–a introduction: waves and phasors - higher education · ﬁrst electric battery, by alessandro...

1Introduction:

Waves and Phasors

Overview

1-1 Dimensions

1-2 T

1-3 T

1-4 T

1-5 Review

1-6 Review

Historical Timeline

1-1 Dimensions, Units, and Notation

1-2 The Nature of Electromagnetism

1-3 Traveling Waves

1-4 The Electromagnetic Spectrum

1-5 Review of Complex Numbers

1-6 Review of Phasors

C H A P T E Rφ0 = π/4 φ0 = –π/4

TT2

3T2

y

–A

A

Leads ahead ofreference wave Lags behind reference wave

Reference wave (φ0 = 0)

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OVERVIEW

Liquid crystal displays have become integral parts ofmany electronic gadgets, from alarm clocks and cellphones to laptop computers and television systems.LCD technology relies on special electrical and opticalproperties of a class of materials known as liquidcrystals, which are neither pure solids, nor pure liquids,but rather a hybrid of both. The molecular structure ofthese materials is such that when light travels throughthe material, the wave polarization of the emerging lightdepends on whether or not a voltage exists across thematerial. Consequently, when no voltage is applied,the exit surface appears bright, and conversely, whena voltage of a certain level is applied across the LCDmaterial, no light passes through it, resulting in a darkpixel. The in-between voltage range translates into arange of grey levels. By controlling the voltage acrosseach individual pixel in a two-dimensional array ofpixels, a complete image can be displayed (Fig. 1-1).Color displays are composed of three subpixels with red,green, and blue filters. The wave-polarization behaviorin a LCD is a prime example of how electromagnetics isat the heart of electrical and computer engineering.

The subject of this book is applied electromagnetics,which encompasses the study of electric and magneticphenomena and their engineering applications, underboth static and dynamic conditions. Primary emphasisis placed on the fundamental properties of time-varying(dynamic) electromagnetic fields because of theirgreater relevance to practical problems in manyengineering disciplines, including microwave and opticalcommunications, radar systems, bioelectromagnetics,and high-speed microelectronics, among others. We shall

C. 2-D array

UnpolarizedLight

EntrancePolarizer

ExitPolarizer

Two-DimensionalPixel Array

Figure 1-1: Wave-polarization principle in a liquid crystaldisplay (LCD).

study wave propagation in guided media, such as coaxialtransmission lines, optical fibers and waveguides; wavereflection and transmission at the interface betweendissimilar media; radiation by antennas, and severalother related topics. The concluding chapter is intendedto illustrate a few aspects of applied electromagneticsthrough an examination of design considerationsassociated with the use and operation of radar sensorsand satellite communication systems.

3




4 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS

We begin this chapter with a historical chronologyof electricity and magnetism. Next, we introduce thefundamental electric and magnetic field quantities weuse in electromagnetics, as well as their relationships toeach other and to the electric charges and currents thatgenerate them. The laws governing these relationshipsconstitute the basic infrastructure we use in the study ofelectromagnetic phenomena. Then, in preparation for thematerial presented in Chapter 2, we provide short reviewsof three topics: traveling waves, complex numbers, andphasor analysis. Although the reader most likely hasencountered these topics in circuit analysis or otherengineering disciplines, short reviews of the properties oftraveling waves and of the convenience phasor notationprovides should prove useful in solving time-harmonicproblems.

Historical TimelineThe history of electromagnetics may be divided into twooverlapping eras. In the classical era, the fundamentallaws of electricity and magnetism were discovered andformulated. Building on these fundamental formulations,the modern era of the past 100 years, characterizedby the introduction of a wide range of engineeringapplications, ushered the birth of the field of appliedelectromagnetics, the topic of this book.

EM in the Classical EraChronology 1-1 (pages 6 and 7) provides a timeline forthe classical era. It highlights those inventions and dis-coveries that have impacted the historical development ofelectromagnetics in a very significant way, albeit that thediscoveries selected for inclusion represent only a smallfraction of the many scientific explorations responsiblefor our current understanding of electromagnetics. As weproceed through the book, we will observe that some ofthe names highlighted in Chronology 1-1, such as thoseof Coulomb and Faraday, appear again later as we discussthe laws and formulations named after them.

The attractive force of magnetite was reported by theGreeks some 2800 years ago. It was also a Greek, Thalesof Miletus, who first wrote about what we now call staticelectricity; he described how rubbing amber caused itto develop a force that could pick up light objects suchas feathers. The term electric first appeared in print in∼1600 in a treatise on the (electric) force generated byfriction, authored by the physician to Queen Elizabeth I,William Gilbert.

About a century later, in 1733, Charles-Francois duFay introduced the concept that electricity consists oftwo types of “fluids,” one positive and the other negative,and that like-fluids repel and opposite-fluids attract. Hisnotion of fluid is what we today call electric charge.Invention of the capacitor in 1745, originally called theLeyden jar, made it possible to store significant amountsof electric charge in a single device. A few years later, in1752, Benjamin Franklin demonstrated that lightningis a form of electricity. He transferred electric chargefrom a cloud to a Leyden jar via a silk kite flown ina thunderstorm. The collective 18th century knowledgeabout electricity was integrated in 1785 by Charles-Augustin de Coulomb, in the form of a mathematicalformulation characterizing the electrical force betweentwo charges in terms of the strengths and polarities of thecharges and the distance between them.

The year 1800 is noted for the development of thefirst electric battery, by Alessandro Volta, and 1820 wasa banner year for discoveries about how magnetism isinduced by electric currents. This knowledge was put togood use by Joseph Henry, who developed one of theearliest designs for electromagnets and electric motors.Shortly thereafter, Michael Faraday built the first electricgenerator (the converse of the electric motor). Faraday,in essence, demonstrated that a changing magnetic fieldinduces an electric field (and hence a voltage). Theconverse relation, namely that a changing electric fieldinduces a magnetic field, was proposed by James ClerkMaxwell in 1873 when he introduced his four (now)




1-1 DIMENSIONS, UNITS, AND NOTATION 5

famous equations. Maxwell’s equations represent thefoundation of classical electromagnetic theory.

Maxwell’s theory, which predicted a number ofproperties for electromagnetic waves, was not fullyaccepted by the scientific community at that time, notuntil those properties were verified experimentally withradio waves by Heinrich Hertz in the 1880s. X-rays,another member of the EM family, were discovered in1895 by Wilhelm Roentgen. On the applied side, NikolaTesla was the first to develop the a-c motor, considered amajor advance over its predecessor, the d-c motor.

Despite the advances made in the 19th century in learn-ing about electricity and magnetism and how to put themto practical use, it was not until 1897 that the fundamentalparticle of electric charge, the electron, was identifiedand its properties quantified (by J. J. Thomson). Theability to eject electrons from a material by shiningelectromagnetic energy, such as light, on it is known asthe photoelectric effect. To explain this effect, AlbertEinstein adopted the quantum concept of energy that hadbeen advanced a few years earlier (1900) by Max Planckin his formulation of the quantum theory of matter. Byso doing, Einstein symbolizes the bridge between theclassical and modern eras of electromagnetics.

EM in the Modern Era

In terms of engineering applications, electromagneticsplays a role in the design and operation of everyconceivable electronic device, including diodes, tran-sistors, integrated circuits, lasers, display screens, bar-code readers, cell phones, and microwave ovens, toname but a few. Given the breadth and diversity ofthese applications, it is far more difficult to constructa meaningful timeline for the modern era than waspossible earlier for the classical era. However, it is quitepossible to develop timelines for specific technologiesand to use them as educational tools by linking theirmilestone innovations to electromagnetics. Chronologies1-2 (pages 8–9) and 1-3 (pages 10–11) present timelines

Table 1-1: Fundamental SI units.

Dimension Unit SymbolLength meter mMass kilogram kgTime second sElectric Current ampere ATemperature kelvin KAmount of substance mole mol

for telecommunications and computers, respectively,representing technologies that have become integral partsof today’s societal infrastructure. Some of the entriesin the tables refer to specific inventions, such as thetelegraph, the transistor, and the laser. The operationalprinciples and capabilities of some of these technologiesare highlighted in special sections called TechnologyBriefs, scattered throughout the book.

1-1 Dimensions, Units, and Notation

The International System of Units, abbreviated SIafter its French name Systeme Internationale, is thestandard system used in today’s scientific literature forexpressing the units of physical quantities. Length is adimension and meter is the unit by which it is expressedrelative to a reference standard. The SI system is basedon the units for the six fundamental dimensions listedin Table 1-1. The units for all other dimensions areregarded as secondary because they are based on andcan be expressed in terms of the six fundamental units.Appendix A contains a list of quantities used in thisbook, together with their symbols and units.

