201489647 Wireless Receiver Design for Digital Communications

Decades of at the bench experience are collected within this book, providing readers with practical lessons and approaches in radio receiver design for wireless communication systems. Starting with the Basics of RF Engineering this is a virtual replacement for a mentor sharing his knowledge and walking his apprentice along the path toward the design of real world systems.

Wireless Receiver Design for Digital Communications, 2nd Edition is a complete modernization (and necessary re-titling) of the bestselling Radio Receiver Design, published over a decade ago and authored by Kevin McClaning and Tom Vito. It serves as a reference for professional electrical engineers; while chapter-end exercises for each chapter facilitates use as a Masters level textbook in Communication Engineering courses. It is replete with proven concepts, illustrations, design examples, and exercises that help clarify the role of each component within the system design.

KEY FEATURES Extensive chapters on mixers, oscillators, filters, and amplifiers Details all major components related to receiver design including

cascade interaction Provides excellent introductions and technical background on basic

and well as advanced component characteristics Provides a clear pathway from Basics of RF Engineering to real

world system design Exercise solutions and PowerPoint slides available for adopting

instructors

Rather than drag the reader through a lifeless rehash of concepts smothered in equations, McClaning artfully guides the reader through the fundamental aspects, provides a generous collection of practical examples, entertains and illuminates with his war stories. Simply put, this book is a treasure in my canonical library. I never part with it! - James Gitre, Motorola Mobility

Microwave and RF Design: A Systems Approach - by Michael Steer ISBN 9781891121883

Transceiver and System Design for Digital Communications, 3rd Ed. - by Scott BullockISBN 9781891121722

Kevin McClaning graduated from the University of Akron in 1982 with a Bachelors Degree in Electrical Engineering and continued his education at Johns Hopkins, attaining an MSEE in 1986.

He has been the lead architect on five large receiving systems for various government customers and consults on RF matters on a regular basis. He has taught two electrical engineering courses at JHU - Microwave Systems and Components and Wireless Communication Circuits.

No mentoring would be complete without hearing all the war stories. This book is replete with stories (many humorous) of how problems were solved and catastrophes avoided. Find out why the Professor on Gilligans Island almost got it right; how well-intentioned cleaning crews can cause bursts of bit errors; and how a pterodactyl was taken down by radar.

McClaning-7200021 Mccl7200021fm ISBN : 978-1-891121-80-7 July 22, 2011 14:58 i

Wireless Receiver Designfor

Digital Communications

McClaning-7200021 Mccl7200021fm ISBN : 978-1-891121-80-7 July 22, 2011 14:58 ii

McClaning-7200021 Mccl7200021fm ISBN : 978-1-891121-80-7 July 22, 2011 14:58 iii

Wireless Receiver Designfor

Digital Communications

Second Edition

Kevin McClaning

Raleigh, NCscitechpub.com

McClaning-7200021 Mccl7200021fm ISBN : 978-1-891121-80-7 July 22, 2011 14:58 iv

Published by SciTech Publishing, Inc.911 Paverstone Drive, Suite BRaleigh, NC 27615(919) 847-2434, fax (919) 847-2568scitechpublishing.com

Copyright 2012 by SciTech Publishing, Raleigh, NC. All rights reserved.

No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by anymeans, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections107 or 108 of the 1976 United Stated Copyright Act, without either the prior written permission of the Publisher,or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 RosewoodDrive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600, or on the web at copyright.com. Requests to thePublisher for permission should be addressed to the Publisher, SciTech Publishing, Inc., 911 Paverstone Drive, SuiteB, Raleigh, NC 27615, (919) 847-2434, fax (919) 847-2568, or email [email protected].

The publisher and the author make no representations or warranties with respect to the accuracy or completeness ofthe contents of this work and specically disclaim all warranties, including without limitation warranties of tnessfor a particular purpose.

Editor: Dudley R. KayProduction Manager: Robert LawlessTypesetting: MPS Limited, a Macmillan CompanyCover Design: Brent Beckley and Jennifer McClaningPrinter: Sheridan Books, Inc., Chelsea, MI

This book is available at special quantity discounts to use as premiums and sales promotions, or for use in corporatetraining programs. For more information and quotes, please contact the publisher.

Printed in the United States of America10 9 8 7 6 5 4 3 2 1

ISBN: 978-1-891121-80-7

Library of Congress Cataloging-in-Publication Data

McClaning, Kevin, 1959-Wireless receiver design for digital communications / Kevin McClaning. 2nd ed.

p. cm.Rev. ed. of Radio receiver design / Kevin McClaning, Tom Vito. 2000.Includes bibliographical references and index.ISBN 978-1-891121-80-7 (hardcover : alk. paper)

1. RadioReceivers and receptionDesign and construction. 2. Digital communications. I.McClaning, Kevin, 1959- Radio receiver design. II. Title.

TK6563.M38 2012384dc23

2011027275

McClaning-7200021 Mccl7200021fm ISBN : 978-1-891121-80-7 July 22, 2011 14:58 v

Contents

Preface to the Second Edition xiAcknowledgments xiii

1 Radio Frequency Basics 1

1.1 Introduction 11.2 Nomenclature 11.3 Decibels 21.4 Signal Standards 91.5 Frequency, Wavelength, and Propagation Velocity 141.6 Transmission Lines 171.7 Descriptions of Impedance 221.8 S-Parameters 381.9 Matching and Maximum Power Transfer 411.10 Introduction to Radio Frequency Components 461.11 Bibliography 571.12 Problems 58

2 Signals, Noise, and Modulation 63

2.1 Introduction 632.2 A Real-Valued, Ideal Cosine Wave 632.3 Single-Sided Spectra and Complex Basebanding 732.4 Two Noiseless Sine Waves 872.5 Band-Limited Additive White Gaussian Noise 992.6 An Ideal Sine Wave and Band-Limited AWGN 1042.7 The Quadrature Modulator 1092.8 Analog Modulation 1152.9 Digital Modulation 1302.10 Quadrature Modulators, Baseband Filtering,

and Spectrum Control 1622.11 General Characteristics of Signals 1722.12 Summary 174

v

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vi Contents

2.13 Bibliography 1752.14 Problems 175

3 Propagation 179

3.1 Introduction 1793.2 Types of Propagation 1793.3 Propagation Through Free Space 1803.4 Propagation Through a Homogenous Medium 1843.5 Propagation Through a Nonhomogenous Medium 1873.6 Multipath Propagation 1893.7 Bibliography 2053.8 Problems 205

4 Antennas 207

4.1 Introduction 2074.2 Antenna Equivalent Circuits 2084.3 Aperture 2154.4 The Isotropic Radiator 2164.5 Antenna Gain, Beamwidth, and Aperture 2174.6 Bibliography 2264.7 Problems 227

5 Filters 229

5.1 Introduction 2295.2 Linear Systems Review 2305.3 Filters and Systems 2425.4 Filter Types and Terminology 2435.5 Generic Filter Responses 2455.6 Classes of Low-Pass Filters 2525.7 Low-Pass Filter Comparison 2645.8 Filter Input/Output Impedances 2675.9 Transient Response of Filters 2735.10 Band-Pass Filters 2755.11 Noise Bandwidth 2805.12 Butterworth Filters in Detail 2845.13 Miscellaneous Items 2965.14 Matched Filters 297

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Contents vii

5.15 Bibliography 3035.16 Problems 303

6 Noise 309

6.1 Introduction 3096.2 Equivalent Model for a Radio Frequency Device 3096.3 Noise Fundamentals 3116.4 One Noisy Resistor 3196.5 System Model: Two Noisy Resistors 3216.6 Amplifier Noise Model 3276.7 Signal-to-Noise Ratio 3296.8 Noise Factor/Noise Figure 3346.9 Cascade Performance 3406.10 Examining the Cascade Equations 3466.11 Minimum Detectable Signal 3466.12 Noise Performance of Lossy Devices 3476.13 Bibliography 3566.14 Problems 356

7 Linearity 363

7.1 Introduction 3637.2 Linear and Nonlinear Systems 3647.3 Amplifier Transfer Curve 3657.4 Polynomial Approximations 3697.5 Single-Tone Analysis 3717.6 Two-Tone Analysis 3737.7 Distortion Summary 3807.8 Preselection 3827.9 Second-Order Distortion 3837.10 Third-Order Distortion 3867.11 Narrowband and Wideband Systems 3887.12 Higher-Order Effects 3897.13 Second-Order Intercept Point 3907.14 Third-Order Intercept Point 3967.15 Measuring Amplifier Nonlinearity 4027.16 Gain Compression/Output Saturation 4057.17 Comparison of Nonlinear Specifications 4087.18 Nonlinearities in Cascade 410

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viii Contents

7.19 Compression Point 4237.20 Distortion Notes 4237.21 Nonlinearities and Modulated Signals 4267.22 Bibliography 4307.23 Problems 431

8 Mixers 443

8.1 Introduction 4438.2 Frequency Translation Mechanisms 4458.3 Nomenclature 4478.4 Block versus Channelized Systems 4598.5 Conversion Scheme Design 4608.6 Frequency Inversion 4658.7 Image Frequencies 4688.8 Other Mixer Products 4718.9 Spurious Calculations 4758.10 Mixer Realizations 4798.11 General Mixer Notes 4908.12 Bibliography 4938.13 Problems 493

9 Oscillators 507

9.1 Introduction 5079.2 Ideal and Real-World Oscillators 5079.3 Phase Noise 5139.4 Effects of Oscillator Spurious Components 5369.5 Frequency Accuracy 5399.6 Other Considerations 5469.7 Oscillator Realizations 5479.8 Bibliography 5669.9 Problems 567

