class2
Class2
Class1
Class1
Class5
Class5
Class5
Class5
Class5
A
B
C
D
E
F
AA
BB
Multiplexingof
andotherspeech
Along-distancecallin
EndOffice TollCenter
TollCenter
TollCenter
TollCenter
TollCenter
TollCenter
Primary
Primary
Primary
Primary
EndOffice
EndOffice
EndOffice
EndOffice
EndOffice
Mona'sspeech
Figure2.11
POTS
Telecommunication Networks  ◆  21 
First,it’sconnectedtoanumberofendoffices(A,B,C,andDinFigure2.11),
eachendofficeliketheonethatMona’scallcamefrom.IfMona’scallwasintended
foraphonethatwasconnectedtoendofficeCforexample,thenMona’sspeechwould
besenttoendofficeCandfromtheretotheintendedphone.
ThesecondthingtheClass4switchingcenter(tollcenter)isconnectedtois
otherClass4switchingcenters(BBinFigure2.11).Ifthecallisintendedforaphone
connectedtoendofficeE,thenitwouldlikelybesenttotollcenterBB,thentoend
officeE,andthentotheintendedtelephone.
ThethirdthingaClass4switchingcenterisconnectedtoisanevenbigger
switchingcenter,whichiscalled(nobigsurprise)aClass 3 switching center(also
calledaprimary center).IfMona’scallisgoingtoaphonenotconnectedtoanyofthe
endofficesavailablefromtollcentersAAorBB,thenMona’sspeechmovesalongto
theclass3switchingcenter.
ThegoodnewsisthattheClass3switchingcenterworksinprettymuchthe
samewayastheClass4switchingcenter.Basically,fromtheClass3switchingcenter,
Mona’scallcangoto:(1)anyClass3,4,or5switchingcenterthatisconnectedtothe
Class3switchingcenterholdingMona’sspeech—itwillgothatrouteiftheintended
phoneisconnectedtooneoftheseClass3,4,or5switchingcenters;otherwise,(2)
thecallwillbeswitchedtoanevenbiggerswitchingcenter,calledtheClass 2 switching 
center(andalsocalledthesectional center).
Let’ssayMona’scallheadsouttotheClass2switchingcenter.Fromhereitgets
toCarlinoneoftwoways:(1)ifCarl’sphoneisconnectedtoaClass2,3,4,or5
switchingcenterthatisdirectlyconnectedtotheClass2switchingcentercontaining
Mona’svoice,thenMona’svoicegetssenttothatswitchingcenter;otherwise,(2)
Mona’scallwillgotothelast,biggestswitchingcenter,theClass 1 switching center(or
regional center).TheClass1switchingcenterwillthensendthecalltoeitheraClass
1,2,3,4,or5switchingcenterthatit’sconnectedto,dependingonwhichcenteris
mostdirectlyconnectedtoCarl’sphone.
Andthat,myfriends,ishowMona’sconcernaboutCarl’seatinghabitsgetsfrom
MonainMontrealtoCarlinColorado.It’sa5-levelhierarchyofswitchingcenters.
Therearesome13,000tollcenters,265primarycenters,75sectionalcenters,and12
regionalcenters.
2.2.3 TheSignalsSentfromSwitchingCentertoSwitchingCenter
WenowunderstandthatPOTSismadeupoffivestagesofswitchingcenters.We
understandhowaperson’sspeechgetsfromoneplacetoanotherusingthephone
system.Beforewemoveon,Iwanttodiscusswhatsignalsaresentfromoneswitch-
ingcentertoanother.
22  ◆  Chapter Two 
Atelephonecallstartsoutwiththephoneinyourhand.ThatcallgoestoaClass
5switchingcenter.Ifthecallislocal,itgoesfromthatClass5switchingcenterrightto
thepersonyou’retalkingto.Ifthecallisnotlocal,theClass5switchingcenterputs
thecalltogetherwithotherlong-distancecalls,andsendsthemtogethertoaClass4
switchingcenter.Let’slookatthesignalcreatedataClass5switchingcenterthatis
senttoaClass4switchingcenter.
Class 5 to Class 4 
TheClass5switchingcentergetsthecallyouplace.Italsogetsalotofothercalls
fromyourneighborhood.WhentheClass5switchingcenterrealizestherearea
bunchofcallsthatarenotinitsarea(andinsteadarelong-distancecalls),itputsthese
signalstogetherandsendsthemout.Specifically,itputsthesignalstogetherasshown
inFigure2.12:
Class5SwitchingCenter
A A
Yourcall 
#1 
t
T=1/8000=0.125ms
8000samples/sec
The
Big
Switch
#2
Symbol
Add
to
1
*
Quantizer
bit
bit Toclass4
mapper
#3
Forceseachamplitude
tothenearestone
outof256permitted
outputamplitudes
. . 
. . 
. . 
#24
*
A
Piecefrom
line#1
t
...

line#2
line#24
Piecefrom
Piecefrom
T/24
T=0.125ms
Figure2.12ThesignalcreatedataClass5switchingcenter(headedtoClass4)
Telecommunication Networks  ◆  23 
1.First,atthelinemarked#1,isyourcall.Butthatisonlyonethingcominginto
theClass5switchingcenter.Thecentertakesyourcallandatthesametime
takes23others,foratotalof24calls.ThosecallsincomingtotheClass5switch-
ingcenterarethelinesmarked#1to#24.
2.Theswitchingcenterputsthese24callsonasinglelineusingTDM.Thatidea
isexplainedinSection2.1.3,butherearesomemoredetailsexplainingexactly
whattheClass5switchingcenterdoes.
2a.Eachvoicecallissampled.Specifically,samplesofeachvoicecallare
takenatarateof8000samples/second(i.e.,at8000Hz).Thatmeansthat
eachsampletakenfromavoicecalllastsatotaltimeofT=1/8000=0.125ms.
2b.Eachsampledvoicesignalmeets“thebigswitch.”Thebigswitchmakes
contactwitheachdigitalvoicesignalbrieflyonceevery0.125ms.Asaresult,
thesignalthatendsuponthewirefollowing“thebigswitch”isacollectionof
all24voicesignals.ThisisshowninFigure2.12betterthanmywordscan
explain.
2c.Onthewirefollowingthebigswitch,wehaveall24voicesamples
smooshedtogetherinthetimeT=0.125ms.Thenumberofsamplesineach
secondisnow24×8,000samples/second=192,000samples/second.
3.The192,000samples/secondonourwirenowenterthroughadevicecalleda
quantizer.Thequantizersimplychangestheamplitudeoftheincomingsamples.
Eachincomingsample,whichhassomeamplitudeA,ismappedtoanoutgoing
samplewhoseamplitudeisoneof256possiblevalues.
4.Eachsample,withanamplitudethatisnowoneof256levels,canbefully
representedbyasetof8bits.(Thisisbecausewith8bitswecanrepresentall
integersbetween1and256.)Adevicecalledasymbol-to-bitmappertakeseach
samplewithoneof256possibleamplitudes,andrepresentsitwith8bits.While
beforewehad24samplesineach0.125ms,wenowhave24×8=192bitsineach
0.125ms.
5.Totellpeoplewhereeachsetof192bitsbeginsandends,anextrabit(a0)is
squeezedinsothatwenowhave193bitsineach0.125ms.Thatmeanswehavea
bitrateof193bits/0.125ms=1.544Mb/s.
Thesebits,withabitrateof1.544Mb/s,aresentfromtheClass5switching
centertotheClass4switchingcenter.ThesignalsentbetweentheClass5andClass4
switchingcentersisnamedDS-1.ThewirewhichconnectstheClass5totheClass4
switchingcenteriscalledatrunk line,specificallyaT-1 trunk line.
24  ◆  Chapter Two 
Other Signals between Switching Centers 
Figure2.13showsthedifferentsignalssentbetweendifferentswitchingcenters.
ComingintoaClass4centerisaDS-1signal.Itislikelythatthisincomingsignalwill
needtobesenttotheClass3center.Ifthat’sthecase,theClass4centercreatesa
veryspecialsignalfortransmissionuptoClass3.Specifically,ittakesfourDS-1
signalsthatarecomingintoit(fromClass5centers)andputsthesesignalstogether
ontoasinglewireusingTDM.ItaddsafewextrabitstohelptheClass3center
identifythebeginningofwhatitgetsandtheendofit.Whenit’sallsaidanddone,the
signalthatmovesfromClass4toClass3isastreamofbitswithabitrateof6.312
Mb/s.ThatsignaliscalledaDS-2signal.
AsignalentersintoaClass3switchingcenter.Thissignalmightneedtomoveup
toaClass2switchingcenter.Inthatcase,theClass3switchingcenterputstogethera
specialsignaljustforClass2.IttakessevenDS-2signalsthatarecomingintoit(from
Class4centers),andputsthemtogetheronasinglewireusingTDM.Itaddsafew
extrabitstohelptheClass2officeidentifyplaceswherebitstreamsbeginandend.
Ultimately,itsendsasignaltoClass2thatisastreamofbitswithabitrateof44.736
Mb/s.ThissignaliscalledaDS-3signal.
Finally,itispossibleinPOTSthatasignalarrivingataClass2switchingcenter
mightbesentuptoaClass1switchingcenter.Ifthat’sthecase,Class2putstogethera
specialsignalforClass1.Specifically,aClass2centertakesfiveoftheDS-3signalsthat
comeintoit(fromClass3centers),andpackagesthemtogetheronasinglewireusing
TDM.IncludingafewextrabitstomakethepackagelookniceforClass1,thestreamof
bitssenttoClass1hasabitrateof274.1746Mb/s.ThissignaliscalledDS-4.
WhatI’vesaidsofaristrue,butnotthecompletetruth.Ingeneral,itispossible
inPOTSthatDS-1,DS-2,DS-3,andDS-4signalscouldbefoundbetweenanytwo
switchingcenters.Forexample,IpresentedDS-3asthesignalClass3sendstoClass
2switchingcenters.ItisalsopossibleinPOTSthatClass3sendsaDS-2signalto
Class2.
2.3 CommunicationChannels
Sofarinourdiscussion,communicationsignalshavebeensentonwires.However,
thereareanumberofdifferentwaysinwhichacommunicationsignalcanbesent
fromonepointtoanother.Usingawireisjustoneway—POTSlikesandembracesthis
way.Inthissection,I’lloutlinethedifferentwaysyoucansendasignalfromonepoint
toanother—thatis,I’lloutlinedifferentcommunicationchannels.
2.3.1TransmissionLines(Wires)
Therearetwotypesoftransmissionlines(wires)overwhichcommunicationsignals
arecommonlysent.Thesearetwisted-pair cableandcoaxial cable.
Telecommunication Networks  ◆  25 
Class4SwitchingCenter
bits
added
...toclass3
DS-2
Class5S.C.DS-1
Class5S.C.DS-1
Class5S.C.DS-1
Class5S.C.DS-1
extra
(a)
Class3SwitchingCenter
Class 4S.C.
Class4S.C.
Class4S.C.
.
.
Class4S.C.
.
bits
added
...toclass2
DS-3
extra
(b)
Class2SwitchingCenter
Class3S.C.
Class3S.C.
Class3S.C.
bits
added
extra
...toclass1
Class3S.C. DS-4
Class3S.C.
(c)
Figure2.13
(a)CreationofDS-2signal,(b)CreationofDS-3signal,(c)CreationofDS-4signal
26  ◆  Chapter Two 
Intwisted-paircable,asignalissentfromonepointtoanotherasacurrentalong
awire.Signalsthataresentinthefrequencyrangeof0to1MHzcanbesupported.
Themostcommonusefortwisted-paircablesisinPOTS.Itformsmostofthelocal
loopconnections.Specifically,inPOTS,aninsulatedtwisted-paircableleavesfroma
homeandiscombinedwithmanyothertwisted-paircablesfromneighboringhomes.
Youendupwithone,bigfatsetoftwisted-paircablessenttotheendoffice(Class5).
Coaxial cablesarethesecondtypeof“wire”usedtosendcommunicationsignals.
Incoaxialcables,thecommunicationinformationissentasacurrentalongawire.
Coaxialcablescansupportsignalsinthe100kHzto400MHzrange(amuchlarger
rangeoffrequenciesthanthetwisted-paircablecansupport).Perhapsthemost
commonuseofcoaxialcableisinconnectionsfromTVcableproviderstoyourhome.
Otherusesincludelong-distancelinesinPOTSandlocalareanetworks(LANs),
discussedalittlelaterinthischapter.
2.3.2 TerrestrialMicrowave
Anotherwayinwhichinformationcanbesentfromonepointtoanotherisbyuseof
(1)amodulator,whichturnstheincominginformationsignalintoahigh-frequency
electricalsignalonawire;and(2)anantenna,whichturnsthehigh-frequencysignal
intoanelectromagneticwavesentthroughtheatmosphere.Atthereceiverside,you
use(1)areceiverantenna,whichpicksuptheincomingelectromagneticwaveand
turnsitbackintothehigh-frequencyelectricalsignal,and(2)ademodulator,which
returnsthehigh-frequencyelectricalsignalbacktotheoriginalinformationsignals.
Someexamplesofcommunicationsystemswhichsendinformationinthiswayare
radiostations,wirelesscommunicationsystems(laterinthischapter),andterrestrial
microwave,whichI’llexplainrightnowsoyoucangetabetterunderstandingofthis
idea.
AterrestrialmicrowavetransmitterisshowninFigure2.14.I’llexplainitswork-
ingshereinthreepoints.
1.InFigure2.14youseetheincominginformationsignal.Inthisexample,the
incominginformationsignalcontainsvoicesignals;specifically,itcontainstwo
DS-3signalscombinedonasinglewireusingTDMmethods.
2.Theincominginformationsignalentersamodulator,andthemodulatormaps
theincomingsignalintoahigh-frequencyelectricalsignal.Forexample,the
modulatormaymaptheincomingsignalsothatitisnowcenteredaround3GHz,
11GHz,or23GHz.
3.Theantennatakestheincomingelectricalsignalandmapsitintoanelectro-
magneticwaveoffrequency3GHz,11GHz,or23GHz(forexample).These
frequenciescorrespondtomicrowavefrequenciesontheEM(electromagnetic)
spectrum,sothesystemiscalledamicrowavesystem.
Telecommunication Networks  ◆  27 
Betweenthetransmitterandreceiverweplacesomedevicescalledrepeaters,
showninFigure2.15.Repeatersareplacedevery26miles(40kilometers).There-
peatermaydooneoftwothings:(1)itmaysimplyamplifyandretransmitthesignalat
ahigherpower(non-regenerativerepeater);or(2)itmayreceivetheincomingsignal,
removenoiseasbestitcanthroughademodulation/remodulationprocess,andthen
retransmitthesignalatahighpower(regenerativerepeater).Weuserepeatersfortwo
reasons.First,becauseofthecurvatureoftheearth,thetransmitantennawillbehidden
fromthereceiverifwedonotplacearepeaterbetweenthetransmitterandreceiver
(Figure2.15).Second,repeaterscanbeusefulinreducingtheimpactofchannelnoise.
...
M
o
d
u
l
a
t
o
r

antenna
signal
2DS-3
signals
systemtransmitter
Parabolicdish
Highfrequency
(centeredaround
3GHz,forexample)
EMwave
Figure2.14
Terrestrialmicrowave
2
6
m
i.
26mi.
2
6
m
i.
#1 #2
Repeater
Repeater
Transmitter Receiver
Figure2.15Repeaters
28  ◆  Chapter Two 
Finally,aftertravellingthroughtherepeaters,thesignalarrivesatthereceiver,
showninFigure2.16.First,areceiverantennaisappliedtoreturnthesignaltoan
electricalsignalofafrequencyof3GHz.Then,ademodulatorisappliedthatreturns
thehigh-frequencysignaltotheoriginalinformationsignal.
Theterrestrialmicrowavesystemjust
describedtypicallyoperatesinfrequency
Receiver
rangesof1GHzto50GHz,andithas
electricalsignal
Demodulator
antenna
Highfrequency
beenappliedinanumberofdifferent
communicationsystems.For
example,ithasbeenusedasa
partofPOTS,toconnecttwo
EMwave
Information-bearing
Class2switchingcenters
separatedbyterrainsuchas
signal
swampwhereitisverydifficult Figure2.16Terrestrialmicrowavereceiver
tolaywire.Thisterrestrial
microwavesystemhasalsobeensetuptoconnectlargebranchesofbigcompanies.It
hasalsobeenimplementedasabackuptofiber-opticlinks(laterinthischapter).
2.3.3 SatelliteConnections
Withsatelliteconnections,acommunicationsystemissetupasshowninFigure2.17.
Here,atransmittertakestheincominginformationsignal,usesamodulatortoturnit
intoahigh-frequencysignal,andthenusesanantennatoturnthesignalintoanelec-
tromagneticwavesentthroughtheatmosphere(justasinterrestrialmicrowave).
Figure2.17 
Satellitecommunication 
system Satellite 
Solarpanels
generatingelectricity
tooperatesatellite
M
o
d
u
la
t
o
r

D
e
m
o
d
u
la
t
o
r
signal signal
6GHz 4GHz Information Information
Ocean
High-frequency High-frequency
electricalsignal electricalsignal
Telecommunication Networks  ◆  29 
Thissignalisthensentuptoasatelliteinorbitaroundtheearth.Specifically,the
satelliteisplacedat36,000km(22,300miles),andatthataltitudeitorbitstheEarthat
6,870mph.Movingatthisspeed,thesatelliteappearstobestationarywhenlookedat
fromtheequator,andsoitissaidtobeinageo-stationaryorbit.Thesatellitepicksup
thesignalanddoestwothings:
(1)itactsasarepeater;and
(2)itshiftsthefrequency(forexample,from6GHzto4GHz)andsendsitback
downtoearthinthedirectionofthereceiver.Modernsatellitesarebeingbuilt
whichcandomoresignalprocessingatthesatelliteitself,butwewon’tgointo
thosedetailshere.
ThesignalleavesthesatelliteandheadsbacktothereceiveronEarth.Atthat
receiver,anantennaisappliedthatturnstheincomingelectromagneticwavebackto
anelectricalsignal,andthatelectricalsignalisreturnedtotheoriginalinformation
signalbyadevicecalledademodulator.
Satellitecommunicationsoperateinanumberoffrequencybands.Herearesome
ofthem:(1)C-band,whichrefersto6-GHzuplink(“up”tothesatellite)and4-GHz
downlink(“down”tothereceiver);(2)Ku-band,whichis14-GHzuplink/12-GHz
downlink;and(3)ACTSwhichrefersto30-GHzuplink/20-GHzdownlink.
SomeofthemanyusesofsatellitecommunicationsincludesatelliteTVdistribu-
tion,liveTVtransoceaniclinks,telephonecommunicationsovertheoceans,backupto
fiber-opticlinks,andGPS(GlobalPositioningSystem).GPSreferstoasystemof
satelliteswhichenablesanyone,usingahand-helddevice,todeterminetheirexact
positiononourplanet(veryusefulforships,backpackers,andcompanieswithlarge
fleetsoftrucks/cars/aircraft).
2.3.4 Fiber-opticLinks
Fiber-opticcableisarevolutioninconnectingtransmitterstoreceivers.Itseems
possibletosupportincredible—andIdomeanincredible—ratesofinformationalong
thiscable.Usingfiber-opticcable,itappearsthatwecansupport10
14
bits/s,atthe
veryleast.Rightnow,wedon’tknowhowtosendinformationatthishigharate
(althoughwe’removinginthatdirection)—wealsodon’thavethatmuchinformation
tosendyet!
Inafiber-opticcablesystem,youhaveaninformationsignal,whichisanincom-
ingelectricalsignal.Touseafiber-opticcable,youhavetoturnthiselectricalsignal
intolight.Figure2.18showshowyoumightdothisatthetransmitterside,usinga
devicecalledanemitter.Theemittermightbe,forexample,aLED(light-emitting
diode)oranFP(FarbyParot)laserdiode.
30  ◆  Chapter Two 
Incominginformation
Cladding
Originalinformation
onelectricalsignal
onelectricalsignal
Light
Emitter
Detector
Core
Outgoinginformation
signal
onalightwave
Figure2.18Fiber-opticlink
Thefiber-opticcable,whichyoucanalsoseeinFigure2.18,ismadeupoftwo
parts:acoreandacladding.Thelighttravelsdownthecore.Itremainsinthecore
andneverleavesit(neverenteringthecladding).Thisisbecausethetwomaterials
chosenforcoreandcladdingarecarefullyselectedtoensurethattotalinternalrefrac-
tionoccurs—thatmeansthataslightenterstheboundarybetweencoreandcladding,
itisbentinsuchawaythatitremainsinthecore.(Formoreonthis,checkouta
physicsbook.)
Atthereceiverside,thelightsignal,whichhasmadeitthroughthefiber-optic
cable,mustbereturnedtoanelectricalsignal.Thatjobisdoneusingadetector,as
seeninFigure2.17.Thedetectormightbe,forexample,aPINdiodeoranAPD
(avalanchephotodiode).
Thelightsentdownthefiber-opticcablecorrespondstoanelectromagneticwave
withafrequencyintherangeof10
14
to10
15
Hz.Asmentionedalready,thesystem
appearscapableofsendinginformationatratesof10
14
bits/s.
Becauseoftheincrediblepromiseoffiber-opticcable,thesecablesarebeing
placedeverywhere.TheyhaveevenbeenplacedontheoceanfloorbetweenNorth
AmericaandEurope,competingwithandinmanycasesreplacingsatellitelinks.Fiber-
opticlinksarebeingusedtoconnectswitchingcentersinPOTS.Theyarelimitingthe
useofcoaxialcableandtwisted-paircabletoareaswheretheyhavealreadybeenput
inplace.
Telecommunication Networks  ◆  31 
2.4 DataCommunicationNetworks
We’veseenPOTS,andwe’vestudiedthedifferentmediumsthroughwhichacommuni-
cationsystemcansenditsinformation.Let’stakesometimeheretoexploresomeofthe
differentwaysandnetworks(otherthanPOTS)whichareinplacetohelppeoplecom-
municatewithoneanother.We’llstartwithadiscussionofdatacommunication.Sofar,in
alltheexampleswe’veconsidered,we’vediscussedvoicecommunicationbetween
differentlocations.Butnowwe’lltalkaboutthetransmissionofdata,whichherewe’ll
definetobecomputercommunication(thatis,communicationbetweencomputers).
Oneofthemostcommonwaysinwhichpeople’scomputerscommunicateis
throughtheuseofamodemandPOTS(plainoldtelephonesystem),showninFigure
2.19.Hereweseeacomputerreadytosenddataonthefarleft.Thedataitwantsto
communicateissenttoadevicecalledamodem,whichisshortformodulator/
demodular.Themodemnexttothecomputerontheleft(sendingcomputer)actslike
amodulator,whichmeansthatitturnsthedigitalsignal(data)intoawaveformready
fortransmissionoverthetelephone—specifically,itturnsthedataintoawaveformthat
lookslikeaspeechsignalasfarasthetelephoneisconcerned(butitsuredoesn’t
soundlikeone).Then,themodemisconnectedtothetwisted-paircablethatleaves
thehouse.Thedatasignalentersintothefive-layertelephonesystemPOTS,and
travelsthroughthissystemofwiresandswitchingcentersuntilitarrivesatthe
receiver’shome.There,thesignaltravelsintothehomealongatwisted-paircableand
entersintothemodem.Themodeminthiscaseactsasademodulator,anditturnsthe
signalthatlookslikeaspeechsignal(butdoesn’tsoundlikeone)backintotheoriginal
datasignal.Thatentersintothereceiver,andinthisway,thetwocomputerscantalk.
actslike
actslike
thetelephonenetwork
(adigitalsignal)
Asignalwhichlooks
Modem Modem
ThroughPOTS,
Datatobesent
likeavoicesignal Datasent
amodulator
ademodulator
Computersource Computersink
Figure2.19DatacommunicationviaamodemandPOTS
Currently,becauseoftheinternetandthefactthatpeoplewantdataathigherand
higherrates(higherratesthanyoucangetusingamodem),somealternativestothe
modemarenowbeingused.OneinvolvesusingthecoaxialcablethatsendsTVsignals
intoyourhousetoalsosenddatatoandfromyourcomputer.Asecondalternativeisto
useDSL(digitalsubscriberline).Inthiscase,thedatasignalthatyourcomputerwants
tosendiscombinedwithanytelephonevoicesignalsthatareleavingthehouse(using
FDM)andtogetherthesearesentoutthetelephonelinetotheClass5switchingcenter.
Iwon’tgetintothedetailsofeithersystemhere.
32  ◆  Chapter Two 
Beyondthesewaysofcommunicatingdata,whichbasicallyareallaboutupdating
existingvoice/TVlinkssothattheyalsocarrycomputerdata,thereisanother(a
better)way.Thatwayisbasedonusingwhatiscalledapacketswitchednetwork.
Apacket switched networkisacommunicationnetworkdesignedspecificallyfor
thecommunicationofdata.Theunderlyingideabehindthenetworkisthis.Datais
very“bursty,”whichmeansthatdatasentbetweencomputersisusuallyfilledwith
timeswhenlotsofdataissent,followedbytimeswhennodataissent,followedby
timeswhenlotsofdataissent,andsoon.Asaresult,peoplewantedtobuildacommu-
nicationnetworkthatallowedfortheconnectionbetweentwocomputerstobeturned
“on”quicklywhendatawastobesent,andtobeturned“off”whentherewasnodata
tobesent(sothatthecommunicationlinkcouldbeusedforothercomputerconnec-
tions).Hereiswhatengineerscameupwith.
Figure2.20representsapackedswitchedcommunicationnetwork.Let’ssaythe
computermarked“A”wantstosenddatatothecomputermarked“B.”Whathappens
isthis.
1.ThedatafromAisbrokendownintosmall,equallysizedpacketsofdata.Each
packethasonitthenameofthecomputeritwantstogoto(inourcasecomputerB).
2.ThefirstpacketfromcomputerAissentouttothenearestnode.Thenode,
whichrepresentsaspecialprocessor,basicallyactslikeatrafficpoliceman.It
looksatthedataanddoestwothings.
2a.Itcheckstoseeifthepackethaserrorsinitthatoccurredintransmis-
sion.Ifitseestoomanyerrors,itissentback;ifhasonlyaveryfewerrorsit
issenton.
Figure2.20
Computer Computer
Datacommunicationthrougha
A B
packet-switchednetwork
Node Node
Node Node
Datapacket Datapacket
Telecommunication Networks  ◆  33 
2b.Ifthepacketistobesenton,thenodedecideswhichisthebestnodeto
senditto,basedonitsfinaldestination.
3.Thismovementfromnodetonodecontinuesuntilthedatapacketarrivesat
computerB.EachpacketfromcomputerAtocomputerBmayfollowadifferent
pathofnodes.
Andthis,myreaders,ishowyoubuildnetworksintendedonlyforthetransmis-
sionofdata.
2.5 MobileCommunications
Oneofthefastestgrowingmarketsintheworld(andtheoneIamactivelyresearch-
ing)isthefieldofwirelesscommunications,alsocalledmobilecommunications.
Peopleonthemove(intheircar,ontheiryacht,walkingthroughbusydowntown
streets)believethattheymustbeabletoreachotherpeoplethroughvoice(and
possiblydata)communications.Thesepeoplewantanytime/anywhere/with-anyone
communications,andtheyarewillingtospendtheirhard-earneddollarstohaveit.As
aresult,manyengineersworkhardtogivethistothem.
TakingalookatFigure2.21,weseetheideaunderlyingamobilecommunication
system.Wehaveapersoninhiscarwithamobilephone,whichismadeupofthree
parts.Whenhetalksintoit:
1.thephoneturnsthevoicesignalintoadigitalsignalusingadevicecalleda
codec(coder/decoder);
M
O
D
E
M

C
O
D
E
C

M
O
D
E
M

C
O
D
E
C

EMwave
EMwave
825MHz
875MHz
Base 
station 
MTSO 
Highfrequency Highfrequency
signal signal
Digitalversion Digitalversion
ofvoicesignal ofvoicesignal
Electricalsignal Electricalversion
representingvoice ofvoice
signal signal
Figure2.21 Mobilecommunicationsystemfundamentals
34  ◆  Chapter Two 
2.thephonetakesthedigitalsignalitjustcreatedandturnsitintoahigh-fre-
quencysignal(forexample,centeredatabout825MHz)usingadevicecalleda
modem(modulator/demodulator);
3.thehigh-frequencysignalfeedsanantenna,andthatantennaturnsthehigh-
frequencyelectricalsignalintoanelectromagneticwave(offrequency825MHz).
Theelectromagneticwavethentravelsthroughtheairandispickedupbyan
antennaonatowercalledabasestation.Theantennaonthebasestationturnsthe
signalintoanelectricsignal.Thebasestation,withhelpfromaswitchingcentercalled
theMTSO(MobileTelephoneSwitchingOffice),figuresoutwherethecallisgoing,
andthenresendsthesignaldowntothereceivingmobilephoneatadifferentfre-
quency(forexample,atabout875MHz).
Thereceivingmobilephonehasanantennaonit,andthisantennapicksupthe
electromagneticsignalintendedforthephone,andturnsitintoanelectricalsignal
(withfrequency875MHz).Thishigh-frequencyelectricalsignalisturnedintoalow-
frequencydigitalsignalbythemodem.Finally,thedigitalsignalisreturnedtothe
originalspeechsignalbythecodec.Thissignaltravelsfromphonetoperson’sear—
andthereyouhaveit,amobilecommunicationsystem.
Whenengineersfirstbuiltsuchasystem,
theyhadaproblem.Thegovernmentonly
gavethemasmallbandoffrequenciesto
usefortheirwirelesscommunication
systems(forexample,824MHz–849
MHzand869MHz–894MHz).The
problemwiththatwasthatmanypeople
wantedtousethewirelesscommunica-
tionsystemsbutwiththelimited
frequencies,thesystemscouldonly
supportafewusers.Then,anengi-
neerhadanideahecalledthecellular
concept,andwirelesstelecommunica-
tionswaschangedforever.The
cellularconceptideaisshownin
Figure2.22.
1. Youcanseethattheentirearea 
isdividedupintosmall,funny- 
shapedregions.Eachregionis 
calledacell(asincellsinthebody). 
Eachcellistypicallybetween1mileto 
12mileslong. 
CellA
CellB
Figure2.22 Cellularconcept
Telecommunication Networks  ◆  35 
2. Eachcellhasitsownbasestation.Mobilephonesinacellcommunicate
throughthebasestationinitscell.
3.Thetransmitpowerofamobilephoneinacelliskeptlow.Anytransmissions
inthe cellAthattravelinthedirectionofcellBareeffectivelyzerobythetime
theygettocellB.
4.CellAandBusetheexactsamefrequencies.Theycandothisbecausethey
areseparatedinspacebyadistancethatmakescellA’stransmissionsunnotice-
ableincellB(andvice-versa).
Inthisway,phonecompanieswereabletosupportalargenumberofwireless
users.WhenIteachthisinmyclass,therearetwocommonquestions.I’llaskthese
questionshereandthenprovidetheanswers.
Q:WhathappensifyouareincellAandwanttocallsomeoneincellB?
A:IfyouaremakingacallwhileincellA,andtheuseryouwanttotalktoisin
cellB,thenyourcallgoestothebasestationincellA.Itissentfromtheretothe
MTSO(MobileTelephoneSwitchingCenter);fromtherethecallissenttothe
basestationincellB,whichsendsitdowntotheintendedlistenerincellB.
Q:Whathappensif,whendrivinginyourcar,youheadoutofonecellandinto
another?
A:Ifyoumoveoutofonecellandintoanother,thenthewirelesscommunication
systemswitchesyourcallfromcommunicationwiththebasestationintheold
celltocommunicationwiththebasestationinthenewcell.
2.6 LocalAreaNetworks(LANs)
Thefinalsectionofthischapterdescribesatypeofcommunicationsystemknownasa
localareanetwork,orLANforshort.Asthenamesuggests,thisisanetworkintended
toallowanumberofdifferentusers(usuallycomputers)locatedclosetooneanother
tocommunicatetogether.Itmightbeusedinanofficebuilding,orinauniversity
setting,toallowpeople’scomputerstotalktooneanother.Typically,LANsprovide
veryhigh-bit-ratecommunications,enablingmultipleuserstocommunicatelotsof
informationveryquickly.
TwotypesofLANshavebecomepopular.ThefirstiscalledEthernet,andthe
secondisknownasTokenring.
EthernetisshowninFigure2.23.Here,youseethreecomputersconnected
togetherbyasinglecoaxialcable.Theseuserscancommunicateat10Mb/s(tradi-
tionalspeed)or100Mb/s(highspeed).Let’ssaythatcomputerAwantstotalkwith
computerB.Here’swhathappens:
36  ◆  Chapter Two 
Computer Computer
A B
Looks-
Seesitisnot
Looks-
Data
Data
forhim.
Seesdataisforhim.
Readsdata.
Data
Figure2.23 LANusingEthernet
1.ComputerAlistenstothecabletoseeifthereisacommunicationtransmis-
sion.Ifthereisnotransmission,thencomputerAsendsitsinformationonthe
cable.Ifthereisatransmission,thencomputerAwaitsuntilthetransmissionis
done,andthenitbeginstosenditssignal.
2.WhencomputerAfirstsendsoutitsinformation,itcheckstoseeifanother
computerhassentatransmissionatthesametime.Ifnot,ComputerAcontinues
tosenditsdata.IfcomputerAandanothercomputerweretryingtosenddataat
thesametime,a“collision”isdetected.ComputerAstopsitstransmission,waits
arandomamountoftime,andthentriestosenditagain.
3.WhencomputerAisabletosendinformationontheline(andno“collision”is
detected),itsendsoutthenameofthedestinationcomputer,ComputerB.Allthe
computerslookatthedata,toseeifitisforthem.
4.ComputerBseesitsnameandreadstheinformation. 
Thatis,briefly,allthereistoEthernetconnections. 
ThesecondtypeofcommunicationsystemiscalledaTokenringnetwork.The 
TokenringnetworkisshowninFigure2.24,anditoperatesasfollows:
1.Whenthereisnocomputertransmitting,thebits11111111aresentaround
theringconnectingthecomputers.These8bitsarecalledthetoken.
2.WhencomputerAwantstosendinformation,itlistenstotheringtomakesure
thetokenisonthering.Ifitis,itinvertsthelastbitofthetoken(puttingout111
11110)andthenfollowsthiswithitsinformation.ComputerAmakessurethat
itputsthenameoftheintendedcomputerinitsinformation—forexample,ifthe
dataisheadedtocomputerB,computerAmakessureitputs“computerB”inits
sentdata.
Telecommunication Networks  ◆  37 
Computer
A
Computer
Coaxial
cable
Data
11111110

1
1
1
1
1
1
1

Figure2.24
Token
Tokenringnetwork
B
3.EverycomputerpicksupthedatasentbycomputerA,andreadsthenameto
seeifitistheintendedaddressee.Ifitisnot,thecomputerputstheinformation
backontheringwithoutreadingit.Ifthedataisintendedforthecomputer,the
computerreadsthedata,andthenputsitbackonthering.
4.Eventually,thedatagetsbacktothecomputerwhosentit.Thiscomputerpicks
upitsdata,andputsthetoken11111111backonthering.
AndthatishowcomputerscommunicateusingTokenringlinks.
2.7 Conclusion
Thischapterintroducedyoutotheworldofcommunicationsystems.Fromhereon,
we’llfocusonthedetailsofhowthesesystemswork.Forexample,wewilllookintoa
cellularphone.Wewilldescribehowthecodec(coder/decoder)worksinChapter4.
Wewillunderstandhowthemodem(modulator/demodulator)worksinChapter5.We
willintroduceanimportantpartofdigitalcommunicationsystems,calledthechannel
coder/decoder,inChapter6.Basically,inwhatfollows,wegetintowhatiscalledthe
“physicallayer”—howalltheinsideswork.ButfirstisChapter3,areviewofthe
statisticsandsignalsandsystemsdetailsthatyou’llwanttomakesureyouknow
beforeforgingahead.
38  ◆  Chapter Two 
Problems
1. Brieflydescribethefollowing:
(a) howtoconnect1000userstooneanotherwithoutdirectly 
connectingeveryusertoeveryotherone. 
(b)twowaystoput1000usersonasinglewire.
(c) thecellularconcept.
2. DescribethreecommunicationsystemsthatsendtheirsignalasanEMwave
throughtheatmosphere.
3.Onyourown,findontheweborinthelibraryatelecommunicationsystemnot
describedinthischapter(e.g.,HDTV,theInternet)andprovideatwo-page
overview,highlightinghowthesystemworks.
3 
Chapter 
A Review of Some Important 
Math, Stats, and Systems 
M
yintentioninwritingthischapteristogiveaquickreviewofkeymathematical,
statistical,andengineeringconceptswhichwe’lluseinourstudyoftelecommu-
nications.Ibasicallyprovideabriefreviewofrandomvariablesandrandomprocesses,
andthentalkinpassingabouttheFouriertransformandlineartimeinvariant(LTI)
systems.
Ifthismaterialisnewtoyou,readwhatfollowscarefullyandusethereferencesto
fillinanyblanksyoufind;ifthisstuffisoldhat,thengiveitapassingglance,ifforno
otherreasonthantofamiliarizeyourselfwiththenotationfoundthroughoutthebook.
3.1RandomVariables
First,I’llbrieflyreviewrandomvariables,limitingmydiscussiontocontinuousrandom
variables.
3.1.1Definitions
First,asusual,somedefinitions.Arandom event,A,referssimplytoaneventwithan
unknownoutcome.Anexampleofarandomeventistomorrow’sweather.
Next,arandom variable,x,isanumberwhosevalueisdeterminedbyarandom
event,A.Forexample,itmaybetomorrow’soutdoortemperatureinFortCollins,
Colorado(whereIlive).
3.1.2TheDistributionFunction:OneWaytoDescribex 
Let’ssayyou’vegotarandomvariable,x,whichistomorrow’stemperature.Youwant
tosomehowbeabletodescribex =(tomorrow’stemperature)tosomeone.Youdon’t
knowexactlywhatthisvaluewillbe,butyoudoknowit’sthemiddleofsummer,so
youknowit’salotmorelikelytobe80degrees(Fahrenheit)thanitistobe0degrees.
Thissectionisallabouthowyoucandescribearandomvariable,liketomorrow’s
temperature,tosomeonewithoutusingalotofwords.
: (
N
N

N


40  ◆  Chapter Three 
Onewaytofullycharacterizeourrandomvariablex isbyafunctioncalledthe
( ) .Thefunction F X probability distribution function, F X ( ) isdefinedinwordsas
x x
follows: F X ( ) isthelikelihoodthattherandomvariable xislessthanthenumberX.
x
Inanice,neatequation, F X ( ) isdefinedas
x
. ( ) N 2 < : )
(3.1)
whereP(__)isshorthandforthewords“probabilitythat __happens.”
Letmeclarifythemeaningof F X ( ) byanexample.Let’sagaingobacktox
x
beingtomorrow’stemperature,andlet’ssayit’sthemiddleofsummer.Then:
(1) F (0)=P(x<0)=(theprobabilitythatthetemperatureislessthan0),andthismay
x
be1/1000000(1inamillion),and(2) F (70)=P(x<70)=(theprobabilitythatthe
x
temperatureislessthan70),andthismaybe½(1in2).Byproviding F X ( ) forall
x
possiblevaluesofX,youcompletelycharacterizeyourrandomvariable.
Herearefoursimplepropertiesof F X ( ) :
x
( ) <1:thatis,since F X (1)0< F X ( ) representstheprobabilitythat x<X,it,like
x x
allprobabilities,mustbebetween0(neverhappens)and1(alwayshappens).
(2) F (– ∞ )=0:thatis, F (–∞ )=P(x <–∞ )=(theprobabilitythatx islessthan
x x
−∞ )=0(sincenonumbercanbesmallerthan–∞ ).
(3) F (∞ )=1:thatis, F (∞ )=P(x < ∞ )=(theprobabilitythatx islessthan ∞ )
x x
=1(sinceeveryvaluemustbesmallerthan ∞ ).
(4) F (x
1
)≥ F (x
2
) if x >x
2
:thatis,forexample,theprobabilitythatx islessthan20
x x 1
isatleastasbigastheprobabilitythatx islessthan10.
3.1.3TheDensityFunction:ASecondWaytoDescribex 
Asecondwaytodescribeourrandomvariablex istouseadifferentfunctioncalledthe
probability density function(pdfforshort).Let’stakealookagainatourrandomvari-
able x,whichrepresentstomorrow’ssummertemperature.Thepdfforthisvariableis
denoted p (x)orp(x),anditsmeaningwillbedescribedfirstinanequation,thenin
x
words,theninagraph,and,finally(phew),usingsomeintuition.
(1)Inanequation
( ( N 2 ≤ N ≤ N )
∫
N F )@N
(3.2)
A Review of Some Important Math, Stats, and Systems  ◆  41 
(2)Inwords,ifyouwanttoknowhowlikelyitisthattomorrow’stemperaturex 
willbebetween70degreesand80degrees,allyouhavetodoisintegratep(x) 
overtherangeof70to80.
(3)Inagraph,anexampleofapossiblep(x)isshowninFigure3.1.Ifyouwantto
figureouttheprobabilitythattomorrow’stemperaturexwillbebetween70and80,
allyouhavetodoisfigureouttheareaunderthep(x)curvebetween70and80.
(4)Andnow,weturnontheintuition.Intuitivelyspeaking,p(x)at x =70givesyou
anideahowlikelyitisthattomorrow’stemperaturewillbeabout70degrees.
Valuesofx atwhichp(x)isbiggestarethosevaluesmostlikelytohappen.
p(x)
x
75
70 80
Figure3.1Possiblep(x)forx=tomorrow’stemperature
3.1.4TheMeanandtheVariance
Ifyoutellme F X ( ) orp(x),thenIknoweverythingthereistoknowaboutx,but
x
whatifIdon’twanttoknoweverythingabout x,orwhatifit’shardforyoutotellme
everythingabout x.Ineithercase,therearesomecommonwaysforprovidingpartial
(butimportant)informationabout x.

(


42  ◆  Chapter Three 
(1)the mean,  x
m
(alsoknownasE(x)):Onethingyoucantellmeistheaverage
(ormean)valueof x.Ifx istomorrow’stemperature,then x
m
istheaverage
(consideringalltheyearsgoneby),oftomorrow’stemperature.Ifyouknowp(x),
thenyoucaneasilycompute x likethis:
m
∞
x
m

∫
x p x dx
(3.3)
( )
−∞
(2)the variance, σ
2
:Anotherimportantpieceofinformationabouttherandom
n
variablex ishowmuchx varies.That’smeasuredbyawell-namedterm,variance.
Ifx changesalot(forexample,iftomorrow’stemperaturecouldbeanywhere
between40and120degrees,andyoureallycan’tbesurewhereinthatrangeit
willfall),thenthevarianceisabignumber.Ifx changesverylittle(forexample,
you’reverysuretomorrow’stemperaturewillbebetween73and77degrees),
thenthevarianceisasmallnumber.Ifyouknowp(x),youcancomputevariance
usingthefollowingintegralequation:
∞
σ
∫
(N − N ) N F )@N
(3.4)
∞ −
Onelastthingaboutvariance.Youcanusuallygetafeelifitwillbebigorsmall
bylookingatagraphofp(x).Ifthisgraphiskindofflatandspreadoutoveralotofx 
values,variancewillbebig;ifthisgraphispeaky,ornotspreadoutoveralotofx 
values,thenvarianceissmall.Figure3.2showsyouwhatImean.
p(x) p(x)
x x
(a) (b)
Figure3.2
(a)Randomvariablexwithlargevariance
(b)Randomvariablexwithsmallvariance




N

"
N



(
N
A Review of Some Important Math, Stats, and Systems  ◆  43 
Example 3.1 
Givenarandomvariablexandtoldthatxhasaprobabilitydistributionfunction
p(x)showninFigureE3.1,determineitsmeananditsvariance.
p(x)
1
0 2
2
FigureE3.1Thep(x)forourexample
We’lluseequation(3.3)togetthemeanlikethis:
∞
N
∫
N F N

( ) @N
(E3.1)
∞ −
∞
N


∫
N ⋅


@N +
∫
N ⋅ @N + N ⋅ @N
(E3.2)
∫
∞ −
N


∫
N ⋅


@N
(E3.3)
(E3.4)
N (E3.5)
We’lluseequation(3.4)togetthevariance(it’splug-and-chug!)
∞
σ


∫
(N − N ) N F ) @N
(E3.6)
∞ −
N

N

O

O


N O
$ $


44  ◆  Chapter Three 
0 ∞
2
2
2
1
2
2 2
σ
∫
(
x −1
)
⋅ dx +
∫
(
x −1
)
⋅ 0 dx +
∫
(
x −1
)
⋅ 0 @N
(E3.7) x
0 −∞ 2
2
2
2
σ
∫
(x −1) ⋅
x
1
2
dx
(E3.8)
0
3
2
2
⋅
1
(
x −1
)
σ
3
2 (E3.9)
x
0
−
− σ
N
( )
(E3.10)
!

(E3.11)
σ
N
3.1.5MultipleRandomVariables
Whatdoyoudoifyouhavemorethanonerandomvariable?Let’ssayyouhavetwo
randomvariables,x whichistomorrow’stemperature,andywhichisthetemperature
onthedayaftertomorrow.Howdoyoudescribethesetworandomvariables?That’s
usuallyhandledusingajointprobabilitydensityfunctionp(x,y).Thisisjustasimple
extensionoftheprobabilitydensityfunctionp(x).Mathematically,p(x,y)isdefinedby
theintegralequation
2 (N ≤ ≤ O N ≤ ≤ O )
∫ ∫
F ( O N ) @N @O
(3.5)
Inwords,youcanusep(x,y)totellhowlikelyitisthat(x,y)fallswithinanyrange
ofpossiblevalues.
Butwhystopattworandomvariables?Let’ssayyougiveme26randomvariables
a,b,c,righttoz,representingthetemperatureforthenext26days.Ifyouwantto
describethesestatistically,youcancharacterizethemwiththejointprobabilitydensity
functionp(a,b,c,...,z).Thisisdefinedbytheintegralequation
z
2
b
2
a
2
a
1
b P
(
a ≤ ≤ a
2
, b ≤ ≤ b
2
,…, z ≤ ≤ z
2
)
… p
(
a b ,…, z
)
da db …dz
1 1
z
∫ ∫ ∫
,
z
1
b
1
a
1
(3.6)
Andso,justlikethep(x)orp(x,y)beforeit,youcanusep(a,b,c,...,z)totellyou
howlikelyitisthatyourvaluesfallwithinanyrangeofpossiblevalues.
A Review of Some Important Math, Stats, and Systems  ◆  45 
3.2 RandomProcesses
3.2.1ADefinition
Randomprocessesaresortoflikerandomvariables,onlythere’salittlemoretoknow.
We’llworkourwaytodefiningarandomprocesslikethis:
(1)First,let’ssaywehave A,arandomevent(aneventwithanunknownout-
come).We’llmake A therandomevent:“Willitbesunnytomorrow?”
(2)Now,arandomvariablex isa
x(t,A )
1
numberwhoseexactvaluede-
pendsontherandomevent A.For
example,x maybetomorrow’s
hightemperature,andthatwill
dependonwhetherit’ssunnyor
A=A
1
not.Sowecanwritex = x(A)— 
thatis,x isafunctionof A.
(3)Now,wejumptoarandom
t
process.Arandom process,x(t),
isafunctionoftimet,wherethe
exactfunctionthatoccursde-
pendsonarandomevent A.For
2
A=A
2
A=A
N
t
t
N
.
.
.

x(t,A )
x(t,A )
example,letx(t)betomorrow’s
temperatureasitchangeswith
timeovertheday;thevaluesof
x(t)willdependonA (whether
it’ssunnyornot).So,wecan
writex(t) =x(t,A)toindicatethat
thetimefunctiondependsonthe
randomevent A.Here,x(t,A)isa
randomprocess.
Aplotofarandomprocess
isshowninFigure3.3.Wesee
thatthereisatimefunctionx(t) =
x(t,A),andtheexactvalueofthe
timefunctionx(t) =x(t,A)de-
pendsontherandomevent A.If
Figure3.3
Plotofarandom
processx(t,A)
46  ◆  Chapter Three 
therandomevent Acorrespondstoitsfirstpossibleoutcome—thatis,itissunny
tomorrow—thenwegetthetimefunctionx(t,A
1
)tellingustomorrow’stemperature
throughouttheday.Iftherandomevent Acorrespondstoitssecondpossibleoutcome,
A —itispartlycloudytomorrow—thenwegettimefunctionx(t,A
2
)tellingushow
2
tomorrow’stemperaturechangesthroughouttheday.Andsoon.
There’soneveryimportantthingtonoteaboutarandomprocessx(t,A).Takea
lookatFigure3.4.ThereI’vedrawnx(t,A)andI’vedrawnadottedlinethroughtime
t = t .Attimet=t
1
,wehavex(t
1
,A),whichisanumberwhoseexactvaluedependson A.
1
That’sjustarandomvariable!So,thesampleofa
x(t,A )
randomprocessx(t,A)att = t isarandom
1 1
variable.We’llcallitx
1
 = x
1
(A).
A=A
1
A=A
2
t
t
t
2
N
1 1
1 2
1 N
t=t
1
t=t
1
t=t
1
.
.
.

x(t,A )
x(t,A )
x(t,A)
x(t,A )
x(t ,A )
Figure3.4 
Arandomprocess, 
highlightingtimet=t
1
A=A
N
A Review of Some Important Math, Stats, and Systems  ◆  47 
3.2.2 ExpressingYourself,oraCompleteStatisticalDescription
Now,yousay,Iunderstandwhatarandomprocessis.It’ssimplyafunctionoftime
whoseexactfunctiondependsonarandomevent A.ButhowdoIcharacterizeit?Let’s
sayIwanttotellmymathematicianfriendsallaboutmyrandomprocess—howdoIdo
that?
Tototallyandfullydescribeyourrandomprocess,you’llhavetoprovidethis
information:
(1)Atanytimet = t
1
,you’llhavex(t,A) = x(t
1
,A) = x
1
,arandomvariable.Your
mathematicallyinclinedfriendswillneedtoknowallaboutthisrandomvariable
x
1
,whichmeansyou’llwanttoprovidep(x
1
).
(2)Atanytwotimest = t andt = t
2
,you’llhavex(t
1
,A) = x
1
andx(t
2
,A) = x
2
,both
1
ofwhicharerandomvariables.Youwillneedacompletecharacterizationofthese
tworandomvariables,andasyouknow,that’sdonebyprovidingthejointprob-
abilitydensityfunctionp(x
1
,x
2
).
(3)AtanyKtimes,t = t
1
,  t = t
2
,andsoonuptot = t
K
,you’llhavex(t
1
,A) = x
1
, 
x(t
2
,A) = x
2
andsoonuptox(t
K
,A) = x
K
,allofwhicharerandomvariables.This
time,you’llhavetoprovideacompletecharacterizationoftheserandomvari-
ables,whichmeansyou’llhavetoprovidep(x
1
,x
2
,...,x
K
).
Actually,allyouhavetogiveyourfriendsiswhat’sin(3),sinceyoucanuse
p(x
1
,x
2
,...,x
K
)togettheinformationin(1)and(2),namelyp(x
1
)andp(x
1
,x
2
).However,
providingtheinformationin(3)isoftentoughstuff.Generally,it’sdifficulttoprovide
p(x
1
,x
2
,...,x
K
)forarandomprocess.
3.2.3 ExpressingSomeofYourself,oraPartialDescription
Whatmostpeopledoisprovidepartialinformationaboutarandomprocess—just
enoughinformationtobeabletoproceedwithatelecommunicationsstudy.Whatthey
provideiscalledasecond-ordercharacterizationoftherandomprocess.Inasecond-
ordercharacterization,youprovidetwothings:
(1)the mean of x(t
1
,A) = x
1
:thisnumber(whichmaybedifferentatdifferent
timest
1
)tellsyoutheaveragevalueofx(t,A)att = t
1
.Thisvaluecanbegenerated
usingtheequation
∞
m
( )
∫
x
1
p (x
1
)dx
1
(3.7)
x t
1
−∞


48  ◆  Chapter Three 
(2)the autocovariance, R
x
(t
1
,t
2
):thisnumber(whichmaybedifferentfordifferent
t
1
andt
2
values)describestherelationshipbetweentherandomvariablex(t
1
,A) = 
x andtherandomvariablex(t
2
,A) = x
2
.Thelargerthisnumber,themoreclosely
1
relatedx(t
1
,A)istox(t
2
,A).Thisvaluecanbegeneratedmathematicallythrough
theequation
∞ ∞
4 ( J J )
∫ ∫
( N − N ( J )) ( N − N ( J )) F ( N N ) @N @N
N (3.8)
∞ − ∞ −
3.2.4 AndinTelecommunications…
Now,youknowwhatarandomprocessis,andyouknowhowtofullydescribeitand
howtopartiallydescribeit.Inmostcommunicationsystems,randomprocesseshavea
rathernicepropertythatmakesthemsimpletodealwith.Inmosttelecommunication
applications,randomprocessesarewidesensestationary(WSS).Thismeansjusttwo
things:
(1)the mean of x(t
1
,A) = x
1
:thisvalueisthesameasthemeanoftherandom
variablex(t
2
,A)=x
2
,andthesameasthemeanoftherandomvariablex(t
3
,A)=x
3
,
andsoon.Mathematically,thatmeansthatthevalue
( ) x t
2
) x t
3
) … x = IECA K>AH )
(3.9)
m
x t
1 m
(
m
(
m
(2)the autocovariance, R
x
(t
1
,t
2
):thisvaluedependsonlyonthetimedifference
betweent
1
andt
2
.Thatis,mathematically,
R t , t ) R t
1
− t
2
) R
x
(τ) (3.10)
x
(
x
(
1 2
Example 3.2 
Determinethemeanandtheautocovarianceofarandomprocessdescribedby
(
j \
F ( J N ) )) F ( N )
π
ANF
(
,
,−
N


,
(
(
(E3.12)
foralltimest
1
,and
2
x
1
p x t A x t A ))
p x x ) p x )⋅ p x )
2
1
π
exp
(
2
2
,
( (
1
, ) , (
2
, (
1
,
2
(
1
(
2 ,
j − − x
2
(
\
(E3.13)
J



A Review of Some Important Math, Stats, and Systems  ◆  49 
Solution:Tosolveforthemean,wetestoutequation(3.7):
∞
N ( )
∫
N

N F ) @N

(
(E3.14)
∞ −
ANF
,
,
j − N \
(
( @N
(E3.15)

∞
∫
N
π
(


, ∞ −

∞
j − N

\
N ANF
,
,
(
( @N
(E3.16)

π
∫

(

, ∞ −
∞



π
(
,
,
j
− ANF
,
,
j − N

\
(
(
(
\
(E3.17)
(
∞ −
(

,
,

(E3.18)
Tofigureouttheautocovariance,weplugandplaywithequation(3.8)likethis:
∞ ∞
( )

J J 4
N
(
∫ ∫

N ) ( −

N − ) ( )

N N F

@N @N
(E3.19)
∞ − ∞ −
∞ ∞


N N ⋅
∫ ∫
⋅ (

N N F )

@N @N
(E3.20)
∞ − ∞ −
∞ ∞

∫ ∫
( )
1 2 1
x x p x ⋅ ⋅ ⋅ ( )
2 1 2
p x dx dx
(E3.21)
−∞ −∞
∞ ∞

∫

N ( ) ⋅

@N N F
∫

N ( )

@N N F

N ⋅

N
(E3.22)
∞ − ∞ −
J
50  ◆  Chapter Three 
3.3 SignalsandSystems:AQuickPeek
3.3.1 AFewSignals
Asawaytobeginourbriefreviewofsignalsandsystems,I’lldescribetwosignalsthat
we’llusethroughouttheupcomingchapters.
First,there’sδ (t),whichiscalledtheimpulsefunction(ordeltafunction,orjust
impulse).Thisfunctionisverytallandveryskinny,andcenteredaround0.Tobeabit
morespecificregardingwhatitlookslike,takealookatFigure3.5.There,you’llseea
functionofheight1/TandofdurationT.AsTgoesto0,thisfunctionbecomesvery,
veryskinnyandvery,verytall.WhenTgoesto0,theplotinFigure3.5correspondsto
δ (t).So,lookingcloselyatFigure3.5asTgoesto0,wecansaythreethingsaboutδ (t):
(1) δ (t)isinfinitelytall;
(2)δ (t)isinfinitelyskinny;and
(3)theareaundertheδ (t)functionis1;thatis,mathematically,
∞
∫
δ ( ) @J
−∞
1/T
T
δ → (t)=thisplotasT 0
t
Figure3.5 Describingδ δδ δδ(t)
Themostcommonwaytorepresentδ (t)usingtheplotisshowninFigure3.6.
Thenextfunctionthatwe’llusefrequentlyisasquarewavefunction,whichI’ll
call π ( ) t .Thisfunctionisofheight1anddurationT.Mathematically,youcanwrite
π ( ) usingthesimpleequation t
A Review of Some Important Math, Stats, and Systems  ◆  51 
π( )
¦
¦
≤ J ≤ 6
J
(3.11)
AIA
¦
Youcanalsoseewhat π ( ) t lookslikegraphicallyinFigure3.7.
δ (t)
Figure3.6
1 Themostcommonwayto
representδ δδ δδ(t)graphically
t
t
π (t)
1
T
Figure3.7
Thesignalπ ππ ππ(t)
3.3.2AnotherWaytoRepresentaSignal:TheFourierTransform
Therearelotsofwaystorepresentasignals(t).Oneeasywayistowriteamathemati-
calequationsuchas
s t (
c
t ( ) cos 2π f t )⋅ π( ) (3.12)
B ( )
52  ◆  Chapter Three 
s(t)=cos(2 f t)• π (t) π
c
Figure3.8 Plotofs(t)
t
Anotherwayistodrawapicture,whichshowsyouwhats(t)lookslikeandhowit
changeswithtime.Forexample,forthes(t)ofequation(3.12),youcanseeitsplotin
Figure3.8.
Butthere’sanotherwaytodescribesignals,discoveredbyafellownamedFou-
rier.Herealizedthatanytimesignals(t)canbecreatedbyaddingtogethercosineand
sinewaveformswithdifferentfrequencies.Youcandescribes(t)byindicatinghow
muchofeachfrequencyfyouneedtoputtogethertomakeyoursignals(t).Tofigure
thisout,allyouhavetodoisthesimpleintegral
(
∞
− πBJ
5 ( ) . ¦ ¦
∫
J I ) A @J
(3.13)
B J I ) (
∞ −
ThisiscalledtheFouriertransformofs(t).Forthes(t)ofequation(3.12),the
S(f)iscomputedtobe
−
c
)
−
c
)
+
c
)
+
c
)
1
sin
(
π(
f f T
)
− j 2π( f f T 2
+
1
sin
(
π(
f f T
)
− j 2π( f f T 2
S f ( )
2T π( f f T 2T π( f f T
e
−
c
)
e
+
c
)
(3.14)
Whilethislooksmessy,youcanseeaplotofpartofS(f)(the|S(f)|)inFigure3.9.
So,nowyoucanplotyoursignalintime,showinghowitchangesintime,oryoucan
plotyoursignalinfrequency,showinghowmuchofeachfrequencyyouneedtomake
upyourtimesignals(t).Yourchoice.
Example 3.3 
DeterminetheFouriertransformofthesignals(t)=δ(t).
We’llturntoequation(3.13)forhelp,wherewefind: 
∞
− πρJ
5 ( )
∫
A J I @J
(E3.23)
∞ −
A Review of Some Important Math, Stats, and Systems  ◆  53 
∞
− π BJ
( )
∫
δ ( A J @J
(E3.24)
B 5 )
∞ −
A
− π B
(E3.25)
A

(E3.26)

(E3.27)
ThisS(f) isplottedinFigureE3.2. Interestingly,whatweseehereisthat
thesignals(t)=δ (t) ismadeupofanequalamountofallthefrequencies.
S(f)
1
f
FigureE3.2 TheFouriertransformofδ δδ δδ(t)
f
–fc f
c
–1/T f
c
f
c
+1/T
S(f)
... ...
Figure3.9
Plotof, ,, ,,S(f) , ,, ,,
3.3.3 Bandwidth
Oftentimes,engineerswanttoknowonethingaboutasignals(t),itsbandwidth.The
bandwidth,s(t), isawaytotellsomeonehowmanyfrequencycomponentsyouneedto
makeupyoursignals(t).Thisdefinitionof“bandwidthofasignal”isarathervague
definition.Engineershavetriedtocomeupwithamorespecificdefinition,but,for
somereason,theyjustcouldn’tagreeononewaytoexactlydefinethewords“band-
widthofasignal.”Theydecided,justtopleaseeveryone,they’dusemanydifferent
definitions.Herearesomeofthese:
54  ◆  Chapter Three 
f
| | S(f)
... ...
absoluteBWis ∞
–∞ +∞
(a)
f
| | S(f)
... ...
f
c
–1/T f
c
+1/T
null-to-null
BW=2/T
peakvalueofS(f)
(b)
f
| | S(f)
... ...
3dBBW
peakvalue=P
value=P/2 value=P/2
(c)
Figure3.10 
ForaparticularS(f),plotshows(a)absolutebandwidth, 
(b)null-to-nullbandwidth,and(c)3-dBbandwidth 
A Review of Some Important Math, Stats, and Systems  ◆  55 
Absolute bandwidth: therangeoffrequenciesoverwhichS(f) (or|S(f)|)isnot
zero(Figure3.10(a)).Forexample,forthe|S(f)| ofFigure3.9,redrawninFigure
3.10(a),thisvalueisinfinity.
Null-to-null bandwidth: therangeoffrequenciesinS(f) (or|S(f)|)asshownin
Figure3.10(b).Inwords,andit’samouthful,thisbandwidthistherangeoffrequen-
ciesfrom[thefrequency,totherightofthepeakvalueof|S(f)|,atwhichthefirst0
occurs]to[thefrequency,totheleftofthepeakvalueof|S(f)|,atwhichthefirst0
occurs].So,forexample,forthe|S(f)| ofFigure3.9redrawninFigure3.10(b),thefirst
zerototherightofthepeakisatf = fc –1/T,andthefirstzerototheleftofthepeakis
atf = fc + 1/T,sothenull-to-nullbandwidthis2/T. 
3-dB bandwidth: therangeoffrequenciesinS(f) (or|S(f)|)asshowninFigure
3.10(c).Inwordsand,again,it’samouthful:therangeoffrequenciesfrom[thefre-
quencytotherightofthepeakvalueof|S(f)|where|S(f)|ishalfitspeakvalue]to[the
frequencyontheleftofthepeakwhere|S(f)| ishalfitspeakvalue].
Nowyouknowhowtofigureoutthebandwidthofasignal.Typically,whenIuse
thewordbandwidth,oftenwrittenBWforshort,I’llbereferringtothenull-to-null
bandwidth.
3.3.4 ALinearTimeInvariant(LTI)System
Figure3.11showsasystemwithaninputx(t)andanoutputy(t).You’llnoticethe
systemiscalledLTI,whichmeans
(1) it’s linear:ifIputinax
1
(t)+bx
2
(t),I’llgetoutay
1
(t)+by
2
(t).Here,y
1
(t)isthe
system’sresponsetox
1
(t),andy
2
(t)isthesystem’sresponsetox
2
(t);
(2) it’s time invariant:ifIputinx(t–T),I’llgetouty(t–T)—thatis,adelayinthe
inputcreatesadelayintheoutput.
ThereisjustonethingIwantyoutoknowaboutanLTIsystem.Let’ssayyou
findoutthatthesystemoutputisy(t) = h(t)whenthesysteminputistheimpulse
x(t)=δ(t).Justbyknowingthisonelittlething,h(t), calledthe impulse response, you
cangettheoutputy(t)foranyinputx(t).Allyouhavetodotogettheoutput,giventhe
inputx(t),iscalculateanintegralcalledconvolution.Itgoeslikethis:
LinearTimeInvariant
Input
(LTI)
Output
x(t)
System
y(t)
Figure3.11LTIsystem
56  ◆  Chapter Three 
∞
J O ( ) ( )
∫
D (τ ) J N − τ τ
(3.15)
( ) J N J D ( ) @
−∞
Sometimescalculatingthisintegralcanbeapain.Inthatcase,youcaninstead
figureouttheoutputy(t),giventheinputx(t),byusingtherelationship
( ) 0 ( ) ( ) (3.16) B ; B : B
whereY(f), X(f),andH(f)aretheFouriertransformsofy(t),x(t),andh(t),
respectively.
3.3.5 SomeSpecialLinearTimeInvariant(LTI)Systems
Earlier,Italkedaboutsomespecialsignals.Now,I’lltalkaboutsomespecialsystems.
I’llusethemalotthroughoutthisbook,soit’sratherimportantyouknowwhatthese
systemsare.
First,there’salineartimeinvariantsystemcalledalow-passfilter,orLPFfor
short.ThisreferstoasystemwhereH(f)isasshowninFigure3.12.Inotherwords,
it’sasystemwheretheH(f) ismademostlyoffrequencieslessthanf
m 
.Takealookat
Figure3.13.Here,youcanseethesignalx(t)(i.e.,X(f))comingintothesystem.The
systemhasanH(f) thatmakesitalow-passfilter.Whatcomesoutofthesystemis
x(t)
LTI
System
h(t) H(f)
characterizedby
y(t)
H(f) H(f)
f
f
f
m
OR
f
m
Figure3.12LTIsystemcorrespondingtoLPF
A Review of Some Important Math, Stats, and Systems  ◆  57 
Y(f) = H(f)X(f), andthisisalsodrawninFigure3.13.Youcanseetheoutputcorre-
spondssimplytothelow-frequencycomponentsoftheinput.Hence,thename
low-passfilter(LPF),becauseonlythelowfrequenciespassthroughtotheoutput.
Next,there’salineartimeinvariantsystemcalledabandpassfilter(BPFfor
short).YoucanunderstandtheBPFbystudyingFigure3.14.IthasanH(f) asshown
inthefigure;inwords,theH(f) isnot0onlyaroundsomefrequencyf .Now,lookat
c 
whathappenswhenaninputx(t)(X(f))entersintotheBPF.InFigure3.14,theoutput
ismadeupofjustthefrequenciesaroundf
c
.Thatis,theBPFonlyallowsfrequencies
aroundsomef  togettotheoutput.
c
H(f)
1
1
1
10
f
f
f
f
m
y(t) x(t)
X(f) Y(f) H(f)
LTISystem
Figure3.13WhathappensinanLPF
H(f)
y(t) x(t)
LTISystem
X(f) H(f) Y(f)
1
10
10
f
f f
f
c
Figure3.14WhathappensinaBPF
58  ◆  Chapter Three 
h(t) y(t) x(t) h
–1
(t)
x(t)
LTISystem
InverseLTISystem
Figure3.15DescribinginverseLTIsystem
Finally,we’vegotaninverse system. ForagivenLTIsystemwithimpulseresponse
h(t) (H(f)),theinversesystemisdefinedasfollows:
(1)inpictures,takealookatFigure3.15andyou’llgetafeelforwhattheinverse
systemdoes;
(2)inwords,ifItakeasignalx(t),andpassitthroughtheLTIsystemwith
impulseresponseh(t),andthenpassitthroughtheinversesystem,Igetback
myoriginalx(t); 
(3)mathematically,theinversesystemischaracterizedasthesystemwith
impulseresponseh
–1
(t) (H 
–1
(f))suchthat:
( ) J D
−
( ) ( ) J J D δ
(3.17)
( )
−
B 0

( ) ⋅ B 0
(3.18)
3.4 Onward
We’vecometotheendofourratherbriefreview.I’llseeyouaswemoveaheadinto
theheartoftelecommunications,startingwithChapter4.