For quantities ranging in value between 10−18 and1018, a set of prefixes, arranged in steps of 103, arecommonly used to denote multiples and submultiplesof units. These prefixes, all of which were derived fromGreek, Latin, Spanish, and Danish terms, are listed in





ca. 900 Legend has it that while walking across a field in northernGreece, a shepherd named Magnus experiences a pullon the iron nails in his sandals by the black rock he isstanding on. The region was later named Magnesia andthe rock became known as magnetite [a form of iron withpermanent magnetism].

ca. 600 Greek philosopher Thalesdescribes how amber, after beingrubbed with cat fur, can pick up feathers [static electricity].

ca. 1000 Magnetic compass used as anavigational device.

1600 William Gilbert (English) coins the term electric after theGreek word for amber (elektron), and observes that acompass needle points north-south because the Earthacts as a bar magnet.

1671 Isaac Newton (English) demonstrates that white light is amixture of all the colors.

1733 Charles-Francois du Fay (French) discovers thatelectric charges are of two forms, and that like chargesrepel and unlike charges attract.

1745 Pieter van Musschenbroek (Dutch) invents the Leydenjar, the first electrical capacitor.

1752 Benjamin Franklin(American) inventsthe lightning rod anddemonstrates that lightning is electricity.

1785

Charles-Augustin de Coulomb (French)demonstrates that the electrical force betweencharges is proportional tothe inverse of the squareof the distance between them.

1800

Alessandro Volta (Italian)develops the first electric battery.

1820

Hans Christian Oersted(Danish) demonstrates theinterconnection betweenelectricity and magnetism through his discovery thatan electric current in a wire causes a compass needle to orient itself perpendicular to the wire.

1820 Andre-Marie Ampere (French)notes that parallel currents inwires attract each other and opposite currents repel.

1820

Jean-Baptiste Biote (French)and Felix Savart (French) develop the Biot-Savart law relating the magnetic field induced by a wire segment to the current flowing through it.

Chronology 1-1: TIMELINE FOR ELECTROMAGNETICS IN THE CLASSICAL ERA

Electromagnetics in the Classical Era

BC

BC





1888 Nikola Tesla(Croation American)invents the ac(alternating current) electric motor.

1895 Wilhelm Roentgen (German)discovers X-rays. One ofhis first X-ray images wasof the bones in his wife'shands. [1901 Nobel prize in physics.]

1897 Joseph John Thomson (English) discovers the electronand measures its charge-to-mass ratio. [1906 Nobel prizein physics.]

1905 Albert Einstein (German American) explains thephotoelectric effect discovered earlier by Hertz in 1887.[1921 Nobel prize in physics.]

1827 Georg Simon Ohm (German) formulates Ohm's lawrelating electric potential to current and resistance.

1827 Joseph Henry (American) introduces the concept of inductance, and builds one of the earliest electric motors. He also assisted Samual Morse in the development of the telegraph.

1831 Michael Faraday (English) discovers that a changing magnetic flux can inducean electromotive force.

1873 James Clerk Maxwell (Scottish) publishes his Treatise on Electricity and Magnetism in which he unites the discoveries of Coulomb, Oersted, Ampere, Faraday, and others into four elegantly constructed mathematicalequations, now known as Maxwell’s Equations.

1887

Chronology 1-1: TIMELINE FOR ELECTROMAGNETICS IN THE CLASSICAL ERA (continued)

Electromagnetics in the Classical Era

Heinrich Hertz (German) builds a system that can generate electromagnetic waves (at radio frequencies) and detect them.

1835 Carl Friedrich Gauss (German) formulates Gauss's lawrelating the electric flux flowing through an enclosed surface to the enclosed electric charge.





Chronology 1-2: TIMELINE FOR TELECOMMUNICATIONS

Telecommunications

1825

1837 Samuel Morse(American) patents the electromagnetic telegraph, using a code of dots and dashes to represent letters and numbers.

1872 Thomas Edison (American) patents the electric typewriter.

1876 Alexander Bell (Scottish-American) invents the telephone, the rotary dial becomes available in 1890, and by 1900, telephone systems are installed in many communities.

1887 Heinrich Hertz (German) generates radio waves and demonstrates that they share the same propertiesas light.

1887 Emil Berliner (American) invents the flat gramophonedisc, or record.

1893 Valdemar Poulsen(Danish) invents thefirst magnetic sound recorder using steelwire as recording medium.

Guglielmo Marconi (Italian) files his first of many patentson wireless transmission by radio. In 1901, he demonstrates radio telegraphy across the Atlantic Ocean. [1909 Nobel prize in physics, shared with Karl Braun (German).]

1897 Karl Braun (German) invents the cathode ray tube (CRT). [1909 Nobel prize with Marconi.]

1902 Reginald Fessenden (American) invents amplitude modulation for telephone transmission. In 1906, he introduces AM radio broadcasting of speech and music on Christmas Eve.

1912 Lee De Forest(American) develops the triode tube amplifier forwireless telegraphy. Also in 1912, the wireless distress call issued by the Titanic was heard 58 miles away by the ocean liner Carpathia, which managed to rescue 705 Titanic passengers 3.5 hours later.

1919 Edwin Armstong (American) invents the superheterodyne radio receiver.

1920 Birth of commercial radio broadcasting; Westinghouse Corporation establishes radio station KDKA in Pittsburgh, Pennsylvania.

1896William Sturgeon(English) develops the multiturn electromagnet.





1958 Jack Kilby (American) builds first integrated circuit (IC) on germanium and, independently, Robert Noyce (American)builds first IC on silicon.

Echo, the first passive communication satellite is launched, and successfully reflects radio signals back to Earth. In 1963, the first communication satellite is placed in geosynchronous orbit.

1969 ARPANET is established by the U.S. Department of Defense, to evolve later into the Internet.

1979 Japan builds the first cellular telephone network:• 1983 cellular phone networks start in the United States.• 1990 electronic beepers become common.• 1995 cell phones become widely available.• 2002 cell phone supports video and Internet.

1984 Worldwide Internet becomes operational.

1988 First transatlantic optical fiber cable between the U.S. and Europe.

1997 Mars Pathfinder sends images to Earth.

2004 Wireless communication supported by many airports, university campuses, and other facilities.

Vladimir Zworykin (Russian-American) invents television. In 1926, John Baird (Scottish) transmits TV images over telephone wires from London to Glasgow. Regular TV broadcasting began in Germany (1935), England (1936), and the United States (1939).

1926 Transatlantic telephone service between London and New York.

1932 First microwave telephone link, installed (by Marconi) between Vatican City and the Pope’s summer residence.

1933 Edwin Armstrong (American) invents frequency modulation (FM) for radio transmission.

1935 Robert Watson Watt(Scottish) invents radar.

1938 H. A. Reeves (American) invents pulse code modulation (PCM).

1947 William Schockley, Walter Brattain, and John Bardeen (all Americans) invent the junction transistor at Bell Labs. [1956 Nobel prize in physics.]

1955 Pager is introduced as a radio communication product in hospitals and factories.

1955 Navender Kapany (Indian American) demonstrates the optical fiber as a low-loss, light-transmission medium.

1923

1960

Chronology 1-2: TIMELINE FOR TELECOMMUNICATIONS (continued)

Telecommunications





1941 Konrad Zuze (German) develops the first programmable digital computer, using binary arithmetic and electric relays.

1945 John Mauchly and J. Presper Eckert develop the ENIAC, the first all-electronic computer.

1950 Yoshiro Nakama (Japanese) patents the floppy disk as amagnetic medium for storing data.

1956 John Backus (American) develops FORTRAN, thefirst major programminglanguage.

1958 Bell Labs develops the modem.

1960 Digital Equipment Corporationintroduces the first minicomputer, the PDP-1, to be followed with the PDP-8 in 1965.

1964 IBM’s 360 mainframe becomes the standard computer for major businesses.

1965 John Kemeny and Thomas Kurtz(both American)develop the BASICcomputer language.

Chronology 1-3: TIMELINE FOR COMPUTER TECHNOLOGY

Computer Technology

ca 1100 Abacus is the earliest known calculating device.

1614 John Napier (Scottish) develops the logarithm system.

Blaise Pascal(French) buildsthe first addingmachine using multiple dials.