10 Cascade Design 577

10.1 Introduction 57710.2 Minimum Detectable Signal 57910.3 Dynamic Range 58010.4 Gain DistributionNoise and Linearity in Cascade 58810.5 System Nonlinearities 610

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Contents ix

10.6 TOI Tone Placement 61210.7 Automatic Gain Control 61410.8 Frequency Planning and IF Selection 62110.9 A Typical System 64010.10 Design Examples 64310.11 Bibliography 65210.12 Problems 652

11 Digitizing 667

11.1 Introduction 66711.2 Nyquist-Shannon Theorem 66711.3 Sampling at Discrete Instants in Time 66811.4 Sampling with Discrete Resolution 67611.5 Sources of Spurious Signals 67911.6 Analog-to-Digital Converters 68311.7 Using an ADC in an RF System 68811.8 Bibliography 69311.9 Problems 693

12 Demodulation 695

12.1 Introduction 69512.2 A Transmitter Model 69612.3 The Pulse-Shaping Filter 69712.4 A 16QAM Modulator 69912.5 A Receiver Model 70312.6 Estimation of Carrier Frequency 70712.7 Estimation of Baud Rate 71812.8 Constellation Impairments 72112.9 Bibliography 73012.10 Problems 730

Appendix 733A.1 Miscellaneous Trigonometric Relationships 733A.2 Euler Identities 734A.3 Law of Cosines 734

Selected Answers 735

Index 745

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McClaning-7200021 Mccl7200021fm ISBN : 978-1-891121-80-7 July 22, 2011 14:58 xi

Preface to the Second Edition

Avoid studies of which the result dies with the worker.

Leonardo da Vinci

There have been two paradigm shifts since we wrote the first edition of this book. Thefirst major shift is that the world is now full of digital signals. Commercial televisionand radio have changed from analog to digital formats. Cellular telephone signals aredigital, and even video baby monitors transmit digital signals. Im not sure you can evenfind an analog signal on the air anymore. In a way, this is sad. In the either you get aperfect signal or you get nothing world of the digital broadcast, our children will neverknow the distinct pleasure of watching a snowy ultrahigh frequency (UHF) broadcastor listening to a crackling AM radio station after it has traveled from the other side ofthe nation.

The second paradigm shift since the publication of the first edition is the availabilityof inexpensive digital processing power. As of this writing (early 2011), processing poweris essentially free. In the past, radio receivers performed filtering and demodulation inthe analog world. Many old-school receivers performed demodulation using analog tech-niques that were finicky and required large amounts of circuit board real estate. Controlfunctions such as automatic gain control (AGC) and automatic frequency control (AFC)were realized in the analog world using individual diodes, operational (op) amps, resistorsand capacitors. Today, it makes technical and economic sense to perform these functionsin the digital domain.

We find ourselves in a world where the receiver is primarily a downconverter whosesole purpose is to translate the signal of interest to a frequency and power level that issuitable to be sampled by an analog-to-digital converter (ADC). Similarly, transmittersconvert the users information into complex waveforms using digital signal processing(DSP) techniques. A digital-to-analog converter (DAC) then converts these signals to theanalog domain. The transmitters remaining task is to convert the modulated signal to itsfinal frequency and power level.

However, it is still an analog world, and we must address analog concepts. Ourreceivers require analog filters, and, although filter realizations have changed, filteringconcepts of previous years have remained valid and useful. Oscillator phase noise isessentially low-level, accidental analog phase modulation, and it still limits receiver per-formance in many areas. Linearity, as measured by component compression points andsecond- and third-order intercept points, is very important in a world that contains manysignals existing in proximity.

The format of this book reflects the state of receiver design as it exists in early 2011.I have only lightly updated many chapters from the previous edition as the material therecontinues to be relevant and useful. I added chapters on ADCs and included an overviewof the demodulation of digital signals as it is performed in the digital domain.

xi

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xii Preface to the Second Edition

As in the first edition, I leave you with a quote from Groucho Marx: Why, a 4-year-oldchild could understand this report. Run out and find me a 4-year-old child. I cant makehead or tail out of it (from Duck Soup).

Kevin McClaningAugust 2011

Preface to the First EditionMy sister, the lawyer, said that, if we wanted this book to sell, we needed to put in abloodthirsty clown who lives in the sewers. So far, though, we havent been able to workhim in.

We write as engineers. We dont pretend to be heavy theoretical types, but we dowrite in the hope that this book will be useful. We didnt concentrate on technology buton useful and proven concepts. Some of the most useful books we own were written inthe 1940s by fellows named Frederick Terman, John Kraus, and Mischa Schwartz. Sure,the books contain a lot of information on electron tubes and are a little light on modernfilter design, statistical decision theory, and quadrature modulation, but they are clear andwell written. They were formulated by engineers for engineers, and as such they are stilluseful. We hope that in 20 years people will say, The McClaning/Vito book is a littledated, but its clear and well written.

We made up our minds early to follow one cardinal rule: to be clear. If we havesucceeded, errors should be easy to detect. If we hear about errors in this book (and I hopewe do), I consider it a good thing. It means we have been clear enough to bring doubts intothe readers mind. The reader is thinking about the material and understands it enough tofind inconsistencies.

Wed like to thank our normally noisy children, Chris and Jenny (McClaning) andMandy, Nick, Steve, and James (Vito) for being quiet while we wrote this. Wed also liketo thank our wives, Kitty and Terri, for putting up with us (and not just while we werewriting this book).

Well leave you with a quote from Groucho Marx: From the moment I picked upyour book until I laid it down, I was convulsed with laughter. Someday, I intend readingit (quoted in Life, February 9, 1962).

Enjoy.Kevin McClaning and Tom Vito

August 1998

BibliographyTerman, Frederick Emmons, Radio Engineering, Third Edition, McGraw-Hill Book Company, Inc.,

1947.Kraus, John D., Antennas, McGraw-Hill Book Company, Inc., 1950.Schwartz, Mischa, Information Transmission, Modulation, and Noise, McGraw-Hill Book Com-

pany, Inc., 1959.

McClaning-7200021 Mccl7200021fm ISBN : 978-1-891121-80-7 July 22, 2011 14:58 xiii

Acknowledgments

No book is an island, to horribly paraphrase the poet John Donne. I would like to thankthe following individuals for their contributions to this ponderous tome:

James McPherson and Tom Vito, who both contributed mightily in terms of schedul-ing, reviews, and occasional material

Jeff Houser, Roger Kaul, and Ron Tobin, from Johns Hopkins University, friends withwhom I voyaged through the teaching experience

Edward Sheriff and Brian Sherlock, buddies from work and two of the smartest peopleI know.

Christopher and Jennifer McClaning, my two children and my world

Kathy Sessions, the true love of my life who is always there with support and laughter

I would also like to thank the following brilliant and eccentric people:

Mark Bennett, Roger Gilbert, Paul Hughes, Dan Loveday, and Philip Pring, fromDetica, United Kingdom

Dr. David Waymont and Philip Morris of Waymont Consulting in the United Kingdom;

Terry Fry, of Zeus Technology Systems

xiii

McClaning-7200021 book ISBN : 978-1-891121-80-7 June 14, 2011 16:6 1

C H A P T E R

1Radio Frequency BasicsBegin at the beginning, and go on til you come to the end; then stop.

The King of Hearts, Alice in WonderlandThe journey of a thousand miles begins with a single step.

Loa Tsu

Chapter Outline

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.3 Decibels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.4 Signal Standards . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.5 Frequency, Wavelength, and Propagation Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.6 Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.7 Descriptions of Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.8 S-Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

1.9 Matching and Maximum Power Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

1.10 Introduction to Radio Frequency Components. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

1.11 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

1.12 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

1.1 INTRODUCTION

There are entire books that deal with statistics, logarithms, significant figures, transmissionlines, and s-parameters. We are interested in understanding and designing radio receiversand systems, which requires at least a passing knowledge of all these topics and more.This chapter serves as an introduction to many of the basic concepts we will need in laterchapters.

This is an optional chapter. Many readers will find the information presented herevery basic and may choose to pass this chapter over. For others, it will be required reading.

Some of the purists out there may argue that the information presented here is impre-cise, incomplete, and greatly simplified. Again, the purpose of this chapter is to acquaintthe reader with some of the fundamental material well need later on. For a more completetreatment of any of the topics listed here, refer to the bibliography at the end of the chapter.

1.2 NOMENCLATURE

Throughout this chapter and the following chapters, we will use quantities commonlyexpressed in both linear and logarithmic (or dB) formats. In this book, anytime we express

1


2 C H A P T E R 1 Radio Frequency Basics

a quantity in dB, we will give the quantity a dB subscript. For example, noise figureexpressed in decibels is

FdB (1.1)while noise figure expressed as a linear quantity is

F (1.2)If we really want to make the point, we may write a linear quantity as

FLin (1.3)A major source of mistakes in receiver design is improperly swapping decibel and linearquantities.

1.3 DECIBELS

Engineers originally used decibels to make measurements of human hearing more under-standable. Human hearing covers a 1:10,000,000,000,000 (1:1013) range from the thresh-old of hearing to the threshold of pain. This enormous range makes it hard to plot andanalyze data. Expressing quantities in decibel units makes data comparisons easier andgraphical data more manageable.

Figure 1-1 shows the range of powers that exist on a typical INMARSAT satellite link.The power levels fall over a 1026 range. Since Figure 1-1 uses decibels and a logarithmicscale, the data are very readable.

Other examples of the large ranges system designers must accommodate are as follows:

A typical communications receiver can easily process signals whose input powers varyby a factor of 1012 or 1 trillion.