AIA
A Review of Some Important Math, Stats, and Systems  ◆  59 
Problems
1. Considerarandomvariablewithprobabilitydistributionfunction
j
(
x a
)
\
p x ( )
1
exp , −
2
−
β
2
2
,
(
(
2πβ
2
,
(
(Q3.1)
(a) Evaluatethemean.
(b)Figureoutthevariance.
2. Considerarandomprocesswhere
2
( ) J N

)
(N N ) F
( )
( )⋅ F
( )
( ) (Q3.2)
J N

( J N

N
J N

N


( )
(Q3.3)
J N

Determinethevalueoftheautocovariance.
3. Showthat
4 ( J J ) σ
( )
(Q3.4)
N J N

4. DeterminetheFouriertransformofthetimesignal
¦
¦
1, −
T
2
≤ ≤
T
t
2
x t ( )
¦
¦
0,
(Q3.5)
¦
4 
Chapter 
Source Coding and Decoding:
Making it Digital 
T
hischaptertalksabouthowtoturnananalogsignalintoadigitalone,aprocess
calledsource coding.WetookabrieflookatitattheendofChapter1,when
Monica’sanalogspeechsignalwasturnedintoadigitalsignal.We’llnowtalkatsome
lengthaboutsourcecoding.
Beforegoingon,justabriefreminderaboutwhywewanttoturnanalogsignalsto
digital.Somanynaturallyoccurringsourcesofinformationareanalog(humanspeech,
forexample),andwewanttomakethemdigitalsignalssowecanuseadigitalcommu-
nicationsystem.
4.1Sampling
Thekeyfirststepinturninganyanalogsignaltoadigitaloneiscalledsampling.
Samplingisthechangingofananalogsignaltosamples(orpieces)ofitself.
4.1.1IdealSampling
Therearethreemethodsofsamplingthatwe’lllookattogether,thefirstofwhichis
calledideal sampling,orimpulse train sampling. Asthenamesuggests,itisasam-
plingthatisimpossibletophysicallycarryout.So,youask,whyareyoutellingme
aboutsomethingthatcanneverbedone? Well,havefaith—therearetwogoodrea-
sons. First,idealsamplingleadstoanunderstandingoftheveryimportantsampling 
theorem. Second,idealsamplingwillhelpyoueasilygraspandunderstandthetwo
practicalsamplingmethodsthatfollow.
The Sampling 
Idealsamplingissimple,andit’sshowninFigure4.1.Here,ananaloginputx(t)—say,
Carl’sspeech—entersthesampler.Thesamplerdoesonething:itmultipliesCarl’s
speechbythesignal
62  ◆  Chapter Four 
X
x(t) x
s
(t)=x(t)· p(t)
=
p(t) =
t t
0 =0 T
s
2T
s
3T
s
4T
s
5T
s
... ...
1
–T
s
0 T
s
Figure4.1Idealsampling
∞
t p )
∑
δ (t − kT ) (4.1) (
s
k −∞
calledanimpulsetrain,showninFigure4.1. Theoutputofthesampleristhen
∞
( ( ( x (t) t x ) ⋅ t p ) t x ) ⋅
∑
δ (t − kT ) (4.2)
s s
k −∞
Themultiplicationofx(t)bytheimpulsetrainp(t)leadstotheoutputshownon
theoutputsideofFigure4.1.Here,weseethattheoutputismadeupofimpulsesat
times kT
s
ofheight x kT ) ;thatis,mathematically, (
s
∞
( x (t)
∑
kT x )δ (t − kT ) (4.3)
s s s
k −∞
Inessence,thisiseverythingthereistoknowaboutidealsampling.Well,almost
everything.
The Information in the Samples 
Afterbuildingthesampler,onetelecommunicationengineerbegantowonder:“How
muchoftheinformationinCarl’sspeechsignalx(t)islostwhenit’ssampledtocreate
s
( ) ?”We’llspendthissectionansweringhisquestion.Theeasiestwaytoanswerit
istoturntothefrequencydomainforhelp.Let’sfirsttakealookatCarl’sincoming
speechsignal,x(t).Whendescribedinthefrequencydomain,asX(f),we’llassumefor
thissectionthatitlookslikeFigure4.2(a).
x t
Next,we’llconsidertheimpulsetrainp(t). TheFouriertransformofp(t)is
∞
P f − ( )
1
∑
δ( f k f ) (4.4)
T
s
s k −∞
wheref  =1/T
s
andf  iscalledthesampling rate.You’llfindapictureofthissignalin
s s
Figure4.2(b).
Source Coding and Decoding: Making it Digital  ◆  63 
( ), theFouriertransformoftheoutputsignal x t Let’snowevaluate X f ( ) .We’ll
s s
thenusethistofigureouthowmuchoftheinformationinCarl’sspeechsignalx(t)is
lostwhenit’ssampled. X f ( ) correspondsto
s
X f ) F x t )} F x(t p t )}. (4.5) ( { ( { ) ⋅ (
s s
Tosimplify X f ( ), we’llusethefollowingproperty:Multiplicationinthetime
s
domainisconvolutioninthefrequencydomain.Withthisinmind,wehave
∞
( ) X f ∗ P f ( ) ∗
1
∑
δ( f kf ), (4.6) X f ( ) ( ) X f −
T
s s
s k−∞
where*denotesconvolution.Next,applyingsimplepropertiesofconvolution,we’re
abletomoveX(f)intothesum;thatis, X f ( ) becomes
s
∞
( )
1
∑
X f − X f ( ) ∗δ( f kf ). (4.7)
T
s s
s k−∞
Finally,wegetanevensimplerequationfor X f ( ) byapplyingtheshifting
s
propertyofthedeltafunction.Thisgetsusto
∞
X f ( − ( )
1
∑
X f kf ), (4.8)
T
s s
s k−∞
Let’swriteoutthesummationtermbytermtohelpusbetterunderstand X f ( ) .
s
Doingthis,weget X f ( ) describedby
s
1
( ( ( ( X f ) {X f ) + X f + f ) + X f + 2 f )+...
s s s
T (4.9)
s
( ( + X f − f ) + X f − 2 f )+...}
s s
f
f
1
–f
M
f
M
... ...
1/T
s
–2f
s
–f
s
f
s
2f
s
0
X(f)
P(f)
(a)
(b)
Figure4.2
(a)X(f),theFouriertransformoftheinputsignalx(t)
(b)P(f),theFouriertransformoftheimpulsetrainp(t)
64  ◆  Chapter Four 
Nowthatwe’vegot X f ( ) describedmathematically,let’sunderstandwhatthis
s
meansusingwordsandapicture.Inwords,thislastequationindicatesthatX f ( )
s
consistsofanumberofX(f )’s,shiftedbymultiplesof f ,andalladdedtogether.
s
Usingpictures,forX(f)showninFigure4.2(a), X f ( ) isshowninFigure4.3.
s
X
s
(f)
X(f+2f
s
) X(f+f
s
) X(f) X(f–2f
s
) X(f–f
s
)
1
T
s
1
T
s
1
T
s
1
T
s
1
T
s
1
T
s
–2f
s
–f
s
–f
s
+f
M
–f
M
0 f
M
f
s
–f
M
f
s
2f
s
D C A B
Figure4.3 X(f),theFouriertransformoftheidealsamplingoutputx
s
(t)
s
Here’swhereitgetsreallyinteresting,soreadoncarefully.TakealookatFigure
4.3.Here,weseethataslongasA<BandD<C,thentheX(f)termin X f ( ) doesnot
s
overlapwithitsneighbors X f f ) and X f f ) . Thismeansthat,aslongas ( − ( +
s s
A<B andD<C,X(f)ispreservedperfectlyin X f ( ) . Inotherwords,aslongasA<B
s
andD<C,thenallofCarl’sincomingspeechsignalx(t)iscompletelysafeinthe
sampledsignal x t ( ) .
s
NowA<Bmeans f
M
< f f
M
—thatis, f > 2 f .Similarly,D<C means
s
−
s M
− +
M
f f
M
< − f ;thatis, f
s
> 2 f .So,aslongas f > 2 f ,thenA<B and D<C,and
s M s M
X(f)ispreservedperfectlyin X f ( ) .
s
WOW.Alloftheinformationofx(t),theincomingspeechsignal,isinthe
sampledsignal x t ( ) ifwejustinsurethat f > 2 f
M
. That’salwaysseemedlikean
s s
incredibleresulttome.Itseemslogicaltothinkthatatleastsomeinformationwould
belostwhenwesampleanincomingsignal,butnot so,aswejustsaw.
Getting Back All the Information from the Samples 
Telecommunicationengineersjumpedupanddownwithjoy,ateicecream,andhada
party,allthewhileshouting,“It’sgreatthatallofCarl’sspeechx(t)isfoundinhis
sampledsignal x t ( ) ,becausethatmeansIcansamplehisspeechandnotloseany
s
information.” Then,oneengineerpipedup,“Now,let’ssayatsomelatertimeIwantto
getCarl’sspeechx(t)backfromthatsampledsignal x t ( ) . HowdoIdothat?”Andall
s
theengineersscratchedtheirheads,wentbacktotheirchalkboards,and5minutes
latercamebackandsaid,“Easy!”
TakealookatFigure4.3,whichshowsus X f ( ) .Fromthisfigure,it’sclearthatif
s
wesimplygetridofeverythingexceptthestuffbetween − f
M
and f ,andaddagainof
M
T ,thenwe’llhaveregainedCarl’sspeech.Howdowegetridofeverythingexceptthe
s
stuffbetween − f
M
and f ? Easyagain—justusealow-passfilter(LPF)tocutoff
M
everythingoutsideof − f and f
M
,addagainof T ,andvoila,Carl’ssoundisback.
M s
Source Coding and Decoding: Making it Digital  ◆  65 
Onelastnote. Wehavesome
choiceregardingexactlywhatfrequen-
f
M
f f f 0 –f
M
X
s
(f)
LPFH(f)
ciestheLPFwillcutout. Takealook
atFigure4.4.Aslongasthecutoff
frequencyofthefilter, f ,isbetween
c
f
M
and f f
M
,thenthisLPFwilldo −
s
afinejobinrecoveringx(t)exactly
f
fromitssampledsignal x t ( ) . One
s
c s s
commonchoiceforthecutofffre-
quencyofthefilterisf
c
=f /2. This
Figure4.4 s 
putsthecutofffrequency f smack
TheuseofanLPFwithfrequency c
dabinthemiddleoftheX(f)andits
responseH(f)torecoverX(f)fromX
s
(f)
shiftedreplica X f f ) . ( −
s
Some Commonly Used Words 
Anumberoftermsarecommonlyusedwhentalkingaboutsampling.First,there’sthe
sampling theorem. Thesamplingtheoremsimplystatesthatasignalcanberecovered
fromitssamplesaslongasitissampledat f > 2 f
M
. Weknowthat,andnowwe’ve
s
gotanameforit.
Next,there’stheNyquist rate, f
N
. TheNyquistrateisthesmallestsamplingrate
f
s
thatcanbeusedifyouwanttorecovertheoriginalsignalfromitssamples.From
whatwejustsaw,weknowthat f 2 f
M
.
N
Finally,there’sthewordaliasing. I’llexplainthiswordwiththehelpofFigure
4.5. Inthisfigure,youseewhathappenswhenwesampleatarateof f < 2 f
M
. As
s
youcansee,inthiscase,in X f ( ) ,thereisanoverlappingoftheX(f)component
s
withthe X f f ) component. Asaresult, ( −
s
X
s
(f)
theoriginalX(f)isnolongerpreserved.
TheoverlappingofX(f)and X f f )
1
T
s
( −
s
iscalledaliasing.
f
Overlappingofterm(1
/
Ts
)X(f) 
and(1
/
T
s
)X(f–f
s
) 
iscalledaliasing 
Figure4.5 X(f)whenf
s
<2f
s M
I I I
I I
B (
I
66  ◆  Chapter Four 
Example 4.1 
Determinetheoutput(inthetimedomainandinthefrequencydomain)ofan
idealsamplerwithsamplingrate2Hz(samplingtime0.5s)whentheinput
correspondsto
( )
IEπJ
J N
(E4.1)
πJ
or,equivalently,inthefrequencydomain
¦ − ≤ B ≤
B : ( )
¦




(E4.2)
AIA
¦
Solution:Turningtoequation(4.3),weknowthattheoutputinthetime
domaincorrespondsto
∞
( )
∑
N (6 )δ (J − 6 )
(E4.3)
J N
−∞
∞
k
j \ j
s
( )

∑
x
, (
δ
,
t −
k
\
x t
2
( (E4.4)
k −∞ ( , (
2
,
whichisplottedinFigureE4.1(a).
Todeterminetheoutputinthefrequencydomain,weturnthepagesback
untilwereachequation(4.8),whichtellsussimply
: ( )
∑
B : −B )
(E4.5)
B

∞
(
6
I
−∞
∞
: ( )
∑
B : − )
(E4.6)
−∞
( ( ( ⋅[…+ B : +)+ B : )+ B : −)+…] (E4.7)
andthisfellowisplottedinFigureE4.1(b).
x
s
(t)
–1
/
2
1
/
–1 2 0 1
1.5
... ...
x(t)=sinπt
πt
X
s
(f)
ω
...
f
1
/
2
3
/
2
2 5
/
2
–5
/
2
–2 –3
/
2
–1
/
2
t
2
(a)
(b)
FigureE4.1(a)Signalintime (b)infrequency
Source Coding and Decoding: Making it Digital  ◆  67 
4.1.2Zero-orderHoldSampling
We’llnowtalkaboutamethodofsamplingthatcanbephysicallybuilt,amethodcalled
zero-order hold sampling.First,I’llprovideyouwithabriefdescriptionusingonlya
picture.Theinputsignal—we’lluseCarl’sspeechsignal,x(t)—andtheresulting
outputsampledsignal, x t ( ) ,arebothshowninFigure4.6.
s
Althoughthistypeofsamplingmethodisnotphysicallybuiltinthisway,Figure
4.7givessomeinsightintowhatgoesoninzero-orderholdsampling.Figure4.7
showstheincomingsignal,Carl’sspeechsignalx(t),firstgoingintoanidealsampler.
Here,itgetsmultipliedbyapulsetrainp(t). Aswesawpreviously,thisleadsto
impulsesofheight kT x ) onceevery kT seconds;thatis,itleadsto (
s s s
∞
x t ( − ( )
∑
x kT )δ(t kT ). (4.10)
i s s
k−∞
Next,theoutputoftheidealsampler, x t ( ) ,entersintoalineartimeinvariant
i
(LTI)system.TheLTIsystemisdescribedbytheimpulseresponseh(t)shownin
Figure4.7(arectangularimpulseresponseofamplitude1andduration T ). This
s
leadstotheoutput:foreachincomingsampleofheight x kT ) (in x t ( ( ) ),theoutput
s i
correspondsto“holdingon”tothevalue x kT ) foradurationof T .Thetotaloutput (
s s
x t ( ) isasignalwith ( ) isshowninFigure4.7,andit’sdescribedbythewords: x t
s s
height kT x ) ineachtimeinterval [kT k + 1)T ) . ( , (
s s s s
The Information in the Samples 
Wesawearlierthatforidealsamplingalltheinformationcontainedintheinputsignal
x(t)ispreservedintheoutputsamples x t ( ) ,aslongaswesampleatarate f > 2 f
M
.
s s
Telecommunicationengineerssuspectedthatthesamethingwastrueforzeroorder
holdsampling.We’llnowseethattheirsuspicionswerecorrect.
hold
x(t)
x
s
(t)
sampling
Zero-order
0 T
s
2T
s
3T
s
Figure4.6 Inputandoutputofzero-orderfieldsampling
68  ◆  Chapter Four 
Idealsampling
X
t
x(t)
p(t)
T
s
1
t
t
x
s
(t) x
i
(t)
h(t)=
h(t)
t
1
–T
s
0 T
s
Figure4.7 Whatgoesoninzero-orderholdsampling
Wewanttofigureoutifwecangetx(t)backfrom x t ( ) ,becauseifwecan,then
s
weknowthatalltheinformationinx(t)issavedin x t ( ) .TakealookatFigure4.7.We
s
canseethat,torecovertheoriginalx(t)fromthesampled x t ( ) ,wewantto(1)undo
s
theeffectsoftheLTIsystemwithresponseh(t),then(2)undotheeffectsoftheideal
sampling(multiplicationbyp(t)).
Figure4.8shows
usasystemthatcan
undotheeffectsofthe
x
s
(t) x(t)
h
–1
(t)
LPF
LTIsystemh(t)and f
c
=f
s
/2
gain=T
s
thentheeffectsofthe
idealsampling.Here,
Figure4.8 Systemundoingtheeffectsof
theeffectsoftheLTI
systemwithresponse
zero-orderholdsampling
h(t)areundonefirstby
applyingtheinverseLTIsystem,thesystemwithimpulseresponse h t
−1
( ) . Then,the
effectsoftheidealsamplingareundoneby(aswesawearlier)usingalow-passfilter
withcutofffrequencyf
c
=f /2.
s 
Here’sanimportantnote—theidealsamplingeffectscanonlybeundoneifwe
sampleat f > 2 f
M
.Soagain,aslongas f > 2 f
M
thenalltheeffectsofthesampling
s  s
canbeundone.
Example 4.2 
Determinetheoutput(inthetimedomain)ofazero-orderholdsamplerwith
samplingrate2Hz(samplingtime0.5s)whentheinputcorrespondsto
( )
IEπJ
J N 
(E4.8)
πJ
Source Coding and Decoding: Making it Digital  ◆  69 
Solution:Inthezero-orderholdsampler,twothingshappen:first,youhave
youridealsampling,whichcreatesimpulsesatthesampletimesof0,0.5s,1s,1.5s,
andsoon.TheoutputoftheidealsamplerisshowninFigureE4.2(a).Then,you
followthatwiththe“hold”circuit,theLTIsystemwhicheffectively“holds”each
sampleforthesamplingtimeof0.5s.ItsoutputisdrawninFigureE4.2(b).
x (t)
i
–1
/
2
–1
/
2 1
/
2
1
/
2
3
/
2
–3
/
2
3
/
2
–3
/
2
–1
–1
2 –2
2
t t
0
0
1
1
x(t)
πt
x
s
(t)
=
sinπt
(a) (b)
FigureE4.2 (a)Aftersampling(b)After“holding”
4.1.3NaturalSampling
AnotherpracticalmethodtosampleCarl’sspeechsignalx(t)isnaturalsampling.The
workingsofnaturalsamplingareshowninFigure4.9.Here,weseeCarl’sincoming
speechsignalx(t)multipliedbyasignalcalledp(t). Thisp(t)ismadeupofabunchof
tall,skinny,rectangularshapesofheight
1
T
andwidthT;thesetallskinnyrectangles
arespaced T
s
secondsapart.Figure4.9showsthismoreclearlythanmydescription.
X
p(t)
0
t
T 2T
x(t)
t
x
s
(t)
=
 
s s
T T T
... ...
1/T
Figure4.9 Naturalsampling
0 T
s
2T
s
70  ◆  Chapter Four 
X(f)
Theoutputofthesamplerissimply
x t ( ) ⋅ p t ) .Thissignalisjustasequenceofpieces
ofx(t).Itconsistsoftall,skinnyshapespiecedtogether,
1
withoneshapecomingevery T
s
seconds;eachshape
lastsforatimeTsecondsandhasaheightshapedby
x(t). Again,Figure4.9showsthisbetterthanmy
description.
s
( ) x t (
f The Information in the Samples 
–f
M
f
M
Again,thequestionarises:howmuch
Figure4.10
informationisinthosesamples?Thesame
X(f),Fouriertransformofinputx(t)
answerapplies:ifyousampleat f > 2 f
M
,you
s
cangetbackCarl’sspeechsignalx(t)fromthe
sampledsignal.Weturntothefrequencydomainasahelpfultooltoshowthatallthe
informationisinthesamplesif f > 2 f
M
. WebeginbyassumingthatX(f),theFou-
s
riertransformofCarl’sspeechsignal,x(t),lookslikeFigure4.10.Next,wewantto
s
( ) , X f figureouttheFouriertransformof x t
s
( ) .Thisisdone,usingsimplemath,as
follows:
s
( ) F x t )∗ p t )} (4.11) X f { ( (
Asawaytosimplifythis,we’llcomeupwithanotherequationforp(t).Because
p(t)isaperiodicsignal,itcanbewrittenusingaFourierseriesaccordingto
∞
p t
j 2πkf t
s
( )

∑
c
k
⋅ e
(4.12)
k −∞
where c
k
aretheFourierseriescoefficients.Forthep(t)athand, c
k
canbecomputed
togettheresult
c
k

1
sinc(
nT
)
1 sin(π nT T )
s
/
. (4.13)
T T T π nT T /
s s s s
Ploppingthisp(t)equationinthe X f
s
( ) ofequation(4.9)getsus
∞
⋅
j kf t
} X f { ( ( ) F x t ) ⋅
∑
c e
2π
s
(4.14)
s k
k−∞
Usingsomewell-knownFouriertransformstuff,wecansimplifythisequationto
get
∞
j kf t
} X f { ( ) ⋅ ( )
∑
c
k
⋅ F x t e
2π
s
(4.15)
s
k−∞
∞
( )
∑
c
k
⋅ X f − kf ) (4.16) X f (
s s
k−∞
Source Coding and Decoding: Making it Digital  ◆  71 
( ) describedmathematically.Letmeexplainthis X f Nowwehave X f ( ) to
s s
youinwordsandpictures.Inwords, X f ( ) consistsofmanycopiesofX(f)added
s
together,wherethe k
th
copyisshiftedby k f
s
andmultipliedby c . Figure4.11
k
showsthisinonesimplepicture.
Now,hereistheimportantpart.Aswithidealsampling,ifyou’lltakealookat
Figure4.11,youcanseethataslongaswekeep f − f
M
> f —i.e., f
s
> 2 f —then
s s M
thesignalX(f)iscontainedperfectlyin X f ( ) .Itcanberecoveredexactlybysimply
s
( ) throughalow-passfilter(LPF)thatgetsridofeverythingbut c X f passing X f ( ) ,
s o
andintroducingagainof1/c intheLPF.
0
X
s(f)
C
0
... ...
C
0
X(f)
X(f+f
s
) X(f–f
s
) C
1
C
1
C
–1
f
s
C
–1
–f
s
–f
s
+f
M
–f
M
f
M
f
s
–f
M
f
s
Figure4.11X(f),Fouriertransformoftheoutputsignalx
s
(t)
s
4.2 Quantization
You’vemadeittothesecondpartofsourcecoding.Takeamomenttopauseand
congratulateyourselfongettingthisfar.Now,let’scontinueourtalkonsourcecoding
(makingananalogsignaladigitalone). Asyousawinthelastsection,thefirstpartof
sourcecodinginvolvesthemappingoftheanalogsignalintosamplesofitself,apro-
cesssuitablynamedsampling.Thenextoperationcarriedoutinsourcecodingis
calledquantization,andthedevicewhichdoesitiscalledaquantizer.
4.2.1MeettheQuantizer
Aquantizerisafancywordforaverysimpledevice.Itisactuallyjustan“amplitude
changer”;ittakestheincomingsignal x t ( ) andchangesitsamplitude(ateverytime)
s
totheclosestofoneofNallowedvalues.
Thisismosteasilyexplainedbyexample,solet’sjumprightintoone.Toillus-
trate,I’llneedyourhelp.EverytimeIsayanumber,youturnitintothenearestinteger
between1and10.SoIyellout7.36,andyougivemebackthenumber7.Iyellout3.9,
youshoutback4.Ipipeout9.22,andyousay9.Ifyouunderstandthis,youtotally
understandquantizers.Alltheydoisoutputtheclosestamplitude(amongNpossible
amplitudes),giventheamplitudeoftheincomingsignal.Whatcouldbesimpler?
3.3322
72  ◆  Chapter Four 
X
s
(t) X
s
(t)
Quantizer
{0,1,...,9}
Allowedoutput
amplitudesare
C
0 C
4
A
A
0
3
B B
0
1
t
t
Figure4.12Illustrationofaquantizerandhowitworks
Tomakethisevenclearer,takealookatFigure4.12.Here,weseeaquantizerwith
input x t ( ) (theoutputofasampler).Let’sassumethatthequantizerallowsoutput
s
amplitudesfromtheset{0,1,2,…,9}.ConsiderthesamplelabeledAonthefigure,with
anamplitudeof3.3322.Itentersintothequantizer.Thequantizerchangestheamplitude
totheclosestallowedamplitude,whichinthecaseofsampleAis3.Soweendupwith
sampleA
o
.ThequantizerchangessampleBinthesameway.
OK,nowthatyou’vegotit,I’mgoingtogiveyouamoretechnicaldescription,
firstinwords,thenusingmath,andfinallyusingapicture.Don’tletanyofthisintimi-
dateyou,becauseyou’vealreadygot
it!In words,aquantizerisadevicethat
X
mapstheamplitudeoftheincoming
signaltothenearestallowedlevel.
Mathematically,aquantizerisadevice
x thatperformsthemapping x Q( ) ,
where x referstothequantizerinput,
x describesthequantizeroutput,and
Q( ) isafunctionwhichmapsthe ⋅
, values (−∞ ∞) totheclosestvaluein
theset C { , y y
2
,..., y },i.e.,
1 N
, Q: (−∞ ∞ ) → C. Graphically,given
aninputofx, thequantizeroutput x
canbeevaluatedusingagraphofthe
formshowninFigure4.13.Thisfigure
showsthattheoutputcorrespondsto
theallowedamplitudeclosesttothe
amplitudeoftheinput.
X
y
5
=2
y
4
=1
y
2
=–1
y
1
=–2
y
3
=0
–0.5 –1.5 0.5 1.5
Figure4.13Figureillustratinghowa
quantizercanbedescribedgraphically
Source Coding and Decoding: Making it Digital  ◆  73 
Example 4.3 
ConsiderthequantizerwiththeinputshowninFigureE4.3(a)andwithaninput
amplitude–outputamplituderelationshipdrawninFigure4.3(b).Drawaplotofits
output.
x(t)
^
x(outputamplitude)
x(inputamplitude)
3
2
1
t
3.9
3.1
1.3
1 2 3 0.5 1.5 2.5
(a) (b)
FigureE4.3 (a)Quantizerinput
(b)inputamplitude-outputamplituderelationshipdescribingquantizer
Solution:Whentheinputbetweentime0and1isofamplitude1.3,then(using
FigureE4.3(b))itsoutputbetweenthosetimesisamplitude1.Youcanseethatin
FigureE4.4.Withtheinputbetweentimes1and2setto3.1,thentheoutput
betweenthosetimesissetat3.Finally,withinputbetweentimes2and3setto3.9,
thentheoutputfromFigure4.3(b)becomes3.Thefinaloutputplotisfoundin
FigureE4.4.
^
x(t)
2
t
1
1
3
3
FigureE4.4 Quantizeroutput
Who wants it? 
Nowthatyouknowwhataquantizeris,youmaybescratchingyourheadwondering
whotheheckwouldwanttouseadevicethatmapsanincomingamplitudetoan
outputamplitudethatmustbeoneofNallowedvalues?Theansweris:almostevery
telecommunicationengineerwhowantstobuildadigitaltelecommunicationsystem/
network.Andthereasonisbestexplainedbyexample,asfollows.
Let’ssayyouareactingasaquantizer:Ishoutoutanumberbetween1and10,
andyoushoutbacktheclosestintegerbetween1and10.SoIshout“1.4”andyou
shoutback“1”.Iscream“5.93”andyousay“6”.Let’ssaywedecidewewanttobuilda
74  ◆  Chapter Four 
digitalcommunicationsystem.TheinformationIshouted,anynumberbetween1and
10(6.984,forexample)cannotberepresentedbyadigitalsignal(becausetherearean
infinitequantityofnumbersbetween1and10).Ontheotherhand,thenumbersyou
shoutedback—1,2,3,…,10—canberepresentedbyadigitalsignal(becausethere
areonly10possiblenumbers)—sothisinformationcanbesentusingadigitalcommu-
nicationsystem.
So,thequantizerperformsaveryimportanttask.Itturnsanincomingsignalinto
adigitalsignal,whichcanbecommunicatedusingadigitalcommunicationsystem.Of
course,someoftheamplitudeinformationislostwhenwequantizeasignal.WhenI
said1.4andyoureplied1,we“lost”the.4.Asyou’llseelater,wetrytobuildquantiz-
ersinsuchawayastominimizethelossofinformation.
Quantizer Terms 
Telecommunicationengineers,withthebestofintentions,havecreatedseveralterms
anddefinitionstomakeiteasiertodescribeaquantizer.
First,there’sthewordcodebook, C.Aquantizerchangestheamplitudeofthe
inputtooneofNallowedoutputamplitudes.ThesetofNallowedoutputamplitudesis
collectivelycalledthecodebook,orC forshort.BecauseCcontainsNamplitudes,itis
oftendenotedmathematicallybywriting C {y
1
, y ,..., y
N
}. IntheexampleofCarl
2
speakinganumber,andyou,thequantizer,speakingbacktheclosestintegerbetween
1and10,thecodebookissimplyC = {1,2,…,10}.
Next,there’sthewordcodeword,  y
i
.Thecodeword,  y
i
,simplyreferstothe i
th
of
theNoutputamplitudesallowedbythequantizer.Intheexampleofthereaderas
quantizer(speakingbacktheclosestintegerbetween1and10),thecodeword y
1
wouldbe1,thecodeword y
2
wouldbe2,andsoon.
Nextisthewordcell, R
i
.Thecell  R
i
referstothesetofinputamplitudesthatare
mappedtothecodeword y
i
.Forexample,considerthecaseofthereader(asquan-
tizer)shoutingbacktheclosestintegerbetween1and10.IfIsay7.6,thereader
screamsback8.IfIsay8.3,thereaderagainshoutsback8.Infact,anynumberinthe
setofnumbers[7.5,8.5)ismappedbythereadertothenumber8.Hence,theset[7.5,
8.5)formsacell.Sincethiscellcorrespondstotheamplitudesmappedto y
8
8 ,this
celliscalled R
8
.Simple.
Therearetwotypesofcells,granular cellsandoverload cells.Agranular cell 
referstoacellthatisbounded.Forexample,considerthecell R
8
fromtheprevious
paragraph,whichyou’llrecallwas R
8
=[7.5,8.5).Becausethiscellconsistsofasetof
numbersthatdonotstretchouttoeitherplusorminusinfinity,itiscalledagranular
cell.
Source Coding and Decoding: Making it Digital  ◆  75 
Anoverload cell,ontheotherhand,referstoacellthatisunbounded.Consider
theexampleofCarlscreaminganumber,andyouscreamingbackthenearestinteger
in1to10.Isay9.7,yousay10.Isay10.2,yousay10.Isay100.67,yousay10.Isay
100,000,yousay10.Fromthis,youcantellthatthecellforthenumber10consistsof
[9.5,∞ ).Becausethiscellendswith ∞ ,itiscalledanoverloadcell.
Types of Quantizers 
Notonlydowetelecommunicationengineershavewordstodescribeanyquantizer,
butwealsohavewordstohelpuscategorizequantizers.
Aquantizeriscalledamid-tread ifithasa0asoneofitscodewords(allowed
outputamplitudes).Inotherwords,aquantizerisamid-treadifitchangesanyampli-
tudeverycloseto0intoa0.Forexample,let’ssayIscreamoutanumber,andyou
mapittotheclosestintegerbetween–5and+5.IfIscream0.3,yousay0.IfIshout
–0.2,youshout0.Inthiscase,you’reactinglikeamid-treadquantizer.Anexampleofa
mid-treadquantizerisshowninFigure4.14(a).
Aquantizeriscalledamid-riserifitdoesnothave0asoneofitscodewords.In
otherwords,aquantizerisamid-riserifitdoesNOTchangeamplitudescloseto0into
0.Forexample,let’ssayIscreamanumber,andyoushoutbackthenearestnumber
endingin.5,between–5and5.Ishoutout“–0.3,”andyoureplywith“–0.5.”Ishoutout
“0.4,”andyouyelp“0.5.”Inthiscase,youareactinglikeamid-riser.Anexampleofa
mid-riserisshowninFigure4.14(b).
X
X
X X
2
1.5
1
0.5
0
0
–1
–0.5
–2
–1.5
–0.5 –1 –1.5 –2 0.5 1 1.5 2
–2.5
2.5
(a)
(b)
Figure4.14
(a)Exampleofmid-treadquantizer (b)Exampleofmid-riserquantizer
76  ◆  Chapter Four 
x
1
1
1
1
1
1
1
1
1
2
3
4
5
6
7
8
1
9
10
1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
x
x
2 4.5 8
10
6
3
1
2
3
4
(a) (b)
Figure4.15
(a)Exampleofuniformquantizer(b)Exampleofnon-uniformquantizer
Aquantizeriscalleduniformifallitscodewordsareequallyspaced.Forexample,
whenIscreamanumberandyoureplywiththeclosestintegerin1to10,thepossible
codewordsare1,2,3,andsoonupto10.Thesecodewordsareallspacedapartbythe
samedistance(of1),andsothisisauniformquantizer.ThistypeisshowninFigure
4.15(a).
Aquantizerisanon-uniformquantizerifallitscodewordsareNOTequally
spaced.Let’ssayIshoutoutanumber,andyoureplywitheither1,3,6,or10,which-
everisclosest.Inthiscase,thecodewordsarenotequallyspaced,andsoyouare
actingasanon-uniformquantizer.ThisexampleisshowninFigure4.15(b).
Example 4.4 
LookingatthequantizersinFigureE4.5,determineiftheyaremid-treadormid-
riseandiftheyareuniformornon-uniform.
Solution:ThequantizerinFigureE4.5(a):
1. haszeroasoneofitscodewords,makingitmid-tread,and
2. hascodewordswhicharenotequallyspaced,makingitnon-uniform.
Meanwhile,thequantizerinFigureE4.5(b):
1. doesnothavezeroasoneofitscodewords,makingitmid-rise,and
2.hasequallyspacedcodewords,makingituniform.
x
Source Coding and Decoding: Making it Digital  ◆  77 
^ ^
x x
x
–1
1
4
–4
–0.5
0.5
1.5
–1.5
FigureE4.5
TwoQuantizers
x
(a) (b)
4.2.2 TheGoodQuantizer
Nowthatyou’vegotanunderstandingofwhataquantizerisanddoes,let’sseehowto
buildagoodquantizerforagivenapplication.
What Is a Good Quantizer? 
Beforewecantalkabouthowonegoesaboutbuildingagoodquantizer,wemust
understandwhatismeantbyagoodquantizer.We’llintroduceameasureofperfor-
mancethatwecanusetotellusifaquantizerisagoodone.
Measures of Performance 
Aquantizermapsaninputamplitudetoanoutputamplitude,andtheoutputamplitude
takesononeofNallowedvalues.Aswementionedearlier,we’dliketokeeptheinput
amplitudeandoutputamplitudeclose,becauseinthiswaylessinformationislostin
thequantizer.
Withthisinmind,atanymomentoftime,wecantellhowwellaquantizeris
doingbylookingatthedifferencebetweentheamplitudeintothequantizerandthe
amplitudecomingoutofthequantizer.Thatis,atanymomentoftime,thequantizer
performancecanbemeasuredbytheerror signal 
( ) N A N X − N (4.17)
wherexistheinputtothequantizerattimet and x istheoutputofthequantizerat
thatsametimet.Agoodquantizerhasasmallerrorterm,andapoorquantizerhasa
largeerrorterm.
Engineers,however,areusuallynotsointerestedinhowwellsomethingisdoingat
amomentintime(unless,ofcourse,it’sanexplosionorsomethingequallydramatic),
butratherathowthingsaredoingoverall,oronaverage.Toprovideoverallmeasuresof
performance,engineersassumethattherearesomethingsabouttheamplitudescoming


)
2 IA
534
N
A

)
N
78  ◆  Chapter Four 
intothequantizerthatareknown(orcanbe“guess-timated”).Specifically,weassume
thatwecandetermine(andweusuallycan)howlikelyitisthataparticularamplitude
comesintothequantizer;thatis,weassumeweknowtheprobabilitydensityfunctionof
theamplitudescomingintothequantizer, p x ( ) ,wherexistheincomingamplitude.
x
Assumingweknow p x ( ) ,thefirstoverallmeasureofquantizerperformanceis
x
mean squared error,ormseforshort.Meansquarederror,asthenamesuggests,is
justtheaverage(ormean)oftheerrore(x)squared;thatis,meansquarederroris
IA -[( N − NX) ]
∞
∫
( N − NX) F ( @N N

N
)
(4.18)
∞ −
Sincewewantaquantizertohaveasmallerror(differencebetweeninputampli-
tudeandoutputamplitude),itmakessensethatengineerscallaquantizerwithasmall
meansquarederrora“goodone.”
Thesecondoverallmeasureofquantizerperformanceissignal to quantization 
noise ratio,orSQNRforshort.(Someothertelecommunicationbookscallthissame
measureSNR,butIfindthatcanbeconfusedwithtermswe’lltalkaboutinChapter5,
soI’llkeepthenotationSQNR.)SQNRreferstotheratioofthesignalinputpowerto
thepoweroftheerror(ornoise)introducedbythequantizer.Mathematically,it’s
describedby
∞
∫
( N − N ) F ( @N N
2
I

∞ −
(4.19)
wherex
m
istheaverage(ormean)xvalue.Becausewewantquantizerstohavea
smallerror,andthesizeoftheerrortermappearsonthedenominator(bottom)ofthe
SQNR term,itmakessensethatengineerssay“goodquantizershavealargeSQNR.”
A “Classic” 
Togiveyouabetterunderstandingoftheoverallmeasuresmse andSQNR,what
followsisa“classic”exampleofhowtocomputethemse andSQNRforaparticular
uniformquantizer.Let’ssayxisauniformlydistributedrandomvariablebetween[a,b];
thatistosay, p x ( ) lookslikeFigure4.16(a).Auniformquantizerisshowninthe
x
graphofFigure4.16(b).Let’sfigureoutthemse andSQNR ofthequantizerforthe
giveninputsignalandquantizer.
First,themse.Asyou’llrecall,theequationformse isgivenby
mse E x x) ] (4.20) [( −
2
∞
IA
∫
( N − NX) F ( @N N
(4.21)
∞ −

)
>
=
N
4

4



N N

Source Coding and Decoding: Making it Digital  ◆  79 
∧
X
X
.
.
.

.
.
.

y
N
∆
∆
∆
b
y =a+
1
∆/2
a a+∆
x
p(x)
a b
1
(b–a)
(a)
=
1
B
(b)
Figure4.16
(a)p(x)ofquantizerinputx (b)Graphicaldescriptionofuniformquantizer
Now,lookingatFigure4.16(a),weseethat p x
x
( ) is0outsidetheinterval[a,b],
andweusethisinformationinequation(4.19)togetto:
IA
∫
(N − NX) F ( @N N
(4.22)
Now,weusetwopiecesofinformationtochangetheintegral.We’llusethe
definitionofacell, R
i
,tohelpusouthere(soifyoudon’tremember,flipbackand
haveaquickpeek). (1)First,theinterval[a,b]canbebrokenupintocells
R R
2
,..., R
N
.(2)Second,forallvaluesofxthatfallintothecell R
i
,theoutputofthe
1
,
quantizer, x ,correspondstothevalue y
i
.Applyingthesetwopiecesofinformationto
theintegralcreates
IA
∫
(N − O ) F ( ) @N N
(4.23)
@N N +…+
∫
(N − O ) F ( )
N
mse
∑
∫
(x − y
E
) p (x)dx
x
(4.24)
i
R
i

N

\
mse
∑
∫
(x − y
E
) dx
(
,
j
KIEC p
x

B
,
(
(4.25)
B
i
R
i
Now,considertheithcell R
i
.Inthiscell,thetermx –y
i
issimplythedifference
betweentheinputtothequantizerxandtheoutputofthequantizer y
i
;thatis,x – y
i
issimplytheerrorofthequantizer,whichwe’llcall err
i
.Usingthesubstitution
err
i
=x – y
i
inthisintegralleadsto:









80  ◆  Chapter Four 
N
err
i,max

mse
∑
∫
(
err
)
d
(
err
)
i
(4.26)
B
i
err
i,min
i
Thelimitsoftheintegral err
i,min
and err
i,max
refertothesmallestvalueofthe
quantizererrorandthelargestvalueofthequantizererrorinacell R
i
.Let’scalculate
thesevalueswiththehelpofthegraphofFigure4.16(b).Takealookatthefirstcell,
whichmapsallthevaluesbetweenaanda+ ∆ intothevalue y
1
+ ∆ /2.Itfollows a
thatthelargestvalueoferroroccurswhentheinputx = a+ ∆ ismappedtotheoutput
y
1
+ ∆ /2;inthiscasetheerroris ∆ /2.Similarly,thesmallestvalueoftheerror a
a occurswhentheinputx = aismappedtotheoutput y
1
+ ∆ /2.Inthiscase,the
errorvalueis–∆ /2.It’seasilyshownthatforanycellthelargestvalueoferroris
∆ /2andthesmallestvalueoferroris–∆ /2.Usingthismaximumvalueoferrorand
minimumvalueoferrorintheintegralleadsto
∆

N

mse
∑
∫
(err ) d (err )
(4.27)
B
i
i ∆
i
−
SinceallNoftheintegralsareidentical,thisequationcanberewrittenas
∆

N

B
mse
∑
∫
(err ) d (err )
(4.28)
i ∆
−
∆
N
mse
∫
(err ) d (err )
(4.29)
B
∆
−
∆

mse
∆
∫
∆
(err ) d (err )
(4.30)
−
wherethelastequationcomesaboutbecause,lookingatFigure4.16(b),weseeN 
steps ∆ intherangeB,whichtellsusN ∆ =B,orN/B=1/∆ .Finally,evaluatingthe
integralandapplyingsomesimplemathleadsusto
∆
1 (err )
3
|
2
mse (4.31)
∆ 3 −
∆
2
∆
mse
1
(
( )
3
−
(−∆)
3
) (4.32)
∆ 24 24

(
2 IA
534

A
N

(
IA
Source Coding and Decoding: Making it Digital  ◆  81 
1 ∆
3
mse (4.33)
∆ 12
∆
2
mse (4.34)
12
Aswe’llnowsee,onceyou’vegotthemseofthequantizer,it’seasytogetthe
SQNR.Inwhatfollows,wecomputetheSQNRassumingthattheintervalof[a,b]
(showninFigure4.16(a))correspondsto[–A, A].StartingwiththebasicSQNR 
equation,wehave
∞
∫
( N − N ) N F ) @N
2
I

∞ −
(4.35)
∞
∫
( ) N F ) @N
534
∞ − (4.36)
A
2
SQNR
∆
3
2
(4.37)
12
Next,we’llgetanewvaluefor ∆ thatallowsustosimplifythisequationeven
further.ReferringtoFigure4.16(b),weseeNstepsofsize ∆ between[a,b]=[–A,A].
ItfollowsthatN ∆ =2A,or,equivalently ∆ =2A/N.PluggingthisintotheSQNR 
equationleadsto
SQNR N
2
(4.38)
Creating the Good Quantizer 
Verysimply,wecancomputethemseorSQNRofaquantizer,andifthemseissmallor
theSQNRislarge,thenwecansaywe’vegotagoodquantizer!Inthissection,we’ll
talkabouthowtobuildtheverybestquantizersforanygiveninput.
Firstoff,letmeexplainwhatImeanbybuildingagoodquantizer.We’lluseFigure
4.17tohelpusout.Noticethatyouknoweverythingaboutaquantizer,ifyouknowtwo
things:(1)thecodewords y y
2
,..., y
N
;thatis,theNvaluesthatthequantizerallowsas
1
,
output;and(2)thecells R R ,..., R
N
;inotherwords,thevaluesofinputxthatare
1
,
2
mappedbythequantizerto y
1
,thevaluesofinputxmappedto y
2
,andsoon.Specifying
thesevaluesiswhatImeanbybuildingaquantizer.
82  ◆  Chapter Four 
Tohelpusinbuildingthebest
X
quantizer,we’llassumethattheamplitude
distributionofthequantizerinputxis
given.Thatis,ifx(t)isthequantizerinput,
y
6
we’llassumeweknowthelikelihoodthat
x(t)hasaparticularamplitudexatany
y
5
timet.Mathematically,we’reassumingwe
know p x
x
( ) ,thelikelihoodthatx(t)has
R
1
R R
2 3 R R
4
R
5 6
amplitudex atanygiventime.
y
X
4
The First Method: Two Rules
and an AlgorithmThismethodfor
y
3
buildingthebestquantizerissoeasy
you’llwhopyourselfontheheadandsay
y
2
“Icouldhavethoughtofthat(andbeen
famous).”Let’ssayIsomehowmagically
knowthecodewords {y
1
,..., y
N
},andI
y
1
nowwanttofigureoutthebestcells
Figure4.17Quantizergraphindicatingthe
{R ,..., R
N
} (bestfromthestandpointof
variablesofthequantizer
1
minimizingthemse).Itsimplymakes
sensethatifI’mtryingtominimizetheerror,thebestcell R
1
(inputsxmapped
toy
1
)isthesetofvaluesofxthatareclosestto y
1
;thebestcell R
2
(inputsx 
mappedto y
2
)consistsofthevaluesofxclosestto y
2
;andsoon.That’sthefirst
rule.Letmestatethisruleformallyforyou:
Rule 1: Nearest Neighbor Rule
Given the codewords { y
1
,…, y
N
}, the best cell  R
i
 is the set of values of x that are 
closer to  y
i
 than any other codeword. 
Tomaketherulelookmoreimpressive,we’llwriteitmathematicallyas
−
2
R
i
{x ∈ | x y |
2
≤| x y
j
| ∀ j ≠ i} −
i
Asaverysimpleexampleofthisrule,considertheearlierexampleofmysayinga
numberbetween1and10andyou,thequantizer,sayingbacktheintegerthatmini-
mizestheerrorbetweenthenumberIshoutandtheintegeryoushoutback.Ishout
8.7andyoushout9(error0.3);Ishout1.2andyouscream1(error0.2).Inthisgame
it’seasytoseethatifwe’retryingbuildaquantizertominimizetheaverageerror(or
averagesquarederror),thevaluesIshoutbetween(forexample)[7.5,8.5)are
mappedto8byyou.Inotherwords,thecell R
8
ismadeupofthevaluesclosestto
y
8
8 .That’sexactlywhatourRule1says.
NowforRule2.Thisruleisallabouthowtodothis:Igiveyouacell R
i
,andyou
givemethe“best”codewordforthatcell, y
i
.(Bybest,Imeanthecodewordthat
minimizesthemse.)Howdoyoudothat?Youchoose y
i
tobetheaveragevalueofall
Source Coding and Decoding: Making it Digital  ◆  83 
thevaluesinthecell R
i
.So,ifmycellis R
8
,andit’smadeupofthexvalues[7.5,8.5),
andallthesexvaluesareequallylikely,thenthebestchoicefor y
i
istheaverage
valueof8(bestforminimizingtheaveragevalueofsquarederror).Let’sdefinethis
rulewithmoremathematicaldetail.
Rule 2: Centroid Rule
Given a cell  R
i
, the best codeword  y
i
 for that cell is the average value of all the x’s in the 
cell; i.e., the best codeword  y
i
 is the centroid of the cell  R
i
. 
Mathematically,
y E x x ∈R ] (4.39)
i
[ |
i
y
i

∫
x p x
x R
i
( ) dx
(4.40)
R
i
x dx
∫
x p
x
( )
R
i
y
i
(4.41)
p x dx
∫
x
( )
R
i
We’llnowusethesetworulestocomeupwithanalgorithmforbuildingthebest
quantizer.Basically,thefirstrulesays:“giventhecodewords{y
1
,…,y
N
},Itellyou
howtogetthebestcells{R
1
,…,R
N
}”.Thesecondrulesays:“Ifyougivemethecells
{R
1
,…,R
N
},thenItellyouhowtogetthebestcodewords{y
1
,…,y
N
}.”AndIsay:
“Hey,ifIputthesetworulestogether,I’llgetaniterativealgorithmthatwillallowme
tocomeupwithboththebest{y
1
,…,y
N
}andthebest{R
1
,…,R
N
}.”It’llworklike
this:
Algorithm for Building the Best Quantizer: Generalized Lloyd Algorithm:
1. a. Setm=1.
b. Choosesomeinitialcodebook{y
m
,…,y
N
m
}.Oftentimesagoodstarting
1
choiceisauniformly(evenly)spacedsetofNvalues.
m m
2. Giventhecodewords{y
m
,…,y
N
m
},computethebestcells{R
1
,..., R }using
1 N
Rule1,theNearestNeighborrule.
m m
3. Giventhecells{R
1
,..., R },computeanewsetofcodewordslabeled
N
{y
m+1 m+
,…,y
N
1
}byusingRule2,theCentroidrule.
1
4. a. Computethemseforthequantizerwithcodewords{y
m
,…,y
N
m
}.
1
m+
b. Computethemseforthequantizerwiththecodewords{y
m+1
,…,y
N
1
}.
1
c. Ifthepercentchangeinthemseisbelowsomesmallnumber(e.g.1%)
thenSTOP;otherwise,replacethevalueofmbym+1andreturntostep2.
x
84  ◆  Chapter Four 
Essentially,thisalgorithmiterates(repeats)Rule2togetcellsandRule1toget
codewords;eventually,thealgorithmstopswhenthecodewordsandthecellstogether
createsuchalowmsethatevenifweupdatethemfurtherthemsereallydoesn’t
improvemuch.It’sworthnotingfromamathematicalstandpointthatthisalgorithm
doesnot,ingeneral,guaranteetheverybestchoiceofcodewordsandcells,butit
usuallyworkssodarnwellthatalmosteverytelecommunicationsengineerusesitto
buildtheirquantizer.
The Second Method: Squeezing and StretchingThesecondmethodforcreating
thebestquantizer(aquantizerthatminimizesthemse)israthercreative.We’lluse
Figure4.18tohelp.Here,weseeadevicethat’smadeupofthreeparts:(1)ablock
G( ) thatmapstheinputxtotheoutputG(x);(2)auniformquantizer;and(3)ablock ⋅
−1
⋅ G
−1
( ) thatmapsthesignal y totheoutput G ( y) .
x=G
–1
(y) x
G(
.
)
y=G(x) y
G
–1
(
.
)
Quantizer
Uniform
Figure4.18Anewwaytocreatethebestquantizer
G(
.
)
y
G
–1
(
.
)
y=3,6,or9
x=2,4,or5
maps
Quantizer
Uniform
you
maps3 2
[2,3.5) [2,4.5) 6 4
9 5
Figure4.19IllustrationexampleoftheworkingsofthenewquantizerofEq.4.18
Source Coding and Decoding: Making it Digital  ◆  85 
ThethreepartsofFigure4.18canworktogethertocreateanytypeofquantizer
wewant.Thishappensbysimplychoosingthe G( ) and G
1
( ) carefully.Letme ⋅
−
⋅
illustratethatbyexample.Let’ssayIcalloutavaluebetween2and10,andyou,a
quantizer,callbackeither2,4,or5,whicheverminimizestheerror(whicheveris
closest).SoIyell2.3andyousay2;Ishout7.9andyousay5.However,wecandothis
differently.Let’ssaythatyou,ourquantizer,aretiredandonlywanttoplaythequan-
tizergameifyoucanbeauniformquantizerandcallbackeither3,6,or9.Well,Istill
wantaquantizerthatmapsinputstooutputsofeither2,4,or5.Canwebothgetwhat
wewant?Yes.TakealookatFigure4.19,whichhelpsexplainhowthiscanhappen.As
weseeinthisfigure,we’llneedtointroducetwonewfunctions:first,lookingatthe
−
⋅ rightofthefigure,we’veintroduced G
1
( ) ,whichmapstheoutputvaluesyousay—3,
6,or9—totheoutputvaluesIwanttohear:2,4,or5.That’shalfofit;theotherhalfis
shownontheleftofFigure4.19,wherewefindthefunction G( ) .Thisfunction ⋅
changestheinputsyouhearmecallingout.Considerthis:ifwhatIwantareoutputsof
either2,4,or5,thenI’llwantinputs[2,3.5)togoto2(becausethosenumbersare
closestto2);but,becausetheoutputsyousayare3,6,or9,thenforthenumbersyou
hearbetween[2,4.5)you’llsay3(becausethesenumbersareclosestto3)—soI’ll
introduceafunctionG( ) thatchangesmyoriginalinputsof[2,3.5)toinputsin[2,4.5). ⋅
Inthisway,youalwayssay3whenIwantyoutosay2(andthat’sgreatbecausethe
−
⋅ function G
1
( ) ontherightofthefigurewillmapyour3tomy2).Byintroducing
thesefunctionsG( ) and G
1
( ) asshowninFigure4.19,I’veturnedyou,Mr.orMs. ⋅
−
⋅
UniformQuantizersaying3,6or9,intoexactlythequantizerIwant.Soyousee,going
backtotheoriginalFigure4.18,wecanusethistogetanyquantizerwewant,withthe
exactchoiceof G( ) and G
1
( ) determiningthespecifictypeofquantizer. ⋅
−
⋅
Inthissectionwe’llfindoutwhatindeedisthebestchoiceof G( ) and G
1
( ) .By ⋅
−
⋅
( ) ,the G( ) and bestchoice,Imean:Giventhedistributionoftheinputamplitude, p x ⋅
x
−
⋅ G
1
( ) thatallowsthequantizerofFigure4.18tohaveaminimummse.
Letmestopheretomakesomebriefpointsandintroducesomecommonlyused
words.Briefly,thefunction G( ) and G
1
( ) arealwaysinversesofoneanother,so ⋅
−
⋅
onceyouknowoneyoucanalwaysspecifytheother.Now,somenotation. G( ) is ⋅
typicallycalleda compressor becauseinmostcasesofpracticalinterestitendsup
−
⋅ smooshingtheoriginalinputintoasmallersetofvalues. G
1
( ) istypicallycalledan
expandor,because,beingtheinverseof G( ) ,itusuallymapsthequantizeroutput ⋅
valuesintoalargersetofvalues.Finally,theentirequantizerofFigure4.18isoften
dubbedthe compandor,withthecomcomingfromcompressorandthepandorcoming
fromexpandor.
Now,howdowecreate G( ) and G
1
( ) insuchawaythatweminimizethemse ⋅
−
⋅
ofthecompandorofFigure4.18?I’mjustgoingtosimplystatethegeneralresultthat
someresearchingengineercameupwith,becausethedetailsdon’treallyaddany
greatinsight.Theansweris:givenaninputwithdistributionp(x),themseisminimized
bychoosing G( ) accordingtotheequation ⋅
IA
:

:


:
86  ◆  Chapter Four 
X
( )

∫
3
Kp x dx
(4.42)
G X
( )
0
Themseofaquantizerusing G( ) ⋅ (asinFigure4.18)isshown(notherethough)
tobe:forinputswithvaluesintherangeof [−x , x ] ,
max max
G

N
=N
( )
@N
∫
N F

(4.43)

( ) N /

−N
=N
whereqisthesizeofonecellintheuniformquantizerand

( ) G ⋅ isthederivativeof
⋅ G( ) .
Example 4.5 
Determinetheoptimalcompressioncharacteristicwhentheinputtoacompandor
isdescribedbyFigureE4.6.
Solution:TofigureouttheG(x),weturntoequation(4.42),whichtellsus
: / ( ( )
∫
!
N F ) @N
(E4.10)
Now,forXintherangeof[0,3],wehavep(x)=1/6.Applyingthisto(E4.10)
leadsusto
( )
∫
( ⋅ : / )
!

$

@N ≤ : ≤ !
(E4.11)
)
$

!

( ) ( ⋅ : / @N ≤ : ≤ !
(E4.12)
∫
p(x)
x
1
/
6
–3 3
FigureE4.6 Thepdfoftheinput

:
! :
!
:

!

:
Source Coding and Decoding: Making it Digital  ◆  87 
( ) ( )
$
!


: ≤ : ≤ ! (E4.13) : /
WhenXislargerthan3,weuseequation(E4.10),whichtellsus
: / ( ( )
∫
!
N F ) @N : > !
(E4.14)
!
⋅

@N +
∫
⋅ @N : > !
(E4.15)

∫
$
( )
$
!


⋅ ! : > ! (E4.16)
WhenXisintherange[–3,0],wehavep(x)=1/6.Applyingthisto(E4.10),weget
( )
∫
( ⋅ : / )
!

$

@N ≤ − : ≤
(E4.17)
K
6
X , 3 ≤ X ≤ 0 (E4.18)
)
1
3
−
(
WhenXislessthan–3,weuse(E4.10)todiscover
: / ( ( )
∫
!
N F ) @N : − < !
(E4.19)
−3
)
1
3
X

∫
(
K ⋅
1
6
dx +
∫
(K ⋅ 0)
1
3
, dx x < − 3
(E4.20)
0 −3
( ) ( )
$
!


− ⋅ ! N − < ! (E4.21)
Puttingitalltogether,weendupwith
¦
)
)
)
1
3
1
3
1
3
6
6
6
K
K
K
, X < − 3
¦
−3 ⋅
(
¦
G X , 3 − ≤ X ≤ 3
( )
¦
¦
X ⋅
(
(E4.22) ¦
¦
3 ⋅
(
, X > + 3
¦
¦
whichisplottedinFigureE4.7.
88  ◆  Chapter Four 
G(x)
x
–3
3
FigureE4.7 Thecompressor
ThevalueofKistraditionallydeterminedbydecidingthatwewantthefollowing:
whentheinputisatitsmaximumvaluex
max
,theoutputshouldbeG(x  )=x
max 
.Inthis
max
case,thatmeanswewantthevalueofG(3)tobe3.Requiringthisleadsto
( )
!

$

(E4.23) ! !⋅
$
(E4.24)
4.2.3 TheQuantizerandtheTelephone
Let’ssaywewanttobuildaquantizerforatelephone,onethat’sgoingtobeappliedto
asampledspeechsignal.Thissectiondiscusseshowwebuildsuchaquantizerand
explainsthestandardquantizer(forasampledspeechsignal)usedintelephone
systemsintheUnitedStatesandEurope.
The Idea 
Sinceallkindsofpeopleusethetelephone—littlepeopleandbigones,fasttalkersand
slowones,loudspeakersandquietones—telecommunicationengineerssatdownand
said,“We’llhavetocomeupwithaquantizerthatdoesagoodjobformostanypos-
sible p x
x
( ) (theinputamplitudedistribution),sincesomanydifferentpeoplewilltalk
intoourphone.”Andwiththis,thequestforthebesttelephonequantizerbegins.
Let’ssaywedecide,justlikethoseearlytelecommunicationsengineersdid,that
we’regoingtobuildourtelephonequantizerusingthecreativecompandor(ofFigure
4.18).Theissuethenbecomeshowtochoosea G( ) thatgetsagoodperformancefor ⋅
mostanypossible p x
x
( ) .We’llbeginbyconsideringtheperformancemeasureof
SQNR.Asyoumayrecall(andyoucancheckbackifyoudon’t),SQNRcorrespondsto
∞
−
2
SQNR
P
s

−∞
(
x x
m
)
p
x
(
x
)
dx
∫
(4.44)
P mse
e
Source Coding and Decoding: Making it Digital  ◆  89 
Inthecaseofacompandor,withinputsbetween [−x , x ] ,wecanuseequa-
max max
tion(4.43)formse,andnowtheSQNRequationbecomes
∞
−
2
∫
(x x
m
) p
x
(x) dx
SQNR
P
s

−∞
P
q
2
x
max
p x
e
∫
x
( )
dx
(4.45)
12
2
−x
max
G x

( )
Now,fromthisequation,istheresomewaywecanmaketheperformance(SQNR)
independentof p x
x
( ) ?IfwecansomehowchooseG(x)sothattheintegralinthe
denominatorequalsaconstanttimestheintegralinthenumerator,thentheintegrals
cancel,andIgetSQNRindependentof p x
x
( ) .That’sjustwhatwewanted.Togetthe
topintegralandbottomintegraltocancel,wejustset G( ) ⋅ accordingto(assumingx
m
=0)
| G x
2
K

( )| | |
2
(4.46)
x
K

( ) (4.47) G x
x
G x K ( ) ⋅ log (x) + C (4.48)
e
Whilethisresultmakessensemathematically,I’mnowgoingtotakeafew
linestoshowyouthatthisresultalsomakesgoodcommonsense,too.Wewantto
keeptheSQNR(ratioofinputpowertoerrorpower)constant.Sowhenbigvalues
ofx(highpower)comein,we
wanttohavebigvaluesofquan-
getsmooshedto
getmappedto
Largevaluesofx
smallerrange
Smallvaluesofx
thisverybigrange
G(x)=logx
tizationerror;andwhensmall
valuesofxcomein(lowinput
power)wewanttohavesmall
valuesofquantizationerror.This
willkeepaconstantSQNR.Let’s
seehowtheG(x)wechose
above,alogfunction,creates
this.We’lluseFigure4.20,a
graphofG(x),tohelp.First,
considerbigvaluesof x (high
power);whenbigvaluesof x 
comein,alookattheG(x)of
Figure4.20tellsusthatbig
valuesof x aremappedtoasmall
Smallvalues Largevalues
ofx ofx
rangeofvaluesinG(x).Inthe
overallquantizer/compandor,
Figure4.20 GraphillustratingG(x)=logx
x
90  ◆  Chapter Four 
thissmallrangeofG(x)thenpassesthroughauniformquantizer,whereitprobably
getsmappedtoonlyoneortwoquantizationlevels.Asaresult,wehavealargerangeof
x gettingmappedtoonlyoneortwolevels,whichcreatesalotofquantizationerror(high
errorpower).So,largeinputs(highinputpower)createlotsofquantizationerror(high
errorpower).Similarly,lookingatFigure4.20,smallinputs(lowinputpower)get
stretchedoutintoalargerrangeofvaluesbythefunctionG(x);therefore,considering
theoverallcompandor/quantizer,whenthislargerangeofG(x)outputsgoesthrough
theuniformquantizer,itismappedtoalargenumberoflevels,whichresultsinonlylittle
quantizationerror(lowerrorpower);sowegetlowinputpowercreatinglowerror
power.ThisisperfectforcreatingconstantSQNR.
However,whentelecommunicationsengineerstookacloselookattheG(x)of
Figure4.20,oneofthemhadatroublingthought:“Whathappenswhenanegative
inputarrivesattheG(x)functioninthecompandor?”Indeed,thatwasaverygood
point,forasyou’llprobablynotice,theG(x)functionofFigure4.20doesnottellyou
whatG(x)willoutputfornegativeinputs.Butanotheroftheengineersofferedasimple
solution:“Wewantbiginputs(eitherpositiveornegative)togivebigerrors,andwe
wantsmallinputs(eitherpositiveornegative)togivesmallerrors.Sowhatwe’lldois
simple:we’llusethis!”andhepointedtoafigurejustlikeFigure4.21.Yousee,inthis
figure,whatG(x)doestothenegativevaluesisidenticaltowhatitdoestopositive
values:bignegativevaluesarecompressedbyG(x)creatingbigerrorsintheuniform
quantizer(thatfollowsG(x));smallnegativevaluesareexpandedbytheG(x),creating
smallerrorsintheuniformquantizer(thatfollowsG(x)).So,theproblemofwhattodo
withnegativeinputswasquicklysolved.
x
G(x)
G(x)=logx
G(x)=–log(–x)
–x
ofG(x)
–x
Smallvaluesof
mappedtolargerange
Bigvaluesof
mappedtosmallerrangeofG(x)
Largevalues Smallvalues
of–x of–x
Figure4.21 GraphillustratingaG(x)withconsiderationofnegativeinputs
x
Source Coding and Decoding: Making it Digital  ◆  91 
But,alas,thatsame
eagle-eyedengineernow
spottedanewproblemin
theG(x)curveofFigure
4.21:“Whatshallwedo
whenazero(ornumber
closetoit)comesintothe
G(x)endofthe
compandor?”Anothergood
point,becausealookat
G(x)inFigure4.21indi-
catesthatitdoesn’ttellus
whattodowhentheinput
of x equalorcloseto0
comesintothequantizer.
“Well,”pipedoutanother
engineer,“I’vegotasimple
G(x)
solution.Let’smakeG(x) 
G(x)=–log(–x) G(x)=x G(x)=log(x)
mapaninputofx=0toan
outputof0,andhaveitmap
Figure4.22 ShapeofG(x)usedintelephone
inputscloseto0tooutputs
systemtomaintainconstantSQNR
closeto0.Andwe’llletthe 
othertwocurvesofFigure4.21applywhenever x isnotsocloseto0.ThenewG(x)
willlooklikethis.”AndhepointedtoafigurethatlookedlikeFigure4.22. 
Next,engineersdecidedtheyneededacarefulmathematicaldescriptiontofully
describethegoings-onintheG(x)ofatelephonecompandor(Figure4.22).Itwashere
thattheEuropeansandtheAmericanshadafallingout.TheAmericansusedamath-
ematicaldescriptionofG(x)calledtheµ-law description,whiletheEuropeansmadeup
theirowndescriptionofthisG(x)calledtheA-lawdescription.Sincebotharejust
mathematicaldescriptionsofthesameG(x)ofFigure4.22,I’llpresentjustoneof
thesedescriptions.BeingNorthAmerican,I’llprovidethe µ -lawdescription.The
µ -lawdescriptionoftheG(x)ofFigure4.22isthelongequation
\
log
,
j
1+
µ x
e (
G x
(
x
max ,
⋅ sgn ( )
(4.49)
x
max
( ) G
log
(
1 + µ
)
e
wheretypically µ =255,log
e
(x)isthenaturallogarithmofx(alsowrittenasln(x)),and
sgn( x) is+1if x ispositiveand–1if x isnegative.
Now,we’llseehowtheaboveG(x)equationdescribesthecurveofFigure4.22.
92  ◆  Chapter Four 
Attimeswhen x iscloseto0(specificallyµ|x|/x  <<1),theG(x)equationsimplifiesto
max
µ x
j
µ
\
x (
max
x
max
( )
G
log
e
( )
⋅ sgn
( )

,
,
G
max
G x
max
x
µ
,
log
e
( )
(
⋅ x
(4.50)
µ
(
( ,
So,attimeswhen x iscloseto0,G(x)isjustavalueclosetox,asinFigure4.22
(specifically,G(x)isaconstantmultipliedbyx).Attimeswhen x isfarfrom0(specifi-
callyµ|x| /x  <<1),G(x)inthiscasereducesto
max
log
e
j
,
µ x \
(
G x
(
x
max ,
⋅ sgn ( )
(4.51)
x
max
( ) G
log
e
( )
µ
So,when x isfarfrom0,G(x)isavalueproportionaltothelogarithmofx,asinFigure
4.22.The µ -lawequationforG(x),then,isjustanotherwayofsaying“seeFigure4.22.”
Telephonequantizersarebuiltusingthecompandor(Figure4.18),andwe’vejust
seentheG(x)thattelecommunicationengineersdecidedtouse.Thefinaldecisionwas
howmanylevelstouseintheuniformquantizerofthecompandor.Aftersomelong
daysofworkandonetiredslide-rulelateritwasdecidedthat256levelswouldbeused
intheuniformquantizer.BoththeEuropeansandAmericansagreed,andwiththat,
thecompandorusedintelephoneswasspecified.
4.3 SourceCoding:PulseCodeModulator(PCM)
Younowhaveasolidunderstandingofsampling,thefirstpartofsourcecoding.You
alsohaveagoodunderstandingofquantizers,thesecondpartofsourcecoders.Inthis
sectionwe’llputsamplersandquantizerstogether,andthrowinathirddevice,tobuild
asourcecoder.Becausethereareotherwaystobuildsourcecoders,aswe’llseelater,
thissourcecoderisgivenaveryparticularname—thepulse code modulator (PCM).
4.3.1IntroducingthePCM
You’llprobablyrecallthatthesourcecoderisadevicethatmapsananaloginputintoa
digitaloutput.Onewaytobuildit,calledthePCM,isshowninFigure4.23whereasam-
plerisused,followedbyaquantizer,whichisfollowedbyathirddevicecalleda
symbol-to-bitmapper.Here,ananalogsignal,whichwecallx(t),comesinattheinputside
ofFigure4.23.Asamplercreatessamplesofthisoriginalsignal,whichwe’llcall x t ( ) ;you
s
canseethisinFigure4.23.Aquantizertakeseachsamplethatcomesinandcreatesanew
samplethatcomesout;thisnewsamplehasanamplitudethattakesononeofNallowed
levels.We’llcallthissignal x t ( ) ,andyoucanseeanexampleofitinFigure4.23.
s
Source Coding and Decoding: Making it Digital  ◆  93 
x(t)
t t t
x
s
(t)
x
s
(t) x
s
(t)
T
s
2T
s
3T
s
0.7
0.3
1.8
2
1
0
01 10
====
Quantizer
Symbol-to-bit
mapper
x(t)
∧
10 00
Sampler
0 T
s
2T
s
3T
s
0 T
s
2T
s
3T
s
1 2 2
Figure4.23 PCMandhowitworks
Finally,adevicecalledasymbol-to-bitmappertakesinthequantizedsamples 
( ) and,foreachsamplein x t x t ( ) thatcomesin,itoutputsasetofbits,0’sand1’s.A
s s
0mayberepresentedbyashortpulseof–5Vanda1byashortpulseof+5V.Letme
explainhowthisdeviceworksbyexample.Let’ssaythequantizeroutputssamples
whichtakeononeoffourlevels—forexample,theoutputsamplesofthequantizerare
valuesintheset{0,1,2,3}.Thesymbol-to-bitmapperassociatesauniquesetofbits
witheachsample;forexample,itassociatesthebits00withthesymbol0,itassociates
thebits01withsymbol1,...,anditlinksthebits11withsymbol3.Whenagiven
samplecomesin,itputsoutthebitsithasassociatedwiththatsample.It’sreallyquite
asimpledevice,andyoucanlookatFigure4.23togetabetterfeelforhowitworks.
Tosumup,thetag-teamcombinationofsampler,quantizer,andsymbol-to-bit
mappertogethertakeananalogsignal x(t)andmapittoadigitalsignal,inthiscasea
setofbits.
4.3.2 PCMTalk
Telecommunicationengineersassociateanumberoftermswiththepulsecodemodu-
latorofFigure4.23,asawaytohelpdescribeitsoperation.I’lldiscussthreeofthese
keywordshere.First,there’ssampling rate,orhowmanysamplespersecondthe
samplercreates.Asyou’llprobablyrecall,thesamplingrateisusuallychosentobeat
leasttwotimesthemaximumfrequencyoftheinput,becauseifyoudothis,thenall
theinformationintheoriginalsignaliskeptinthesamples.
Next,there’sthetermsymbol rate.Thisisthenumberofsamplespersecondthat
leavethequantizer.Sincethequantizercreatesonesampleoutforeachsamplethat
comesin,thesymbolrateisalsotherateofthesymbolsthatcomeintothequantizer.
But,ifyoutakeaquickpeekatFigure4.23,you’llnoticethatthenumberofsamples
thatcomeintothequantizerexactlymatchesthenumberofsamplesthatcomeoutof
thesampler,sothisnumberisalwaysequaltothesamplingrate.
0
94  ◆  Chapter Four 
Finally,there’sthebit rate.Thisindicateshowmanybitspersecondcomeoutof
thesymbol-to-bitmapper.Thisnumbercanbeevaluatedbythesimplecomputation
bit rate symbol rate ×
# of bits
symbol
(4.52)
4.3.3 The“Good”PCM
Telecommunicationengineersneededawaytoevaluatehowwellasourcecoder,like
thepulsecodemodulator,wasworking.Ultimately,theydecidedthata“good”source
coderwasonethathadtwothingsgoingforit.First,theamountoferrorinthequan-
tizerpartshouldbesmall;thatis,theywantedalargeSQNR.Theycalledthis“good”
becauseitmeantonlyaverylittlebitofinformationwasbeinglostatthequantizer
part.Second,thebitrateofthesourcecodershouldbesmall.Theseengineerscalled
smallbitrates“good”becausetheydiscoveredthatasmallerbitratemeansasmaller
bandwidthforthesourcecoderoutputsignal(showninFigure4.24),andthatwas
goodbecausealotofcommunicationchannelswouldonlytransmitsignalswitha
smallbandwidth.
... ...
...
...
Highbitrate=manybits/sec Lowbitrate=fewbits/sec
T
1
(small) T
2
(big)
–
1
–
2 2 1
(big) (small)
infrequencydomain infrequencydomain
1/T 1/T 1/T 1/T
BW 2/T
1
BW 2/T
1
(a) (b)
Figure4.24 Illustratingthathighbitrateleadstolargesignalbandwidth
Source Coding and Decoding: Making it Digital  ◆  95 
However,onebrightengineersawaproblem.“Waitaminute!Youtelecommuni-
cationguyswantoppositethings.Yournumber-onewant(largeSQNR)andyour
number-twowant(smallbitrate)areopposites.Let’ssayyouwantaquantizerwitha
highSQNR(lowerror).Thenyou’llneedaquantizerwithalotofallowedoutput
levels,forexample,1024.Butthismeansyou’vegottohave10bits(2
10
1024 )for
eachsymbol,whichwillmeanaHIGHbitrate(asweseefromequation(4.52)).”
Hewasright.Sincewhatwewantareoppositethings,wehavetodefinea“good”
sourcecoderlikethis:
1. If theSQNRisfixed,wegetaverysmallbitrate(comparedtoothersource
coders);or, 
2. Ifthebitrateisfixed,wegetaverylargeSQNR (comparedtoothersource
coders).
Allthetelecommunicationengineersnoddedtheirheadsincollectiveagreement
withthisnotionof“good,”andsoitwas.
4.3.4 SourceDecoder:PCMDecoder
Ifyou’vemadeitthisfar,it’sverylikelythatyouunderstandhowthesourcecoder,the
PCM,transformsanincominganalogsignalintoadigitalone,andhowtodecideona
“good”PCM.TakingalookatFigure4.25,weseewhathappenstothedigitalsignal
outputbythePCMinthecommunicationsystem:it’stransformedbyamodulator,sent
acrossthechannel,andpickedupbyareceiver.Continuingtoexplorethisfigure,you
canseethereceiverhardatwork:ittriestoreconstructtheoriginalsignalx(t).Basi-
cally,it’sthereceiver’sjobtoundotheeffectsofthetransmitterandthechannel,as
bestitcan.AsyoucanseeinFigure4.25,apartofwhatthereceiverdoesisundothe
effectsofsourcecoding,aprocesssuitablynamedsource decoding,andwe’lltalkhere
abouthowitworks(whenthesourcecodingisPCM).
ThesourcedecoderwhichundoestheeffectsofPCMisshowninFigure4.26.
Thefirstthingitundoesisthesymbol-to-bitmapping,usingabit-to-symbol mapping,
whichyou’llfindinFigure4.26.Thisdevice,foreachincomingsetofbits,recreates
thesample,withoneofN possibleamplitudes,thatwasoutputbythequantizer.
x(t)
x(t)
Quantizer
Symbol-to-bit
Mapper
bits bits
Channel
... ...
Modulator Demodulator
Source
Decoder
PCM
UndoesPCMeffects
Transmitter Receiver
Figure4.25 WhathappenstothePCMsignalinthecommunicationsystem
96  ◆  Chapter Four 
Bit-to-symbol
Mapping
LPF
... ...
t t
T
s
T
s
01
==

1 2
1
2
0
undosymbol-to-bitmapper
toundo
10
nothinghere
undosamplereffects
quantizereffects
Figure4.26 ThesourcedecoderforPCM
Thesourcedecoderwouldnextliketoundotheeffectsofthequantizer,buta
quicklookatFigure4.26showsthatitdoesn’tdothat.Letmeexplainwhat’sgoingon
here.Aquantizer,asyouknow,mapsinputswithanyamplitude,forexample,6.345,to
anoutputwithonlyoneofNpossibleamplitudes.Forexample,forinput6.345,the
outputis6.So,whenavalueof6ismadeavailabletothesourcedecoder,ithasnoway
ofknowingexactlywhatinputcameintothequantizer.Wastheinput6.001?Howabout
6.212?Alltheseinputswouldcreateanoutputof6.
LookingagainatFigure4.26,you’llseealow-passfilter(LPF),whichisusedto
undotheeffectsofsampling.That’sbecause,asyou’llrecall,andyoucancheckbackif
youdon’t,thattheeffectsofthesampleraretotallyundonebyalow-passfiltering.
Sothereyouhaveit.Insummary,thesourcedecoderforPCMismadeupoftwo
parts,apiecethatundoesthesymbol-to-bitmapping,followedbyapartthatremoves
thesamplingeffects.
4.4 PredictiveCoding
Pulsecodemodulation,whilepopular,isnottheonlytypeofsourcecodingouttherein
thetelecommunicationworldtoday.Withacleardefinitionofwhatconstitutesagood
sourcecoder,telecommunicationengineerssetouttomakereallygoodsourcecoders.
Thischaptersharessomeofwhattheyfound,whichwe’llcallpredictivecoding.
Source Coding and Decoding: Making it Digital  ◆  97 
4.4.1TheIdeaBehindPredictiveCoding
I’llexplaintheideabehindthespecialtypeofsourcecodingcalledpredictivecoding
withthehelpofFigure4.27.Thisfigureusesdiscrete-timesignalnotation,andI’ll
explainthatasIgothroughastep-by-stepdescriptionofthefigure.Ataglance,Figure
4.27showsastrikingsimilaritytoFigure4.23(thePCM),withonlyonemaindiffer-
ence:thereisasubtractionafterthesampler.I’lltakeyouinforacloselookandwe’ll
seewhat’sgoingon.First,we’vegotasignalcomingin,whichwe’llcallx(t).Itgoes
rightintothesampler,whichoutputssamplesoftheincomingx(t).InFigure4.27,we
usethenotation x toindicatethe n
th
sampleoutputbythesampler.Here’swhere
n
somethingnewandexcitinghappens.Ratherthanpassthis x righttothequantizer,
n
wefirstdoasubtraction.Imaginethatyoucouldsomehowmagicallypredictthevalue
ofthesample x
n
,creatingapredictedvalue x
P
.Well,infact,asI’llshowyoualittle
n
later,wecancreatejustsuchasignalwiththemagicofengineering.Whathappensin
Figure4.27is,oncethesample x isgenerated,apredictedvalue x
P
isimmediately
n n
created,andsubtractedfrom x
n
.Theoutputforaninputsample x isthesample E
n
,
n
theerrorbetweentheactualsamplevalue x andthepredictedvalue x
P
.
n n
x(t)
Quantizer
Symbol-to-bit
Mapper
+
+
x
n
E
n
x
n
P
sampler
E
n
–
bits
Maindifferencebetween
predictivecoderandPCM Figure4.27
Thepredictivecoder
Theerrorsample E =x
n 
–x
P
(andnottheactualsamplevalue x
n
)nowenters
n n 
intothequantizer.Thequantizermapsthiserrorsampletoanoutputsamplemadeup

ofoneofNpossibleamplitudes.We’llcallitsoutput E
n
.Finally,eachofthequantized
samplesisturnedtoasetofbitsbyasimplemappingdevicecalledthesymbol-to-bit
mapper.ItworksinjustthesamewayasdescribedinthePCMsection.
4.4.2 Why?
Youmightbesaying,“That’sallfineanddandy,Ifollowyou,butwhywouldanyone
wanttousethisdeviceinsteadofthePCM?”I’lltakesometimeouthereandanswer
thatimportantquestion.
98  ◆  Chapter Four 
Let’ssaywe’vegotawaytogetareallygoodpredictedvalue x
P
,onethat’sreally
n
closeto x
n
.Inthatcase,theerrorsignal E = x –x
P
isaverysmallnumbercloseto0.
n n n
Now,considerthis:useaquantizerthat,foreachinputsample E
n
,outputsasample
withanamplitudethattakesononeoftwopossiblelevels,either–δor+δ,whereδisa
verysmallnumbercloseto0.Thiscreatestwothingsthattelecommunicationengineers
getexcitedabout:(1)becausethequantizerinput E wasasmallnumbercloseto0to
n

beginwith,theerrorintroducedbythequantizer(E –E
n
)issmall,andwegetalarge
n
SQNR;(2)also,becausethequantizeronlycreatestwopossibleoutputlevelsforeach
sample,thenumberofbitspersampleislow;thisleadstoalowbitrate.
Inaword,WOW—usingthepredictivecoder,itmaywellbepossibletogethigh
SQNRandlowbitrate,everythingthetelecommunicationengineerwantsfroma
sourcecoder!
4.4.3 ThePredictedValueandthePredictiveDecoder
Thekeytothegoodworkingofthepredictivecoderiscomingupwithagoodpre-
dictedvalue x
P
.Infact,therearetwodifferenttypesofpredictivecoders,andwe’ll
n
talkatsomelengthabouthoweachcomesupwiththepredictedvalue x
P
.
n
Butbeforewegointothat,let’stakealookatthesourcedecoderforapredictive
coder,thedevicelocatedinthereceiverthatundoestheeffectsofthepredictivecoder
(asbestitcan).Thisdevicehelpsthereceiveroutputtheoriginalinformationsignal
thatwassent.
ThesourcedecoderisshowninFigure4.28.Itworkstoundotheeffectsofthe
predictivecoderinreverseorder.Itbeginswithabit-to-symbolmapper,adevicewhich
undoestheeffectsofthesymbol-to-bitmapperinthesourcecoder.Next,itwouldlike
tohaveadevicethatundoestheeffectsofthequantizer;butsincethereisnosuch
device(aswesawinsection4.3.4),itmoveson.Thenexteffecttotrytoundoisthe
subtractionof x
P
attheencoder.Thisisundoneatthesourcedecoderbyaddingback
n
thevalue x
P
.Finally,theeffectsofsamplingareundonebyalow-passfilter;andvoila,
n
you’vegotasourcedecoderwhichundoespredictivecodereffects.
LPF
Bit-to-symbol
Mapper
+
+
+
bits
E
n
x
n
x
n
P
received
afterbeing
sentacrosschannel
undoessymbol-to-bit nothinghereto undoremovalof undosamplereffects
mapper undoquantizereffects predictedvalue
Figure4.28 Thedecoderforapredictivecoder
2