Gottfried von Leibniz (German) builds calculator that can do both addition and multiplication.

Charles de Colmar (French) builds the Arithometer, the first mass-produced calculator.

1642

1671

1820

1885 Dorr Felt (American) invents and markets a key-operated adding machine (and adds a printer in 1889).

1930 Vannevar Bush (American) develops the differential analyzer, an analog computer for solving differential equations.

BC

PRINTFOR Counter = 1 TO Items PRINT USING “##.”; Counter; LOCATE , ItemColumn PRINT Item$(Counter); LOCATE , PriceColumn PRINT Price$(Counter)NEXT Counter





Chronology 1-3: TIMELINE FOR COMPUTER TECHNOLOGY (continued)

Computer Technology

1989 Tim Berners Lee (British) invents the World Wide Web by introducing a networked hypertext system.

1991 Internet connects to 600,000 hosts in more than 100 countries.

1995 Sun Microsystems introduces the Java programminglanguage.

1996 Sabeer Bhatia (Indian American) and Jack Smith (American) launch Hotmail, the first webmail service.

1997 IBM’s Deep Blue computer defeats World Chess Champion Garry Kasparov.

1997 Palm Pilot becomes widely available.

1968

1971 Texas Instruments introduces the pocket calculator.

1971 Ted Hoff (American) invents the Intel 4004, the first computer microprocessor.

1976 IBM introduces the laser printer.

1976 Apple Computer sells Apple Iin kit form, followed by the fully assembledApple II in 1977 and the Macintosh in 1984.

1980 Microsoft introduces theMS-DOS computer disk operating system. Microsoft Windows is marketed in 1985.

1981 IBM introduces the PC.

IBM

pal

mO

ne In

c.

IBM

Ap

ple

IBM

Tom

How

e

Knn

ight

-Rid

der

Texa

s In

stru

men

ts

Douglas Engelbart (American) demonstrates a word-processor system, the mouse pointing device and the use of “windows.”





Table 1-2: Multiple and submultiple prefixes.Prefix Symbol Magnitude

exa E 1018

peta P 1015

tera T 1012

giga G 109

mega M 106

kilo k 103

milli m 10−3

micro µ 10−6

nano n 10−9

pico p 10−12

femto f 10−15

atto a 10−18

Table 1-2. A length of 5 × 10−9 m, for example, may bewritten as 5 nm.

In electromagnetics we work with scalar and vectorquantities. In this book we use a medium-weight italicfont for symbols (other than Greek letters) denoting scalarquantities, such as R for resistance, whereas we use aboldface roman font for symbols denoting vectors, suchas E for the electric field vector. A vector consists ofa magnitude (scalar) and a direction, with the directionusually denoted by a unit vector. For example,

E = xE, (1.1)

where E is the magnitude of E and x is its direction.Unit vectors are printed in boldface with a circumflex ( ˆ )above the letter.

Throughout this book, we make extensive use ofphasor representation in solving problems involvingelectromagnetic quantities that vary sinusoidally in time.Letters denoting phasor quantities are printed with atilde (∼) over the letter. Thus, E is the phasor electricfield vector corresponding to the instantaneous electricfield vector E(t). This notation is discussed in moredetail in Section 1-6.

1-2 The Nature of Electromagnetism

Our physical universe is governed by four fundamentalforces of nature:

• The nuclear force, which is the strongest of the four,but its range is limited to submicroscopic systems,such as nuclei.

• The weak-interaction force, whose strength isonly 10−14 that of the nuclear force. Its primaryrole is in interactions involving certain radioactiveelementary particles.

• The electromagnetic force, which exists betweenall charged particles. It is the dominant force inmicroscopic systems, such as atoms and molecules,and its strength is on the order of 10−2 that of thenuclear force.

• The gravitational force, which is the weakest ofall four forces, having a strength on the order of10−41 that of the nuclear force. However, it is thedominant force in macroscopic systems, such as thesolar system.

Our interest in this book is with the electromagnetic forceand its consequences. Even though the electromagneticforce operates at the atomic scale, its effects can betransmitted in the form of electromagnetic waves thatcan propagate through both free space and materialmedia. The purpose of this section is to provide anoverview of the basic framework of electromagnetism,which consists of certain fundamental laws governing theelectric and magnetic fields induced by static and movingelectric charges, respectively, the relations between theelectric and magnetic fields, and how these fields interactwith matter. As a precursor, however, we will takeadvantage of our familiarity with the gravitational forceby describing some of its properties because they providea useful analogue to those of the electromagnetic force.




1-2 THE NATURE OF ELECTROMAGNETISM 13

m1

m2

Fg12

Fg21

R12^

R12

Figure 1-2: Gravitational forces between two masses.

1-2.1 The Gravitational Force: A UsefulAnalogue

According to Newton’s law of gravity, the gravitationalforce Fg21

acting on massm2 due to a massm1 at a distanceR12 from m2, as depicted in Fig. 1-2, is given by

Fg21= −R12

Gm1m2

R212

(N), (1.2)

where G is the universal gravitational constant, R12 isa unit vector that points from m1 to m2, and the unitfor force is newton (N). The negative sign in Eq. (1.2)accounts for the fact that the gravitational force isattractive. Conversely, Fg12

= −Fg21, where Fg12

is theforce acting on mass m1 due to the gravitational pullof mass m2. Note that the first subscript of Fg denotesthe mass experiencing the force and the second subscriptdenotes the source of the force.

The force of gravitation acts at a distance; that is, thetwo objects do not have to be in direct contact for eachto experience the pull by the other. This phenomenonof direct action at a distance has led to the concept offields. An object of mass m1 induces a gravitationalfieldψψψ1 (Fig. 1-3) that does not physically emanate fromthe object, but its influence exists at every point in spacesuch that if another object of mass m2 were to exist at adistance R12 from object m1 then the second object m2

m1

–R

1ψ ψ ψ ψ ψ ψ

Figure 1-3: Gravitational fieldψψψ1 induced by a massm1.

would experience a force acting on it equal in strengthto that given by Eq. (1.2). At a distance R from m1, thefieldψψψ1 is a vector defined as

ψψψ1 = −RGm1

R2(N/kg), (1.3)

where R is a unit vector that points in the radial directionaway from objectm1, and therefore −R points towardm1.The force due to ψψψ1 acting on a mass m2 at a distanceR = R12 along the direction R = R12 is

Fg21= ψψψ1m2 = −R12

Gm1m2

R212

. (1.4)

The field concept may be generalized by defining thegravitational fieldψψψ at any point in space such that, whena test mass m is placed at that point, the force Fg actingon m is related toψψψ by

ψψψ = Fg

m. (1.5)

The force Fg may be due to a single mass or a distributionof many masses.





1-2.2 Electric Fields

The electromagnetic force consists of an electrical forceFe and a magnetic force Fm. The electrical force Fe

is similar to the gravitational force, but with a majordifference. The source of the gravitational field is mass,and the source of the electrical field is electric charge, andwhereas both types of fields vary inversely as the squareof the distance from their respective sources, electriccharge may have positive or negative polarity, whereasmass does not exhibit such a property.

We know from atomic physics that all matter containsa mixture of neutrons, positively charged protons, andnegatively charged electrons, with the fundamentalquantity of charge being that of a single electron, usuallydenoted by the letter e. The unit by which electric chargeis measured is the coulomb (C), named in honor of theeighteenth-century French scientist Charles Augustin deCoulomb (1736–1806). The magnitude of e is

e = 1.6 × 10−19 (C). (1.6)

The charge of a single electronis qe = −e and that ofa proton is equal in magnitude but opposite in polarity:qp = e. Coulomb’s experiments demonstrated that:

(1) two like charges repel one another, whereas twocharges of opposite polarity attract,

(2) the force acts along the line joining the charges, and

(3) its strength is proportional to the product ofthe magnitudes of the two charges and inverselyproportional to the square of the distance betweenthem.

These properties constitute what today is calledCoulomb’s law, which can be expressed mathematicallyby the following equation:

Fe21 = R12q1q2

4πε0R212

(N) (in free space), (1.7)

+q1

+q2

Fe12

Fe21

R12^

R12

Figure 1-4: Electric forces on two positive point chargesin free space.

where Fe21 is the electrical force acting on charge q2 dueto charge q1,R12 is the distance between the two charges,R12 is a unit vector pointing from charge q1 to charge q2

(Fig.1-4), and ε0 is a universal constant called theelectrical permittivity of free space [ε0 = 8.854 × 10−12

farad per meter (F/m)]. The two charges are assumedto be in free space (vacuum) and isolated from allother charges. The force Fe12 acting on charge q1 due tocharge q2 is equal to force Fe21 in magnitude, but oppositein direction; Fe12 = −Fe21 .