On its journey to a geosynchronous satellite, a signal might be attenuated by a factorof 1020.

Filters routinely provide attenuation factors of 106 or 1 million.

1.3.1 Definitions

The Bel is defined as

Bel = B = log(

P2P1

)(1.4)

The decibel (or dB) is ten Bels or

dB = 10 log(

P2P1

)(1.5)

where P2 and P1 are signal powers.These are the fundamental definitions of the Bel and decibel. All other decibel-like

quantities are derived from these two definitions. These are power ratios (not voltage orcurrent ratioswell derive those shortly). A quantity described in decibels is fundamen-tally a dimensionless number, although well often label the quantity with a remindersubscript.

McClaning-7200021 book ISBN : 978-1-891121-80-7 July 11, 2011 11:52 3

1.3 Decibels 3

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1018

1016

1014

1012

1010

108

106

104

102

1

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106

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180

160

140

120

100

80

60

40

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100 Mod

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INMARSATTerminal(Transmit)

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FIGURE 1-1 Thepower budget of anINMARSAT satellitelink.

EXAMPLE

dB Power Conversions

A radio frequency (RF) amplifier accepts 3.16E-6 watts from a signal source and supplies its loadresistor with 2.2E-3 watts. Whats the power gain of the RF amplifier? Give your answer in dB.



Solution

Using equation (1.5), we set P2 equal to the amplifiers output power and P1 equal to the amplifiersinput power. So, P1 = 3.16E-6 watts, and P2 = 2.2E-3 watts and

GaindB = 10 log(

P2P1

)

= 10 log(

2.2 1033.16 106

)= 10 log (696.2)= 28.4 dB

(1.6)

The power gain or simply the gain of the amplifier is about 28.4 dB.

1.3.2 dB Math

The logarithmic nature of the decibel allows us to calculate quickly. For example:

A doubling of power is a 3 dB increase:

10 log(2Pin

Pin

)= 3.01 dB 3 dB

(1.7)

Similarly, halving the power is a 3 dB decrease:

10 log(

Pin2Pin

)= 3.01 dB 3 dB

(1.8)

A 4-times increase in power:

10 log(4Pin

Pin

)= 6.02 dB 6 dB

(1.9)

A 4-times power increase is like doubling the power twice and 3 dB + 3 dB = 6 dB. A 4-times decrease in power:

10 log(

Pin4Pin

)= 6.02 dB 6 dB

(1.10)

A 4-times power decrease is like halving the power twice and (3 dB) + (3 dB) =6 dB. A 10-times increase in power:

10 log(10Pin

Pin

)= 10 dB (1.11)

Table 1-1 summarizes these results.


1.3 Decibels 5

TABLE 1-1 Power ratios and their decibelequivalents.

Power Ratio Decibels

106 60.0 dB0.001 30.0 dB0.01 20.0 dB0.1 10.0 dB0.5 3.0 dB1.0 0.0 dB2 3.0 dB3 4.8 dB4 6.0 dB5 7.0 dB8 9.0 dB

10 10.0 dB100 20.0 dB

1,000 30.0 dB106 60.0 dB

TABLE 1-2 Power levels in dBm and their linear equivalents.

Power Level in dBm Linear Power Level

30 dBm 0.001 mW = 1 W20 dBm 0.01 mW = 10 W10 dBm 0.1 mW = 100 W

0 dBm 1.0 mW = 1000 W10 dBm 10 mW20 dBm 100 mW30 dBm 1,000 mW = 1 W40 dBm 10 W50 dBm 20 W

1.3.2.1 Orders of Magnitude1

When we compare a power level to a 1 milliwatt reference, we use the symbol dBm tomean dB above or below a milliwatt. Similarly, we use dBW to describe a power levelthat is reference to 1 watt or dB above or below a watt. Table 1-2 shows that every 10 dBincrease in a quantity represents a linear factor of 10 increase in that quantity (in otherwords, you gain an order of magnitude for every 10 dB increase). Similarly, a decrease of10 dB represents multiplying the quantity by 1/10 (or you lose an order of magnitude).

Given a number in the format of

ABCD.EF dB

1Herein lies a story. The term decade undisputedly refers to a factor of 10. The term order of magnitudewas first used in the world of astronomy, where it refers to the brightness of a star. An astronomical orderof magnitude works out to be the fifth root of 100, and five orders of magnitude add up to 20 dB. In the1960s, engineers stole the term order of magnitude for their own nefarious uses. In the world of RF andin this book, the term order of magnitude means a factor of 10.



the numbers represented by the ABC tell you the order of magnitude, while the D.EFtells you the position in that order of magnitude.

1.3.2.2 Amplifiers, Attenuators, and DecibelsIn wireless system design, we often encounter situations where signals pass through ampli-fiers and attenuators. These calculations are easy to perform if we express all our quantitiesin decibels.

When we multiply in the linear domain, we add in the decibel (or logarithmic) domain.When we divide in the linear domain, we subtract in the decibel domain. Mathematically,the relationship

C = A B CdB = AdB + BdB

(1.12)

is always true.For example, if a signal with 82 dBm of power passes through an amplifier with

15 dB of power gain, the output signal power isPout, dBm = Pin, dBm + Gp, dB

= 82 + 15= 67 dBm

(1.13)

Using the linear quantities, we first convert the input signal power and gain from decibelto linear terms:

82 dBm = 10 log (Pin, mW) Pin, mW = 6.31 109 mW

15 dB = 10 log (G P) G p = 31.6

(1.14)

We multiply and convert back to decibels:Pout = Pin G P

Pout, mW =(6.31 109mW) (31.6)

= 199.5 109mWPout, dBm = 10 log

(199.5 109)

= 67.0 dBm

(1.15)

1.3.2.3 Gains and LossesThe terms gain and loss come up quite a bit when we discuss signals and systems. Itsworth taking a little time to clarify these terms and their concepts.

For example, when a device has a 6 dB gain, the device will accept a signal, multiplyits power by 4 times (or add 6 dB), and present that power to the outside world. If theinput signal power is Pin, then the output signal power is

Pout = 4Pin (1.16)or, in decibels,

Pout, dBm = Pin, dBm + 6 dB (1.17)If a devices has a 6 dB loss, the device accepts a signal, divides the signal power by 4 (orsubtracts 6 dB from the input signal power) and then presents that power to the outsideworld. If the input signal power is Pin, the output signal power will be

Pout = Pin4 (1.18)


1.3 Decibels 7

or

Pout, dBm = Pin, dBm 6 dB (1.19)So, a loss of x dB is equivalent to a gain of x dB. A negative gain is equivalent to apositive loss. We relate gains and losses in equation form as

GaindB = LossdB (1.20)and

GainLin = 1LossLin (1.21)

EXAMPLE

Gains and Losses in Cascades

Figure 1-2 shows a cascade made up of blocks with gain and losses. Find the cascade gain.

Pad Amp BPF Mixer Pad BPF

Gain (dB)

Gain

6

0.251

15

31.6

3

0.501

10

0.100

4.5

0.355

3.2

0.479

Amp

9.5

8.91

FIGURE 1-2 Calculating cascadegain.

Solution

The cascade gain is

G p, cas, dB = G1,dB + G2,dB + G3,dB + G4,dB + G5,dB + G6,dB + G7,dB= (6) + 15 + (3) + (10) + 9.5 + (4.5) + (3.2)= 2.2 dB

(1.22)

The cascade has a gain of 2.2 dB or a loss of 2.2 dB. We can find the cascade gain using only thelinear terms:

G p,cas = G1G2G3G4G5G6G7=

(1

3.98

)(31.6)

(12

) (1

10

)(8.91)

(1

2.82

) (1

2.09

)= 0.60= 2.2 dB

(1.23)

1.3.2.4 Measurement Accuracy, Significant Figures, and DecibelsSuppose we make a power measurement and our meter reads 0.000000 dBm. Is thisnumber of decimal places reasonable? If not, then how many are reasonable?

Assume that the power level is 0.000000 dBm (or 1 mW) exactly. If the power levelincreases by 1 dB to +1 dBm, then

1 dBm = 10 log(PMeasured, mW) PMeasured, mW = 1.2589 mW

(1.24)



which is a 26% change. This is a universal trutha change of 1 dB in any quantityrepresents a change of about 26%. Any reasonable meter would have a good chance ofmeasuring this amount of change accurately.

A 0.1 dB increase in a measured quantity results in

0.1 dB = 10 log (PMeasured, mW) PMeasured, mW = 1.0233 mW

(1.25)

which is about a 2% change. This is measurable with care. The 10th dB digit is meaningfulin a carefully controlled environment.

Looking at the 100th (or 0.0x) dB digit, we find

0.01 dB = 10 log (PMeasured, mW) PMeasured, mW = 1.00233 mW (1.26)

which is a 0.2% change. Under normal engineering circumstances, this amount of accuracyis unusual unless we exercise extreme measures.

Were interested in the answers to several questions regarding significant figures:

Can I reasonably measure the quantity expressed to the number of digits reported?This is a complex question involving the accuracy of the equipment used to make themeasurement, the experiment itself, and the calculations performed with the data.

Do I really care about the least significant digits? For example, often we need enoughgain to overcome the system noise or enough noise figure to see a 125 dBm signal.We do not care about the precise value. The situation is a little like the amount of moneyyou haveit doesnt matter exactly how much you have as long as you have enough.

Components with very tight tolerances tend to be expensive. Testing to tight tolerancesis also expensive. It is important to understand how many significant digits are reallyneeded to avoid requiring more accuracy than is necessary.