Source Coding and Decoding: Making it Digital  ◆  99 
Now,here’saveryimportantpoint.Tomakethesourcedecoderdescribedabove
(andseeninFigure4.28)possible,telecommunicationengineersarealwaysvery
carefultomakesurethatanysignal x
P
theyremoveduringsourcecodingisasignal
n
thatcanbecreatedandaddedbackatthereceiversideduringsourcedecoding.Inthe
nextsections,we’llexploretwopredictivecoders,onepersection.We’lldetail(among
P
otherthings)howthesedevicesget x
P
,andhowtheyuseonly x valuesthatcanbe
n n
createdandaddedbackatthereceiverduringsourcedecoding.
4.4.4 TheDeltaModulator(DM)
TheDeltaModulatoristhefirstoftwopredictivecodersthatwe’lltalkaboutinthis
book.Beingapredictivecoder,itworksinjustthewaywesawinFigure4.27.Allwe’ll
dohereisdetailhowthissystemcreatesits x
P
,andunderstandingthat,we’llexplore
n
inabitofmathematicaldetailitsinnerworkings.
How the DM creates an  N
Thispredictivecodercreatesthepredictedvalue x
P
usingareallysimpleidea:Ifyou
n
sampleanincomingsignalveryquickly,thenagoodestimateofthecurrentsample
x issimplytheprevioussamplevalue x
n−1
.Thatis,agoodpredictedvalueofthe
n
sample x is x
P
=x
n−1
.
n n
However,asonetelecommunicationengineerwasquicktopointout,therewasa
problemwiththisidea.“Ifyouuse x
P
=x
n−1
,”sheargued,“there’snowaytoaddback
n
P
x atthedecoder(Figure4.28),because x
n−1
isasampleoftheinputsignal,andthe
n
decoder(inthereceiverendofthechannel)hasnowaytocreateexactlythat.”
Andwiththatthetelecommunicationengineersscratchedtheircollectiveheads
forawhile,untilonedayoneofthemscreamed,“I’vegotit!Havealookatthis.”She
pointedtothesourcedecoderinFigure4.28.“Whilethesourcedecoderdoesn’thave
accesstothe x (or x
n−1
),itdoeshaveaccessto x (and x
n−1
).So,wecanuse x
n−1
as
n n
thepredictedvalue,ratherthan x
n−1
.”Thatseemedagoodidea,andeveryonewasin
agreement.Aftersometimeandabitofhardwork,engineersdecidedtouse x
n−1
as
thepredictedvalue,thatis, x
P
=x
n−1
.
n
Someengineersdecidedtofinetunethisidea,andastheyplayedaroundwithit,they
P
foundthataslightlybetterchoiceof x
P
is x =a x
n−1
,whereaisavalueverycloseto,but
n n
justalittlelessthan,1;specifically,theyfiguredoutthattheoptimalvalueofawas
1
a
R
x
( )
(4.53)
R ( ) 0
x
whereR (k)=E[x
n 
⋅  x
n – k
].
x
Inwhatfollows,we’llusethepredictedvalue x
P
=a x
n−1
.
n
100  ◆  Chapter Four 
∧ ∧ ∧
Quantizer
Symbol-to-bit
Mapper
+
×
+
+
×
+
x
n
x
n
x
n
E
n
E
n
E
n
sampler
∧
bits
∧
x
n
p
=ax
n–1
∧
x
n
p
=ax
n–1
Z
–1
Z
–1
bits Bit-to-symbol
Mapper
channel,
...
LPF
+
+
a
–
received
modulator,
demodulator
a
Deltamodulator(DM)
Sourcedecoder
forDM
Figure4.29 Thedeltamodulatorandsourcedecoderblockdiagram
The Block Diagram of the DM 
Let’snowtakealookattheblockdiagramoftheDMtogetanunderstandingofhow
thisdeviceisbuilt.BoththeDMandthesourcedecoderareshownintheblock
diagramofFigure4.29.ThesolidlineshowsthatthisDMandsourcedecoderhasthe
sameformasthegeneralpredictivecoderandsourcedecoderofFigures4.27and
4.28.Thedashedlinesshowthecreationofthepredictivevalue x
P
=a x
n−1
attheDM
n
anddecoder.
Let’stakesometimeoutnowanddescribehow,inthisblockdiagram,thevalue
P
x =a x
n−1
iscreatedattheDManddecoder.We’llstartatthesourcedecoder,because
n
it’seasier.Atthesourcedecoder,thevalue x
n
isavailableattheoutput.So,allwedo—
andthedashedlinesofFigure4.29showthis—istaketheavailable x
n
,delayitbyone
P
sampletime(withthe z
−1
block)andmultiplyitbya;andthenwe’vegotit: x = a  x
n−1
.
n
Atthesourcecoder,thevalueof x
n
isnotreadilyavailable,sowhatthedashed
linesonthecodersideofFigure4.29doisthis:(1)firstmake x byadding x
P
and
n n

E (thisishow x ismadeatthedecoder),then(2)delaythe x that’sjustmadeand
n n n
multiplyitbyatoget x
P
=a  x
n−1
.
n
The Sampler and the Quantizer in the DM 
WhiletheblockdiagramoftheDMiswell-describedbyFigure4.29,andadescription
ofhowitworksisalsogiven,thereareafewthingsabouttheDMthathavebeenleft
unsaid,andinthissectionIwanttosaythem.Thesearethingsregardingthesampler
andthequantizer.
The Sampler: First,wesaidearlierthattheideabehindtheDMwasthatifyou
sampledasignalfastenough,thentheprevioussample x
n−1
wasagoodpredicted
valuefor x
n
,andwemodifiedthatslightlyandcameupwiththepredictedvalueof
P
x =a x
n−1
,whereaisavaluecloseto1.WhatIwanttoexplorehereis:howfastdoes
n
thesamplerinaDMactuallysamplethesignal x(t) suchthatthesample x
n−1
isavery
goodpredictedvalueof x ?Idon’tknowexactlyhowtelecommunicationengineers
n
cameupwiththis,andIexpectthatitwastrialanderror,butultimatelywhen x(t) isa
speechsignalthesamplertendstoworkatfourtimestheNyquistrate,oreighttimes
Source Coding and Decoding: Making it Digital  ◆  101 
theNyquistrate;thatis,thesamplertendstoworkateighttimesorsixteentimesthe
maximumfrequencyoftheincomingsignalx(t). 
The Quantizer: Next,let’sconsiderthequantizer.Thequantizermapsthepredicted

error, E
n
=x –x
P
,toanoutput E withoneofNlevels.Howmanyoutputlevels(N)
n n n
x
doesthequantizeruse?ThesamplingrateinaDMissohighthattheprevioussample
n−1
(andthevaluea x
n−1
)isaverygoodpredictedvaluefor x ,andsotheerrorterm
n
E =x –x
P
=x –a x
n−1
isverycloseto0.Becauseofthis,aquantizercanbebuiltthatis
n n n n
verygoodandverysimple:thequantizermapstheinput E
n
(closeto0)toanoutputwith
oneofonlytwolevels,either+δ or–δwhereδ isavaluecloseto0.Theexactvalueofδ
dependsonthestatisticsoftheinputsamples x .
n
4.4.5 TheSignalsintheDM
Nowwe’llseewhatthedifferentsignalsintheDMlooklike,givenaparticularinputto
theDM.HavealookatFigure4.30;thistellsyouwhatsignalswe’llbestudying.We
willassumethata = 1tokeepthisanalysisniceandsimple.Let’ssaytheinputtothe
DM,pointAinFigure4.30,looksjustlikeFigure4.31(a).Then,aftersampling,the x
n
samplesthatpopoutofthesampler(atpointB)areshowninFigure4.31(b).
x(t)
Quantizer
Symbol-to-bit
Mapper
+
×
+
+
×
+
–
x
n
E
n
E
n
E
n
bits
Z
–1
Z
–1
x
n
Bit-to-symbol
Mapper
channel,
...
LPF
+
+
a
C D B E F
x
n
p
x
n
p modulator,
demodulator
∧ ∧ ∧
a=1
A
Figure4.30 BlockdiagramoftheDMshowingsignalsofinterest
We’llstudyalltheothersignalsofFigure4.30,startingourstudyatthefirst
P
sampletimen =0.We’llassumethattheinitialvalueof x
P
is0,i.e., x
0
=0.Then,
n
followingFigure4.30,wediscoverthefollowing:(1)At C:thesignal
E
0
x
0
− x
P
x
0
− 0 x
0
ispositive.(2)At D:withapositivevalueenteringthe
0

quantizer,theoutputofthequantizeris+∂ .(3)At E:oncethevalue E =+∂ issent
n
acrossthechannel,thevalueatpointEis+∂ .(4)At F:theoutput x
0
isthen
P
x
0
x
0
+ ∂ 0 + ∂ ∂.
Let’sconsiderthenextsampletimen =1:(1)At C:wehavethesignal
P P P
E
1
x
1
− x
1
P
.Let’sevaluatethevalueof x
1
: x
1
a x
n−1
1 x
0
+∂ .Usingthis x
1
value,wehave E x
1
− x
P
x
1
− ∂.Assuming x
1
ofFigure4.31(b)islargerthan ∂ ,
1 1
thenthevalue E
1
ispositive.(2)At D:theinputtothequantizerispositive,soit

followsthatthequantizeroutput E is+∂ .(3)At E:assumingthevalue+∂ issafely
n
sentacrossthechannel,thenthevalueatpointEisalso+∂ .(4)At F:finally,the
P
outputvalueiscomputedas x
1
x
1
+ (value at E) x
1
P
+ ∂ ∂ + ∂ 2∂ .
102  ◆  Chapter Four 
x(t)
x
n
x
0
x
1
x
2
x
3
n t
...
0 1 2 3
(a) (b)
∧
E
n
=x–x
n
p
E
n
=x–x
n
p
x
2
–2δ 
x
1
–δ 
x
3
–3δ
x
0
–0=x
0
...
n
0 1 2 3
δ
...
0 1 2 3
n
(c) (d)
∧
x
n
4δ
3δ
2δ
δ
...
n
0 1 2 3
(e)
Figure4.31 FollowingthesignalsinFigure4.30:(a)atpointA;(b)atpointB;
(c)atpointC;(d)atpointDandE;(e)atpointF
Source Coding and Decoding: Making it Digital  ◆  103 
WecancontinuethecomputationofthesignalsinFigure4.30ateachandevery
timeinstant(andifyoutryproblem4.7you’llgetjustthatopportunity).Ultimately,we
endupwiththevaluesatpointCofFigure4.31(c),thevaluesatpointDandEof
Figure4.25(d),andthevaluesatpointFofFigure4.31(e).
Idon’tknowifit’scleartoyouyetfromthedescriptionaboveandFigure4.31,but
there’saveryeasyway(ashortcut)todescribingtheinput-output(x to x
n
)relation-
n
shipoftheDM(assuminga = 1).Iftheinputvalueislargerthanthepredictedvalue,
thenincreasetheoutputbyδ;iftheinputvalueissmallerthanthepredictedvalue,
thendecreasetheoutputbyδ.Andtohelpyoudothis,remember—thepredicted
valueisjustthepreviousoutputvalue.
Aratherneatresultiseasytoseefromtheaboveshortcut.Sinceyou’realways
increasingordecreasingtheoutputbyδ,theniftheoutputis0atstart-up,theoutput
atanylatertimeisalwayssomekδ,wherekisaninteger.
Theinput–output(x to x
n
)relationshipcanalsoberesolvedusingadifferent
n
shortcut,whichusesagraph.Assumea = 1.First,drawtheinputsignalusinga
dashedline,asI’vedoneinFigure4.32.Thenaddlittledashesonthex-axisatthe
sampletimes.I’vealsoaddedthisintoFigure4.32.Now,attime0,thepredictedvalue
is0. Becausethisissmallerthantheinputattime0(seeFigure4.32),increasethe
outputbyδ(whichmeanstheoutputbecomesitsinitialvalueof0plusδ,whichisδ).
ThisisdrawninFigure4.32usingasolidline(seethelinelabeled1).Sincetheoutput
doesn’tchangeuntilthenextsamplecomesin,drawtheoutputasconstantoverthe
timebeforethenextsamplearrives.Again,IshowthisinFigure4.32(seetheline
labeled2).
Now,atsampletime1,anewinput
x(t)
samplearrives;wecandeterminethe
valueofthissamplebylookingatthe
x
n
dottedlineofFigure4.32atsample
time1.Atthissampletime,thepre-
dictedvalue(whichequalstheprevious
outputvalue)isδ.Becauseinthiscase
theinputvalueisbiggerthanthe
predictedvalue,increasetheoutputby
δ.Thisisshowninthesolidlineof
Figure4.32(seethelinelabeled3).
2
Sincetheoutputvaluedoesn’tchange
1
untilthenextinputsamplearrives,
0 1 2 3 4 5
3
4
6
t
n
drawtheoutputasconstantoverthat
time,asseeninFigure4.32(seethe (b)
linelabeled4).
Figure4.32 Graphicalevaluationof
input-outputrelationshipofDM
104  ◆  Chapter Four 
Continuingtodothisforallothersampletimesleadstotheinput-outputrelation-
shipseeninFigure4.32.Weseefromthisfigurethattheoutput x followstheinput
n
x(t) withastaircase-type
waveform,increasingand
x(t)
decreasingbyδtotrytokeep
upwiththechangesinx(t). 
3.1
Example 4.6 
Determinetheoutputofa
deltamodulatorwitha=1 
andδ=1,whentheinput
2.1
showninFigureE4.8is
sampledatarateofone
samplepersecond.
1.1
Solution:Todeter- 
minetheoutput,we 
returntotheprophetic 
0.1
wordsutteredinSection
0 1 2 3 4 5
t
4.4.5:“iftheinputvalueis
largerthanthepredicted
FigureE4.8InputtotheDM 
value,increasetheoutput 
ˆ
byδ ;iftheinputvalueis
x(t)
smallerthanthepredicted
ˆ
x(t)
value,thendecreasethe
outputbyδ.”FigureE4.9
showsyouhowtoapply
theruleandtheoutput
thatpopsoutwhenyou
do.
input
value
t
1
0.1
input
x
2 3 4 5
output
predicted
value
predicted
value
value
FigureE4.9OutputoftheDM
Source Coding and Decoding: Making it Digital  ◆  105 
4.4.6 OverloadandGranularNoise
Telecommunicationengineerssawsomeinterestingthingsgoingonintheinput–output
relationshipofaDM,andtheymadeupsomewordstodescribewhattheysaw.Thefirst
interestingthingstheysawwereat
x(t)
timeswhentheinputishardly
changing.Let’sconsideraninput
x
n
that’snearconstantatthevalue0.5.
We’llassumethattheoutputis
initially0,andwe’llassumed=1,as
showninFigure4.33(a).What’s
happeningisthis:attime0,the
1
outputincreasestod=1;thenat
time1,theoutputdecreasesbackto
...
0;then,atthenexttimetheoutput
0.5
increasesbackupto1.Ultimately,
...
wehavetheoutputjumpingback
t
andforthbetween0and1whilethe
0 1 2 3 4 5
n
inputisconstantatabout0.5.(You
(a)
cancheckthisoutyourselfusing
eitheroftheshortcutsdescribedin
theprevioussection.)Telecommuni-
cationsengineerscalledthis
x(t)
phenomena—theoutputjumping
x
n betweentwovalueswhentheinput
...
issmall—granular noise. 
...
Thenextinterestingthing
happenedattimeswhentheinput
changedveryquickly.Takealook,
forexample,attheinputofFigure
4.33(b).Here,theoutputincreases
by+ δateachtime. (Again,youcan
checkthatthisiswhattheoutput
doesbyusingoneoftheshortcutsI
toldyouaboutearlier.)Butyou’ll
noticefromFigure4.33bthateven
withtheoutputincreasingby+δat
t
n eachsampletime,itstillfallsfurther
0 1 2 3 4 5
andfurtherbehindtheinput,andit’s
(b)
unabletokeeppacewiththequickly
Figure4.33
changinginput.Telecommunication
(a)Granularnoise;(b)Overloadnoise
engineerslabeledthisphenomena
overload noise. 
106  ◆  Chapter Four 
x(t)
x
n
x
n
t
x(t)
n
0.5
...
...
t
0 1 2 3
n
...
...
0 1 2 3 4 5 6
(a) (b)
Figure4.34 (a)Demonstratingtheeffectofdecreasingδ δδ δδ ongranularnoise
(b)Demonstratingtheeffectofincreasingδ δδ δδonoverloadnoise
Becausegranularnoiseandoverloadnoisewereexperiencestelecommunication
engineerswantedtoavoid,theyfoundthemselvescontemplatingthefollowing:“We’re
freetochangeδ;howcanwechangeδtodecreasegranularnoise?And,howcanwe
changeδtogetridofthatunwantedoverloadnoise?”
Let’sfirstconsiderthepossibilityofchangingδtodecreasetheamountofgranu-
larnoiseaDMexperiences.Granularnoiseoccurswhentheoutputoscillatesabove
andbelowaslowlychanginginput,asinFigure4.33(a).Let’sseewhathappensifwe
decreaseδ(Figure4.34(a)).Weseeinthiscasethattheamountofoscillatingabove
andbelowtheinputhasdecreasedsignificantly;so,decreasingδdecreasesthegranu-
larnoise.
Let’snextconsiderthecaseofoverloadnoise,aneventwhichoccurswhena
quicklychanginginputcan’tbefollowedbytheoutput,asshowninFigure4.33(b).
Let’sseewhathappensinthiscasewithanincreaseinδ.Figure4.34(b)showsusjust
this.Increasingtheδallowstheoutputtofollowtheinputmoreclosely,decreasingthe
amountofoverloadnoise. 
So,decreasingδdecreasesthegranularnoise,whileincreasingδhelpscutdown
ontheoverloadnoise.Giventhat,howdotelecommunicationengineerschooseδ?
Typically,theydooneoftwothings.Thepickavalueofδthattheyfindcreatesa
compromisebetweentheamountofgranularnoiseandtheamountofoverloadnoise;
or,ifthey’regettingfancy,thevalueofδisupdatedwhiletheDMisrunning,increas-
ingattimesofoverloadnoiseanddecreasingattimesofgranularnoise.
Source Coding and Decoding: Making it Digital  ◆  107 
4.4.7 DifferentialPCM(DPCM)
I’vetalkedatlengthaboutonetypeofpredictivecoder,theDM,andyougotthroughit
all.Waytogo.Now,I’mgoingtotalk(but,thankgoodness,notasmuch)aboutasecond
typeofpredictivecoder,calledthedifferentialPCM,orDPCMforshort.Beingapredic-
tivecoder,theDPCMisbuiltasshowninFigure4.27.WhatmakestheDPCMdifferent
fromtheDMisthepredictedvalue, x
P
,explainedindetailinthenextsection.
n
The Predicted Value 
Thepredictedvalue x
P
usedintheDPCMcameaboutlikethis.Atelecommunica-
n
tionsengineeronedayhadtheideatousethepreviousKsamplestocomeupwitha
predictedvalue. Specifically,herecommendedthatwegetthepredictedvalueaccord-
ingtotheequation
K
P
x
∑
a x
−
(4.54)
n k n k
k1
x
n
P
a x
n−1
+ a x
n−2
+...+a x
n K
(4.55)
1 2 K −
Now,thisengineerhadtwoproblemstoworkout.First,asyou’llrecall(andifyou
don’t,aquickpeekatFigures4.27and4.28willworktoremindyou),thevalue x
P
is
n
neededatboththepredictivecoderandthepredictivedecoder.But,thevaluesof x
n−1
,
..., x usedinthe x
P
equationareonlyavailableatthepredictivecoder(andaren’t
n N n −
availabletothepredictivedecoder).
Ourengineer,lookingatthedecoderofFigure4.28,asked,“Whatinformationis
thereatthedecoderthatcanbeusedtocreateapredictedvalue x
P
?”Heansweredby
n
realizingthat,while x
−1
,...,x
−
isn’tavailablethere, x
n−1
,..., x
n K
canbefoundthere.
n n K −
“I’lljustchangemyequationslightlytothis,”hesaidandwrote
K
P
x
∑
a x
−
(4.56)
n k n k
k1
x
n
P
a x
n−1
+ a x +...+a x
n K
(4.57)
1 2 n−2 K −
Thefinalquestionthatremainstobeansweredisthis:Whatdoweuseforthe
values a
1
, a
2
,upto a
K
?Thistakesabitofmath.Tostart,we’lldecidewewantto
choose a
1
, a
2
,..., a
K
suchthat x
P
isascloseaspossibleto x
n
;specifically,we’ll
n
decidewewanttochoose a a
2
,..., a suchthat
1
,
K
1
, [(
P
a a
2
,..., a arg min E x − x )
2
] (4.58)
N n n
a a , ,..., a
1 2 N
whereargminmeans“thevalueofa
1
,a
2
,...,a thatminimizes.”
N
a
1
,a
2
,...,a
N 
⋅
⋅
⋅
⋅
⋅
⋅
⋅ ⋅
⋅ ⋅
⋅ ⋅
4 ( )
4
=
=
4 4 4
4 4 4
N
N
N N N
N N N




108  ◆  Chapter Four 
Substitutinginthe x
P
equationin(4.56)leadsto
n
K
arg min E x [(
n ∑
− a x
k n k −
2
) ] (4.59) a a a
1
, ,...,
N 2
a a ,
1 2
a
k
,...,
N 1
iscloseto x
n k −
,wecanrewritethisas

Assumingthat
n k −
x
K
∑
− a x
k n k −
)
2
] (4.60) arg min E x [( a a a
1
, ,...,
N 2 n
a a ,
1 2
a
k 1
Fromnowon,wewillusemathematicstocomeupwithamoreexplicitequation
forthe a
1
to a
K
values.We’llstartbytryingtofindthebest a
1
,bytakingthederiva-
tiveoftheabovetermwithrespectto a
1
andsettingitto0.Thisleadsto
K

,...,
N
∑
− E[2 ( ) ( )] 0 (4.61)  − x a x
k n k −
x ⋅ ⋅
1
1
Andnowwejustdosomesimplemathtocomeupwith
K
−

n n
k
1 ∑
−
−
− ⋅ E x x 2 [
n n
]
1 −
0 (4.62) a x x
k n k n −

k 1
K
∑
− E x x [
n n−
a E x x [
n k −
] ] 0 (4.63) 
k 1 1
1
−
Usingcorrelationnotation,wecanrewritethisaccordingto
K

n
k
∑
− R
x
a R
k x
k) 0 (4.64) ( ) 1 (1−
k 1
K
∑
a R
k x
− k) R
x
(1 (1) (4.65)
k 1
That’stheequationwegetwhenwetrytogetthebest a
1
.Whenwerepeatthisto
getanequationforthebest a
2
upto a
K
,wegetatotalofKequations.Puttingthese

equationstogetherbyusingsimplematrixnotation,wegetthefollowing:
,
,
,
,
,
,

(
(
(
(
(
(
,
,
,
,
,
,
,
(
\
(
(
(
(
(
(
,
(

⋅
⋅
⋅
( )
( ) (
( ) − ( ) ( ) − j
,
,
,
,
,
,
j \
(
(
(
(
(
(
j \ …
(4.66)
) )
,
− −
( (

N

N N

N

N N
N
Source Coding and Decoding: Making it Digital  ◆  109 
R R K) ,yougetthebest So,ifsomeonegivesyouthestatistics R (0), (1),..., (
x x x
a a
2
,..., a bysimplysolvingtheabovematrix.
1
,
K
Example 4.7 
Figureoutthebestchoiceforthevaluesofa anda
2
ina2-tappredictorwhenthe
1
inputhasacorrelationfunctiongivenby
4 ( ) (E4.25)
4 ( ) 4 (− ) # (E4.26)
4 ( ) # (E4.27)
Solution:Weturntoequation(4.66),anduseK=2,whichleadsusto
j 4
N
( ) 4 (− )\ j =

\ j 4
N
( )\
( , ( , (
(
,
,
4 ( )
,
(
(E4.28)
,

(
4 ( ) 4 ( )
(
,
,
(
=
, ( N

j # \ j =

\ j # \
,
,
( (
(
(E4.29)
(
#
(
(
,
,
(
,
=
,
(
(
,
,
#
,
j =

\ j # \
−
j # \
,
(
(
,
,
(
(
,
,
(
(
(E4.30)
,
=
, (
#
, (
#
, (
j =

\ " j − # \ j # \
, (
(

, (
(
,
,
(
(
(E4.31)
(
=
,
!
(
,
,
− #
, (
#
,
"
j ! \

,
,
&
(
(
(E4.32)
!
(

,
j \

,
,

(
(
(E4.33)
(

,
The Block Diagram 
ThissectionprovidesyouwithablockdiagramoftheDPCM,anditssourcedecoder,
sothatyoucangetanunderstandingofhowpeoplebuildthisdevice.Figure4.35
showsthatblockdiagram.ThesolidlineshowshowtheDPCManditsdecoder
maintainthegeneralpredictivecoder–decoderstructureofFigures4.27and4.28,and
thedashedlineshowsyouhowthepredictedvalueisgenerated.Here,theblock
calledN-tappredictorreceives x
n
asinputandoutputs
110  ◆  Chapter Four 
K
P
n

∑
a x
k n k
. (4.67)
−
k1
It’snotimmediatelyapparentfromFigure4.35justhowthepredictedvalueof
equation(4.67)isbeinggeneratedatthesourcecoder.Thesourcecoderiscreating

intwosteps:(1)first,itcreatesthevalue
n
byadding x

E
P
n
x
x
P
n
P
n
;then(2)it x to
n
createsthepredictorvalue x usingthis .
n
Quantizer
Symbol-to-bit
Mapper
+
+
+
+
E
n
E
n
E
n
sampler
∧
∧
Bit-to-symbol
Mapper
LPF
channel,
...
+
+
x
n
x
n
x
n
x
n
p
–
x
n
p
modulator,
demodulator
K-tapPredictor
K-tapPredictor
∧
∧
x(t)
Figure4.35 DPCManditssourcedecoder
Finally,thereareacoupleofdetailsregardingthesourcecoderthatIwantto
brieflytouchon.ThesearedetailsaboutthesamplerandthequantizerintheDPCMof
Figure4.35.First,there’sthequestionofhowfastthesamplersamples.Inthreewords,
notveryfast(whencomparedtotheDM).TypicallyinDPCM,thesamplingisatthe
NyquistrateorperhapstwotimestheNyquistrate;thatis,thesamplerissamplingat
twicethemaximumfrequencyoftheinput,x(t),orfourtimesthatmaximumfrequency.
Thenthereisaquestionaboutthequantizer—howmanyallowedoutputlevelsNdoesit
use?Typically,thequantizeroperateswith8,16,or32allowedoutputlevels.
4.5 CongratsandConclusion
Congrats!Theendofa(ratherlengthy)chapter.Torecap,andI’llbebrief,welearned
indetailabouttwopartsofthesourcecoder,thesamplerandthequantizer.First,we
learnedaboutthreetypesofsamplersandsawthecoolestofresults:aslongasyou
sampleat(atleast)twotimesthemaximumfrequency,youcangetyouroriginalsignal
backfromyoursamples.Thenwelearnedaboutquantizers,afancywordforan
“amplitudechanger”—itmapstheinputamplitudetooneofNallowedoutputampli-
tudes.Youalsosawtwowaystobuildaquantizerthatminimizetheaverageerror
betweeninputandoutput.Notonlythat,butyousawthequantizermostcommonly
usedintelephonecommunications.
Thenwedecidedtogetadventurous,andweputthesamplerandquantizerto-
getherandbuiltasourcecodercalledaPCM.Next,weconsideredadifferentsource
codercalledthepredictivecoder.ItlookedalotlikethePCM,theonlydifferencebeing
that,beforeyougottothequantizer,youremovedapredictedvaluefromyoursample.
Wetalkedatlengthabouttwodifferenttypesofpredictivecoders,theDMandthe
DPCM,andfinally,youcameheretothesummary,wherewewrappeditallup.
Ifyoufeelyouwantmoreonthismaterial,andbeassuredtherearebooksand
booksonthisstuff,havealookatthereferences.
x

( J

Source Coding and Decoding: Making it Digital  ◆  111 
Problems
1. Ifyouknowthatthesignalx(t)iscompletelycharacterizedbythesamples
takenatarateof5,000Hz,what(ifanything)canyousayaboutX(f)?
2. DeterminetheNyquistsamplingrateforthefollowingsignals
( )
IE(" π J )
J N
(a)
π J
(Q4.1)
( )
IE (" π J )
(b)
J N

(Q4.2)
π J
3. Considerthesignal
J O ) N ( )∗ N ( ) J (Q4.3)
where
1
( ) 0, X f f >1000 0 (Q4.4)
2
( ) 0, X f f > 2000 0 (Q4.5)
Whatistheminimumsamplingperiodthatensuresthaty(t)iscompletelyrecov-
erablefromitssamples?
4. Assumethesignalx(t)hasthefrequencyrepresentationshowninFigureQ4.1.
(a) Whatdoestheoutputofanidealsamplerlooklikeinthefrequency 
domain? 
(b)WhatistheminimumsamplingratethatIcanuseandstillrecovermy
signalfromitssamples?
X(f)
–f
1
–∆f –f
1
–f
1
+∆f f
1
–∆f f
1
f
1
+∆f
FigureQ4.1Theinput
f
112  ◆  Chapter Four 
5. Considerzero-orderholdsampling.
(a) Iftheinputx(t)hasaFouriertransformshowninFigureQ4.2,what 
doestheoutputwaveformlooklike(1)inthefrequencydomainand 
(2)inthetimedomain?AssumesamplingattheNyquistrate.
(b)Iftheinputx(t)hasaFouriertransformshowninFigureQ4.2,what 
doestheoutputwaveformlooklike(1)inthefrequencydomainand 
(2)inthetimedomain?AssumesamplingatTWICEtheNyquistrate.
x(f)
FigureQ4.2 Theinput
f
–f
m
f
m
6. Plottheoutputofasamplerinfrequencygiven:
• Theinputsignalhasmaximumfrequency5,300Hz.
• Idealsamplingisused.
• Samplingisatarateof5,300Hz.
• Theinputsignalinthefrequencydomainistriangular(i.e.,itisamaximum
at0Hzanddegradesto0linearlyasfrequencyincreasesto5,300Hz(and
to–5,300Hz).
7. ConsideraquantizerwithaninputdescribedinFigureQ4.3.
(a) Drawaquantizerwith7levels.Makeitmid-tread,letithave–3asits 
smallestoutputvalue,andmakesurethatthestepsizeis1. 
(b)Evaluatethemseofyourquantizergiventheinput.
(c) EvaluatetheSQNR.
(d)Iftheinputtothequantizerhasanamplitudewithaprobabilitydistribution
uniformbetween–3.5and+3.5,whatistheSQNRofthequantizer?
Source Coding and Decoding: Making it Digital  ◆  113 
1
/
4
1
/
6
–3 2
p(x)
FigureQ4.3 Theinputpdf
x
8. FindouthowmanylevelsaquantizermustusetoachieveanSQNRgreater
than30dBgiven:
• TheincomingaudiosignalissampledatitsNyquistrateof8,000samples/
sec.
• Theamplitudesoutputfromthesamplerhaveauniformprobabilitydistri-
butionfunction.
• Auniformquantizerisused.
9. Determinetheoptimalcompressioncharacteristicfortheinputx whoseprob-
abilitydensityfunctionisprovidedinFigureQ4.4.
10. (a) Plottheµ =10compressorcharacteristicgiventhattheinputvaluesare
intherange[–2.5,2.5].
(b)Plotthecorrespondingexpander.
p(x)
FigureQ4.4 Theinputpdf
1
/
6
1
/
12
x 
–4 –2  2 4
114  ◆  Chapter Four 
11.EvaluatethesymbolrateandthebitrateofthePCMsystemdescribedbythe
following:
• Thesamplingrateis5,300Hz
• Thequantizerisan8-levelquantizer.
12.Acomputersends:
• 100lettersevery4seconds
• 8bitstorepresenteachletter
• thebitsenteraspecialcodingdevicethattakesinasetofbitsandputsout
oneof32possiblesymbols.
Whatisthebitrateandwhatisthesymbolrateoutofthespecialcodingdevice?
13.Overthetime0sto2s,determine(1)theinputtotheDM,(2)theoutputof
theDM,and(3)thetimesofgranularandoverloadnoisegiven:
• TheinputtotheDMisx(t)=t
2
• Thesampleroffers10samples/second
• ThestepsizeoftheDMis0.1V
14. (a)DrawtheoutputoftheDMgiven:
• Theinputcorrespondstox(t)=1.1t+0.05
• Theinputissampledattimest=0,1,2,3,4
• Thestepsizeis1anda =1.
(b)Repeat(a),thistimeusinga =0.5.
15.Atwo-tappredictivefilterisbeingdesignedtooperateinaDPCMsystem.The
predictorisoftheform
(Q4.6) N

2
=

N
−
+ =

N
−
(a) Provideanexplicitequationfortheoptimalselectionofa anda
2
(interms
1
ofautocorrelationfunctions)whichminimizesthemeansquaredprediction
error.
(b)Provideageneralequationforthemeansquaredpredictionerrorusing
thevaluesdeterminedin(a).
(c) Determinethevaluesofthepredictortapsin(a)andthepredictionerror
in(b)given:
¦ n
( )
¦
¦
1 − , n 0,1, 2, 3
3 R n
x
(Q4.7)
¦
0 , else
¦
5 
Chapter 
Getting It from Here to There:
Modulators and Demodulators 
I
nmanyways,thisisthemostimportantchapterofthebook,becausethere’dbeno
telecommunicatingwithoutthedevicesdescribedinthischapter.
5.1AnIntroduction
Thischapterisreallyabouttwosimplethings:themodulator,anditsopposite,the
demodulator.Thebestwaytoexplainthemistoimagineyourselfstuckwithaparticu-
larcommunicationproblem.Let’ssayafteryoureadChapter4,yougotexcitedand
builtasourcecoder—aPCM—whichturnsyourvoicesignalintoadigitalbitstream
(Figure5.1).YouthencalleduptheFCC(FederalCommunicationCommission)and
toldthemyouwanttouseyoursourcecodertosendadigitalvoicemessagetoa
friend.Theyrespond,“Sure,justbesureyousenditovertheFamilyRadioService
band,whichis462.5625–462.7125MHz.”
Youbuiltthis!
Chapter5
PCM
Channel
Thestuffof
thischapter
Alsothestuffof
mustlieinthe
462.5625MHzto
462.7125MHz
Bits
...10110...
Bits
...10110...
inputspeech
SourceCoder
Modulator
Sourcedecoder
forDCM
Demodulator
Youbuiltthis!
FCCsaysyoursignal
range
Figure5.1Introducingthemodulatoranddemodulator
116  ◆  Chapter Five 
Youthink,“HowdoIdothat?”
Theanswerisfoundinthischapter.Amodulator isadevicethatturnsyourdigital
bitstreamintoasignalthatisreadytobesentoverthecommunicationchannel.
Generallyspeaking,you’lltransmityourinformationsignaloverachannel(forex-
ample,copperwire,coaxialcable,orasanEMwavethroughtheatmosphere).That
channelwillonlyallowcertainfrequenciestomakeittothereceiver(thatis,itwillact
asabandpassfilter(BPF)).Themodulator’sjobistoturntheinformationbitsintoa
waveformthatcanmakeittothereceiver.
Ibetyoucanguesswhatademodulatordoes.Itsitsatthereceiversideandturns
theincomingwaveformcreatedbythemodulatorbacktotheoriginalbitstream.
5.2 Modulators
Therearetwotypesofmodulators:basebandmodulatorsandbandpass
modulators.
5.2.1BasebandModulators
Basebandmodulatorsaredevicesthatturnyourbitstreamintoawaveformcentered
around0Hz(Figure5.2).You’dusethistypewhenyourchannelallowsfrequencies
around0Hztogetsafelytothereceiver.Therearemanybasebandmodulators,each
onecreatingitsownuniquewaveformaround0Hztosendacrossthechannel.
Bits
s(t)
s(t)
t
Baseband
modulator
...10110... S(f)
f
0Hz
Figure5.2Abasebandmodulator
Getting It from Here to There: Modulators and Demodulators  ◆  117 
NRZ Modulators 
OnekindofbasebandmodulatoriscalledanNRZmodulator,shortfornon-return-to-
zero.Asyoumayhavealreadyguessedfromthename,withanNRZmodulatorthe
waveformcreatedneverreturnsto0(zero)volts.Let’slookatacoupleofmodulators
fromthisfamily.Thefirst-born,whichwentontogreatsuccessinmanyacademicand
businesscircles,wasthepopularNRZ-L.YoucanseewhattheNRZ-Lmodulatordoes
toa0andtoa1inFigure5.3.A0staysat+Vvoltsforthebitduration T
b
,anda1stays
at–Vforabitduration T
b
.NotnearlyaspopularistheNRZ-M(theblacksheepofthe
family,Isuspect).Here,wesendeither+Vvoltsforabittime T
b
,or–Vvoltsforbit
time T
b
.Ifthebitisa0,thesignalleveldoesn’tchange.Ifthebitisa1,thesignallevel
changes(e.g.,from+Vto–V).Figure5.4showsyouwhatImean.
1
mappedto mappedto
+V
-V
V
V
t t
T
b
T
b
(a)
0
in 1 0 1 1 0 0
+V
-V
out
t
(b)
Figure5.3NRZ-L
118  ◆  Chapter Five 
+V
-V
V
V
t t
T
b
T
b
Sendeither
or
1
mappedto
changeinsignallevel
nochangeinsignallevel 0
mappedto
(a)
+V
-V
out
in
1 0 1 1 0 0
(b)
Figure5.4 NRZ-Mdescribed
RZ Modulators 
LeavingtheNRZfamily,wecometoitswell-knownrivals,theRZs.Asyou’lllikely
guess,RZisshortforreturn-to-zero.Thesemodulatorsmakesurethat,foratleast
someofthetime,thetransmittedsignalsitsat0.FirstisunipolarRZ.Figure5.5shows
youwhathappenstoa1andwhathappensto0afteritpassesthroughthistypeofRZ
modulator.Next,there’sbipolarRZ,anotherstraightforwardmodulatorwhoseoutput
waveformisshowninFigure5.6.Andfinally,we’vegotRZ-AMI(alternativemark
inversion),inwhicha1correspondstoanalternatingsymbol,anda0issentas0
(Figure5.7).
1
Getting It from Here to There: Modulators and Demodulators  ◆  119 
+V
V
V
t
t
0 1
T
b
/2 T
b
/2
T
b
(a)
+V
0
out
in
1 0 1 1 0 0
(b)
Figure5.5 UnipolarRZ
V V
+V
0
T
b
/2T
b
/2
t t
T
b
/2T
b
/2
(a)
-V
+V
0
out
in
1 0 1 1 0 0
(b)
-V
Figure5.6 BipolarRZ
120  ◆  Chapter Five 
V V 
+V
1 or
(alternates)
T
b
/2 T
b
/2
t
t
T
b
/2 T
b
/2 
V
-V 
+V
0
t
T
b
(a)
+V
0
out
in
1 0 1 1 0 0
-V
(b)
Figure5.7 RZ-AMI
Phase-encoded Modulators 
Finally,wecometothemodulatorfamilyknownasthephase-encodedgroup.Among
themostpopularistheManchesterCodingmodulator,inwhichbit1ismappedtoa
waveformthatstartsat+Vandendsupat0,andwherebit0becomesthewaveform
startingoutat0andendingat+V(Figure5.8).Anotherpopularmemberofthephase-
encodedgroupistheMillerCodingmodulator.Thisonebelievesinchoice,soitlets
bit0becomeoneoftwowaveformsandalsoletsbit1beoneoftwopossiblewave-
forms(Figure5.9).Thedecisionastowhichofthetwopossiblewaveformstooutput
isbasedonthissimplerule:alwaysmakesurethatthere’satransition(e.g.,from+V
to–V)betweenbits(Figure5.9).
Getting It from Here to There: Modulators and Demodulators  ◆  121 
+V
V
V
t
(a)
1 0
-V
+V
T
b b
/2
T
b b
/2
/2T
/2T
Figure5.8
Manchester Coding
t
out
in
1 0 1 1 0 0
(b)
V V
1
Figure5.9
MillerCoding
0
+V
-V
V V
t t
t t
T
b
T
b
or
+V +V
-V -V
or
T
b
/2 T
b
/2 T
b
/2 T
b
/2
(a)
+V
0
out
in 1 0 1 1 0 0
-V
Alwaysa+Vto-Vor-Vto+V
transitionateachnewbit
(b)
122  ◆  Chapter Five 
Which Modulator to Use? 
Choicecanbeawonderfulthing,butitcanalsobeoverwhelming.I’vejustgivenyou
sevenpossiblechoicesforyourbasebandmodulator,butifyouwanttobuildacommu-
nicationsystem,you’llhavetochoosejustoneofthem.Aswithallthings,whatyou
choosedependsonwhatyouwant.Therearesixthingsmostpeoplewantfromtheir
basebandmodulator.Ultimately,you’llhavetodecidewhichmodulatorisbestsuited
foryourcommunicationneeds.
(1)No DC component:Insomecommunicationsystems,verylow-frequency
components(thatis,frequenciesaround0Hz)don’tmakeittothereceiver.In
thesecases,wewanttousemodulatorsthatoutputwaveformswithNOfre-
quencycomponentrightat0Hz(Figure5.10).Suchmodulatorsaresaidtohave
noDCcomponent.
Channel Modulator Receiver
Frequencies
Avoidsendingasignal
withacomponentat0Hz
around0Hz
don'tmakeit
Figure5.10ModulatorswithnoDCcomponents
It’seasytotellifamodulatorcreatesawaveformwithacomponentat0Hz.Just
lookatwhathappensifyousendasequenceof0101010101...Iftheaverageis0,then
there’snoDCcomponent;iftheaverageisnot0,thenthereisaDCcomponent.For
example,unipolarRZhasaDCcomponent,NRZ-Ldoesnot.
(2)Self-Clocking: Inmostcommunication
+V 
systems,wewanttohelpthereceiverout. 
...
Onewaytodothisistomakeiteasyforthe
0
...
receivertocreateaclocksignalwithdura-
T
b
T
b
tion T
b
(Figure5.11).Themodulatorcan
helpthereceivercreateaclocksignalby Figure5.11
sendingawaveformthatalwaystransitions Clocksignalwithduration
(forexample,+Vto0or+Vto–V)once T wantedatthereceiver
b
everybitduration T
b
.Suchamodulator
helpsthereceivercreateaclockandpeople(andreceivers)appreciateit.For
example,theManchestercodehelpswithself-clocking,buttheNRZ-Ldoesn’t
(considerwhathappensifwesendall1’s—therearenotransitions).
(3)Error Detection: Receiversappreciateitifyoucanhelpthemdetectanerrorin
transmission.Modulatorscanhelpreceiversdetecterrorsbysendingawaveform
Getting It from Here to There: Modulators and Demodulators  ◆  123 
wheresomewaveshapesare
Receiver
picksup:
notallowed.Apicturewillhelp,
T
b
sotakealookatFigure5.12. 
There,anRZ-AMImodulatoris
if1sent 
used,andthereceiverseesthe
Impossible:mustbeeither 
if0sent
waveformshowninFigure5.12.
Youknowfromhavinglookedat
Figure5.12ErroneousrecemptioninRZ-AMI
theRZ-AMImodulatorthatthe
waveforminFigure5.12couldnotbetheonesent,sothereceivercaneasilyspot
atransmissionerror.
(4) BW compression: Inmostcommunicationsystems,you’dliketosendyour
signalwithassmallabandwidthaspossible.TakealookatFigure5.13(a).For
everysignalsentofduration T
b
,theBWisproportionalto1/T
b
.So,consider
NRZ-Landbipolar-RZ(Figure5.13(b),(c)).InNRZ-L,thewaveformistwiceas
longasinbipolar-RZ,sothebandwidthofNRZ-Lishalfthatofbipolar-RZ.
s(t) S(f)
s(t)
s(t)
S(f)
S(f)
t
t
t
T
T
b
-1/T
-1/T
b
1/T
1/T
b
NRZ-Lsignal
(a)
(b)
Mostofthefrequency
componentsarein
thisrange
f
1
f
1
f
T
b
/2 -2/T
b
2/T
b
BipolarRZ
Figure5.13
(c)
Bandwidth considerations
124  ◆  Chapter Five 
(5) Inversion Insensitive: Mostpeopledon’tlikeinsensitive,butinthiscase,
insensitiveisagoodthing.Sometimes,whenyousendyourwaveformacrossthe
channel,itgetsturnedupside-downalongtheway(+Vbecomes–Vand–V
becomes+V).Inversioninsensitivemeansthatevenwheneverythingisturned
upside-down,thereceivercanstillfigureout(correctly)whichbitisa0and
whichisa1.Forexample,NRZ-Lisnotinversioninsensitive,becauseifthings
getturnedupside-down,thewaveformforbit0becomesthewaveformforbit1,
andthereceiverwillmakemistakes.
(6) Noise immunity: Thisshouldcomeasnosurprise—peoplewantmodulators
thatarerelativelyimmunetonoiseonthechannel.Evenwhenthechanneladds
noise,itshouldstillbeeasyforthereceivertotellthedifferencebetweenthe
waveformforbit0andthewaveformforbit1.Forexample,NRZ-Lisconsidered
morenoiseimmunethanunipolar-RZbecause,simplyput,thewaveformsin
NRZ-Laremore“different,”soittakesmoreaddednoisetocreateanerror.
Thereyouhaveit—thesixthingspeoplelookforinbasebandmodulators.The
finalchoiceisbasedonwhichofthesecriteriaismostimportanttoyou.
Example 5.1 
NameabasebandmodulationschemewhichprovidesbothzeroDCcomponent
andlowbandwidth.
Solution:Let’sstartbytakingalookatNRZ-M.
InNRZ-M,a1issentasa+Vforthebitdurationanda0issentasa–Vfor
thebitduration.Onaverage,whenanequalnumberof0’sand1’saresent,an
equalnumberof+V’sand–V’saresent.Asaresult,theaveragevalueofthe
signalsentis0.TheNRZ-MhasazeroDCcomponent.
InNRZ-M,thesignalsent(when1or0isinput)isconstantfortheentire
durationT.Asaresult,thissignalhasasmallbandwidthasdiscussedearlier.
Hence,NRZ-Mmeetsbothourrequirements:ithasnoDCcomponentandit
employsasmallbandwidth.
5.2.2 BandpassModulators
Abandpassmodulatortakesincomingbitsandoutputsawaveformcenteredaround
frequency ω
c
.Thisisthemodulatoryouwanttousewhenyourcommunication
channelwillprovidesafepassagetofrequenciesaround ω
c
(thatis,whenthechannel
letsfrequenciesaround ω
c
gettothereceiverwithlittledistortion).Basically,these
modulatorsareprettystraightforward.Themodulatorcreatesthewaveform
s t t + ( ) Acos(ω θ) (5.1)
Getting It from Here to There: Modulators and Demodulators  ◆  125 
where ω isafrequencyatornear ω .Themodulatorstorestheinformationbitsin
c
eithertheamplitude(A),thefrequency(ω ),orthephase(θ ).Forexample,bit1may
t + t + besentas +Acos(ω θ) andbit0maybesentas −Acos(ω θ) .Inthiscasethe
informationisintheamplitude—thereceivercheckstoseeifithasa+Aor–Ato
figureoutwhatbitwassent.
Now,let’stalkinmoredetailabouthowbandpassmodulatorswork.
ASK 
Thefirstbandpassmodulatorswe’lllookatarecalledASKmodulators,shortfor
amplitudeshift-keyingmodulators.Thisreferstothemodulatorsthat,giventheinput
bits,createthewaveform s t t + ( ) Acos(ω θ) ,wheretheinputbitsarestuffedinthe
amplitude(A).We’llstartwiththesimplestoftheASKmodulators,calledBinaryASK
orB-ASKforshort.Figure5.14showsyouwhathappenswhenbits010arriveattheB-
ASKmodulator.Asyoucansee,wheneverabit0isinput,themodulatorpopsout
−Acos(ω t ) forthebitduration.Wheneverabit1arrives,themodulatorthrowsout
c
+Acos(ω t ) forthebitduration.TakeapeekatTable5.1forasummaryofhowB-
c
ASKworks.Inthistable,thetimesiT to(i+1)Trefertothetimedurationofthe
incomingbit.
Inputbits
1 0 0
Output 
waveform 
-Acosω
c
t +Acosω
c
t -Acosω
c
t
0
T 2 T
3T
Figure5.14B-ASKmodulator
O p t u t u w r o f e v a m
n I p t u bi s t O p t u t u w r o f e v a m
s ( h h t r o and r o f m)
0 s = ) t ( –A s o c ω , t  iT≤ ( < t   i T ) 1 + – s o c  A ω t · π t (  –iT)
0 c  c 
1  s
1
( s o c  A + = ) t ω , t  iT≤ ( < t   i T ) 1 + s o c  A +  ω t · π (t–i ) T
c c 
π(t-iT)
1
t
Table5.1BASK
iT (i+1)T
126  ◆  Chapter Five 
1 0 1 1 Inputbits
Output
waveform
Acosω
c
t 3Acosω
c
t
T
Figure5.154-ASKmodulator
Nextup:4-ASK.Here,welettwobitsenterthemodulatoratthesametime.
Ratherthansimplymapbit0tooneamplitudeandbit1toanotheramplitude,welet
themodulatorgrabtwoinformationbitsatatime,sothatthemodulatorinputisnow
either00,01,10,or11.The4-ASKmodulatormapseachsetoftwobitstoawaveform
withadifferentamplitude.Sometimespicturesareeasier,soglanceatFigure5.15.
Hereweseetheinputbitsare1011.Thatfirstbitpair10comesintothemodulator,
whichpopsouttheoutputwaveformwithamplitudeA.Thenbitpair11jumpsintothe
modulator,andthemodulatorrespondsbythrowingouttheoutputwaveformwith
amplitude3A.Table5.2providesasummaryof4-ASK.Ontheleft,youcanseethe
possibletwobitsthatcanenterintothemodulator.Ontherightaretheoutputwave-
forms,whicharedifferentforeachpairoftwoinputbits.Thedifferentmodulator
outputsdifferonlyinamplitude.There’sonethingtopointoutinthistableaboutthe
timesiTto(i+1)T,thetimedurationoftheoutputwaveform.Theoutputofamodula-
torlastsaslongastheinputtothemodulator.So,in4-ASK,theoutputofthe
modulatoristwobitdurationslong.You’llwanttonoticethateachwaveformcreatedin
a4-ASKmodulatorlaststwiceaslongasawaveformcreatedinaB-ASKmodulator.
That’sbecauseeachwaveformin4-ASKisrepresentingtwobits(eachofduration T
b
),
andsowilllastatotaltimeofT = 2 T
b
;meanwhile,eachwaveforminB-ASKisrepre-
sentingonlyonebit(ofduration T
b
),andsoitwilllastforatimeT =  T
b
.
s t i b t u p n I m r o f e v a w t u p t u O
m r o f e v a w t u p t u O
) m r o f d n a h t r o h s (
0 0  s
0
=  ) t (  – s o c  A 3  ω
c 
T i , t  ≤ T ) 1 + i ( < t  – s o c  A 3  ω
c 
t · π t (  – ) T i 
1 0  s
1
=  ) t (  – s o c  A  ω
c 
T i , t  ≤ T ) 1 + i ( < t  – s o c  A  ω
c 
t · π t (  – ) T i 
0 1  s
2
s o c  A  =  ) t (  ω
c 
T i , t  ≤ T ) 1 + i ( < t  s o c  A  ω
c 
t · π t (  – ) T i 
1 1  s
3
s o c  A 3  =  ) t (  ω
c 
T i , t  ≤ T ) 1 + i ( < t  s o c  A 3  ω
c 
t · π t (  – ) T i 
Table5.24-ASK
Getting It from Here to There: Modulators and Demodulators  ◆  127 
Somesay“sizematters”or“biggerisbetter.”Tothosewhodo,weofferyou8-
ASK.AsimpleextensionoftheideasofB-ASKand4-ASK,in8-ASKthreebitsareinput
tothemodulatoratthesametime,andthemodulatoroutputsoneofeightpossible
waveforms.Table5.3summarizeswhat’sgoingon.Asyoucanseehere,whenbits000
enterthemodulator,itoutputs −7 Acos(ω t ) ;ifthenext3bitsenteringthemodulator
c
are111,thenitoutputs +7 Acos(ω t ) .Andsoon.
c
t u p t u O m r o f e v a w
I b t u p n i s t t u p t u O m r o f e v a w
) m r o f d n a h t r o h s (
0 0 0  s =  ) t (  – s o c  A 7 ω , t  iT≤ ( < t   i T ) 1 + – s o c  A 7 ω t · π t ( –i ) T
0 c  c 
1 0 0  s =  ) t (  – s o c  A 5 ω , t  iT≤ ( < t   i T ) 1 + – s o c  A 5 ω t · π t ( –i ) T
1 c  c 
0 1 0  s =  ) t (  – s o c  A 3 ω , t  iT≤ ( < t   i T ) 1 + – s o c  A 3 ω t · π t ( –i ) T
2 c  c 
1 1 0   s =  ) t (  – s o c  A ω , t  iT≤ ( < t   i T ) 1 + – s o c  A ω t · π t ( –i ) T
3 c  c 
0 0 1   s
4
s o c  A  =  ) t (  ω , t  iT≤ ( < t   i T ) 1 + s o c  A  ω t · π t ( –i ) T
c c 
1 0 1  s
5
s o c  A 3  =  ) t (  ω , t  iT≤ ( < t   i T ) 1 + s o c  A 3  ω t · π t ( –i ) T
c c 
0 1 1   s
6
s o c  A 5  =  ) t (  ω , t  iT≤ ( < t   i T ) 1 + s o c  A 5  ω t · π t ( –i ) T
c c 
1 1 1   s
7
s o c  A 7  =  ) t (  ω , t  iT≤ ( < t   i T ) 1 + s o c  A 7  ω t · π t ( –i ) T
c c 
Table5.38-ASK
Ofcourse,thingsdon’tstopwith8-ASK.Youcouldmakea16-ASKmodulator,a
32-ASKmodulator,orevena5092-ASKmodulator.Thetablesjustgetbigger,butthe
ideastaysthesame.
PSK 
ThemostpopularofthebandpassmodulatorsarethePSKmodulators,shortforphase
shift-keyingmodulators.Withthese,inputbitsaremappedintooutputwaveformsof
theform s t t + ( ) Acos(ω θ) ,andtheinformationbitsarestuffedinthephase θ .
We’llstartwiththesimplestcasefirst,BPSK(binaryPSK).
InBPSK,whenbit0goesintothemodulator,themodulatorspitsoutthewave-
o o
form Acos(ω t + 0 ) .Ifbit1strutsintothemodulator,itpopsout Acos(ω t +180 ) .
c c
Figure5.16showsyouwhathappenswhenbits010strollintoaBPSKmodulator.
Table5.4(lookatthetoppartmarkedBPSK)summarizestheBPSKideainaneat
fashion.
128  ◆  Chapter Five 
0 1 0 Inputbits
Output
wav orm ef
Acos(ω t+0 °)
Acos(ω t+180 °)
Acos(ω
c
t+0 °)
c
c
Figure5.16BPSKmodulator
O t u p t u w r o f e v a m
I t u p n bi s t O t u p t u w r o f e v a m
n a h t r o h s ( d r o f m)
K S P B  0  s
0
(  s o c  A  =  ) t (  ω 0 + t  ° , ) iT≤ ( < t   i T ) 1 + (   s o c  A  ω 0 + t  °) · π t ( –i ) T
c c 
1  s
1
(  s o c  A  =  ) t (  ω 0 8 1 + t  ° , ) iT≤ ( < t   i T ) 1 + (   s o c  A  ω 0 8 1 + t  °) · π t ( –i ) T
c c 
K S P - 4  0 0  s
0
(  s o c  A  =  ) t (  ω 0 + t  ° , ) iT≤ ( < t   i T ) 1 + (   s o c  A  ω 0 + t  °) · π t ( –i ) T
c c 
1 0   s
1
(  s o c  A  =  ) t (  ω 0 9 + t  ° , ) iT≤ ( < t   i T ) 1 + (   s o c  A  ω 0 9 + t  °) · π t ( –i ) T
c c 
0 1  s
2
(  s o c  A  =  ) t (  ω 0 8 1 + t  ° , ) iT≤ ( < t   i T ) 1 + (   s o c  A  ω 0 8 1 + t  °) · π t ( –i ) T
c c 
1 1   s
3
) t ( =   (   s o c  A  ω t 0 7 2 +  ° , ) iT≤ ( < t  i T ) 1 + (  s o c  A  ω 0 7 2 + t  °) · π t ( –i ) T
c c 
K S P - 8   0 0 0   s
0
(  s o c  A  =  ) t (  ω 0 + t  ° , ) iT≤ ( < t  i T ) 1 +
c 
1 0 0  s
1
(  s o c  A  =  ) t (  ω 5 4 + t  ° , ) iT≤ ( < t  i T ) 1 +
c 
0 1 0  s
2
(  s o c  A  =  ) t (  ω 0 9 + t  ° , ) iT≤ ( < t  i T ) 1 +
c 
1 1 0   s
3
(  s o c  A  =  ) t (  ω 5 3 1 + t  ° , ) iT≤ ( < t  i T ) 1 +
c 
0 0 1  s
4
(  s o c  A  =  ) t (  ω 0 8 1 + t  ° , ) iT≤ ( < t  i T ) 1 +
c 
1 0 1  s
5
(  s o c  A  =  ) t (  ω 5 2 2 + t  ° , ) iT≤ ( < t  i T ) 1 +
c 
0 1 1  s
6
(  s o c  A  =  ) t (  ω 0 7 2 + t  ° , ) iT≤ ( < t  i T ) 1 +
c 
1 1 1  s
7
(  s o c  A  =  ) t (  ω 5 1 3 + t  ° , ) iT≤ ( < t  i T ) 1 +
c 
Table5.4PSKexplained
Getting It from Here to There: Modulators and Demodulators  ◆  129 
Nextupis4-PSK.Here,themodulatorworksontwobitsatatime.Foreverytwo
bitsthatcomeintothemodulator,themodulatoroutputsadifferentwaveform.Con-
siderthis.Theincomingbitsare0011.Themodulatorfirstgetsholdofthebitpair00
o
anditmapsthisinto A cos(ω t + 0 ) .Next,itgrabsthebitpair11,mullsitover,and
c
o
popsout A cos(ω t + 270 ) .YoucanseealltheactioninFigure5.17.Table5.4(the
c
partmarked4-PSK)summarizestheworkingsof4-PSK.Asyoucanseehere,foreach
inputbitpair,yougetadifferentwaveformoutput.It’sverycommonfor4-PSKtobe
calledbythenamequadraturePSK(that’sQPSKforshort).
Next,you’vegot8-PSK.It’sthat“biggerisbetter”philosophy,again.(Gottabe
carefulwiththatone—it’llgetyouintrouble,aswe’llseelateron.)Atanyrate,in8-
PSK,threebitsareinputatthesametime,andthemodulatorthinksitover,andpops
outoneofeightpossiblewaveforms.TakealookatthebottomofTable5.4.Forevery
possiblesetofthreeincomingbits,adifferentwaveformisoutput,andthesewave-
formsdifferfromoneanotherinphase.
Ofcourse,youdon’thavetostopthere.There’s16-PSK,64-PSK,and,hey,why
not10192-PSK?
1 1 Inputbits 0 0
Output 
waveform 
Acos(ω
c
t+0 ) ° Acos(ω t+270° )
c
Figure5.174-PSKmodulation
FSK 
Nextstop,theFSKmodulators,shortforfrequencyshift-keyingmodulators.Asthename
suggests,herewestufftheinformationbitsintothefrequency.We’lllookatthesimplest
first,whichisBFSK(binaryFSK).Here,ifbit0goesintothemodulator,itsendsout
∆ ∆
A cos((ω + ω )t +θ) .Ifbit1goesin,thenoutcomes A cos((ω + ω )t +θ) .The
∆
ω
0 c 0 c 1
isthefrequencyoffsetusedtorepresenta0,and
∆
ω
1
isthefrequencyoffsetusedtorepre-
senta1.YoucanseeanillustrationoftheBFSKmodulatorinactioninFigure5.18.Table
5.5providesasummaryofthisBFSKmodulator.
Nextup,4-FSK.Here,twobitsareinputtothemodulatoratthesametime,and
the4-FSKmodulatoroutputsoneoffourpossiblewaveforms.TakealookatTable5.5,
whereyoucanseeallthe4-FSKmodulatoraction.
Ofcourse,those“bigger-is-better”guysalsointroduced8-FSK,16-FSK,andsoon
...samegreatidea,justabiggertabletodescribeit.
130  ◆  Chapter Five 
Inputbits
Output
wav orm
0 1 0
ef
Figure5.18BFSKmodulation
s t i b t u p n I m r o f e v a w t u p t u O
m r o f e v a w t u p t u O
) d n a h t r o h s (
K S F B   0  s
0
( ( s o c  A  =  ) t (  ω
c 
+
∆
ω
0 
T i , ) t )  ≤ T ) 1 + i ( < t   ( (  s o c  A  ω
c 
+
∆
ω
0
) t )  · π ) T i t (  –
1  s
1
( ( s o c  A  =  ) t (  ω
c 
+
∆
ω
1 
T i , ) t )  ≤ T ) 1 + i ( < t   ( (  s o c  A  ω
c 
+
∆
ω
1
) t )  · π ) T i t (  –
K S F - 4   0 0   s
0
( ( s o c  A  =  ) t (  ω
c 
+
∆
ω
0 
T i , ) t )  ≤ T ) 1 + i ( < t   ( (  s o c  A  ω
c 
+
∆
ω
0
) t )  · π ) T i t (  –
1 0   s
1
( ( s o c  A  =  ) t (  ω
c 
+
∆
ω
1 
T i , ) t )  ≤ T ) 1 + i ( < t   ( (  s o c  A  ω
c 
+
∆
ω
1
) t )  · π ) T i t (  –
0 1   s
2
( ( s o c  A  =  ) t (  ω
c 
+
∆
ω
2 
T i , ) t )  ≤ T ) 1 + i ( < t   ( (  s o c  A  ω
c 
+
∆
ω
2
) t )  · π ) T i t (  –
1 1  s
3
( ( s o c  A  =  ) t (  ω
c 
+
∆
ω
3 
T i , ) t )  ≤ T ) 1 + i ( < t   ( (  s o c  A  ω
c 
+
∆
ω
3
) t )  · π ) T i t (  –
Table5.5FSKmodulator
QAM 
QAMmodulators(shortforquadratureamplitudemodulationmodulators)takeincom-
ingbitsandpopoutwaveformsoftheusualform s t ( ) Acos(ωt +θ) .Whatmakes
QAMmodulatorsuniqueisthattheinputbitsarestoredinboththeamplitude(A)and
thephase(θ )oftheoutputwaveform.
Iknowthisisgoingtoseemrathershort,butI’mgoingtostoptalkingabout
QAMfornow.I’llprovidemoredetailsalittlelater,afterI’vehadachancetotellyou
aboutorthonormalbasis.
Getting It from Here to There: Modulators and Demodulators  ◆  131 
Example 5.2 
Sketchtheoutputofa4-ASKand4-PSKsystemwhentheinputbitsare001110.
Solution:TurningtoTable5.2and5.4,weseetheoutputwhendifferentbits
areinput.Usingthesetables,withinput001110,wefindtheoutputshownin
FigureE5.1.
00 1 1 10
4-ASK
–3Acos(ω
c
t)
Acos(ω
c
t+0°) Acos(ω
c
t+270°) Acos(ω
c
t+180°)
3Acos(ω
c
t) Acos(ω
c
t)
4-PSK
A
3A
A
FigureE5.1Outputof4-ASKand4-PSK
Choosing a Modulation Method 
Bandpassmodulationoffersalotofchoices.Youfirsthavetochoosefromamong
ASK,PSK,FSK,andQAM.Onceyou’vemadethatchoice,you’vegottopickanum-
ber:willitbe,forexample,4-ASK,8-ASK,or16-ASK?Inthissection,I’llbrieflydiscuss
themostcommoncriteriaforchoosingthemodulationschemethat’sbestforyour
communicationsystem.Let’sstartbytakingalookatthedifferentmodulationtypes
(ASK,PSK,FSK).Ratherthantellyouwhattouse,I’mgoingtotellyouwhattoavoid 
using.
InASK,alltheinformationisintheamplitude.Let’ssayyourcommunication
channelcreatesamplitudedistortions.Inthatcase,ASKisnotthemodulationyou
want.That’sbecausethechannelmessesupthepartofyoursignalcontainingthe
information.Youdon’twantthat.
132  ◆  Chapter Five 
InPSK,alltheinformationisinthephase.Let’ssayyourchannelintroduces
phasedistortionsandyouareunabletoremovethesephasedistortionsatthereceiver.
Inthiscase,youdon’twanttousePSK,becausethechannelisdistortingthepartof
thesignalwithinformation.
Finally,inFSK,you’vegotinformationinthefrequencies.Ifthechannelcauses
significantfrequencydistortionsthatyoucan’tcorrect,thenyoudon’twanttouse
FSK.FSKhasanotherdrawback.IttakesupmorebandwidththaneitherPSKorASK.
InASKandPSK,youalwayssendtheinformationsignalatthesamefrequency—it’s
alwayssentaroundω
c 
(e.g.,Table5.4).ButinFSK,yousendthesignalatalotof
differentfrequencies(e.g.,aroundω +
∆
ω orω +
∆
ω
1
(Table5.5)).Thismeansyou’ve
c 0 c
gottohavemorefrequenciesavailabletoyouifyouwanttosendanFSKsignal.
Let’ssayyou’vedecidedonPSK,butshouldyouuseBPSKor4-PSKor8-PSK?
Let’stalkalittleabouthowtodecide.Tothoseofyouwhoarethinking“biggeris
better”(forexample,4-PSKbeatsBPSK),thereisonepointthatsupportsyourthink-
ing.Whenyouusebiggersizes,yoursignalhasasmallerbandwidth.IfyouuseBPSK,
theneachsignalthatleavesthemodulatorisofdurationT =  T
b
.Ifyouinsteaduse4-
PSK,theneachsignalthatleavesyourmodulatorisofdurationT=2 T
b
.InFigure5.19,
youseethattheBPSKsignalhasanull-to-nullbandwidthof2/ T
b
,whilethe4-PSK
signalhasanull-to-nullbandwidthofonly1/ T
b
. So,biggermodulationsize,smaller
bandwidth.
s(t) S(f)
-1/T
b
1/T
b
1/2T
b
-1/2T
b
t
t
T=T
b
f
2/T
b
(a)
...
...
f
T=2T
b
1/T
b
(b)
Figure5.19Signalsofdifferentdurationsinfrequencyandtime
c
Getting It from Here to There: Modulators and Demodulators  ◆  133 
But,forthoseofyousay“Life’snotallaboutsize,”well,youtooarecorrect.
Whilebeingbiggermeansthatyouuseasmallerbandwidth,italsomeansthatyouare
morevulnerabletochannelnoise.ConsidersendingBPSKor4-PSK.Ifthechannel
noiseisgoingtocauseanerrorinBPSK,it’sgottomake A cos(ω t ) looklike
c
o
A cos(ω t + 180 ) .Meanwhile,in4-PSK,ifthechannelnoiseisgoingtocausean
o
error,itonlyhastomake A cos(ω t ) looklike A cos(ω t + 90 ) .It’seasiertomakea
c c
signallooklikeithasa90-degreephaseshiftthanitistomakeitlooklikeithasa180-
degreephaseshift.Soit’seasiertomakesymbolerrorsatthereceiverin4-PSKthanit
iswithBPSK.We’lltalkmoreaboutthiswhenwelookatdemodulators,comingsoon
toasubsectionnearyou.
5.3 Just-in-TimeMath,orHowtoMakeaModulatorSignal
LookFunny
WhenIwastakingmyPh.D.comprehensiveoralexam,oneofmyprofessorsaskedhow
IwouldgetbywiththelittlemathIknew.Iexplainedtohimtheconcept,whichhelater
called“just-in-timemath.”Basically,thisistheideathatwhenmyresearchrequiredthat
Ilearnsomemoremath,I’dreadthemathbooksthen.Inkeepingwiththespiritof“just-
in-timemath,”I’mgoingtosharewithyousomemathtoolsthatwe’llneedinorderto
understanddemodulators(comingupnext).Theparticularmathskillyou’regoingto
learnishowtorepresentasignalasapointinspace.TakealookatFigure5.20.Inpart
(a)ofthefigure,we’vegotapointinspace.Inpart(b),youseethatwecancharacterize
thatpointas(1,1)byintroducinganx-axisanday-axis.Itturnsoutthatyoucandothe
samethingwithsignals.Takealookatpart(c)ofthefigure.Thereyouseeasquare
wave.Inpart(d),youseethatwecanrepresentthissquarewaveasapoint(1,1),which
meansthatthesignalismadeupofonepartofthex-axissignalandonepartofthey-axis
signal.Allkindsofdetailwillfollow.
y
Figure5.20
P P=(1,1)
1
Representingpointsandsignals
x
1
(a) (b)
2
s(t)
φ
2
(t)=
1/2 1
s(t)=(1,1)
1 2
2
t
φ
1
(t)=
1 1
(c) (d)
1/2
134  ◆  Chapter Five 
5.3.1TheIdea
Theideaisthis.Let’ssayyouthrowmeasetofsignals{s (t), s
2
(t), ..., s (t)},andyou
1 M
makesuretheyhavefiniteenergy(andallthesignalswelookatwillbefiniteenergy
signals).ThenIcanalwaysrepresentthesesignalslikethis:
( ) s ϕ ( ) + s ϕ (t )+... +s
1N
ϕ
N
(t ) (5.2)
1
s t
11 1
t
12 2
…
( ) s ϕ ( ) + s ϕ (t )+... +s
MN
ϕ
N
(t ) (5.3)
M
s t
M1 1
t
M2 2
ϕ
1
t ),..., ϕ
N
(t )} arecalledanorthonormal basis.Thenumberof ϕ
j
( ) ’s Here, { ( t
(N)isalwayslessthanorequaltothenumberof s t
i
( ) ’s(M)yousentme.The
{ ( ϕ
1
t ),..., ϕ
N
(t )} havetwoparticularlynoteworthyproperties,bothofwhichhaveto
dowithintegrating.Andheretheyare.Ifyouintegratetwodifferent ϕ
j
( ) ’stogether t
youget0,andifyouintegratea ϕ
j
( ) withitselfyougeta1.Thatis,inaniceconcise t
equationform:
∞
t t dt 0, i ≠ j
(5.4)
∫
ϕ
i
( )ϕ
j
( )
−∞
∞
∫
ϕ
j
( )ϕ
j
(t dt 1
(5.5)
t )
−∞
AlsoinEquations(5.2)to(5.3),thecoefficients{s
ij
, i = 1, 2, ..., M, j = 1, 2, ..., N}
arejustnumbers.Thesenumbers,like s forexample,tellyouhowmuch ϕ
2
( ) t
12
thereisin s t
1
( ) .
Now,Equations(5.2)to(5.3)tellmethereisanewwayIcanexpresssignal
s t
1
( ) .Inthepast,Ieitherhadtowriteoutamathematicalequation(forexample,
1
( ) t )orIhadtodrawyouapicture.Now,Icanwrite s (s
11
,..., s ) ,andyou s t
1 1N
knowthattomakes t
1
( ) —allyoudoisthemathofEquation(5.2).
Thisisallverynice,buttherearetwoveryimportantthingsIhaven’tyettold
you.Howdowefigureoutthecoefficients,{s
ij
, i = 1, 2, ..., M, j = 1, 2, ..., N},and
howdowefigureoutthesignals { ( ϕ
1
t ),..., ϕ
N
(t )}?
Thecoefficents,from s
11
to s
MN
,areapieceofcake.Allyouhavetodoisthis
integration:
∞
s
ij

∫
s t t dt
(5.6)
( )ϕ
j
( )
i
−∞

Getting It from Here to There: Modulators and Demodulators  ◆  135 
Theorthonormalset { ( ϕ
1
t ),...,ϕ
N
(t )}isalittlebitmorework,butit’sstillquite
easy.YousimplyfollowthemathematicalalgorithmthatI’mgoingtogiveyounext,
calledtheGram-Schmidtorthogonalizationprocedure(bigwords,simplealgorithm):
(1)Toget ϕ
1
( ) t ,justcompute
s t
t
1
ϕ ( )
( )
1
E
1
∞
where -
∫
I ( ) ( ) I J @J J .
∞ −
(2)Toget ϕ
2
( ) t ,compute
ϕ ( )
θ ( ) t
t
2
2
E
θ 2
∞
t
2
( ) − s ϕ ( ) and -
θ

∫
θ ( ) ( ) where θ ( ) s t t J θ @J J .
2 21 1
∞ −
(3)Toget ϕ
3
( ) t ,justcompute
ϕ ( )
θ ( ) t
t
3
3
E
θ3
∞
where θ ( ) s t
31 1
t
32 2
t J θ @J J . t
3
( ) − s ϕ ( ) − s ϕ ( ) and -
θ !