The expression given by Eq. (1.7) for the electricalforce is analogous to that given by Eq. (1.2) for thegravitational force, and we can extend the analogy furtherby defining the existence of an electric field intensity Edue to any charge q as follows:

E = Rq

4πε0R2(V/m) (in free space), (1.8)

where R is the distance between the charge and theobservation point, and R is the radial unit vectorpointing away from the charge. Figure 1-5 depicts theelectric-field lines due to a positive charge. For reasonsthat will become apparent in later chapters, the unit for Eis volt per meter (V/m).





+q

E

R

Figure 1-5: Electric field E due to charge q.

Electric charge exhibits two important properties. Thefirst is the law of conservation of electric charge, whichstates that the (net) electric charge can neither be creatednor destroyed. If a volume contains np protons and ne

electrons, then the total charge is

q = npe − nee = (np − ne)e (C). (1.9)

Even if some of the protons were to combine with anequal number of electrons to produce neutrons or other el-ementary particles, the net charge q remains unchanged.In matter, the quantum mechanical laws governing thebehavior of the protons inside the atom’s nucleus and theelectrons outside it do not allow them to combine.

The second important property of electric charge isthe principle of linear superposition, which states thatthe total vector electric field at a point in space due to asystem of point charges is equal to the vector sum of theelectric fields at that point due to the individual charges.This seemingly simple concept will allow us in futurechapters to compute the electric field due to complexdistributions of charge without having to be concernedwith the forces acting on each individual charge due tothe fields by all of the other charges.

The expression given by Eq. (1.8) describes the fieldinduced by an electric charge when in free space. Letus now consider what happens when we place a positive

+–

+–

+–+–+–

+–+–+–+

–

+–+–+

–+

–

+–

+–

+–

+–

+ – + – + –+

–

+ – + –

+–

+–

+–

+–

+–

+–

+–

+–

+–

+–+–+–+

–

+–+–

+–

+–

+–

+–

+–

+–

+ – + – + –+

–

+ – + –+

–+–

+–

+–

+–

+–

+–

+–

q++ – + – + – + –

Figure 1-6: Polarization of the atoms of a dielectricmaterial by a positive charge q.

point charge in a material composed of atoms. In theabsence of the point charge, the material is electricallyneutral, with each atom having a positively chargednucleus surrounded by a cloud of electrons of equal butopposite polarity. Hence, at any point in the materialnot occupied by an atom the electric field E is zero.Upon placing a point charge in the material, as shownin Fig. 1-6, the atoms experience forces that cause themto become distorted. The center of symmetry of theelectron cloud is altered with respect to the nucleus, withone pole of the atom becoming more positively chargedand the other pole becoming more negatively charged.Such a polarized atom is called an electric dipole, andthe distortion process is called polarization. The degreeof polarization depends on the distance between theatom and the isolated point charge, and the orientationof the dipole is such that the dipole axis connectingits two poles is directed toward the point charge, asillustrated schematically in Fig. 1-6. The net result ofthis polarization process is that the electric dipoles ofthe atoms (or molecules) tend to counteract the fielddue to the point charge. Consequently, the electric fieldat any point in the material would be different fromthe field that would have been induced by the point





charge in the absence of the material. To extend Eq. (1.8)from the free-space case to any medium, we replace thepermittivity of free space ε0 with ε, where ε is now thepermittivity of the material in which the electric field ismeasured and is therefore characteristic of that particularmaterial. Thus,

E = Rq

4πεR2(V/m). (1.10)

Often, ε is expressed in the form

ε = εrε0 (F/m), (1.11)

where εr is a dimensionless quantity called the relativepermittivity or dielectric constant of the material. Forvacuum, εr = 1; for air near Earth’s surface, εr = 1.0006;and for materials that we will have occasion to use in thisbook, their values of εr are tabulated in Appendix B.

In addition to the electric field intensity E, we will oftenfind it convenient to also use a related quantity called theelectric flux density D, given by

D = εE (C/m2), (1.12)

and its unit is coulomb per square meter (C/m2). Thesetwo electrical quantities, E and D, constitute one of twofundamental pairs of electromagnetic fields. The secondpair consists of the magnetic fields discussed next.

1-2.3 Magnetic Fields

As early as 800 B.C., the Greeks discovered that certainkinds of stones exhibit a force that attracts pieces of iron.These stones are now called magnetite (Fe3O4) and thephenomenon they exhibit is magnetism. In the thirteenthcentury, French scientists discovered that, when a needlewas placed on the surface of a spherical natural magnet,the needle oriented itself along different directions fordifferent locations on the magnet. By mapping thedirections taken by the needle, it was determined that the

S

N

B

Figure 1-7: Pattern of magnetic field lines around a barmagnet.

magnetic force formed magnetic-field lines that encircledthe sphere and appeared to pass through two pointsdiametrically opposite each other. These points, calledthe north and south poles of the magnet, were foundto exist for every magnet, regardless of its shape. Themagnetic-field pattern of a bar magnet is displayed inFig. 1-7. It was also observed that like poles of differentmagnets repel each other and unlike poles attract eachother. This attraction–repulsion property is similar to theelectric force between electric charges, except for oneimportant difference: electric charges can be isolated,but magnetic poles always exist in pairs. If a permanentmagnet is cut into small pieces, no matter how small eachpiece is, it will always have a north and a south pole.

The magnetic lines encircling a magnet are calledmagnetic-field lines and represent the existence of amagnetic field called the magnetic flux density B.A magnetic field not only exists around permanentmagnets but can also be created by electric current.This connection between electricity and magnetism wasdiscovered in 1819 by the Danish scientist Hans Oersted





B

BB

B^

z

y

x

r

B

BB

B

Iφ φ φ φ φ

Figure 1-8: The magnetic field induced by a steadycurrent flowing in the z-direction.

(1777–1851), who found that an electric current in a wirecaused a compass needle placed in its vicinity to deflectand that the needle turned so that its direction was alwaysperpendicular to the wire and to the radial line connectingthe wire to the needle. From these observations, itwas deduced that the current-carrying wire induced amagnetic field that formed closed circular loops aroundthe wire, as illustrated in Fig. 1-8. Shortly after Oersted’sdiscovery, French scientists Jean Baptiste Biot and FelixSavart developed an expression that relates the magneticflux density B at a point in space to the current I inthe conductor. Application of their formulation, knowntoday as the Biot–Savart law, to the situation depicted inFig. 1-8 for a very long wire leads to the result that themagnetic flux density B induced by a constant current Iflowing in the z-direction is given by

B = φφφµ0I

2πr(T), (1.13)

where r is the radial distance from the current and φφφ is anazimuthal unit vector denoting the fact that the magnetic

field direction is tangential to the circle surrounding thecurrent, as shown in Fig. 1-8. The magnetic field ismeasured in tesla (T), named in honor of Nikola Tesla(1856–1943), a Croatian-American electrical engineerwhose work on transformers made it possible to transportelectricity over long wires without too much loss. Thequantity µ0 is called the magnetic permeability of freespace [µ0 = 4π × 10−7 henry per meter (H/m)], and itis analogous to the electric permittivity ε0. In fact, as wewill see in Chapter 2, the product of ε0 andµ0 specifies c,the velocity of light in free space, as follows:

c = 1√µ0ε0

= 3 × 108 (m/s). (1.14)

The majority of natural materials are nonmagnetic,meaning that they exhibit a magnetic permeabilityµ = µ0. For ferromagnetic materials, such as iron andnickel, µ can be much larger than µ0. The magneticpermeability µ accounts for magnetization properties ofa material. In analogy with Eq. (1.11), µ of a particularmaterial can be defined as

µ = µrµ0 (H/m), (1.15)

where µr is a dimensionless quantity called the relativemagnetic permeability of the material. The values of µr

for commonly used ferromagnetic materials are given inAppendix B.