Given the component specifications with which I have built my system, can I reallyjustify the number of significant digits Im using? For example, the typical gain andnoise figure specifications for a typical RF amplifier might be

Power Gain: 11.0 dB (0.5 dB)Noise Figure: 2.5 dB (0.5 dB)

Given these specifications (along with the topology of the system), it might be hard tojustify stating that the power gain of the system is 24.22 dB when the gain variation ofjust one of the amplifiers is 0.5 dB.

1.3.3 Decibels, Current, and Voltage

Figure 1-3 shows a simplified schematic diagram of an amplifier. We apply a voltage Vinto the amplifier. The power being dissipated in the input resistor Rsys is

Pin = V2

inRsys

(1.27)


1.4 Signal Standards 9

Vin+

Rsys

Rout

Vout+

RL

AmplifierFIGURE 1-3Simplified schematicof an RF amplifier.

The power delivered by the amplifier to the load resistor RL is

Pout = V2

out

RL(1.28)

Using equation (1.5), the gain of the amplifier in decibels (dB) is

GaindB = 10 log(

PoutPin

)

= 10 log(

V 2out/RLV 2in/Rsys

)

= 20 log( Vout

Vin

)+ 10 log

(RsysRL

)(1.29)

Equation (1.29) is exactly correct. However, electrical engineers arent known especiallyfor their rigor, and last term of equation (1.29) involving Rsys and RL is often ignored,even in systems with wildly differing impedance levels.

Most RF systems have the same impedances on the input and on the output. InFigure 1-3, for example, Rsys, Rout, and RL are the same value (usually 50 or 75 ohms). IfRsys = RL , then

10 log(

RLRsys

)= 0 whenRL = Rsys (1.30)

and

GdB = 20 log( Vout

Vin

)when RL = Rsys (1.31)

Similarly, we can show

GdB = 20 log(

IoutIin

)when RL = Rsys (1.32)

where Iin and Iout are the signal currents flowing in Rsys and RL of Figure 1-3.Equations (1.31) and (1.32) are true only if the source and load resistors are the same.

1.4 SIGNAL STANDARDS

Engineers commonly use the decibel format to express absolute power or voltage levels.We replace P1 in equation (1.5) with some standard, agreed on value. For example, assumewe are building a system where milliwatt (or mW) power levels are common. We might



choose P1 to be 1 mW, and equation (1.5) would becomePReferenced to 1mW = 10 log

(PMeasured, mW

1 mW

)= 10 log (PMeasured, mW)= 10 log

(PMeasured, W

0.001

) (1.33)

where

PMeasured, mW = the power level measured in mWPMeasured, W = the same power level measured in watts

Equation (1.33) defines the dBm standard.Here are several of the most common standard ways of expressing various RF

quantities:

1.4.1 dBm

This is the oldest standard. The telephone company originally set this one up to measuresignal levels on their lines, but now almost everyone has adapted it.

The reference power for dBm is 1 milliwatt (thats where the m in dBm comes fromitstands for power referenced to 1 mW). A quantity expressed in dBm represents an absolutepower level (e.g., Pout = 20 dBm).

0 dBm = 1 mWPdBm = 10 log

(Pwatts0.001

)= 10 log(PmW)

(1.34)

The impedance of a telephone line is about 600 , so 1 mW (or 0 dBm) measures a littleless than 0.775 VRMS. In a 50 system, a 1 mW or 0 dBm sine wave has a root meansquare (RMS) voltage of 0.224 VRMS.

1.4.2 dBW

This is a power measurement referenced to 1 watt. Quantities expressed in dBW are usefulin the world of transmitting equipment. dBW is an absolute unit for expressing power level(e.g., The transmitter was putting out +30 dBW).

0 dBW = 1 wattPdBW = 10 log(Pwatts)

(1.35)

EXAMPLE

dBm and dBW

Find an expression relating dBm and dBW.

Solution

Solving equation (1.34) for Pwatts producesPwatts = (0.001)(10PdBm/10) (1.36)



Equating this to equation (1.35) producesPdBW = 10 log [0.001(10PdBm/10)]

= 10 log(0.001) + 10 log(10PdBm/10) PdBw = PdBm 30

(1.37)

So to convert dBm to dBW, we add 30 dB. For example, 0 dBW is +30 dBmboth equal 1 watt.

1.4.3 dBf

This standard often refers to the sensitivity of consumer receiving equipment. The refer-ence power is 1 fW = 1 femtowatt = 1015 watts. dBf is an absolute power level (usuallya rather small one).

0 dBf = 1 fW = 1 femtowatt = 1015 wattPdBf = 10 log(PfW)

(1.38)

1.4.4 dBV

This standard is dB referenced to 1 voltRMS. Quantities expressed as dBV are not powerreferences unless we specify some impedance (or assumed one from the syntax).

0 dBV = 1 voltRMSVdBV = 20 log(Vvolts,RMS)

(1.39)

Note the multiplier of 20 (which arises because voltage is squared to produce power) andthe missing

10 log(

RLRsys

)(1.40)

term. A quantity expressed in dBV represents an absolute voltage level in a system (forexample, 2 VoltsRMS = 6 dBV) and can represent an absolute power level if the systemimpedance is specified or understood.

1.4.5 dBmV

This is another voltage standard. dBmV is decibels referenced to one millivoltRMS. Thisstandard is common in the video and cable TV industries.

0 dBmV = 1 mVRMSVdBmV = 20 log

( Vvolts,RMS0.001

)= 20 log (VmV,RMS)

(1.41)

Like dBV, dBmV represents an exact voltage measurement (1 voltRMS = 60 dBmV).dBmV can express power if we know the system impedance.

Unfortunately, the previous definitions are often written simply as dB without identi-fying subscripts. This leaves the reader to ascertain just what reference the author had inmind.



EXAMPLE

Power Measurements

Express 24 mW in dBm, dBW, dBf, dBV. and dBmV. Assume a 50 system.

Solution

dBm: Equation (1.34) producesPdBm = 10 log(PmW)

= 10 log(24) = 13.8 dBm (1.42)

dBW: Using equation (1.35), PdBW = 10 log(Pwatts). Since 24 mW = 24 103 watts thenPdBW = 10 log(24 103) = 16.2 dBW.dBf: Using equation (1.38), we know

24 mW = 24 103 watts= 24 1012 fW = 133.8 dBf (1.43)

To find the dBV and dBmV solutions, we must determine the voltage across a 50 resistor thatsdissipating 24 mW. Since P = V2RMS/R with P = 24 103 watts and R = 50 , solving forVRMS = 1.1 VRMS = 1100 mVRMS.dBV: Using equation (1.39), we know

VdBV = 20 log(Vvolts, RMS)= 20 log(1.1) = 0.83 dBV (1.44)

dBmV: Using equation (1.41), we knowVdBmV = 20 log(VmV, RMS)

= 20 log(1100) = 60.8 dBmV (1.45)

EXAMPLE

Power Measurements

A Yamaha RX-700U stereo receiver has a usable sensitivity specification of 9.3 dBf or 0.8 VRMS atthe antenna terminals. What is the input impedance of this stereo receiver?

Solution

Using equation (1.38), we know PdBf = 9.3 dBf, which implies9.3 dBf = 10 log(PfW)

PdBf = 8.51 fWPWatt = 8.51 1015 watts

(1.46)

Since

Pwatts =V 2in,RMSRRcvr,in

(1.47)



and were given that V2in, RMS = 0.8E-6 volts, then8.51 106 watts = (0.8 10

6 VRMS)2

RRcvr, inRRcvr, in = 75

(1.48)

EXAMPLE

Power Amplifiers and Decibels

a. An amplifier is generating an output signal of 30 dBm. If we increase the output power by 3 dB,how much more power (in linear terms) does the amplifier produce?

b. A power amplifier is generating an output signal of +40 dBm. If we increase the output power by3 dB, how much more power (in linear terms) does the amplifier produce?

Solution

a. Equation (1.34) relates the amplifiers output power in dBm to the power in mW:PdBm = 10 log(PmW)

30 dBm = 10 log(PmW) Pout, mW = 103 mW

(1.49)

Increasing this power by 3 dB is equivalent to doubling the power, so the increase is 103 mW (asmall number).

b. Using equation (1.34) again produces+40 dBm = 10 log(Pout, mW)

Pout, mW = 104 mW Pout, W = 10 watts

(1.50)

Doubling the output power represents a 10 watt increase in the amplifiers output power (a bignumber).

The lesson is that a 3 dB power increase is easy to accomplish at relatively small power levelsbecause it doesnt represent much power. At high output power levels, unit decibel increases amountto large increases in output power.

1.4.6 Other Standards

Later, we will need other quantities expressed as decibels. These quantities dont fit thedefinition of a decibel because they arent power ratios, but they are still useful.

1.4.6.1 dBKThis is a temperature measurement. The reference is 1K.2

0 dBK = 1 kelvindBK = 10 log(Temperature inK) (1.51)

2A temperature expressed in kelvin is unitless, and 1K = 273.15C. However, there is the opportunityfor confusion as 100K may refer to 100,000 or a temperature of 100 Kelvin in the absence of context. Toaddress possible confusion, we will refer to temperature expressed in Kelvin as K, while the kilo prefixwill be expressed as k (e.g., 100 kHz).



We will use dBK when we discuss amplifier noise, satellite communications and linkbudgets.

EXAMPLE

Convert 430K to dBK.

Solution

Using equation (1.51)dBK = 10 log(Temperature in K)

= 10 log (430)= 26.3 dBK

(1.52)

1.4.6.2 dBHzThis is a measure of bandwidth. The reference is 1 hertz.

0 dBHz = 1 hertzdBHz = 10 log(Bandwidth in hertz) (1.53)

We will need dBHz for discussions of noise bandwidth and receiver sensitivity.