∫
θ ( ) ( )
3 ! !
∞ −
( ) ,andifyouget ϕ
k
t (4)Keepgoing,upto ϕ
M
t ( ) 0alongtheway,justthrow
thatoneout,becauseyoudon’tneedit.
Well,nowyouknowthisratherneattrick.Ifsomeonegivesyouasetofsignals,
youcanfigureoutthesetofsignals { ( ϕ
1
t ),...,ϕ
N
(t )},andyoucanfigureoutasetof
coefficients(values){s
ij
, i = 1, 2, ..., M, j = 1, 2, ..., N},andyoucanwritetheirsetof
signalsintermsofyour ϕ
j
( ) ’sand s
ij
’s.Moreover,youcanrepresenttheirsignalas t
avector,takingtheir s t
1
( ) andwritingitas s ( ,..., s ) .
1
s
11 1N

J


-
J
I ( )


J
I ( )


J


J
J


-
J
136  ◆  Chapter Five 
Example 5.3 
DeterminetheorthonormalbasisforthetwosignalsshowninFigureE5.2.
Solution:We’llusetheorthogonalizationprocedure,whichtellsus
ϕ ( )
I ( )
(E5.1)

∞
j


(E5.2)
,
∫
I

( )
(
\
,
@J J
(
( ∞ − ,

j \


,
∫
" @J
(
(E5.3)
, (
( ,

I ( )
(E5.4)
s
1
(t) s
2
(t)
2
1
t t
2 1
FigureE5.2 Twosignals
ThisfirstelementoftheorthonormalbasisisdrawninFigureE5.3.Return-
ingtotheorthogonalizationproceduretogetthesecondelement,wediscover
ϕ ( )
θ ( )
(E5.5)
θ


-



-
J J


-
J J
J

J


-
J

J
Getting It from Here to There: Modulators and Demodulators  ◆  137 
( )
− t s t s ϕ
1
( )
2 21

E
θ
2
(E5.6)
∞
I (J)−
∫
I

( ) ( ) J J ϕ @J J ⋅ϕ ( )

∞ −
(E5.7)
θ

I ( )−
∫
@J ⋅ϕ ( )
(E5.8)
θ

ϕ ( )
I ( )−ϕ ( )
(E5.9)
θ

ϕ
1
(t)
t
1
1
FigureE5.3 Theϕ ϕϕ ϕϕ
1
(t)
Tohelpgetabetterfeelingofwhatthenumeratorlookslike,you’llfindit
drawninFigureE5.4.Usingthis,we’llfinishsolvingforthesecondorthonormal
basiselementlikethis:
ϕ ( )
θ ( )
(E5.10)
θ

θ ( )

∞
j



(E5.11)
,
J
,
∫
θ ( ) @J
(
\

(
( ∞ − ,
138  ◆  Chapter Five 
t
ϕ
2
( )
θ
2
( )
t
2
j
1 dt
\
(
1
2
(E5.12)
,
∫
( 1 ,
θ
2
( )
t (E5.13)
ThetwoelementsoftheorthonormalbasisaredrawninFigureE5.5.
θ
2
(t)=s
2
(t)–ϕ
1
(t)
t
1
1 2
FigureE5.4Aplotofθ
2
(t)=s
2
(t)– ϕ
1
(t)
t
1
ϕ
1
(t)
1
t
1
ϕ
2
(t)
1 2
FigureE5.5 Theorthonormalbasis
5.3.2 RepresentingModulatedSignals
Nextwe’llseesomesimpleexamplesofwhatwasdiscussedabove.Ifyoutotally
understoodwhatIsaidinthelastsection,thenthispartwillbeapieceofcake.Ifyou
don’tfullyunderstandwhatIsaidearlier,thenthissectionwillhelpyounailit.
First,we’lllookatthesignalsthatleavethePSKmodulator,thenlookatthe
signalsthatleavetheASKmodulator,andfinallylookatthesignalsthatleavethe
QAMmodulator.I’mgoingtotellyouabouttheorthonormalbasis(the ϕ
j
( ) ’s)and t
thecoefficients(s
ij
’s)correspondingtothesesignals.BeforeIdothis,letmeempha-
sizethatthereasonI’mdoingthishasnothingtodowithagreatlovefororthonormal
basis,butitactuallyturnsouttobeimportantwhenwelookatdemodulators.
6
J
?
Getting It from Here to There: Modulators and Demodulators  ◆  139 
BPSK 
Let’slookatthesignalsthatleavetheBPSKmodulator.Youcanseethesetwosignals,
called s t
1
( ) ,atthetopofTable5.4.
0
( ) and s t
The Orthonormal Basis:Giventhetwo-signalset { ( ), ( s t s t )},we’llstartwith
0 1
theorthonormalbasis { ( ϕ
1
t ),...,ϕ
N
(t )} forthesetwosignals.First,youknowyou’ve
beengiventwosignals { ( ), ( s t s t )},sotheorthonormalbasis {ϕ
1
(t ),...,ϕ
N
(t )} willbe
0 1
madeupoflessthanorequaltotwo ϕ
j
( ) ’s. t
YoucangoaheadanddothatalgorithmIgaveyouinSection5.3.1,andyou’llget
theorthonormalbasis(actually,it’sgoodpractice,sotakeamomentandtryithere).
OK,here’swhatyou’llfind(ifyoudidn’t,checkyourmath).The{ϕ
1
(t), ..., ϕ (t)} for
N
ϕ
BPSKisjusttheone-elementset
1
t )},where ϕ ( )

?I( )⋅π(J −EJ).That’sright,there’sonlyoneelementin ω J
{ (

theorthonormalbasis.ThismeansyoucanwritethePSKoutputsignalsas
t ( ) s ϕ ( ) (5.7)
0
s t
01 1
t ( ) s ϕ ( ) (5.8)
1
s t
11 1
s
The Coef ficients:Next,dotheintegralIshowedyou(Equation(5.6))togetthe
01
and s
11
(Goahead.Again,it’sgoodpractice).Here’swhatyoushouldget:for
0
( ) ,the s A
T
2
andfor s t
1
( ) ,the s −A
T
2
. s t
01 11
A Plot:Onenicethingaboutanorthonormalbasisisthatitletsyourepresent
theBPSKsignalsasapointinspace.Onceyou’vegotthe { ( ϕ
1
t ),...,ϕ
N
(t )} (inthis
casethe { ( ϕ
1
t )}),andyou’vegotthecoefficients(inthiscasethe s
01
A
T
2
and
A
T
2
tellsyouhowmuch ϕ
),thenyoucanplot s t
1
( ) .Figure5.21showsyouwhatImean.It
0
( ) and s t
s
11
−
t youneedtoget s t t youneedtoget
1
( )
0
( ) ,andhowmany ϕ
1
( )
1
( ) .Andthat’sit. s t
representss (t) representss (t)
1 0
x x φ √ ω π
1
t) (t-iT)
c
• (t)= (2/T)cos(
S =-A√(T/2) S =A √(T/2)
11 01
Figure5.21 PlotofBPSKsignalss
0
(t)ands
1
(t)
1
140  ◆  Chapter Five 
4-PSK 
Herewe’llbeaskingandansweringthefollowingquestion:“Howmany ϕ
j
( ) ’s(or- t
thonormalbasisfunctions)doesittaketorepresentthe4-PSKsignals(andwhatare
those ϕ
j
( ) ’s)?”Theanswerisreallyquiteeasy. t
The Orthonormal Basis:Ifwe’regoingtorepresentthe4-PSKsignalsonan
orthonormalbasis,we’vefirstgottorememberwhattheylooklike.Takeaquickpeek
atTable5.4whereyoucanseethe4-PSKsignals {s t s t s t s t )} onceagain. ( ), ( ), ( ), (
0 1 2 3
Now,lookatTable5.6.There,I’vewrittenthe4-PSKsignalsasasumofasineanda
cosine(usingasimpletrigidentity).I’vedonethisbecauseinfactitsuggestsaparticu-
larorthonormalbasis,asI’llshowyounext.
IfyoucarryoutthealgorithmofSection5.3.1,you’llgetanorthonormalbasis
ϕ
1
t ),...,ϕ
N
(t )},andyou’dfindthebasismadeupofacosineandsine.Specifically,
you’dfindtheorthonormalbasisis { (t ),ϕ (t )} where
{ (
ϕ
1 2
t ϕ ( )
2
T
cos(ω t ) ⋅ π(t iT) and ϕ (t ) −
2
T
sin(ω t ) ⋅ π(t iT) . −
c
−
1 c
Now,thisisintuitivelypleasing,because,lookingatTable5.6,it’sobviousthat4-
PSKsignalsaremadeupofasumofsinesandcosines,andthat’swhatthe
orthonormalbasistellsus:youcanwrite4-PSKsignalsasasumofsinesandcosines.
t u p n I
s t i b
s m r o f e v a w t u p t u O
4 K S P - 0 0  s
0 
) t ( =   (   s o c  A  ω
c 
0 + t  °)π ) T i t (  – =  ( s o c  A  ω
c
) t  · π ) T i - t (   +  0 
1 0  s
1 
) t ( =   (   s o c  A  ω
c 
+ t  0 9  °) · π ) T i t (  – =  0  ( n i – s  A  ω
c
) t  · π ) T i - t ( 
0 1  s
2 
) t ( =   (   s o c  A  ω
c 
+ t  0 8 1  °) · π ) T i t (  – =  – ( s o c  A  ω
c
) t  · π ) T i - t (  +  0 
1 1  s
3 
) t ( =   (   s o c  A  ω
c 
+ t  0 7 2  °) · π ) T i t (  – =  0  +  ( n i s  A  ω
c
) t  · π ) T i - t ( 
Table5.64-PSKwrittenasasumofsinesandcosines
Getting It from Here to There: Modulators and Demodulators  ◆  141 
The Coefficients:Nowthatyouhavetheorthonormalbasis { (t ),ϕ (t )} ,let’s ϕ
1 2
s s
02
) ,then ( , moveaheadandtakealookatthevalues ( , s s
12
) ,andsoon.Bydoing 
01 11
thesimpleintegrationthatgivesthesevalues(Equation(5.6)),wequicklyandeasily
find
( , s s
02
) ( A
T
2
,0)
01
( , s s
12
) (0, A
T
2
)
11
( , ( s s
22
) −A
T
2
,0)
21
( , s s
32
) (0,−A
T
2
)
31
The Plot:Nowwecanrepresent4-PSKsignalsonaplot,whichwilltellushow
t andhowmuch ϕ
2
( ) much ϕ
1
( ) t mustbeaddedtogethertocreatetheQPSKsignal.
TakealookatFigure5.22andsee4-PSKinawholenewlight.
φ
2
(t)
φ
1
(t)
x
x
x
x
0
(s ,s
01
√
(0,-A√ (t)
3
(0,A√
1
(-A√
2
representss (t)
)=(A
02
(T/2),0)
(T/2))representss
(T/2))representss (t)
(T/2),0)representss (t)
Figure5.22 Plotof4-PSKsignals{s
0
(t),s
1
(t),s
2
(t),s
3
(t)}
usingorthonormalbasis
8-PSK 
“Mustwe?”IthinkasIwritethistitle.“Imeanwe’vealreadydoneBPSKand
4-PSK,dowereallyhavetodo8-PSK?”Ithinkabouthowniceashortbookwouldfeel
asIwritelate,lateintothenight.But,don’tworry,after4-PSK,thisone’sabreeze,so
we’lljustdoitanditwillbeoverinnotime.
7
1
2
142  ◆  Chapter Five 
Table5.7showsyouthe8-PSKsignalsonceagain, {s t s t ),..., s t )} ,and ( ), ( (
0 1
addsalittletoitbyshowinghoweach8-PSKsignalcanbewrittenasasumofsines
andcosines.You’vereallygottwowaystofindthe { (t ),..., ϕ
N
(t )} (orthonormal ϕ
1
basisfunctions)for8-PSK.Firstoff,youcouldjumpaheadandusethealgorithmof
Subsection5.3.1.Or,ifyou’vegotagoodgut,youcoulduseyourintuition,usingthe
factthatevery8-PSKsignalcanbewrittenasasumofsinesandcosines.Eitherway,
youcancometothisresult:the8-PSKsignals {s t s t ),..., s t )} canberepresented ( ), ( (
0 1 7
onthesameorthonormalbasisas4-PSK,namely { (t ),ϕ (t )} where ϕ
1 2
ϕ ( ) −
1
t
T c
− t
T
cos(ω t ) ⋅ π(t iT) and ϕ ( ) −
2
sin(ω t ) ⋅ π(t iT) .
c
Allthat’sleftarethevalues ( , s s
12
) ,andsoonupto ( , s s
02
) ,then ( , s s
72
) .You
01 11 71
cangetthisbyusingtheintegralofEquation(5.6),orifyouprefer,youcanagainuse
thatoldgutofyours,turnontheintuition,andlookingatTable5.7,seehowmuch
t andhowmuch ϕ
2
( ) ( ) (thisgivesyou ( , of ϕ
1
( ) t youneedtocreate s t s s
i 2
) ).Either
i i1
way,you’llcometothesamevalues,whichyou’llfindinTable5.8.
I b t u p n i s t t u p t u O m r o f e v a w s
K S P - 8  0 0 0  s 
0 
) t ( =   ( s o c  A  ω
c 
· °) 0 + t  π ) T  i  – t (  =  ( s o c  A  ω
c 
· ) t  π ) T  i - t (   +  0 
1 0 0  s 
1 
) t ( =   ( s o c  A  ω
c 
· °) 5 4 + t  π ) T  i  – t (  =  ( s o c  ω
c 
· ) t  π ) T  i - t (
A
– ( n  i  s  ω
c 
· ) t  π ) T  i - t ( 
A
0 1 0  s 
2 
) t ( =   ( s o c  A  ω
c 
· °) 0 9 + t  π ) T  i  – t (  =  0  ( n  i  s – A  ω
c 
· ) t  π ) T  i  – t ( 
1  1 0  s 
3 
) t ( =   ( s o c  A  ω
c 
· °) 5 3 1 + t  π ) T  i  – t (  =  ( s o c  ω
c 
· ) t  π ) T  i  – t (  
A -
– ( n  i  s  ω
c 
· ) t  π ) T  i  – t ( 
A
0 0 1  s 
4 
) t ( =   ( s o c  A  ω
c 
· °) 0 8 1 + t  π ) T  i  – t (  =  – ( s o c  A  ω
c 
· ) t  π ) T  i  – t (   +  0 
1 0 1  s 
5 
) t ( =   ( s o c  A  ω
c 
· °) 5 2 2 + t  π ) T  i  – t (  =  ( s o c  ω
c 
· ) t  π ) T  i  – t ( 
A -
+  ( n  i  s  ω
c 
· ) t  π ) T  i - t ( 
A
0 1  1  s 
6 
) t ( =   ( s o c  A  ω
c 
· °) 0 7 2 + t  π ) T  i  – t (  =  0  ( n  i  s A +  ω
c 
· ) t  π ) T  i  – t ( 
1  1  1  s 
7 
) t ( =   ( s o c  A  ω
c 
· °) 5 1 3 + t  π ) T  i  – t (  =  ( s o c  ω
c 
· ) t  π ) T  i - t ( 
A
+ 
A
( n  i  s  ω
c 
· ) t  π ) T  i  – t ( 
2  2 
2  2 
2  2 
2  2 
Table5.78-PSKwrittenasasumofsinesandcosines
Getting It from Here to There: Modulators and Demodulators  ◆  143 
t u p t u O t u p t u O m r o f e v a w d e t n e s e r p e r n o
m r o f e v a w m r o n o h t r o al s a b is
8-PSK s  ) t (  s = (s 
1 0
, s
02
) =  (A 
2 

T
0  )
0 0 
s  ) t (  s
1
= (s 
1  1
, s 
2 1 
) = 
,
,
j
A
2
T 

A
2
T 
(
,
\
(
1 
(
2
A 
T
)
s  ) t (  s = (s 
1 2
, s 
2 2
) = 
(0,
2 2 
j
−A T   A T  
\
s  ) t (  s = (s 
1 3
, s
32
) = 
(
,
,
2 
,
(
(
2 
,
s  ) t (  s = (s 
1 4
, s 
2 4
) =  (−
3 3 
A 
2 

T
0  )
4 4 
j
−A T   −A T  
\
s  ) t (  s = (s 
1 5
, s 
2 5
) = 
(
,
,
2 
,
(
( 5 5 
2 
,
s  ) t (  s = (s 
1 6
, s
62
) =  (0,  −
2
A 
T
)
6 6 
j
A  T −A T 
\
s  ) t (  s = (s
71
, s  ) = 
(
,
,
2
, 
(
( 7  7 72 
2 
,
Table5.8 8-PSKwrittenasasumofsinesandcosines
ASK 
Goodnews.ASKiseasy.LookatTables5.1to5.3.There,you’llseetheASKoutput
signals,andyou’llquicklynoticeonething:allASKoutputsignalsaresimply
cos(ω t ) ⋅ π(t iT) terms;theonlydifferenceistheamplitudes. −
c
TogettheorthonormalbasisforeitherBASK,4-ASK,or8-ASK,youcanperform
thealgorithmofSection5.3.1usingtheASKsymbols { (
M
( s t ),..., s t )} ,or,youcan
0
simplyrealizethis:IfIconsidertheone-elementorthonormalbasis { (t )},where ϕ
1
t
T
− ϕ ( )
2
cos(ω t ) ⋅ π(t iT) ,thenIcanrepresentallASKsignalsasaconstant
c
timesthis ϕ
1
( ) —sothisfellowservesasmyorthonormalbasis. t
Next,weturnourattentiontocomputingthecoefficients s
01
,then s
11
,andsoon
upto s
1
.YoucanperformtheintegralofEquation(5.6),oryoucanuseyourintu-
M
ition,lookingat ϕ
1
( ) ( ) ,andyoucanfigureouthowmany ϕ
1
( ) t ,thenlookingat s t t ’s
i
youneedtoget s t ( ) .Eitherway,you’llgetthevaluesshowninTable5.9.
i
1
144  ◆  Chapter Five 
t u p t u O t u p t u O m r o f e v a w d e t n e s e r p e r n o
m r o f e v a w m r o n o h t r o al s a b is
K S A B  s 
0 
) t ( 
s 
1 
) t ( 
K S A - 4  s 
0 
) t ( 
s 
1 
) t ( 
s 
2 
) t ( 
s 
3 
) t ( 
K S A - 8  s 
0 
) t ( 
s 
1 
) t ( 
s 
2 
) t ( 
s 
3 
) t ( 
s 
4 
) t ( 
s 
5 
) t ( 
s 
6 
) t ( 
s 
7 
) t ( 
T
s = s  = –A 2
0  1 0 
T
s= s  = A  2
1  1  1 
s = s  = –3A
0  1 0 
T
s
1
= s
1  1
= –A
2 
T
s
2
= s 
1 2
= A  2 
T
s
3
= s 
1 3
= 3A 2 
s = s  = –7A
0  1 0 
s= s  = –5A
1  1  1 
s = s  = –3A
2  1 2 
T
s
3
= s 
1 3
= –A 2 
T
2 s = s  = A
4  1 4 
T
2 s = s  = 3A
5  1 5 
T
s = s  = 5A 2
6  1 6 
T
s = s  = 7A 2
7  1 7 
2
T
2
T
2
T
2
T
Table5.9ASKsignalsrepresentedonorthonormalbasis{φ
1
(t)}
Getting It from Here to There: Modulators and Demodulators  ◆  145 
QAM 
Thelastone!InQAM,theinformationbitsarestuffedintoboththephase(θ )andthe
amplitude(A)ofthecosinewaveform.Thatis,atypicaloutputwaveformforQAM
lookslikethis:
j
( ) A
j
cos(ω t +θ
j
) ⋅ π(t iT) (5.9) s t −
c
Now,applyingalittlebitoftrig,wecanrewritethisas
s (t) = A
j
cos(θ )cos(ω t) ⋅ π(t – iT) – A sin(θ )sin(ω t) ⋅ π(t – iT) (5.10)
j j c j j c
Youcanplainlyseethatthe s t
j
( ) canbeexpressedasasumofonesineandone
cosineterm.Soitjustmakessensethattheorthonormalbasisconsistsofacosineand
asineterm.Specifically,oneorthonormalbasismadeupofasineandacosinetermis
ϕ
1
t thisone: { (t ),ϕ (t )} where ϕ ( )
2
T
cos(ω t ) ⋅ π(t iT) and −
2 1 c
ϕ ( ) −
2
T
sin(ω t ) ⋅ π(t iT)
. This will work wonderfully as the orthonormal basis
t −
1 c
forQAM.Usingthisorthonormalbasis,wecanwritetheQAMoutputsignal s t
j
( )
aboveas:
j
( ) s
j1 1
t s t ϕ (t ) + s
j 2
ϕ ( ) (5.11)
2
Now,youcanusetheintegralinEquation(5.6)togetthevalues s
j1
and s
j 2
,or
youcanfiguretheseoutbycomparingtheabove s t
s
j
( ) equations(namely(5.10)and
(5.11))andseeingwhatthese s
j1
and s
j 2
mustbe.Eitherway,you’llcomeupwith
j1
A
j
T
2
sinθ
j
,whichmeanstheQAMsignalcanbewritten
accordingto
cosθ
j
and s
j 2
A
j
T
2
s t
j1
,
j
( ) ↔ s
j
(s s
j 2
) ( A
j
T
2
cos θ
j
, A
j
T
2
sin θ
j
) (5.12)
Onenicewaytolookatthis s t
j
( ) istoplotitasapointinspace,whichIdoin
Figure5.23.
φ
2
(t)
(A
j
√(T/2)cosθ
j
,A
j
√(T/2)sinθ
j
)
representss(t)
j
φ
1
(t)
Figure5.23 AplotofasingleQAMpoint
146  ◆  Chapter Five 
Beforeweheadoffintothebrave φ
2
(t)
newworldofdemodulators,onelast
thingaboutQAM.Atypical16-QAM x
3A
x
√(T/2)
x x
constellationlookslikeFigure5.24. 
There,nowyou’vehadachancetosee 
all16outputsignals,andIdidn’thaveto A√(T/2) 
x x
makeatableofall16inputbitpairs
x x
A√(T/2) 3A√(T/2)
(0000,0001,…,1111)andallthecorre-
-3A√(T/2) -A√(T/2)
φ
1
(t)
spondingoutputsignals.This
orthonormalbasisstuffsuredoescome
x x x x
inhandysometimes.
-A√(T/2)
5.4 BringitHome,Baby,or
x x x x
Demodulators
-3A√(T/2)
Figure5.24 Plotof4-PSKsignals
WhatIwanttodohereisbuildade-
{s
0
(t),s
1
(t),...s
15
(t)}for16-QAM
modulatorforyou,andteachyouhowto
builditforyourself.But,first,we’llwant
todefinethedemodulator.Whatisit?Thedemodulatorisadevicethatgetsthesignal
sentacrossthechannel,andturnsitbackintobits.
5.4.1WhatDemodulatorsDo
Let’sstartwithapicture.TakealookatFigure5.25.There,youcanseehowthesignal
( ) leavesthemodulator,packingitsbagsandtakingajourneyacrossthechannel. s t
m
Finally,afteritslongjourney,itarrivesatthereceiverside.Justlikeallustravellers,
afteralongtripourhairisamessandourstomachsmaybeturnedupside-down.Well,
thesignal s t ( ) isnodifferentthanweare—itlooksalittleshakenup.Specifically,by
m
thetimethesignalarrivesatthereceiver,it’snolongeritsold s t ( ) self,butratherit’s
m
become
( ( ) +η t r t ) s t ( ) (5.13)
m
where η( ) ( ) headedalongfrom t isthenoiseaddedbythechannelasoursignal s t
m
transmittertoreceiver.
Thedemodulatordoesn’tknow s t ( ) wassentacrossthechannel.Allitknowsis
m
that,forexample,a4-PSKsignalfromtheset {s t s t s t s t )} wassent.Itsjob ( ), ( ), ( ), (
0 1 2 3
istodothebestitcantofigureoutwhich s t ( ) wassentacrossthechannel,given
m
( ) .Onceitfiguresthisout,itusesalook-uptabletofindoutwhatinformationbits
arestuffedinthesignalitthinkswassent.Itthenoutputsthosebits.
r t
Thekeytobuildingagooddemodulatoristominimizetheeffectsofnoiseand
givethehighestprobabilityofguessingthecorrectsentsignal.Thiswillmakesure
Getting It from Here to There: Modulators and Demodulators  ◆  147 
Channel
bits
b
output
bits
b
s
m
(t)
r(t)
s
Modulator
Demodulator
information
Figure5.25 Themodulatoranddemodulator
thatthebitsthatleavethedemodulator(receiverside)areascloseaspossibletothe
bitsthatcomeintothemodulator(transmitterside).
5.4.2 TheChannelandItsNoise
We’vemodeledthechannelasanunpleasantfellowthattakesthesignalyousent,tires
itout,andultimatelyaddsanoisetoit.Thisisnottheonlychannelmodel,andchan-
nelscandoalotofotherthings(someratheruglythings),butthisnoisemodelisone
ofthemostcommonones.Tocharacterizethenoiseterm η t ,themostcommon ( )
noisemodel,andtheonewe’llworkwith,iscalledAWGN,shorthandforadditive 
white gaussian noise.Eachwordhasameaning:
Additive:tellsusthenoisesignal η t isaddedtothesentsignal s t ( ) ( ) ;
m
Gaussian: tellsus,first,thatthenoisesignal η t isarandomprocess.Second,if ( )
youconsiderasampleofthatnoise η t ,itisarandomvariablewithagaussian ( )
distribution.Thatis,intheconciselanguageofmath,
1
p t )| ) p(η ) exp(
−η
i
2
) ; (5.14)

(η(
t t
i
i
2πσ
2
2σ
n
2
n
White:tellsusthatthenoise η t ,arandomprocess,hastheautocorrelation ( )
function:
2
R
η
τ
n
( ) ( ) σ ∂ τ .
148  ◆  Chapter Five 
Whatthisfunctionmeansinwordsisthatsamplesof η t arecompletelyindependent ( )
fromoneanother.
Noise: thislittlewordmeansthat η t isanunwantedsignalintroducedbythe ( )
channel.
Puttingitalltogether,weknowthat η t isaddedtothesentsignal,it’sarandom ( )
process,andweknowitsfirst-andsecond-orderstatistics(p( ) ( ) ). η and R
η
τ
i
5.4.3 BuildingaDemodulator,PartI—theReceiverFrontEnd
Icandescribethedemodulatorintwoparts,eachparthavinganimportantroletoplay
indetectingthereceivedsignal.Thefirstpartiscalledthereceiver front end.
What it does 
Thesignal
( ( ) +η t r t ) s t ( ) (5.15)
m
jumpsintothereceiverfrontend.Thereceiverfrontendsaystoitself:“Itwouldbe
easierfortherestofthereceivertoworkwithnumbers(oravectorofnumbers),
( ) .”Soitdecidestorepresent r t ratherthanthiscontinuoustimefunction r t ( ) asa
vector.Onewaytodothisistofindanorthonormalbasisfor r t ( ) can ( ) ,suchthat r t
bewrittenas
( ) rϕ (t ) + r ϕ ( ) ... t r t
1 1 2 2
t + + r ϕ
N
( ) (5.16)
N
Inthiscase, r t ( ) canberepresentedas r (r ,..., r ) .Therestofthereceiver
1 N
couldthenworkwiththisvector,whichiseasierthanworkingwith r t ( ) .Andthat’s
whatitdoes…thereceiverfrontendturns r t ( ) into r ,atermeasierfortherestof
thereceivertoworkwith.Thedetailsfollow.
An orthonormal basis for r(t) 
Thefirstthingthereceiverfrontendisaskedtofigureoutistheorthonormalbasisfor
r t ) s t ( ) ϕ
1
t ),...,ϕ
N
(t )} inEquation(5.16) ( ( ) +η t .Inotherwords,whatwillthat { (
m
( ) :itismadeupoftwosignals, s t ( ) looklike?Let’stakealookat r t ( ) and η t ,so
m
we’lltakealookattheorthonormalbasisforeachofthesesignals,andwe’llusethat
togetanorthonormalbasisfor r t ( ) .
Let’sstartwith s t ( ) isa4-PSK ( ) .We’llpickoneoutofahatandsaythat s t
m m
basisfor
2
T
signal.Thatmeans,fromwhatwesawintheprevioussection,thattheorthonormal
s t ϕ
1 2
t − ( ) is { (t ),ϕ (t )}where ϕ ( ) cos(ω t ) ⋅ π(t iT) and
m 1 c
ϕ (t ) −
2
T
− sin(ω t ) ⋅ π(t iT) .
c 2


J

Getting It from Here to There: Modulators and Demodulators  ◆  149 
Next,let’slookat η t ,arandomprocessdescribedasAWGN.Howdoyouget ( )
theorthonormalbasis { ( ( ) ϕ
1
t ),...,ϕ
N
(t )}forarandomprocesslike η t ?Answering
thatwilljustintroducealotofmathandstats,whichultimatelywillleadustoasimple
point—soI’mjustgoingtotellyouthesimplepoint.Forarandomprocess,specifically
onedescribedasAWGN,anyinfinitelylongorthonormalbasiswilldo.Sowe’llchoose
thisone.Ipick {ϕ (t ),ϕ (t ),ϕ (t ),ϕ (t ),...}where { (t ),ϕ (t )}formstheorthonor- ϕ
1 1 2 3 4 2
malbasisfor s t
1
t
T c
− ( ) (i.e., ϕ ( )
2
cos(ω t ) ⋅ π(t iT) and
m
t ϕ ( ) −
2
T
sin(ω t ) ⋅ π(t iT) ), and ϕ (t ),ϕ (t ),... are some other functions which
2 c
−
3 4
whencombinedwith { (t ),ϕ (t )}formaninfinitelylongorthonormalbasis(we’ll ϕ
1 2
seelaterthatitwon’tmatterwhattheyare).
( ) andanorthonormalbasisfor η t , Nowwehaveanorthonormalbasisfor s t ( )
m
soyousay:what’stheorthonormalbasisfor r t ) s t ( ) ?Well, η t canbe ( ( ) +η t ( )
m
( ) canberepresentedon representedon {ϕ (t ),ϕ (t ),ϕ (t ),ϕ (t ),...} and s t
1 2 3 4 m
{ (t ),ϕ (t )}.Itfollowsthat r t ϕ
1 2
( ) canberepresentedon {ϕ (t ),ϕ (t ),ϕ (t ),ϕ (t ),...}
1 2 3 4
since:thisbasisisadequatetorepresent s t ( ) (usingthefirsttwoelements)and
m
adequatetorepresent η t (usingalltheelements),soit’sgottobeenoughtorepre- ( )
( ) s t ) +η t sent r t ( ( ) .
m
Representing r(t) as a vector using the orthonormal basis 
( ) ,let’sfigureoutthe r (r r r Nowthatwe’vegottheorthonormalbasisfor r t , , ... )
1 2 3
in r t ) rϕ ( ) + r ϕ ( ) + rϕ ( ) ... (
1 1
t
2 2
t
3 3
t +
This r isthevectorthatthereceiverfrontendwantstofigureoutandhandoffto
therestofthereceiver.First, r
1
.Togetit,allwehavetodoisusetheintegralequation
(5.6).Thenextfourlinesshowtheintegralcomputationtogetthat r
1
:
H
∫
( ) ( ) @J (5.17) J H ϕ J
J ( ))ϕ (J )@J (5.18) H
∫
(I ( )+η J
J ϕ (J )@J +
∫
( ) ( ) @J (5.19) H
∫
I ( ) ϕ η J
r
1
s +η
1 m1
( ) along ϕ
1
( ) Here, s
m1
isjustthecoefficientof s t t . (InFigure5.22,it’sthex-
m
axispartofthe4-PSKsignal.)Whataboutthe η
1
?Usingsomestatisticalarguments
whichIwon’tpresent, η
1
turnsouttobeaGaussianrandomvariablewith0meanand
variance σ
2
.Hownice.
n
Next,let’stakealookat r
2
.Again,wecanfigurethisoneoutbyusingtheinte-
gralequation(5.6).Thenextlineshowsyouwhathappenswhenyoudoit:
! !
! !
J
! ! !
J
J
2
150  ◆  Chapter Five 
r s
m2
+η
2
(5.20)
( ) along ϕ
2
( ) Thistime, s
m2
isthecoefficientof s t t .(InFigure5.22,it’sthey-
m
axiscomponent.)Here, η
2
isaGaussianrandomvariablewith0meanandvariance
σ
2
.Notonlythat,butaftersomestatisticalwrangling(about12lineslong),youcan
n
showthat η
2
and η
1
arestatisticallyindependent.Thatis,knowing η
1
tellsyou
nothingabout η
2
,andknowing η
2
tellsyouzippoabout η
1
.
Onto r
3
.Pullingequation(5.6)outofourpocketandpoppingitdownonpaper
leadsusto
H
∫
( ) ( ) J H ϕ @J J (5.21)
J ( ))ϕ ( @J J (5.22) H
∫
(I ( )+η J )
H
∫
I (J )ϕ ( ) ( ) ) @J J +
∫
ϕ η ( @J J (5.23)
H
∫
(I ϕ ( )+ I ϕ ( ))ϕ ( ) ϕ η @J J (5.24) J J @J J +
∫
( ) ( )
! ! !
J ϕ @J J + I

∫
ϕ ( ) ( ) ϕ η @J J (5.25) H I

∫
ϕ ( ) ( ) J ϕ @J J +
∫
( ) ( )
! ! ! !
t and ϕ
3
( ) Nowbecause ϕ
1
( ) t arepartsofanorthonormalbasis,thenbydefini-
tion(orbylookingbackupatEquation(5.4)),weseethattheirintegralis0;andthat
t and ϕ
3
( ) alsohappensfor ϕ
2
( ) t .Thisrealizationleadsusto:
r s
m1
⋅ 0 + s ⋅ 0 +η `(5.26)
3 m2 3
r
3
η
3
(5.27)
Now, η
3
is(usingafewlinesofstatisticalanalysis)aGaussianrandomvariable,
independentof η
1
andindependentof η
2
.
Similarly,wecancompute r
4
, r
5
,andsoon,andwe’llfind
r
4
η
4
(5.28)
r
5
η
5
(5.29)
Andsoon.Herethe η
4
, η
5
,yadayadayada,areallGaussianrandomvariables
independentofalltheother η
i
terms.
Thereceiverfrontendlooksatthisandsays:“Youknow,Ireallyonlywanttogive
therestofthereceiverinformationthat’susefulfordetection.CertainlyI’llgiveit r
1


Getting It from Here to There: Modulators and Demodulators  ◆  151 
and r
2
,becausethesecontainaninformationterm.Buttherereallyisnopointin
passingon r
3
, r
4
,andsoon,becausetheyjustcontainnoiseterms,andthesenoise
termsareindependentofalltheothernoises.AndIdon’tseeanyuseinpassingnoise
totherestofthereceiver.”
Andso,wisely,thereceiverfrontendonlyprovidestherestofthereceiverwith
thepartsof r thatcontainasignalterm.Inthecaseof4-PSK,thatis r
1
and r
2
.Inthe
caseofASKorBPSK,itseasytoshowthatitwouldsimplybe r
1
.
Building the Receiver Front End 
Nowthatthereceiverfrontendhasdecidedtogivetherestofthereceiverthepartof
r containingthesignalterms,thissectionisallaboutbuildingareceiverfrontend.
Therearemanydifferentwaystobuildareceiverfrontend,sothatitcandoitsjoband
doitwell.Thereceiverfrontendpresentedhereiscalledthecorrelatorreceiverfront
end.GlanceatFigure5.26,andyou’llseewhatthecorrelatorreceiverfrontendlooks
like.
LookingatFigure5.26,youseethattheincomingsignalisr(t).It’swellknown
thatifyouwant r
1
,allyouhavetodoistheintegral H
∫
( ) ( ) @J .InFigure5.26, J H ϕ J
youcanseethatthereceiverfrontenddoesthisonitstopbranch.Ifthereceiverfront
endalsowantstoprovide r
N
(anditwillifthere’sasignalcomponentin r
N
),thenit
justcomputestheintegral H
∫
( ) ( ) @J .Andthat’sjustwhatthebottombranchof J H ϕ J
Figure5.26does.And,voila—thereceiverfrontend.Lifeisgood.
φ
1
(t)
r(t)
x
x
.
.
.
.
.
.
∫
∫
(i+1)T
(i+1)T
iT
iT
r
1
r
N
φ
N
(t)
Figure5.26 Correlatorreceiverfrontend
J
?


152  ◆  Chapter Five 
Example 5.4 
DescribethereceiverfrontendinaBPSKreceiver.
Solution:Turningbackthepagestosection5.3.2,you’llfindthattherewe
discoveredtheorthonormalbasisforaBPSKsignalissimply
ϕ ( )
6
?Iω J ⋅π(J −E6 ) (E5.14)
Now,allthatthereceiverfrontenddoesismapthereceivedsignalr(t) toits
valuealongtheorthonormalbasis,r
1
.FortheBPSKcase,twowaysofdoingthis
areshowninFigureE5.6.
r(t)
x
r
1
r(t)
x
r
1
(i+1)T
iT
∫
∞
–∞
∫
( ) ( )
1
2
cos
c
t t T
T
ϕ ω −
2
cos
c
t
T
ω t i ⋅ π
(a) (b)
FigureE5.6 ThereceiverfrontendforBPSK—twoimplementations
5.4.4 TheRestoftheDemodulator,PartII—TheDecisionMakers
Sothereceiverfrontendhasdoneitspart—it’sturnedr(t)into r .Nowit’supto
therestofthedemodulator,whichwecallthedecisiondevice.
What It Does 
Thedecisiondevicereceives r ,which,for4-PSK,correspondsto
r ( , r ) (5.30) r
1 2
r s
m1
+η
1
(5.31)
1
r s
m2
+η
2
(5.32)
2
Inshorthandnotation,wecanwrite
r s
m
+η . (5.33)
Thedecisiondevicethinkstoitself:“AllIseeis ( , r r
1 2
) ,whichisanoisyversion
of ( , ( ) ,orcorrespondingly,whichvector ( , s s
2
) .Iwonderwhichsignal s t s s
m2
)
m1 m m m1
s s
(andtherearefourpossibilitiesin4-PSK),wassenttomebythemodulator?”The
m2
) (i.e.,which s t decisiondevicespendsitstimefiguringoutwhich ( , ( ) )the
m1 m
modulatortriedtosend.
Getting It from Here to There: Modulators and Demodulators  ◆  153 
Oncethedecisiondevicefigures φ
2
(t)
s s
2
) ,orwhich s t outwhich ( , ( ) is
m1 m m
x (s ,s )
11 12
sent,itsjobisnearlydone.Itdoesone
moresimplething:itlookstoseewhich
(r,r )
1 2
bitsthemodulatorstoresinthat s t ( ) ,
m
anditoutputsthosebits,and,then,the
demodulatorhascompleteditsjob.
x x φ
1
(t)
Forexample,takealookatFigure
(s ,s
22
) (s
01
,s
02
)
21
5.27.There,thex’smarkfourpossible
s s
m2
) valuesthatmaybeoutputby ( ,
m1
a4-PSKmodulator.Theomarks
1 2
) ,thereceivedvaluesthatcome ( , r r
x(s ,s
32
)
31
intothedecisiondevicefromthe
receiverfrontend.Thedecisiondevice
hasthejoboffiguringoutwhich
Figure5.27 Possible4-PSKsymbols
s s
m2
) wassent,giventhat ( , ( , r r )
andthereceivedvector
m1 1 2
wasreceived.Oncethedecisiondevice
decideswhich ( , s s
m2
) wassent,itthengoestoTable5.4.There,itcantellwhich
m1
s s
2
) .Forexample,ifthedemodulatordecides ( , bitsarestoredinthe ( , s s ) was
m1 m 11 12
sent(correspondingtosending s t ( ) ),thenitknowsthebitssentwere01.Thisisthe
1
demodulatoroutput.
How It Works 
“Well”,saysthedecisiondevice,“IunderstandwhatitisI’mgoingtodo,butI’m
unclearaboutexactlyhowtodoit.Specifically,howdoIfigureoutwhich s =( ,
m1 m
s s
m2
)
wassentwhenIlookandsee r ( , r r
1 2
) ?”Inthissection,we’lltellthedecisiondevice
exactlyhowtodothat.Tocomeupwithouranswer,we’regoingtodosomestatistical
work,whichwillultimatelyleadustoanintuitivelypleasinganswer.
Firstoff,wewanttomakesurethedecisiondevicemakesasfewerrorsas
possible.We’llstarttobuildthedecisiondevicebyrequiringthatitminimizethe
ε ,theprobabilityofitmakinganerror.Anotherwaytosaythisis:we’llrequire
that,giventhedecisiondevicesees r ,itmustchoosethe s
i
thatismostlikelytohave
occurred.Statistically,thisisstatedsimplyas
P( )
s
i
argmax p( | r) (5.34) s
i
s
i
Letmeexplaintoyouwhatthatmath/statsequationmeans.Itstatestwothings:
(1)the  s
i
argmax  part: thisisshorthandforthewords“Havethedecision
s
i
devicechoosetooutput s
i
,thevalueof s
i
thatmaximizes...”;
154  ◆  Chapter Five 
(2)the p ( ) part: thisisshorthandforthewords“howlikely s
i
istooccur,given
Isee r .”
Puttogether,thisequationsays“Havethedecisiondevicechoosetheoutput s
i
,
thevalueof s
i
thatismostlikelytooccur,givenIsee r .”
Forthenextlittlebit,we’regoingtomanipulateEquation(5.34)untilwegetit
intoaniceandeasyform.First,weapplyBayes’Rule,whichletsusrewritetheequa-
tionaccordingto
i
) ( | s p(s )
( )
(5.35)
i
s
i
argmax
p r
p r
s
i
Next,lookatthattermonthebottom(inthedenominator).It’snotafunctionof
s
i
.Thattermwillbethesameregardlessof s .Itplaysnoroleinchoosingwhich s
i i
i
)
i
makes
p r ( | s p(s )
( )
biggest.Itcanbedropped.Sowedropit,andendupwith
p r
r
i
) s
i
argmax p( | s p(s ) . (5.36)
i
s
i
Now,theterm p r ( | s ) istheprobabilityofhavingaparticular r givenan s
i
was
i
sent.Now r =s
i
+η ,soif r =(10,10)and s
i
=(6,7),then η = r –s
i
=(10,10)–(6,7)=
(4,3).Inotherwords,theprobabilityofhaving r given s
i
istheprobabilitythat η =r –s ,
i
ormathematically, p r ( | s ) =p(η =r –s ).Stuffingthisequalityinourequationleadsto
i i
−
i
) s
i
argmax p(η r s p(s ) (5.37)
i
s
i
Let’slookatthatnoiseterm η .Itcorrespondsto η (η ,..., η
N
) ,and,forconve-
1
nience,we’llsayherethatitconsistsoftwoterms,thatis, η ( , η η ) .InSection
1 2
5.4.3,wetalkedabouthow η
1
and η
2
areindependentofoneanother.Thismeansthat
wecanwrite p( ) p(η ) p(η ) .Throwingthisintoourequationleadsto η
1 2
s
i
argmax p(η r s p (η r s p(s ) (5.38)
1 1
−
i1
)
2 2
−
i 2
)
i
s
i
Next,weknowthatthenoisetermsareGaussian,with0mean,whichmeanswe
1 −η
i
2
write p(η
i
) exp( ) .Plugginginthisinformationleadsustoanew 
2πσ
2
2σ
n
2
n
equation,namely
2
−
2
\
exp , −
1
2
−
σ
n
2
(
,
\
2πσ
2
exp , −
2
2σ
2
(
,
( )
(5.39)
sˆ
i
arg max
1
j
(r s
i1
)
(
1
j
(r s
i 2
)
( p s
i
,
s
i 2πσ
2
(
,
n (
n
n
Getting It from Here to There: Modulators and Demodulators  ◆  155 
Now,herecomesaneatstatementthatyoumaynothaveheardbefore.Ifyou’re
lookingforthevalueofxthatmaximizesf(x),thenyoucanalternativelylookforthe
valueofxthatmaximizeslnf(x).Applyingthisideatoourequationgetsusto
−
2
−
2
\ \
] ,
1
j
(
r s
i1
)
(
1
j
(
r s
i 2
)
( p
( )
]
sˆ
i
argmax ln
,
,
2πσ
2
exp
(
, −
1
2σ
n
2
(
,
2πσ
2
exp , −
2
2σ
2
(
,
s
i
]
, ,
s
i
¸
n n (
n
]
(5.40)
−
2
−
2
1 2
sˆ
i
argmax ln
,
,
1
]
]
−
(
r s
i1
)
+
,
,
1
]
]
−
(
r s
i 2
)
+ ln
,
p
( )
]
s
i
]
s
i
,
¸
2πσ
2
]
]
2σ
n
2
,
¸
2πσ
2
]
]
2σ
n
2
¸
n n
(5.41)
Togetthisequation,Ijustusedtwosimplerulesofmath:(1)lnxy=lnx +lnyand
(2)lnexpx=x.Next,Icanremovethetermsthatarenotafunctionof s ,sincethey’ll
i
bethesameforeachandevery s ,andwon’thelpmeindecidingwhich s
i
makesthe
i
termbiggest.Doingthatgetsmeto
)
2
)
2
1
−
i1
−
(r s
+ ln[ p s )] (5.42) s
i
argmax −
(r s
2
−
i 2
(
s
i
2σ
2
2σ
2
i
n n
Then,I’llmultiplythoughbytheconstant 2σ
2
whichleavesmewith
n
)
2
)
2 2
s
i
argmax − (r s − (r s + 2σ ln[ p(s )] (5.43)
1
−
i1 2
−
i 2 n i
s
i
Now,here’salittlemathematicaltrick:Icanchoosethevaluethatmaximizesx,
orequivalently,Icanchoosethevaluethatminimizes–x.Usingthis,Icanrewritethe
equationaccordingto
)
2
)
2 2
s
i
argmin(r s + (r s − 2σ ln[ p(s )] (5.44)
1
−
i1 2
−
i 2 n i
s
i
Thatis,inshorthandnotation
2
s
i
argmin | r s | −2σ ln[ p(s )] (5.45) −
i
2
n i
s
i
Andthat,atlast(phew),isthefinalequationforhowthedecisiondevicegoes
aboutpicking s .Now,intheverycommoncasewhereallthe s ’sareequallylikely—
i i
thatis,p(s )isaconstant—thenthisequationcanberewrittenas
i
s
i
argmin | r s |
2
(5.46) −
i
s
i
156  ◆  Chapter Five 
Andhereliesthisreallyniceandeasyinterpretation.Thisequationsimplysaysto
tellthedecisiondevicetochoosethe s
i
closestto r .TakealookatFigure5.27.All
Equation(5.46)saysinthiscaseischoose s
1
,becauseit’sclosestto r .Nowisn’tthat
anintuitivelypleasingresult?Allthatmath,anditallcomesdownto:“Given r ,choose
the s
i
that’sclosestontheplot.”Isn’titnicewhenmathandcommonsensemeet?
Waitaminute...isn’tthatjustengineering?
5.4.5 HowtoBuildIt
Let’slookatwhatwenowknow.First,weknowthatthedecisiondevicereceives r ,
andtriestofigureoutwhich s
m
wassent.Itmakesitsdecisionbychoosingthe s
m
basedonequation(5.45)(or(5.46)ifallsymbolsareequallylikely—thatis,p(s )
m
equalsaconstant).Oncethedecisiondevicedecidesonwhich s
m
wassent,itthen
figuresoutwhichbitsthemodulatorstoredinthat s
m
(or,equivalently,that s t ( ) ),
m
anditoutputsthosebits.That’sprettymucheverythingthereistoknowaboutdeci-
siondevices.Theonlythinglefttotalkaboutishowtobuildthem.
The Correlator Receiver 
Figure5.28showsthecompletedemodulator,beginningwiththereceiverfrontend
andfollowingwiththedecisiondevice.Whenwebuildthedemodulatoraccordingto
Figure5.28,peoplecallitthecorrelatorreceiver.Thereceiverfrontendpartisthe
partbefore(totheleftof)thedottedlineandlooksjustliketheimplementationIdrew
earlierinFigure5.26.Aswe’dexpect,ittakesther(t)andhandsoutthe r .Tothe
rightofthedottedlineinthefigure,I’vedrawnthedecisiondevice.Itworkslikethis:
(1)processor: theprocessorreceives r =(r r ,..., r ) .ItoutputsMvalues,where ,
1 2 N
Misthenumberofpossible s t ( ) signals(forexample,for4-PSK,itoutputsfour
m
values).Specifically,foreach s t ( ) ,theprocessoroutputs|r –s |
2
.
m m
(2)adders: thereareMadders,oneforeachoutputfromtheprocessor.Thejth
adderreceivestheprocessoroutput|r –s
j
|
2
,andaddstoit–2σ
2
lnp(s
j
).So,to
n
statetheobvious,thejthadder’soutputis | r s
j
|
2
−2σ ln[ p(s
j
)] . −
2
n
(3)choose min: the“choosemin”devicereceivesMinputs, 
| r s
j
|
2
−2σ ln[ p(s
j
)] forj=1,2,...,M.Itoutputsonevalue,the s
j
,whichhas  −
2
n
thesmallestvalueof | r s
j
|
2
−2σ ln[ p(s
j
)] . −
2
n
2
Combined,thesethreepiecescarryout s
i
argmin | r s | −2σ ln[ p(s )] . −
i
2
n i
s
i
Getting It from Here to There: Modulators and Demodulators  ◆  157 
φ
1
(t)
x
x
∫
∫
(i+1)T
(i+1)T
iT
iT
r
1
. .
. .
r(t)
. .
r
N
φ
N
(t)
receiverfrontend
P
r
o
c
e
s
s
o
r
-2σ
n
2
lnp(s )
1
r s
1

2
-
Choose
min.
+
+
∧ ∧
. .
s
i
Look
b
i
. .
up
table
(bits) . .
r s
m

2
-
-2σ
n
2
lnp(s )
M
decisiondevice
Figure5.28 CorrelatorReceiver
Example 5.5 
DescribethedecisiondeviceinaBPSKreceiver(assumingeachtransmitted
signalisequallylikely).
Solution:TurningbackthepagestoExample5.4,you’llseewealready
decidedthatthereceiverfrontendisasshownontheleftofFigureE5.7.Onthe
rightsideofFigureE5.7isthedecisiondevice.Aswejustsaidinthetext,it
consistsof
1.Theprocessor,whichinthiscaseinputsr = r
1
,andoutputsM=2branches.
Thetopbranchputsoutthevalueof|r
1
–s |
2
andthebottombranchoutputs
01
2
thevalue|r –s | .
1  11
2.Theadders(orlackofthem!):withallsymbolsbeingequallylikely,
p(s
0
)=p(s
1
).Hence,thereisnoneedfortheaddersheresincealltheywould
doisaddthesamenumbertoeachbranch.
The“choosemin”device:ThechoosemindeviceoutputstheBPSKsymbol
closesttothereceivedsignal.
J J
?
158  ◆  Chapter Five 
ϕ ( )

6
?I( )π ω (J −6 )
p
r
o
c
e
s
s
o
r
2
x
∞
–∞
∫
r
1
–s
01
r
1
r(t)
2
r
1
–s
11
Choose
min.
receiverfrontend decisiondevice
FigureE5.7 ThereceiverfordeletingBPSKsymbols
The Matched Filter Receiver—Version 1 
Engineers,beingcreativetypes,cameupwithasecondwaytobuildthedemodulator,
showninFigure5.29.Let’slookatthisfigure,andcompareittoouroriginalfigure,
Figure5.28.First,looktotherightofthedottedline(atthedecisiondevice):onthe
rightofthedottedline,youcanseethatthetwofiguresareidentical,meaningthatthe
twodecisiondevicesareidentical.Next,glancetotheleftofthedottedline(the
receiverfrontend):thetwolookquitedifferent.ButI’mgoingtoshowthatinfact
thesetworeceiverfrontendsdoexactlythesamething,bydemonstratingthatthe
inputandoutputofthereceiverfrontendinthenewfigure(Figure5.29)areidentical
totheinputandoutputofthereceiverfrontendintheoriginalreceiverfigure(Figure
5.28).
Thismeansafilter 
withimpulseresponseh(t)=φ
1
(T-t)
-2σ
n
2
lnp(s
1
) 
r(t)
.
.
.
.
.
.
.
.
.
P
r
o
c
e
s
s
o
r
 
1
2
r s -
 
2
r s -
m
Choose
min.
Look
up
table
s
i
∧ ∧
b
i
(bits)
+
+
-2σ
n M
2
lnp( ) s
decisiondevice
φ
1
φ
N
t=T
t=T
A
B
receiverfrontend
(T-t)
(T-t)
Figure5.29 Thematchedfilterreceiver—Version1
J 6
K (

Getting It from Here to There: Modulators and Demodulators  ◆  159 
Boththereceiverfrontendsreceiveasinputr(t),so,they’veobviouslygotthe
sameinput.Now,ifIcanshowyouthatthevalueAinFigure5.29isequalto r
1
,and
thevalueBinFigure5.29isequaltor
N
,thenthesereceiverfrontendswillalsohave
thesameoutput,andsotheydoexactlythesamething.
It’sallmath,showingthatAisequalto r
1
,andherewego:
( ) * r t ) | where h t ) ϕ (T t ) (5.47) A h t ( ( −
t T 1
A
∫
τ ( h( ) ⋅ r t −τ ) dt (5.48)
t T
( )
∫
ϕ (6 −τ )⋅ J H −τ )@J (5.49)
A ϕ
1
(T − τ ⋅ r T − τ) dt (5.50)
∫
) (
)
∫
ϕ ( )⋅ K H )@K (5.51)
A r
1
(5.52)
Followingtheexactsamesetofmathematicalarguments,wecanshowthat
B=r
N
.Sothereyouhaveit,thereceiverfrontendofFigure5.29hasthesameinput
andoutputasthereceiverfrontendinFigure5.28,andsotheydotheexactsame
thing.Wecanreplaceonebytheother,andstillbuildtheequivalentreceiver.And
that’sreallyallFigure5.29says.
The Matched Filter Receiver —Version 2 
Yep,theymadeanotherone,andit’ssomewhatpopular,soweneedtolookatit.It’s
goingtotakesomemathtofigureouthowtobuildthisreceiver,andabitofmathto
explaintheactualimplementation,buthanginthere.
Thisnextreceiverimplementationstartsoutbyrewritingthewayadecision
devicemakesitsdecisions.Whenlastwelooked,wesawdecisiondevicesmaking
decisionsaccordingto
2
s
i
argmin | r s | −2σ ln[ p s )] (5.53) −
i
2
n
(
i
s
i
Now,we’regoingtowriteoutthefirstterm,whichleadsusto
2 2
( s
i
argmin(r
1
− s
i1
)
2
+ (r
2
− s
i 2
) − 2σ ln[ p s )] (5.54)
n i
s
i
)
2
( , assuming r r r ) .Next,we’regoingtoevaluate (r s (r
1
− s
i1
)(r s ) ,which
1 2 1
−
i1 1
−
i1
thistimeleadsusto(afteracoupleoflinesofmath)
⋅ ⋅
⋅ ⋅
J J J J J
160  ◆  Chapter Five 
s
i
argmin | |
2
−2(r s + r s
i 2
)+| s | −2σ ln[ p s )] (5.55) ⋅ ⋅
2 2
( r
1 i1 2 i n i
s
i
Now,the|r|termisthesameforall s
i
,andasaresultitwon’taffectourdecision
onwhich s
i
toselect.Sowecandropit,whichleadsusto
s
i
argmin − 2(r s + r s
i 2
)+| s | −2σ ln[ p s )] (5.56) ⋅ ⋅
2 2
(
1 i1 2 i n i
s
i
Now,foramathtrickwe’veusedoncebefore:Choosingthevalueminimizing–x 
isthesameaschoosingthevaluemaximizingx.Usingthislittletrick,let’swrite:
s
i
argmax 2(r s + r s
i 2
) + (| s | −2σ ln[ p s )]) (5.57) ⋅ ⋅
2 2
(
1 i1 2 i n i
s
i
s
i
argmax (r s + r s
i 2
) +
1
2
(| s
i
|
2
−2σ ln[ p s )]) (5.58) ⋅ ⋅
2
(
1 i1 2 n i
s
i
s
i
argmax (r s
i1
+ r s
i 2
) + c (5.59)
1 2 i
s
i
⋅ s
i
argmax r s
i
+ c
i
(5.60)
s
i
2
( where c
i

1
2
(| s
i
|
2
−2σ ln[ p s )]) .
n i
Nowwe’vegotanewequationdescribinghowthedecisiondevice,given r ,
makesachoiceonwhich s
i
tooutput.I’llshowyouhowengineersbuildademodula-
torthatusesadecisiondevicebasedonEquation(5.60).There’sreallyjustone
popularwaytodoit,showninFigure5.30.ThevaluemarkedAinFigure5.30is
actuallyequalto r s
1
,andthevaluemarkedBisactuallyequalto r s .Withjusta
M
littlelookingandalittlethinking,Ibelieveitbecomesapparentthatthedemodulatorof
Figure5.30carriesouttheoptimizationofEquation(5.60).
Let’sdosomemaththatshowsthatindeedAisequalto r s
1
: ⋅
∞
)
∫
r t s t dt
(5.61)
( ) ( )
1
−∞
∞
)
∫
(Hϕ ( )+H ϕ ( )+H ϕ ( )+…)⋅ (I ϕ ( )+ I ϕ ( ))@J
(5.62) ! !
∞ −
) ( I H

∫
ϕ ( ) ( ) J ϕ @J J )+ J ϕ @J J + I H

∫
ϕ ( ) ( )

( I H

∫
ϕ ( ) ( ) @J J +…) J ϕ @J J + I H

∫
ϕ (J)ϕ ( )
(5.63)

Getting It from Here to There: Modulators and Demodulators  ◆  161 
s(t) c
1 1
A
. . .
. . .
r(t)
. . .
x
x
∫
∫
Choose
max.
+
+
∧
s
i
Look
up
table
∧
b
i
B
s
M
(t) c
M
Figure5.30 Thematchedfilterreceiver—Version2
A (r s ⋅1 + r s ⋅1) + (r s ⋅ 0 + r s
21
⋅ 0+... ) (5.64)
1 11 2 12 1 12 2
A ⋅ r s
1
(5.65)
Inanidenticalmanner,wecanshowthatB is r s ⋅ .Andthat’sit.Takethis,add
M
somelooking,throwinsomethinking,and,voila,Figure5.30doesindeedimplement
thedecisionshowninEquation(5.60).
5.5 HowGoodIsItAnyway(PerformanceMeasures)
Ofcourse,anytimeyoubuildsomething,youwanttobeabletotellpeoplehowgoodit
is.“It’sgreat!”youmightdeclare.“Afour-starperformance!”Whilethesedescriptions
workwellformoviereviewers,they’rerarelysufficientfornumber-orientedengineers.
“Exactlyhowwonderfulisit?”thoseengineerswillask.“Givemeanumber.”You
mightthinktoreply“Aperfect10!”,butthey’llwantamoredescriptivenumber.This
sectionisallaboutprovidingsuchanumber.
5.5.1APerformanceMeasure
First,we’vegottodecidewhatnumberwewanttogivethoseengineers,andwhatthat
numberwilldescribe.Let’scomeupwithawaytonumericallymeasuretheperfor-
manceofamodulator-demodulatorpair.Thebestwayistoaskasimplequestion:
What’sthemostimportantthinginamodulator-demodulatorpair?Well,themost
importantthingisthatthedemodulatorcorrectlydecidewhatthemodulatorsent.So,
ifyouwanttotellsomeonehowyourmodulator-demodulatorpairisdoing,you’llwant
totellthem P( ) ,theprobabilitythatthedemodulatormakesanerrorwhenit’s ε
decidingwhatsignalwassent.Thesmallerthe P( ) ,thebetteryourdemodulator. ε
162  ◆  Chapter Five 
5.5.2 Evaluationof P( ) ε forSimpleCases
WhatI’lldohereisteachbyexample.We’lltakealookatBPSKandstudythe P( ) ε
foraBPSKmodulator–demodulatorpair.
The BPSK Modulator Remembered 
TakealookatTable5.4,whichsummarizestheBPSKmodulator.There,youcansee
thata0ismappedtos
0
(t)anda1mappedtos
1
(t).Also,peekingbackatSection5.3.2,
youcanuncoverthattheorthonormalbasisforBPSKissimplytheonesignal{ϕ
1
( ) }, t
ands
0
(t)ands
1
(t)areeasilyrepresentedon ϕ
1
( ) bys ands
11
.Figure5.21showsyou t
01
avisualplotofthetwoBPSKsymbols,plottedontheirbasis{ϕ
1
( ) }. t
The BPSK Demodulator: A Summary 
Aswesawinourearlierwork,ademodulatorconsistsoftwoparts:(1)areceiverfront
end,and(2)adecisiondevice.
Thereceiverfrontendtakesr(t)andmapsitto r .Thedetailsofhowitworks
andhowyoubuilditareallinSection5.4.Ifyoujustfollowthat,you’llfindinthecase
ofBPSKthatthereceiverfrontendisjustlikethepictureinFigure5.31(looktothe
leftofthedottedline).
-2σ
n
2
lnp(s
01
)
P
r
o
c
e
s
s
o
r
r s 
2
-
1 01
φ
1
(t)
x
∫
r
1
r s 
2
-
1 11
Choose
min.
+
+
∧
s
0i
Look
up
table
∧
b
i
(bits)
r(t)
-2σ
n
2
lnp(s
11
)
decisiondevice receiverfrontend
Figure5.31 BPSKdemodulator
Getting It from Here to There: Modulators and Demodulators  ◆  163 
Thedecisiondevicetakes r andfiguresoutwhichsymbol s wassent.Howit
m
worksandhowtobuilditareonceagainthetopicsofSection5.4,andifyoulookthat
overandapplyittoBPSK,you’lldiscoverthatthedecisiondevicelooksjustasit’s
drawnontherightsideofFigure5.31.Also,you’llfindthattheequationthatthe
decisiondeviceiscarryingoutissimply
)
2 2
s
i
argmin (r s − 2σ ln[ p s )] (5.66)
1
−
i1 n
(
i1
s
i
We’llassumeequallylikelysymbols(thatis,p(s
1
)=constant),whichmeansthat
i
thedecisiondeviceiscarryingouttheequation
)
2
s
i
argmin (r s (5.67)
1
−
i1
s
i
Thatis,thedecisiondevicecarriesoutthesimplerule:outputthesymbolwhich
isclosesttothereceived r
1
.
Evaluating the P(ε εε εε) 
Nowthatwe’verecappedthemodulatorandthedemodulatorforBPSK,we’rereadyto
movefull-steamaheadandfindthe P( ) ε .InBPSK,therearetwowaysanerrorcan
happen:Thedemodulatorthinksyousent s t
0
( ) ,orthe
1
( ) ,butinfactyousent s t
demodulatorthinksyousent s t
1
( ) .Writingthisstatistically,
0
( ) ,butinfactyousent s t
wehave
P(ε) = P(output s
1
(t)|send s
0
(t)) ⋅ P(send s
0
(t)) + P(output s
0
(t)|send s
1
(t)) ⋅ P(send s
1
(t))
(5.68)
Wecansimplifythisequationbymakingasimplerealization:thesymbolsare
equallylikely,whichmeanstheP(sent s t
0
( ) )=0.5.Pluggingthis
1
( ) )=P(sent s t
informationintoour P( ) ε equationleadsusto
P(ε) = 0.5 · [P(output s
1
(t)|send s
0
(t)) + P(output s
0
(t)|send s
1
(t))] (5.69)
Now,let’sseeifwecanfigureoutanothervalueforP(output s t
1
( ) ).
0
( ) |sent s t
TakealookatFigure5.32(a).There,wesent s t
1
( ) ,whichcorrespondstosending s
11
.
Thedecisiondevicepicksup r s
11
+η .Thisdecisiondevicemakesadecision
1 1
accordingtotherule“Choosethesymbolwhichthereceived r
1
isclosestto.”
LookatwhathappensinFigure5.32(b),where η
1
isbiggerthan
2
T
A .Inthat
case,the r
1
willbeclosestto s
01
,andthedemodulatorwilldecide s (thatis,
01
s t
0
( ) ).If η
1
isbiggerthan A
T
2
,thedemodulatorwillmakeanerror.So,the 
0
( ) |sent s t P(output s t
1
( ) )isequalto p
(
η > A
T
)
.
1 2




)
1
164  ◆  Chapter Five 
x
x
x
x
sentthis
sentthis
S =–A
11
√
11
√
01
√
S =A
01
√
φ
1
(t)
φ
1
(t)
1 1
(a)
(T/2)
S =-A (T/2)
S =A (T/2)
(T/2)
r =S +n
11
r =S +n
1 1 11
when 
n >  
1
A√(T/2)
Figure5.32
(b) SentandreceivedsignalsinBPSK
Similarly,wecanshowthatP(output s t
0
( ) )isequalto
1
( ) |sent s t
p
(
η > A
T
2 )
.Puttingthisinformationtogooduse,weplugitintoour P( ) ε
equation,whichleadsusto
T T
ε =0.5[p(η > A ) + p(η < − A ) ] (5.70) P( )
1
2
1
2
Thegoodnewshereisthattheequationjustgetssimpler.Oneoftheneatproper-
tiesofaGaussianrandomvariableisthat,ifxisaGaussianrandomvariable,p(x>A)and
p (x<–A)arethesame.Puttingthisinformationtouseinthe P( ) ε equationleadsusto
T
ε
1
) (5.71) P( ) =p(η > A
2
∞
η \
2( )
∫
ANF ,
j
−
σ


(
,
(
@η ε
6
πσ

(
,
(5.72)

N


6
E
2
Getting It from Here to There: Modulators and Demodulators  ◆  165 
Now,wejustmakeasimplesubstitution,namelytolet u
η
1
σ
.Substitutingthis
n
intotheintegrationbringsusto
j \
ε 2( )
∞
∫
6
π
ANF
(
,
,
−
K


,
(
( @K
)
(5.73)
σ

P( ) Q(
A
T
2
) (5.74) ε
σ
n
wheretheQ(x)functionisjustshorthandnotationfortheintegral
j \
N 3 ( )
∞
∫
π
ANF
(
,
,
−
K


,
(
( @K
.
Iwanttorewritethisequationinmorecommonnotation,sothatifyouhappento
openanothertelecommunicationsbook,youcaneasilyfollowtheirnotation.First,
there’s E :it’seasytoshowthattheenergyoftheBPSKsignal, E ,isgivenby
s s
@J J ) -
I

∫
I ( )
.Additionally,it’scommontoexpressthenoisevarianceof η
1
as σ N / 2 .PluggingthisintoourP( ) ε equationgivesustheendresult:
n o
2E
s
) P( ) ε Q( (5.75)
N
o
TheQ()functioncanbelookedupinatable.
P(
10
–1
10
–2
10
–3
10
–4
)
So,ifyougivemetheenergyoftheBPSKsignal(oritsAandT 
variance,thenI’llgiveyouthe P
fromwhichIcangetitsenergy),andyougivemethenoise
doisgivemetheratiooftheenergyoftheBPSKsignalto
( ) ε .Actually,allyou’vegotto
thechannelnoisevariance(thatis,justgiveme
E
s
),whichisoftencalledthesignal-to-noise
N
o
ratio orSNR,andI’llgiveyouthe P( ) ε .
TakealookatFigure5.33.Thereyoucan
seeaplotof P( ) ε forBPSKthatwasgenerated
s 0
E /N (dB)
usingEquation(5.75).Whatyou’llnoticeisthat,
asthesignal-to-noiseratiogetshigher,the
Figure5.33BPSKperformance P( ) ε decreasesrapidly—veryrapidly.
166  ◆  Chapter Five 
5.5.3 SomeWell-knownP(ε εε εε)’s 
WhilewecouldcalculatealltheP( ) ’sforallthemodulationschemesinamanner ε
similartothatinthelastsection,wesimplydon’thaveto.Peoplehavedonealotofit
forus.I’mgoingtogiveyousomecommonlyusedP( ) ’sforyoutouseasyouplease. ε
Justlookatthelasttableofthechapter,Table5.10.
M u d o l t a i n o m r o f r e P e c n a
P ( )
j
2E π
\
s
K S P  - M  ∑ ≈ 2  Q
,
,
sin 
(
(
N
0 
M
(
,
K S F  - B  P( )
j
∑ Q
,
,
E
s
(
\
(
N
0 
,
(
K S F  - M  P( ) ≈ ∑ (M −1  ) Q
j
,
E
s
(
\
,
(
N
0 
,
(
Table5.10 Modulationschemesandperformance
5.6 WhatWeJustDid
Anotheronebitesthedust.Let’stakeamomenttorecapandsummarizewhatwejust
wentthrough.Modulatorstaketheincomingbitsandmapthemtoawaveformready
tobesentoverthechannel.You’vegotbasebandmodulators(NRZ,RZ,andsome
others),andyou’vegotbandpassmodulators(ASK,PSK,FSK,andQAM).
Thesemodulatoroutputsignalsflyacrossthechannel.Thechannelintroducesan
unwantedelementcalledanoise.
Thedemodulatoratthereceiverendthenpicksupthisnoisyversionofwhatwas
sent.Ithasthetaskoffiguringoutwhatthemodulatorsentfromthisnoisysignal.It
doesthisbybreakingitsjobupintotwotasks:areceiverfrontend,whichmapsthe
receivedsignalr(t)into r ,andadecisiondevice,which,lookingintenselyat r ,
selectswhichsignalwasmostlikelytohavebeensent.
Finally,wediscussedhowtogoabouttellingsomeonehowamodulator–demodu-
latorpairperforms.Wedecidedonthemeasurecalledprobabilityoferror,or P( ) ε ,
andIexplainedhowyoucouldcalculateit.
Andsohereweare,atanothercrossroads,anotherend.But,asyou’lllearnwhenyou
turnyourattentiontoChapter6—everyendingisjustanewbeginning.Seeyouthere.
Getting It from Here to There: Modulators and Demodulators  ◆  167 
Problems 
1. DiscussthebenefitsanddrawbacksofManchestercodingintermsof(1)DC
component;(2)self-clocking;(3)bandwidthusage;(4)inversioninsensitivity;and
(5)noiseimmunity.
2. Ifyouwantabasebandmodulationschemethatis
(1) insensitivetoamplitudeinversionsonthetransmissionline,and
(2) insensitivetothe60-Hzspike(considerthisaDCspike)intheEM 
spectrumcausedbydevicesdrivenbyACcurrent,whatwouldbeagood 
choiceforthebasebandmodulator? 
3. A16-levelquantizerhasoutputlevels–7,–6,...,6,7,8.Itisfollowedbyasymbol
tobitmapperthatmaps–7to0000,–6to0001,andsoon.Theoutputofthebit
mapperfeedsan8-PSKmodulator.
(a) Assumethattheoutputofthesampler(thatfeedsthequantizer)is6.32
(amplitudeoffirstsample),4.16(amplitudeofsecondsample),and1.12
(amplitudeofthirdsample).Drawtheoutputofthemodulator.
(b)Ifthesamplingrateis10,000Hz,whatisthesymbolrateoutofthe 
modulator? 
4. Giventhattheinputbitsare000110101001,providetheoutputofaBPSK,
QPSK,8-PSKand16-PSKmodulator.
5. GiventhefoursignalsinFigureQ5.1,findanorthonormalbasisforthesesignals.
s
0
(t) s
1
(t)
2 2
1
t t
1 2 3 0 1 2 3 0
FigureQ5.1
Foursignals
s
2
(t) s
3
(t)
2
0
3
1
–1
1
1
0 1 2 3
t t
–1
–2
168  ◆  Chapter Five 
6. Oneoftwoequallylikelysignalsissentacrossachannel.Thechanneladdsan
additivewhitegaussiannoise(AWGN).Thesignalsentiseither
( ) ( )
0
s t p t (Q5.1)
or
( ) (
1
s t p t ) 2 − (Q5.2)
wherep(t)correspondstothesignaldrawninFigureQ5.2.
(a) Provideanequationforthereceivedsignalr(t).
(b)Determineanorthonormalbasisforthetransmittedsignals.
(c) Sketchthereceiverfrontend.
(d)Provideanequationfortheoutputofthereceiverfrontend,r.
(e) Sketchthedecisiondeviceforthereceiver.
(f) Sketchablockdiagramfortheentireoptimaldemodulator.
p(t)
2
t
0
p(t)= 2
1
FigureQ5.2 p(t)
7. Youareaskedtobuildademodulatorfor8-ASK.Youaretoldallsignalsare
equallylikely.
(a) Provideareceiverfrontendforthedemodulator.
(b)Sketchthedecisiondevice.
(c) Sketchthecompletedemodulator.
8. AnFSKmodulatorsendsoneofthefollowingsignals
( )
1
s t ( ) ( ) cos , 1
c
A t iT t i T ≤ < + ω (Q5.3)
( )
2
s t
(
cos
c
A t ω
)
, t iT
∆
+ ω t ≤ < ( ) 1 i T + (Q5.4)
( )
3
s t
(
cos
c
A t ω
) ( ) 2 , 1 t iT t i
∆
+ ≤ < + ω T (Q5.5)