We stated earlier that E and D constitute one of twopairs of electromagnetic field quantities. The second pairis B and the magnetic field intensity H, which are relatedto each other through µ:

B = µH. (1.16)





1-2.4 Static and Dynamic Fields

Because the electric field E is governed by the charge qand the magnetic field H is governed by I = dq/dt , andsince q and dq/dt are independent variables, the inducedelectric and magnetic fields are independent of oneanother as long as I remains constant. To demonstratethe validity of this statement, consider for example asmall section of a beam of charged particles that aremoving at a constant velocity. The moving chargesconstitute a d-c current. The electric field due to thatsection of the beam is determined by the total charge qcontained in that section. The magnetic field does notdepend on q, but rather on the rate of charge (current)flowing through that section. Few charges moving veryfast can constitute the same current as many chargesmoving slowly. In these two cases the induced magneticfield will be the same because the current I is the same,but the induced electric field will be quite differentbecause the numbers of charges are not the same.

Electrostatics and magnetostatics, corresponding tostationary charges and steady currents, respectively,are special cases of electromagnetics. They representtwo independent branches, so characterized becausethe induced electric and magnetic fields are uncoupledto each other. Dynamics, the third and more generalbranch of electromagnetics, involves time-varying fieldsinduced by time-varying sources, that is, currents andcharge densities. If the current associated with the beamof moving charged particles varies with time, then theamount of charge present in a given section of the beamalso varies with time, and vice versa. As we will seein Chapter 6, the electric and magnetic fields becomecoupled to each other in that case. In fact, a time-varyingelectric field will generate a time-varying magnetic field,and vice versa. Table 1-3 provides a summary of the threebranches of electromagnetics.

The electric and magnetic properties of materials arecharacterized by the two parameters ε andµ, respectively.

A third fundamental parameter is also needed, theconductivity of a material σ , which is measured insiemens per meter (S/m). The conductivity characterizesthe ease with which charges (electrons) can move freelyin a material. If σ = 0, the charges do not movemore than atomic distances and the material is said tobe a perfect dielectric, and if σ = ∞, the chargescan move very freely throughout the material, which isthen called a perfect conductor. The material parametersε, µ, and σ are often referred to as the constitutiveparameters of a material (Table 1-4). A medium is saidto be homogeneous if its constitutive parameters areconstant throughout the medium.

REVIEW QUESTIONS

Q1.1 What are the four fundamental forces of natureand what are their relative strengths?

Q1.2 What is Coulomb’s law? State its properties.

Q1.3 What are the two important properties of electriccharge?

Q1.4 What do the electrical permittivity and magneticpermeability of a material account for?

Q1.5 What are the three branches and associatedconditions of electromagnetics?

1-3 Traveling WavesWaves are a natural consequence of many physicalprocesses: waves and ripples on oceans and lakes; soundwaves that travel through air; mechanical waves onstretched strings; electromagnetic waves that constitutelight; earthquake waves; and many others. All thesevarious types of waves exhibit a number of commonproperties, including the following:

• Moving waves carry energy from one point toanother.




1-3 TRAVELING WAVES 19

Table 1-3: The three branches of electromagnetics.

Branch Condition Field Quantities (Units)

Electrostatics Stationary charges Electric field intensity E (V/m)(∂q/∂t = 0) Electric flux density D (C/m2)

D = εEMagnetostatics Steady currents Magnetic flux density B (T)

(∂I/∂t = 0) Magnetic field intensity H (A/m)B = µH

Dynamics Time-varying currents E, D, B, and H(Time-varying fields) (∂I/∂t = 0) (E,D) coupled to (B,H)

Table 1-4: Constitutive parameters of materials.

Parameter Units Free-space Value

Electrical permittivity ε F/m ε0 = 8.854 × 10−12 (F/m)

1

36π× 10−9 (F/m)

Magnetic permeability µ H/m µ0 = 4π × 10−7 (H/m)

Conductivity σ S/m 0

• Waves have velocity; it takes time for a wave to travelfrom one point to another. In vacuum, light wavestravel at a speed of 3 × 108 m/s and sound waves inair travel at a speed approximately a million timesslower, specifically 330 m/s.

• Some waves exhibit a property called linearity.Waves that do not affect the passage of other wavesare called linear because they pass right througheach other, and the total of two linear waves issimply the sum of the two waves as they wouldexist separately. Electromagnetic waves are linear,as are sound waves. When two people speak to oneanother, their sound waves do not reflect from oneanother, but simply pass through independently of

each other. Water waves are approximately linear;the expanding circles of ripples caused by twopebbles thrown into two locations on a lake surfacedo not affect each other. Although the interactionof the two circles may exhibit a complicatedpattern, it is simply the linear superposition of twoindependent expanding circles.

Waves are of two types: transient waves caused bya short-duration disturbance and continuous harmonicwaves generated by an oscillating source. We willencounter both types of waves in this book, but mostof our discussion will deal with the propagation ofcontinuous waves that vary sinusoidally with time.





(a) Circular waves (c) Spherical wave(b) Plane and cylindrical waves

Plane wavefrontTwo-dimensional wave

Cylindrical wavefront Spherical wavefront

Figure 1-10: Examples of two-dimensional and three-dimensional waves: (a) circular waves on a pond, (b) a plane light waveexciting a cylindrical light wave through the use of a long narrow slit in an opaque screen, and (c) a sliced section of a sphericalwave.

u

Figure 1-9: A one-dimensional wave traveling on a string.

An essential feature of a propagating wave is that itis a self-sustaining disturbance of the medium throughwhich it travels. If this disturbance varies as a functionof one space variable, such as the vertical displacementof the string shown in Fig. 1-9, we call the wave aone-dimensional wave. The vertical displacement varieswith time and with the location along the length ofthe string. Even though the string rises up into asecond dimension, the wave is only one-dimensionalbecause the disturbance varies with only one spacevariable. A two-dimensional wave propagates out acrossa surface, like the ripples on a pond [Fig. 1-10(a)], andits disturbance can be described by two space variables.And by extension, a three-dimensional wave propagatesthrough a volume and its disturbance may be a functionof all three space variables. Three-dimensional wavesmay take on many different shapes; they include planewaves, cylindrical waves, and spherical waves. A planewave is characterized by a disturbance that at a givenpoint in time has uniform properties across an infiniteplane perpendicular to the direction of wave propagation





[Fig. 1-10(b)] and, similarly, for cylindrical and sphericalwaves the disturbances are uniform across cylindrical andspherical surfaces, as shown in Figs. 1-10(b) and (c).

In the material that follows, we will examine some ofthe basic properties of waves by developing mathematicalformulations that describe their functional dependence ontime and space variables. To keep the presentation simple,we will limit our present discussion to sinusoidallyvarying waves whose disturbances are functions of onlyone space variable, and we will defer discussion of morecomplicated waves to later chapters.

1-3.1 Sinusoidal Wave in a Lossless Medium

Regardless of the mechanism responsible for generatingthem, all waves can be described mathematically incommon terms. By way of an example, let us considera wave traveling on a lake surface. A medium is said tobe lossless if it does not attenuate the amplitude of thewave traveling within it or on its surface. Let us assumefor the time being that frictional forces can be ignored,thereby allowing a wave generated on the water surface totravel indefinitely with no loss in energy. If y denotes theheight of the water surface relative to the mean height(undisturbed condition) and x denotes the distance ofwave travel, the functional dependence of y on time tand the spatial coordinate x has the general form

y(x, t) = A cos

(2πt

T− 2πx

λ+ φ0

)(m), (1.17)

where A is the amplitude of the wave, T is its timeperiod, λ is its spatial wavelength, and φ0 is a referencephase. The quantity y(x, t) can also be expressed in theform

y(x, t) = A cosφ(x, t), (1.18)

where

φ(x, t) =(

2πt

T− 2πx

λ+ φ0

)(rad). (1.19)

–A

02

3λ 2

A

y(x, 0)

x

(a) y(x, t) versus x at t = 0

–A

0 T2

T 3T2

A

T

y(0, t)

t

(b) y(x, t) versus t at x = 0

λ

λ λ

Figure 1-11: Plots of y(x, t) = A cos(

2πtT

− 2πxλ

)as a

function of (a) x at t = 0 and (b) t at x = 0.

The angle φ(x, t) is called the phase of the wave, andit should not be confused with the reference phase φ0,which is constant with respect to both time and space.Phase is measured by the same units as angles, that is,radians (rad) or degrees, with 2π radians = 360.

Let us first analyze the simple case when φ0 = 0:

y(x, t) = A cos

(2πt

T− 2πx

λ

)(m). (1.20)

The plots in Fig. 1-11 show the variation of y(x, t) withx at t = 0 and with t at x = 0. The wave pattern repeatsitself at a spatial period λ along x and at a temporalperiod T along t .