EXAMPLE

Convert 30 kHz to dBHz.

Solution

Using equation (1.53)dBHz = 10 log(Frequency in hertz)

= 10 log (30, 000)= 44.8 dBHz

(1.54)

1.5 FREQUENCY, WAVELENGTH,AND PROPAGATION VELOCITY

1.5.1 General Case

We relate frequency, wavelength, and propagation velocity by f = v (1.55)

where

= the wavelength of the signalv = the velocity of propagationf = the frequency of the signal


1.5 Frequency, Wavelength, and Propagation Velocity 15

Equation (1.55) is the most general form of the equations relating wavelength to propa-gation velocity. The period T of a wave is

T = 1f (1.56)

1.5.2 Free Space

When a wave is traveling through empty space or through air, the velocity of propagation[v in equation (2.36)] is the speed of light, commonly denoted as c and

c = 2.9979 108 meterssecond

3 108 meterssecond

(1.57)

so equation (1.56) becomes0 f = c (1.58)

where

0 = the wavelength GHz in free space.

EXAMPLE

Wavelength and Frequency

Find the wavelengths of the following frequencies in the atmosphere: 1 MHz (low high-frequency[HF] band), 20 MHz (upper HF band), 100 MHz (middle very high frequency [VHF] band), 150 MHz(the 2 meter band), 300 MHz, 600 MHz, 1 GHz, 3 GHz, 10 GHz, 30 GHz, and 100 GHz.

Solution

Since we are discussing signals that are propagating through the atmosphere, we can assume the waveis moving at the speed of light and that equation (1.58) applies.

TABLE 1-3 The relationship between frequency and wavelength.

Wavelength

Frequency (MHz) Meters Yards Feet Inches

1 300.0 328.0 984.0 11,800.020 15.0 16.4 49.2 591.0100 3.0 3.28 9.84 118.0150 2.0 2.19 6.56 78.7300 1.0 1.10 3.28 39.4600 0.50 0.547 1.64 19.7

1,000 0.30 0.328 0.948 11.83,000 0.10 0.109 0.328 3.94

10,000 0.03 0.0328 0.0948 1.1830,000 0.01 0.0109 0.0328 0.394

100,000 0.003 0.00328 0.00948 0.118

It is convenient to remember that 300 MHz equates to 1 meter of wavelength.



1.5.3 The Speed of Light

The speed of light is 2.998E8 meters/second, and there are 39.37 inches in a meter; thus,we can write

c =(

2.998 108 meterssecond

) (39.37

inchesmeter

)= 11.80 inches

nanosecond(1.59)

or

c 1 ftnanosecond

(1.60)

With less than a 2% error, we can say that light travels about 1 foot in 1 nanosecond (orconversely, it takes about 1 nanosecond for light to travel 1 foot).

EXAMPLE

Satellite Delay Time

A geostationary satellite is 35.863E6 meters (= 22,284 miles or 117.66 106 feet) above the earth.How long does it take a signal to travel from the surface of the earth to the satellite and back to theearth again?

Solution

Since the speed of light = c = 2.9979 108 meters/second, it takes(2)(35.863 106) meters

2.9979 108 meters/second = 0.239 seconds (1.61)to travel the distance. Using our 1 nsec/foot rule of thumb, we find

(2)(35.863 106 meters)(

1 foot0.304 meters

) (1 nsecfoot

)= 0.236 sec (1.62)

1.5.4 Physical Size

The physical size of a system or device expressed in wavelengths is an important quantity.For example:

Wires begin to act like antennas when their physical sizes approach wavelength dimen-sions (typically when their physical dimensions get larger than /10 or so).

Wires begin to look like transmission lines when the physical length of the wire is ofthe same order of magnitude as a wavelength (/15 is the rule of thumb).

If two wires carrying a signal are physically separated by /5, they begin to act like anantenna and will radiate energy into space.

When the manufacturing tolerances of a device approach a tenth of a wavelength, thedevice will begin to misbehave. Connectors will exhibit excessive losses; antennas willexhibit gain and sidelobe variations.

We will encounter this effect many times as we discuss wireless systems. The behaviorof an antenna depends strongly on its physical size in wavelengths. Table 1-4 relatesfrequency to wavelength and physical size:


1.6 Transmission Lines 17

TABLE 1-4 The relationship between frequencyand physical dimensions of an object.

f /2 /20

10 Hz 30,000 km 4,800 km 1,500 km60 Hz 5,000 km 800 km 250 km100 Hz 3,000 km 480 km 150 km400 Hz 750 km 120 km 38 km1 kHz 300 km 48 km 15 km

10 kHz 30 km 4.8 km 1.5 km100 kHz 3 km 480 m 150 m1 MHz 300 m 48 m 15 m

10 MHz 30 m 4.8 m 1.5 m100 MHz 3.0 m 0.48 m 15 cm

1 GHz 30 cm 4.8 cm 1.5 cm10 GHz 3.0 cm 4.8 mm 1.5 mm

where

f = frequency = wavelength/2 = the boundary between antenna near and far fields/20 = antenna effects begin to occur in wires and slots

1.6 TRANSMISSION LINES

Transmission line effects occur when the physical size of a system approaches a wavelengthat the frequency of operation. An equivalent description is that the time a signal takes totravel between components is nonzero.

Figure 1-4 illustrates the point. We drive a 300 MHz signal to two receivers: onelocated 1 inch away from the driver and the second located 7 inches away.

Signals travel more slowly in a transmission line than they travel in free space. Wewill assume the signal travels at 2E8 meters/second so it requires 0.127 ns to travel 1 inch.The signal at point B arrives at t = 0.127 ns. The signal at point C arrives at t = 0.889 ns.The difference in propagation time between A-B and A-C is 0.762 ns. This time is anappreciable fraction of the 3.33 ns period of the 300 MHz clock. The digital logic at pointA and point B would not see the rising clock edge at the same instant. This propagationdelay problem (termed clock skew) is very important in high-speed computers.War StoryTransmission Line Propagation DelayOpen up an old computer and examine the memory printed circuit boards (PCBs). Mostmemory PCBs contain several integrated circuits (ICs) that must be clocked at the sametime. The time-critical signals will enter the memory PCB from the motherboard and thenwill be run to the individual memory ICs. Some lines run in a serpentine, back-and-forthmanner so that the propagation delay to each IC is identical.

When they were first building radars early in World War 2, the designers neededto generate precisely timed, very high-voltage pulses to power their transmitters. Forexample, a system might need a 20 kV, 1 sec pulse. The engineers built special high-voltage transmission lines whose propagation delay times were 1 sec. They would charge



FIGURE 1-4Clock skew in digitallogic caused byunequal line lengths.

F = 300 MHz => T = 3.33 nsec

7 Inches

1 Inch

A

Receiver 1

Transmitter

t (nsec) 3.331.66

0.127 nsec

Receiver 2

B

C

A

Bt (nsec)

t (nsec)

0.889 nsec

C

FIGURE 1-5Transmission linereflection for amatched load. Thesignal incident onthe matched load iscompletelyabsorbed by thematched load.

RL = Z0

Reflected Wave(Nothing)

Z0 = Transmission LineCharacteristic Impedence

Incident Wave

the line up to 20 kV and then connect the line to the high-voltage sink at the instant thatthey needed the pulse. The sink would pull the energy from the transmission line until theline dissipated, that is, for exactly 1 sec.

1.6.1 Characteristic Impedance

Characteristic impedance, denoted Z0, is the most important property of a transmissionline. Well come to think of characteristic impedance as the impedance at which our RFsystems operate.

Terminating a transmission line with a resistor whose value is the characteristicimpedance of the line allows the resistor to absorb completely any signal we push into theline, as in Figure 1-5. If the transmission line is terminated with a resistor whose value isnot Z0, then the load resistor will not dissipate the entire signal. The resistor will absorbsome of the energy, and the rest will be reflected back into the line, as shown in Figure 1-6.These reflections have profound effects on system performance.

A transmission lines characteristic impedance is a function of its physical geometryand the materials used to build it. Well look at two common examples.



RL Z0

Z0 = Transmission LineCharacteristic Impedence

Reflected Wave

Incident Wave

FIGURE 1-6 Transmission line reflection for a mismatched load. Part of the signal incidenton the mismatched load is absorbed by the load, but some of the signal reflects from themismatch and returns to the source.

Z0

d

r

D

FIGURE 1-7 Theconstruction detailsof a coaxial cable.

1.6.1.1 Coaxial CableFigure 1-7 shows the physical configuration of a coaxial cable. The characteristic impedanceis

Z0 = 138R

log(

Dd

)(1.63)

where

R is the permittivity or dielectric constant of the insulatord is the outer diameter of the inner conductorD is the inner diameter of the outer conductor

The characteristic impedance of the cable depends on the permittivity of the insulatorand the relative diameters of the outer conductor and the center conductor.

War StoryEquation (1.63) tells us that the ratio of diameters of the outer conductor to the innerconductor strongly determines the characteristic impedance of a coaxial cable. The char-acteristic impedance of a cable will change if someone steps, kinks, or otherwise physicallydamages the cable. Small changes in the D/d ratio result in large changes in Z0.

1.6.1.2 Twin Lead CableFigure 1-8 shows the physical configuration of a twin lead cable. The characteristicimpedance is

Z0 = 276R

log(2D

d

)(1.64)

whereR is the permittivity or dielectric constant of the insulatord is the diameter of each wireD is the center-to-center distance between the conductors



FIGURE 1-8 Theconstruction detailsof a twin lead cable.

d

Dr

Again, the characteristic impedance of the cable depends on the permittivity of the insulatorand the physical configuration of the two wires.