Getting It from Here to There: Modulators and Demodulators  ◆  169 
s t
∆
)
, ≤ <
4
( ) Acos
(
ω t +3 ωt iT t (i +1)T (Q5.6)
c
where
∆
2π
ω
T
(Q5.7)
(a) Findanorthonormalbasisforthesefoursignals;
(b)BuildanoptimaldemodulatorwhentheFSKsignalisreceivedinthe
presenceofAWGN.
9. ConsiderabinaryFSKmodulatorwhichtransmitsoneofthefollowingsignals:
s t
0
t
0
( ) Acos (ω + θ
0
) (Q5.8)
s t
1
t
1
( ) Acos (ω + θ
1
) (Q5.9)
where
θ θ KEBH H=@ L=KA E [ π] (Q5.10)
(a) Findanequationrelatingω andω
1 
suchthatthetwotransmittedsignals
0
areorthogonal,i.e.,suchthat
(i T +1)
∫
0
( ) ( ) s t s t dt 0
(Q5.11) 1
iT
Fortheremainderofthisproblem,assumethatthetwotransmittedsignalsare
orthogonal.
(b)Findanorthonormalbasisforthetwotransmittedsignals.
(c) Plotthetwotransmittedsignalsontheorthonormalbasis.
(d)AssumethatthesignalsoutofabinaryFSKmodulatorareequallylikely,
andthattheyaresentoveranAWGNchannel.Drawablockdiagramofthe
optimaldemodulator.
(e) Express,ontheplotyoudrewin(c),theoperationofthedecisiondevicein
thedemodulator.
10. Evaluatetheprobabilityoferrorwhen
• amodulatorsendsoneoftwoequallylikelysymbols;
• themodulatoroutputsareeither0or1,asshowninFigureQ5.3;
• thechanneladdsAWGN
• anoptimaldemodulatorisused.
170  ◆  Chapter Five 
s
0
=0
X X
s
1
=1
ϕ
1
(t)
FigureQ5.3
Modulatoroutputsonorthonormalbasis
11. Determinetheoutputsymbolrateofthe8-PSKmodulatorgiven
• Ananaloginputentersasampler–theanaloginputhasamaximum
frequencyof12kHz.
• ThesignalissampledattheNyquistrate.
• Thesampledsignalentersintoan8-levelquantizer.
• Thequantizeroutputpassesthroughasymbol-to-bitmapper.
• Thebitsfromthemapperenterintothe8-PSKmodulator.
6 
Chapter 
Channel Coding and Decoding:
Part 1–Block Coding and Decoding 
W
e’renownearthemidpointofthebook.Thischapterisallaboutwhat’sknownas
channelcoding,anditspartner,channeldecoding.We’llstartwithabrief
overviewofthemeaningofthewordschannel codingand channel decoding.
First,inFigure6.1you’llseeadevicecalledthechannelcoder,locatedrightin
themiddleofthesourceencoderandthemodulator.(Itisthisdevicethatperforms
channelcoding.)Ifyoulookattheinputtothechannelcoder,you’llseethatitisa
streamofbits,anditsoutputisalsoastreamofbits.Thisleadsustoanintriguing
question.Whywouldsomebodyintroduceadevicethattakesinastreamofbitsand
putsoutastreamofbits?Toanswerthisquestion,weneedtolookalittlemoreclosely
atwhatthechannelcoderdoes:Ittakeseachsetofkincomingbitsandmapsitintoa
setofnoutgoingbits,wherenisgreaterthank.Theextran – kbitsintroducedinto
thebitstreambythechannelcoderareaddedsothatwecandetecttransmission
errorsand/orremovetransmissionerrorsatthereceiverside.
TRANSMITTER
encoder
Channel
coder
bits bits
x(t)
speechsignal
...
sent
channel
s(t)
Source
Modulator
forexample, forexample,
forexample, across
101 101110
Figure6.1Introducingthechannelcoder
172  ‹  Chapter Six 
NowtakealookatFigure6.2,whereyou’llseeapictureofanupdatedreceiver
side.Thisreceiverincludes,betweenthedemodulatorandthesourcedecoder,a
deviceknownasthechanneldecoder.Thechanneldecoderundoestheoperationof
thechannelcoder.Thatis,thechanneldecodermapseachsetofincomingnbitsinto
itsbestguessontheoriginalsetofkbits.Specifically,whatitdoesisusetheextra
n – kbitsintroducedatthechannelcodertocorrect/detectanyerrorsthatmighthave
occurredduringtransmission.
Therearetwotypesofchannelcoding:blockcodingandconvolutionalcoding.
Thischapterwillfocusonblockcoding,aswewillsaveconvolutionalcodingfor
Chapter7.
RECEIVER
decoder
Channel
decoder
bits bits
...
r(t)=s(t)+n(t)
(t)
speech
x
Source
Demodulator
forexample, forexample,
received
fromchannel
101110 101
signal
Figure6.2 Introducingthechanneldecoder
6.1 SimpleBlockCoding
6.1.1TheSingleParityCheckBitCoder
Inthissimpleexampleofchannelcoding,calledthesingle parity check bit coder,each
setofk incomingbitsismappedtok+1outgoingbits.TakealookatFigure6.3,which
clearlyexplainswhatisgoingon.Theadditionalbitisaddedtoensurethat,ifyou
addedallthebitstogether,youwouldgetatotalof0(usingmodulo2addition:0+0=0,
1+0=1,0+1=1,1+1=0).
Channelcodingwhereonebitisaddedtocreatea
totalsumof0iscalledeven parity. Youcan
insteadaddonemorebitsothatthetotal
Channel
coder
1 0 1 1 0 1 0
whenaddingallbitsis1,andthisis
calledodd parity.Youcandecide
whichyoulikebetter,buttokeep
add1bit
toensuresumof thingssimpleinthisbook,unless
allbitsis0
otherwisestated,youcanassume
Figure6.3 Singleparitycheckbitcoder
I’malwaysreferringtoevenparity.
Channel Coding and Decoding: Part 1–Block Coding and Decoding  ‹  173 
Nowweknowwhatthechannelcoderdoes.Whataboutthechanneldecoder?
UsingtheexampleofFigure6.3,thedecodertakesthefourbitsreceivedandadds
themalltogether(modulo2).Ifthetotalofthesumis0,thenitreportsthattherewas
noerrordetected.Ifthetotalhappenstobe1,thenitreportsthatinfactanerrorhas
occurred.
Forexample,lookatFigure6.4.InFigure6.4(a)youseethefourtransmittedbits.
InFigure6.4(b)youseethefourreceivedbits.Addingthereceivedbitstogether
modulo2yougetatotalof1.Seeingthat1tellsyouanerrorisdetected.
Let’sseewhetherornotwecandetecttwoerrorswithsingleparitycheckbits.In
Figure6.4(c),youseethereceivedbits,withtwoerrorsinthem.Addingallthere-
ceivedbitstogether,yougetasumof0,whichindicates“Noerror.”Butweknow
thereareinfacttwoerrors.So,singleparitycheckbitscan’tbeusedtodetectthecase
oftwobiterrorsoccurring.Actually,ifyoutakeafewminutesandlookatdifferent
scenarioscarefully,you’llfindthatasingleparitycheckbitcanalwaysdetectanodd
numberoferrors,butcanneverdetectanevennumberoferrors.
1 0 1 0
sumis0
(a)
1 1 1 0
sumis1
ERROR! 
bitin 
error 
(b)
1 1 1 1 sumis0
Doesnotdetect
2errors
bitin bitin
error error
(c)
Figure6.4 Errordetectionatchanneldecoder
(a)sentbits (b)1bitreceivedinerror
(c)2bitsreceivedinerror
174  ‹  Chapter Six 
Let’sconsiderwhetherornotwe’reabletocorrect(andnotjustdetect)anytrans-
missionerrorsusingsingleparitycheckbits.LookatFigure6.5(a),toseethesentbits.
InFigure6.5(b)and6.5(c),youseetwocasesofreceivedbits,eachwithadifferentbitin
error.Now,computethesumforthecaseofFigure6.5(b)andthecaseofFigure
6.5(c)—inbothcaseswegetasumof1.Thistellsusthere’sanerror,butwehaveno
wayofknowingwhichbitisinerror—wecan’ttellifit’sthecaseofFigure6.5(b)or
Figure6.5(c).Therefore,singleparitycheckbitscannotbeusedtocorrectanyerrors.
sumis1
ERROR!
bitin
error
(a) (b)
sumis1
ERROR!
1 0 1 0
1 1 1 0
0 0 1 0
bitin 
error 
(c)
Figure6.5 Thelackoferrorcorrectionatthechanneldecoder
(a)sentbits (b)1bitreceivedinerror(Case1) (c)1bitreceivedinerror(Case2)
Example 6.1 
Determinetheoutputofaparitycheckbitcoderwhich,forevery3bitsin,puts4
bitsout.Assumetheinputis001110.Ifbiterrorsoccurredinthefirsttwobits,
couldadecoderdetectthis?
Solution:Withinput001110,anewbitisaddedaftereach3bits.Thatbit
makessurethatthesumofeachsetisnow0.So,with001110comingin,0011
110 0 comesout.A1isaddedafterthefirstthreebits,witha0addedafterthe
finalthreebits.
Ifthefirsttwobitsareinerror,wereceive1111110 0. Atthereceiver,we
formasumovereachsetof4bitsandmakesurethesumaddsto0.Ifitdoes,we
say“noerrordetected”;ifitdoesn’taddto0,wesay“errordetected.” Sointhis
case,wecreate1+1+1+1(modulo2sumoffirst4bits)=0,andoverthenextsetof
4bitswecreate1+1+0+0(modulo2)=0.So,eventhoughtwoerrorsoccurred,
nonearedetected.
Channel Coding and Decoding: Part 1–Block Coding and Decoding  ‹  175 
6.1.2SomeTerminology
Beforewemoveontoconsiderthemoresophisticatedformsofblockcoding,I’dlike
tointroducefourkeytermstoremember.Considerablockcodethatmapseach
incomingsetofkbitstoanoutgoingsetofnbits:
1.First,inshorthandnotation,thisblockcodewillbecalledan(n,k) code.
2.Thisblockcodeissaidtohaveacode rate of k/n.
3.Thisblockcodeisalsosaidtohavearedundancyof(n–k)/k.
4.Andfinally,thiscodeissaidtohave(n–k)redundant bits(thatis,checkbitsor
paritybits),whichrefertotheadded(n–k)bits.
6.1.3RectangularCodes
Inrectangularcodes,eachsetofM·Nbitsaremappedtoasetof(M +1)·(N +1)bits.
Channel Coders for Rectangular Codes 
Letmestartbyelaboratingonwhatgoesonatthechannelcoder.InFigure6.6(a),you
willseethatthebitstreamismappedfromaserialformintoamatrixform.Inthis
case,wehaveeachsetof9bitsmappedtoa3by3matrix.Figure6.6(b)showsyou
whathappensnext—namely,aparitybitiscreatedforeachrowandforeachcolumn.
Withtheadditionofthisbittoeachrow,thetotalsumofeachrowisnow0.Withthe
additionofanextrabittoeachcolumn,thetotalmodulo2sumofeachcolumnis0.
Thesebits,nowina4by4matrix,aresentseriallyacrossthechannelasasetof16
bits.Andthat’sit—that’sallthechannelcodingforarectangularcode.
paritycheck
bitforrow1
1 1 1 1
0 1 0 1
1 1 0 0
0 1 1 0
0 1 0
1 1 0
1 1 1
111 010 110
1111 0101 1100 0110
paritycheck
bitforcolumn1
(a) (b)
Figure6.6 Theworkingsofthechannelcoderforrectangularcodes—in2parts
176  ‹  Chapter Six 
Channel Decoders for Rectangular Codes 
We’llseewhat(ifany)biterrorswecancorrectanddetectusingthechanneldecoder.
Forsimplicity,I’lltakeyouonadecoderjourneyusingtheexampleshowninFigure6.6.
Thechanneldecoderstartsbyreturningtheincomingserialbitsbacktomatrix
form.YoucanseethisongoinginFigure6.7,whereoneofthereceivedbitsisinerror.
Next,thechanneldecodercomputesthesumforeachcolumnandthesumforeach
row.Asyoucansee,ifthereisanerrorinoneofthecolumns,thenthesumwillbe1.
Ifthereisanerrorinoneoftherows,thenthesumwillbe1forthatrow.Youcansee
clearlyfromthesumsformedinFigure6.7thatyoucanlocatetheexactcolumnand
theexactrowindicatingwherethesingleerrorhasoccurred,soyouknowwhatbitis
inerror.Knowingthebitinerror,youcancorrectthebitbychangingitsvalue.There-
fore,fromFigure6.7,weseethatrectangularcodescaneasilycorrectonebiterror.
Nowthisisanicething,especiallywhencomparedtothecaseofasingleparitycheck
bit,whereyoucouldonlydetectanerrorbutnotcorrectit.Asyoucansee,weare
buildinguptomoreandmoresophisticatedblockcoders.
sum= 
1111 0001  1 1 1 1
0 0 0 1
1 1 0 0
0 1 1 0
1100 0110
bitin
error
0
1 errorinrow!
0
0
sum= 0 1 0 0
error
in
column!
Figure6.7 Channeldecoderforrectangularcodes
Example 6.2 
Givenarate(2×2)/(3×3)rectangularcoder,determinetheoutputforinputbits1101.
Solution:Firstthefourbitsaremappedtoa2×2matrix,andthenparitybits
areaddedalongthecolumnsandrowstomakethesumofeachcolumnandthe
sumofeachrow0.ThisisshowninFigureE6.1.Then,thesebitsareoutput
serially(thatis,asonelongset),leadingtotheoutput110011101.
Channel Coding and Decoding: Part 1–Block Coding and Decoding  ‹  177 
FigureE6.1
Determiningrectangularcodeoutput
1 1 0
0 1 1
1 0 1
6.2 Linearblockcodes
6.2.1Introduction
Ingeneral,achannelcodergrabseachk-bitsetandthrowsoutann-bitset.Let’ssayit
takesina3-bitsetanditthrowsouta6-bitset.Thereareeightpossible3-bitsetsthat
cancomein,whereasthereare64possible6-bitsetsthatcancomeoutofthechannel
coder.
Figure6.8(a)showsonepossible3-bitto6-bitchannelcoder.Figure6.8(b)shows
asecondpossible3-bitto6-bitchannelcoder.Ineithercase,weseethateach3-bitset
inputtoachannelcoderismappedtoa6-bitsetattheoutput.Butthedifference
betweenthetwoisthatthe6bitsoutputwhen,forexample,111comesin,aredifferent
forthetwocoders.Ingeneral,howwouldwedecidewhich6-bitsetstooutputfromthe
channelcoder?
Linear block codersareagroupofblockcodersthatfollowaspecialsetofrules
whenchoosingwhichsetofoutputstouse.Therulesareasfollows,usinga(6,3)
codeforillustrativepurposes:
Let
V  =thesetofallpossible646-bitsequences
n
U=thesetofeight6-bitsequencesoutputatthechannelcoder
Usingthisnotation,theruleisthis:
UmustbeasubspaceofV
n 
.
Thismeanstwoverysimplethings:
1.Umustcontain{000000}
2.Adding(modulo2)anytwoelementsinUmustcreateanotherelementinU.
Ofcourse,examplesmakethismuchclearertounderstand.LookatFigure
6.8(b)—isthisalinearblockcode?
178  ‹  Chapter Six 
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Channel
coder
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
1 1 1
1 1 0
1 0 1
1 0 0
0 1 1
0 1 0
0 0 1
0 0 0
6bitoutput
forinput000
possible3-bit 6-bitoutputsfor
inputs
(a)
given3-bitinput
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Channel
coder
6bitoutput
forinput000
(b)
Figure6.8 Whichchannelcoderdoyouchoose?
First,lookingatthechannelcoderoutputs,weseethattheelement000000isin
theoutputofthechannelcoder(thesetU).Thatsatisfiesthefirstpartofourrule.
Now,let’stryaddinganytwoelementsinUandseeifwegetanotherelementinU.
Theaddition,bytheway,ismodulo2.Here,110110(7thelement)+011010(3rd
element)=101100(5thelement).Yes,wedecide,itdefinitelyisablockcode.
Example 6.3 
Determinewhetherornottheinputbit–outputbitsinTableE6.1couldrepresent
alinearblockcoder.
Solution:Foralinearblockcode,youmustmakesure0000isinyouroutput,
andthattheadditionofanytwooutputelements(modulo2)leadstoanother
outputelement.
Weimmediatelysee0000asanoutputelement,soweknowthatwe’reOK
withthefirstrequirement.Next,we’vegottomakesureaddinganytwoelements
(modulo2)leadstoanotherelement.Tryingthisout,wefind,forexample,0101
0 0 0 0 0 0
1 1 0 0 0 1
0 1 1 0 1 0
1 0 1 0 1 1
1 0 1 1 0 0
0 1 1 1 0 1
1 1 0 1 1 0
0 0 0 1 1 1
Channel Coding and Decoding: Part 1–Block Coding and Decoding  ‹  179 
(element2)+1010(element3)=1111(element inputbits
4);and0000(element1)+0101(element2)=
0101(element2);and1010(element3)+1111 0 0
(element4)=0101(element2);andsoon.Yes,
0 1
everyelementsumleadstoanewelement.
1 0
Withthetworulessatisfied,wecansafely
1 1
saythatTableE6.1canrepresentalinear
blockcode.
outputbits
0 0 0 0
0 1 0 1
1 0 1 0
1 1 1 1
TableE6.1Alinearblockcode?
6.2.2 UnderstandingWhy
Thisunusualrulethatalinearblockcodemustsatisfymightseemlikearandomly
chosenrule.But,asyouwillnowsee,thisruleactuallymakessense.Tounderstandit,
let’sconsiderhowwecanbuildablockcoder.Themostobviouswaytodoitisshown
inFigure6.9.Theblockcodergetsthe3-bitinput,usesalook-uptable,andselectsa6-
bitoutputusingitslook-uptable.Howwonderfullysimple.Until...
Therearesomeblock
codersthatmapeachincom-
Channelcoder:useslook-uptable
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0 0 0
1 1 0
0 1 1
1 0 1
1 0 1
0 1 1
1 1 0
0 0 0
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Input Output
1 10110
ing92-bitsetintoanoutput
1 1 0
127-bitset—a(127,92)code.
Ifwewanttousealook-up
tableforthis,we’restuckwith
havingtoconstructalook-up
tablewithoneinput-output
pairforeachpossible92-bit
input,andthereareabout
10
28
possible92-bitpairs.That
makesforanunbelievably
Figure6.9 Channelcoderbuiltusinglook-uptable
largeandexpensivelook-up
table!
Withlinearblockcodes,thereisanotherwaytogeneratetheoutputbitsgiven
thoseinputbits,andit’saneasymethodrequiringalotlessmemory.Itworkslikethis:
yougivemetheinputbits(let’ssay3inputbits)intheformofavectorm(101),and
I’llgiveyoutheoutputbits(let’ssay6outputbits)intheformofavectoru=(011101)
byusingthesimplematrixmultiplication
u=m G. (6.1)
Gisakbynmatrixof0’sand1’scalledthegenerator matrix.Also,wheneveryou
doanadditionwhencomputingm G,you’llwanttodoamodulo2additionforthe
equationtowork.Youcanonlyusethissimplewaytocreateanoutputfromaninputif
yousatisfythelinearblockcoderulewesawearlier.Thisisahandywaytogenerate
/


180  ‹  Chapter Six 
outputbitsfrominputbits,becauseinthiscaseallyouhavetostoreisG,andthatis
onlyn·k bits.
Let’sseeifthisreallyworks,ratherthanjusttakingmywordforit.Considerthe
generatormatrixGcorrespondingto
j \
(
(
(


,
,
,
(

(6.2)

,
Thisisthegeneratormatrixforthe(6,3)codeseeninFigure6.10.Let’ssaythe
inputism =(101).Thentheoutputis
u=mG  (6.3)
j
,
,
,
\
(
(
(

( )
(6.4) 
(

,
)
Ifwecomparethattotheoutputweexpectfromthelook-uptableofFigure6.9,
thenweseewe’vegotamatch.
inputtolinearblockcoder outputoflinearblockcoder
( (6.5) 
0 0 0 0 0 0 0 0 0
0 0 1 1 1 0 0 0 1
0 1 0 0 1 1 0 1 0
0 1 1 1 0 1 0 1 1
1 0 0 1 0 1 1 0 0
1 0 1 0 1 1 1 0 1
1 1 0 1 1 0 1 1 0
1 1 1 0 0 0 1 1 1
last3bitsofoutput
Figure6.10
Systematiclinearblockcodes =3inputbits

Channel Coding and Decoding: Part 1–Block Coding and Decoding  ‹  181 
6.2.3 SystematicLinearBlockCodes
LookatFigure6.10again,whichshowstheinputandtheoutputofalinearblock
coder.Specifically,takealookattherightsideofthisfigure,whichshowstheoutput.
Lookcarefullyatthelastthreebitsoftheblockcoderoutput,andnotethatthelast
threebitsintheoutput6-bitsetmatchthe3inputbits.
Notalllinearblockcodessatisfythisproperty,butiftheyhappento,theyare
calledsystematic linear block codes.People—mostlyengineers—liketousesystematic
linearblockcodesbecauseithelpsthemsavememory.First,recallthatforlinear
blockcodesyoucangetthechannelcoderoutputbyapplyingthesimpleequation
u =mG  (6.6)
Ifyouknowthatthelastbitsoftheoutputmatchtheinputbits,thenyoucan
easilyshowthatGwilllooklike(forthecaseof3-bitsetsmappedto6-bitsets)
/ ( 2
!× !
1
!× !
) (6.7)
2 2 \ j 2
!
, (

,
2

2 2
!

(
(6.8)
,
2 2 2
,
(
( ! ! !!
Thatmeansallyouneedtostoreinthememoryofthechannelcoderisthe
matrixP.Itwillbemadeupofk·(n–k)elementsthatareeither0or1,soitreallywon’t
requirealotofmemorytostoreit.
Thatisaboutallthereistoknowaboutchannelcodingforblockcodes.Therereally
isonlyonerulethatmakesablockcodealinearblockcode,andit’sonlysixwordslong:
maketheoutputbitsasubspace.Onceyou’vedonethat,thenyou’vegotaneasywayto
getoutputbitsfrominputbits,usingG.Ifyouwanttomakeiteveneasier,addtherule
thatthelastbitsoftheoutputhavegottomatchtheinputbits,andyou’vemadethe
matrixGevensimpler(andyougettocallyourcodeasystematiclinearblockcode).
Example 6.4 
TrueorFalse:ThematrixinEquation(E6.1)isthegeneratormatrixforthelinear
blockcodeofTableE6.1inExample6.3.
j \
/
,
,
(
(
(E6.1)

, (
Solution:IfGisthegeneratormatrixforthelinearblockcodeofTableE6.1,
then:forevery2-bitinputm(e.g.,m =(00)),theoutputmustbetheu(e.g.,u =
(0000))showninTableE6.1.



� /




� /




� /






� /
182  ‹  Chapter Six 
Let’sseeifthat’strue:
(
\

j
( K
( (E6.2) 
(E6.3) 
(E6.4) 
(E6.5) 
K
K
)
(
(
,
,
,
(
)
)
)
)
(
\

(
(
,
(
\
(
\

(
(
,
j
)
,
,
(
j
,
,
(
j
)
)
(
(




) (
) (
)
Yep.Foreveryinputm,wegettheoutputuofTableE6.1.Itfollowsthatthe
Gofequation(E6.1)isindeedthegeneratormatrix.
6.2.4 TheDecoding
Whatyoudoatthetransmitter,you’llwanttoundoatthereceiver.Atthetransmitter,
withalinearblockcoder,wemapped,forexample,3-bitsetsinto6-bitsets.Forthe
sakeofsimplicityinpresentation,let’sassumethatourlinearblockcodermapsinput
bitstooutputbitsasdrawninFigure6.10(alsoseeninFigure6.9).
Now,thelinearblockdecoderseesa6-bitsetcomingin.Let’ssaywesentfrom
thecoder(011101)(theoriginal3bitswere(101)).Thismayarriveattheblock
decoderinputwithoutanybiterrors(thatis,wesee(011101))oritmightcometo
thechanneldecoderinputwithanerrorinit(forexample,wesee(111101)).The
jobofthelinearblockcoderistodothebestitcantofigureoutthe3-bitsetthatwas
inputtothechannelcoder.Inourcase,thedecoderwilltrytofigureoutthat(101)
wasindeedtheinput.Iftherearenoerrorsinthe6-bitsetsent,thenthisshouldbea
simpletask.Ifthereareerrors,thenthelinearblockdecoderwillhopefullycorrect
thoseerrorsandthenfigureoutthat(101)wassent.Let’sseeexactlyhowitworks.
Insimple,almostnontechnicallanguage,youlookatwhatyou’vegot,youcorrect
errorsasbestyoucan(ifthereareany),andthenyoudecideonwhat3-bitinputwassent.
So,forexample,let’ssayyoupickupatthedecoder(111101).Inthatcase,looking
overatFigure6.10,youseethatthissimplyisn’toneoftheeightpossiblechannelcoder
outputs.Anerrormusthavehappenedsomewherealongtheway.Soyouask:whichof
theeightchannelcoderoutputsofFigure6.10isclosestto(111101)?Inlookingatall

(
(
,
,
,
(
(
)
K
(
Channel Coding and Decoding: Part 1–Block Coding and Decoding  ‹  183 
theeightpossiblecoderoutputs,youdecidethattheclosestoneis(011101),because
thisdiffersfromthereceived6-bitsetbyonlyonebit.Thisiserrorcorrection,because,
asbestyoucan,you’vecorrectedthechannelerror.Now,withtheerrorcorrected,and
(011101)inhand,youusethelook-uptableofFigure6.10anddecidethat(101)was
inputtothecoder,soyououtput(101).
Thechannelcoderdoesexactlyasdescribedabove.Only,beingathingrather
thanaperson,itdoesn’thavecommonsensetoreferto,soitmustbegivensome
mathematicalrulesthatallowittoworkasifitdemonstratedcommonsense.Hereare
detailsonhowwecangetthechanneldecodertoworkinamannerthatallowssimple
implementation.
Forstarters,wehavetointroducesomemathematicalterms.Thefirsttermis
calledtheparity check matrix,H, anditworkslikethis:ifyougivemeachannelcoder
thathasageneratormatrixG,thentheparitycheckmatrixHisdefinedasthematrix
thatsatisfiestheequation:
G H=0 (6.9)
where0referstotheall-zerosmatrix.Inotherwords,Histhematrixthatwhen
multipliedbyGproduceszip,zero,nada.Forexample,forthegeneratormatrixof
Equation(6.7),thecorrespondingparitycheckmatrixissimply
! !
0
,
,
j 1
×
\
(
(
(6.10)
! ! , (
2
×
which,fortheexampleofGinEquation(6.2),means
,
j
(
\
,

(

0 ,
,


(
(

,
,




(
(


(6.11)
,
,
(

(
(
,

YoucancheckthisoutyourselfbysimplymultiplyingtheG H,becausewhen
you’lldothisyou’llget0,asprophesied.
Ifyoulackcommonsense,asachanneldecoderdoes,youcanusethe Hatthe
decodersideinitsplace.Let’ssayyousendoutfromthechannelcoderthe6-bitset
u=m G=(011101).


/




0
184  ‹  Chapter Six 
Let’sconsiderCase1,whichisthecasewhenyoureceiveatthechanneldecoder
the6-bitsetvthatmatchestheusent(i.e.,v=u=m G=(011101)).Let’sseewhat
happensifwetakethereceivedvandmultiplyitbytheparitycheckmatrixH.Inthis
case,weget
v H=u H=m G H=m 0=0. (6.12)
Nowlet’sconsiderCase2,thecasewherethevwereceivedoesnotmatchwhat
wesentfromthechannelcoder,becauseanerroroccurredalongtheway.We’ll
considerthecasewherev=(111101).Inthiscase,wecandescribewhat’sgoingon
accordingtov=u+e,whereuisthesentsequence(011101)andeistheerror
thatoccurred,representedas(100000).Let’sseewhathappenswhenwemultiply
thereceivedvbytheparitycheckmatrixH.Inthiscaseweget
v H =u H+e H=m G H+e H=m 0+e H=0 + e H=e H=(100000)H 
(6.13)
whichisnot0.Infact,doingthemathfortheparitycheckmatrixofEquation(6.11),
you’llfindthatv H=(100000)H=(100).
Let’sinterprettheseresults.First,lookatCase1(noerrorinreceived6-bitset,
andv H=0),thenlookatCase2(anerrorinreceived6-bitset,andv H=(100)).
Fromthesetwocases,wecandetermineaverysimpleresult.Anytimethereisno
error,multiplyingthereceived6-bitvectorbyHgives0,andifthereisanerror,
multiplyingthereceived6-biterrorbyHdoesnotgive0.So,usingH,thechannel
decoderhasaneasywaytodetecterrors.
Example 6.5 
TrueorFalse:theparitycheckmatrixHforthegeneratormatrix
j \
,
,
(

(
(
,
(E6.6)
is
j
,
,
\
(
(
,

(
(
(
,
(E6.7) 
,
,
(
Solution:Tofindout,wesimplymultiplyG Handseeifwegetzero.Let’sfindout:





 


/0
Channel Coding and Decoding: Part 1–Block Coding and Decoding  ‹  185 
\
(
j
,
\ j
( ,
\ j

(E6.8)
Yep.Wegetzero,confirmingthattheHofequation(E6.7)isindeedthegenerator
matrixforGin(E6.6).
Nowlet’slookintocorrectionoferrors.ConsiderCase2.Wesentu=
(011100),theerrordescribedbye =(100000)arose,whichledtothedecoder
inputv =(111100).Weendedupatthedecoderreceiving(100)aftermultiplying
byH.Ifwecouldtellthedecoderthat,ifaftermultiplyingbyH,ifyousee(100),
assumetheerrorise =(100000),thenwecancorrectforthiserror.Inotherwords,
ifwecanmatchwhatwegetfromv Htoerrorse,thenwecancorrecttheerrors.
Let’sfollowengineeringconventionandcallwhatwegetfromvHthevectorS
(i.e.,vH =S),whichiscalledsyndrome.
So,you’vereceivedv,andyoumultiplieditbyHandyougotS.Nowlookatthis:
v H=e H=S. (6.14)
(Weknowv H=e HfromEquation(6.13).)So,foreveryerrorethereisasyn-
dromeS.
Infact,it’seasytoshowthattherearemoree’sthanthereareS’s,somorethan
oneewillsharethesameS.Forexample,inthe3-bitsetto6-bitsetcaseofFigure
6.10,thereare8S’swhilethereare63e’s.
Here’swhatthechanneldecodermustdo(anexamplefollowstheexplanation):
1)ForeachpossiblevalueofS,determinewhicherroreyouthinkhasoccurred.
Dothisasfollows,usingour(6,3)codeasanexample:
a) RealizethevalueofS=(000)=0indicatesnoerror.Thatmeansthereare
8–1=7possibleSvaluesthatwecanconsider.
b) Startwiththemostcommonerrors,whicharetheone-biterrors,i.e.,the
errorsrepresentedbythevectorse1 =(100000),e2 =(010000),
e3=(001000)toe6 =(000001).Foreachoftheseerrorse,figureout
theSusinge H=S. Soyou’llgetS1,S2,S3,uptoS6,oneforeacherror
vector.Thataccountsforatotalof6ofthe7remainingSvalues.
c) ThereisoneSvalueleftunused.Starttoconsidertheevaluescorrespond-
ingtotwo-biterrors(e.g.,e=(101000)),andfindanevaluesuchthat
eHleadstotheremainingS.
(
(
,
,
,
(
(
(
(
,
,
,
,
(
(
(
,
,
,
(


/
186  ‹  Chapter Six 
e S = e H
Step1.a.
0 0 0 0 0 0 0 0 0
1 0 1 0 0 0 0 0 1
Step1.b.
0 1 1 0 0 0 0 1 0
1 1 0 0 0 0 1 0 0
0 0 1 0 0 1 0 0 0
0 1 0 0 1 0 0 0 0
1 0 0 1 0 0 0 0 0
Step1.c. 1 1 1 0 1 0 0 0 1
Figure6.11 MappingerrorsetosyndromesS
2)Now,whenavectorvarrives,createv H=e H,andyou’llgetS.Fromthe
resultsofpart(1),decidewhicherrorethisScorrespondsto,andcorrectthe
erroryouthinkhappened.
Here’sanexampletoillustratethepoint.ConsiderthechannelcoderofFigure
6.10,whichhasGasshowninEquation(6.2)andHasgiveninEquation(6.11).
Followingstep1,youcancreateatablelinkingpossibleerrorvectorseto S=e H.
Whenyoudothis,you’llgetresultsmatchingFigure6.11.
Now,let’ssayyousentu =(101110),andyoureceivedv =u +e=u +
(100000)=(001110).Computingv H,weget
v H=e H=S =(100). (6.15)
NowlookingtoFigure6.11,weseethissyndromecorrespondstoerrore =
(100000),sowecorrectforthiserror.Withtheerrorcorrected,wethenfigureout
thatthe3-bitinputwassimply(110)usingFigure6.10.Goodwork!
Example 6.6 
Determineatableoferrorsandcorrespondingsyndromesforthe2-bitto4-bit
linearblockcoderdescribedinTableE6.2anddescribedbygeneratormatrix
j \
,
,
(

(
(
,
(E6.9) 
Channel Coding and Decoding: Part 1–Block Coding and Decoding  ‹  187 
m
TableE6.2 Linearblockcode
0 0
0 1
1 0
1 1
u
0 0 0 0
0 1 0 1
1 1 1 0
1 0 1 1
andwithparitycheckmatrix
j \
, (
,

(
0
,
,
,

(
(
(
(E6.10)

, (
Solution:
1. Thefirsterrortoconsideristheno-errorcaseofe =(0000).Inthiscase,wehave
j
1 0
\
, (
5
A=0
eH ( 0 0 0 0)
,
0 1
(
( 0 0)
,
1 0
( (E6.11) 
, (
, (
(
0 1
,
2. Thesecondcasestoconsideraretheone-biterrors,startingwithe =(0001),
inwhichcasewehavethesyndrome
j \
, (

(
5
A( )
A0 ( )
,
,
,


(
(
(
( )
(E6.12)
,


, (
3. Next,we’llconsidertheone-biterrore =(0010),inwhichcasethesyndrome
correspondsto
j \
, (
,

5
A( )
( )
,


(
(
( )
,
,

(
(
(E6.13)

, (
188  ‹  Chapter Six 
4. Continuingon,weconsidertheonebiterrore=(0100),inwhichcase
j \
, (
,

5
A( )
( )
,


(
(
( )
,
,

(
(
(E6.14)

, (
5. That’sit.We’reoutofsyndromes.So,ourtableoferrorstosyndromescorre-
spondstoTableE6.3.
So,wheneverwereceiveasetoffourbitsv,wemultiplyitbyH,andweget
vH=eH=S.Wethenusethesyndrometabletodeterminewhicherrorethat
correspondsto,andcorrectthaterror.
S e
_ _
TableE6.3Mappingsyndromesanderrors
0 0 0 0 0 0
0 1 0 1 0 1
1 1 1 0 1 0
1 0 1 1 1 1
6.3 PerformanceoftheBlockCoders
Nowthatyouknowhowtheseblockscoderswork,we’llcharacterizetheirperfor-
manceinordertotellwhichblockcodersarebetterthanothers.We’llfirst
think—hmmm,whatdowewanttheseblockcoderstodo?Wewantthemtodetectbit
errorsandwewantthemtocorrectbiterrors.So,wewillmeasuretheperformanceof
blockcodersbydetermininghowwelltheyareabletodoboth.Specifically,we’llmake
ourperformancemeasures:
1.P  ,theprobabilitythatachanneldecoderfailedtodetect(ormissed)abit
m
error;and/or
2.P,theprobabilitythatachanneldecoderfailedtocorrectabiterror.
Withourmeasuringstickinhand,let’srevisitthechannelcodersweoriginally
lookedat,anddetermineP
m
and/orP.
6.3.1PerformancesofSingleParityCheckBitCoders/Decoders
Forsingleparitycheckbitcoders/decoders,wereceiven=(k +1)bitsandcanalways
detectanoddnumberoferrorsinthesebits.Butwealwaysfailtodetectaneven


Channel Coding and Decoding: Part 1–Block Coding and Decoding  ‹  189 
numberoferrors.So,theprobabilityP
m
(theprobabilitywemissedanerror)isjustthe
probabilitythatanevennumberoferrorsoccurred.Thatis,
n
P
∑
P ( j,n)
(6.16)
m
j 2
j∈ even
whereP(j,n)referstothelikelihoodofhavingjbiterrorsoccurinablockofnbits.
Thisvaluecanbefoundfromsomestatisticsliterature,butratherthanmakeyoulook
itup,I’lljusttellyouthatitrefersto
−
( )
j
,
,
\
(
( F

( − F)
(6.17)
2

, (
where
j \
1.
,
,
(
(
referstothevalue
( − )
and
(

,
2.pistheprobabilitythatabiterroroccurswhenabittravelsfromchannelcoder
outputtochanneldecoderinput(i.e.,theprobabilitythatabiterroroccurswhen
abitissentthroughthemodulator,acrossthechannel,andfinallythroughthe
demodulator—lookatFigures6.1and6.2topicturethis).
Also,sincethesesingleparitycheckbitscan’tcorrectanyerrors,weknowthat
P=1(i.e.,you’re100%surethatyouwon’tcorrectanerror).
6.3.2 ThePerformanceofRectangularCodes
Wesawthatrectangularcodesmapk=M·Nbitston=(M+1)·(N+1)bits,andbydoing
thistheycancorrectonebiterror.OnethingIdidn’tshowyouearlieristhatthe
rectangularcodescannotcorrecttwoormoreerrors,justthatone.(Youcanshowthis
toyourselfbyputtingtwoerrorsintherectangularcodeandseeingthatyoucan’t
correcttheseerrors.)So,theprobabilitythatyoufailtocorrectanerror,P,issimply
theprobabilitythatmorethanonebiterroroccurs.Knowingthis,youcaneasilyopen
statisticsbooksandfigurethisouttobe(oryoucantakemywordthatthisPis)
2
∑
2( )
(6.18)
6.3.3 ThePerformanceofLinearBlockCodes
Inlinearblockcodesyoutakeachunkofkbitsandmapitintoachunkofnbits. In
thewordsofallfamousengineeringtextbooks,letmejustsay“itcanbeshownthat”
N
J +

190  ‹  Chapter Six 
thenumberofbitsinerror,ineachchunkofincomingnbits,thatcanalwaysbe
correctedist,wheretisthenumberyoucalculatefromtheequation
− ] , @
E
J
(6.19)
,
¸

]
]
Here
¸ ]
referstotheintegeryougetbyroundingxdowntothenearestinteger,
andd
min
isthenumberyougetbylookingatallthechannelcoderoutputsandcounting
thenumberof1sinthechannelcoderoutputwiththefewest1s(excludingtheoutput
(000000)).Forexample,considerthechannelcoderdescribedbyFigure6.10.In
thischannelcoder,theoutputwiththefewestnumberof1s(notcounting
(000000))istheoutput(000111).Thishasthree1sinit,sod
min 
=3.Then,com-
putingt,wegett=
, !− ]
=1.Thistellsusthat,forthechannelcoderinFigure6.10,you
,
¸
]
]
canbuildachanneldecoderandalwayscorrectall1-biterrors.Youmightbeableto
correctafew2-biterrorsaswell,butt = 1tellsusthatthechanneldecodercannot 
alwayscorrectall2-biterrors.
Theprobabilitythatthelinearblockcoderwillnotcorrectanerror,P,iswell
approximatedbytheprobabilitythatmorethanterrorsoccurintheincomingchunk
ofnbits.So,usingaweebitofstatistics,thatmeansmathematicallythat
2
∑
2 ( )
(6.20)
Example 6.7 
If,inthemodulator-channel-demodulatorpartofthecommunicationsystem,the
probabilitythatabitisinerroris1%,whatis:
1. theP  forarate¾paritycheckbitdecoder?
m
2. thePforarate5/10linearblockdecoder(assumet=2)?
Solution:Fortheparitycheckbitdecoder,turningtoequation(6.16),wefind
n
P
∑
P ( j, n)
m
j 2 (E6.15)
j∈ even
4

∑
P j, 4)
j 2
(
(E6.16)
j∈ even
!







Channel Coding and Decoding: Part 1–Block Coding and Decoding  ‹  191 
4
4
j \
j
4⋅ j

∑ , (
p (1− p )
j
j 2
( ,
(E6.17)
j even ∈
4
4
j \ j 4⋅ j

∑ , (
(0.01) (1−0.01)
j
j 2
( ,
(E6.18)
j∈even

j " \
"

j
,
,
" \
(
( ( ) ( '') +
,
,
(
( ( ) ( '')
(E6.19)

,
"
, ( (
≅ $ ⋅
−"
(E6.20)
Forthelinearblockdecoder,weturntoequation(6.20),whichleadsusto
2
∑
2 ( )
(E6.21)

∑
2 ( )−
∑
2 ( )
(E6.22)
−
∑
2 ( )
(E6.23)

⋅

,
, −
∑
j \
(
( F ( − F)
(E6.24)
(

,
,
j \ 0 10 j \ 1 9 j \ 2
10 10 10
−
,, (
(0.01) (1−0.01) +
, (
(0.01) (0.99) +
, (
(0.01) (0.99)
8
]
]
1
0 1 2
¸
( , ( , ( ,
]
(E6.25)
0.000114
(E6.26)
192  ‹  Chapter Six 
6.4 BenefitsandCostsofBlockCoders
Sofar,allwe’vetalkedaboutarethewonderfulbenefitsofblockcoders.Bymappingk 
bitstonbits(n > k),theyareabletodetectbiterrorsthatoccurintransmission,and,
evenbetter,theycancorrectsucherrors.But,alas,asisthecaseinallthingsengi-
neering(bigandsmall),thereisalwaysatradeoff.Weknowwhatwegainfromblock
codes—howaboutwhatwelose?
TakealookatFigure6.12(a).Thereyouseethreebits,mappedbyabaseband
modulatortoanon-return-to-zeroformat.EachbitsentisofdurationT.Nowlookat
Figure6.12(b).Thereyou’llseethatablockcoderwasintroduced,andittookthree
bitsandmappedittosixbits.Tooperateinrealtime,thechannelcoderhadtosmoosh
(squeeze)allsixbitsinthetimeintervalthattheoriginalthreebitsfitinto.So,the
samemodulatormapsthesixbitsintoanon-return-to-zeroformat,whereyoucan
clearlyseethateachbitisofdurationT/2.
Baseband
modulator
011101
101 101
Channel
coder
Baseband
modulator
1 0 1 0 1 1 1 0 1
T T T T
/
2
T
/
2
T
/
2
T
/
T
/
2
T
/
2 2
3T 3T
(a) (b)
Figure6.12Illustratingthecostofchannelcoding
We’reataplacewhere,withoutthechannelcoder,thebitssentacrossthechan-
nelareofdurationT,andwiththechannelcoding,thebitssentacrossthechannelare
ofdurationT/2.Itiseasilyshown(andI’veshownitinChapter5),thatthebandwidth
(frequencyrange)occupiedbyasentbit,BW,variesinverselywithbitduration.So,
before(orwithout)thechannelcoding,wehaveabandwidthof1/T,andwithchannel
codingwehaveabandwidthof2/T.Thismeansthatchannelcodingcomesatthecost
ofsignificantlyincreasedbandwidth.
Channel Coding and Decoding: Part 1–Block Coding and Decoding  ‹  193 
Specifically,thetradeoffisthis.Youdecidetouseablockcoder.Yougettodetect
andcorrecterrorsatthereceiverside.Youpaythepriceofrequiringmorebandwidth
tosendthesameamountofinformation.
6.5 Conclusion
Anotherchaptercomeandgone.Welearnedaboutsomeinterestingdevicescalled
blockcoders.Theyaddextrabitstothebitstream,andthatcostsyoubandwidth,but
theygiveyouthepowertodetectandcorrecterrorsatthereceiverside.
Yousawthreewaystoimplementthis.Firstandsimplestwasthesingleparity
checkbitcoder,whichaddedonebittoeachblockofkbits,puttingoutk+1bits.With
thisaddedbit,whichmadesureyourk+1bitsaddedto0,youcoulddetectanodd
numberoferrors.
Rectangularcodersdidonebetter.BymappingM·Nbitsto(M+1)·(N+1)bits,you
wereabletocorrectonebiterror.
Thenlinearblockcoderswereintroduced.Theyrequiredthatyouroutput
codewordsbeasubspace,andbymakingthatsimplerequirementtheygaveyoua
powerfulwaytobuildtheblockcoder.Theblockdecoderusesasyndromeinorderto
correcterrors.
Finally,wegaveyouawaytocompareblockcoders,byprovidingtwoperfor-
mancemeasuresandtellingyouhowtocomputethemforthedifferentblockcoders
availabletoyou.Theend,fornow.
/
194  ‹  Chapter Six 
Problems
1. Considerarate6/7channelcoderusingasingleparitycheckbit.
(a) Whatistheoutputforinput100101110011?
(b)Doesthechanneldecoderdetecterrorsif(1)bit2isinerror?(2)bit2and
bit8areinerror?(3)bit2andbit4areinerror?(Explain.)
2. Considerarate9/16rectangularcode.
(a) Whatdoesthechannelcoderoutputforinputbits100110001?
(b)Ifanerroroccursatbit3,explainhowtheerroriscorrected.
(c) Ifanerroroccursatbit3andatbit7,explainifandhowanerroriscorrected.
3. (a)Whatarethetworulesthatablockcodemustfollowinordertobealinear
blockcode?
(b)Usingthesetworules,makeupa(4,2)linearblockcode,drawingatableto
describeit.
(c) Verifythatthe(4,2)codeyoumadein(b)isindeedalinearblockcodeby
showingthatitsatisfiesthetworulesin(a).
(d)Usingtrialanderror(oranyothermethodyoucanthinkof),findthe 
generatormatrixGthatdescribesthelinearblockcode. 
(e) VerifythatGisindeedthegeneratormatrixbyinsuringthatforeveryinput
mitsatisfies
u=mG  (Q6.1)
(f) FindtheparitycheckmatrixH,andshowthatGH =0.
(g)Buildasyndrometableforcorrectingerrors(atablewitherrorsinone
columnandsyndromesintheother).
(h)Demonstrate(usingyoursyndrometable)whathappenswhenv = u + e =
(0000)+(0001)enterstheblockcoder.
4. Considerthe3-bitto6-bitlinearblockcoderdescribedbygeneratormatrix
j

\
(
(
(
,
,
,
(

(Q6.2)

,
Channel Coding and Decoding: Part 1–Block Coding and Decoding  ‹  195 
(a) Plotatableofinputbitsandoutputbitsthatdescribethelinearblockcoder.
(b)DeterminetheparitycheckmatrixH.
(c) Createasyndrometable,witherrorsinonecolumnandsyndromesinthe
other.
(d)Explainifandhowthechanneldecodercorrects e =(001000).
5. Imagineyouhaveachannelcoderanddecoder.Inbetweenthemis
• aBPSKmodulatorwithA=1andT=2
• anAWGNchannelwithN  =2
o
• anoptimalBPSKdemodulator(Chapter5)
(a) WhatistheprobabilitypthatabiterroroccursattheBPSKdemodulator?
(b)Ifarate2/3paritycheckbitchannelcoder/decoderisused,whatisthe
probabilitythatthechannelcoderfailstodetectanerror?
(c) Ifarate4/9rectangularcodeisused,determinetheprobabilitythatyou
failtocorrectanerror.
(d)Givena3/6linearblockcodewitht=1,findouthowlikelyitisthatitfails
tocorrectanerror.
6. Nameachannelcoder/decoderthatcan
(a) Detectall1and3biterrors.(Explain.)
(b)Detectall1,3,and5biterrors.(Explain.)
(c) Correctall1biterrorsinevery4bits.(Explain.)
(d)For(a),(b),and(c),provideanexampleofareceivedsignalwithanerror
onit,andshowhowthechannelcoder/decoderdetectsorcorrectsit.
(e) Giventheinputbits111000101001,providetheoutputbitsforthethree
channelcodersyouprovidedin(a),(b),and(c).
7.StudyFigureQ6.1.Nowsketchtheoutputofthemodulatorwhen(1)the
modulatorisBPSKand(2)themodulatorisQPSK.
Channel
111101
Modulator
Modulator
Singleparitycheckbit
FigureQ6.1
3bit 4bit
Achannelcoderandmodulator
7 
Chapter 
Channel Coding and Decoding:
Part 2–Convolutional Coding and Decoding 
I
nChapter6,wetookacarefullookatblockcodersanddecoders.We’renowgoing
tolookatanotherclassofchannelcoderanddecoder,somethingcalledaconvolu-
tionalcoderanddecoder.Justlikethenameindicates,convolutionalcodersand
decodersarealittlebitmorecomplicatedthanblockcodersanddecoders.
7.1ConvolutionalCoders
Atthetransmitterside,convolutionalcoders,likeblockcoders,takeeachsetofkbits
andputoutasetofnbits.Thatmeansanextra(n –k)bitsareintroduced.Atthe
receiverside,aconvolutionaldecoderusestheextrabitstodetectandcorrectbit
errors.
Thereisonemaindifferencebetweenconvolutionalcodersandblockcoders.In
blockcoders,youtakeablockofk inputbitsandyouputoutablockofnoutputbits.
Thenoutputbitsonlydependontheincomingkinputbits.Convolutionalcodershave
whatcouldbedescribedasagreaterappreciationofhistory.Inconvolutionalcoders,
eachk-bitblockthatcomesinismappedtoann-bitoutput.Butwithconvolutional
coders,then-bitoutputdependsonthecurrentk-bitblockthatcamein,andalso on
thepreviousKblocksofkbitsthatcamein.ThisisseeninFigure7.1.Ratherthan
elaborateingeneral,let’sgotoanexample.
7.1.1OurExample
We’lltakealookataconvolutionalcoderwhichtakesinblocksofk=1bitatatime
andputsoutachunkofn=2bits.Now,ifthiswereablockcoder,wemightsay,“Ifa0
comesin,mapittoa(00),andifa1comesin,mapittoa(11).”Butit’snotablock
coder.Thisconvolutionalcodermakesitsdecisionnotjustconsideringthecurrentk=
1bitbutalsobyconsideringthepreviousK=2k=2bits.
Apicturewillmakethismuchclearer,sotakealookatFigure7.2.Thearrowon
theleftshowswherethebitscomein.Withk=1,bitscomeinonebitatatime.The
boxwithlinesthatthek=1bitwalksintoismeanttorepresenta3-bitshiftregister.
198  ‹  Chapter Seven 
1 1 1 1 0 0 1 1 1
Coder
. .
Convolutional
remembers
Blockof
Blockofn=6
k=3bits
1 1 0 1 1 1
bitsoutput
comesin
PreviousK=2sets 
ofk=3bits. 
Usesthisplusinputto 
determineoutput. 
Figure7.1Convolutionalcoderidea
Convolutionalcoder
1
1 1
0 1
1
+
+
Blockof
k=1bit
comesin
Blockof
n=2bits
comesout
1
0
1 2 3
holds stores
current K=2
incoming previousbit
bit setsofk=1bit
Figure7.2 Exampleofconvolutionalcoder
Channel Coding and Decoding: Part 2–Convolutional Coding and Decoding  ‹  199 
Whenthek =1bitcomesin,itisstoredinthepositionlabeled“1.”Itmovesthebit
thatwasinposition1overtoposition2,andwhatisinposition2overtoposition3.
Now,weknowwhatishappeningtotheinputs,butwhatabouttheoutputs?You
canseetwoadders,oneatthetopofthepictureandoneatthebottom.Thetopadder
takesallthreebitsintheshiftregister,addsthemtogether(modulo2),andusesthat
tocreatethefirstbitoftheoutput.Thebottomaddertakesthefirstandthirdbits,adds
themtogether(modulo2),andthisformsthesecondbitoftheoutput.Theswitchat
theveryrightismeanttoshowtheconversionofthetwooutputbitsfromparallel
formtoserialform.
Inthisway,k =1bitcomesin,andusingthisonebitandtheK=2previousbits,two
outputbitsarecreated.Wegetonebitin,twobitsout,andthisisaconvolutionalcoder.
7.1.2MakingSureWe’veGotIt
Figure7.3showsasimpleexampledescribingtheworkingsoftheconvolutional
channelcoderinFigure7.2.Thebitscominginare1,then0,then100.Let’stogether
figureoutwhatiscomingoutoftheconvolutionalcodergiventhisinput.Attime0,
beforethingsgetstarted,wehaveall0sintheshiftregister.Now,let’smovetotime1,
whenbit1comesin,andseewhatcomesout.LookingatFigure7.3,weseethatwhen
a1comesintotheshiftregister,itentersintoposition1anda00endsupbumpedinto
position2and3.Sotheshiftregistercontentsare100.Thatmeansthatoutput1is
1+0+0(modulo2)=1,andtheoutput2is1+0(modulo2)=1.Sotheoutputatthis
timeis11.
Time Inputbit Shiftregister Outputbits
contents bit1 bit2
1  2  3 
0 -
1 1
0  0  0 
1  0  0 
0  1  0 
1  0  1 
0  1  0 
0 0 1 
1 1 
2 0  1 0 
3 1  0 0 
4 0  1 0 
5 0  1 1 
Figure7.3 Showingtheworkingsoftheconvolutionalcoder
Similarly,attime2a0comesin,anditbumpsthe10intoposition2and3inthe
shiftregister.Asaresult,thetwooutputbitsare:0+1+0=1(forthefirstoutputbit),
and0+0=0forthesecondoutputbit.Wecancontinueinthiswayforalltheoutput
bits,andsoonweendupwiththeoutputsshowninFigure7.3.
_____________________________________________________________________
200  ‹  Chapter Seven 
7.1.3PolynomialRepresentation
1+X+X
2
Anotherwaytoexpresstheoperations
oftheconvolutionalcoderisthrough
theuseofpolynomials.Letmeexplain
byexample.Considertheconvolutional
codershowninFigure7.4,looking
specificallyatthelinesconnectingthe
shiftregistertothetopadder,(which
outputsthefirstoutputbit).Wehavea
lineconnectingshiftregisterposition1
totheadder,andwerepresentthisasa
1+X
2
1.Wealsohavealineconnectingshift
registerposition2totheadder,andwe
Figure7.4 Polynomialrepresentationof
representthisasX.Finally,wehavea
convolutional coder
lineconnectingshiftregisterposition3totheadder,andwerepresentthisasX
2
.We
denotealltheseconnectionsthatfeedthetopadderoutputtingbit1bythepolynomial
g
1
(X) = 1 + X + X
2
(7.1)
Similarly,wedescribethebottomadder,outputtingbit2,inthisway.Asthis
adderisconnectedtoshiftregisterposition1,thisgetsrepresentedbya1.Asitisalso
connectedtobitregisterposition3,thisgetsrepresentedusinganX
2
.Puttingthis
together,thebottomadderoutputtingbit2isdescribedbythepolynomial
g
2
(X) = 1 + X
2
(7.2)
Usingg
1
(X)andg
2
(X),wehaveafulldescriptionofthechannelcoder,andwecan
usethisdescriptiontodeterminetheoutputgiventheinput.Let’ssaytheinputsare
thebits1,then0,then1,then0,then0,whichwecanwriteasm=10100.We
representthisinpolynomialformasm(X)=1+0X+1X
2
+0X
3
+0X
4
=1+X
2
.Now,we
cangettheoutputbysimplepolynomialmultiplicationlikethis:
outputbit1atdifferenttimes:m(X)⋅g
1
(X)=(1+X
2
)(1+X+X
2
)=1+X+ + X
3
+X
4
outputbit2atdifferenttimes:m(X)⋅g
2
(X)=(1+X
2
)(1+ X
2
)=1+ +X
4
totaloutputbitsatdifferenttimes 11 10 00 10 11
(Note:Togetthecorrectresultwhenmultiplyingm(X)⋅g
i
(X)usemodulo2
addition,e.g.,X
2
+X
2
=(1+1)X
2
=0X
2
=0.)
1
1
1
1 0
1 1
0
+
+
1 2 3
X
2
X
2
1
1
X
Channel Coding and Decoding: Part 2–Convolutional Coding and Decoding  ‹  201 
IfyoucomparethistoourworkinSection7.1.2,youcanconfirmthatwegotthe
correctoutputbitsforthegiveninputbits.So,ifyouhavealoveofpolynomials,you
canalwaysrepresentyourchannelcoderusingpolynomials.Therearemanyother
possiblerepresentationsfortheconvolutionalcoder,suchasimpulseresponseand
statetransitiondiagrams,butratherthangoingthroughallofthese,we’lljumptothe
mostusefulofthemall,thetrellisdiagram.
7.1.4TheTrellisDiagram
Thetrellisdiagram,showninFigure7.5,isawayofrepresentingwhatgoesonatthe
convolutionalcoderfromonetimetoanother.
time0 time1 time2
1 2
00
00 
11 
01
10
11
(a)
time0 time1 time2 time3
1 2
00
00
01
10
11
11
00 00
00 00
10 10
10
10
10
10
01
01
01 01
01
01
00
11
11
11
11
11
(b)
Figure7.5 Trellisrepresentationofconvolutionalcoder
(a)partial(b)complete
202  ‹  Chapter Seven 
Figure7.5(a)showsyouthebasicsofatrellisdiagram.Toconstructatrellis
diagram,drawasetoftimes,saytime0,thentime1,thentime2,andsoon,going
fromlefttorightonthepage.Beloweachtime,drawabigdot,oneforeachpossible
state ornode.Thestate(node)isashortwordforsaying“whatwillstayintheshift
registerafterthenewbitcomesin.”Forexample,lookatthecoderofFigure7.2.Ifa
newbitcomesin,whatwillstayintheshiftregisteriswhatiscurrentlyinposition1
andposition2(whatisinposition3willgetbumpedout).Sointhatcasethestate=
(bitinposition1,bitinposition2).Possiblestatesare00,01,10,and11.Sobeloweach
timewedrawthestates.
Nextweadddottedlinesandsolidlinesleavingeachstateandenteringanew
state.Adottedlineisusedtodescribewhathappenswhena1entersattheinput.So
forexample,forthecoderofFigure7.2,saya1comesinandweareatstate00(0isin
position1,and0isinposition2).Then,when1comesin,theshiftregisterwillnow
contain(100),whichmeanstheoutputbitswillbe11andthenewstate=(what’sin
position1,what’sinposition2)=(10). ThiseventisshowninFigure7.5(a),onthe
trellisdiagram.
Weaddadottedlineandasolidlinetoeachandeverydot(state)ateachand
everytime,whichleadsustothediagramofFigure7.5(b).Thisiscalledthetrellis
diagram,anditfullydescribesallthepossibleongoingsoftheconvolutionalcoder.It
tellsyouwhatcomesout(bylookingatthetopoftheline)givenwhatwasintheshift
registerposition1andposition2(thedot)andtheinputbit(theconnectingline).
Example 7.1 
Determinethepolynomialrepresentationandthetrellisdiagramfortheconvolu-
tionalcoderdescribedbyFigureE7.1.
Solution:UsingtherulesoutlinedinSection7.1.3,thepolynomialrepresen-
tationcorrespondsto
g
1
(X) = 1 + X
2
(E7.1)
g
2
(X) = X
2
(E7.2)
UsingtherulesoutlinedinSection7.1.4,thetrellisdiagramrepresentationis
showninFigureE7.2.
+
FigureE7.1
Convolutional Coder
in out
Channel Coding and Decoding: Part 2–Convolutional Coding and Decoding  ‹  203 
00
01
10
10
11
11
00
00
FigureE7.2
Thetrellisdiagram
01
10
11
01
7.2 ChannelDecoding
Let’snowexplorewhatthechanneldecoderdoes.First,we’lldefinethetaskofthe
channeldecoder.InFigure7.6,atthechannelcoder,abitsequencemcomesin;we’ll
considerthechannelcoderofFigure7.2withm=(000)comingin.Thisismappedto
theoutput,whichwe’llcallu;intheexamplewe’reconsidering,u=(000000).Thesesix
bitsaresentacrossthechannelbythemodulatorandreturnedtosixbitsbythede-
modulator.Thebitsthatcomeoutofthedemodulator,whichfeedthechanneldecoder,
willbereferredtoasv.Thesebitsv mayormaynotbeequaltou.Anerrormayhave
occurredintransmission,resultinginv =u+e,whereerepresentsthebiterror.Forthe
examplewe’reconsidering,we’llsaywereceivev =u +e =(000000)+(000001)=
(000001).Inthiscase,onebiterrorhasoccurredintransmissionatposition6.
Thegoalofthe
channeldecoder,
then,istotakethe
u=(000000) m=(000)
bitsvthatithas
received,andcome
upwiththebest
C
Coder
andoutputs2bits)
Convolutional
ofFigure7.2
(takesin1bitatatime
Modulator
h
guessatm,the a
n
originalinformation n
e
bits.We’llcallthe
l
channeldecoder’s
guessatmthevector
Convolutional
Decoder
m =bestguessonm
m′.Wewanttofinda
waytomakem′as
v=(000001)
abiterror
Demodulator
occurredhere
closetomaspos-
Figure7.6 Theroleofthechannel(convolutional)decoder
sible.
204  ‹  Chapter Seven 
7.2.1UsingaTrellisDiagram
Howdoesthechanneldecoderusea“trellisdiagram”tohelpitfigureouthowtoput
outanm′thatmatchesm?ThetrellisdiagramofFigure7.5(b)showsthepossible
outputbitsofthechannelcoder.Ifyoulookatthetopofthedottedanddashedlinesof
thetrellisdiagram,youseethosepossibleoutputs.
Let’sreturntoourexampleofFigure7.6andconsiderthetransmittedoutput
u=(000000).Wecanusethetrellisdiagramtoverifythatthisuisapossibleoutput
ofthechannelcoder.Lookingatthetrellisdiagram,weseethat(000000)isapossible
outputbyfollowingthechainof00outputsthatsitabovethesolidlinesatthetopof
thetrellisdiagram.Whataboutthereceivedinputv=(000001)?Isthatapossible
outputofthechannelcoder?Ifyoulookthroughthetrellisdiagram,youcanfind00
outputattime1,followedbya00outputattime2,butyouseeitcan’tbefollowedby
theoutput01attime3.Sothereceivedv=(000001)isnotapossibleoutputofthe
channelcoder.
Therefore,onethingthechanneldecodercandoislookatthereceivedv(for
example,v =(000001))andaskitselfifitmatchesanoutputpaththroughthetrellis.
Iftheanswerisno,thentheremustbeanerrorintransmission.That’saneasywayto
detecterrors.
Aconvolutionaldecodercanalsocorrecterrorsusingthetrellisdiagram.Todo
this,itsimplyasksitselfthefollowing:Givenv=(000001),whatisthepathinthe
trellisclosesttothisv(intermsoffewestbitsdifferent)?Exhaustivelysearchingall
pathsinthetrellis,theclosestpathtov=(000001)isu′=(000000),asseenin
Figure7.7(a).Thechanneldecoderdecidesthatthesentbitsmusthavebeenu′=(00
0000).“Ifthesearethesentbits,thentheinputthatcreatedthemism′=(000),”
saysthechanneldecoder,“andsowehavedecidedonouroutput.”
Thereisasimpletoolthatwecanusetofindm′oncewehavefiguredoutu′,the
closestpathinthetrellistothereceivedbitsv.Allwehavetodoislookatthetrellis—
attheseriesofsolidanddashedlinesthatcorrespondstou′.Asolidlinetellsusa0is
thebitinm′andadashedlinetellsusthata1isthebitinm′.Letmeexplainby
example,usingFigure7.7(b)tohelp.InFigure7.7(b),youseethatu′=(000000)is
theoutputpathcorrespondingtothetoplinesofthetrellis.Now,ifyoulookatthefirst
branchofthepath,thereyouseebelowthe00asolidline—thattellsyouthefirst
outputbitinm′isa0.Ifyoulookbelowthesecond00,youseeasecondsolidline—
thistellsyouthesecondoutputbitisa0,andsoon.
So,thefunctionofaconvolutionaldecoderisreallyquitesimple.Itlooksatv,
thenlooksatthetrellisdiagramandsearchesituntilit’sfoundtheoutputpathinthe
trellisu′closesttov.Fromthisu′,itdecidesonandoutputsm′.
Channel Coding and Decoding: Part 2–Convolutional Coding and Decoding  ‹  205 
time0 time1 time2 time3
.  .  . 
00
.  .  . 
01) received: v =(00
0difference 0difference 1difference
1 2
onepossiblepath
00
a2
nd
possiblepath
01
10
a3
rd
possiblepath
11
00
00 00 00
time0 time1 time2
(a)
11
11 11 11
00 00
00 00
10 10
10 10
10
10 10 10
10
10
10
10
01 01
01 01
01
01 01 01
01
01 01 01
00
00
00 00
11 11
11 11
11
11 11 11
time3
u
. . . . . .
00 00)
=bitssent=(000000) u
inputbitis0 inputbitis0 inputbitis0
=(00
1difference 1moredifference 1moredifference
solidlineindicates solidlineindicates
Total1different
Total2different
Total3different
Decide:
1 2
00 
01 
10
11
Decide: =(000) m
(b)
Figure7.7
(a)Decidingonclosestpathintrellis(3pathsshown)
(b)determiningoutputbitsatconvolutionaldecoder
206  ‹  Chapter Seven 
Example 7.2 
FortheconvolutionalcoderdescribedbythetrellisdiagramofFigureE7.2,
determinetheoutputofthechanneldecoderwhenthechanneldecoderreceives
v =(1111).
Solution:ThetrellisdiagramisdrawnovertwotimesinFigureE7.3.Look-
ingoveratthisfigure,weseethattheoutputofthetrelliscodercanbe(1111)
whenitfollowstheshadedpath.Thistellsusthattheinputbitsthatweresent
musthavebeenm =(00),since:thebestpathcorrespondstotwosolidlines,
eachsolidlineindicatinganinputbitof0.
00
00
01
01
01
10
11
00
10
11
11
10
00
01
01
00
10
11
11
10
FigureE7.3 Trellisdiagramover2times
7.2.2TheViterbiAlgorithm
OnethingIhaven’tyetmentioned—ifyouhavereceived,forexample,v =(00000111
10001000001111110110010101111101),thenyouhavetosearchthroughthe
trellistofindthepathclosesttothisv.Butthatsearchisalongone.Luckily,afellow
namedViterbipresentedasimplewaytolookthroughthetrellis.TheViterbiAlgo-
rithm(VAforshort)letsyou(orbetteryet,yourcomputerorDSPchip),findthe
closestpathtovinthetrelliseasilyandeffectively.
Getting the Basic IdeaTheViterbiAlgorithmisbasedonaverysimpleidea:
Startatthetime0inthetrellis,andmovethroughthetrellisfromlefttoright.Ateach
time,youcansystematicallyeliminatesomeofthepathsinthetrellisasbeingclosest
tov.Infact,accordingtotheViterbiAlgorithm,youcaneliminate(ateverytime)all
Channel Coding and Decoding: Part 2–Convolutional Coding and Decoding  ‹  207 
butfourofthepathsthroughthetrellisasbeingclosesttov,forthetrellisofFigure
7.5.(Ingeneral,youcaneliminateallbut“s”pathsthroughatrellisatanyonetime,
where“s”isthenumberofstates.)
ThispatheliminationisbrieflyexplainedwiththehelpofFigure7.8.AttimeL+1,
lookatthetopnode.Therearetwopathsthatheadintothisnode,oneoriginatingfrom
parentnodeAandtheotheroriginatingfromparentnodeB.AttimeL+1,youcan
makeadecisionastowhoisthebetterparentandchoosebetweenAandB.You
repeatthisforeverynodeattimeL+1,thusleavingyouwithfourpathsthroughthe
trellisateverytime.
Understanding by ExampleLet’ssayweinputtothechannelcoderofFigure
7.2thebitsm=(00),inwhichcaseitoutputsthebitsu=(0000).Let’salsosaythe
channeldecoderreceivesv=(0000).Let’susetheViterbiAlgorithmtosearchforu′,
thepaththroughthetrellisclosesttov.Oncewe’vefoundu′,wecanoutputm′,our
guessatthebitssentintothechannelcoder.
First,wedrawthetrellis,andweassociatewitheachstartnodethenumber0.
Thismeansthatwearenotbiasedtowardanyonestartnodeoveranyother.Thisis
showninFigure7.9(a).Then,westartbyexaminingthetopnodeattime1,whichyou
canseeinFigure7.9(b).Forthisnode,therearetwopossibleparentnodes,node0at
time0andnode1attime0.Wearegoingtodecidewhichisthebestparentnode,
usingthisprocedure:
1.Ifwestartedatnode0(time0)andmovedtonode0(time1),thefirstoutput
bitswouldbe00.Comparingthistovwherethefirsttwooutputbitsare00,we
say“0biterrorsifparentisnode0(time0).”Weaddthis0tothe0 numberthat
wegavenode0(time0)inFigure7.9(a),foragrandtotalof0.
2.Ifwestartedatnode1(time0)andmovedtonode0(time1),thefirstoutput
bitswouldbe11.Comparingthistovwherethefirsttwooutputbitsare00,we
say“2biterrorsifparentisnode1(time0).”Weaddthis2tothe0 numberthat
wegavenode1(time0)in
Figure7.8(a),foragrand
totalof2. timeL timeL+1
Now,sincestartingatnode0
(time0)andmovingtonode0
(time1)createsthefewesttotalbit
errors(0),weproclaimthatthe
00
01
nodeA
nodeB
Candecideonbest
parentnode(AorB)
parentnodefornode0(time1)is 10
node0(time0),andthatitcarries
withitthenumber0(forzerototal 11
errorswiththisselection).We
Figure7.8 UnderlyingideaofVA
208  ‹  Chapter Seven 
time0 time1
00 0
01 0
10
0
11 0
(a)
Readthisfirst
v
= ( 00
00
)
difference:
00
11
0
node0
Total
2difference:
Total
Readthislast
0 0= 0 + 
time0
time1 
3
node0 Bestparent:
00 0
0
node0
01 0 lowesttotal
+
node1
0 2 = 2
10
0
11 0
(b)
Figure7.9
(a)Settinginitialvaluesto0
(b)pickingbestparentnode(withlowesttotal)fornode0
Channel Coding and Decoding: Part 2–Convolutional Coding and Decoding  ‹  209 
00
10
01
10
11
01
time0 time1
v 00
0
0
0
0
= ( 00 )
node2
node3
node1 
1
node2
difference1:
Total0+1=1
difference1:
Total0+1=1
lowesttotal
Bestparent:
(c)
time0 time1 
node2
node3
node1
lowesttotal0
lowesttotal1
Bestparent:
Bestparent:
(d)
node2 
Figure7.9
(c)pickingbestparentnodefornode1
(d)bestparentnodefornodes243
210  ‹  Chapter Seven 
repeatthisfornode1(time1).YoucanseethisongoinginFigure7.9(c).There,node
2(time0)andnode3(time0)arepossibleparentnodes.Wedecidedbetweenthese,
asfollows:
1.Fornode2(time0)asstartingnodemovingtonode1(time1),theoutputis
10.Comparingthistothefirsttwobitsofv,whichare00,wesay“1biterrorif
parentnodeisnode2(time0).”Weaddthis1tothe0 numberthatwegavenode
2(time0)inFigure7.9(a),foragrandtotalof1.
2.Fornode3(time0)asstartingnodemovingtonode1(time1),theoutputis
01.Comparingthistothefirsttwobitsofv,whichare00,wesay“1biterrorif
parentnodeisnode3(time0).”Weaddthis1tothe0 numberthatwegavenode
3(time0)inFigure7.9(a),foragrandtotalof1.
Sincestartingatnode2(time0)ornode3(time0)andmovingtonode1(time1)
createsthesametotalbiterrors,weproclaimthattheparentnodefornode1(time1)
isnode2(time0)(weuseatie-breakerruleof“alwayschoosethetopnodeincaseofa
tie”).Thatcarrieswithitthenumber1(foronetotalerrorwiththisselection).
Werepeatthisfornode2(time1)andnode3(time1),andtheresultofdoingthis
isshowninFigure7.9(d).Thatisallwedoforourfirstmovefromlefttoright
throughthetrellis.
Atthenexttime,wedoaverysimilarthing.Westartagainatthetopnode,this
timestartingwithnode0(time2).LookingatFigure7.10(a),wecanseethatthisnode
hastwopossibleparentnodes,whicharenode0(time1)andnode1(time1).We
decidebetweenthesetwonodesasfollows:
1.Ifwestartedatnode0(time1)andmovedtonode0(time2),thesecondsetof
outputbitswouldbe00.Comparingthistov wherethesecondsetofoutputbits
are00,wesay“0biterrorsifparentisnode0(time0).”Weaddthis0tothe0 
numberthatwegavenode0(time1)inFigure7.9(b),foragrandtotalof0.
2.Ifwestartedatnode1(time1)andmovedtonode0(time2),thesecondtwo
outputbitswouldbe11.Comparingthistovwherethesecondtwooutputbitsare
00,wesay“2biterrorsifparentisnode1(time1).”Weaddthis2tothe1 num-
berthatwegavenode1(time1)inFigure7.9(c),foragrandtotalof3.
Sincestartingatnode0(time1)andmovingtonode0(time2)createsthefewest
totalbiterrors(0),weproclaimthattheparentnodefornode0(time2)isnode0(time
1),andthatitcarrieswithitthenumber0(forzerototalerrorswiththisselection).We
repeatthisfornode1(time2),node2(time2)andnode3(time2);theresultsare
showninFigure7.10(b).
3
Channel Coding and Decoding: Part 2–Convolutional Coding and Decoding  ‹  211 
node1
v = ( 00
00
)
difference:0
Total0 0 0 + =
time0 time1
time2 Bestparent:
00
node0
node0
node0
0lowesttotal
node1
1
11
time2
(a)
1
0
1
1
difference:2
Total1+2=3
lowesttotal
lowesttotal
lowesttotal
node1 Bestparent
2
node2 Bestparent
1
node3 Bestparent
1
2
(b)
Figure7.10(a)bestparentfornode0(time2)
(b)bestparentfornodes1,2,3(time2)
Wecontinuethisprocessuntilwehaverunthroughthetrellisforthelengthof
timecorrespondingtothelengthofv.Inourcase,sincevconsistsonlyoftwosetsof2
bits(v=(0000)),wearedoneaftertwotimes.Attheend,wehavefourendnodeswith
fournumbers.Forexample,endnode0(time2)comeswithvalue0,whileendnode2
(time2)comeswiththevalue1.Wechoosetheendnodewiththesmallestvalue.In
ourexample,lookingatFigure7.10,wechoosetheendnode0.
212  ‹  Chapter Seven 
Fromhere,weknowthehistoryofparentnodes,sowecan“backtrack”through
thetrellis,anddeterminetheu′andthem′.Andwe’redone.Inourcase,wechoose
node0(time2)withafinalvalueof0,andwe“backtrack”asshowninFigure7.11.
Thisleadsustotheoutputm′=(00).
Nowyouknowjusthowthechanneldecoderworkstoundotheeffectsofchannel
coding,andalongthewaycorrectbiterrors.
00
01
time 0 time1 time2
node0
backto
node0
4thitem
Bestparent=
startyour
readinghere
Followpathback
Bestparent
isnode0
Followline
toread
tonode 0=
bestparentnode
10
11
5 4
,youhavepaththroughtrellis.
Thispathtellsyouu
From
1
to
=(0000)andm =(00)
Figure7.11Explaining“backtracking”throughtrellistogetoutput
Example 7.3 
UsetheViterbiAlgorithmtodeterminetheoutputofaconvolutionaldecoder,
given
• theinputbitstothedecoderarev=(1111)and
• theconvolutionalcoderisdescribedbythetrellisdiagraminFigureE7.2.
Solution:FigureE7.4showsthetrellisdiagramwhentheViterbiAlgorithm
isperformed.Itshows(1)thenumber-of-errorscomputationforeachbranch,(2)
thebestparentselectionateachnode,(3)theverybestfinalnodeselection;and
(4)backtrackingtodeterminethebestpaththroughthetrellis.Fromthisbest
path,thedecoderdecidesthatthebestoutputis(00).