If we take time snapshots of the water surface, theheight profile y(x) would exhibit the sinusoidal patternsshown in Fig. 1-12. For each plot, corresponding to aspecific value of t , the spacing between peaks is equal





λ λ 2

3λ 2

3λ 2

3λ 2

λ λ 2

λ λ 2

y(x, 0)

y(x, T/4)

y(x, T/2)

A

–A

A

–A

A

–A

(a) t = 0

(b) t = T/4

(c) t = T/2

x

x

x

P

P

P

up

Figure 1-12: Plots of y(x, t) = A cos(

2πtT

− 2πxλ

)as

a function of x at (a) t = 0, (b) t = T/4, and (c) t =T/2. Note that the wave moves in the +x-direction witha velocity up = λ/T .

to the wavelength λ, but the patterns are shifted relativeto one another because they correspond to differentobservation times. Because the pattern advances alongthe +x-direction at progressively increasing values of t ,the height profile behaves like a wave traveling in thatdirection. If we choose any height level, such as thepeak P , and follow it in time, we can measure the phasevelocity of the wave. The peak corresponds to when the

phase φ(x, t) of the wave is equal to zero or multiplesof 2π radians. Thus,

φ(x, t)= 2πt

T− 2πx

λ=2nπ, n = 0, 1, 2, . . . (1.21)

Had we chosen any other fixed height of the wave, say y0,and monitored its movement as a function of t and x, thisagain is equivalent to setting the phase φ(x, t) constantsuch that

y(x, t) = y0 = A cos

(2πt

T− 2πx

λ

), (1.22)

or

2πt

T− 2πx

λ= cos−1

(y0

A

)= constant. (1.23)

The apparent velocity of that fixed height is obtained bytaking the time derivative of Eq. (1.23),

2π

T− 2π

λ

dx

dt= 0, (1.24)

which gives the phase velocity up as

up = dx

dt= λ

T(m/s). (1.25)

The phase velocity, also called the propagation velocity,is the velocity of the wave pattern as it moves acrossthe water surface. The water itself mostly moves up anddown; when the wave moves from one point to another,the water does not move physically along with it.

The direction of wave propagation is easily determinedby inspecting the signs of the t and x terms in theexpression for the phaseφ(x, t)given by Eq. (1.19): if oneof the signs is positive and the other is negative, then thewave is traveling in the positive x-direction, and if bothsigns are positive or both are negative, then the wave istraveling in the negative x-direction. The constant phasereference φ0 has no influence on either the speed or thedirection of wave propagation.





The frequency of a sinusoidal wave, f , is thereciprocal of its time period T :

M 1.9D 1.1

M1.ϒ-1.81-3.2 Sinusoidal Wave in a Lossy MediumIf awave is traveling in thex-direction in alossymedium,its amplitude will decrease ase

αx.Thisfactoriscalled

the

attenuationfactor

,andαiscalledthe

attenuation

constant

ofthemediumanditsunitisneperpermeter

(Np/m).Thus,ingeneral,

y(x,t)

/Ae

αxcos

(ωt

βx+φ0).(1.33)

Thewaveamplitudeisnow

Aeαx,andnot just

A. Fig-

ure1-14showsaplotof

y(x,t)

asafunctionof

xatt/0for

A/10m,

λ/2m,

α/0.2Np/m, and

φ0/0.Note

thattheenvelopeofthewavepatterndecreasesas

eαx.

Therealunitof

αis(1/m),theneper(Np)partis

adimensionless, articialadjective traditionallyusedasa

reminderthattheunit(Np/m)referstotheattenuation

constantofthemedium,

α. Asimilarpracticeisapplied

tothephaseconstant

βbyassigningittheunit(rad/m)

insteadofjust(l/m).

M 9 3 - 1 0 6 a n d D - 3


φ0 = π/4 φ0 = –π/4

Tt

T2

3T2

y

–A

A

Leads ahead ofreference wave Lags behind reference wave

Reference wave (φ0 = 0)

Figure 1-13: Plots of y(0, t) = A cos [(2πt/T )+ φ0] for three different values of the reference phase φ0.

–10 m

–5 m

0

5 m

10 m

y(x)

y(x)

10e–0.2x

x (m)1 2 3 4 5 6 7 8

Figure 1-14: Plot of y(x) = (10e−0.2x cosπx)meters. Note that the envelope is bounded between the curve given by 10e−0.2x

and its mirror image.

REVIEW QUESTIONS

Q1.6 How can you tell if a wave is traveling in thepositive x-direction or the negative x-direction?

Q1.7 How does the envelope of the wave pattern vary

with distance in (a) a lossless medium and (b) a lossymedium?

Q1.8 Why does a negative value of φ0 signify a phaselag?





Example 1-1 Sound Wave in Water

An acoustic wave traveling in the x-direction in a fluid(liquid or gas) is characterized by a differential pressurep(x, t). The unit for pressure is newton per square meter


travel before its amplitude will have been reduced to (a)10 V/m, (b) 1 V/m, (c) 1 µV/m?

Ans. (a) 4.6 m, (b) 9.2 m, (c) 37 m. (See C

DROM )

1-4 The Electromagnetic Spectrum

Visible light belongs to a family of waves called theelectromagnetic spectrum (Fig. 1-15). Other membersof this family include gamma rays, X rays, infraredwaves, and radio waves. Generically, they all are calledelectromagnetic (EM) waves because they share thefollowing fundamental properties:

• An EM wave consists of electric and magnetic fieldintensities that oscillate at the same frequency f .

• The phase velocity of an EM wave propagating invacuum is a universal constant given by the velocityof light c, defined earlier by Eq. (1.14).

• In vacuum, the wavelength λ of an EM wave isrelated to its oscillation frequency f by

λ = c

f. (1.34)

Whereas all EM waves share these properties, each isdistinguished by its own wavelength λ, or equivalentlyby its own oscillation frequency f .

The visible part of the EM spectrum shown in Fig. 1-15covers a very narrow wavelength range extendingbetween λ = 0.4 µm (violet) and λ = 0.7 µm (red). Aswe move progressively toward shorter wavelengths, weencounter the ultraviolet, X-ray, and gamma-ray bands,each so named because of historical reasons associatedwith the discovery of waves with those wavelengths. Onthe other side of the visible spectrum lie the infrared bandand then the radio region. Because of the link between λand f given by Eq. (1.34), each of these spectral ranges

may be specified in terms of its wavelength range oralternatively in terms of its frequency range. In practice,however, a wave is specified in terms of its wavelength λif λ < 1 mm, which encompasses all parts of the EMspectrum except for the radio region, and the wave isspecified in terms of its frequency f if λ > 1 mm (i.e.,in the radio region). A wavelength of 1 mm correspondsto a frequency of 3 × 1011 Hz = 300 GHz in free space.

The radio spectrum consists of several individualbands, as shown in the chart of Fig. 1-16. Each bandcovers one decade of the radio spectrum and has aletter designation based on a nomenclature defined bythe International Telecommunication Union. Differentfrequencies have different applications because they areexcited by different mechanisms, and the propertiesof an EM wave propagating in a material may varyconsiderably from one band to another. The extremelylow frequency (ELF) band from 3 to 30 Hz is usedprimarily for the detection of buried metal objects. Lowerfrequencies down to 0.1 Hz are used in magnetotelluricsensing of the structure of the earth, and frequenciesin the range from 1 Hz to 1 kHz sometimes are usedfor communications with submerged submarines and forcertain kinds of sensing of Earth’s ionosphere. The verylow frequency (VLF) region from 3 to 30 kHz is used bothfor submarine communications and for position locationby the Omega navigation system. The low-frequency(LF) band, from 30 to 300 kHz, is used for some formsof communication and for the Loran C position-locationsystem. Some radio beacons and weather broadcaststations used in air navigation operate at frequencies inthe higher end of the LF band. The medium-frequency(MF) region from 300 kHz to 3 MHz contains thestandard AM broadcast band from 0.5 to 1.5 MHz.