War StoryEquation (1.64) produces the Z0 of the cable shown in Figure 1-8. The conductors in twinlead are not shielded from the outside world, as is the inner conductor of coaxial cable.Equation (1.64) is entirely different if the cable is near a metal sheet.

One of the authors worked in a popular electronics store in college, where they soldboth coaxial and twin lead TV antenna cable. The return rate on the twin lead cable wasmuch higher than the return rate on the coaxial cable. We eventually reasoned that peoplewould run the cable along their aluminum siding and that the proximity of the metal wouldchange the characteristic impedance of the twin lead more than it would change the Z0of the coaxial cable. Since the antenna was operating into an unmatched characteristicimpedance, signal loss occurred, and the cable was judged to be of poor quality.

1.6.2 Transmission Lines and Pulsed Input Signals

Figure 1-9 shows a transmission line experiment. We are feeding a lossless transmissionline, whose characteristic impedance is Z0, with a very narrow pulse while we vary theload resistor RL . We will plot Vin, the voltage at the input of the transmission line, overtime.

1.6.2.1 Open-Circuit LinePlot (b) of Figure 1-9 shows Vin when the load is an open circuit (RL = ). We observetwo pulses at point A: the source-generated pulse at t = 0 and a second pulse. The secondpulse has the same magnitude as the first pulse but appears at t = td .

We explain the second pulse by considering the transmission line as a simple timedelay. The initial pulse travels through the cable until it encounters the open circuit at theload end. The pulse reflects off the impedance discontinuity and travels back in the directionit came. The second pulse is traveling from the load end of the cable toward the generator.

The time td is the time required to travel down the cable and back again. We can usethis technique to determine the cables propagation velocity.

When the pulse first enters the transmission line, but before the return pulse hashad time to return from the open circuit, the transmission line appears to be a resistorwhose value is its characteristic impedance. Voltage division between RS and the cablescharacteristic impedance Z0 causes the initial input voltage to be one-half the input voltage.

1.6.2.2 Short-Circuit LinePlot (c) of Figure 1-9 shows Vin when we short-circuit the load end of the transmissionline. Everything is the same except that the return pulse is in the opposite polarity. Theshort circuit at the end of the cable causes the pulse to come back inverted.



VS

VS

VS

RS

Vin

+

RL

RL = Infinity

RL = 0

RL = Z0

Z0

A

Vin

Vin

Vin

(a)+2 V

+1 V

+1 V

+1 V

(b)

1 V

+1 V

td

td

td

(No Pulse)

Time

(c)

(d)

Let RS = Z0

PulseGenerator

Time

Time

Time

FIGURE 1-9Transmission lineunder pulsedconditions withvarious values ofloading. We will setVS equal to 2 volts,and we will set thesource impedance,RS, equal to thecharacteristicimpedance of thecable.

1.6.2.3 Z0 Terminated LinePlot (d) of Figure 1-9 shows the effect of terminating the cable with a resistor whosevalue equals the characteristic impedance of the cable. No pulse ever returns because thematched load resistor absorbs all of the energy in the pulse. This is a desirable situationand is one of the reasons we like to keep our system impedances matched.

1.6.3 Propagation Velocity or Velocity Factor

Waves travel slower in a transmission line than they travel in free space. The propagationvelocity in any medium is v.

Velocity factor is the ratio of the propagation velocity of a wave in a transmission lineto the waves velocity in free space or

v f = vc

= t0td

1

(1.65)

wherev f = the velocity factor of the transmission line (0 v f 1)v = the propagation velocity in the transmission linec = the speed of lightt0 = the propagation time in free spacetd = the propagation time in the transmission line



EXAMPLE

Measuring Velocity Factor

We launch an 8 nsec long pulse into a 6 foot long piece of RG-223 coaxial cable. The load end of thecable is an open circuit. The return pulse came back 19.1 nsec after we launched the first pulse into thecable. What is the velocity factor of this piece of cable?

Solution

The 19.1 nsec time between pulses is the time required for the pulse to travel down the cable andreturn. The one-way travel time is 19.1/2 = 9.55 ns. Our pulse traveled 6 feet in 9.55 ns. Usingequation (1.60), our pulse would travel the 6 feet in 6 ns if it were traveling in free space, so thevelocity factor of the cable is

v f = t0td

= 6 nsec9.55 nsec

= 0.63(1.66)

1.6.4 Guide Wavelength

Waves move slower than the speed of light inside a transmission line, so the wavelengthinside the transmission line (the guide wavelength = g) will be shorter than the wave-length in free space. The guide wavelength is

v f = g0

(1.67)where

v f = the velocity factor of the transmission line (0 v f 1)g = the guide wavelength in the transmission line0 = the wavelength in free space

Equation (1.58) relates 0 to the frequency.

1.7 DESCRIPTIONS OF IMPEDANCE

The notion of impedance permeates radio frequency design. Many of the properties of oursystems are related directly to the terminal impedances of our components, referenced tothe system impedance. The environment in which were operating determines the systemimpedance, and it is always some standardoften 50 or 75 .

The system impedance is set by the impedances of the test gear we have availableand by the characteristic impedance of the transmission lines with which we connect thecomponents.

We have many ways to describe impedance.

1.7.1 Reflection Coefficient

Open and short circuits are extreme conditions. If we place an arbitrary complex load atthe load end of a transmission line, the magnitude of the reflected pulse will be

VReflected = VIncident (1.68)


1.7 Descriptions of Impedance 23

Rs

Z0

Vs

PulseGenerator

Launched byPulse

GeneratorVIncident

encounters theload and

generates areflectionReflected back

to the source VReflected =VIncident

VIncident

Rs RL

FIGURE 1-10Transmission linereflection coefficient.

where = complex reflection coefficient. The reflection coefficient will be a complex

number if the load impedance is complex.VIncident = the complex voltage incident on the load. In other words, its the magnitude

of the pulse we sent into the transmission line. VIncident travels from thesource to the load.

VReflected = the complex voltage reflected off the load and back into the transmissionline. VReflected travels from the load to the source.

See Figure 1-10. We can show

= ZL Z0ZL + Z0 1 || +1 (1.69)

where = the complex reflection coefficientZL = the complex impedance that we are comparing to the characteristic impedance Z0Z0 = the characteristic impedance of the system in which we are operatingFigure 1-11 shows a graph of real load impedance vs. the reflection coefficient in a

50 system. At low values of RL , the reflection coefficient approaches 1; most of thevoltage we send down the cable reflects back toward the source (although the magnitudeof the pulse is reversed). When RL is large, approaches +1, and most of the voltagereflects off the load but the sign stays the same.

When the value of the load resistor equals the transmission lines characteristicimpedance, the reflection coefficient equals zero. None of the energy we send into thecable comes backall of the energy dissipates in the load resistor.

1.7.1.1 Transmitting SystemsIn transmitting systems, we send power into an antenna so the power will radiate away intofree space. If the impedance of the transmitting antenna isnt matched to the impedance ofthe transmission line, some of the energy we are sending into the cable reflects off antennaand is not radiated into space. Weve lost efficiency.



FIGURE 1-11Transmission linereflection coefficientversus loadresistance for a 50 system. Loadresistance is realvalued loadimpedance.

1001

0.8

0.6

0.4

0.2

0

Load Resistance (Ohms)

Reflection Coefficient vs. Load Resistance (Z0 = 50 )

Ref

lect

ion

Coe

ffic

ien

t0.2

0.4

0.6

0.8

1

101 102 103Z0

1.7.1.2 Receiving SystemsIn receiving systems, we feed a receiver with an antenna. Any impedance mismatchbetween the antenna and the receiver will cause energy to reflect off the mismatch. Signalenergy will be lost, and our receiving system will lose sensitivity.

EXAMPLE

Reflection Coefficient

Find the reflection coefficient for the following load impedances in a 50 system:

a. 10 b. 125 c. 50 d. 20 j40 e. 50 + j20

Solution

Liberally applying equation (1.69) producesa.

= 10 5010 + 50 = 0.667 (1.70)

A real impedance less than Z0 produces a real, negative reflection coefficient.b.

= 125 50125 + 50 = 0.429 (1.71)

A real impedance greater than Z0 produces a real, positive reflection coefficient.



c. = 50 50

50 + 50 = 0 (1.72)

A real impedance equal to Z0 produces a zero valued reflection coefficient.d.

= (20 j40) 50(20 j40) + 50

= 30 j4070 j40

= 0.077 j0.615= 0.620 97

(1.73)

A complex impedance produces a complex reflection coefficient.e.

= (50 + j20) 50(50 + j20) + 50

= j20100 + j20

= 0.039 + j0.192= 0.196 79

(1.74)

EXAMPLE

Mismatched Transmission Line

Figure 1-12 shows a signal source driving a transmission line.

Rs = 15 Z0 = 50

Vs = Step Function= 0 for t < 0= 1 for t > 0

RL = 125

FIGURE 1-12 Example of a mismatchedtransmission line. The transmission line ismismatch on both the source and load ends.

The signal source is a step function, that is,VS = 0 volts for t < 0VS = 1 volt for t 0

(1.75)

and Z0 = 50 , RS = 15 , and RL = 125 . The time for a wave to propagate from one end of thetransmission line to the other is seconds. Plot the voltage versus distance over the length of the cablefor t = 0.3, 1.3, 2.3 , etc.