E
@
A
1
Channel Coding and Decoding: Part 2–Convolutional Coding and Decoding  ‹  213 
receive: 11 11
00 error=2(total2) 00 error=2(total2)
0 (00)
(01)
2 (10)
3 (11)
(total0)
=node1(total0)
(total0)
e
r
r
o
r

=

1

(
t
o
t
a
l

2
)
e
r
r
e
r
r
o
r

=

0

(
t
o
t
a
l
0
)

e
r
r
e
r
r
o
r

=

0

(
t
o
t
a
l
0
)

e
r
r
o
r

=

1

(
t
o
t
a
l
1
)

e
r
r
o
r

=

1

(
t
o
t
a
l

1
)
e
r
r
o
r

=

1

(
t
o
t
a
l

1
)

bestparent
=node 1
bestparent
bestparent=3
10 10
11 11
e
r
r
o
r
=
1(
t
o
t
a
l1)
01
e
rro
r=
1(t
o
t
a
l1)
01
bestparent=3
(total1)
00
e
rro
r=
2
(t
o
t
a
l3
)
00
e
r
r
o
r
=
2
(
t
o
t
a
l2
)

bestparent=0
bestparent=0 10
(total1)
10
(total1)
11
o
r

=

0

(
t
o
t
a
l

0
)

11
o
r

=

0

(
t
o
t
a
l

1
)

bestparent=2
01 error=1(total1) 01 error=1(total2)
(total2)
bestparent=2
(total1)
FigureE7.4 UsingtheVAtodeterminebestpath
7.3 PerformanceoftheConvolutionalCoder
Nowthatweknowhowthecoderandthedecoderinconvolutionalcoding/decoding
work,wewanttoevaluatetheirperformance.Channelcodersanddecodersareintro-
ducedwiththeintentionofcorrectingbiterrors,sowe’lldeterminehowwella
convolutionalcoderisabletocorrectbiterrors.Ingeneral,aconvolutionalcoder
cancorrecte biterrorsinaboutevery4kbits,wherethevaluekisthesolutionto
2
k
=(numberofstates(nodes))andeisthevaluecomputedaccordingto:
− , ]
,
¸

(7.3) 
]
]
d
Thekeytodetermininghowwellthetrellisdecoderisabletocorrectbiterrorsis
min
.Ittellsyouhoweasyitcouldbetomistakeonepathforanotherinthetrellis.
Specifically,d
min
isthesmallestdistancebetweenanytwopathswiththesamestart
andendnodeinatrellis.Alternatively,thisd
min
valueisequaltothesmallestdistance
betweentheall-zeroespathandanotherpaththroughthetrellisthatstartsatnode0
(time0)andendsatnode0(anylatertime).
214  ‹  Chapter Seven 
Forexample,considerFigure7.12.Thereyouseeatrellisandyouseetheall-
zeroespath(thepaththatwhenfollowedgivesyouanoutputthatisall0bits).Yousee
anotherpathhighlightedthatstartsatnode0andendsatnode0.Aboveeachbranch
ofthatpath,youseetheoutputbits,andanumber.Thatnumbertellsyouhowmany
1’sareoutputbyfollowingthispath(andthereforehowfaritisfromtheall-zeroes
path).Youcombineallthesenumbers.Yourepeatthisforeachpaththatstartsatnode
0andendsatnode0.Thesmallestnumberyougetisd
min
.InFigure7.12,d = 5.
min
Apaththatstarts
atnode0 
andendsat
Allzeroespath 
node0
time0 time1 time2 time3
00 00 00 00
01
10
11
11
2
10
1
11
2
5
Totaldistancefrom
allzeroes:
(2+1+2)
Figure7.12Computingd
min
7.4 CatastrophicCodes
Therearesomeconvolutionalcodesthatareverybadtouseinacommunication
system—infact,theyaresobadthattheyareknowninengineeringcirclesascata-
strophiccodes.Aconvolutionalcodeiscalledacatastrophiconewheneverthe
followingispossible.LookatFigure7.13.Therem=(0000000...)isinputtothe
convolutionalcoder,andu=(00000000000000...)isoutputbytheconvolutional
coder.Thisissentacrossthechannel,whereonlythreebiterrorsoccurforalltime,
andwereceivev=(11010000000000...).Inresponse,theconvolutionalcoder
outputsm′=(1111111...).Thatis,onlyafinitenumberofbiterrorswereintro-
ducedinthechannel,buttheconvolutionaldecoder,inseeingthefinitenumberof
errors,madeaninfinitenumberoferrors.Yikes—catastrophic!Canthisreally
happen?Theansweristhatinabadlydesignedconvolutionalcoderitcan!
Let’slookathowaconvolutionalcodercanmakethistypeoferror,soyoucanbe
suretoavoidbuildingonelikethis.LookattheconvolutionalcoderdrawninFigure
7.14.Itcanbewritteninpolynomialformasfollows:
g
1
(X)=1+X  (7.4)
g
2
(X)=1+X
2
(7.5)
1
Channel Coding and Decoding: Part 2–Convolutional Coding and Decoding  ‹  215 
m=(00000...)
Coder
1bitin-2bitsout
Convolutional
u=(00000000...)
Modulator
C
h
a
n
n
e
l
Demodulator Coder:
Convolutional
UsesVA
m=(11111...)
v=(11010000...)
Figure7.13Acatastrophiccode
Convolutionalcoder Inmodulo2multiplicationand
addition,wecanexpressg
2
(X)=
1+X
2
=(1+X)⋅(1+X).Youcansee
thatinthiscaseg
1
(X)isafactorof
g
2
(X).Wheneverthishappens,the
convolutionalcodeiscatastrophic.If
itdoesn’thappen,thenyourcodeis
okay.
0
1 1
0
1
+ 
+ 
Letmeelaborateonthecata-
strophiccodeideawithasimple
Figure7.14Convolutionalcoder
example.TakethecoderofFigure
illustratingcatastrophiccode
7.14.Ifyoutakeafewmomentsto
representitbyatrellisdiagram,
you’llendupwiththediagramshowninFigure7.15.Usingthistrellisdiagram,con-
siderthis:youinputm=(0000...)whichmeansthechannelcoderoutputsu=(0000
0000...).Thechannelmakesonly3biterrors,andyoureceivev=(1101000000...).
Thedecodersearchesforthepathinthetrellisclosesttothisreceivedsignal,andit
findsthatthereisapathwithnoerrorsthroughthetrellis,asshowninFigure7.15.
Usingthispath,theconvolutionalcoderoutputsm′=(11111..).Wow—catastrophic.
216  ‹  Chapter Seven 
Sentthispathu
00
01
10
11
00 00 00 00
time0 time1 time2
11 11 11
11
00 00 00 00
10
10 10 10
11 11 11 11
00
01 01 01
10
10 10 10
01 01 01 01
time3 time4
Receivethispathv
whenonly3
channelerrors
Figure7.15Explainingacatastrophicerrorinacatastrophiccode
7.5 BuildingYourOwn
Let’ssaythatyousetouttobuildyourownconvolutionalcoder.Herearethecriteriato
followtomakeitagoodone(mostimportantcriteriafirst):
1.Makesureit’sNOTcatastrophic.
2.Maked
min
(thesmallestdistancebetweentheall-zeroespathandanyotherpath
startingandendingatnode0)asbigaspossible.Thatmeansyourcodecan
correctmoreerrorse.
3.Makesurethereisonlyonepathwithadistanced
min
inthetrellis.
4.Buildthecodersothatthetrellisdiagramhasasfewpathsaspossiblewith
distanced
min
+1.
5.Buildthecodersothatthetrellisdiagramhasasfewpathsaspossiblewith
distanced
min
+2.
Nowyou’rereadytobuildyourownconvolutionalcoder(although,infairness,
youmayjustenduppurchasingachipwithitsownready-to-gochannelcodersonit,
alreadymeetingtheabovecriteria).
Channel Coding and Decoding: Part 2–Convolutional Coding and Decoding  ‹  217 
Problems 
1. Drawthetrellisdiagramforthek =1,n =3,K =3convolutionalcoderde-
scribedby
g
1
(X) = 1 + X + X
2
(Q7.1)
g
2
(X) = 1 + X (Q7.2)
g
3
(X) = 1 + X
2
(Q7.3)
2. DrawthetrellisdiagramfortheconvolutionalcoderdescribedbyFigureQ7.1.
+
FigureQ7.1
Convolutional coder
out
in
3. GiventhetrellisdiagramofFigureQ7.2,determineitsblockdiagram.
00
0
10
11
0 FigureQ7.2
Trellisdiagram
01 0 0
0 10 0
11 0 0
00
4. Considertheconvolutionalcoder(k =1,n =2,K =3)showninFigureQ7.3.
UsetheViterbiAlgorithmtodeterminetheoutputoftheconvolutionaldecoder
whenitreceives(1000001011).
218  ‹  Chapter Seven 
FigureQ7.3
Convolutional coder
out
+
+
in
5. ConsidertheconvolutionalcoderinFigureQ7.4.UsetheViterbiAlgorithmto
determinetheconvolutionaldecoderoutputgivenitsinputis(1101011001).
+
FigureQ7.4
Convolutional coder
out
in
6. Considertheconvolutionalcoder(k=1)ofFigureQ7.5.
(a) DeterminenandK.
(b)Describetheconvolutionalcoderusingatrellisdiagram.
(c) Assumethesignalreceivedbytheconvolutionaldecoderis00010001.
Determine,usingtheViterbiAlgorithm,theoutputofthedecoder.(In
casesoftiesatanode,alwaysassumethetoppathwins.)
+
FigureQ7.5
Convolutional coder
out
in
7.DrawthetrellisdiagramfortheconvolutionalcoderinFigureQ7.6.Determine
ifthisconvolutionalcoderiscatastrophic.Ifitis,show(usingthetrellisdiagram)
howafinitenumberoferrorsinthechannelcancauseaninfinitenumberof
errorsattheconvolutionalcoder.
Channel Coding and Decoding: Part 2–Convolutional Coding and Decoding  ‹  219 
+
+
in
FigureQ7.6
Convolutional coder
out
8. (a) Buildanyrate1/2convolutionalcoder(donotuseoneoftheonesalready
usedintheproblemsetortheoneusedinthetext).
(b)Describeitusingatrellisdiagram.
(c) Explaininthreetofoursentenceshowtheconvolutionaldecoderworks.
(d)Determineiftheconvolutionalcoderyoubuiltwascatastrophic.
8 
Chapter 
Trellis-Coded Modulation (TCM)
The Wisdom of Modulator and 
Coder Togetherness 
F
igure8.1showsacommunicationsystemwithallthepartsthatwehavedrawnfor
itsofar.Thesourcecoderturnstheinformationtobits;thechannelcoderadds
extra(redundant)bitstohelpthereceivercorrectbiterrors;andthemodulatorturns
bitsintosignalsreadytobesentoverthechannel.Aftertravelingthelengthofthe
channel,anoisysignalarrivesatthedemodulator.Thedemodulatorreturnsthesignal
tobits,thechanneldecodercorrectsthebiterrors,andthesourcedecoderreturns
thebitstotheoriginalanaloginformation.
Intrellis-codedmodulation,orTCMforshort,theideaistoconsidersomethings
together.Specifically,atthetransmitterside,thereisathoughtfulmatchingofthe
modulatortothechannelcoder.Atthereceiverside,theoperationofthemodulator
andthechanneldecoderareactuallycombined.Let’slookatwhythisisdoneandthen
athowitallworks.
r(t)=s(t)+n(t)
Channelcoder
Channeldecoder
x(t)
signal
m
bits
u
bits
s(t)
channel +
n(t)
v
bits
m
bits
x(t)
Modulator Sourcecoder
Demodulator Sourcedecoder
Information
TCMconsidersthistogether
TCMcombinesthis
Figure8.1Thedigitalcommunicationsystem(showingwhereTCMcomesin)
222  ‹  Chapter Eight 
8.1TheIdea
Considerthetransmittersideofacommunicationsystem,whichconsistsofasource
coder,achannelcoder,andamodulator.First,imaginethatthechannelcoderisgone,
andwehavethetransmittershowninFigure8.2(a).Wehaveasourcecoderoutput-
tingbitswithabitdurationofT(abitrateofR=1/T),andaBPSKmodulatorwhich,
foreachbit(0or1)thatcomesin,outputsasignalasshowninFigure8.2(a).Figure
8.2(a)alsoshowsthesignalsoutputbythemodulatorinthefrequencydomain.From
thisfigure,wecanseethattheoutputsignalshaveabandwidth(null-to-null)of
BW=2/T.
Now,let’sconsiderthetransmissionsystemwiththechannelcoderbackinplace,
asshowninFigure8.2(b).Here,thesourcecodermapstheinformationintobitsof
durationT.Theconvolutionalchannelcodertakesthesebits,andforeachbitthat
comesinitputsouttwobits.Tooperateinrealtime,thetwobitsthatcomeoutmust
BPSK
Modulator
Sourcecoder
x(t)
Information
signal 1 0 1
T T T
s(t)=
−Acos(ωt)+Acos(ωt)−Acos(ωt)
c c  c 
T T T
s(t)=−Acos(ω π t) (t T) t) (t 2T) t) (t)+Acos( ω π − − Acos(ω π −
c c c
s(f)
f
f (1/T)
c
− f
c c
f+(1/T)
BW=2/T
Figure8.2(a) Systemwithoutchannelcoder
Trellis-Coded Modulation (TCM): The Wisdom of Modulator and Coder Togetherness  ‹  223 
fit(intime)intothesameamountoftimethattheincomingonebitfitsinto—soeach
bitoutputfromthechannelcoderisofwidthT/2.SeeFigure8.2(b)togetaclearer
pictureofthis.Finally,thebitsthatleavetheconvolutionalcoderarepassedthrough
themodulator,andeach0and1aremappedtothesignalasshownbytheBPSK
modulatorofFigure8.2(b).
ThisfigurealsoshowstheoutputoftheBPSKmodulatorinthefrequencydo-
main.Asyoucansee,thissignalhasabandwidth(null-to-nullbandwidth)ofBW=4/T.
Ifwebrieflycomparethecaseofasystemwithachannelcoderandthecaseofa
systemwithoutone—Figure8.2(b)vs.Figure8.2(a)—youcanseeonecleardiffer-
ence:youbenefitfromachannelcoderbecauseitcorrectserrors,butwhenyouusea
channelcoderyoupaythepriceofsendingsignalsthatrequirealargerbandwidth.
Oneday,afellownamedUngerboecksaidtohimself,“Theremustbesomeway
togetthebenefitofachannelcoderwithoutthecostofincreasedbandwidth.”He
Sourcecoder
BPSK
Channel
Coder
Modulator
Convolutional
s(t) x(t)
Informational
1 0 1 1 1 0 1 0 1 
signal 
T T T
T T T T T T 
2 2 2 2 2 2 
s(t)=
−Acos(ω
c
t) −Acos(ω
c 
t) −Acos(ω
c 
t)
Acos(ω
c
t)
T T T T T T
2 2 2 2 2 2
S(f)
f
f (2/T)
c
− f
c c
f+(2/T)
BW=4/T
Figure8.2(b) Systemwithchannelcoder
224  ‹  Chapter Eight 
cameupwiththeideashowninFigure8.3.First,comparingFigure8.3toFigure
8.2(b)showsusthatthesetwofiguresareverycloselyrelated.Infact,theonlydiffer-
enceisthatthesystemofFigure8.3usesaQPSKmodulator,whilethesystemof
Figure8.2(b)usesBPSK.Let’sseewhatdifferencethismakes.First,thetransmitter
usesthesamesourcecodertomaptheinformationtobitsofdurationT.Then,a
channelcoderisused,whichforeveryonebitthatcomesincreatestwobits,eachof
duration T/2.Finally,themodulator,aQPSKmodulator,takeseverytwobitsandmaps
themtoasymbolasshowninFigure8.3.
Figure8.3showstheoutputofthesysteminthetimeandfrequencydomain.
Lookingatthisoutputinthefrequencydomain,weseethatthesignaloutputhasa
bandwidth(null-to-nullbandwidth)ofBW=2/T.Thisbandwidthisthesameband-
widthasthatoftheoriginal(nochannelcodingsystem)ofFigure8.2(a).
Basically,whatUngerboeckrealized(andwhatFigures8.2and8.3confirm)is
thatifyouintroduceachannelcoder,andfollowitbyamodulatorthattakesinmore
bits,thenyoucanusechannelcodingwithoutincreasingthebandwidthofthetrans-
mittedsignal.Thatturnedouttobeaverygoodidea.
Channel
Coder
Sourcecoder
Convolutional
1 0 1 1 1 0 1 0 1
T T T
T T T T T T
2 2 2 2 2 2
* 
* 
QPSK
Modulator
11 01 01
s(t)=
Acos(ω
c
t+3 /2) π Acos(ω
c
t+ /2) π Acos(ω
c
t+π/2)
T T T
c c
π π
c
π π s(t)=Acos(ω t+3 /2) (t)+Acos( ω t+π/2) (t− T)+Acos(ω t+π/2) (t− 2T)
S(f)
f
f 1/T
c
− f
c c
f+1/T
BW=2/T
Figure8.3 Ungerboeck’sideaforanewtransmitter
Trellis-Coded Modulation (TCM): The Wisdom of Modulator and Coder Togetherness  ‹  225 
8.2 ImprovingontheIdea
Therewerestillanumberofdetailsthathadtoberesolvedbeforethisideawasem-
bracedbytheengineeringcommunity.InFigure8.4(a)youseetheoutputsofaBPSK
modulator,whichareAcos(ω
c
t)π(t – iT)andAcos(ω t +180°)π(t – iT).Anerrorwillonly
c 
occurifthenoiseofthechannelisbigenoughtomakea180-degreephaseshiftoccur.In
Figure8.4(b)youseeadrawingoftheoutputsoftheQPSKmodulator,whichshowsthat
thesignalssentbythemodulatorareAcos(ω t)π(t – iT),Acos(ω
c
t +90°)π(t – iT),
c 
Acos(ω
c
t +180°)π(t – iT)andAcos(ω t +270°)π(t – iT).Inthiscase,anerroroccursif
c 
thenoiseisbigenoughtomakeitlooklikea90-degreephaseshifthasoccurred.By
addingaQPSKmodulatorinplaceofaBPSKmodulator,thenoiseinthesystemwill
introducemoreerrors—possiblysomanymorethatthesystemofFigure8.3willbe
lousywhencomparedtothesystemsofFigure8.2(a)and(b).
BPSK
Modulator
In Out
Outputinsignalspace
0
1
s(t)=Acos( t) (t-iT)
0 c
ω π
s(t)=Acos( t+ ) (t-iT)
1 c
ω π π
s
1
s
0
–A√(T/2) A√(T/2)
ϕ √ ω π
1 c
(t)= (2/T)cos( t)+ (t-iT)
(a)
2A√(T/2)
QPSK
Modulator
In Out Outputinsignalspace
0 0 s(t)=Acos( t) (t-iT) ω
c
π
0 1
s(t)=Acos( ω t+π/2 π
0
) (t-iT)
1 0 s(t)=Acos( ω t+π π
1 c
) (t-iT)
s
1
A√T/2
A√T/2
√
ϕ
2
(t)
1 1 s(t)=Acos( ω t+3π/2 π
2 c
) (t-iT)
2 A√T/2 3 c
s
0
ϕ
1
(t)
s
2
s
3
(b)
Figure8.4 ComparingBPSKtoQPSK
226  ‹  Chapter Eight 
Ungerboeckresolvedthisproblembycomingupwith“mappingbysetpartition-
ing,”showninFigure8.5.Thereyouseeasourcecoder,anewconvolutionalcoder
whichtakestwobitsandmapsintothreebits,andamodulatorusingan8-PSKconstel-
lation.Let’sexaminethisfigure.
+
+
1 2 3
coder
8-PSK
mm
1 2
m
2
m
1
u
1
u
2
u
3
1 2 3
s(t)
Uses
Source
modulator u u u
mappingby
setpartitioning
ConvolutionalCoder
2bitsin 3bitsout
Figure8.5 Ungerboeck’snewtransmitter
First,thechannelcodertakesintwobits(m
1
,m
2
)ateachtimeandoutputsthree
bits(u
1
,u
2
,u
3
)ateachtime.Tobetterunderstanditsworkings,we’lldrawatrellis
diagramforthechannelcoderasshowninFigure8.6.First,thenodes(dots):there
arefournodes,oneforeachpossible(bit at position 1, bit at position 2)pair.Thenthe
branches:thesolidbranchindicatesthatincomingbitm
2
is0;thedottedbranch
indicatesthattheincomingbitm
2
is1.Ifthebranchisdrawnontop(i.e.,ifitisthetop
branchofapair)thenthatmeanstheincomingbitm
1
is0;ifitcorrespondstothe
bottombranch,thenthatmeanstheincomingbitm
1
is1.Foragivenstate(node)and
agiveninput(branch),thethree-bitoutputthatcomesoutofthechannelcoderis
drawnabovethebranch.
Nowletmeexplainthe
time0 time1 time2
modulatorandhowUngerboeck
1 2
000 000
saiditwouldwork.Hisideawas
00 100
100
this:First,drawprettypicturesof
1
1
1
0
11
0
1
1

1
1
1

1
1
1
0
11
0
1
1

1
1
1

themodulatoroutputs.Lookat
01 0
0
0

0
0
0

Figure8.7,whereweseeontop
10
0
0
1
0

110

10
0
0
1
0

110
theeightpossibleoutputsofan8- 10 0
0
1
0
0
1
PSKconstellation,drawnonthe
orthonormalbasis(lookbackto
Chapter5forarefresherifyou’d
11
1
0
1
0
0
1
1
0
1
010
110
1
0
1
0
0
1
1
0
1
010
110
like). Figure8.6 Trellisdiagramdescribingchannelcoder
4
Trellis-Coded Modulation (TCM): The Wisdom of Modulator and Coder Togetherness  ‹  227 
iT) ϕ −√ ω π −
2 c
(t) = (2/T)sin(
s
3
s
4
s
5
s
6
s
7
s
0
s
1
s
2
1
0
7
6
5
4
3
2
t) (t
φ
1 
(t) = √(2/T)cos(ω π −iT) t) (t
c
2 
3  1 
4  0
7  5 
6 
2 
1 3 
0
7  5 
6 
1 3 
2 
4 0 
6 
5  7 
Figure8.7 Splittingthe8-PSKconstellation
228  ‹  Chapter Eight 
Next,separatethesepoints,byputtinghalfontheleftandhalfontheright—
makingsureyouseparatethemsothateverypointhasanewneighbor,andsothat
eachpointinanewsetofpointsisasfarawayaspossiblefromitsnewneighbors.You
canseethisintheseconddrawingofFigure8.7.Then,takethesepointsandsplit
themagain,justasyousplitthefirstset.Youcanseethisinthethirddrawingof
Figure8.7.Dothisagain,andagain,untilallthat’sleftisasetofsinglepoints,justas
onthebottompictureofFigure8.7.
AtthispointyoumightthinkthatUngerboeckhadgivenupengineeringalto-
getherandtakenanunusualinterestinmodernart,butfearnot,hehadgood
engineeringintentions.Next,hesaid,youbuildamodulatorwiththiscool-looking
picture,whereatthebottomofthepicturearesinglepoints(representingoutput
modulationsignals).Youmatchthebitscomingintothemodulatortotheoutput
modulationsignalsshownatthebottomofthepictureinFigure8.7,usingthetrellis
diagram.Ungerboeckcameupwithjusttworules:
(1)Allparalleltransitions(transitionsthatstartandendatthesamenode)are
separatedbythemaximumpossibledistance;and
(2)Alltransitionsdivergingfromormergingtoasinglenodeareassignedthe
nextmaximumdistance.
It’shardtoexplainthese
time0 time1
rulesingeneral,solet’sgotoour
example,wheretheywillmake
sense.TakealookatFigure8.8.
Lookfirstatthetopbranches.
000
100
0
1
1
1
1
1

0
1
1

111
0
0
0
1
0
0

0
1
0

110

0
0
1
1
0
1

0
0
1

1
0
1

010
a
b
c
0
4
2
6
2
6
0
4
3
7
1
5
1
5
3
7
1 2
00
Therearetwobranchesstarting
atnode0andendingatnode0—
thesearecalledparallelbranches.
Ungerboecksaidinrule(1)that
01
thesebranchesshouldbeas-
signedpointsthatareasfarapart
aspossible.LookingatFigure8.7,
Iwillassignonebranchpoint4
andtheotherbranchpoint0.You
10
canseethisinFigure8.8,marked
(a).Youcanalsoseethatour
selectiontellsusthatmodulator
inputbits(000)aremappedby
11
themodulatortopoint0,and(10
110
d
0)ismappedbythemodulatorto
Figure8.8 Assigningmodulationsymbolsto
themodulatoroutputlabeled4
incomingbitsusingthetrellis
(seeFigure8.8,marked(b)).
Trellis-Coded Modulation (TCM): The Wisdom of Modulator and Coder Togetherness  ‹  229 
Nextwe’
—givingthebrancheseitherthepoints1and5,or2and6,or3
andittellsuswhichofthesetochoose.Rule(2)saysthatallbranchesleavingthe
samenodemustbegiventhenextmaximumdistance—inourcase,since“0and4”
’llwanttochoose“2and6”tobetheotherpointsleaving
“3
and7”or“1and5”
In
000
001
010
011
100
101
110
111
Outputsymbolnumber
0
1
3
2
4
5
7
6
Outputsymbol
s
A
/
s
s
0
1
3
2
4
5
7
6
√
√
√ √
− √ √
√
− √
−
√
−
√
√
−
√
−
T/2,0)
2
T/2, T/2)
T/2, T/2)
T/2)
T/2,0)
T/2, T/2)
T/2, T/2)
)
Outputsignal
iT)
0 c
ω π −
ω π π −
ω π π −
ω π π −
ω −
ω π π −
ω π π −
ω π π −
/ iT)
iT)
iT)
)  (t iT)
iT)
iT)
iT)
1 c
3 c
2 c
4 c
5 c
7 c
6 c
A
/
√2
A
/
√2
A
/
√2
A
/
√2
A
/
√2
A
/
√2
A
/
√2
A
/
√2
Figure8.9 8-PSKmodulatorinputandoutputcreatedusingtrellisofFigure8.8
Forthisexample,we’
llmovetothebranchesgoingfromnode0tonode2.Therearetwo
brancheshere,sofollowingrule(1),wewanttoassigntothesebranchestwopointsas
farapartaspossible
and7(Figure8.7).Weseemtohaveachoicehere.Butnowrule(2)comesintoplay,
are
alreadyleavingthenode0,we
thenode0(sincethesepointsarefartherfrom0and4thanourotherchoicesof
).ThisisshowninFigure8.8(marked(c)).Fromhere,wecansee
thattheinputbits(011)aremappedtothemodulatoroutputlabeled2,andtheinput
bits(111)aremappedtothemodulatoroutputlabeled6(seeFigure8.8,marked(d)).
Wecancontinueinthisway,movingthroughourentiretrellis,andwequickly
endupwiththetrellisshowninFigure8.8.Thisindicatesthatthemodulatormaps
inputbitstooutputsymbolsasyoucanseeinFigure8.9.
Andwiththat,Ungerboecksmiled,forhehadjustdevisedasmartwaytobuild
modulatorsthatwerematchedtochannelcoders.
=(A
s =(
s =(
=(0,A
s =( A
s =(
s =(
=(0,
s(t)=Acos( t) (t
π π
s(t)=Acos( t+ 4) (t
s(t)=Acos( t+3 /4) (t
s(t)=Acos( t+ /2) (t
s(t)=Acos( t+
s(t)=Acos( t+5 /4) (t
s(t)=Acos( t+7 /4) (t
s(t)=Acos( t+3 /2) (t
Example 8.1 
llusetheconvolutionalcoderofFigure8.5.Butthistime,an
8-ASKmodulatorwillbeused,inplaceofan8-PSKmodulator.Matchtheoutputsof
the8-ASKmodulatortothedifferentbranchesofthetrellisshowninFigure8.6.
230  ‹  Chapter Eight 
Solution:First,the8-ASKconstellationisseparatedintoparts,separatingthe
constellationintopointsfurtherandfurtherapart,asshowninFigureE8.1.
Next,weassigneachpointinthe8-ASKconstellationtoabranchinthe
trellisaccordingtothetworulesspelledoutbyUngerboeck:allparalleltransi-
tionsseparatedbyamaximumdistance,andalltransitionsdivergingormerging
getthenextmaximumdistance.Applyingthisruleleadstothefirsttwocolumns
ofFigure8.9,wherethistimethe0’s,1’s,…,7’srefertothepointsmarkedin
FigureE8.1.
The8PSKconstellationdrawninitsorthonormalbasis
x x x x x x x x
1
(t)
0 1 2 3 4 5 6 7
x x x x x x x x
0 2 4 6 1 3 5 7
x x x x x x x x
0 4 2 6 1 5 3 7
x x x x x x x x
0 4 2 6 1 5 3 7
FigureE8.1Separatingthe8-PSKpoints
8.3 TheReceiverEndofThings
Atthereceiverside,Ungerboeckhadanunusualidea:hewouldcombinethedemodu-
latorandchanneldecodertogetherandtheywouldoperateasasingledevice,as
showninFigure8.10.HewouldcallhisnewunittheTCMdecoder.We’llexplainhow
itworksusinganexample,showninFigure8.11.
Trellis-Coded Modulation (TCM): The Wisdom of Modulator and Coder Togetherness  ‹  231 
Channel
decoder
r(t)=s(t)+n(t) bits
v
bits
m
Demodulator
Combine
toget
T C M
Decoder
bits
m
r(t)=s(t)+n(t)
Figure8.10Ungerboeck’sideaatthereceiverside
8.3.1TheInput
First,we’llfigureoutwhatiscomingintothedecoder(itsinput).Figure8.11(a)
showsthecommunicationsystemunderconsideration.Theinputtothechannel
coderism =(00000000),whichmakesitsoutputu =(000000000000).Thesebits
enterintothemodulator.Lookingbackatthemodulatorintheprevioussection(we
studieditinSection8.2,anddetaileditinFigure8.9),weknowthatforeachinputof
000,themodulatoroutputsthesignalcorrespondingto“0”,whichcorrespondsto
Acos(ω
c
t)π(t – iT).Moreimportantlytoourcaseinhand,fortheinputu =(000000
000000),theentireoutputcorrespondsto(asseeninFigure8.11):
s t ) Acos (ω t )π(t ) + Acos (ω π(t T ) + Acos (ω π(t − 2T ) + Acos (ω π(t − 3T ) (
c
t ) − t ) t )
c c c
(8.1)
s t
1
( ) + s t
3
( ) + s t ( ) = s t
2
( ) + s t
4
( )
(8.2)
wheres
i
(t)=Acos(ω
c
t)π(t –(i – 1)T).Graphically,wehaves(t)correspondingtothe
plotshowninFigure8.11(a).Thissignals(t)leavesthemodulatorandissentout
acrossthechannel,whichaddsanoisetothetransmittedsignals(t).Theresulting
signalis
( ) ( )+ J ) (8.3) J H J I (
232  ‹  Chapter Eight 
Channel
coder
8-PSK
s(t)
modulator
m=(00000000) u=(000000000000)
Outputsymbol#
0 0 0 0
Outputsymbol
s
0
s
0
s
0
s
0
Outputsignal s(t)=
0 T 2T 3T 4T
Acos(ω
c
t) Acos(ω
c
t) Acos(ω
c
t) Acos(ω
c
t)
1 2 3
s(t) s(t) s(t) s(t)
(a)
s(t)
+
r(t)=s(t)+n(t)
n(t)
0 T 2T 3T 4T
1 1 1
2 2 2
4 4 4
3 3 3
r(t)=s(t)+n(t)
r(t)=s(t)+n(t)
r(t)=s (t)+n(t)
r(t)=s(t)+n(t)
1 2 3 4
r(t)=r(t)+r (t)+r (t)+r (t)
(b)
Figure8.11GettingtheinputtotheTCMdecoder
(a)modulatoroutputs(t)
(b)channeloutput=TCMdecoderinput=r(t)=s(t)+n(t)
4
!
J
E E
J
E
J
?
E E E

E E

H
E

E E
Trellis-Coded Modulation (TCM): The Wisdom of Modulator and Coder Togetherness  ‹  233 
whichisdrawninFigure8.11(b).Asshowninthisfigure,anotherwaytoexpressr(t),
moreconvenientinourupcomingpresentation,istowriter(t)asasumoftime-sepa-
ratecomponents.Thisleadstoanr(t)writtenas:
J H

J
"
( ) H (J )+H (H )+H ( )+H ( ) J
(8.4)
wherer
1
(t)=s
1
(t)+n
1
(t),r
2
(t)=s
2
(t)+n
2
(t)andsoon.Thisr(t)feedstheTCM
decoder.
8.3.2 TheTCMDecoderFrontEnd
Withr(t)comingin,let’sseewhattheTCMdecoderdoes.Itgetsstartedwitha
decoderfrontend.Lookingateachtimeintervalseparately,wesee
H ( ) I ( )+ ( ) J
(8.5)
wheres
i
(t)=Acos(ω
c
t)⋅ π(t –(i – 1) T) .
Weknow,fromourreadinginChapter5,thatther
i
(t)canberepresented
fullyontheorthonormalbasis φ
1
( )
2
cos (ω π
(
t −(i −1)T
)
, φ
2
( )
π ω (J −(E −)6 ).Specifically,r
i
(t)canberepresentedas:
t
T c
t ) t
−

6
IE( )
H ( H H ) (8.6)
where
H I
E
+
(8.7)
I
E
+
E
(8.8)
Here, (I I ) ()
6
) andn
i 
andn
i 
representindependentGaussianrandom
1 2
variables.
Now,Ungerboeckknewthistoo.Hesaidtohimself:Sincer
i
(t)canbefully
representedasavectoroftwovalues,whatI’lldotostartouttheconstructionofthe
TCMdecoderisbuildadevicethatmapsr
i
(t)intothevectorr
i
=(r
i
,r
i
).
1 2
Withthatinmind,UngerboeckdrewtheTCMdecoderfrontendshowninFigure
8.12.Thissimplymapsr
i
(t)intoitsalternativerepresentationofr
i 
=(r
i
,r
i
).
1 2
Whenthedecoderfrontendwasdonewiththeincomingr(t),thisiswhatcame
outofthereceiverfrontend:
IN: r
1
(t), r
2
(t), r
3
(t), r
4
(t)
OUT:r
1
, r
2
, r
3
, r
4
" "
+
! !
+

+

8
234  ‹  Chapter Eight 
×
×
i i i
S
S
ϕ
1
(t)
r
1
i
r
2
i r(t)=s(t)+n(t)
ϕ
2
(t)
Figure8.12TheTCMdecoderfrontend
8.3.3 TheRestoftheTCMDecoder
UngerboecknotedthathehadinhisTCMdecodersomethinghecouldwriteasone
bigvector:
H (

H H

H
!
H
"
) (8.9)
Hecouldwritethisas
H � I+
(8.10)
i
wheres=(s
1
,s
2
, s
3
, s
4
)(ands
i
=(s
1
,s
2
i
))representsthesentsignalfromthemodula-
i
torandn=(n
1
,n
2
, n
3
, n
4
)(andn
i
=(n
1
,n
2
i
))representsthechannelnoise.
Hewantedtogofromthislongvectorrtowhathecalledm′,adecisiononthe
incomingbitsm.Theideaisthis.Youhaver=(r
1
,r
2
, r
3
, r
4
).You’dliketodecideon
thefoursentsymbolss=(s
1
,s
2
, s
3
, s
4
).Ifyouknews,thesignalssentfromthe
modulator,thenyoucould,bylookingatthetableofFigure8.9,figureoutm,thesent
bits.
So,givenr,youwanttofigureoutwhatI’llcalls′,yourbestguessonthetransmit-
teds.Todothis,youlookatthetrellisdiagram,whichyou’llfindredrawninFigure
8.13.Youfindtheonepaththroughthetrellisthatcreatesansascloseaspossibleto
thesentr.Bythis,Imeanyoufindthepaththroughthetrellisthatminimizesthe
value
H H H H − I − I − I − I (8.11)
! "
Trellis-Coded Modulation (TCM): The Wisdom of Modulator and Coder Togetherness  ‹  235 
2 3
RECEIVE r = (r
1
r r r
4
)
time0 time1 time2 time3 time4
00
01
10
11
s
s
0
s
4
s
2
s
6
s
2
s
6
s
0
s
4
s
3
s
7
s
1
s
5
s
1
s
5
s
3
s
0
s
4
s
2
s
6
s
2
s
6
s
0
s
4
s
3
s
7
s
1
s
5
s
1
s
5
s
3
s
0
s
4
s
2
s
6
s
2
s
6
s
0
s
4
s
3
s
7
s
1
s
5
s
1
s
5
s
3
s
0
s
4
s
2
s
6
s
2
s
6
s
0
s
4
s
3
s
7
s
1
s
5
s
1
s
5
s
3
7
s
7
s
7
s
7
Figure8.13ThetrellisdiagramofFigure8.8–
representingchannelcodingandmodulation
wheres
i 
referstotheithoutputsymboloftheselectedpathinthetrellisand
i i
|r
i
–s
i
| referstothedistancebetweenr
i
=(r
1
,r
2
i
)ands
i
=(s
1
,s
2
i
).Theonepaththrough
thetrellisclosesttorindicatesthesents,andfromthisyoucanthendeterminethem.
Let’slookatanexample.Intheexamplewe’vebeenconsideringthroughout
Section8.3,thesentscorrespondsto
I (I I I I ) (8.12)
i i
where
s
i

(
s s
2
)

(
A
T
2
, 0
)
.Let’sassumeA =1VandT=2sec,whichleadsus
1
,
i
tos
i
=(s
1
,s
2
i
)=(1,0).We’llalsoassumethatthenoisenisanoisewhichleadsto
! " "
H I+ � (I + �

I + �

I + �
!
I + � ) (8.13)
where
i
s + n
i
(1, 0 ) + (0.1, 0.1 ) (1.1, 0.1 ) (8.14)
Wecaneasilyshow,byanexhaustivesearchofthetrellisdiagramofFigure8.13,
thatthepaththroughthetrellisthathasacorrespondings=(s
1
,s
2
, s
3
, s
4
)closesttor
isthatshowninFigure8.14.Now,lookingatthebranchesofthispaththroughthe
trellis(solidbranchindicatesm
2
=0,topbranchindicatesm
1
=0),weimmediately
figureoutthatthedecodershouldoutputm′=(0000000000).
236  ‹  Chapter Eight 
s
0
=(1,0)
s
0
=(1,0)
s
0
=(1,0)
s
0
=(1,0)
Figure8.14Bestpaththroughtrellisgiven
r=((1.1,0.1),(1.1,0.1),(1.1,0.1),(1.1,0.1))
Example 8.2 
Assume,asinthesectionjustbeforethisexample,thatarate2/3convolutional
coderisfollowedbyan8-PSKmodulator.Letthetrellisdiagramdescribingthese
operationscorrespondtoFigure8.8.Assumetheinputbitsarem =(00000000),
whichmeanstheoutputfromtheconvolutionalcoderisthenv =(000000000000),
andtheoutputfromthemodulatoristhusthe8-PSKoutputthatcanbeexpressed
onanorthonormalbasisas
s=(s
1
,s
2
,s
3
,s
4
)=((1,0),(1,0),(1,0),(1,0)) (E8.1)
Now,assumethatthechannelisparticularlynoisy,andsomehowthede-
modulatorreceives
r =s+n=((1,0),(1,0),(1,0),(1,0))+((0.9,0),(0.9,0),(0.9,0),(0.9,0))
=((1.9,0),(1.9,0),(1.9,0),(1.9,0))
(E8.2)
Determinetheoutputofanoptimaldecoder.
Solution:Atthispointinthechapter,anexhaustivesearchthroughthe
trelliswouldhavetobeperformedtodeterminethebestpath.However,inthis
case,alittlebitofthinkinglet’susavoidanexhaustivesearch.
AsyoucanseeinFigureE8.2,thereceivedvalue(1.9,0)iscloserto(1,0)
thananyotherpointinthe8-PSKconstellation.Thistellsusthatifthepathcorre-
spondingtooutputsof(1,0)exists,thiswouldbetheclosestpathtoinputs(1.9,0).
! "
Trellis-Coded Modulation (TCM): The Wisdom of Modulator and Coder Togetherness  ‹  237 
x
x
x x
x
x
x
x
(1.9,0) (1,0)
closestpointinan
8-PSKconstellation
FigureE8.2 Closestpointto(1.9,0)inour8-PSKconstellation
That(1,0)’spathdoesindeedexist—it’sthetoplinethroughthetrellis.So,
followingthistopline,whichcorrespondstothebestpaththroughthetrellis,we
decidetooutputthebits(00000000).
8.3.4 SearchingfortheBestPath
Onethingthatthedecodermustdoissearchforthebestpaththroughthetrellis—
thatis,searchforthepaththroughthetrellisthathasansclosesttor.Onewayto
figureoutthebestpathistoperformanexhaustivesearchofallpathsandfigureout
whichonedemonstratesansthatisclosesttor.Butthisiscomputationallyvery
expensive.
Thereisabetterandeasierwaytosearchthroughthetrellis,usingamethod
we’veseenearlier,whenwediscussedchanneldecodersforconvolutionalcodes.It’s
calledtheViterbiAlgorithm.
Let’slookatanexampletoshowhowtoapplytheViterbiAlgorithmtofindthe
bestpaththroughthetrellisinthecaseoftrellis-codedmodulation.We’llusethesame
examplefromtheprevioussubsection,where
H (H H H H ) (8.15)
and
r
i
=(1.1,0.1).
Let’susetheViterbiAlgorithmtosearchfors′,thepaththroughthetrellis
closesttor.Oncewe’vefounds′,wecanoutputm′,ourguessatthebitssentbythe
transmitterside.First,wedrawthetrellis,andweassociatewitheachstartnodethe
number0.Thismeansthatwearenotbiasedtowardanyonestartnodeoverany
other.ThisisshowninFigure8.15(a).Then,westartbyexaminingthetopnodeat
238  ‹  Chapter Eight 
00
0
01
0
10
0
11
0
(a)
1
r =(1.1,0.1)
time0 time1
00
01
s
0
=(1,0)
s
4
=(–1,0)
s2
=
(
0
,1
)

s
6
=
(0
,–
1)
nodeunderconsideration=node0(time1)
Best: 
bestparentnode=node0(time0) 
branch=topbranch 
Lowesttotal=0.02
10
11
(b)
Figure8.15UsingtheVAtomovefromtime0totime1throughtrellis
Trellis-Coded Modulation (TCM): The Wisdom of Modulator and Coder Togetherness  ‹  239 
time1
Node0(time0),top
0.02
Node3(time0),top
0.5
Node1(time0),top
0.02
Node2(time0),top
0.02
(c)
Figure8.15(continued)UsingtheVAtomovefromtime0totime1throughtrellis
time1,whichyoucanseeinFigure8.15(b).Forthisnode,therearetwopossible
parentnodes,node0attime0andnode1attime0.Eachnodecomeswithtwopos-
siblebranches,atopbranchandabottombranch.Wearegoingtodecidewhichisthe
bestparentnodeandthebestbranchfornode0(time1),likethis:
1.Ifwestartedatnode0(time0)andmovedtonode0(time1)usingthe top 
branch,thefirstoutputsymbolwouldbe(1,0).Comparingthistorwherethe
firstsymbolis(1.1,0.1),wesay“d
1
=(1.1–1)
2
+(0.1–0)
2
=0.02distanceifparent
isnode0(time0),topbranch.”Weaddthis0.02tothe0numberthatwegave
node0(time0)inFigure8.15(a),foragrandtotalof0.02.
2.Ifwestartedatnode0(time0)andmovedtonode0(time1)usingthebottom 
branch,thefirstoutputsymbols
1
wouldbe(–1,0).Comparingthistorwherethe
firstsymbolis(1.1,0.1),wesay“d
2
=(1.1– (–1))
2
+(0.1–0)
2
=4.85distanceif
parentisnode0(time0),bottombranch.”Weaddthis4.85tothe0numberthat
wegavenode0(time0)inFigure8.15(a),foragrandtotalof4.85.
240  ‹  Chapter Eight 
3.Ifwestartedatnode1(time0)andmovedtonode0(time1)usingthetop
branch,thefirstoutputsymbols
1
wouldbe(0,1).Comparingthistorwherethe
firstoutputsymbolis(1.1,0.1),wesay“d
3
=(1.1–0)
2
+(0.1–1)
2
=2.02distanceif
parentisnode1(time0),topbranch.”Weaddthis2.02tothe0numberthatwe
gavenode1(time0)inFigure8.15(a),foragrandtotalof2.02.
4.Ifwestartedatnode1(time0)andmovedtonode0(time1)usingthebottom
branch,thefirstoutputsymbols
1
wouldbe(0,–1).Comparingthistorwherethe
firstoutputsymbolis(1.1,0.1),wesay“d
4
=(1.1–0)
2
+(0.1–(–1))
2
=2.42
distanceifparentisnode1(time0),bottombranch.”Weaddthis2.42tothe0
numberthatwegavenode1(time0)inFigure8.15(a),foragrandtotalof2.42.
Sincestartingatnode0(time0)andmovingtonode0(time1)alongthetop
branchcreatesthesmallestdistance(0.02),weproclaimthattheparentnodefornode
0(time1)isnode0(time0)andthebestbranchisthetopbranch,andthatitcarries
withitthenumber0.02(fortotaldistancewiththisselection).
Werepeatthisfornode1(time1),node2(time1)andnode3(time1),andthe
resultsareshowninFigure8.15(c).Thatisallwedoforourfirstmovefromleftto
rightthroughthetrellis.
Atthenexttime,wedoasimilarthing.Westartagainatthetopnode,thistime
startingwithnode0(time2).LookingatFigure8.16,wecanseethatthisnodehas
twopossibleparentnodes,whicharenode0(time1)andnode1(time1).Eachparent
nodecomeswithtwopossiblebranches.Wedecidebetweenthesenodesandbranches
asfollows:
1.Ifwestartedatnode0(time1)andmovedtonode0(time2)usingthe top 
branch,thesecondoutputsymbolwouldbe(1,0).Comparingthistorwherethe
secondsignalis(1.1,0.1),wesay“d
1
=(1.1–1)
2
+(0.1–0)
2
=0.02 distanceif
parentisnode0(time1),topbranch.”Weaddthis0.02tothe0.02numberthat
wegavenode0(time1)inFigure8.15(b),foragrandtotalof0.04.
2.Ifwestartedatnode0(time1)andmovedtonode0(time2)usingthebottom 
branch,thesecondoutputsymbolwouldbe(–1,0).Comparingthistorwhere
thesecondsignalis(1.1,0.1),wesay“d  = 4.85distanceifparentisnode0(time
2
1),bottombranch.”Weaddthis4.85tothe0.02numberthatwegavenode0
(time1)inFigure8.15(b),foragrandtotalof4.87.
3.Ifwestartedatnode1(time1)andmovedtonode0(time2)usingthetop
branch,thesecondoutputsymbolwouldbe(0,1).Comparingthistorwherethe
secondsentsignalisr
2
=(1.1,0.1),wesay“d = 2.02distanceifparentisnode1
3
(time1),topbranch.”Weaddthis2.02tothe0.5numberthatwegavenode1
(time1)inFigure8.15(c),foragrandtotalof2.52.
Trellis-Coded Modulation (TCM): The Wisdom of Modulator and Coder Togetherness  ‹  241 
2
r =(1.1,0.1)
time1 time2
0.02
0.5
s
0
=(1,0)
s
4
=(–1,0)
s2
=
(
0
,1
)

s
6
=
(0
,–
1)
node0(time2)
0.02
0.5
Figure8.16 MovingthroughthetrellisusingtheVA(fromtime1totime2).
Onlymovementtotopnodeisdrawn.
4.Ifwestartedatnode1(time1)andmovedtonode0(time2)usingthebottom
branch,thesecondsentsymbolwouldbe(0,–1).Comparingthistorwherethe
secondoutputsymbolis(1.1,0.1),wesay“d = 2.42distanceifparentisnode1
4
(time1),bottombranch.”Weaddthis2.42tothe0.5numberthatwegavenode0
(time1)inFigure8.15(b),foragrandtotalof2.92.
Sincenode0(time1)tonode0(time2),alongthetopbranch,createsthesmall-
esttotal(0.04),weproclaimnode0(time1)andthe“top”branchthebestparentand
branch.
Wecontinuethisprocessfornode1(time2),node2(time2)andnode3(time2).
Andthatendsournextmovefromlefttorightthroughthetrellis.
Werepeattheprocessattime3,andthenattime4.Now,attheend,wehavefour
endnodeswithfournumbers.Forexample,endnode0(time4)comeswithvalue
0.08.Wechoosetheendnodewiththesmallestvalue.
Fromhere,weknowthehistoryofparentnodesandbranches,sowecan“back-
track”throughthetrellis,anddeterminethes′andthem′.Andwe’redone.
Inourcase,wechoosenode0(time4)withafinalvalueof0.08,andwe“back-
track,”leadingustothepathshowninFigure8.14.Thisleadsustotheoutputm′=
(0000000000).
NowyouknowhowtheTCMdecoderworkstoundotheeffectsatthetransmit-
tersideandcorrectbiterrorsalongtheway.
242  ‹  Chapter Eight 
Problems 
1. YouwanttobuildaTCMsystemthatistousetheconvolutionalcoderofFigure
Q8.1andthemodulationschemeofFigureQ8.2.
(a) Drawthetrellisdiagramrepresentingtheconvolutionalcoder.
(b)Assignmodulationsymbolstooutputbitsbyusingmapping-by-set 
-partitioning.Provideacompletetrellisdiagramincludingmodulation 
outputs. 
(c) DrawtheTCMdecoderfrontend.
(d)GivenA =1andthattheoutputoftheTCMdecoderfrontendis(0.7,0),
(1.2,0),(3.3,0),(1.2,0)usetheViterbiAlgorithmtodeterminetheoutput
oftheTCMdecoder.
m
1
u
1
FigureQ8.1
Convolutional coder
u
2
m
2
u
3
+
+
2bitsinput 3bitsoutput
φ
2
(t)
FigureQ8.2
8outputsofmodulator
φ
1
(t) x x x x
x
x
x
x
5 4
3A
6
0 1
2
3
A
7
Trellis-Coded Modulation (TCM): The Wisdom of Modulator and Coder Togetherness  ‹  243 
2.ConsidertheconvolutionalcoderofFigureQ8.3.Assumethatthemodulation
is32-ASK.Usingyourunderstandingoftrellisdiagramsandmapping-by-set
partitioning,drawatrellisdiagramthatincludestheoutputofthemodulator.
m
1
u
1
FigureQ8.3
Convolutional coder
+
m
4
m
3
u
4
u
3
m
2
u
2
+
u
5
inputbits outputbits
3. ProvideatrellisdiagramandablockdiagramthatfullydescribesaTCMcoder
meetingthefollowingcriteria:
• Thechannelcodermusthave1bitinputand2bitsoutput;
• ThemodulationschememustbeQPSK.
4.DescribethebenefitsofusingtheTCMofFigureQ8.1inplaceofusingthe
sameconvolutionalcodingfollowedbya4-PSKmodulator.
9 
Chapter 
Channel Filtering and 
Equalizers 
I
nChapter5,wetalkedaboutusingmodulatorstomapbitsmtoasignals(t)that
couldbesentoverthechannel.Wesawthatthechanneladdedanoisen(t)tothe
sentsignal,givingthereceiverthenoisyr(t)=s(t)+n(t).Wetookthisastepfurther
whenwetalkedaboutdemodulators,which,givenr(t)=s(t)+n(t),didtheirbestto
regeneratetheoriginalbitsequencem.
Nowwetakethisevenfurther.Thistimewe’llconsidersomethingcalledpulse 
shapingatthemodulator.Modulatorsstilltakebitsmandturnthemtoasignals(t)—
wejustconsideramoregeneralwaytodothis.Wecannowtalkofchannelsthatdo
twothings—addanoiseanddoafilteringc(t),givingthereceiverr(t)=c(t)∗ s(t) + 
n(t).We’llalsospendsometimetalkingaboutwhatgoesonatthereceiver,whichis
called equalization.
9.1ModulatorsandPulseShaping
ConsiderthemodulatorswetalkedaboutinChapter5.WetalkedofASK,PSK,and
QAM(aswellasFSK,butwe’llleavethatoneoutfornow).LookovertotheQPSK
modulatorinFigure9.1(a)forarefresher.InFigure9.1(b),weseethatincomes
m =(1000101000),andoutgoesthewaveforms(t).Wecanexpressthemodulator
output,suchasthisQPSKoutput,mathematicallyaccordingto
( ) s t ( ) + s t ( ) + s t s t
0
( ) + s t ( ) + s t
4
( ) (9.1)
1 2 3
( ) A
0
cos (ω + θ
0
)π( ) + A cos (ω + θ π(t T ) + A cos (ω + θ )π(t − 2T ) s t
c
t t
1 c
t
1
) −
2 c
t
2
+ A
3
cos (ω + θ π(t − 3T ) + A cos (ω + θ )π(t − 4T )
c
t
3
)
4 c
t
4
(9.2)
4
s t
c i
)
−
( )

∑
A
i
cos
(
ω t +θ π
(
t iT
)
(9.3)
i 0
246  ‹  Chapter Nine 
QPSK
in out
0 0 Acos(ω
c
t+0),iT<t<(i+1)T=Acos(ω
c
t+0)π(t–iT)
0 1 Acos(ω
c
t+π/2),iT<t<(i+1)T=Acos(ω
c
t+π/2)π(t–iT)
1 0 Acos(ω
c
t+π),iT<t<(i+1)T=Acos(ω
c
t+π)
.
π(t–iT)
1 1 Acos(ω
c
t+3π/2),iT<t<(i+1)T=Acos(ω
c
t+3π/2)
.
π(t–iT)
(a)
QPSK
in: out:
m =10 00 10 10 00 10 00 10 10
s(t)=
A
c
cos(ω
c
t+π) A
c
cos(ω
c
t) A
c
cos(ω
c
t+π) A
c
cos(ω
c
t+π) A
c
cos(ω
c
t)
s
0
(t) s
1
(t) s
2
(t) s
3
(t) s
4
(t)
0 T 2T 3T 4T
(b)
Figure9.1DescribingamodulatorusingQPSKasanexample
(a)input-outputrelationship
(b)input-outputexample
00
Channel Filtering and Equalizers  ‹  247 
4
s t
i
jθ
i
jω t
( )

∑
Re
¦
Ae e
c
¦
π
(
t − iT
)
(9.4)
i 0
Whenyouwritewhatcomesoutinthisway,itindicatesasimplewaytobuildthe
modulator.Forexample,theQPSKmodulatorcanbebuiltasshowninFigure9.2.
Here,bitscomeinandtheyaremappedbyacodertoacomplexamplitudeindicating
j
i
theAe
θ
value.Thisisthenpassedthroughwhatiscalledthepulseshaper,which
i
j
i
j
i
turnstheinformationAe
θ
intotheshape Ae
θ
π (t −iT ) .Finally,wemultiplythis
e
i i
j t
by
ω
c
andtaketherealpart,and,voila,foreverytwobitsthatcomeinwe’vegot
jθ
i
jω t
Re
¦
Ae e
c
¦
π (t −iT ) comingout.
i
e
jω
c
t
Symbol,I
i
=A
i
e
jθ
i
A
i
e
jθ
i
π(t–iT)
bits
0 0 Ae
j0
0 1 Ae
jπ/2
1 0 Ae
jπ
1 1 Ae
i3π/2
m=(10 00) symbols=(Ae
jπ
Ae
j0
)
signals=Ae
jπ
π(t)+Ae
j0
π(t–1T) s(t)= ∑ Re{Ae
1
coder x Re{
.
}
pulse
shaper
π(t) s(t)
jθ
i
e
jω
c
t
}
.
π(t–iT)
i=0
Figure9.2 Constructionofmodulator(e.g.QPSK)
Now,examinethemodulatorshowninFigure9.3.Itisidenticalineverywayto
themodulatorinFigure9.2,withoneexception.Thepulse-shapingfilter,whichwas
previouslyπ(t),ischangedtog(t).Asaresult,theoutputofthemodulatorisnow
L−1
jθ
i
jω t
(
s t
i
( )

∑
Re
¦
Ae e
c
¦
g t − iT
)
(9.5)
i 0
L−1
j t
( s t
i
ω
c
¦
g t − iT )
(9.6)
( )
∑
Re
¦
I e
i 0
j
i
whereg(t)maybeanyshapeyou’dlike,andI isshorthandfor Ae
θ
. Sofarinallthe
i i
modulatorswe’velookedat,we’veusedg(t)=π(t).Soon,asyoureadon,you’llseethat
willbeareasonforusingadifferentg(t).Butfornow,simplyknowthatyouhavea
newwaytobuildmodulatorsandaneasywaytogeneralizethem.
248  ‹  Chapter Nine 
jω
c
t
pulse
shaper
new
g(t)
e
coder
A
i
e
jθ
i
g(t–iT) bits Symbol,I
i
=A
i
e
jθ
i
Re{
.
} x
s(t)
00
Ae
j0
Ae
jπ/2
0 1
Ae
jπ
1 0 
Ae
j3π/2 
1 1
Figure9.3 Constructionofnewmodulatorwithdifferentpulseshaper
Example 9.1 
ConsideraBPSKmodulator.ExpresstheoutputofaBPSKmodulatorinan
equationthatlookssimilartoEq.(9.6).
Solution:Forstarters,let’sconsidertheoutputofaBPSKmodulatoratone
time.Itlookslike
s t t
i
)  − ( ) Acos (ω + θ π(t iT)  (E9.1)
i c
where

θ 0

or 180  (E9.2)
i
Now,inamoregeneralform(wecangeneralizethepulseshape),wewrite
thisequationaccordingto
s t t
i
(  − ( ) Acos (ω + θ ) g t iT )  (E9.3)
i c
ApplyingalittlebitofmathtotheoutputofaBPSKmodulator,weendupwith
s t
jθ
i
jω t
( − ( ) Re
¦
Ae e
c
¦
g t iT )  (E9.4)
i
Now,ifweconsidertheoutputatawholebunchoftimes,wehave
L−1
s t
jθ
i
jω t
( − ( )
∑
Re
¦
Ae e
c
¦
g t iT ) 
(E9.5)
i0
L−1
ω
c
s t
i
j t
(
−
( )

∑
Re
¦
I e
¦
g t iT
) 
(E9.6)
i 0
Channel Filtering and Equalizers  ‹  249 
where
I
i
Ae
jθ
i
Ae
j 0

or Ae
j180

(E9.7)
9.2 TheChannelThatThoughtItWasaFilter
Wehavejustre-expressedthemodulatorsthatwesawinChapter5—infact,we’ve
foundawaytogeneralizethemsomewhat.InChapter5wealsosawchannels.Those
channelsaddedanoise,andthechannelwasmodeledasshowninFigure9.4(a).We
arenowgoingtopresentanewchannel,amoregeneralmodel.Achannelmaydo
morethanaddanoise—itmayactasafilterc(t)andaddanoisen(t).Inthiscase,the
channelismodeledasshowninFigure9.4(b).
channel
filter
+
+
new
r(t)=s(t)+n(t)
r(t)=s(t)
*
c(t)+n(t) s(t)
s(t)
n(t)
c(t)
(a)
(b)
The Channel
TheChannel
n(t)
Figure9.4
(a)Channelmodelseensofar(b)Newchannelmodel
Let’ssaywehaveournewmodulatorfromFigure9.3,andthesignals(t)outof
thismodulatorissentacrossthechannel,whichactsasafilter,asshowninFigure
9.4(b).WethereforehavethesystemshowninFigure9.5.Let’sevaluatethechannel
outputr(t):
r t ) s t ∗c t ( ( ( ) ( ) + n t ) (9.7)
250  ‹  Chapter Nine 
coder Re{
.
}
bits
pulse
shaper
x
n(t)
r(t)
g(t)
channel
filter
c(t)
+
I
i
=A
i
e
jθ
i
e
jω
c
t
newmodulator newchannel new
received
signal
Figure9.5 Newmodulatorandnewchannel
Here,r(t)isthesentsignalpassedthroughthechannelfilterwithanoiseadded
ontoit.Plugginginthevaluefors(t)comingoutofthemodulatorinEquation(9.6)
leadsusto
L−1
ω
c
r t
i
j t
(
−
( )
+ n t
( )

∑
Re
¦
I e
¦
g t iT
)
∗c t
( )
(9.8)
i0
or,usingpropertiesofconvolution,
L−1
ω
c
( )

∑
Re
¦
I e
¦
h t iT
)
+ n t
)
(9.9)
r t
i
j t
(
−
(
i 0
where h t ( ) ∗c t
E
( )
channelfilterc(t)shiftedinfrequencysothatitiscenteredaround0Hz.Thatis,the
channel,throughitsfilterc(t),reshapesthetransmittedpulses,changingtheshape
fromg(t)toh(t),anditaddsanoisen(t).
( ) g t
E
( ) and c t isthebasebandfiltercorrespondingtothe
Example 9.2 
Determinetheoutputofachannelthatperformsafilteringandaddsanoisegiven
•thechannelfilterisdescribedaccordingto
( ) δ( ) + δ(t − τ) (E9.8) c t t
E
•themodulatorusedisaBPSKmodulator.
Solution:Usingequation(9.9),wehave
L−1
ω
c
( )

∑
Re
¦
I e
¦
h t iT
)
+ n t
)
(E9.9)
r t
i
j t
(
−
(
i0
Channel Filtering and Equalizers  ‹  251 
where
h t
( )
∗c t
( )
g t
E
( ) (E9.10)
UsingtheinformationaboutthechannelinEquation(E9.8)leadsustothe
result
( ) g t ) ∗ δ( ) + δ(t − τ)]
]
(E9.11) h t ( ,
¸
t
h t ( ( ( ) g t ) + g t − τ) (E9.12)
Applyingthistoequation(E9.9)leadsto
L−1
r t
i
j t
¸
( )
− g t − τ
)]
]
+ n t
( )

∑
Re
¦
I e
ω
c
¦
⋅ ,g t
( ( )
(E9.13)
i0
KnowingthatthemodulatorisBPSKfurthertellsus(fromExample9.1)that
jθ
i

I
i
Ae where θ 0 or 180 (E9.14)
i
Equation(E9.13)and(E9.14)togetherdescribetheoutputofthechannelfilter.
9.3 Receivers:AFirstTry
Wenowhaveageneralformforthemodulatorswe’vebeenstudying.Weknowifwe
sendthesignals(t)acrossachannelthatdoesafiltering,wegetout(atthereceiver
side)ther(t)ofEquation(9.9).Atthereceiverside,thejobathandisthis:takether(t)
comingin,andfigureoutm′,abestguessoftheoriginalbitsmthatweresent.Alter-
natively,ifwecancomeupwithI′,agoodguessonthecoderoutputsI=(I
0
,I
1
,...),we
canfigureoutm′.Sowe’llsaythatthetaskofthereceiveristocomeupwithI′,as
goodaguessaspossibleonthevalueofI.
ThereceiverI’lldescribeherewon’tturnouttobethebest receiveryoucanuse,
butitwillbeausefuloneinhelpingusunderstandtheproblemsareceiverfaceswhen
itpicksupareceivedsignalr(t)ofEquation(9.9),asignalthatincludeschannel
filtering.
9.3.1TheProposedReceiver
TakealookatthereceiverinFigure9.6.Ittakestheincomingsignalr(t)andfirst
multipliesitontopbyacosineandonthebottombyasine.Thisleadstotwosignals
whicharefedintoafilterf(t),whichpassesfrequenciesaround0Hzbutcutsout
higherfrequencyterms.
252  ‹  Chapter Nine 
cos(ω
c
t)
r
c
(t) 
kT 
r(t)
x
f(t)
x
device
Decision
∧
I
k
or
r′(t) r″(t)
r
k
bits
r
s
(t) 
sin(ω
c
t) 
Figure9.6 Proposedreceiverfornewr(t)
First,themultiplicationbyacosineleadsto(afteraboutthreelinesofmath):
L−1
r t I h t iT ) + n t ( )
∑
Re¦ ¦ ( −
c
( )
i 0
c i
( ))
(9.10)
+
(
high frequency terms that will be cut out by f t
andthemultiplicationbyasineleadsto(afteradifferentthreelinesofmath)
L−1
r t I h t iT
)
+ n t
( )

∑
Im
¦ ¦ (
−
s
( )
i 0
s i
( ))
(9.11)
+
(
high frequency terms that will be cut out by f t
where n t t
c
( )
n
( )
sinω t . ( )
n
( )
cos ω t and n t t
c s c
Tosimplifythepresentationandnothavetobringtwodifferentsignalswithusin
therestofourdiscussion,wecanrepresentthisasasinglecomplexsignal,namely
′
( ) r t
s
( ) (9.12) r t
c
( ) + jr t
whichleadsto
L−1
r t I
i
I h t iT ) +
¸
,n t ( )]
]
′
( )
∑
,
¸
Re¦ ¦+ j Im ¦ ¦] ( −
c
( ) + jn t
i
]
s
i 0
(9.13)
+ (
high frequency terms
)
Channel Filtering and Equalizers  ‹  253 
L−1
r t
i
( − ′
( )
′
( )
∑
I h t iT ) + n t
i 0
( ))
(9.14)
+
(
high frequency terms that will be cut out by f t
where n t
c
( ) + jn t ′
( ) n t ( ).Thisisthesignalweshowenteringthechannelfilter
s
f(t).Themultiplicationbysineandcosineinessence:(1)removesthe ω t termfrom
c
thereceivedsignal(thatis,returnsthesignaltobaseband);and(2)allowsustowrite
thesignalinahandycomplexnotation.
Next,weapplythefilterf(t).Thisleadsto
L−1
r t
i
(
−
(
′
( )
∗ f t ′′
( )

∑
I h t iT
)
∗ f t
)
+ n t
( )
(9.15)
i 0
L−1
′′
( )
∑
I x t iT ) + n t r t
i
( − ′′
( )
(9.16)
i 0
( ) h t ) ∗ f t ) g t c t ( ) and n t ′
( )∗ f t where x t ( ( ( )∗ ( ) ∗ f t ′′
( ) n t ( ) .Thistellsus
E
thatoursentsignal,afterpassingthroughachannelfilterandhavingnoiseaddedon
toit,andthenpassingthroughareceiverfilterf(t),basicallyisthesentsignalwiththe
combinedeffectsofthefilteringandthenoise.
WenowsamplethesignalattimeskT(k =0,1,2,...).AttimekT,weendupwith
theoutputr
k
whichcorrespondsto
L−1
r
k

∑
I x kT iT
)
+ n
i
(
−
k (9.17)
i 0
where n
k
n kT ′′
( ) ;or,takingthekthtermoutofthesum,wecanwritethisas
L−1
0
i
( − r
k
I x ( ) +
∑
I x kT iT ) + n
(9.18)
k k
i 0
i k ≠
Desired Intersymbol Noise
Information Interference (ISI)
(ThedecisiondeviceisexactlythesameonewebuiltinChapter5,takinginanr
k
andputtingoutaguessonthesymbolsent.)
254  ‹  Chapter Nine 
9.3.2 MakingtheReceiveraGoodOne
We’dliketohavethereceiverworksothatitcreates r
k
I
k
;inotherwords,we’dlike
forthekthreceivedsamplefromthesamplerinFigure9.6tocorrespondexactlytothe
kthinformationsymbolsentatthetransmitter.Ifwecoulddothat,thenthedecision
devicewould,given r
k
I
k
,alwaysmakethecorrectdecisionon I .LookingatEqua-
k
tion(9.18),let’sseewhatitwouldtaketomake r
k
I
k
.Itwouldtakethreethings:
1. x(0) = 1
2. x(kT – iT) = 0, k ≠ i
3. n
k
= 0.
Wecan’tmakethenoisen
k
=0,butrecallthat x t ( ) ∗ ( )∗ f t ( ) g t c t ( ) ,andg(t)
E
isafilterputinatthetransmitter,whilef(t)isafilterweputinatthereceiver.Sincewe
controlg(t)andf(t),wecancontrolx(t).So,wecouldchooseg(t)andf(t)suchthat
x t ( )∗ ( )∗ f t
¦
1, t

0
− ( ≠
E (9.19)
( ) g t c t ( )
¦
¦
0, t kT iT k i )
whichmeanswewouldsatisfypoints1and2above.Iff(t)andg(t)areselectedsuch
thatx(t)satisfiesEquation(9.19),x(t)and/orthecommunicationsystemissaidto
satisfytheNyquist criteria for zero ISI (intersymbolinterference).Herearesome
choicesforx(t)thatsatisfyEquation(9.19).
The sinc scenario: Thefirstwell-knownchoiceforx(t)thatsatisfiesEq.(9.19)is
j
πt
\
j \
sin
,
(
T
,
(
t
x t ( ) sinc
, (

πt
( ,
(9.20)
T
T
Aplotofthisx(t)isshowninFigure9.7(a).Inthefrequencydomain,thissignalis
showninFigure9.7(b).
The raised cosine scenario:Anotherwell-knownchoiceforx(t)iscalledtheraised
cosinefunction,inwhichcasex(t)correspondsto
j
πt
\ j
t
\
α
( )

sin
π
(
,
t
T
,
(
⋅
cos
,
πα
( )
RC t
(
T
t
2
,
(
x t
(9.21)
− α
2
1 4
T
T
2
whereαiscalledtheroll-offfactorandcorrespondstoavaluebetween0and1.Three
raisedcosinefunctionsareshowninFigure9.8(a),andtheirfrequencyresponseis
showninFigure9.8(b).
Channel Filtering and Equalizers  ‹  255 
sin(πt
/
T
)
0 –T –2T T
1
2T 3T
t
x(t)=sinc(t/T)=
πt
/
T
(a)
X(f)=F{sinc(t
/
T
)}
f
–
1
/
2T
1
/
2T
(b)
Figure9.7 x(t)=sinc(t/T):(a)intimeand(b)infrequency
–2T 2T –T T 0
t
α=1(
....
)
α=0.5(--)
α=0(
__
)
x(t)=RC
α
(t)
(a)
x(f)=F{RC
α
(t)}
α=1
α=0
α=α
–1
/
T
–1
/
2T
(1+α) –1
/
2T
1
/
2T
1
/
2T
(1+α) 1
/
T
(b)
Figure9.8 x(t)=RC
α αα αα
(t):(a)intimeand(b)infrequency
256  ‹  Chapter Nine 
9.3.3 TheProposedReceiver:ProblemsandUsefulness
Let’slookatthereceiverthatwe’vebeendiscussingsofar.It’sshowninFigure9.6,
andwefiguredoutthatther 
k
valueis:
L−1
0
k
r
k
I
k
x ( ) +
∑
I
k
x (kT − iT ) + n
(9.22)
i 0
i k ≠
t t where n
k
n′′
(kT ) n′
( )∗ f ( ) .Wealsosawthatifwechosef(t)andg(t)
t kT
carefully,wecouldmakesurethatthex(t)wassuchthatx(0)=1andx(kT – iT)=0,
whichwouldleadto
r
k
I
k
+ n
k
(9.23)
t t where n
k
n′′
(
kT
)
n′
( )
∗ f
( ) .Thislooksgreat.We’venowbeenabletoget
t kT
thereceivertoproduceakthsamplethathasinitthekthinformationsymbolI 
k
and
somenoisen .
k
Butthereareproblemswiththenoisen
k
.Thisnoisecorrespondsto
n
k
n′′
(kT ) n′
( )∗ f ( ) t t .Now,f(t)waschosentomakex(t)satisfytheNyquist
t kT
criteria;wemadenoefforttoconsiderhowf(t)wouldaffectthenoise.Indeed,f(t)
couldcausedramaticnoiseamplification,makingtheinformationI
k
lostintheseaof
loudnoisen
k
.So,generallyspeaking,thisreceiverofFigure9.6isnotusedinits
currentform,except...itdoeshaveonereallyniceuse:whenthechannelfilterc(t)is
flatoverthetransmissionfrequency—thatis,whenc(t)hasafrequencyresponseC(f)
suchasthatshowninFigure9.9.Inthiscase,thechannelC(f)=1forallintentsand
purposes,aslongaswemakesureallthetransmitfrequenciesfallintheflatpartof
C(f).Forallpracticalpurposes,thechannelcanbemodeledas c t ( ). ( ) δ t
E
C(f)=frequencyresponse 
ofchannelfilter 
f
1
f
c
flatoverrange
oftransmission
Figure9.9 Apossiblechannelfilterc(t)showninthefrequencydomain
Channel Filtering and Equalizers  ‹  257 
Inthiscase,forthemodulatorandreceiverwe’reconsidering,wehaveanr 
k
equalto
L−1
r
k
I x
( )
+
∑
I x kT iT
)
+ n
k
0
i
(
−
k (9.24)
i 0
( ) g t ( )∗ f t ( )∗ f t wherethistime x t ( )∗c t ( ) g t ( ),and,asbefore,
E
n
k
n kT ′′
( ( )
∗ f t ′′
( ) with n kT
)
n t
( )
t kT
.