Long-distance communications and short-wave broad-casting over long distances use frequencies in thehigh-frequency (HF) band from 3 to 30 MHz becausewaves in this band are strongly affected by reflectionsby the ionosphere and least affected by absorption in




1-4 THE ELECTROMAGNETIC SPECTRUM 27

1 fm 1 pm 1 nm1 Å

1 EHz 1 PHz 1 THz 1 GHz 1 MHz 1 kHz 1 Hz

1 µm 1 mm 1 km 1 Mm1 m

10–15

1023 1021 1018 1015 1012 109 106 103 1

10–12 10–10 10–9 10–6 10–3 103 106 1081

Frequency (Hz)

Wavelength (m)

visible

Gamma raysCancer therapy

X-rays

Medical diagnosis

Ultraviolet

Sterilization

InfraredHeating,

Night vision

Radio spectrumCommunication, radar, radio and TV broadcasting,

radio astronomy

Atmospheric opacity

100%

0

Atmosphere opaque

Opticalwindow

Infraredwindows Radio window

Ionosphere opaque

Figure 1-15: The electromagnetic spectrum.

the ionosphere. The next frequency region, the veryhigh frequency (VHF) band from 30 to 300 MHz, isused primarily for television and FM broadcasting overline-of-sight distances and also for communicating withaircraft and other vehicles. Some early radio-astronomyresearch was also conducted in this range. The ultrahighfrequency (UHF) region from 300 MHz to 3 GHz isextensively populated with radars, although part of thisband also is used for television broadcasting and mobilecommunications with aircraft and surface vehicles. Theradars in this region of the spectrum are normally used foraircraft detection and tracking. Some parts of this regionhave been reserved for radio astronomical observation.

Many point-to-point radio communication systemsand various kinds of ground-based radars and ship radarsoperate at frequencies in the superhigh frequency (SHF)range from 3 to 30 GHz. Some aircraft navigation systemsoperate in this range as well.

Most of the extremely high frequency (EHF) bandfrom 30 to 300 GHz is used less extensively, primarilybecause the technology is not as well developed andbecause of excessive absorption by the atmosphere insome parts of this band. Some advanced communicationsystems are being developed for operation at frequenciesin the “atmospheric windows,” where atmosphericabsorption is not a serious problem, as are automobilecollision-avoidance radars and some military imagingradar systems. These atmospheric windows include theranges from 30 to 35 GHz, 70 to 75 GHz, 90 to 95 GHz,and 135 to 145 GHz.

Although no precise definition exists for the extentof the microwave band, it is conventionally regardedto cover the full ranges of the UHF, SHF, and EHFbands, with the EHF band sometimes referred to as themillimeter-wave band, because the wavelength rangecovered by this band extends from 1 mm (300 GHz) to1 cm (30 GHz).





Radar, advanced communication systems,remote sensing, radio astronomy

Extremely High FrequencyEHF (30 - 300 GHz)

Radar, satellite communication systems, aircraftnavigation, radio astronomy, remote sensing

Super High FrequencySHF (3 - 30 GHz)

TV broadcasting, radar, radio astronomy,microwave ovens, cellular telephone

Ultra High FrequencyUHF (300 MHz - 3 GHz)

TV and FM broadcasting, mobile radiocommunication, air traffic control

Very High FrequencyVHF (30 - 300 MHz)

Short wave broadcastingHigh FrequencyHF (3 - 30 MHz)

AM broadcastingMedium FrequencyMF (300 kHz - 3 MHz)

Radio beacons, weather broadcast stationsfor air navigation

Low FrequencyLF (30 - 300 kHz)

Navigation and position locationVery Low FrequencyVLF (3 - 30 kHz)

Audio signals on telephoneUltra Low FrequencyULF (300 Hz - 3 kHz)

Ionospheric sensing, electric powerdistribution, submarine communication

Super Low FrequencySLF (30 - 300 Hz)

Detection of buried metal objectsExtremely Low FrequencyELF (3 - 30 Hz)

f < 3 Hz) Magnetotelluric sensing of theearth's structure

1012

109

106

103

300 GHz

1 GHz

1 MHz

1 kHz

1 Hz

Microwave

Frequency (Hz)

Band Applications

Figure 1-16: Individual bands of the radio spectrum and their primary applications.

REVIEW QUESTIONSQ1.9 What are the three fundamental properties of EMwaves?Q1.10 What is the range of frequencies covered by themicrowave band?Q1.11 What is the wavelength range of the visiblespectrum? What are some of the applications of theinfrared band?

1-5 Review of Complex Numbers

A complex number z is written in the form

z = x + jy, (1.35)

where x and y are the real (Re) and imaginary (Im) partsof z, respectively, and j = √−1. That is,

x = Re(z), y = Im(z). (1.36)




1-5 REVIEW OF COMPLEX NUMBERS 29

θ ℜe(z)

ℑm(z)

y z

x

|z|

x = |z| cos θ y = |z| sin θ

θ = tan–1 (y/x)

|z| = x2 + y2+

Figure 1-17: Relation between rectangular and polar rep-resentations of a complex number z = x + jy = |z|ejθ .

Alternatively, z may be written in polar form as

z = |z|ejθ = |z|∠θ (1.37)

where |z| is the magnitude of z, θ is its phase angle,and the form ∠θ is a useful shorthand representationcommonly used in numerical calculations. ApplyingEuler’s identity,

ejθ = cos θ + j sin θ, (1.38)

we can convert z from polar form, as in Eq. (1.37), intorectangular form, as in Eq. (1.35),

z = |z|ejθ = |z| cos θ + j |z| sin θ, (1.39)

which leads to the relations

x = |z| cos θ, y = |z| sin θ, (1.40)

|z| = +√x2 + y2 , θ = tan−1(y/x). (1.41)

The two forms are illustrated graphically in Fig. 1-17.When using Eq. (1.41), care should be taken to ensurethat θ is in the proper quadrant. Also note that, since |z|is a positive quantity, only the positive root in Eq. (1.41)is applicable. This is denoted by the + sign above thesquare-root sign.

The complex conjugate of z, denoted with a starsuperscript (or asterisk), is obtained by replacing j

(wherever it appears) with −j , so that

z∗ = (x + jy)∗ = x − jy = |z|e−jθ = |z|∠−θ . (1.42)

The magnitude |z| is equal to the positive square root ofthe product of z and its complex conjugate:

|z| = +√z z∗ . (1.43)

We now highlight some of the properties of complexalgebra that we will likely encounter in future chapters.

Equality: If two complex numbers z1 and z2 are given by

z1 = x1 + jy1 = |z1|ejθ1, (1.44)

z2 = x2 + jy2 = |z2|ejθ2, (1.45)

then z1 = z2 if and only if x1 = x2 and y1 = y2 or,equivalently, |z1| = |z2| and θ1 = θ2.

Addition:

z1 + z2 = (x1 + x2)+ j (y1 + y2). (1.46)

Multiplication:

z1z2 = (x1 + jy1)(x2 + jy2)

= (x1x2 − y1y2)+ j (x1y2 + x2y1), (1.47a)

or

z1z2 = |z1|ejθ1 · |z2|ejθ2

= |z1||z2|ej (θ1+θ2)

= |z1||z2|[cos(θ1 + θ2)+ j sin(θ1 + θ2)]. (1.47b)




1-6 REVIEW OF PHASORS 31

(c) V I ∗ = 5e−j53.1 × 3.61e−j236.3

= 18.05e−j289.4 = 18.05ej70.6.

(d)V

I= 5e−j53.1

3.61ej236.3

= 1.39e−j289.4 = 1.39ej70.6.

(e)√I = √

3.61ej236.3

= ±√3.61 ej236.3/2 = ±1.90ej118.15

.

EXERCISE 1.3 Express the following complex functionsin polar form:

z1 = (4 − j3)2,

z2 = (4 − j3)1/2.

Ans. z1 = 25∠−73.7 , z2 = ±√5 ∠−18.4 . (See C

DROM )

EXERCISE 1.4 Show that√

2j = ±(1 + j). (See C

DROM )

1-6 Review of Phasors

Phasor analysis is a useful mathematical tool forsolving problems involving linear systems in which theexcitation is a periodic time function. Many engineeringproblems are cast in the form of linear integro-differentialequations. If the excitation, more commonly known as theforcing function, varies sinusoidally with time, the useof phasor notation to represent time-dependent variablesallows us to convert the integro-differential equation intoa linear equation with no sinusoidal functions, therebysimplifying the method of solution. After solving forthe desired variable, such as the voltage or current ina circuit, conversion from the phasor domain back to thetime domain provides the desired result.

The phasor technique can also be used for analyzinglinear systems when the forcing function is any arbitrary(nonsinusoidal) periodic time function, such as a squarewave or a sequence of pulses. By expanding the

C

R

ivs(t)

+

–

Figure 1-19: RC circuit connected to a voltage sourcevs(t).

forcing function into a Fourier series of sinusoidalcomponents, we can solve for the desired variable usingphasor analysis for each Fourier component of theforcing function separately. According to the principleof superposition, the sum of the solutions due to all ofthe Fourier components gives the same result as onewould obtain had the problem been solved entirely in thetime domain without the aid of Fourier representation.The obvious advantage of the phasor–Fourier approachis simplicity. Moreover, in the case of nonperiodic sourcefunctions, such as a single pulse, the functions can beexpressed as Fourier integrals, and a similar applicationof the principle of superposition can be used as well.