Solution

We use equation (1.69) to find the source and load reflection coefficients. The source reflectioncoefficient is

S = RL Z0RL + Z0 =15 5015 + 50 = 0.538 (1.76)



The load reflection coefficient is

L = RL Z0RL + Z0= 125 50

125 + 50 (1.77)= 0.429

Before any reflections propagate down the cable and back again, the transmission line looks like aresistive element with a value of Z0. So when the generator turns on at t = 0, the transmission linelooks like a 50 resistor to the source. Using voltage division, the magnitude of the incident wavepropagating down the transmission line is

VIncident = VS Z0RS + Z0= (1 volt) 50

15 + 50 (1.78)= 0.770 volts

The graph labeled t = 0.3 in Figure 1-13 shows the initial pulse traveling from the sourceto the load.

At t = , the incident pulse encounters the load resistor. The load resistor absorbssome of the incident pulses energy and reflects the rest. Equation (1.68) relates the incidentand reflected voltage, and we know the magnitude of the incident voltage is 0.770 V whilethe load reflection coefficient is 0.429. The reflected voltage is

VReflected = L VIncident= (0.429)(0.770V ) = 0.330 V (1.79)

Figure 1-13 shows the state of the transmission line at time t = 1.3 .

FIGURE 1-13Snapshots ofvoltages present ona mismatchedtransmission line attimes t = 0.3,t = 1.3, t = 2.3and t = 3.3 .

Z0 = 50

Vs Rs = 15 RL = 125

s = 0.538 L= 0.429

t = 0.3

t = 1.3

t = 2.3

t = 3.3

0.330 V

0.178 V

0.076 V

0.770 V

0.770 V

0.922 V 1.10 V

0.846 V0.922 V



The next interesting event occurs at t = 2 . The 0.330 V wave traveling from theload encounters the source. At this discontinuity, the incident voltage is 0.330 volts, andthe reflection coefficient is 0.538. Equation (1.68) gives us the reflected voltage

VReflected = S VIncident= (0.538)(0.330V ) (1.80)= 0.178 V

Figure 1-13 shows the state of the transmission line at time t = 2.3 .At t = 3 , the 0.178 volt waveform transient is incident on the load. The value of

the reflected wave isVReflected = L VIncident

= (0.429)(0.178V ) (1.81)= 0.076 V

Figure 1-13 shows the state of the transmission line at time t = 3.3 .If we continue this process, we will find that the voltage on the transmission line

eventually settles down to what we would expect from a simple direct current (DC)analysis. In other words, after all the transients have died down, the final voltage on thetransmission line will be

VFinal = VDC = VS RLRL + RS= (1 V)

( 12515 + 125

)(1.82)

= 0.893 VWe can arrive at this result either through simple DC analysis or by summing up thereflections on the transmission line until they die out. The result is the same.

1.7.1.3 Complex Reflection CoefficientWe have spent our time examining resistive cable terminations. The astute reader maywonder what happens if we decided to terminate a cable with an inductor or capacitor.

The answer is nothing special. Everything we have discussed still applies, except thatthe reflection coefficient has now become a complex number; it now has a magnitude and aphase angle. With a resistive load, the reflection coefficient is always real. With a complexload, the time-domain examples have certainly become harder to visualize because of thefrequency-dependent nature of the load, but the effects are identical.

Since the impedance of a capacitor or inductor changes with frequency, the complexreflection coefficient will also change with frequency. Different frequency componentsof a signal will experience different phase and amplitude changes as they reflect off thecomplex load.

For a complex load, ZL = RL + jXL , the reflection coefficient is = ||

= ZL Z0ZL + Z0

(1.83)

whereZL = RL + jXL = the complex load (resistance and reactance) = the complex reflection coefficient



|| = the magnitude of the complex reflection coefficient = the angle of the complex reflection coefficient

A complex means that the incident wave suffers a phase change as well as a magnitudechange when it encounters the mismatch.

The equations we are developing for return loss and voltage standing wave ratio(VSWR), for example, will apply for complex reflection coefficient. We have been carefulto specify the magnitude wherever it was required. If the magnitude is not specified, usethe complex reflection coefficient and be prepared for a complex result.

1.7.2 VSWR

The notion of VSWR as a description of impedance arises directly from the effects ofmismatched impedances on signals in transmission lines. The term evolved into a commonexpression of impedance even when there is no transmission line immediately present.

1.7.2.1 Transmission Lines and Sine Wave Input SignalsWe have discussed the behavior of transmission lines under pulsed input signal conditions.Our discussions have centered on the behavior of pulses as they move up and down thetransmission line and when they encounter impedance discontinuities.

Since a pulse is made up of a series of sine waves, we can argue that we will observethe same effects if we launch a continuous RF carrier (i.e., a sine wave) down the trans-mission line. Reflections occur, and the mathematical description is identical to the pulsedconditions. However, because of the continuous nature of the sine wave, the effects willnot be as intuitive as with the pulsed case.

Figure 1-14 shows an example. The voltage generator VS is a sine wave source thatturns on at time t = 0. The system is completely mismatched (i.e. RS = Z0 = RL). Whenwe turn the signal generator on at t = 0, the sine wave first encounters the mismatchbetween the signal generators source impedance and the impedance of the transmissionline. This effect acts to reduce the voltage incident on the line.

The sine wave travels down the line until it encounters the load resistor. The loadabsorbs some of the signal energy and reflects the rest according to the equation

VReflected = VIncident (1.84)The energy that reflects off the load travels back along the line toward the source. If theload is complex, then is complex and the sine wave experiences a phase shift and amagnitude change.

At the source end of the line, the signal from the load splits into two pieces again asit encounters the mismatched source impedance. Some energy is absorbed by the sourcewhile the rest is reflected and sent back again toward the load.

This process of absorption and reflection at each end of the transmission line continuesuntil the transients die out and the line reaches a steady-state condition. The voltage presentat any particular point on the transmission line is the sum of the initial incident wave andall of the reflections.

FIGURE 1-14 Amismatchedtransmission linewith a sinusoidalinput voltage.

Rs

RL 0 for t < 0

sin (0t) for t 0=

Z0

Vs

Rs Z0 RL



Rs = Z0Z0

Vs

Sine WaveSource

Vi

Vr = Vi

RL

FIGURE 1-15 A mismatched transmission line under sinusoidal drive showing the incidentand reflected voltages. The incident sine wave, Vi , travels toward the load and reflects from themismatched load resistor, forming Vr = Vi . The reflected signal, Vr , is completely absorbedby the matched source resistor Rs.

The mathematical expression describing this process is quite complicated, involvingan infinite series. To simplify our analysis, we will examine a transmission line with anarbitrary load resistance but with a matched source resistance (Figure 1-15). When weassume RS = Z0, we eliminate the reflections caused by a mismatch at the source end ofthe cable. The voltage at any physical point on the line is due to only the incident waveand one reflection from the load.

1.7.2.2 Voltage Minimums, Voltage MaximumsFigure 1-16 and Figure 1-17 show the magnitudes of the sine waves we will observe alonga length of transmission lines for several load mismatch conditions. Figure 1-16 describesloads that are greater than Z0 (or 50 , in this case), while Figure 1-17 describes thesituation when the loads are less than 50 .

0.5

1

1.5

2

2.5

1.00

3

0.9 0.8 0.7 0.6 0.5

Wavelengths From Load

VSWR = 10 (RL = 5 )

VSWR = 5 (RL = 10 )

VSWR = 2 (RL = 25 )

VSWR = 1 (RL = 50 )

Rel

ativ

e A

mpl

itu

de

0.4 0.3 0.2 0.1 Load

Voltage vs. Physical Position (Lo-Z Load) FIGURE 1-16VSWR. Themagnitude of thevoltages presenton a mismatchedtransmission linewhen ZL > Z0.



FIGURE 1-17VSWR. Themagnitude of thevoltages present ona mismatchedtransmission linewhen ZL < Z0.

EXAMPLE

Transmission Line Voltages (Hi-Z Load)

Assume we are operating a 50 transmission line with a 250 load resistance. What are thetime-domain voltages we would observe at the following distances from the load?

a. 0.10 Gb. 0.25 Gc. 0.40 Gd. 0.50 Ge. 0.60 Gf. 0.65 G

Solution

Reading from the RL = 250 (or VSWR = 5) curve of Figure 1-16, we finda. 1.37 cos(t + 1)b. 0.37 cos(t + 2)c. 1.37 cos(t + 3)d. 1.65 cos(t + 4)e. 1.37 cos(t + 5)f. 1.00 cos(t + 6)where

= 2 f = the frequency of operationn = a phase constant depending on the frequency and upon the distance from the source to the load



EXAMPLE

Transmission Line Voltages (Low-Z Load)

Assume we are operating a 50 transmission line with a 10 load resistance. What are thetime-domain voltages we would observe at the following distances from the load?

a. 0.10 Gb. 0.25 Gc. 0.40 Gd. 0.50 Ge. 0.60 Gf. 0.65 G

Solution

Reading from the RL = 10 (or VSWR = 5) curve of Figure 1-17, we finda. 0.90 cos(t + 1)b. 1.35 cos(t + 2)c. 0.90 cos(t + 3)d. 0.65 cos(t + 4)e. 0.90 cos(t + 5)f. 1.12 cos(t + 6)where

= 2 f = the frequency of operationn = a phase constant depending on the frequency and upon the distance from the source to load

Given a line with a mismatched load, well define the largest voltage present on the lineas Vmax and the smallest voltage present on the line as Vmin or

Largest voltage on the T-Line Vmax cos(t + 1)Smallest voltage on the T-Line Vmin cos(t + 2) (1.85)

The values of Vmax and Vmin change with the value of the load resistor.When the line is matched (when RL = Z0), then Vmax = Vmin. When the line is not

matched, Vmax = Vmin. As the loads reflection coefficient increases (i.e., as the match getsworse), Vmax and Vmin become increasingly different.1.7.2.3 VSWRWe refer to the ratio of the voltage maximum to the voltage minimum as the voltagestanding wave ratio of the line:

VSWR = VmaxVmin

1 (1.86)

The concept of VSWR comes about naturally from our observations of physical transmis-sion lines and from theory. At one time, VSWR was a very common unit of impedance



FIGURE 1-18Measuring VSWR ona transmission line.

measurement because it was relatively easy to determine with simple equipment and with-out disturbing the system under test. Although it is derived from transmission line effects,VSWR is simply a measure of impedancejust like the reflection coefficient ().1.7.2.4 VSWR RelationshipsA little algebra with Vmax, Vmin, , Z0, and RL reveals the following relationships:

VSWR = 1 + ||1 ||1 VSWR 0 || 1 (1.87)

and

|| = VSWR 1VSWR + 1 (1.88)

Figure 1-16 and Figure 1-17 show another feature of the voltage peaks and valleys. AsFigure 1-18 shows, the distance between two adjacent voltage maximums or two adjacentvoltage minimum is G/2 where G is the wavelength in the transmission line or theguide wavelength. Also, the distance between a voltage minimum and its closest voltagemaximum is G/4.