Ifwewanttotrytomaker 
k
ascloseaspossibletoI  ,thenwe’llmakesurethat
k
x t ( )∗ f t
¦
1, t

0
− ( ≠
(9.25)
( ) g t ( )
¦
¦
0, t kT iT k i )
Ifwedothisandalsomakesureg(t)= f(t),itcanbeshown(withabitofstatistical
wrangling)thatwenotonlymakex(0)=1andx(kT – iT)=0,butalsothisf(t)doesnot
causenoiseamplification.
Thisreceiveristhereforeapopularchoicewheneverthechannelis“flat”overthe
rangeoftransmissionfrequencies.
Example 9.3 
Assumingaflatchannel,figureoutapossibleg(t)andf(t)sothatthereisnoISI
inthereceivedsignalr 
k
.
Solution:Equation(9.24)showsusthereceivedr 
k
whenthechannelisflat.If
wewanttogetridofISI(themiddleterm),allwehavetodoismakesurethat
x(t)=g(t)∗ h(t)satisfiesEquation(9.25).
Theeasiestwaytofindag(t)andh(t)thatsatisfyEquation(9.25),theeasiest
wayistofindanx(t)=g(t)∗ h(t)thatsatisfies
πt
sin
( ) g t ∗ h t
(E9.15)
x t ( ) ( )
πt
T
T
Tosolvefortheg(t)andh(t)fromhere,turntothefrequencydomain,where
wefindoutthatwecanwriteEquation(9.25)accordingto
¦
−1 1
X f ( )⋅ F f )
¦
f
( ) G f (
¦
1,
2T
≤ ≤
2T
(E9.16)
¦
0, else
¦
258  ‹  Chapter Nine 
Onesolutiontothisequationistolet
¦
−1 1
G f (
¦
1, ≤ ≤
( ) F f )
¦
2T
f
2T
(E9.17)
¦
0, else
¦
Turningthisbacktothetimedomainweendupwith
πt
sin
( ) f t g t ( )
πt
T
(E9.18)
T
So,hereisonepossibleg(t)andh(t)thatsatisfiesEquation(9.25),insuring
thatinaflatchannelwegetnoISI.
9.4 OptimalReceiverFrontEnd
Sofar,weconsideredoneexampleofapossiblereceiver,showninFigure9.6.This
receiverexplainstheideaofareceiverfilterf(t)andhowitcanbeusedtoremove
intersymbolinterference(ISI).Inthecasewherethechannelis“flat”overthefrequen-
ciesoftransmission,thatreceiveractuallyturnsouttobetheoptimalreceiver.Butin
othercases,thatreceivercausesnoiseamplification,andmaynotbeaverygoodone.
Youneedalternatives!Youneedchoices!You’llwanttoknowhowtobuildtheverybest
receiverpossible.That’swhythissectionishere—totellyouexactlyhowtobuildthe
optimalreceiverfrontend,whichisthefirstpartoftheoptimalreceiver.
Westartatourreceiverwiththeinput
L−1
ω
c
( )
∑
Re
¦
I e
¦
h t iT ) + n t )
(9.26)
r t
i
j t
( − (
i 0
( ) g t c t where h t ( ) ∗
E
( ) .Withoutanylossofinformation,wecanmapthisr(t)signal
j t
c
toasignalwithoutthecarrier e
ω
.Wedothisbymultiplyingther(t)byacosineand
thenbyasine,asshowninFigure9.10.Thisleadstotheoutputs(whichwesawearlier)
L−1
r t I h t iT ) + n t
c
( )
∑
Re¦ ¦ ( −
c
( )
i
i0
(9.27)
+ (
high frequency terms which will be cut out in a moment
)
L−1
r t I h t iT ) + n t ( )
∑
Im¦ ¦ ( −
s
( )
s i
i0
(9.28)
+ (
high frequency terms which will be cut out in a moment
)
Channel Filtering and Equalizers  ‹  259 
cos(ω
c
t)
r
c
(t)
r(t)
r′(t)
r
s
(t)
Figure9.10
sin(ω
c
t) Thefirstpartofanoptimalreceiverfrontend
x
x
where n t t
c
( ) n( )sinω t ( ) n( )cos ω t and n t t
c s c
Tosimplifythepresentationandkeepusfromhavingtobringtwodifferent
signalswithusintherestofourdiscussion,wecanrepresentthisasasinglecomplex
signal,namely
r t
c
( ) + jr t ′
( ) r t
s
( ) (9.29)
whichleadsto
L−1
r t I
i
I h t iT )
+
¸
,n t ( )]
]
′
( )

∑
,
¸
Re¦ ¦
+ j Im ¦ ¦] (
−
c
( )
+ jn t
i
]
s
i −0
(9.30)
+(high frequency terms, cut out in a moment )
L−1
r t
i
(
− ′
( )
+
(
high frequency terms, cut out in a moment
)
′
( )

∑
I h t iT
)
+ n t
i −0
(9.31)
′
( ) = s t
′
( ) ÷(high frequency terms, cut out in a moment ) r t
′
( ) ÷n t
(9.32)
L−1
where n t
c
( ) + jn t ′
( )

∑
I
i
h t iT
)
.
′
( ) n t ( ) and s t
(
−
s
i0
Now,each I
i
A
i
e
jθ
i
isoneofafinitenumberofvalues(oneofMvalues,tobe
exact,becauseI
i 
representstheoutputfromthecoder,whichtakesinnbitsand
outputsoneofM=2
n 
possibleI
i 
’s.SeeFigure9.3foranillustrativeexample.)That
L−1
meansthat s t
(
− ′
( )

∑
I
i
h t iT
)
isatimefunctionthatisoneofafinitenumber
i0
oftimefunctions(oneofM
L
possibletimefunctionstobeexact).Sowehave
L−1
r′(t) = s′(t) + n′(t),where s t
(
− ′
( )

∑
I
i
h t iT
) isoneofafinitenumberoftimefunctions.
i0
260  ‹  Chapter Nine 
IfyoulookbacktoChapter5,wesawtherethat:
ifyoureceiver(t)= s(t)+ n(t),wheres(t)takesononeofafinitenumberof
values,M values,tobeexact,
then anoptimalreceiverfrontendistheoneshowninFigure9.11(a).
So,nowthatwehaver′ (t)= s′(t)+ n′(t),wheres′(t)takesononeofafinitenum-
berofvalues(oneofM
L
values),thenanoptimalreceiverfrontendforr′ (t)mustbe
theoneshowninFigure9.11(b).
s
1
(t)
x
r(t)=s(t)+n(t)
x
s
M
(t)
∫
∫
u
1
u
M
(a)
cosω
c
t s′
1
(t)
x
r′(t)
x
x
x
=s′(t)+n′(t)
∫
∫
u
r(t)
u
M
L
sinω
c
t
s′
M
L(t)
(b)
Figure9.11
(a)OptimalreceiverfrontendforreceivedsignalofChapter5,namelyr(t)=s(t)+n(t)
(b)CorrespondingoptimalreceiverfrontendforreceivedsignalofChapter9:
(1)firstcreatingr′ ′′ ′′(t)=s′ ′′ ′′(t)+n′ ′′ ′′(t),then(2)usingreceiverfrontendanalogousto(a)
1
Channel Filtering and Equalizers  ‹  261 
Ofcourse,nobodywantstobuildthereceiverfrontendshowninFigure9.11(b)
becausethereareM
L
branches,whichcanbequitealot.But,thisdidgivesomesmart
engineeranidea—namely,totakeacloserlookatthatu 
k
comingoutofthereceiver
frontendofFigure9.11(b).Thatu 
k
correspondsto:
t s
k
′ t dt (9.33) u
k

∫
r′
( ) ( )
L−1
u
k
r′
( )
∑
I h t − iT dt
(9.34)
∫
t
i k
( )
,
i 0
L−1
u
k

∑
I
i k
∫
r′
( ) ( ) t h t − iT dt
(9.35) ,
i0
L−1
u
k

∑
I
,
r
i
′
(9.36) i k
i0
t h t −iT dt .Ifwehave r
i
′ (fori =0,1,...,L –1),wecouldgenerate
everyu
k
wewant.Wedon’thavetobuildareceiverfrontendasbigastheoneof
Figure9.11(b)—allwehavetodoisbuildonethatprovidesthevalueof r
i
′ for
i =0,1,2,..., L–1.Ifwecandothat,thenfromthisreceiverfrontendwehavethe
valuesneededtobuildthereceiverfrontendofFigure9.11(b)ifwewantit—orany
otherimportantreceiverwewanttobuild,forthatmatter.
TakealookatFigure9.12.Fromthisreceiverfrontend,r(t)ismappedtor′ (t)by
thecosineandsinemultiplication;thenwithr′ (t)comingintothefilterandsampler,r
i 
′
comesout—likethis:
LetO
i
=outputoffilterandsampleratsampletimei;then
where r
i
′
∫
r′
( ) ( )
O
i
=r
′
( )∗ h(−t) t (9.37)
t=iT
O
i
= r
′
( )h(J−t d J
∫
J ) (9.38)
t=iT
O
i
= r
′
( )h(J−iT d J
(9.39)
∫
J )
O
i
=r
i
′
(9.40)
Figure9.12,then,isourreceiverfrontend.Ittakesinr(t)anditputsout r
i
′ ,
i=0,1,2,..., L –1.Withthesevalues r
i
′ ,wehaveallthenumbersweneedtogenerate
anyvalueswemightwantinourreceiver.
262  ‹  Chapter Nine 
cosω
c
t 
r
c
(t) 
r(t)
x
r′(t)
x
h(–t)
kT
r
i
′
r
s
(t)
sin(ω
c
t)
Figure9.12Optimalreceiverfrontend
9.5 OptimalRest-of-the-Receiver
We’vegonefromthenewmodulator,throughthenewchannelthatputoutthenew
r(t),andthroughareceiverfrontendwhichoutputs r
i
′ .Youcanseeitalltogetherin
Figure9.13.Nowlet’sbuildtheoptimal“rest-of-the-receiver,”whichputsoutm′ ,the
bestpossibleguessonthetransmittedbitsm.
RECEIVER
TRANSMITTER CHANNEL FRONTEND
coder Re{
.
}
(bits)
pulse
shaper x
n(t)
g(t)
c(t) +
s(t)
r(t)
r
k
′
h(–t)
m
e
jω
c
t
cosω
c
t
sinω
c
t
kT
x
x
Figure9.13Thenewmodulator(transmitter)withpulseshaping,followedbythenew
channel(withachannelfilter),followedbytheoptimalreceiverfrontend
9.5.1TheInput
ComingintowhatI’mcallingtherest-of-the-receiveristhevalue:
r
k
′
= r
′
( )h (t −k T )d t
(9.41)
t
∫
Let’stakeamorecarefullookat r
k
′ ,andseewhatwe’vegottoworkwith.Here
comesthemaththatallowsustobetterunderstand r
k
′ .Takingthe r
k
′ ofEquation
Channel Filtering and Equalizers  ‹  263
L −1
(9.41),andsubstitutingin r t t t
i
(
t ′
( )
s′
( )
+ n′
( )

∑
I h t − iT
)
+ n′
( )
,weendupwith
i0
L−1
l
r
k
′
=

I h t −iT )÷n
′
(t)
l
l
⋅ h t −kT dt
(9.42)
∫

∑ i
( ( )

i=0
l
L−1
i
( ) ( )
∫
) ( ) r
′
=
∫
∑
I h t −iT h t −kT dt ÷ n
′
(t h t −kT dt
(9.43) k
i=0
L−1
k
( ) ( ) r
′
=
∑
I
i
∫
h t −iT h t −kT dt ÷n
k (9.44)
i=0
L−1
r
′
=
∑
I
i
∫
h t h t −(k −i T dt ÷n
k
( ) ( ) )
k (9.45)
i=0
t h t − kT dt . LookattheintegralinEquation(9.45).Let’s
( ) g t t ( )∗h (−t ) h (τ )h t −τ τ .Froma
where n
k

∫
n′
( ) ( )
compareitto x t ( )∗c ( )∗h (−t ) h t
∫
( )d
E
brieflookattheintegralofEquation(9.45)andx(t),you’llbeabletounderstandwhatI
meanwhenIwrite:
L−1
r
′
=
∑
I
i
x
((k −i T
)
÷n
k
)
k (9.46)
i=0
which,letting x
k i
x
((k − i T
) canberewrittenas
−
)
L−1
r
k
′
=
∑
I
i
x
k −i
÷n
k (9.47)
i=0
Therearetwootherwayswecanexpressthis r
k
′ .Thefirstisashorthandnota-
tionfromdigitalsignalsandsystemsliterature,whichtellsuswecanwrite r
k
′
accordingto
r
′
= I
k
∗ x
k
÷n
k
(9.48)
k
where I
k
∗ x
k
denotesthesumofEquation(9.47)andrepresentsadiscrete-time
convolution.Thesecondwaywecanwritether
k 
′ ofEquation(9.47)istowriteoutthe
sumlonghand,leadingto
r 
′
=( I  x  + I
1 
x
k−1 
+ … + I
k−1 
x )+ I
k
x
0 
+
(
I
k +1 
x
− 1 
+ I
k + 2 
x
− 2 
+ … + I
L−1 
x
k−( L−−1)
)
+ n
k k  0  k  1 
(9.49)
264  ‹  Chapter Nine 
Thatistosay,theoutputweareseeing, r
k
′ ,correspondstotheoutputgenerated
bytheshiftregister,multiplier,andaddershowninFigure9.14.There,theI ’sare
i 
storedinashiftregisteroflengthL,andeachI
i 
ismultipliedby x
k i −
;thentheyareall
addedtogetherandanoiseisaddedontothem.
I
0
I
1
I
k
I
L–2
I
L–1
+ 
+ 
r′
k
x
k
x
0
x
k–(L–2)
x
k–(L–1)
x
k–1
... ...
... ...
x x x x x
n
k
Figure9.14 Illustratingpictoriallythevalueofr
k
′ ′′ ′′ comingoutofthereceiverfrontend
andintothe“restofthereceiver”
9.5.2 AProblemwiththeInput,andaSolution
Thereisaproblemwith r
k
′
,somethingthatitmakesitveryhardtobuildanoptimal
rest-of-the-receiver.Wehavethat
L−1
r
k
′
=
∑
I
k
x
k−i
÷n
k (9.50)
i=0
where n
k
n t h t − kT )dt. Hereinliesourproblem.Then(t)inthen 
k
n 
integralisadditivewhiteGaussiannoise,anditisfilteredbyh(–t)(thatis,integrated
withh(t – kT)).Theoutput,n 
k
,canbeshowntobeaGaussianrandomvariablethatis
correlatedwithnoisesamplesatothertimes—thatis,n 
k
iscorrelatedwithn 
k –1 
and
k–2 
,andsoon,aswellasbeingcorrelatedwithn  andn 
k +2 
,andsoon.Thisnoise
∫
( ) (
k +1 
correlationmakesbuildinganoptimalreceiveraverydifficultandcomplicatedtask.
It’saproblemengineersarestillworkingtoresolve.
Channel Filtering and Equalizers  ‹  265 
Butthereisawayaroundthisproblem.Wecanintroduceanotherfilter,adigital
filtercalledawhiteningfilter,w 
k
,sothat,afterpassing r′ throughw 
k
,weendupwith
k
r
k
= ∗ =(I ∗ ÷ ∗ w (9.51) r
k
′
w
k k
x
k
n
k
)
k
r
k
= I
k
∗ ∗ w
k
)÷ ∗ w (9.52) (x
k
n
k k
r
k
= I
k
∗ ÷ n
′
(9.53) v
k k
where n′ n
k
∗ w
k
isnowaGaussianrandomvariablethatisindependentofallother
k
Gaussianrandomvariablesn
k
′
+1
, n′ ,andsoon.Thatis,wecanfilteroutthedepen-
k+2
denceoftherandomnoiseonotherrandomnoisesamplesusingafilterw 
k
.
Itreallydoesn’tserveustogothroughthedetailsofhowweconstructthe
whiteningfilterw 
k
,asitallcomesoutofstatisticalliterature,andothertextsdelightin
sharingthatwithyou.Let’sjustgetrighttothekeyresult(howtobuildthewhitening
filter!),sowehavewhatweneedtokeepsurgingaheadwithourconstructionofthe
optimalrest-of-the-receiver.
Forthisoneparagraph,I’mgoingtoassumesomeknowledgeofdiscretetime
processing.Ifyoudon’thaveit,thisparagraphmaybehardtounderstand,andinthat
case(ifthisparagraphisimportanttoyou)you’llwanttotakealookatabookon
discretetimeprocessing.Thewhiteningfilterw
k
isobtainedfromtheequation
2
1
jω
)
W e
(
jω
)
(9.54)
X e
(
j
AnyW(e 
ω
)thatsatisfiesEquation(9.54)willdo,althoughitisconventional
j
wisdomtochoosetheW(e 
ω
)thatisalsocausalandstable(thatis,allpolesfallwithin
theunitcircle).
So,thefirststepintheoptimalrest-of-the-receiverisaddingawhiteningfilterw 
k
.
Thisfilterchangesthenoisesamplesforus—itmakeseachnoisesampleindependent
ofothernoisesamples,whichiskeyinbuildinganoptimalreceiver.(Engineersstill
areunsureofwaystobuildoptimalreceiversotherwise.)
9.5.3 TheFinalPartoftheOptimalReceiver
Sofar,we’vebuilttheoptimalreceiverfrontend,whichreceivesr(t)andputsout
r
′
= I
k
∗ ÷ n
k
(9.55)
k
x
k
266  ‹  Chapter Nine 
Werealizedwehadaproblemwith r′ ,withthenoisesamplesattimek,n 
k
.This
k
noiseisarandomvariablethatdependsonothernoisesamplesatothertimes(for
example,n 
k+1
).Soweputinafiltercalledawhiteningfilterwhichtookinr ′ andput
k 
outr  ,where
k
L−1
r
k
∗ +
∑
I v
k i
+ n′ I
k
v
k
n
k
′
i − k (9.56)
i0
and n′ isarandomvariablethatisindependentofothernoisesamplesatothertimes
k
(forexample, n′ ).Fromhere,wearereadytobuildthefinalpartoftheoptimal
k+1
receiver.Tostart,wetakeabetterlookatr 
k
.First,ifwewriteoutthesuminlong-
hand,wefindthatwecanrewriter 
k
inthefollowingform:
r
k
(I v
k
+ I v
k−1
+ + I v ) + I v
0 1
…
k−1 1 k 0
(9.57)
+
(
I v
k+ − 1
+ I
k+2
v
−2
+ + I
L−1
v
k L
)
+ n
k
′
1
…
− −1) (
Now,thatmeanswecanvisualizer 
k
beingbuiltasshowninFigure9.15.There,
wehavetheI
i 
’sbeingstoredinashiftregister;eachI
i
ismultipliedbyv
n – i
,andthen
theseareaddedtogether.Finally,thenoiseisaddedon.
I
0
I
1
I
k
I
L–2
I
L–1
+ 
+ 
v
k
v
k–1
v
0
v
k–(L–2)
r
k
...
x x x x x
v
k–(L–1)
...
n
k
′
Figure9.15Illustratingthecreationofr
k
(comingoutofthewhiteningfilter)
Channel Filtering and Equalizers  ‹  267 
Tokeeptheconstructionoftherestofthereceivereasytounderstand,let’s
considerasimpleexample.Let’ssay
r
k
I v
0
+ I
k−1
v
1
+ n
k
′ (9.58)
k
Thatis,we’reassumingthatalltheotherv 
k–i
termsare0,sowehaveasimpler
equationtoworkwith.We’llalsoassume(1)v
0
=1andv
1
=0.5,and(2)thatwehave
BPSKmodulation,inwhichcaseI 
k
iseitherAor–A.Ther inthissimplercasecorre-
k
spondstoFigure9.16.
Now,asmarkedinthatfigure,wecanseer 
k
createdasfollows:An“input”I 
k
comesin,itmovesthe“state”(what’satposition1)overtoposition2(creatingI 
k–1
).
Multiplicationbyv andv
1
,andtheadditionofnoise,createsthefinaloutputr  .
0 k
Wecandrawatrellisdiagramdescribingthecreationoftheoutputr 
k
ifthenoise
werenotpresent.ThisdiagramisshowninFigure9.17:
1.The“state”(inFigure9.16)isrepresentedbynodes,thedotsdrawnateach
time—atanyonetime,thisstateiseither–Aor+A.
2.The“input”I 
k
(inFigure9.16)isrepresentedasabranch—iftheinputis
I  =– A,thebranchissolid;iftheinputisI 
k
=A,thebranchisdotted.
k
3.Whenaninputcomesin(thatis,abranchisdrawn),itchangesthestate.You
canseethisindicatedbythestateabranchbeginsat,andthestateitendsin.
4.Finally,ifweknowthe“state”andthe“input,”thenwehavealltheinformation
weneedtofigureouttheoutputr 
k
(assumingnonoise).Thatoutputisdrawn
aboveeachbranch.
"state"
I
k
v
0
=1
v
1
=0.5
"input"
r
k
x x
+
+
I
k–1
1 2
n
k
′
Figure9.16Illustratingthecreationofr
k
inasimplercase
268  ‹  Chapter Nine 
time0 time1 time2
–1.5A –1.5A
–A
0.5A
–0.5A
0.5A
–0.5A
…
+A 
1.5A 1.5A 
Figure9.17Atrellisdescribingthecreationofr
k
(withoutthenoise)
Here,I’mgoingtousea“commonsense”argument(ratherthanalengthystatis-
ticalone)toexplainhowtheoptimaldecoderusesthetrellisofFigure9.17togetthe
outputI′ :
1.Sincethenoise n
k
′ iseasilyshowntobeaGaussianrandomvariablewithzero
mean,thenthat,bydefinition,meansthattheaveragenoiseis0.
2.Itmakessense,then,thatgiventhesequenceofreceivedr 
k
(k=0,1,2,..., L–1),
whichonaveragecontainnoisen 
k
thatequals0,youcouldtrytomatchther ’sto
k
thebestpathofnoise-freeoutputsinthetrellis.Thatis,youwouldwanttofind
thepathinthetrelliswhoseoutputs(drawnwithnonoise)areclosesttother
k 
(k=0,1,2,..., L–1)thatyoureceived.Onceyou’vefoundthispath,thenyoucan
figureout,byfollowingthebranches,whatinputsI 
k
(k=0,1,2,..., L–1)toputout
fromthereceiver.
Here’sanexample.Let’ssaythatwereceivethetwovaluesr
1
=1.7Aandr
2
=
1.7A.Wewantthereceivertooutputthebestguessonthesentbits.Todothis,we
turntoourtrellis,andwefindthepaththroughthetrelliswithoutputsthatareclosest
tor
1
=1.7Aandr
2
=1.7A.Lookingateachpaththroughthetrellis,wefindouthow
closeeachpathis—seeFigure9.18(a)through(d).Wethendecidethatthe“bottom
lines”pathofFigure9.18(d)istheclosest.Thebranchesonthispathindicatethatthe
outputsareI
1
=A andI
2
=A,whichtellsusthattheinputbitsarem =(1,1).Our
receiveroutputsthesebits.
Thisoptimalrest-of-the-receiveriscommonlycalledanMLSE(MaximumLikeli-
hoodSequenceEstimator).
Ingeneral,searchingthroughatrellisfortheclosestpathcanbeadifficulttask.
But,inChapters7and8,wesawahandywaytodoit—awaycalledtheViterbialgo-
rithm.WhenyouwanttousetheViterbialgorithmtofindthebestpath,justuseitin
exactlythesamewayexplainedinSection8.3.
Channel Filtering and Equalizers  ‹  269 
r
1
=1.7A r
2
=1.7A
time0 time1 time2
–1.5A –1.5A
PATH 1
(a)
Distance
2
=(1.7A–(–1.5A))
2
+(1.7A–(–1.5A))
2
=20.48A
2
time0 time1 time2
–1.5A
0.5A
PATH 2
(b)
Distance
2
=(1.7A–(–1.5A))
2
+(1.7A–0.5A)
2
=11.72A
2
time0 time1 time2
1.5A
–0.5A
PATH 3
(c)
Distance
2
=(1.7A–1.5A)
2
+(1.7A–(–0.5A))
2
=10.32A
2
time0 time1 time2
PATH 4
(d)
1.5A 1.5A
Distance
2
=(1.7A–1.5A)
2
+(1.7A–1.5A)
2
=0.08A
2
Figure9.18Lookingfortheclosestpaththroughthetrellis
270  ‹  Chapter Nine 
Example 9.4 
If,afterawhiteningfilter,thereceivedsignalisdescribedby
r
k
I
k
v
0
+ I
k −1
v
1
+ n
k
(E9.19)
where
v
0
v
1
1 (E9.20)
I
k
+1 or I
k
−1 (E9.21)
thenfigureouttheoutputofanMLSE(usingtheViterbialgorithm)whenthe
inputtothedecodercorrespondsto
r
1
−2.1 and r
2
0.1 (E9.22)
Solution:FigureE9.1showsthetrellisdiagramcharacterizingthereceived
signalinEquation(E9.19).ThisdiagramiscreatedfromEquation(E9.19)using
thefourstepsoutlinedinthewritingyoujustread.
AlsoshowninFigureE9.1istheapplicationoftheViterbialgorithmto
determinethebestpaththroughthetrellis.Foreachnode,thebestparentis
foundbychoosingtheparentwiththesmallesttotaldistance.Whenwegettothe
lastnodes,welookforthenodewiththesmallesttotalandwebacktrackthrough
thetrellisasmarkedbythearrows.Thesearrowstellusthattheoutputofthe
trellisshouldbe(–11)(basedonthebranchlinesofthebestpath).
received –2.1 0.1
–1
–2 dist
2
=0.01 total=0.01
–2
dist
2
=4.41 total=4.42
0
d
i
s
t2
=

4
.
4
1

bestparent=–1
d
i
s
t2
=

0
.
0
1

bestparent
=–1
2
=

4
.
4
1
t
o
t
a
l

=

4
.
4
1

(total0.01)
t
o
t
a
l

=

0
.
0
2 2
=

0
.
0
1
(total4.42)
0
d
i
s
t
t
o
t
a
l

=

4
.
4
1

0
d
i
s
t
t
o
t
a
l

=

4
.
4
2
+1
2 dist
2
=16.81 total=16.81 2
bestparent bestparent
=–1 =–1
(total4.41) (total0.02)
FigureE9.1
Trellisdescribingreceivedsignal+ViterbiAlgorithmusedtogetbestoutput
Channel Filtering and Equalizers  ‹  271 
9.5.4 AnIssuewithUsingtheWhiteningFilterandMLSE
First,congratulations.WhenyoubuildareceiverasshowninFigure9.19you’vegotan
optimalone.Youpasstheincomingsignalthroughanoptimalreceiverfrontendtoget
r
k
′ ;thenyoupassitthroughawhiteningfiltertogetr .Finally,youpassitthroughan
k
MLSE(whichfindstheclosestpaththroughanoiselesstrellis),andthatputsoutthe
bestguessonbitsm′.
However,thereisaproblemwiththeMLSE.Inmanycommunicationsystems,
thenumberofstatesandthenumberofbranchesinthetrelliscanbeverylarge.In
fact,thetrelliscanbesolargethatitbecomesveryexpensivetobuildreceiversthat
implementtheMLSE.Peoplebegantolookforinexpensivealternatives.Inessence,
therestofthischapterisallaboutcheaperoptions.Wewanttofindalternativesto
usingthewhiteningfilterfollowedbytheMLSE—alternativesthatwillperformwell,
andwillbefarmorecosteffective.
cosω
c
t
x
r(t)
r
′
(t)
x
h(–t)
r
k
kT r
k
′
ω
k
MLSE
m
′
sinω
c
t
Figure9.19Theoptimalreceiver
9.6 LinearEqualizers
WestartherewithFigure9.20(a):we’vegotthenewmodulatorputtingouts(t),the
newchannelputtingoutr(t)= s(t)*c(t)+n(t),andtheoptimalreceiverfrontendputting
out r
k
′ .Weknowhowtobuildanoptimalrest-of-the-receiver,butit’ssodarnexpensive
thatwe’relookingforcheaperalternatives.Thefirstalternativeiscalledthelinear
equalizer.
TheuseofthelinearequalizerinthereceiverisshowninFigure9.20(b).Itis
simplyadiscretefilterwithimpulseresponsec thattakestheinput r
k
′ andturnsitinto
k
anewvariablecalledr
k
;thisr isthenpassedthroughadecisiondevicewhichoutputs
k
I
k
′ ,thebestguessonthedatasymbolI .Thedecisiondevicealwaysworksinthe
k
sameway,exactlythewaydecisiondevicesweredescribedinChapter5.
272  ‹  Chapter Nine 
Whatisthisimpulseresponsec
k
,andhowdowechooseit?Therearemany
differentpossiblechoicesforc
k
,andwe’llspendthenextfewsectionsdiscussing
thesechoices(onesectionperchoice).
cosω
c
t
coder Re{
.
}
bits
pulse
shaper
x
n(t)
c(t) +
x
x
kT
r
k
′
e
j(ω
c
t)
sinω
c
t
h(–t)
m
symbol
I
k
NEWOPTIMAL
NEWMODULATOR NEWCHANNEL
RECEIVERFRONT
END
(a)
device
Decision
r
k
′
c
k
r
k I
k
′
LINEAR
EQUALIZER
(b)
Figure9.20 (a)Themodulator,channel,andreceiverfrontend
(b)Thenewrest-of-the-receiver(alinearequalizerfollowedbyadecisiondevice)
9.6.1ZeroForcingLinearEqualizer
Tounderstandthiscase,let’sstartwithalookatwhatiscomingintothereceiver,
namelyr ′ :
k
r′ I
k
∗ x
k
+ n
k
(9.59)
k
TheintentioninthezeroforcinglinearequalizeristoforcetheISItozero—togetthe
outputr
k
tohavetheform
r
k
I
k
+ n′′ (9.60)
k
That’seasy.Allwedointhiscaseismakec
k
undox
k
.Thatis,wechoosec
k
tobe
theinversefilterforthefilterx
k
.Mathematically,wechoose(usingthez-transform
domain)
C(z) = 1/X(z) (9.61)
Channel Filtering and Equalizers  ‹  273 
There’soneproblemwithbuildingthelinearequalizerinthisway:itfocuseson
gettingridofthefilteringeffect(thex
k
),butitpaysnoattentiontowhathappenstothe
noise.Thenoisepassesthroughthefilterc
k
,anditmaygetmuchbigger.Thisphe-
nomenon,knownasnoiseamplification,cancausetheI
k
togetlostinthemuchlarger
noise n
k
′′ .Forthisreason,thistypeoflinearequalizerisrarelyused.
Instead,engineershaveafondnessforthe...
9.6.2 MMSE(MinimumMeanSquaredError)Equalizer
Withthistypeofequalizer,incomes
r′ I
k
∗ x
k
+ n
k
(9.62)
k
or,writtendifferently,
L−1
r
k
′
∑
I
i
x
k −i
+ n
k (9.63)
i 0
] ,
L−1
]
r′ I
k
x
0
+
,
∑
I
i
x
k −i
+ n
k
]
k
,
(9.64)
i 0
, i k ]
] ¸
≠
We’dliketochoosetheequalizer(thefilter)c
k
sothatwhatcomesout,namely
r
k
∗c
k
isasclosetoI
k
aspossible.Withthatinmindwesay,mathematically,that r
k
′
wewanttochoosec
k
tominimizethefunction:
2
( ) E
,
f c
k
¸
r′ ∗c
k
− I
k
]
(9.65)
k
]
whereE[x]istheexpectedvalueofx.Thatis,inwords:onaverage,wewantthe
outputofthefiltertobeascloseaspossibletoI
k
.Afteraboutapageofstatistical
wrangling,youcanshowthatthisrequirementismetbychoosingc
k
tosatisfy(inthe
z-transformdomain)
C(z)=1/[X(z)+ N
o
] (9.66)
whereN  /2isthevarianceofthenoiseintroducedinthechannelandX(z)isthe
o 
z-transformofx
k
.
Example 9.5 
Ifareceiverseestheinput
r′ I
k
+ 0.5 I
k −1
+ 0.25 I
k−2
+ n
k
(E9.23)
k
274  ‹  Chapter Nine 
provideanequationforthezeroforcinglinearequalizerandanequationforthe
MMSElinearequalizer.AssumeN
o
=0.5.
Solution:Writingthereceivedsignalintheformofaconvolution,wehave
r′ I
k
∗ δ + 0.5 δ + 0.25 δ
k−2
)
+ n (E9.24) (
k k k−1 k
ComparingthiswithEquation(9.59),werecognizethat
x
k
δ + 0.5 δ + 0.25 δ
k−2
(E9.25)
k k −1
which,usingz-transforms,correspondsto
( ) + 0.5 z + 0.25 z
−2
(E9.26) x z 1
−1
Now,forthezeroforcinglinearequalizer,weturntoEquation(9.61)which
inthiscaseleadsusto
C z
−2 (E9.27)
( )
1 + 0.5 z
−1
1
+ 0.25 z
and,fortheMMSElinearequalizer,weturntoEquation(9.66),whichthistime
leadsusto
C z
−2 (E9.28)
( )
1.5 + 0.5 z
−
1
1
+ 0.25 z
9.7 OtherEqualizers:theFSEandtheDFE
Inadditiontothelinearequalizer,sometimescalledLEforshort,engineersdesigned
othercheapalternativestotheoptimalwhiteningfilterfollowedbyMLSE.I’llprovide
justabriefoverviewhere,andI’llletyou,ingraduatecoursesoroutofgeneralinter-
est,readotherbooksthatdescribetheworkingsoftheotheralternativestothe
whiteningfilterandtheMLSE.
First,thereisthefractionallyspacedequalizer,orFSE.Ifsomeoneintroducesyou
tothisequalizer,don’tbeatallintimidated.Itis,intruth,justalinearequalizerin
disguise,forwhichtheyhavefoundawaytodoallofthefilteringinthedigitaldomain,
whereitischeaperandeasieranyway.
Secondly,thereisthedifferentialfeedbackequalizer,orDFE.Inthiscase,what
youdoonceyouhave r′ fromthereceiverfrontendisshowninFigure9.21.Basically,
k
youhavewhatlookslikealinearequalizerfollowedbyadecisiondevice.Themain
differencehereisthatafeedbackfilterhasbeenaddedbelowthedecisiondevice.
Channel Filtering and Equalizers  ‹  275 
Verybriefly,youreceive
r′ ∗ + n
k
(9.67)
k
I
k
x
k
L−1
r
k
′
∑
I x
k i
+ n
k (9.68) i −
i 0
k −1 L−1
\ j
r′ I x
0
+
,
j
∑
I x +
k k i k i ( , ∑
I x
−
\
+ n
k (9.69) − i k i (
( i0 , ( i k +1 ,
Youusethec
k
togetridofthesecondtermandthed
k
togetridofthethirdterm.
+
d
k
c
k
r
k
′
device
I
n
′ r
k
+
–
Decision
FEEDBACK
EQUALIZER
FEEDFORWARD
EQUALIZER
Figure9.21 TheDFE
9.8 Conclusion
Thechannelwasdifferent.Insteadofgivingyouwhatyousentwithanoise—i.e.,r(t)=
s(t)+ n(t)—itgaveyousomethingelse.Itgaveyouwhatyousent,filteredbyachannel
filter,plusanoise—thatis,r(t)= c(t)∗ s(t)+ n(t).Becauseofthat,wehadtointroduce
awholechapter.
Wefirstfoundanewwaytoexpressthemodulator,andwewereabletogeneral-
izeitsothatitdidwhatwecalled“pulseshaping.”Thenwelookedatreceivers.First,
wetriedoneout,andwefoundoutthatithadproblemswithnoiseamplification,except
inthecasewhenwehadachannelfilterthatwasflatoverthefrequenciesoftransmis-
sion.
Thenwefoundtheoptimalreceiverfrontend.Fromthere,webuiltanoptimal
receiver,whichwasmadeupofawhiteningfilterandwhatiscalledanMLSE.Finding
thatcostlytobuild,engineerscameupwithanewdesign.Theybuiltthelinearequal-
izerintwoforms(zeroforcingandMMSE).
That’sthemeatofthischapter,inanutshell.
276  ‹  Chapter Nine 
Problems 
1.(a) Provideanequations(t)describingtheoutputofa16-PSKmodulatorwith
pulseshapingg(t).Makesurethatyours(t)isinaformsimilarto
Equation(9.6).
(b)Repeat(a)fora16-ASKmodulator.
2. Considerasystemwherewehaveatransmitfilterg(t),achannelfiltering
c
E
(t)=δ(t),andthereceiverofFigure9.6withfilterf(t).Thetotalfilteringeffect
isx(t)(theconvolutionofallthefiltereffects).
(a) Drawablockdiagramofthecommunicationsystem,includingthemodula-
torandthereceiverfrontend.Indicatethevalueofthesignalbeforethe
decisiondevice.
(b)Itisdecidedthatx(t)willbearaisedcosinefunction.Useatableofvalues
toplottheraisedcosinefunctionx(t)ofequation(9.21). Ononecolumn,
usetvaluesspacedbyT/4,andontheothercolumnprovidethevalueof
x(t).Dothisovertherange[–2T,2T].Providetwoplots,forroll-offfactors
of0.5and1.0.
(c) Useyourplotandyouranswerto(a)toexplainhowtheraisedcosine 
functionx(t)allowsyoutoavoidISI. 
(d)Provideapossibleselectionforthefilterg(t)andthefilterf(t)sothatthe
totalfilterx(t)istheraisedcosinefilter.
3. Findthreepossiblechoicesforf(t)∗ g(t)(thecombiningofthetransmitand
receivefilter)whenyouaretold
•  thechannelimpulseresponsecanbemodeledasc
E
(t)=δ(t).
•  youwillusethereceiverofFigure9.6.
•  youwanttozeroISI.
•  youwanttotransmitatasymbolratecorrespondingtoasymbolduration
T=9600symbols/sec.
•  youwantthetotalbandwidthofthetransmissiontocorrespondtolessthan
6000Hz.
4. Youaretoldthatthereceivedsignalafterthereceiverfrontendandwhitening
filtercorrespondsto
1 1
r
k 
I
k 
+ I
k −1
+ n
k
′ 
(Q9.1)
2 2
Channel Filtering and Equalizers  ‹  277 
where
I
k
+1 or I
k
− 1 (Q9.2)
(a) AssuminganMLSEisappliedtothissignal,plotthetrellisdiagramthatthe
MLSEusestomakeadecisiononthefinaloutputsymbols.
(b)Ifyoureceivethevalues+1,+1,+1,0,usetheViterbialgorithmtodetermine
whatsymbolstheMLSEdecidesweresent.
5. Yourmanagertellsyouthathisreceiverfrontendisoutputtingthesignal:
1
0.9 0.3
k k k k
r I I n
−
′ + + (Q9.3)
where
2 or I
k
− 1 − or 1 or 2 (Q9.4)
(a) Assumingthenoisen
k
iswhite,heasksyoutodesigntheoptimalreceiver
fordatadetection.Beascompleteanddetailedaspossibleinyourreply–
includethetrellisdiagram,thevaluesonallthebranches,andadescription
oftheViterbialgorithmtobeusedfordetection.
(b)Designtheoptimallinearequalizerto(1)removeISI;(2)minimizeMSE.
(AssumeN
o
=0.7).
6. Outofareceiverfrontendcomesthesignal
r
k
I
k
+ 0.5 I
k −1
+ n
k
′ (Q9.5)
where
I
k
−3A or − A or A or 3A (Q9.6)
andthenoisesareindependentGaussianrandomvariables(i.e.,awhiteningfilter
hasalreadybeenapplied).
(a) DescribetheMLSE.
(b)Giveninputs1.5A,2.5A,and–1.5A,whatwouldyourMLSEoutput?
278  ‹  Chapter Nine 
n=1
shaping
I
n
g(t–nT) I
k
N
c
E
(t)
+
n(t)
f(t)
kT+τ
Pulse
FigureQ9.1Acommunicationsystem
7. YouaregiventhecommunicationsystemofFigureQ9.1.
(a) Determinetheoutputaftereachblockinthefigure.
(b)SpecifythecriteriaonthecommunicationsystemforzeroISIattheoutput
ofthesampler.
(c) IfIselect
¦
1, t 0
x t ( )∗c t ( )
¦
− ( ≠
E (Q9.7)
( ) g t ( )∗ f t
¦
0, t kT iT k i )
forwhatvaluesofτ doIhavezeroISIoutofthereceiverfrontend?
10 
Chapter 
Estimation and
Synchronization
S
ofarinthisbook,wehaveconsideredtwochanneleffects.PriortoChapter9,we
consideredchannelsaddinganoisen(t)tothesentsignal.InChapter9,wesaid
thechannelinfactaddedmorethanjustanoise;italsoactedasafilterwithimpulse
responsec(t).Inthischapter,weagainaddtowhatchannelsdo—wemakethemmore
realistic.
10.1Introduction
Inadditiontonoiseand/orfiltering,channelsalsoaddunwantedparameterstothe
sentsignal.Forexample,takealookatFigure10.1.Thereyouseethatthemodulator
sentthesignal
( ) Acos (ω +θ ), iT ≤ < (i +1)T (10.1) s t t
i
t
c
whichmight,forexample,beaQPSKsignal.Manychannelsarewell-modeledas
follows:
CHANNEL
n(t)
r(t)
s(t)=Acos(ω
c
t+θ
i
),iT≤ t< (i+1)T
r(t)=Acos((ω
c
+∆ω)(t– τ)+θ
i
+θ)+n(t),iT<t–τ < (i+1)T
+
phase
offset
θ
offset
∆ω
timing
offset
τ
Modulator
frequency
Figure10.1Channeleffects
280 ‹ Chapter Ten 
(1)thechanneladdsatimingoffsetτ,aphaseoffsetθ,afrequencyoffset∆ω;and
(2)thechanneladdsanoisen(t).Combiningtheseeffectsleadstothereceivedsignal
( )
Acos
((
ω + ∆ω
)
t +θ θ
)
+ n
( )
, ≤ − τ <
(
i +1
)
T (10.2) + t iT t r t
c i
Manyreceivers,giventhisreceivedsignal,tryfirsttoestimateandremovethe
timingoffsetτ,thephaseoffsetθ,andthefrequencyoffset∆ω.Then,allthatisleftis
thesentsignalinthepresenceofnoisen(t).Theyusethedemodulatorswesawin
Chapter5todetectthesignalinthepresenceofnoisen(t).
Thischapterisallabouthowareceiverestimatesτ,θ,∆ω andanyotherunknown
parameter,sothatitcanremovetheseparametersfromthereceivedsignal(leaving
justthesignalinthepresenceofnoise).Thisprocessofestimatingunknownparam-
etersatreceiversiscalledestimationorsynchronization.Wewillbeginwithageneral
explanationofhowtoestimateanyparameterthatachannelmightintroducetoa
signal,andthenwewillexplaindifferenttechniquesthatthereceivermightuseto
estimateandremoveachannelphaseθ.
10.2 Estimation:Part1
10.2.1OurGoal
Inthissection,weconsiderthefollowingproblem.Wereceiveasignalr(t)witha
randomvaluea(withprobabilitydensityfunctionp(a))containedinit.Forexample,
wemayreceivethesignal
r t
c
t
i
)
, ≤ <
(
i +1
)
T (10.3) ( )
Acos
(
ω +θ + a
)
+ n
(
t iT t
wherea representsarandomphaseoffsetintroducedbythechannel.Wewanttofinda
waytoestimatethevalueofainr(t).Thereisawaytoexpressthisproblemmathemati-
callytomakeiteasier.Anyr(t)canberepresentedonanorthonormalbasis,andcanbe
writteninsteadasavectorr.WesawthisideainChapter5.Forexample,thesignalr(t)
inEquation(10.3)canbefullyrepresentedasthevector r (r r
2
) usingthe
1
,
c
), ≤ < (i +1)T cos (ω t iT t
orthonormalbasis
¦
φ
1
( )
t
t , φ
2
(t )¦
where
φ
1
( )
2
T
and
φ
2
( ) −
2
t sin (ω t iT t
c
), ≤ < (i +1)T
.Specifically,thesignalr(t)in
T
Equation(10.3)canbeexpressedasthevectorrdescribedby
r (r r
2
) (10.4)
1
,
where
r
1
A
T
2
cos (θ + a) + n
(10.5)
i 1
Estimation and Synchronization ‹ 281 
r
2
A
T
2
sin (θ + a) + n
(10.6)
i 2
Theproblemwewanttosolvecannowbestatedas:Givenr,wewanttofindan
estimateofa.WedonotjustwanttoestimateainthespecialcaseofEquations(10.4)
to(10.6),butwewanttodothisforanyrcontainingarandomvaluea.Ingeneral,we
haveanrequation,andwehavearandomvalueacontainedinthatrequation.We
wanttoestimatethata.
10.2.2 WhatWeNeedtoGetanEstimateofaGivenr
Therearetwoorthreethingsrequiredtocreateanestimateofagivenr.Theyareall
statisticalquantities.
1.p(r rr rr|a): Theexactvalueofrdependson:therandomvaluea;thevalueofthe
signalsent;andtherandomvalueofthechannelnoise.Thevalueofr,then,isa
randomvaluethatdependsonthreevalues.Foragivena,rbecomesarandom
variablethatdependsonlyonthesignalsentandthenoise.Thatis,foragivena,
rcanbedescribedasarandomvariablewithadistributionthatwewilllabel
p(r rr rr|a).Werequireknowingorbeingabletoevaluatethisdistributionbeforewe
cancreateanestimateofa,givenr.
2.p(a): a,beingarandomvalue,ischaracterizedbyaprobabilitydensityfunction
p(a).Wealsoneedtoknowthisinordertocomputeanestimateofa.
3.p(r rr rr):AsIexplainedinpoint1,risarandomvalue,andthatmeansitcanbe
describedbyaprobabilitydensityfunctionp(r rr rr).Insomecases,generatingthe
valueofawillrequireweknowthefunctionp(r rr rr). 
Next,we’llseehowtousethesevaluestocreateanestimateofa,givenr.Thereare
threemainwaystocomeupwiththisestimate,andthesearedescribedinthenext
threesections.
10.2.3 EstimatingaGivenr,theFirstWay
Thefirstwayiscalledtheminimummeansquarederror(MMSE)method.Itgener-
atesanestimate aˆ calledtheminimummeansquarederrorestimate.Theidea
mmse
hereistooutputtheestimateofa, aˆ ,thatis,onaverage,asclosetoa aspossible.
mmse
Mathematically,theideaistooutputthe aˆ thatsatisfiesthefollowingstate-
mmse
2
ment:Thevalue aˆ minimizesthevalue E a − aˆ
mmse
)
]
]
(whereE[x]isthe
mmse
¸
,
(
expectedvalueofx).
282 ‹ Chapter Ten 
Aftersomework,youcanshowthattheminimummeansquaredestimate aˆ
mmse
canbecalculatedwithasimplemathematicalequationifyouknowtheprobabilities
p(r rr rr|a), p(a),andp(r rr rr):
∞
ˆ a
mmse

∫
a p
(
a r
)
da
(10.7)
−∞
r
)

p
(
r a
where
p a
a
)
p ( )
( .
r p ( )
10.2.4 EstimatingaGivenr,theSecondWay
Often,themathematicalcomputationoftheestimate aˆ fromEquation(10.7)is
mmse
rathermessy.So,engineers,whohaveneverbeenbigon“messymath,”foundanew
waytogetanestimateofa.Theycalledtheirnewwaythemaximum a posteriori,or
MAP,estimateofa,andlabeledit aˆ
MAP
.Thisistheestimatethatsatisfiesthefollowing
, mathematicalstatement:Choosethevalue aˆ
MAP
thatminimizes E C (a, aˆ
MAP
)]
] ¸
where C a a
MAP
) =1if aˆ
MAP
isnotveryclosetoa,and C a a
MAP
) =0if aˆ
MAP
is (
, ˆ
(
, ˆ
veryclosetoa.
Thismathematicalstatementbasicallysays,inwords,findthe aˆ
MAP
thatisveryclose
toa.
Thevalueof aˆ
MAP
canbefoundifyouknowthetwoprobabilitiesp(r rr rr|a)andp(a).
Ifyouknowthesetwothings,thenthevalueof aˆ
MAP
canbefoundaccordingtothe
mathematicalstatement: aˆ
MAP
isthevaluethatmaximizesthefunctionp(r rr rr|a)· p(a);or,
equivalentlythroughthemathematicalequation
aˆ
MAP
argmax p
(
r a a
)
p ( )
(10.8)
a
or
∂
p
(
r a a
)
p ( )
ˆ
0
(10.9)
a a
MAP
∂a
Generallyspeaking,theestimates aˆ and aˆ
MAP
workouttothesamevalue.In
mmse
fact,inalmosteverycaseofpracticalinterest,thesevaluesareidentical.
Estimation and Synchronization ‹ 283 
10.2.5 EstimatingaGivenr,theThirdWay
Thereisonemorewaytogenerateanestimateforagivenr.Thismethodisused
ifyouhaveaslightlydifferentcasethanthatwehavebeenstudyingsofar.Itisusedif
thechannelintroducesanunknownvaluea—thistimeaisnotdescribedasarandom
valuebutratheraisjustsomeunknownvalue.Forexample,thechannelmayadda
constantatothesentsignal,asshowninFigure10.2,andthevalueaaddedonisnot
well-describedasarandomvalue,butjustsomeunknown,fixednumber.
Engineerswantedtocreateanestimatethatwouldbeasclosetoaasstatistically
possible.Theyfoundawaytocreatethisestimatemathematically,andcalledthe
estimationmethodtheyinventedmaximumlikelihood(orML)estimation.Theesti-
mateforathatthismethodprovidesiscalled aˆ
ML
.
The aˆ
ML
estimateiscreatedaccordingtothefollowingmathematicalequation:
aˆ
ML
argmax p
(
ra
)
,
(10.10)
a
or,equivalently,
∂
p
(
ra
)
ˆ
0
(10.11)
a a
ML
∂a
modulator
receiver
s(t)
+ +
r(t)=s(t)+a+n(t)
a n(t)
someunknown
value
Figure10.2 Achannelintroducesanunknownvalue
Example 10.1 
Let’ssaythatyoupickupsomeunknownvalueacorruptedbyazero-meanunit-
varianceGaussiannoise.Thatis,let’ssayyoufind:
+ n
(E10.1)
where
r a
j −n \
( )
1
exp
2
p n
2π
,
(
2
(
,
(E10.2)
284 ‹ Chapter Ten 
Determinethebestestimateofa.
Solution:We’redealingwithanunknownvalueofahere,sointhiscasewe
turntoEquation(10.11).Specifically,thisequationtellsusthatwerequirep(r|a) 
ifwe’regoingtocomeupwithanestimatefora,solet’sstartbygettingp(r|a).
Thekeyisrealizingthatp(r|a)isthelikelihoodofhavingr (forexample,
r = 1)showup,givena (forexample,a = 0.2).Now,sincer = a + n,I’llgetr 
whena issentonlyifn = r – a (soforourexamplevalues, n =1–0.2=0.8). 
Mathematically,
p r a
)
p n r − a) (E10.3) ( (
Now,pluggingintheknowndistributionfornoise(EquationE10.2),this
means
2
(r a)
\
( p r a
)

1
exp
j
,
− −
(
(E10.4)
2π
,
2
(
, (
Okay,nowwehavep(r|a). We’rereadytocomeupwithourestimateusing
Equation(10.11).
p r a
)
∂
(
a a
ML
0
(E10.5)
ˆ
∂a
(r a)
2
\
(
\
(
∂
j
,
1
exp ,
j
− −
ˆ a a
ML
0
(E10.6)
∂a ,
2π
,
(
2
(
(
(
,
,
2
−
2
(r a)
\
(
j
− −
1
−2 (r a)
⋅ −1 exp ,
ˆ a a
ML
0
(E10.7)
2π
2
( )⋅
,
2
(
, (
2
j
− −
\
− (r aˆ
ML
)⋅ exp ,
(r aˆ
ML
)
( 0
,
2
(
,
(E10.8)
(
a r (E10.9) ˆ
ML
Therefore,ifIseer = a + n = 0.7,withndefinedaccordingtoequation(E10.2),I’ll
guess(estimate)that0.7isinfactthevalueofa.
Estimation and Synchronization ‹ 285 
10.3 EvaluatingChannelPhase:APracticalExample
10.3.1OurExampleandItsTheoreticallyComputedEstimate
Let’snowlookatapracticalexampleofhowtocomeupwithanestimateofagivenr.
TakealookatFigure10.3,wherewehaveatransmitterthatissendingacosinewave
( ) Acos (ω t )overashorttimeinterval.Thechannelintroducesaphaseoffseta s t
c
andaddsanoisen(t).Asaresult,thereceivedsignalis
( )
Acos
(
ω +
( )
, 0 < < r t t a
)
+ n t t T
E
(10.12)
c
wherethechannelphasea isknowntobearandomvalueuniformintherange[0,2π),
andthenoisen(t)representsanadditivewhiteGaussiannoise.
Thereceivergreetstheincomingsignalwithareceiverfrontend,whichmaps
r(t)tothevectorr =(r
1
,r
2
)bymappingitontotheorthonormalbasis ¦
φ
1
( ) t , φ
2
(t )¦
sin (ω
c
t ), 0 < < t T
E .
t cos (ω
c
t ), 0 < < t
where
φ
1
( )
T
2
E
t T
E and
φ
2
( ) −
2
T
E
Specifically,asseeninChapter5,thereceiverfrontendcomputes
r (r r
2
) (10.13)
1
,
where
∞ T
E
T
E
T
E
( ) ( )
∫
( ) ( )
∫
t a )φ (t dt + n t φ
1
t dt r
1

∫
r t φ
1
t dt r t φ
1
t dt Acos (ω +
1
)
∫
( ) ( )
c
−∞ 0 0 0
T
E
Acos (ω +
1
)
∫
c
t a )φ (t dt + n
1
0
(10.14)
CHANNEL
r(t)=Acos(ω
c
t+a)+n(t),0< t< T
E
n(t)
s(t)=Acos(ω
+
phase
offset
a
aisuniformrandomvalue 
between[0,2π). 
c
t),0< t< T
E
Figure10.3 Channelintroducingphaseoffsetandnoise
286 ‹ Chapter Ten 
∞ T
E
T
E
T
E
(
t dt
∫
r t φ
2
t dt
∫
Acos
(ω +
2
) (
t dt r
2

∫
r t
)φ
2
( ) ( ) ( )
t a
)φ (
t dt +
∫
n t
)φ
2
( )
c
−∞ 0 0 0
T
E
Acos
(
ω +
2
)

∫
c
t a
)
φ
(
t dt + n
2
0
(10.15)
Usingalittlebitofmath,thissimplifiesto
r (r r
2
) (10.16)
1
,
where
r
1
A
T
E
cos ( ) + n a
1
(10.17)
2
r
2
A
T
E
sin ( ) + n a
2
(10.18)
2
We’llassumeT
E
=2tosimplifyourpresentation.Itcanbeshown(butyou’llhave
totakemywordforit)thatn
1
andn
2
arebothGaussianrandomvariableswithmean0
andvarianceσ
2
,andn
1
andn
2
areindependent.We’llnowfigureouttheMAPestimate
n 
a giventher=(r
1
,r
2
)ofequations(10.16)to(10.18).Togetthisestimatewewill ˆ
MAP
needtwothings,p(a)andp(r rr rr|a). Let’sstartoutbyevaluatingthese.
1.p(a):Weweretoldthataisarandomvariable,uniformintherangeof[0,2π).
Writingthisstatistically,weknowthatthedistributionofa, p(a),correspondsto
¦
1
( )

¦
2π
, a ∈
[
0, 2π) ¦
p a
(10.19)
¦
0 , otherwise
¦
2.p(r rr rr|a): Wewillneedtofigurethisoneoutstatistically,asfollows:Westartwith
p
(
ra
)
p r r
(
1
, a
) (10.20)
2
Thatis,p(r rr rr|a)istheprobabilitythatr
1
takesonaparticularvaluegivena,andthe
probabilitythatr
2
takesonaparticularvaluegivena.
1
2
Estimation and Synchronization ‹ 287 
Now,weknowthatr
1
=Acos(a)+n
1
. So,givena,thelikelihoodthatr =xisthe
likelihoodthatthenoisen
1
 = x – Acos(a).Similarly,withr  = Asin(a)+ n
2
,thelikeli-
2 
hoodthatr
2
=ygivenaisthelikelihoodthatn =y –Asin(a). Withthesetwothingsin
2
mind,wecanwrite
(
1
, p r r a
)
p n r
1
− Acos ( ), n
2
r
2
− Asin (a)) (10.21) (
1
a
Now,withthenoisesn
1
andn
2
independentofoneanother,thenwecanexpress
theequationaboveaccordingto
(
1
, p r r a
)
p n r
1
− Acos ( ))
⋅ p n r
2
− Asin ( )) (10.22) (
1
a
(
2
a
2
Next,wecancontinuetoapplywhatweknowtothisequation.Weknowthatthe
noisen
1
andn
2
areGaussianrandomvariables,withmean0andvarianceσ
2
—that
n 
is,mathematically, p n n) p n n)
1
exp
(
2σ
2
(
.Substitutingthisinto
,
Equation(10.22),wehave
2πσ
2
n ,
(
1
(
2
j
−n
2
\
n
a
)

1
exp
j
,
−
(
r Acos ( ))
2
\
1
j
−
(
r − Asin ( ))
(
1
, p r r
1
− a
(
⋅ exp
,
2
a
2
(
\
2
2πσ
2
(
, 2σ
n
2
(
,
2πσ
2
(
, 2σ
n
2
(
,
n n
(10.23)
Usingthesimplemathematicalpropertyexp(a)exp(b)=exp(a + b)weendupwith
2 2
a
)

1
exp
j
,
−
(
r Acos ( ))
−
(
r
2
− Asin ( ))
− a a
\
1
(
(
1
, p r r
2
(
(10.24)
2πσ
2
,
2σ
2
n
(
n
,
Thisequationrepresentsp(r rr r r|a),whichisallwe r|a).Wenowhavep(a)andp(r rr r
needtocomputetheMAPestimatep(a). 
Theestimate aˆ
MAP
correspondstothevaluecomputedfromtheequation
aˆ
MAP
argmax p
(
ra p a
) ( )
(10.25)
a
1
From(10.19),p(a)isaconstant
2π
foranyphaseain[0,2π). Beingconstant,it
doesnotaffectthechoiceofmaximumvalue aˆ
MAP
(thatis,thevaluemaximizing
p
(
r a
)
⋅
1
isthesamevaluemaximizingp(r rr rr|a)). Hence,wecanwriteourequationas
2π
aˆ
MAP
argmax p
(
r a
)
(10.26)
a
288 ‹ Chapter Ten 
SubstitutingthevalueswehavefromEquation(10.24)forp(r rr rr|a)weendupwith:
2 2
− a a
\
=ˆ
MAP
argmax
1
exp
,
j
−
(
r Acos ( ))
−
(
r
2
− Asin ( ))
1
(
(
(10.27)
a 2πσ
2
,
2σ
2
n
(
n
,
2 2 2 2 2
exp
,
j
−
(
r
1
+ r
2
)
− A
(
cos ( ) +sin ( ))
+
(
2r Acos ( ) + 2r Asin ( ))
\
a a
1
a
=ˆ
MAP
argmax
1
a
2
(
a 2πσ
n
2
,
(
2σ
n
2
,
(
(10.28)
1
a
2
a
1
j
−
(
r + r
2
)
\
j
−A
j 2r Acos ( ) + 2r Asin ( )
\
=ˆ
MAP
argmax exp ,
1
2 2
( exp
(
,
,
2σ
2
2
,
\
(
(
exp
, (
, (
a 2πσ
n
2
(
,
2σ
2
(
,
n (
2σ
2
, n n
(10.29)
Wecanremoveallmultipliertermsthatarenotafunctionofa becausethese
termswon’taffecttheoptimization.Thisleadsusto
aˆ
MAP
=argmax exp
·
·
í
r Acos( )÷r Asin (a)
1





1
a
2
2 (10.30)
a
·
·
(
s
n
)
Now,weusealittlemathematicaltrick.Wecanfindthevaluethatmaximizesxor
wecanfindthevaluethatmaximizesln(x). Eitherway,wewillendupwiththesame
value.Thistellsusthatourequationcanberewrittenaccordingto
a a
2
=argmax ln


exp·
í
·
·
(
r Acos( )÷r Asin ( )
1





l
l
aˆ
MAP

·
1
2
)
l
l
l (10.31)
a

s

n
a
ˆ a
MAP
1
a
2
argmax
r Acos
( )
+ r Asin
( )
argmax
[
r Acos a + r Asin a
]
1 2
a σ
2
a
n
(10.32)
Estimation and Synchronization ‹ 289 
Borrowingfromabitofmathnotshownhere—takemywordforit—youcan
rewritetheequationaccordingto
T
E
aˆ
MAP
argmax
∫
r ( ) A cos (ω + a )dt
(10.33)
t t
c
a
0
(Thiscomesaboutbecause ∑a
i
b
i
isthesameas
∫
a t b t dt wherea
i 
’sandb
i 
’s ( ) ( )
i
representa(t)andb(t)onanorthonormalbasis.)Now,sinceaisthevaluethatmaxi-
mizestherighthandterm,itisequivalenttothevalueathatsatisfiestheequation
(borrowingfromfirst-yearcalculus)
∂
T
E
( ) A cos (ω + a)dt r t t
c a aˆ
MAP
0
(10.34)
∂a
∫
0
Takingthisderivativeleadsusto
T
E
∫
( )sin (ω + aˆ
MAP
)dt 0
(10.35)
r t t
c
0
Thatmeansthatourestimate aˆ
MAP
istheonethat,givenr(t),forcestheintegral
ofEquation(10.35)to0.Aphysicaldevicethatfindsthis aˆ
MAP
givenr(t)isshownin
Figure10.4.Here,youseeaterm, sin (ω + t aˆ
),where aˆ isthecurrentestimateof
c
thephasea,multipliesr(t).Thisproductispassedthroughanintegrator,which
outputsthevalue
T
E
( )sin (ω + aˆ
)dt . (10.36) r t t
∫
c
0
T
E
0
∫
x
decision
device
closetoφ
notclosetoφ
sin(ω
c
t+a)
∧
output
∧
r(t) sin(ω
c
t+a)
=Acos(ω
c
t+a)+n(t),0< t< T
E
∧
updatea
andsend
itback
Figure10.4 Adevicetoevaluate
^
a
MAP 
290 ‹ Chapter Ten 
Thisvalueentersintoadecisiondevice(typicallyimplementedusingavoltage-
controlledoscillator,orVCOforshort).Iftheintegralis0(orveryclosetoit),thenthe
decisiondevicedecidesthatthecurrent aˆ estimateisthedesiredestimate aˆ
MAP
.If
thevalueisnotcloseto0,thenthedecisiondeviceupdatesthevalueof aˆ andsendsit
backtoseeifitnowmakestheintegralofEquation(10.35)equalto0(orverycloseto
( ) Acos (ω + a) + n t it).Inthisway,given r t t ( ), t ∈
[
0, T
] ,wecangeneratean
c E
estimateofa. 
10.3.2 ThePracticalEstimator:thePLL
Inpractice,engineersestimatethephaseausingadeviceveryclosetotheoneshown
inFigure10.4,namelythedeviceshowninFigure10.5,calledaphase-lockedloop,or
PLLforshort.
LookingatFigure10.5,weseethatthePLLisveryclosetotheestimatorof
Figure10.4.Herearethedifferences:(1)theintegratorhasbeenreplacedbyafilter
F(f);and(2)thedecisiondevicehasbeenimplementedusingaVCO.
=Acos(ω
c
t+a)+n(t),0< t< T
E
x
decision
devicebuilt
sin(ω
c
t+a)
∧
∧
sin(ω
c
t+a)
output
F(f)
r(t)
usingVCO
Figure10.5 Phase-lockedloop
Figure10.6helpstoexplaintheworkingsofthePLL.Attheinputwehave
r t t
( ) (10.37) ( )
Acos
(
ω + a
)
+ n t
c
TokeepthewrittenexplanationofthePLLworkingssimple,I’llsimplyignorethe
noiseforthesakeoftheverbaldescription.We’llbeassumingthattheinputis
( ) r t ( cos A
c
t ω + a) (10.38)
Now,asthisr(t)comesin,itgetsmultipliedbythesinusoid
( )
ˆ sin
c
t a ω + (10.39)
Estimation and Synchronization ‹ 291 
∧
Acos(ω
c
t+a)
.
sin(ω
c
t+a)
∧
∧ ∧
r(t)
=Acos(ω
c
t+a),0< t< T
E
x
sin(ω
c
t+a)
∧
sin(ω
c
t+a)
∧
∧
F(f)
=
A
sin(2ω
c
t+a+a)+
A
sin(a–a)
∧
2 2
2
2
A
sin(a–a)
*
f(t)
≈
A
sin(a–a)
∧
2 ∂t
∂a
∝
A
sin(a–a)
VCO
Figure10.6 ThePLLexplained
where aˆ isthecurrentestimateofourchannelphase.Theoutputsignalfromthat
productis
′
( ) Acos (ω + t a ˆ
) (10.40) r t t a ) sin (ω +
c c
r t (
c
t a a ˆ
) +
A
sin (a a ˆ
) (10.41)
′
( )
A
sin 2 ω + + −
2 2
ThissignalthengoesthroughthefilterF( f ),whichactsasalow-passfilter,
cuttingoutthehigh-frequencyterminr′(t)(thefirstterminr′(t)).Asaresult,the
outputcorrespondsto
A
r t − (
′
( )
2
sin
(
a a ˆ
)∗ f t
) (10.42)
WewillassumethatF( f )iscloseto1(f (t)iscloseto δ(t))inthelow-frequency
range,andhencether ′(t)iswell-approximatedby
A
′
( )
2
sin (a a ) (10.43)
r t − ˆ
ThissignalthenentersintotheVCO,whichactsasadecisiondevice,decidingbased
onitsinputwhethertoupdatethecurrentestimate aˆ ,orkeepitandbedone.Specifi-
cally,theVCOisadevicethatdeterminestherateofchangeofthephaseestimate aˆ ,
accordingto
∂aˆ
K ⋅ (input to VCO ) (10.44)
∂t
292 ‹ Chapter Ten 
ˆ ∂a A
⋅ K sin
(
a a ˆ
) (10.45)
−
∂t 2
Now,ifthephaseestimate ˆ a a ,thenpluggingthisintotheequationabove,
∂aˆ
0
(10.46)
∂t
Thatis,theestimate aˆ isnotchanged.TheVCOdecidestoholdontoitscurrent
estimate.
− ˆ
)
≈ − ˆ
Alternatively,let’ssaythevalueof aˆ isclosetoa.Inthatcase, sin
(
a a a a , 
andwehave
ˆ ∂a A
≈ ⋅ K a a ˆ
) (10.47)
⋅ ( −
∂t 2
∂aˆ
If aˆ issmallerthana,then willbepositive,whichmeansthattheVCOis
∂t
∂aˆ
goingtoincreasetheestimate aˆ .If,ontheotherhand, aˆ islargerthana,then is
∂t
negative,whichmeansthattheVCOwilldecreasethevalueof aˆ .Inthisway,the
a a . VCOactsasadecisiondevice,makingchangesin aˆ until ˆ
Thisis,practicallyspeaking,thewayengineersgenerateanestimateofthe
unknownphasea.ThereareanumberoftweakingstothePLLthataresometimes
performed,andunderdifferentconditionstheymakedifferentchoicesfortheF(f).But
thesedetailsareleftforanothertime—andanotherbook.
10.3.3 UpdatestothePracticalEstimatorinMPSK
Let’sconsiderwhatwe’vedonesofar.Wereceivedthesignal
( ) Acos (ω + ) + t T
E
(10.48) r t t a n(t ), 0 ≤ <
c
wherearepresentsarandomphaseoffsetandn(t)representsanadditivewhite
Gaussiannoise.Wethenbuiltadevicethatcreatestheestimate aˆ ofa.Wealsosaw
analternativeversionofthisestimator,calledthePLL,whichiscommonlydeployed
bymostengineers.
Butinsomecommunicationsystems,wereceive
r t t
)
+ t
( )
Acos
(
ω +θ + a n
( ) (10.49)
c
Estimation and Synchronization ‹ 293 
¦
2π
¦
whereθ isavalueintheset ¦
k k 0, 1, …, M −1
¦ andrepresentsinformation
¦
M
,
¦
bitsstoredinthetransmittedphase.Thisiswhatwereceiveifthetransmittercorre-
spondstoanMPSKmodulator.Youcanseehowthisreceivedsignaliscreatedby
takingalookatFigure10.7.Inthissection,weareinterestedinfindingawaytocreate
anestimatefora,giventhereceivedsignalshowninEquation(10.49),ratherthanthe
r(t)showninEquation(10.48).
CHANNEL
n(t)
=Acos(ω
c
t+θ),
θ∈
2π
k,k=0,1,…,M–1}
r(t)=Acos(ω
c
+θ +a)+n(t)
s(t)
MPSK
+
phase
offset
a
bits
M
modulator
Figure10.7 Apracticalreceivedsignal
Typically,thewayengineersgeneratetheestimateofa,givenr(t),isshownin
Figure10.8.Theidea,asexplainedinthatfigure,isthis:useaprocessorthatgetsrid
ofθ,andthenusethePLLshowninFigure10.5.Onlyonequestionremains:howdo
wegetridofthephaseθ?Onedevicecommonlyemployedtodothisisshownin
Figure10.9.Incomes
( ) Acos (ω +θ + a) + n t r t t ( ) (10.50)
c
r(t)
=Acos(ω
c
t+θ +a)+n(t)
getridofθ∈
2π
k,k=0,1,…,M–1}
M
PLL
sin(ω
c
t+a)
∧
processor:
^
Figure10.8 Creatinganestimateofa,a,givenr(t)
294 ‹ Chapter Ten 
PLL
sin(Mω
c
t+a)
∧
r′′(t)≈ A
M
cos(Mω
c
t+M
.
a)
(
.
)
M
BPF
centered
at
M
.
ω
c
θ∈{
2π
k,k=0,1,…,M–1}
r(t)
=Acos(ω
c
t+θ +a)
r′(t)=[r(t)]
M
M
=A
M
cos(Mω
c
t+
2
M
π
k
.
M+M
.
a)
+lowfreq.terms
Figure10.9 Processortoremoveθ θθ θθ
¦
2π
,
whereθ isavalueintheset
¦
¦
M
k k 0, 1,…, M −1
¦
¦
.Tosimplifythepresentation
¦
oftheworkingsofFigure10.9,we’llignorethenoise.Thatmeanswecanwritethe
inputas
( )

(
j
c
2π
\
t + r t Acos
,
ω + k a
( (10.51)
M
,
wherek =0or1or…orM –1.Afterpassingthisinputthroughthepower-of-M 
device,weendupwith
M
,
j \
]
′
( )
,
Acos
,
ω +
2π
k a r t
¸ (
c
t +
(]
(10.52)
M
,]
\
r t
M
cos
,
M t M ⋅
2
M
π
k M ⋅ a
,
ω + +
(
′
( )
(
j
c
(10.53)
+
(
lower frequency terms
)
Thissignalispassedthroughabandpassfilter,whichcutsoutalltheterms
exceptthetermcenteredaroundfrequencyMω
c 
—thatis,exceptthefirsttermof
Equation(10.53).Asaresult,theoutputofthebandpassfilteris
′′
( ) ω + π + ) ω + r t
M
cos (M t 2 k Ma
M
cos (M t Ma) (10.54)
c c
ThePLLtracksthephaseofthistermand,hence,itendsupwiththephaseesti-
mate ˆ a Ma.Weuseadivide-by-Mtogeneratethedesiredphaseestimateofa.
Estimation and Synchronization ‹ 295 
10.4 Conclusion
Thischapterdescribestheworldofestimation,alsoknownassynchronization.Itisan
importantpartoftelecommunicationsbecause,whensendingasignalfromtransmitter
toreceiver,thechanneloftenaddsunwantedparametersalongtheway,whichwelabel
a.Weneedtobeabletoestimatetheseparameterssowecanremovethemfromthe
receivedsignal.Thischapterprovidedageneralframeworkforcreatingtheestimate
aˆ ofanygivenparametera.Specifically,therearethreewaystogeneratethisesti-
mate.ThefirstmethodisknownasMMSE,thesecondasMAP,andthethirdasML.
Wethenconsideredanimportantpracticalexample,namelytheestimationofchannel
phase.WesawhowtoestimatephaseusingtheMAPmethodandwesawhowengi-
neersimplementadevicesimilartotheestimatorcreatedfromtheMAPmethod.
296 ‹ Chapter Ten 
Problems 
1. Atransmittersendsa.Areceiverpicksupthevaluer,whichischaracterized
by
2
\
a
)

1
exp
j
,
(
r − (a − 1))
(
p r
(
2πσ
2
(
, 2σ (
(Q10.1)
,
DrawablockdiagramofareceiverthatperformsMLdetection.
2. Atransmittersendsavaluea,whichisarandomvariablecharacterizedby
FigureQ10.1.Areceiverpicksupthevaluer.
(a) Ifrischaracterizedby
p r a +1
)

1
2
exp
(
− r
( ) (Q10.2)
1
2
\
p r a −1
)

2
exp
j
−1
r
(
, ( (Q10.3)
(
2
,
drawablockdiagramofareceiverthatperformsMAPdetection.
(b)Ifrischaracterizedby
j −r
2
\
p r a +1
)

1
exp
(
2σ
(
(
(Q10.4)
2πσ
1
2
,
1
2
,
p(a)
Figure Q10.1
Thepdfofa 
1
/
2
δ(t+1)
1
/
2
δ(t–1)
1
/
2
a
–1 +1
Estimation and Synchronization ‹ 297 
j −r
2
\
p r a −1
)

2
1
πσ
2
exp
,
(
2σ
2 ,
(
(Q10.5)
(
2
2
drawablockdiagramofareceiverthatperformsMAPdetection.
3. Areceiverobserves
r a n
(Q10.6)
+
whereaisaGaussianrandomvariablewithmean0andvariances
2
,andnisa
2
Gaussianrandomvariablewithmean0andvariances
1
.
(a) FindtheMMSEestimate.
(b)FindtheMAPestimate.
4. Atransmittersends
( ) Acos (ω +θ ), iT ≤ < (i +1)T (Q10.7) s t t
i
t
whereθ
i
=0°or180°
andreceives
r t t
i
) ( ), iT ≤ < (i +1)T (Q10.8) ( ) Acos (ω +θ + ε + n t t
Drawablockdiagramofasystemthatwillestimatethephaseoffsetandbriefly
describeitsoperation.
11 
Chapter 
Multiple Access Schemes
Teaching Telecommunications Systems to Share 
T
hischapterexploresthedifferentwaysinwhichacommunicationchannelcanbe
sharedbymultipleusers.WelookedatthisbrieflyinChapter2,whereweintro-
ducedtimedivisionmultiplexingandfrequencydivisionmultiplexing.Inthischapter,
weprovidemoredetailssothatyoucanunderstandhowtobuildacommunication
systemthatsharesthechannelamongthemanyuserswhowanttouseit.
11.1 WhatItIs
Let’ssayyouhavebeengivenalicensebytheFCCtobuildawirelesscommunication
systemoperatinginafrequencybandof1.8GHzto1.805GHz.Youdecidethatyou
wouldlikemanyuserstobeabletouseyoursystem.Youalsodecidethateachuser
willcommunicateonlydigitalsignals(thatis,theywilluseasourcecodertomakeall
theinformationtheysenddigital).Youneedtofindawaytoallocateportionsofthe
communicationchanneltoeachuser.Thischapterisabouthowtodothat.
Therearetwogeneralmethodsthatallowmanyuserstoshareasinglecommuni-
cationchannel:
1.Multiplexing schemes. Multiplexingschemesarechannel-sharingschemes
whereportionsofthechannelareassignedtoeachuserbyasystemcontrollerat
acentrallocation.Thesystemcontrollerassigns,inadvance,thechannelpor-
tionsforeachuser,andcontrolseachuser’saccesstothechannel.
2.Multiple access schemes.Multipleaccessschemesrefertochannel-sharing
schemeswhereasystemcontrollerassignsportionsofthechanneltoeachuser
basedoncurrentavailability.Thesystemcontrollercanupdatethesharingof
portionsofthechannelbasedonchangesinsystemdemand.Oncethesystem
controllertellstheuserwhatportionhecanuse,theuserisinchargeofmaking
sureheusestheportionrequested.
Mostmoderncommunicationsystemsusemultipleaccessschemes,sowe’ll
spendourtimeonthem.
300 ‹ Chapter Eleven 
11.2TheUnderlyingIdeas
LookatFigure11.1,sinceitexplainswhatIdescribenext.
userk
signal
s
i
(k)
(t)+s
i
(j)
(t)
sentoverchannel
coder
coder
+ +
n(t)
s
i
(k)
(t)
s
i
(j)
(t)
r(t)=s
i
(k)
(t)+s
i
(j)
(t)+n(t)
channel
userj
source
source
modulator
modulator
(a)
x
x
s
i,M
(t)
s
i,1
(t)
decoder
wantsto
hearuser
k
(notuserj)
DECISIONDEVICE
FORUSERk
R1
r(t)=s
i
(k)
(t)+s
i
(j)
(t)
+n(t)
decision
device
(i+1)T
iT
(k)
(k)
∫
(i+1)T
iT
∫
source
OPTIMALRECEIVER
FRONTENDFORUSERk
DEMODULATORFORUSERk(fromChapter5)
(b)
Figure11.1 (a)Userkanduserjsendtheirsignalsoverthechannel
(b)Areceivertriestopickupuserk’ssignal
Multiple Access Schemes: Teaching Telecommunications Systems to Share ‹ 301 
k
1.InFigure11.1(a),wehaveanewnamingconvention.Weusethenotations
i
( )
( ) t
toindicatethesignalsentoutbythekthuserduringthetime[iT,(i +1)T].Let’ssay
k
thatthekthuser’sinformation,s
i
( )
( )
isoneofthefollowingsignals: t
k k k
i ,1
t
i ,2
t
i M
t
¦
s
( )
( ), s
( )
( ), …, s
( )
( )
¦
.(Let’sassume,forexample,itisasignalfroma
,
QPSKsignalset.)
2.InFigure11.1(a),weassumethattherearetwousers,userkanduserj,using
thecommunicationsystem.Thatmeansthatthesignalbeingsentoverthe
channelisthecombinationofbothsentsignals,
k
3.InFigure11.1(b),wewanttopickupthesignalfromuserk, s
i
( )
( ) t .Wedonot
wantthesignalfromuserj,becausewearenotinterestedinwhatuserjissaying.
FromChapter5,weknowthatthereceiverfrontendtousewhenwewanttopick
k
upthesignals
i
( )
( )
(fromuserk)istheoneshowninFigure11.1(b). t
4.Let’slookatthetopbranchofthisreceiverfrontend.
j
( ) s
( )
( ) + s
( )
( ) + n t ( ) ischannelnoise. 4a.Incomes r t
i
k
t t ( )
,wheren t
i
j
4b.Wewanttomakesurethatthesignals
i
( )
( )
inr(t)doesnotmakeitoutof t
thistopbranch.Ifthissignaldoesnotmakeitoutofthetopbranchthen
j
wehaveeffectivelyeliminatedthepresenceof s
i
( )
( )
(userj)inthispart
ofuserk’sreceiver.
t
4c.Thesignalcomingoutofthetopbranchis
(i +1)T (i +1)T
k k j
R1
∫
s
( )
( ) ( ) t
¸
i
k
t t (
]
t r t dt
∫
s
( )
( )⋅
,
s
( )
( ) + s
( )
( ) + n t )
]
dt
(11.1) i,1 i,1 i
iT iT
(i +1)T (i +1)T
k k k
R1
∫
s
( )
( )s
( )
( )
i,1
t
i
j
t dt +
∫
s
( )
( ) ( ) t
i
k
t dt +
∫
s
( )
( ) s
( )
( ) t n t dt
(11.2) i,1 i,1
iT iT
(i +1)T
k j
Ifwemakesurethat
∫
s
i ,1
t
j
t t
( )
( ) s
i
( )
( )dt 0 ,thenthesignal s
i
( )
( )
willnotmake
iT
(i +1)T
k
t
j
t dt 0
itoutofthetopbranch.So,werequirethat
∫
s
( )
( ) s
i
( )
( )
. i ,1
iT
j
5.Next,wealsowanttomakesurethatthesignal s
i
( )
( )
inr(t)doesnotmakeit t
outofbranch2,branch3,andbranch4.Thatmeanswerequirethat
k j
s
( )
( ) s
( )
( )
∫
i,2
t
i
t dt 0 (11.3)
302 ‹ Chapter Eleven 
k j
s
( )
( ) s
( )
( )
∫
i,3
t
i
t dt 0 (11.4)
k j
s
( )
( ) s
( )
( )
∫
i,4
t
i
t dt 0
6.Generally,wecanstatethatwewanttomakesurethat
k
∫
s
i
( )
( ) s
( )
( ) t
i
j
t dt 0 (11.5)
k
t
j
forallpossibles
i
( )
( )
ands
i
( )
( )
signalssent.ItiseasilyshownusingFourier t
transformsthat,equivalently,wewanttomakesurethat
k
∫
S
i
( )
( ) S
( )
( )df 0 (11.6) f
i
j
f
Inwords,wewanttomakesurethatthesignalsentbyuserjisorthogonaltothe
signalsentbyuserk.Thisistheunderlyingprincipleofmultipleaccesstechniques—
makingsurethatEquation(11.5)or(11.6)issatisfied.Ifyoudothat,thenyoumake
surethatuserj’ssignaldoesnotappearintheoutputofareceiverthatwantstopick
upuserk’ssignal,andviceversa.
Recently,engineershavebecomeamoreforgivinglot.Somesay:“ItwouldbeOK
ifjustatinybitofuserj’ssignalappearedinuserk’sreceiver,aslongasitwassucha
smallamountthatitdidn’taffecttheperformanceofuserk’sreceiver.”Mathematically
whattheyaresayingisthattheywanttomakesurethat
k
∫
s
i
( )
( ) s
( )
( ) t
i
j
t dt < ε (11.7)
whereεisaverysmallnumber.Ifyoudecidetobuildamultipleaccesssystemwhere
thisisthecase,itiscalledinterference-limited,andthesignalssentarecalledpseudo-
orthogonal.
Example 11.1 
Twouserssetouttoshareacommunicationchannel.Oneusersendsherbinary
informationas+x(t)(ifthebittosendis1)or–x(t)(ifthebitis0).Thesecond
usersendshisbitas+y(t)or–y(t).Thex(t)andy(t)areshowninFigureE10.1.
Determineiftheseusersareabletosharethechannelwithoutinterferingwith
oneanother.
Solution:Theusers,we’llcallthemuser1anduser2,willbeabletosharethe
channelifthesignalstheysendsatisfy
∫
1
t
2
s
( )
( ) s
( )
( ) dt 0 (E11.1) t
Multiple Access Schemes: Teaching Telecommunications Systems to Share ‹ 303 
Now,weknowthatuser1sendseither+or–x(t)anduser2sendseither
+or–y(t).So,pluggingthisintoourrequirementleadsto
⋅ ±y t ( )] , ( )]
]
dt 0 (E11.2) ,±x t
] ¸ ∫ ¸
Now,let’susealittlemathandseeifindeedthisequalityholds:
x t ⋅ y t ±
∫
( ) ( ) dt 0 (E11.3)
1
x t y t
∫
( ) ( ) dt 0
(E11.4)
0
1
2
1
1 1 dt +
∫
1⋅ − 1 ⋅ ( ) dt 0
∫ (E11.5)
0
1
2
1
2
+ −
1
2
) 0 (E11.6)
0 0
(E11.7)
(

Yes,theequalityholds.Thetwouserswillbeabletosharethechannel
withoutinterferingwithoneanother.
x(t) y(t)
1
1
1
t
1
/
2
1
-1
FigureE11.1 Signalssentbyuser1anduser2
11.3TDMA
Averycommontypeofmultipleaccesssystem,andonethatsatisfiesEquation(11.5),
iscalledTDMA,shortfortime division multiple access.YoucanseethisideainFigure
11.2.Userkhasaslotoftimeinwhichtosendhisinformation,andthenuserjhasa
differenttimeinwhichtosendhisinformation.Inthisway,thesignalsentbyuserj 
(s
i
( )
( )
j k
t )is t )is0whenuserk’ssignalisnon-zero;andthesignalsentbyuserk(s
i
( )
( )
0whenuserj’ssignalisnon-zero.Thismakessurethattheproductintheintegralof
Equation(11.5)is0,andthereforetheintegralis0.
304 ‹ Chapter Eleven 
s
i
(k)
(t)
coder
coder
t
t
s
i
(j)
(t)
userk
userj
time
align
time
align
source
source
modulator
modulator
Figure11.2TheTDMAidea
ThebasicprincipleofTDMAisalsoshowninFigure11.3.Oneuserusesthe
entirefrequencyrangeofthecommunicationchannelforabriefperiodoftime,then
anotheruserusestheentirefrequencyrangeofthecommunicationchannelforabrief
time.Ananalogycouldbeacocktailpartywhere,toavoidhearingtwoconversationsat
once,onepairofpeopletalkatonetime,thenanotherpairtalkatanothertime,then
anotherpairtalkthetimeafterthat,thenthefirstpairtalkagain,thenthesecondpair
talkagain,thenthethirdpairtalkagain,andsoon.It’sapolitecocktailpartywhere
peoplesharethetimedomain.
Generally,inTDMA,userssendanentiresetofdatasymbolsintheirtimeslot,
andthisentiresetiscalledaburst.Forexample,inawell-acceptedstandardcalled
GSM,userssendaburstof148datasymbolsatonetimeslot.
frequency
time
userj userk userj userk
T
B
T
B
T
B
T
B
Figure11.3TheTDMAideainthefrequency-timedomain
Multiple Access Schemes: Teaching Telecommunications Systems to Share ‹ 305 
Example 11.2 
ProposeaTDMAsystemwhich
• allowstwouserstotransmitdataatarateof1Mb/sand
• allowseachusertosendthreebitsatatime.
Solution:Whatwewanttodoisgiveuser1atimeslotlongenoughforthreebits,
whichisatimeslotoflength
j
3 3
,
1
1 10 bits ⋅ / sec
,
(
\
3 10
−6
sec
(E11.8)
T
SLOT
⋅ T
b
⋅ ⋅
6
(
Wethenwanttogiveuser2aslotlongenoughforhisthreebits,againaslotof
duration
⋅
−6
⋅
−6
3 T
SLOT
⋅ T
b
3 ⋅
(
1 10
)
3 10 sec (E11.9)
We’llrotatebetweengivinguser1aslotforherthreebitsandgivinguser2the
slotforhisthreebits.Intheend,weendupwithaTDMAschemeasshownin
FigureE11.2,wheretheshadedbitsrepresentuser2’sbits,andtheunshaded
onesrepresentuser1’sbits.
t
SLOT1 SLOT2 SLOT3 SLOT4
3
.
T
b
3
.
T
b
=3
.
10
–6
FigureE11.2TheTDMAscheme
11.4FDMA
FDMAisanotherwaytoenablemultipleuserstoshareanentirecommunicationresource.
InFDMA(shortforfrequencydivisionmultipleaccess)eachuserusesadifferentbandof
frequenciestocommunicatehisorherinformation.AnexampleofFDMAisshownin
Figure11.4.There,weseeuserksendinghisinformationatonefrequencyanduserj 
sendingherinformationatadifferentfrequency.Ifyouknowwhoyouwanttolistento,you
tuneyourreceivertopickuptransmissionsatthedesireduser’sfrequency.Thissystem
satisfiesEquation(11.6),becauseuserkis0atthefrequencieswhereuserjistransmit-
ting,anduserjis0atfrequencieswhereuserkistransmitting.Thatmakes
k k
f
i
j
f f
i
j
f S
( )
( ) S
( )
( ) 0 ,andtherefore
∫
S
( )
( ) S
( )
( )df 0 .
i i
306 ‹ Chapter Eleven 
coder
coder
f
f
S
i
(k)
(f)
S
i
(k)
(f)
userk
userj
source
source
modulator
modulator
frequency
shift
frequency
shift
Figure11.4TheFDMAidea
TheuseofFDMAisalsoshowninFigure11.5.Here,weseethateachuseris
givenallthetimetheycouldeverwant,buttheycanonlycommunicateoverasmall
frequencyband.
frequency
userk
userj
time
Figure11.5TheFDMAideainthefrequency-timedomain
11.5CDMA
11.5.1Introduction
CDMAisshortforcodedivisionmultipleaccess.TheideaunderlyingCDMAwasput
forwardbyanattractiveHollywoodactress,HedyLamarr,andcomposerGeorge
Antheil.CDMAitselfwasmadepossiblebyimprovementsintechnologyinthelate
1980sandwasdesignedbyagiantcompanywhichhasafootballstadiumnamedafter
theminSanDiego,Qualcomm.
Multiple Access Schemes: Teaching Telecommunications Systems to Share ‹ 307 
TheideabehindCDMAisthis.Giveuserkauniqueshape,calledasignature 
waveformoracode,andgiveuserjadifferentuniqueshape,alsocalledasignature
waveformoracode.Makesurethatthecodeyougiveuserkandthecodeyougiveuser
jarecarefullychosentoensurethatEquation(11.5)orEquation(11.7)issatisfied.
Forexample:
j
1.Let’ssaythatwegiveuserj theshape x
( )
( )
showninFigure11.6(a).Userj t
j
t
j
sends s
i
( )
( ) +x
( )
( )
tosaythebitsentis1,orhesends s
i
( )
( ) −x
( )
( )
to
saythebitsentis0,asshowninFigure11.6(b).
j
t
j
t t
x
(j)
(t)
(i+1)T
t
iT
+0.5
–0.5
(a) 
+x
(j)
(t) –x
(j)
(t) 
s
i
(j)
(t)=
OR
(i+1)T
(i+1)T
t
iT
iT
0.5
–0.5
+0.5
–0.5
(b)
Figure11.6(a)Code(shape)giventouserj
(b)Possiblesignalssentbyuserj
308 ‹ Chapter Eleven 
k
2.Let’salsosaythatwegiveuserktheshape x
( )
( )
showninFigure11.7(a). t
k
Userksends s
i
( )
( ) +x
( )
( )
tosaythebitsentis1,orhesends t
k
t
k
s
i
( )
( ) −x
( )
( )
tosaythebitsentis0,asshowninFigure11.7(b). t
k
t
k k
t
i
j
t t 3.Now,youcaneasilyshowthat
∫
s
( )
( ) s
( )
( )d t 0 forany s
i
( )
( )
and
i
j
s
i
( )
( )
,whichtellsusthatthischoicesatisfiesEquation(11.5).Thatmeansthat
userjanduserkcansendthesetwosignalsanduserj’ssignalwillnotappearin
userk’sreceiver(andviceversa).
t
x
(k)
(t)
s
i
(k)
(t)=
+x
(k)
(t) –x
(k)
(t)
(a)
(i+1)T
(i+1)T
(i+1)T
t
iT
iT
iT
+1
+1
–1
OR
(b)
Figure11.7(a)Code(shape)giventouserj
(b)Possiblesignalssentbyuserk
Multiple Access Schemes: Teaching Telecommunications Systems to Share ‹ 309 
4.Inthisparticularexample,itiseasilyshownwithagraphichowgivinguserk 
k
t
j
thecode x
( )
( )
andgivinguserjthecode x
( )
( )
allowsustosendbothuserk t
anduserj’ssignalandnotexperienceanyinterference.Considertheexampleof
Figure11.8.Thereyouseeasignalsentbyuserj,andasignalsentbyuserk,and
youseethecombiningofuserkanduserj’ssignal.Youcanseefromthecom-
binedsignalthatyoucanstilltellwhatuserkanduserjsent:(1)userk’s
informationis1ifthecombinedsignalisabovethex-axis,anditis0ifthecom-
binedsignalisbelowthex-axis;(2)userj’sinformationis1ifthecombinedsignal
slopesupward,andhisinformationis0ifthecombinedsignalslopesdownward.
So,bygivingeachuseracarefullychosencode,youcansendsignalsatthesame
timeandatthesamefrequency,butstillhaveawaytoseparateusers.
userk
2T 3T 4T T 0
2T 3T 4T T
0
2T 3T 4T T
1 0 1 1
0 0 1 0
1.5
0.5
0.5
–0.5
–0.5
–1.5
userk
+userj
userj
Figure11.8Howassigningcodes(shapes)letsyoudetermineeachuser’ssignal
310 ‹ Chapter Eleven 
CDMAisagreatidea,becausewithityouhaveuserssendinginformationatthe
sametime,andoverthesamefrequencies,butyoucanstillperfectlyseparateusers.
AnanalogytoCDMAcouldbethatofbeingatacocktailpartyfilledwithtwohumans
andtwoaliens.Thehumansandalienstalkatthesametimeandoverthesamefre-
quencies,butthehumancommunicationiscompletelyundetectabletothealiensand
whatthealiensspeakiscompletelyunnoticeabletothehumans.
TherearetwoclassesofCDMA.Thefirstclassiscalledorthogonal CDMA.In
orthogonalCDMA,userj’sanduserk’ssignalssatisfyEquation(11.5)—thatis
k
t
i
j
t
∫
s
i
( )
( ) s
( )
( )d t 0 . Thesecondclassis pseudo-orthogonal CDMA,whereuserj’s
anduserk’ssignalsinsteadsatisfyEquation(11.7).Inthiscase,alittlebitofuserj’s
signalappearsinuserk’ssignal,butjustaverylittle,notenoughtosignificantlyaffect
theperformanceoftheuserk’sreceiver.Whenyouusepseudo-orthogonalCDMA,
youcansupportmoreusersthanwithorthogonalCDMA,TDMA,orFDMA.
11.5.2DS-CDMA
TherearethreedifferenttypesofCDMA,distinguishedbythecodesgiventoeach
user.Thefirstandmost-usedformofCDMAiscalleddirect sequence CDMA,orDS-
CDMAforshort.
InDS-CDMA,eachuserisgivenacodeliketheoneshowninFigure11.9.Asyou
canseeinthatfigure,eachcodeconsistsofshortpulsesofdurationT
c
,andeachshort
pulsehasaheightofeither+1or–1.InFigure11.9,fourshortpulsescomprisethe
user’scode.Ingeneral,thereareNshortpulsesthatmakeupauser’scode.Userk 
k
takeshercode,saytheoneinFigure11.9,called x
( )
( )
,and,inthesimplestcase, t
k
sends s
i
( )
( ) x
( )
( )
toindicatetheinformationbitis1,andsends t
k
t
k
s
i
( )
( ) −x
( )
( )
toindicatethattheinformationbitis0. t
k
t
x
(k)
(t)
+1
0
T
-1
T
C
2T
C
3T
C
4T
C
Figure11.9Code(shape)assignedtouserkinaDS-CDMAsystem
Multiple Access Schemes: Teaching Telecommunications Systems to Share ‹ 311 
Eachuserisgivenauniquecode.Forexample,Figure11.10showsfourcodes,each
oneofwhichisgiventoadifferentuser.User1usesthetopcode x t
( ) 1
( )
,user2usesthe
secondcode x t
( ) 2
( )
,andsoon.ForthecodesinFigure11.10,itiseasytoshowthat
k j
∫
( )
( )
( )
( )dt 0 for all k, j (k ≠ j ) (11.8) x t x t
k k j j
With s t
( )
( ), ( ) ±x t
( )
( ) ±x t s
i
t
( )
( ) ,andsoon,thisguaranteesthat
i
k j
∫
( )
( )
( )
( )dt 0 for all k, j (k ≠ j ) (11.9) s t s t
x
(1)
(t) 
t 
Figure 11.10
Codeforuser1,2,3,and4
T
C
2T
C
3T
C
4T
C
T
C
2T
C
3T
C
4T
C
T
C
2T
C
3T
C
4T
C
T
T
+1
+1
-1
+1
-1
+1
-1
x
(2)
(t)
x
(3)
(t)
x
(4)
(t)
t
t
t
312 ‹ Chapter Eleven 
Hence,theexampleofFigure11.10representsanexampleoforthogonalCDMA.
Ingeneral,inDS-CDMA,thecodeforuserkcanrepresentedas
N−1
k
k
c
i
(
x t
T
c
( )
( )

∑
(
−1
)
( )
p t − iT
c
)
(11.10)
i 0
k
Here,thec
i
(k)
iseither0or1,andhencethe
( )
( )
iseither+1or–1.Thisterm
−1
c
i
tellsusifthepulsesthatmakeupthecodehaveanamplitudeof–1or+1.The P t ( )
T
c
representsthebasicpulseshapethatthecodeismadeupof—inalltheexamplesI’ve
beendrawing,thispulseshapeisasimplerectangularshape.Ingeneral,thebasic
pulsecantakeonslightlydifferentshapes,butalloftheseare,approximatelyspeak-
ing,rectangular.
11.5.3FH-CDMA
AnothertypeofCDMA,notnearlyascommontodayasDS-CDMAbutstillused,is
calledfrequency-hopping CDMA,orFH-CDMAforshort.InFH-CDMA,thecodesare
notmadeupoflittlepulses,asinDS-CDMA,butinsteadtheyaremadeupoflittle
cosinewaveforms,asshowninFigure11.11.
k
WhathappensinFH-CDMAisthis.Theuserk,usingthecode x t
( )
( )inFigure
k k k k ( )
( ) +x t
( )
( ) −x t 11.11,sendseither s t
( )
( ) torepresentthebit1or s t
( )
( ) to
i i
representthebit0.Tothecommunicationchannel,itlookslikeasignalissentatone
frequency,thensuddenlyitjumps(hops)toanotherfrequency,andthenitsuddenly
hopstoadifferentfrequency.Hencethenamefrequencyhopping.
Tokeepusers’signalsfrominterferingwithoneanother,youmakesurethatthe
frequencyjumpsthatoneusertakesnevercollidewiththefrequencyjumpsthatthe
otheruserstake.Basically,thesystemmustmakesurethattwousersareneverusing
thesamefrequencyatthesametime.
t
T
+1
-1
0
x
(k)
(t)
cos(ω
2
t) cos(ω
1
t) cos(ω
3
t) cos(ω
4
t)
T
c
2T
c
3T
c 4T
c
Figure11.11Code(shape)assignedtouserk
Multiple Access Schemes: Teaching Telecommunications Systems to Share ‹ 313 
InFH-CDMA,thecodeusedbyausercanbedescribedmathematicallybythe
equation
N−1
k
x t
i
k
) T
c
(
−
( )
( )

∑
cos
(
ω
( )
t p t iT
c
)
(11.11)
i0
Here,
cos
(
ω
i
( )
t p t
k
)
( ) representsacosinewaveformwhichexistsoverthevery
ω
T
c
i
k
shortperiodoftime[0,T
c
].Thevalue
( )
indicatesthefrequencyofthiscosinewave-
formandistypicallychosenfromafinitesetofpossiblefrequencies.
11.5.4MC-CDMA
Thefinalmemberoftheexclusivethree-memberCDMAclubisanewcomerthat
announceditspresencein1993,andhasbeenslowlygrowinginpopularitysinceits
latearrival.Itsnameismulticarrier CDMA,orMC-CDMAforshort.InMC-CDMA,
eachuserisgivenacodethatisbestunderstoodinthefrequencydomain.I’llexplain
thiscodeintwoparts.
1.ThecodeforuserkisgeneratedbytheblockdiagramofFigure11.12.Thereyou
seeapulseofdurationTisputatmanydifferentfrequencies.Figure11.13shows
thecodeforuserkinthefrequencydomain.Eachbumpinfrequencyrepresentsa
pulseofdurationTintime.
cos(ω
c
t)
t
T
1
x
(k)
(t)
cos((ω
c
+
∆
ω)t)
cos(ω
c
+(N–1)
∆
ω)t)
x
x
x
x
p(t)
Figure11.12Thecreationofuserk’scodex
(k)
(t)forMC-CDMA(mainidea)
x
(k)
(f)
f
f
c
f
c
+
∆
f f
c
–1)
∆
f +(N
Figure11.13Thefrequencymake-upofuserk’scodex
(k)
(t)
314 ‹ Chapter Eleven 
2.Actually,thecodeisalittledifferentthanthatexplainedinpart1.Figure11.14
showstheactualcodegiventouserk.Thecodeforuserkisnotonlyapulsesent
overthesamefrequenciesoverandoveragain,buta+1or–1isappliedtoeach
frequency.Userksendshisinformationbitof1bysendingthecodeforuserk 
multipliedby+1,andsendstheinformationbit0bysendingthecodeforuserk 
multipliedby–1.
THESE+1and–1values
aredifferentfordifferentusers
t
T
+1
x
(k)
(t)
cos((ω
c
+
∆
ω)t)
cos((ω
c
–1)
∆
ω)t)
x
x
x
x
x
x
p(t)
cos(ω
c
t)
userk
+
–1
–1
+1
+(N
Figure11.14Thecreationofuserk’scodex
(k)
(t)
Userjsendshisinformationinexactlythesameway—theonlydifferenceisthat
hiscodeusesdifferent+1and–1valuestomultiplyeachfrequency.Forexample,user
jmaysendhissignalas+1or–1multipliedbythecodegeneratedasshowninFigure
11.15.IfyoucompareFigure11.15,whichgeneratesthecodeforuserj’ssignal,to
Figure11.14,whichgeneratesthecodeforuserk’ssignal,youcanimmediatelysee
thattheonlydifferenceisinthe+1and–1’sateachfrequencyofthecode.
Bycarefullychoosingthe+1and–1valuesforeachuser’scode,wecanmake
surethattheircodes,andthereforetheirtransmittedsignals,areorthogonalor
pseudo-orthogonal(thatis,theysatisfyEquation(11.5)orEquation(11.7)).
t
T
+1
x
(j)
(t)
cos((ω
c
+
∆
ω)t)
cos((ω
c
–1)
∆
ω)t)
x
x
x
x
x
x
p(t)
cos(ω
c
t)
userj
+
+1
+1
+1
+(N
Figure11.15Thecreationofuserj’scodex
(j)
(t)
Multiple Access Schemes: Teaching Telecommunications Systems to Share ‹ 315 
TheMC-CDMAcodecanbeexpressedmathematicallyaccordingtotheequation
k
t
N−1
c
i
k
∆
)
⋅
( )
x
( )
( )

∑
(
−1
)
( )
cos
((
ω + i ω
)
t p t
c (11.12)
i0
k
wherec
i
(k)
iseither0or1,meaningthat
( )
( )
iseither+1or–1;thistellsusthat
−1
c
i
eachfrequencycomponentismultipliedbyeither+1or–1.Thep(t)representsthe
pulseofdurationTsentateachofthecarrierfrequencies.
11.6CIMA
CIMA,shortforcarrier interferometry multiple access,isanovelsetofmultiple
accesstechniquesunderdevelopmentbyfriendsSteveandArnoldatIdrisCommuni-
cations(whichalsoholdsthepatent).I’vespentthelastyearresearchingit,andmy
graduatestudentsandIthinkitisarevolutionarymultipleaccessscheme,soIde-
cidedtoincludeithere.
TheeasiestwaytounderstandCIMAistoconsideronesimpleexample,an
exampleclosetoMC-CDMA.InCIMA,eachuserisgivenauniquecode.Thecodefor
userkisgeneratedaccordingtoFigure11.16.Here,youcanseethattheCIMAcode
consistsofapulseshapeofdurationT,whichissentoutoveralargenumberoffre-
quencies,justlikeMC-CDMA.Thedifferenceisthateachcarrierfrequencyisnot
multipliedbya+1or–1,asinMC-CDMA,butinsteadthenthcarrierfrequencyis
k
(
( )
multipliedbythephaseoffset e
j n−1)
∆
θ
.
t
T
+1
e
j(ω
c
)∆ω)t
e
j(ω
c
+(N–1)∆ω)t
e
j(N–1)∆θ
(k)
x
x
x
x
x
x
p(t)
e
jω
c
t
e
j∆θ
(k) userk
+
1
takesthe
(makese
j
(
.
)
Re{
.
}
realpart
x
(k)
(t)
cos
(
.
)
).
Figure11.16Thecreationofuserk’scodex
(k)
(t)inCIMA
Let’slookatthecodeforuserkinthetimedomain.ThecodeinCIMAcanbe
expressedmathematicallyaccordingto
N−1
k
x
( )
( )

∑
cos
((
ω + ω
)
t i
∆ k
( )
t i
∆
+ θ
( )
)
p t
c (11.13)
i0
316 ‹ Chapter Eleven 
N−1
∆ k
k
t
,
c
i
∆
)
t i
¦]
( ) x
( )
( )
,
∑
Re
¦
e
j
(
ω + ω +
( )
]
⋅ p t
(11.14)
¸ i 0 ]
N−1
t i
∆ k
k
x
( )
( )
,
,
Re
¦∑
e
j t
(
¦
]
( ) t
¦
ω
c
e
j i
∆
ω +
( )
) ¦
]
⋅ p t
(11.15)
¸
¦i0 ¦
]
N−1
j i
(
∆
ω +
( )
)
]
]
¦
,
t
∆ k
k
x
( )
( ) Re
¦
¦
e
j t
∑
e ( ) t
ω
c ¸
,
]
¦
⋅ p t
(11.16)
¦ i0 ¦
Now,usingthepropertiesofsummationfromourfriendlymathtextbooks(spe-
cificallylookingupgeometricseries),wecanrewritethesuminEquation(11.16)
accordingto
¦ t
∆ k
¦
j t
,
j
1 −e
jN
(
∆
ω +
( )
) \
¦
k
t
ω
c (
¦
¦
,
¦
⋅ p t x
( )
( )
Re
¦
e
(
1 −e
j
(
∆
ω +
( )
) (
¦
( )
t
∆ k
(11.17)
¦
,
¦
Afteraboutfivelinesofaddedmath,wefindoutthatthistermcanberewritten
accordingto









)

N
⋅
D
q
(
=
D k ( )
sin + wt





cos  w t +
D k
q
( )




)

⋅ p t ( )
(11.18)
− N  1 2
k ( )
t ( )
(
D
t + w x 
sin 





1
(
D
2 
wt +
D k
q
( )




)

c 
2 
Drawingthiscodeinthetimedomain,weendupwiththecodeshowninFigure
11.17.Itlookslikeabigpulseinthetimedomainwithlittlelobes(sidelobes)sur-
roundingit.Souserkhasacodewhich
1.isgeneratedaccordingtoFigure11.16;and
2.inthetimedomainlookslikeFigure11.17. 
Userksendsthebit1bymultiplyinghiscode x
( ) k
( ) t by+1andsendingitacross
thechannel;hesendsthebit0bymultiplyingthecode x
( ) k
( ) t by–1andsendingit
acrossthechannel.
Userjhasadifferentcode x
( )
( )
j
t ,generatedasshowninFigure11.18.Asyou
j
canseeinthisfigure,theonlydifferencebetweenthecode x
( )
( ) t foruserjandthe
code x
( ) k
( ) t foruserkisthephasesthatmultiplyeachfrequency.Userj’scodeis
drawninthetimedomaininFigure11.19.
Multiple Access Schemes: Teaching Telecommunications Systems to Share ‹ 317 
T
x
(k)
(t)
t=
∆
θ
(k)
∆
ω
t
Figure11.17 Theuserk’scodex
(k)
(t)(drawnwithoutthecos(ω ωω ωω
c
t)terms)
T
1
x
(j)
(t)
e
j((ωc+∆ω)t)
e
j((ωc+(N–1)∆ω)t)
e
j(N–1)∆θ
(j)
x
x
x
x
x
x
p(t)
e
j
ω
c
t
e
j∆θ
(j)
userk
+
1
Figure11.18 Thecreationofuserj’scodex
(j)
(t)inCIMA
T
x
(j)
(t)
t=
∆
θ
(j)
t
∆
ω
Figure11.19 x
(j)
(t)inthetimedomain(withcos(ω ωω ωω
c
t)notdrawn)
318 ‹ Chapter Eleven 
Whenlookedatinthetimedomain,theonlydifferencebetweenthecodeofuser
kandthecodeofuserjisthattheirpulsesoccuratdifferenttimes.Inthatway,CIMA
lookslikeTDMA—eachusersendsapulsewitha+1or–1onit,andthesepulsesare
separatedintime.However,onemaindifferenceisthis.InTDMA,youalwaysmake
surethatEquation(11.5)issatisfied;inCIMA,youstartoutbymakingsureEquation
(11.5)issatisfied.Whenmoreuserswanttouseyoursystem,somanythatEquation
(11.5)cannolongerbesatisfied,TDMAtellssomeuserstheycannotusethesystem;
CIMAshiftstomakingsurethatEquation(11.7)issatisfiedinstead(thatis,yourusers
shiftautomaticallyfrombeingorthogonaltopseudo-orthogonal)andyoucannow
handleallthenewusers.Notonlythat,butourcurrentresearchisshowingusthat
CIMAisabletoofferbetterperformancethanTDMA.
Inaveryrealway,CIMAisabridgebetweenTDMAandCDMA,andthatisa
nicething.
11.7Conclusion
Thischapterisallaboutthe“how-to”ofsharingacommunicationchannelamong
manyusers.Wesawtheguidingprinciplebehindthisidea,thenwesawhowtoshare
time(TDMA),howtosharefrequencies(FDMA),andfinally,inCDMA,howtoshare
codes.Wealsoproposedanewsetofmultipleaccesspossibilitieswhenweintroduced
CIMA,abrand-newmultipleaccesssystem.
Multiple Access Schemes: Teaching Telecommunications Systems to Share ‹ 319 
Problems 
1. ConsiderthesignalsshowninFigureQ11.1(a),(b),(c),(d).Twouserswantto
usethesystem.User1willsend+s
1
(t)forbit1and–s
1
(t)forbit0.User2will
send+s
2
(t)forbit1and–s
2
(t)forbit0.
(a) Ifthetwouserswanttoexperiencenointerferencewithoneanother,which
ofthefoursignalsetsshouldtheyuse?Explain.
(b)Ifthetwouserswanttosharethesamechannelbutcantoleratealittlebitof
interference,whichofthefoursignalsetswouldyourecommend?Explain.
(c) Whichsignalsetswouldyourecommendtheyavoid?Explain.
s
1
(t) s
2
(t)
Figure Q11.1
Foursignalsets
t
f
t
1
t
1
1
s
2
(t)
s
2
(t)
t
1
s
1
(t)
s
1
(t)
(c)
(b)
(a)
1
f
1
1
S
1
(f)
–f
m
–f
m
f
m
f
m
S
2
(f)
A1
A2
(d)
0.99 1.99
1
/
2
1
/
2
1
/
2
1
/
2
t
1
1
-1
t
1
1
–
A1=A2
320 ‹ Chapter Eleven 
2. ProposeaTDMAsystemwhichallowsfouruserstosend148bitsatatimeata
datarateof9.6kb/s.Explainwhateachusersends,whentheysendit,and
explainwhy.
3. ProposeanFDMAsystemwhichallowssevenuserstosenddataatarateof
9.6kb/s.Explainwhateachusersends,inwhatfrequencybands,andexplain
why.
4. YouareaskedtobuildaDS-CDMAsystemtosupport(a)twoorthogonalusers
and(b)fourorthogonalusers.Usingtrialanderror(oranyothermethod),
determinethecodes(seeequation(11.10))assignedtoeachuser.
5. Considerthethree-userCDMAsystemwithcodesasshowninFigureQ11.2.
Determineifthesecodesareorthogonalorpseudo-orthogonal.Explain.
6. TheCIMAusercodeisgeneratedusingFigure11.16.Show,withoutskipping
anylinesofmath,thattheoutputsignalcorrespondstoequation(11.18).
x
(1)
(t)
1
T
c
2T
c
3T
c
t Figure Q11.2
CDMAcodes
–1
x
(2)
(t)
t
1
–1
T
c
3T
c
x
(3)
(t)
+1
2T
c
t
3T
c
–1
12 
Chapter 
Analog Communications
W
ow,thelastchapter,whichisaboutsomethingthatcamefirst.Beforetherewere
digitalcommunications,beforethereweresourcecodingandchannelcoding
anddigitalmodulationandequalization,therewereanalogcommunications.
Iwillnotgointothedetailsofanalogcommunicationsinthisbook,becausein
mostcasesitisbeingquicklyreplacedbypowerfuldigitalcommunicationtechniques.
But,becausesomeofwhatcurrentlyexistsintheworldtodaywasbuiltwhenthere
wasonlyanalog,itisimportanttohaveabasicunderstandingofit.
12.1Modulation—AnOverview
Ananalogcommunicationsystem,roughlyspeaking,lookslikewhatisdrawnin
Figure12.1.Youcanseethattheinformationsignalx(t)comesinandismappedbya
modulatorintoanewsignals(t)
readytobesentoverthe
channel.Andthatisallthat
happensatthetransmitter
modulator
x(t)
side—nosourcecoder,no
s(t)
signal
channelcoding.Atthereceiver
informationsignal
readytobe
side,thesignalthatarrives
sentoverthechannel
C
h
fromthechannelispickedup, a
anditgoesthroughademodu-
n
n
e
lator,whichreturnsittoabest
l
guessoftheoriginalinforma-
tionsignal.That’sit—no
channeldecoder,nosource
demodulator
x(t)
decoder.So,allwe’lldointhis
chapterisstudythemodulator
anddemodulator.
^
r(t)
bestguessonx(t)
Figure12.1Ananalogcommunicationsystem
322 ‹ Chapter Twelve 
Atthemodulator,theinputsignalx(t)istypicallyabasebandsignal—thatis,a
signalcenteredaround0Hz.YoucanseethisinFigure12.2(a).Thesignalyouwantto
sendoverthechannelmustbesentaroundthefrequencyω
c
,asseeninFigure
12.2(b).Thegoalofthemodulatoristomaptheincomingsignalatbasebandtoa
signalthatiscenteredaroundω
c
.Todothis,themodulatortakesasinusoidatthe
frequencyω
c 
anditshovesx(t)initasitsamplitude,orasvariationsinitsphaseor
frequency.Icanexplainthatlastlinebetterinmath,so:givenx(t),themodulator
outputss(t)accordingto
s t
( )
cos
(
ω +θ
( ))
(12.1) ( )
A t t t
c
wheretheinformationx(t)isputintoeitherA(t)orθ(t).
X(f)
f
(a)
ω
c
f
youranalogsignal
musttravelthechannel
inthisfrequencyband
(b)
Figure12.2 (a)Youranalogsignalx(t)inthefrequencydomain
(b)Thefrequencybandoverwhichyoursignalmustbesent
12.2 AmplitudeModulation(AM)
OneoftheoptionsonyourradioisAM,shorthandforamplitude modulation,asimple
typeofmodulator.
Analog Communications ‹ 323 
12.2.1AMModulators—inTime
ThebasicideainAMmodulatorsistotaketheinformationsignalx(t)andmapitinto
theamplitudeofthesentsignals(t).Specifically,whathappensisthis:
1.Makesurethattheamplitudesofx(t)areintherange[–1,1].Iftheyexceed
thisrange,updatex(t)bymultiplyingbyascalarsothatitfitsintotherange[–1,
1].Inwhatfollows,Iwillassumex(t)hasamplitudesintherange[–1,1].
2.Send
s t
c
( ( )
A
(
1 + mx t
))
cos
(
ω t
) (12.2)
c
wheremisavalueintherange[0,1]andiscalledthemodulationindex.
Let’ssaytheinputcominginisthex(t)showninFigure12.3(a);thatis,the
incomingx(t)isasquarewave.Tocreatetheoutputs(t),let’sassumethatm =1.In
thiscase,wecanfigureouttheoutputasfollows:
x(t)
t
+1
1 2 3
–1
(a)
s(t)
t
2A
c
(b)
Figure12.3 (a)InputtoAMmodulator (b)OutputofAMmodulator(m=1)
324 ‹ Chapter Twelve 
1.Attimeswhenx(t)=–1,wehaveanoutput
s t
c
( ( )
A
(
1 + mx t
))
cos
(
ω t
) (12.3)
c
s t
c
( ( ) A
(
1 + − 1))
cos (ω t ) (12.4)
c
s t ( ) 0 (12.5)
2.Attimeswhenx(t)=+1,wehaveanoutput
s t
c
( ( )
A
(
1 + mx t
))
cos
(
ω t
) (12.6)
c
( ) A (1 1) cos (ω t ) (12.7) s t +
c c
( ) 2 A cos (ω t ) (12.8) s t
c c
Thesetworesultstellusthatforthex(t)ofFigure12.3(a),wehavetheoutput
showninFigure12.3(b).YoucanseefromFigure12.3(b)thattheshape(dottedlines)
ofFigure12.3(b)isthesameasthex(t)inFigure12.3(a).Thex(t)issaidtocreatethe
“envelope”ofthesentsignals(t).
Let’slookatanotherexample.InFigure12.4(a),weseetheinputwaveform
x(t)=cos(Wt),whereWisaverysmallvalue(closeto0).Thatmeansthats(t)hasthe
form(usingm =1again)
( ) A (1 cos Wt ) cos (ω t ) (12.9) s t +
c c
TheplotofthisisshowninFigure12.4(b).Again,youcanseethatthe
x(t)=cos(Wt)shapeformstheenvelopeofs(t).
Ingeneral,toplots(t)givenx(t),youfirstplotx(t).Then,plotmx(t).Next,plot1+
mx(t).Continuingon,plotA
c
(1+ mx(t))—mosttimes,it’seasytogorightfromx(t)to
A (1+mx(t)).Finally,drawindottedlinesatA
c
(1+mx(t))anditsnegative–A (1+
c c
mx(t)).Betweenthesedottedlines,drawasinusoidcos(ω
c
t).That’sit.You’vegotyour
AMsignal.
Analog Communications ‹ 325 
x(t)
s(t)
t
t
(a)
(b)
+1
–1
cos(ω
c
t)
A
c
(1+mx(t))
Figure12.4 (a)InputtoAMmodulator (b)OutputofAMmodulator(m=1)
Example 12.1 
PlottheoutputofanAMmodulatorusingm =1andA
c 
=2,whentheinputisthat
inFigureE12.1.
Solution:Thesentsignal,whichingeneralcorrespondstoEquation(12.2),this
timelookslike
( ) 2 1 + x t s t
( ( ))
cos ω t (E12.1)
c
The2(1+ x(t))lookslikethedottedlinedrawnatthetopofFigureE12.2.This
createstheenvelopeofthesentsignals(t),whichisshowninthesolid“jiggly”
lineofFigureE12.2.
326 ‹ Chapter Twelve 
x(t)
2 4
+1
–1
FigureE12.1InputtoAMmodulator
t
2 4
4
0
2(1+x(t))
FigureE12.2 OutputfromAMmodulator
12.2.2 AMModulation—inFrequency
Let’sseewhatthesentsignals(t)lookslikeinthefrequencydomain—thatis,let’s
lookattheFouriertransformofs(t),S(f ). First,weknow
s t
( ))
cos
(
ω t
) (12.10) ( )
A
(
1 + mx t
c c
s t ( )cos (ω t ) (12.11) ( ) A cos (ω t ) + A mx t
c c c c
Now,turningtothefrequencydomain,andusingbasicpropertiesofFouriertrans-
forms,wehave
S f
( )
A
2
c
(
δ(
f + f
c
) + δ(
f − f
c
))
+
, A m X f
]
¸
1
2
δ
(
f + f
)
+
1
2
δ
(
f − f
)
]
(12.12)
¸
c
( )] ∗
,
c c
]
c c
( )

A
(
δ
(
f + f
)
+ δ
(
f − f
c
))
+
A m
(
X f + f
)
+ X f − f
c
)) (12.13)
S f
c
(
c
(
2 2
Apictureisworthathousandwords,solet’sseewhatS(f )lookslike.Let’ssay
wehaveanX(f )asshowninFigure12.5(a).AccordingtoEquation(12.13),wethen
haveanS(f )asshowninFigure12.5(b).ThedifferentpartsofS(f )inEquation
(12.13)arepointedoutintheplotofFigure12.5(b).
Analog Communications ‹ 327 
X(f)
S(f)
f
(a)
δ(f+f
c
)
f
0
1
A
c
m
2
X(f+f
c
)
A
c
m
2 A
c
.
m
2
A
c
2
A
c
2
δ(f–f
c
)
X(f–f
c
)
f
c
f
c
(b)
Figure12.5
(a)InformationsignalinputtoAMmodulatorinfrequencydomain,X(f)
(b)SignaloutputbyAMmodulatorinfrequencydomain,S(f)(shownwithm=0.5)
Onethingengineersnoticedwhentheylookedatwhatwasgoingoninthe
frequencydomainwasthattransmissionpowerwasbeingspentinsendingtheim-
pulsesinS(f)—theδ(f +f
c 
)andδ(f –f  )—acrossthechannel.Theydecidedtocomeup
c 
withameasuretofigureoutwhatpercentofpowerwasbeingspentsendingthese
impulses,calledmodulation ef ficiency,anddefinedasfollows:
modulationefficiency,η=percentoftotalpowerthatisbeingusedtoconvey
information;
or,moremathematically,
η information power/total power (12.14)
2
m P
x
η
1 + m P
(12.15)
2
x
328 ‹ Chapter Twelve 
whereP  isthepowerinx(t),andiscalculatedusingtheintegral
x
T
1
2
P lim x t dt .
2
( )
x
T →∞
T
∫
T
−
2
Forexample,ifP
x
=1andm =1,thenwehaveamodulationefficiencyη =0.5=50%.
12.2.3 DemodulationofAMSignals—Noise-FreeCase
Inthissection,wewillstudythedemodulatorusedwhenamodulatorsendsanAM
signal.Thedemodulatorreceivesthesignalr(t)thatcomesacrossthechannel,and
putsout x t ,itsbestguessontheoriginalinformationsignalx(t).Tokeepour
ˆ
( )
presentationsimple,I’llassumethattheinputtothedemodulatorisr(t)=s(t).Thatis,
wewillignoreallthechanneleffectsandassumeanidealchannelwherewhatcomes
outisexactlywhatcamein.
Theideabehinddemodulationissimple.Look,forexample,atthesignalinFigure
12.4(b).Let’ssaythatthissignalr(t)= s(t)iscomingintothedemodulator.Thetop
dottedlineinthatpicture,calledtheenvelope,issimplyA (1+mx(t))=A
c
(1+m cosWt).
c
Ifwecanfindawaytogetthattopdottedlineoutofthereceivedr(t)=s(t),thenwein
essencehaveourinformationx(t).(Allwehavetodoonceweget[thetopdottedline]=
A (1+mx(t))=A
c
+A  mx(t)issubtractA
c
andmultiplytheresultbythescalar1/mA .)
c c c
Sothekeyquestionindemodulationis:Canwebuildasimpledevicetogetthat
topdottedlinefromthesignalinFigure12.4(b)?Theanswerisaresoundingyes.
Thereareverycheapandsimpledevices,calledenvelopedetectors,thatcaneasily
extracttheenvelopefromthes(t).So,AMreceiversarecalledenvelopedetectors.
Anexampleofaninexpensiveenvelope
detectorisshowninFigure12.6.Inthis diode
makesallnegative
figure,thelittletriangle,adiode,makesall
negativevalueszero,andleavesallthe
values0
positivevaluesuntouched.Forexample,
withtheinputofFigure12.4(a),redrawnin
Figure12.7(a),theoutputfromthediodeis
Figure12.7(b).Theresistor(R)andthe
capacitor(C)worktogetherasalow-pass
filter.TheRClow-passfiltercutsoutthe
rapidvariationsinthesignalandleavesthe
slowvariationsintact.Thatis,fortheinputof
Figure12.7(b),itcreatesanoutputof
approximatelythatshowninFigure12.7(c).
Giventheincomings(t),we’vegotitsenve-
Low-PassFilter
lopepulledout,whichmeans—good Figure12.6 AnimplementationofanAM
news—we’veprettymuchgotourx(t). demodulator=envelopedetector
R C
Analog Communications ‹ 329
k
(a)
cos(ω
c
r(t)=s(t)
k
A
c
(1+mx(t))
cos(ω
c
t)
k
(c)
(b)
outputof
outputof
diode
A
c
A
c
(1+mx(t))
t)high-frequencyterm
RCLPF
(1+mx(t))low-frequencyterm
Figure12.7 WorkingsofenvelopedetectorofFigure12.6
(a)input (b)diodeoutput (c)LPFoutput
330 ‹ Chapter Twelve 
12.2.4 AnAlternativetoAM—DSB-SC
SomeuseanalternativetoAMcalleddouble sideband suppressed carrier,whichmerci-
fullyiswrittenandspokeninshorthandusingtheacronymDSB-SC.
InAM,wesent
( )
A
(
1 + mx t s t
( ))
cos ω t (12.16)
c c
InDSB-SC,wesendprettymuchthesamething,onlywegetridofthe“1”and
the“m”;thatis,inDSB-SC,wesend
( ) A x t s t ( )cos ω t (12.17)
c c
Insomewaysthisisagoodthing,andinsomewaysitisn’t.Toseeitsbenefits,
let’slookatthefrequencydomain,andstudytheFouriertransformoftheDSB-SCsent
signals(t)—thatis,studyS(f).UsingsimpleFouriertransformproperties,andthes(t)
ofEquation(12.17),wefind
( ) A X f S f
c
( )∗
¸
,
1
2
δ( f − f ) +
1
2
δ( f + f
c
)
]
]
(12.18)
c
c
( )

A
2
¸
, X f − f
)
+ X f + f
c
)
]
]
(12.19)
S f
(
c
(
So,ifX(f)looksthewayit’sdrawninFigure12.8(a),thenitfollowsthatS(f)looks
likeFigure12.8(b).WecanseethatinDSB-SC,wearenotwastinganypowersending
animpulseatω
c
(asinAM,whichwastespowerhere,asshowninFigure12.5).
But,alas,itwasnotallgoodnewsforDSB-SC.Peoplesaid:Well,weneedade-
modulatorforit—let’strytousetheenvelopedetector.Let’ssayx(t)=cos(Wt)(whereW 
isverysmall)asshowninFigure12.9(a).Thatmeansthats(t)= A
c
cos(Wt)cos(ω t),
c 
whichcorrespondstothemultiplicationshowninFigure12.9(b)andleadsto
s(t)= A
c
cos(Wt)cos(ω t)asshowninFigure12.9(c).Now,let’slookattheenvelopeof
c 
s(t),whichisthedottedlineontopofit,showninFigure12.9(d).Youcanclearlysee,
comparingtheenvelopeofFigure12.9(d)withthex(t)ofFigure12.9(a),thatthesehave
completelydifferentshapes.Infact,inthiscase,theenvelopeofs(t)is|x(t)|andnotx(t).
So,theenvelopedetectordoesnotworkfortheDSB-SCsignal.Thisisunfortunate,
becauseenvelopedetectorsaresoinexpensive.
Nevertheless,somedesignersdiddecidetouseDSB-SC.Butfirst,theyhadto
comeupwithademodulatoratthereceiverside,togetx(t),givenr(t).Thedemodula-
torthattheyfoundworkedwellisshowninFigure12.10.Theinputcomingintothe
demodulatoris
r t ( ) A x t ( ) s t ( ) cos (ω t ) (12.20)
c c
Thisgetsmultipliedbyacosineterm,leadingto
Analog Communications ‹ 331 
X(f)
1
f
–f
m
f
m
(a)
S(f)
A
c
2
f
–f
c
– f
m
–f
c
–f
c
+f
m
f
c
– f
m
f
c
f
c
+f
m
(b)
Figure12.8 DSB-SC (a)Informationsignalinfrequencydomain,X(f)
(b)Sentsignalinfrequencydomain,S(f)
′
( ) A x t r t ( ) cos (ω t )⋅ cos (ω t ) (12.21)
c c c
c
r t
( )
,
¸
1 cos
(
2 ω t
)
]
]
(12.22)
′
( )

A
x t +
c
2
Next,thissignalispassedthroughalow-passfilter,whichcutsoutthehigh-
frequencyterm,leadingtotheoutput
′′
( )

A
x t r t
c
( ) (12.23)
2
Finally,amultiplicationisappliedtogeneratex(t).Thisdemodulatorisalittle
morecomplexthanthatofAM.DSB-SCgivesyouachoice—spendlesspowerin
transmissionbyusingDSB-SC,orusealessexpensivereceiverbyusingAM.
332 ‹ Chapter Twelve 
x(t)
(c)
s(t)=x(t)⋅A
c
cosω
c
t
=A
c
cosWtcosω
c
t
t
(a)
(b)
1
–1
x(t)=cosWt
x(t)=cosWt
A
c
cosω
c
t
A
c
–A
c
envelopeofs(t)
t
s(t)=
t
t
t
(d)
Figure12.9 (a)Informationsignalx(t)=cosWt (b)Creatingsentsignals(t)
(c)Sentsignals(t) (d)Envelopeofs(t)
Analog Communications ‹ 333 
x(t)
r(t)=s(t)
x x LPF
r
'
(t) r"(t)
=A
c
x(t)cos(ω
c
t)
cos(ω
c
t) 2
/
A
c
Figure12.10DemodulatorforDSB-SC
Givenr(t)=s(t),itoutputsx(t)
Example 12.2 
AssumingA
c 
=4:
(a) Determinetheoutput(intime)oftheDSB-SCmodulatorwhentheinput
correspondstoFigureE12.3.
(b)Determinetheoutput(infrequency)oftheDSB-SCmodulatorwhenthe
inputcorrespondstoFigureE12.4.
4
x(t)
FigureE12.3
TheinputtoDSB-SCmodulatorfor(a)
+1
t
2 
–1 
X(f)
FigureE12.4
TheinputtoDSB-SCmodulatorfor(b)
5
0
–100Hz 100Hz
Solution: (a) Theoutputsignalcorrespondsto
( ) 4 x t s t ( ) cos ω t (E12.2)
c
334 ‹ Chapter Twelve 
InFigureE12.5,the4x(t)isdrawnasadottedline,andformstheenvelope
forthesignals(t).A– 4x(t)isalsodrawnasadottedline.Thesignals(t)isthe
solid“jiggly”linebetweenthetwodottedlines.
(b)Theoutputsignalcorrespondsto
( )
4
, X f f
) + X f f
)]
(E12.3)
s f
( +
c
]
2
¸
( −
c
FortheX(f )ofFigureE12.4,thisS(f )correspondstoFigureE12.6
s(t)
2 4
4
–4
4x(t)
FigureE12.5
TheDSB-SCoutputfor(a)
t
–4x(t)
S(f)
10
f
–f
c
–100 –f
c
–f
c
+100 f
c
–100 f
c
f
c
+100
FigureE12.6 Outputofmodulator
12.3 FrequencyModulation(FM)
Tosome,FMisadialontheradiowhereyoucanhearlovesongs,classicrock,or
greatesthits.Toengineers,FMisshorthandforfrequency modulation.
Analog Communications ‹ 335 
12.3.1TheModulatorinFM
Theideainfrequencymodulationistomaptheinformationx(t)intothefre-
quencyofthetransmittedsignals(t).Mathematically,whatisdoneisthis:Givenan
informationbearingsignalx(t),yousendoutoverthechannel
( )
A cos
(
ω +θ
( ))
(12.24) s t t t
c c
where
t
θ ( ) K
f
∫
x(τ)dτ . (12.25) t
−∞
Lookingatthisequation,youreallycan’ttellthattheinformationx(t)isplacedin
thefrequencyofs(t).Solet’sdoalittlemaththatwillshowthatx(t)hasindeedbeen
placedinthefrequencyofs(t).Thefrequencyofs(t),atanymomentintimet,isgiven
by
ω( )
d
(
ω +θ ( )) (12.26)
t t t
c
dt
t
t
dt
(
t K
f
∫
x( )dτ
(
\
ω( )
d
j
,
ω + τ
c (12.27)
−∞ ,
ω( ) ω + K x t ) (12.28) t
f
(
c
Thistellsusthatattimet,thefrequencyofthesentsignals(t)isω
c
+K  x(t),which
f
indicatesthatx(t)isdeterminingthefrequencyofs(t).
Let’susepicturestoseewhatisgoingoninFM.Let’ssaywehavetheinforma-
tion-bearingsignalx(t)asshowninFigure12.11(a).Usingthis,wecandetermine
someimportantinformationabouts(t):
1.Attimeswhenx(t)=–1,thefrequencyofs(t)isω(t)=ω
c
 + K  x(t) = ω
c
– K .
f f 
2.Attimeswhenx(t)=+1,thefrequencyofs(t)isω(t)=ω
c
 + K  x(t) = ω
c
 + K .
f f 
Usingthisinformation,wecangettheplotofs(t)showninFigure12.11(b).Here,we
seethevariationinthefrequencyofs(t)asadirectresultofchangestox(t).
Foranotherexample,takealookatFigures12.12(a)and(b).There,weseethe
inputx(t)=cos(Wt)(whereWisaverysmallnumber).Wealsoseetheoutputin
Figure12.12(b)—asx(t)getsbigger,thefrequencyofs(t)getsbigger,andasx(t)gets
smallerthefrequencyofs(t)getssmaller.
336 ‹ Chapter Twelve 
x(t)
+1
. . .
(a)
t
–1
s(t)
A
c
. . .
t
–A
c
(b)
Figure 12.11
(a)Informationsignalx(t)
(b)TransmittedFMsignals(t)
Let’sseeifwecancharacterizethesentsignals(t)inthefrequencydomain—that
is,let’sseeifwecanevaluatetheFouriertransformofs(t),calledS(f ).Tohelpusout,
let’sstartbyrewritings(t):
( ) A cos
(
ω +θ ( ))
(12.29) s t t t
c c
A e
ω +θ( )
¦
s t
j t t
c
( ) Re
¦
(12.30)
c
s t
j t ω
c
( ) Re
¦
,
A e
θ( )
]
e
j t
¦
(12.31)
c
] ¸
Analog Communications ‹ 337 
x(t)=
cosWt
x(t)
t
(a)
t
A
c
–A
c
s(t)
highest
frequency
lowest
frequency
(b)
Figure 12.12
(a)Informationsignalx(t)
(b)Sentsignals(t)inFM
ω
c
j t
¦ ( ) Re
¦
g t e (12.32) s t ( )
whereg(t)=A
c
e
jθ(t)
.TakingtheFouriertransformofs(t)andapplyingpropertiesofthe
Fouriertransform,weendupwith
S f
1
2
G f − f
)
+
1
2
G
(
− − f
c
) (12.33) f
( )

(
c
338 ‹ Chapter Twelve 
t
j t
t whereG(f)istheFouriertransformof A e
θ( )
andθ ( )
∫
x(τ)dτ .Therelationship
c
−∞
betweenG(f)andx(t)issocomplexthatthereisnosimplemathematicalequationto
relatethevalueG(f)tothevalueX(f)—thatmeansthereisnosimpleequationtorelate
S(f)toX(f).
Inthesimplecasewhenx(t)=cos(Wt)andWissmall,wecanderiveanequation
relatingX(f )toS(f ),butthisequationismessy,involvingBesselfunctions,andIjust
wanttoofferanintroductiontoanalogcommunicationshere.Lookinthereferencelist
toseeabookthatcoversthejoysofBesselfunctions.
Example 12.3 
DrawtheoutputofanFMmodulatorwhentheinputcorrespondstoFigureE12.7.
Solution:Theoutputcorrespondstoequation(12.24),whichshowsthatthe
outputcorrespondstoasinusoidwithconstantamplitude.Equation(12.28)tells
usthatthefrequencyoftheFMoutputchangeslinearlywithx(t).Puttingthis
informationtogetherleadstotheoutputplotofFigureE12.8.
x(t)
t
FigureE12.7 InputtoFMmodulator
t
A
c
–A
c
s(t)=A
c
cos(ω
c
t+θ(t))
...
FigureE12.8 OutputofFMmodulator
Analog Communications ‹ 339 
12.3.2 TheDemodulatorinFM
WenowknowabouthowtheFMmodulatorsworkandwhattheydo.Atthereceiver
side,youwanttobuildanFMdemodulatorthatgetsthereceivedsignalr(t)= s(t)and
turnsthatbackintoyourinformationx(t).
Youknowthatthereisnoinformationx(t)intheamplitudeofr(t)= s(t)—allthe
informationisintheinstantaneousfrequencyω(t)=ω
c
+K x(t).Thedemodulator
f
workstogetω(t),theinstantaneousfrequency,outofr(t)= s(t),thereceivedsignal.
AdemodulatorforanFMsignalisshowninFigure12.13.First,alimiteris
applied.Thelimitertakesr(t)= s(t)andgetsridofallamplitudefluctuations,bysimply
forcingallpositivevaluesto+1andallnegativevaluesto–1.Theoutputiscalledr′ (t).
Thelimiter’soperationdoesnotaffecttheabilitytoextracttheinformationsignalx(t),
sincetheinformationisstoredinthefrequency(notintheamplitude).Then,the
signalr ′ (t)ispassedthroughadiscriminator.Thediscriminatoroutputsavaluer″(t),
whichisproportionaltotheinstantaneousfrequencyω(t).Thatis,itoutputs
r t K ′′
( )
ω
(
t
) (12.34)
′′
( ) K
(
ω + K x t r t
f
( )) (12.35)
c
Oncewehavethisoutput,aprocessordoingasimplesubtractionandascalar
multiplicationcreatestheoutput x t .
ˆ
( )
Andthat,myfriends,isFM,itsmodulationanditsdemodulation.
x(t) r'(t) r(t)=s(t) r"(t)
in
1
–1
out
LI MI TER
amplitudeofoutput
ofinput
Simple
&
scalar
^
DI SCRI MI NATOR
frequency
subtraction
multiplication
Figure12.13FMdemodulator
12.4 TheSuperheterodyneReceiver
Thelastsectionofthisbookhasagrandtitle.Thesuperheterodyne receiverisastan-
dardAMreceiverthatyoucanusetopickupanyradiostation.Letmeexplainhow
thisreceiverworks,thenwe’llcallitadonebook.
TakealookatthetopofFigure12.14.Thisshowsapartofthefrequencyspec-
trum.Specifically,eachpeakinthespectrumrepresentsthepresenceofaradiostation
transmittingitssongsandsoundsatthatfrequency.Youwanttobuildaninexpensive
radiowhichallowsyoutopickupanyradiostationthatyou’dliketolistento.
340 ‹ Chapter Twelve 
RF
filter
IF
filter
455kHz
x
f
f
H(f)
H(f)
455kHz
ANTENNA
900kHz
f
AM
900
thislinemeans
cos2πf
L
t
Envelope
Detector
radiostationsat
differentfrequencies
"adjustable"or
"tuneable"
PICKS REMOVESALLBUTDESIRED TURNSAMSIGNAL
UPALL RADIOSTATION TOSOUND
RADIO
STATIONS
Figure12.14Thesuperheterodynereceiver
ThestandardconstructionofsuchareceiverisshowninFigure12.14.Let’ssay
thatyouwanttopickupanAMsignalat900kHz.
1.Tostart,youhaveanantenna,whichpicksupalltheradiosignals.
2.Next,youusethreecomponentstogetridoftheotherradiostationsandleave
youonlywiththesoundsof900kHz.
2a. ThefirstcomponentusedtocutouttheotherradiostationsisatunableRF
(radiofrequency)filter.TheRFfilteristunedto900kHz,andactsasa
bandpassfilter,asshowninFigure12.14.Itcutsdown(butnotout)the
radiosignalsatotherfrequencies—thereasonwedon’tuseaverysharp
bandpassfilterherethatcutsoutalltheotherradiostationsisbecauseitis
expensivetobuildaverysharpfilterthatistunable.
2b. Next,weapplyamixer,whichisjustafancywaytosaythatwemultiply
thesignalbyacosinewaveformwithfrequencyf
L
.Multiplicationbya
cosinecausestheincomingsignaltoshiftinfrequencybyf
L
.Bytuningthe
dialto900MHz,thecosinewaveformfrequencyf issettoshiftthefre-
L
quencyofthe900MHzstationto455MHz.
Analog Communications ‹ 341 
2c. Averysharpfilter,calledtheIF(intermediatefrequency)filter,isappliedto
theincomingsignal.Thisfilterisaverysharpbandpassfilterat455MHz,
asshowninFigure12.14.WiththedesiredAMsignalnowat455MHz,it
makesitthroughtheIFfilter—alltheotherradiofrequenciesarecutout.
3.NowyouhavetheAMsignalyouwantwithallthoseotherradiosignalscut
out.SoyouputyourAMsignalthroughanenvelopedetector,whichextractsthe
soundinformationx(t)fromtheAMsignal.
12.5 Summary
Thisisit.Theendofthechapter.Theendofalongjourneythroughabook.Inthis
chapter,webrieflylookedatanalogcommunicationsystems.WelookedatAMand
FMmodulation,andwesawhowtodetectAMandFMsignals.Weeven“built”a
receiverthatallowsyoutolistentoyourfavoriteradiostation.
Itismysinceresthopethatthisbookprovidedyouwithasolidunderstandingof
thebasicsoftelecommunicationssystemsengineering,withoutthemountainof
“muck”createdbyintimidatingattitudesandbigwords.Lifeissimple,ifit’sjust
explainedthatway.
342 ‹ Chapter Twelve 
Problems
1. FigureQ12.1showstheinputtoamodulator.Determinetheoutputif
(a) themodulatorisanAMmodulator.
(b)themodulatorisanFMmodulator.
(c) themodulatorisaDSB-SCmodulator.
t
3.5 3 2 1
x(t)
...
FigureQ12.1Modulatorinput
2.FigureQ12.2showstheinputtoamodulator.Determinetheoutput(inthe
frequencydomain)if
(a) themodulatorisanAMmodulator.
(b)themodulatorisaDSB-SCmodulator.
X(f)
1
-2
.
f
a
f
a
f
a
2
.
f
a
FigureQ12.2 Modulatorinput
f
3. Giventhatthesignalx(t)inFigureQ12.3isinputtoaDSB-SCmodulator:
(a) Determinetheoutputofthemodulator.
(b)Assumethissignalissentoverachannelwhichisideal(thereceived
signalequalsthesentsignal).
Analog Communications ‹ 343 
(1) Determinetheoutputofthedemodulatorwhenthedemodulatorisa
perfectenvelopedetector.
(2) Determinetheoutputofthedemodulatorwhenthedemodulatorisa
mixer,low-passfilter,andprocessingdeviceasdescribedinthechap-
ter.
x(t)
FigureQ12.3 DSB-SCinput
1
1
–1
2 3
4. Giventheinputin(Q12.1),provideananalyticalequationforthemodulator
outputwhenthemodulatoris
t
¦
t , 0 ≤ <1
¦
x t t ( )
¦
2 −t ,1 < < 2
(Q12.1)
¦
¦
0 , else
(a) anAMmodulator
(b)aDSB-SCmodulator
(c) anFMmodulator
5. AnAMreceiveristunedtopickupthestationat600MHz.Drawablock
diagramoftheAMreceiverusedtodetectthissignal.
Annotated References and 
Bibliography 
Whilethisbookprovidesastreamlinedoverviewoftheworldoftelecommunica-
tionsengineering,I’dliketoreferyoutootherbooksthatcanserveasuseful
resources.Theyoffersomeaddeddetailontopicscoveredinthisbook,andaddress
topicsIdidn’thavethespacetoincludeinthisbook.Here’smyrecommendedlist,
withsomenotesthatIhopeprovevaluable.
1. B.Sklar,Digital Communications: Fundamentals and Applications.Englewood
Cliffs,NJ,Prentice-Hall,1988.
Anolderbook,butinmyopinionaclassic.(Thesecondeditionwillbecomingin
early2001.)Ilearnedtelecommunicationsusingit,andIenjoyedit.Well-written
forthemostpart,andincludessomedetailsnotincludedinthisbook.
2. J.G.Proakis,Digital Communications,4thEdition,Boston,McGraw-Hill,2000.
ThisisTHEclassicbookinthefield.Thebookisalittletoughtoreadinplaces
(intendedforthegraduatestudentaudience)butitisawonderfullydetailedbook
andagreatreference.
ForfurtherreadingonthetopicsinChapter1and2,try:
3. J.G.Nellist,Understanding Telecommunications and Lightwave Systems: An 
Entry-Level Guide. IEEEPress,1995
Anicelittlebookthatgivesanontechnicalintroductiontotelecommunication
systems.Agoodbookifyouwanttointroducesomeonewithoutastrongtechni-
calbackgroundtothefield.
4.RogerB.Hill,“TheEarlyYearsoftheStrowgerSystem,”Bell Laboratories 
Record,Vol.XXXINo.3,p.95.
www.privateline.com/Switching/EarlyYears.html
Thecitedarticleisreprintedonthiswebsite,whichalsohasalotofother
interestingmaterialontelephonehistory.
ForfurtherreadingonthetopicsinChapter3,try:
5. A.Oppenheim,A.WillskyandH.Nawad,Signals and Systems,SecondEdition.
UpperSaddleRiver,NJ,PrenticeHall,1996.
346  ‹  Annotated References and Bibliography 
Anintroductiontosignalsandsystems,whichisrequiredbackgroundreadingif
youwanttothriveintheworldoftelecommunications.Alittlebitwordyinplaces,
butdefinitelyasolidbookinthefield.
6.I’veyettofindagreatbookintheareaofprobabilityandstatisticsforengi-
neers.I’drecommendgoingtoyourfavoriteon-linebooksellerandseewhat
catchesyourattention.OnethatothershaverecommendedisApplied Statistics 
and Probability for Engineers,D.Montgomeryetal.,JohnWiley&Sons,1998.
ForfurtherreadingonthetopicsinChapter4,try:
7. A.GershoandR.M.Gray,Vector Quantization and Signal Compression.Boston,
MA:KluwerAcademicPublishers,1992.
Anincrediblydetailed,nicelywrittenbookintheareaofsourcecoding.Itis
intendedasagraduateleveltextbutifyouwantagreatreferenceonsource
coding,thisisit.
8. TexasInstrumentsApplicationNoteSPRA163A,A-lawandmu-Law
CompandingImplementationsUsingtheTMS320C54x,December1997.
www-s.ti.com/sc/psheets/spra163a/spra163a.pdf
Gooddetailsonµ-lawandA-lawcompanding.
ForfurtherreadingonthetopicsofChapters5and6,try:
9. StephenG.Wilson,Digital Modulation and Coding,PrenticeHall,1995. 
Agraduate-levelengineeringtextthatgivesgoodinsightintothefundamentalsof 
modulationandcoding. 
ForfurtherinsightintothetopicsofChapter7,try: 
10. ViterbiAlgorithmWorkshop,byBrianC.Joseph,www.alantro.com/viterbi/ 
viterbi.htm 
ThisisaniceJavaappletsimulationoftheViterbiAlgorithmasappliedtobinary 
convolutionalcodes. 
ForfurtherreadingonthetopicsofChapter9,checkout:
11. E.LeeandD.Messerschmidt,Digital Communications.Boston,MA:Kluwer
AcademicPublishers,1988.
Atoughbooktoread,butinchapters6-10itdoesofferapresentationtoequaliza-
tionthatissignificantlydifferentthanthatofferedinProakis’sbook.
Annotated References and Bibliography  ‹  347 
ForfurtherreadingonthetopicsofChapter10,try:
12. H.VanTrees,Detection, Estimation and Modulation Theory: Part I.New
York,NY:JohnWileyandSons,1968.
Man,yousay,thisauthorisrecommendinga1968book.Howoldisthisguy
anyway?Actually,thisisTHEclassicinthefieldofdetectionandestimation.It
canbeatoughreadattimes,butifyouwantallthedetailsofdetectionand
estimationdiscussedinChapter10,Irecommendit.It’soutofprint,butyoucan
probablyfindacopyonline.
ForfurtherreadingonthetopicsinChapter11,try:
13. J.S.LeeandL.E.Miller,CDMA Systems Engineering Handbook. Boston,MA:
ArtechHousePublishers,1998.
Thisfatbook,over1,000pageslong,isbyfarthebestandeasiesttoreadbookin
theareaofCDMA.It’swonderfullycomplete,nicelywritten,andjustanall-
aroundgem.
Forfurtherreadingaboutwirelesstelecommunications,consider:
14. T.Rappaport,Wireless Communications.UpperSaddleRiver,NJ:Prentice
Hall,1996.
Ifyouwanttolearnaboutthewirelessworld,Iwouldrecommendinparticular
thefirstfourchaptersinthisbookasanintroduction.Rappaportdoesawonderful
jobofprovidinganeasy-to-readvisionofthewirelesscommunicationchannel,the
firststeptoagoodunderstandingofthewirelessworld.
Index
3-dB bandwidth, 55
4-ASK,  126,  143
4-FSK,  129
4-PSK,  129,  146,  151
4-PSK signals, 140
8-ASK,  127,  143
8-FSK,  129
8-PSK,  129,  141,  142
µ-law description, 91
A
A-law description, 91
absolute bandwidth, 55
additive white gaussian noise, 147
aliasing, 65
alternative mark inversion, 118
AM, 322, 340, 341
AM modulator, 325
AM modulators, 323
AM receiver, 339
amplitude modulation, 322
amplitude shift-keying, 125
analog, 6
analog communication, 7, 8, 321
analog signal, 6
antenna,  26
ASK,  131,  132,  143
ASK modulators, 125
autocovariance,  48
AWGN,  147,  149
B
B-ASK,  125
bandpass filter, 18, 57, 294, 340
bandpass modulator, 124
bandwidth,  53
baseband filter, 250
baseband modulator, 116, 122, 192
baseband signal, 322
BASK,  143
Bayes’ Rule, 154
benefits of block coders, 192
BFSK,  129
Binary ASK, 125
bipolar RZ, 118
bit rate, 94, 95
bit-to-symbol mapper, 98
bit-to-symbol mapping, 95
block coding, 172
BPF, 18, 57
BPSK,  127,  165
BPSK modulator, 139, 162, 222, 225
burst,  304
BW compression, 123
C
carrier interferometry multiple access, 315
catastrophic code, 214, 215
CDMA,  306,  310
cell,  74
Centroid Rule, 83
channel,  3,  147
channel  coder,  171,  175
channel decoder, 171, 172, 176, 183, 203
channel  filter,  250
check bit, 173
CIMA,  315,  318
Class 1 switching center, 21
Class 2 switching center, 21
Class 3 switching center, 21
Class 4 switching center, 20
Class 5 switching center, 19
coaxial cable, 24
code, 307, 311
code division multiple access, 306
code rate, 175
350  ‹  Index 
codebook, 74
codeword, 74, 82
compandor, 85, 89
compressor, 85
convolution, 55, 63, 250
convolutional coder, 199, 200, 201
convolutional coders and decoders, 197
convolutional decoder, 204
correlator receiver, 156
correlator receiver front end, 151
cutoff frequency, 65
D
data communication, 31
decision device, 152, 153, 160, 163, 253, 271,
291,  292
delta function, 50
Delta Modulator, 99, 104
demodulator, 26, 28, 115, 116, 146, 148, 162
DFE,  274
differential feedback equalizer, 274
differential PCM, 107
digital communication, 7, 8
digital signal, 6
digital subscriber line, 31
direct sequence CDMA, 310
discrete time processing, 265
discrete-time  convolution,  263
discrete-time signal, 6
discriminator,  339
DM, 99, 100, 101
double sideband suppressed carrier, 330
DPCM,  107,  109
DS-1, 23, 24
DS-2, 24
DS-3, 24
DS-4, 24
DS-CDMA,  310,  312
DSL, 31
E
end office, 19
envelope,  324
envelope detector, 328, 330
Error Detection, 122
error signal, 77
estimation, 280, 281, 282, 283, 295
Ethernet,  35,  36
even parity, 172
expandor, 85
F
FDM, 16
FDMA,  305,  306
FH-CDMA,  312,  313
fiber-optic cable, 30
fiber-optic links, 29
FM, 334, 339
FM demodulator, 339
Fourier transform, 51, 52, 326, 330, 337
fractionally spaced equalizer, 274
frequency distortions, 132
frequency division multiple access, 305
frequency division multiplexing, 16
frequency domain, 326
frequency  modulation,  334,  335
frequency offset, 280
frequency  shift-keying,  129
frequency-hopping  CDMA,  312
FSE,  274
FSK,  132
FSK modulators, 129
G
Generalized Lloyd Algorithm, 83
generator matrix, 179, 180, 184
Gram-Schmidt orthogonalization procedure,
135
granular cells, 74
granular noise, 106
GSM,  304
I 
ideal sampling, 61, 62, 67
IF,  341
impulse function, 50
impulse response, 58, 272
impulse train, 62
impulse train sampling, 61
L
Index  ‹  351 
instantaneous  frequency,  339
interference-limited,  302
intermediate  frequency,  341
intersymbol  interference,  254,  258
inverse filter, 272
inverse system, 58
Inversion  Insensitive,  124
ISI,  258
J
just-in-time math, 133
LANs, 35
limiter,  339
linear block coder, 177, 178
linear block decoder, 182
linear equalizer, 271, 273
Linear Time Invariant, 55, 56, 67
Local Area Networks, 35
local loop, 20
low-pass filter (LPF), 56, 64, 96, 291, 328
LTI, 55, 56, 58
M
Manchester Coding modulator, 120
MAP,  282
MAP estimate, 286, 287
mapping by set partitioning, 226
Matched Filter Receiver, 158, 159
maximum a posteriori, 282
maximum likelihood, 283
Maximum Likelihood Sequence Estimator, 268
MC-CDMA, 313, 315
mean, 42, 47, 48
mean squared error, 78, 84
mid-riser, 75
mid-tread, 75
Miller Coding modulator, 120
minimum mean squared error, 281
minimum mean squared estimate, 273, 282
mixer, 16, 340
ML, 283
MLSE,  268,  270,  271
MMSE,  273,  281
MMSE linear equalizer, 274
mobile communications, 33
modem, 31
modulation  efficiency,  327
modulator, 26, 115, 116, 322
MPSK modulator, 293
mse, 78, 81, 82, 84, 85, 86
multicarrier CDMA, 313
multiple access, 299, 302
Multiple Random Variables, 44
multiplexing,  16,  299
N
(n,k) code, 175
natural sampling, 69
Nearest Neighbor Rule, 82
node,  202
noise,  249,  279
noise immunity, 124
non-return-to-zero,  117
non-uniform quantizer, 76
NRZ modulator, 117
NRZ-L, 117
NRZ-M, 117
Null-to-null bandwidth, 55
Nyquist criteria for zero ISI, 254
Nyquist rate, 65, 100, 110
O
odd parity, 172
optimal decoder, 268
optimal receiver, 258, 260, 264, 265, 271
orthogonal CDMA, 310, 312
orthogonalization procedure, 136
orthonormal basis, 134, 136, 138, 139, 140,
142,  143,  145,  148,  149,  162,  233,  280,
285,  289
overload, 106
overload cells, 74, 75
overload noise, 106, 107
352  ‹  Index 
P
packet switched network, 32
parity check matrix, 183
PCM, 92, 115
pdf, 40
performance of block coders, 188
performance of linear block codes, 189
performance of rectangular codes, 189
performance of the convolutional coder, 213
phase distortions, 132
phase offset, 280, 285
phase shift-keying, 127
phase-encoded modulators, 120
phase-locked loop, 290
Plain Old Telephone System (POTS), 19
PLL, 290, 292, 293
POTS,  19
predictive coder, 96, 98, 107
predictive decoder, 98
primary center, 21
probability density function, 40, 44
probability distribution function, 40
pseudo-orthogonal,  302
pseudo-orthogonal CDMA, 310
PSK,  132
PSK modulators, 127
pulse code modulator, 92, 93
pulse shaping, 245
pulse-shaping filter, 247
Q
QAM,  145,  146
QAM modulators, 130
quadrature amplitude modulation, 130
quadrature  PSK,  129
quantization,  71
quantizer, 10, 71, 72, 73, 101
quantizer error, 80
R
radio frequency, 340
raised cosine function, 254
random event, 39
random processes, 45
random variable, 39, 45
receiver  antenna,  26,  28
receiver filter, 258
receiver front end, 148, 151, 158, 162, 262
rectangular codes, 175, 176, 189
redundancy,  175
redundant bits, 175
regional center, 21
repeaters,  27
RF,  340
RZ Modulators, 118
RZ-AMI,  118
S
Sampler, 9, 100
sampling, 61
sampling rate, 62, 93
sampling theorem, 65
satellite  connections,  28
sectional center, 21
self-clocking,  122
shifting property, 63
signal to quantization noise ratio, 78
signal-to-noise ratio, 165
signature waveform, 307
sinc scenario, 254
single parity, 173
single parity check bit coder, 172, 188
single parity check bits, 174
SNR,  165
source coder, 9, 95, 110
source coding, 61
source decoding, 95
speech signal, 18
SQNR, 78, 81, 88, 89, 94, 95, 98
square wave function, 50
state,  202
superheterodyne  receiver,  339
switching center, 13, 14, 15
symbol rate, 93
symbol-to-bit mapper, 10, 93
synchronization,  280,  295
syndrome,  185,  188
systematic linear block codes, 181
V
Index  ‹  353 
T
T-1 trunk line, 23 
TCM, 221 
TCM decoder, 230, 233, 241 
TCM Decoder Front End, 233 
TDM, 18 
TDMA, 303, 304, 305, 318 
telecommunication  network,  13 
telephones,  13 
terrestrial microwave, 26, 28 
time division multiple access, 303 
time-division multiplexing, 18 
timing offset, 280 
Token ring, 35, 37 
toll center, 20 
trellis diagram, 201, 202, 204, 226, 234, 267 
trellis-coded modulation, 221, 237 
trunk line, 23 
twisted-pair cable, 24 
U
Ungerboeck,  223 
uniform quantizer, 76 
unipolar RZ, 118, 122 
variance,  42 
VCO, 290, 291, 292 
Viterbi algorithm, 206, 207, 212, 237, 268, 
270 
voltage-controlled oscillator, 290 
W
whitening filter, 265, 271 
wide sense stationary, 48 
WSS, 48 
Z
zero forcing linear equalizer, 272 
zero-order Hold Sampling, 67 
Dem Dem Dem Dem Demystifying ystifying ystifying ystifying ystifying T TT TTec ec ec ec echnolo hnolo hnolo hnolo hnolog gg ggy yy yy
TM TM TM TM TM
series series series series series
Technical publications by engineers, for engineers.
Video Demystified, Third Edition
A Handbook for the Digital Engineer
by Keith Jack
INCLUDES WINDOWS/MAC CD-ROM.
Completely updated edition of the “bible” for
digital video engineers and programmers.
1-878707-56-6 $69.95
NEW!
Short-range Wireless Communication
Fundamentals of RF System Design and Application
by Alan Bensky
INCLUDES WINDOWS CD-ROM. A clearly
written, practical tutorial on short-range RF wireless
design. The CD-ROM contains a number of useful
Mathcad worksheets as well as a full searchable
version of the book.
1-878707-53-1 $49.95
NEW!
PCI Bus Demystified
by Doug Abbott
NEW! INCLUDES WINDOWS CD-ROM with full
searchable version of the text. This concise guide
covers PCI fundamentals, for both hardware and
software designers, including the new PCI Hot-Plug
Specification and new features of the PCI BIOS
spec.
1-878707-54-X $49.95
NEW!
Telecommunications Demystified
A Streamlined Course in Digital (and Some Analog)
Communications for E.E. Students and Practicing
Engineers
by Carl Nassar
NEW! INCLUDES WINDOWS CD-ROM. A
straightforward and readable introduction to the
theory, math, and science behind telecommunica-
tions. The CD-ROM contains useful Matlab tutorials
and a full searchable version of the book.
1-878707-55-8 $59.95
Digital Signal Processing Demystified
by James D. Broesch
INCLUDES WINDOWS 95/98 CD-ROM. A
readable and practical introduction to the fundamen-
tals of digital signal processing, including the
design of digital filters.
1-878707-16-7 $49.95
Digital Frequency Synthesis Demystified
by Bar-Giora Goldberg
INCLUDES WINDOWS CD-ROM. An essential
reference for electronics engineers covering direct
digital synthesis (DDS) and PLL frequency synthe-
sis. The accompanying CD-ROM contains useful
design tools and examples, and a DDS tutorial.
1-878707-47-7 $49.95
Bebop to the Boolean Boogie
An Unconventional Guide to Electronics
Fundamentals, Components, and Processes
by Clive “Max” Maxfield
The essential reference on modern electronics,
written with wit and style. Worth the price for the
glossary alone!
1-878707-22-1 $35.00
Modeling Engineering Systems
PC-Based Techniques and Design Tools
by Jack W. Lewis
INCLUDES WINDOWS CD-ROM.Teaches the
fundamentals of math modeling and shows how to
simulate any engineering system using a PC
spreadsheet.
1-878707-08-6 $29.95
Fibre Channel, Second Edition
Connection to the Future
by the Fibre Channel Industry Association
A concise guide to the fundamentals of the popular
ANSI Fibre Channel standard for high-speed
computer interconnection.
1-878707-45-0 $16.95
Visit www.LLH-Publishing.com for great technical print books, eBooks, and more!