War Story: VSWR and Microwave OvensWe can think of a microwave oven as a transmission line system. The energy source is themagnetron tube inside of the microwave. The transmission line is the cavity in which weplace the food to be heated, and the load is the water in the food.

As long as we have a load on the transmission line (i.e., we have food in the oven),energy transfers from the magnetron tube through the waveguide transmission line andinto the food. We have a low VSWR inside of the microwave oven, and all is well.

However, if we operate the oven without an adequate load (i.e., empty), then the ovenoperates under high VSWR conditions. Areas of high voltages (large electric fields) insidethe cavity will stress the magnetron tube. Also, since the energy from the magnetron meetsa load with a high VSWR (or, equivalently, a high reflection coefficient), all of the energysent out by the magnetron will bounce off the poor load and travel back into the tube. Themagnetron now has to dissipate a large amount of heat and it can fail.

Even when we operate a microwave oven with food in the cavity, the load is stillpoorly characterized. In other words, if we think of a microwave oven and its load as atransmission line with a source and a load, the load is not exactly Z0. After all, what is theimpedance of a bologna sandwich? The system exhibits a nonunity VSWR.



From a practical point of view, this effect causes hot and cold spots in the oven. Thehot spots are areas of high field energy, whereas the cold spots are areas of low field energy.Food that lies in a hot spot heats quickly, while food in a cold spots heats slowly.

To combat this problem, manufacturers often arrange to rotate the food in the oven.Since the rotating food alternately passes through hot and cold spots in the cavity, the foodheats more evenly. Another solution is to install a mode stirrer in the cavity. The modestirrer is a metal propeller-like structure that rotates slowly in the cavity. It reflects theenergy from the magnetron in different directions as it moves. This changes the positionsof the hot and a cold spot in the oven, and the effect is to heat the food more evenly.

War Story: VSWR and Screen Room TestingWe must often perform sensitive electronic tests in a screen room. A screen room usuallytakes the form of a large metal room that shields the inside from the ambient electromag-netic fields on the outside of the room. The room has metal walls, doors, and ceilings withthe appropriate fixtures to get power and air into the room.

Screen rooms are commonly used to test antennas or to measure the electromagneticradiation emanating from a piece of equipment. Such tests are necessary to make sure onepiece of equipment wont interfere with another.

One of the first questions we ask after constructing a screen room is how muchshielding the room supplies. There are many such tests to verify the performance.

In one test, the operator places a transmitter inside the room, closes all the doors, thenchecks for radiation from the transmitter on the outside of the room.

Like the microwave oven example, we have to be concerned with the load that thetransmitter experiences and with the VSWR inside the room. In an empty metal room,the transmitter effectively has no load, and we will experience a high VSWR conditioninside the screen room. The effect of the high VSWR is to place large electromagneticfields in some areas of the room and small electromagnetic fields in other areas of theroom. If an area of small electromagnetic field happens to fall in an area where the screenroom is leaky, then you may not detect the leak. If the area of strong electromagnetic fieldfalls in the neighborhood of a good joint, you may detect that the joint is leaky due to theexcess field placed across the joint by the high VSWR condition in the room.

Commercial gear gets around this problem by sweeping the transmitter slightly infrequency. The positions of the voltage maximums and minimums on a transmission linedepend on the operating frequency (remember that the distances between voltage maximaand minima were G/2). When we change the frequency, we change the position of thevoltage maxima and minima, so we hope to obtain a more realistic measurement of theattenuation of the screen room.

1.7.2.5 Return LossLike VSWR and reflection coefficient, return loss is a measure of impedance with respectto the characteristic impedance of a transmission line.

We have discussed launching pulses down a transmission line. The load absorbedsome of the pulses energy, while some of the energy was reflected back toward the source.Return loss characterizes this energy loss in a very direct manner (see Figure 1-19).

When a known voltage is incident on a load (VIncident), we can calculate an equiva-lent power incident on the load (PIncident). Similarly, when voltage reflects from a load(VReflected), that voltage represents power reflected from the load (PReflected).



FIGURE 1-19Measuring returnloss.

PIncident

PReflected

Radiate a known amount of power intothe cable. That power is incident

on the load

Measure the power which returns fromthe load

RL

The definition for return loss is

Return LossdB = 10 log(

PReflectedPIncident

) 0 dB (1.89)

We often measure return loss directly by launching a known amount of power into atransmission line (PIncident) and measuring the amount of power that reflects off the load(PReflected). Return loss measures the power lost when a signal is launched into a transmis-sion line.

A little algebraic manipulation reveals

Return LossdB = 20 log VReflectedVIncident

= 20 log |L | (1.90)= 20 log

ZL Z0ZL + Z0

Figure 1-20 shows a plot of return loss versus load resistance for a 50 system. Whenthe load resistance is very small or very large, most of the power we launch into thetransmission line reflects off the load. Since PReflected PIncident, the return loss is near0 dB. This is a poorly matched condition.

FIGURE 1-20Return loss versusload resistance in a50 system.

30

25

20

15

10

5

0

Load Resistance (Ohms)

Return Loss vs. Load Resistance (Z0 = 50 )

Ret

urn

Los

s (d

B)

100 101 102 103Z0



When the load resistance is approximately equal to the system characteristic impedance(Z0, which is 50 in this case), most of the power we launch into the line is absorbed bythe load; very little returns. The return loss is a large negative number. A large negativenumber indicates a good match.

EXAMPLE

Reactive Terminations

Figure 1-21 shows a transmission line with a reactive termination. Find the reflection coefficient, returnloss, and VSWR for this load. Assume a 50 system and that the frequency of operation is 4 GHz.

Z0 = 50

10

3 pF

FIGURE 1-21 Transmission line with a complex load.

Solution

The reactance of the load at 4 GHz is 10 j13.3 or 16.6 53. Using equation (1.69) withZL = 10 j13.3 and Z0 = 50 produces

= (10 j13.3) 50(10 j13.3) + 50

= 40 j13.360 j13.3 (1.91)

= 0.589 j0.352= 0.686 149

Note that the reflection coefficient is complex because of the complex load. Since the load impedancechanges with frequency, the complex reflection coefficient changes with frequency.

Equations (1.88) and (1.90) relate the complex reflection coefficient to the VSWR and return loss:

= (10 j13.3) 50(10 j13.3) + 50

= 40 j13.360 j13.3 (1.92)

= 0.589 j0.352= 0.686 149

andReturn LossdB = 20 log(||)

= 20 log(|0.686|) (1.93)= 3.2 dB



1.7.2.6 Mismatch LossThe voltage reflecting off a mismatched load at the end of a transmission line representspower loss. If the load were perfectly matched to the transmission line, we would transferall of the available incident signal power to the load resistor. However, since our loadisnt matched to the transmission lines characteristic impedance, we dont transfer all theavailable power to the load. This is mismatch loss, which we define as

Mismatch Loss = PAvailable PReflectedPAvailable

= PDeliveredPAvailable

(1.94)

where

PAvailable = the power delivered to the matched loadPDelivered = the power delivered to the unmatched loadPReflected = the power reflected off the unmatched load

We know

PAvailable = V2

IncidentZ0

(1.95)

and

PReflected =V 2Reflected

Z0(1.96)

Combining equations (1.94) through (1.96) with equation (1.68) produces

Mismatch LossLoad =V 2Incident/Z0 V 2Reflected/Z0

V 2Incident/Z0

= V2

Incident V 2ReflectedV 2Incident

= V2

Incident (|L |VIncident)2V 2Incident

= 1 |L |2

(1.97)

Mismatch loss is yet another way of describing the relationship between the characteristicimpedance of a transmission line and its load.

We usually specify mismatch loss in decibels

Mismatch LossLoad,dB = 10 log(1 |L |2) (1.98)1.7.2.7 Source and Load MismatchesWe can experience mismatch at both the source and load ends of a transmission line. Whenwe have a source mismatch, we will not transfer the maximum amount of power into theline. The mismatch loss due to source mismatch is

Mismatch LossSource,dB = 10 log(1 |S|2) (1.99)



EXAMPLE

Mismatch Loss

Find the mismatch loss (in decibels) for the following resistors in a 50 system:a. 10 ( = 0.667),b. 25 ( = 0.333),c. 50 ( = 0),d. 100 ( = 0.333)e. 250 ( = 0.667)f. 20 j100 ( = 0.530 j0.671 = 0.855 51.7).

Solution

Liberally applying equation (1.9

201489647 Wireless Receiver Design for Digital Communications

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Transcript of 201489647 Wireless Receiver Design for Digital Communications