The simple RC circuit shown in Fig. 1-19 contains asinusoidally time-varying voltage source given by

vs(t) = V0 sin(ωt + φ0), (1.55)

where V0 is the amplitude, ω is the angular frequency,and φ0 is a reference phase. Application of Kirchhoff’svoltage law gives the following loop equation:

R i(t)+ 1

C

∫i(t) dt = vs(t) (time domain). (1.56)

Our objective is to obtain an expression for the currenti(t). We can do this by solving Eq. (1.56) in the timedomain, which is somewhat cumbersome because the




1-6 REVIEW OF PHASORS 33

The time factor ejωt has disappeared because it wascontained in all three terms. Equation (1.65) is thephasor-domain equivalent of Eq. (1.56).

Step 4: Solve the phasor-domain equation

From Eq. (1.65) the phasor current I is given by

I = Vs

R + 1/(jωC). (1.66)

Before we apply the next step, we need to convert theright-hand side of Eq. (1.66) into the form I0e

jθ with I0

being a real quantity. Thus,

I = V0ej (φ0−π/2)

[jωC

1 + jωRC

]= V0e

j (φ0−π/2)[

ωCejπ/2

+√1 + ω2R2C2 ejφ1

]= V0ωC

+√1 + ω2R2C2ej (φ0−φ1), (1.67)

where we have used the identity j = ejπ/2. The phaseangle φ1 = tan−1(ωRC) and lies in the first quadrant ofthe complex plane.

Step 5: Find the instantaneous value

To find i(t), we simply apply Eq. (1.61). That is, wemultiply the phasor I given by Eq. (1.67) by ejωt andthen take the real part:

i(t)= Re[I ejωt]

= Re

[V0ωC

+√1 + ω2R2C2ej (φ0−φ1)ejωt

]= V0ωC

+√1 + ω2R2C2cos(ωt + φ0 − φ1). (1.68)

In summary, we converted all time-varying quantities intothe phasor domain, solved for the phasor I of the desiredinstantaneous current i(t), and then converted back to the

Table 1-5: Time-domain sinusoidal functions z(t) andtheir cosine-reference phasor-domain equivalents Z, wherez(t) = Re[Zejωt ].

z(t) Z

A cosωt ↔ A

A cos(ωt + φ0) ↔ Aejφ0

A cos(ωt + βx + φ0) ↔ Aej(βx+φ0)

Ae−αx cos(ωt + βx + φ0) ↔ Ae−αxej (βx+φ0)

A sinωt ↔ Ae−jπ/2A sin(ωt + φ0) ↔ Aej(φ0−π/2)

d

dt(z1(t)) ↔ jωZ1

d

dt[A cos(ωt + φ0)] ↔ jωAejφ0

∫z1(t) dt ↔ 1

jωZ1∫

A sin(ωt + φ0) dt ↔ 1

jωAej(φ0−π/2)

time domain to obtain an expression for i(t). Table 1-5provides a summary of some time-domain functions andtheir phasor-domain equivalents.

Example 1-4 RL Circuit

The voltage source of the circuit shown in Fig. 1-20 isgiven by

vs(t) = 5 sin(4 × 104t − 30) (V). (1.69)

Obtain an expression for the voltage across the inductor.

Solution: The voltage loop equation of the RL circuit is

Ri + Ldi

dt= vs(t). (1.70)





R = 6 Ω

L = 0.2 mH

i

vs(t)

+

–

+

–

vL

Figure 1-20: RL circuit (Example 1-4).

Before converting Eq. (1.70) into the phasor domain, weneed to express Eq. (1.69) in terms of a cosine reference:

vs(t)= 5 sin(4 × 104t − 30)

= 5 cos(4 × 104t − 120) (V). (1.71)

The coefficient of t specifies the angular frequency asω = 4 × 104 (rad/s). The voltage phasor correspondingto vs(t) is

Vs = 5e−j120(V),

and the phasor equation corresponding to Eq. (1.70) is

RI + jωLI = Vs. (1.72)

Solving for the current phasor I , we have

I = Vs

R + jωL

= 5e−j120

6 + j4 × 104 × 2 × 10−4

= 5e−j120

6 + j8= 5e−j120

10ej53.1 = 0.5e−j173.1(A).

The voltage phasor across the inductor is related to I by

VL = jωLI

= j4 × 104 × 2 × 10−4 × 0.5e−j173.1

= 4ej (90−173.1) = 4e−j83.1(V),

and the corresponding instantaneous voltage vL(t) istherefore

vL(t)= Re[VLe

jωt]

= Re[4e−j83.1

ej4×104t]

= 4 cos(4 × 104t − 83.1) (V).

REVIEW QUESTIONS

Q1.12 Why is the phasor technique useful? When is itused? Describe the process.

Q1.13 How is the phasor technique used when theforcing function is a non-sinusoidal periodic waveform,such as a train of pulses?

EXERCISE 1.5 A series RL circuit is connected to avoltage source given by vs(t) = 150 cosωt (V). Find(a) the phasor current I and (b) the instantaneous currenti(t) for R = 400 , L = 3 mH, and ω = 105 rad/s.

Ans. (a) I = 150/(R + jωL) = 0.3∠−36.9 (A), (b)i(t) = 0.3 cos(ωt − 36.9) (A). (See C

DROM )

EXERCISE 1.6 A phasor voltage is given by V = j5 V.Find v(t).

Ans. v(t) = 5 cos(ωt + π/2) = −5 sinωt (V).(See C

DROM )

CHAPTER HIGHLIGHTS

• Electromagnetics is the study of electric and mag-netic phenomena and their engineering applications.

• The International System of Units consists of thesix fundamental dimensions listed in Table 1-1.The units of all other physical quantities can beexpressed in terms of the six fundamental units.




GLOSSARY OF IMPORTANT TERMS 35

• The four fundamental forces of nature are thenuclear, weak-interaction, electromagnetic andgravitational forces.

• The source of the electric field quantities Eand D is the electric charge q. In a material, Eand D are related by D = εE, where ε is theelectrical permittivity of the material. In free space,ε = ε0 (1/36π)× 10−9 (F/m).

• The source of the magnetic field quantities Band H is the electric current I . In a material, Band H are related by B = µH, where µ is themagnetic permeability of the medium. In free space,µ = µ0 = 4π × 10−7 (H/m).

• Electromagnetics consists of three branches: (1)electrostatics, which pertains to stationary charges,(2) magnetostatics, which pertains to steadycurrents, and (3) electrodynamics, which pertainsto time-varying currents.

• A traveling wave is characterized by a spatialwavelength λ, a time period T , and a phase velocityup = λ/T .

• An electromagnetic (EM) wave consists of oscillat-ing electric and magnetic field intensities and travelsin free space at the velocity of light c = 1/

√ε0µ0 .

The EM spectrum encompasses gamma rays, Xrays,visible light, infrared waves, and radio waves.

• Phasor analysis is a useful mathematical tool forsolving problems involving time-periodic sources.

GLOSSARY OF IMPORTANT TERMS

Provide definitions or explain the meaning of thefollowing terms:

SI system of unitsfundamental dimensionsCoulomb’s lawelectric field intensity Elaw of conservation of electric charge

principle of linear superpositionelectric dipoleelectric polarizationelectrical permittivity εrelative permittivity or dielectric constant εr

electric flux density Delectrostaticsmagnetostaticsmagnetic flux density Bmagnetic permeability µvelocity of light cnonmagnetic materialsmagnetic field intensity Helectrodynamicsconductivity σperfect dielectricperfect conductorconstitutive parameterstransient wavecontinuous harmonic wavewave amplitudewave period Twavelength λreference phase φ0

phase velocity (propagation velocity) up

wave frequency fangular velocity ωphase constant (wave number) βphase lag and leadattenuation factorattenuation constant αEM spectrummicrowave bandcomplex numberEuler’s identitycomplex conjugateforcing functionphasorinstantaneous function





PROBLEMS

Section 1-3: Traveling Waves

1.1∗ A 2-kHz sound wave traveling in the x-direction inair was observed to have a differential pressure p(x, t) =10 N/m2 at x = 0 and t = 50 µs. If the reference phaseof p(x, t) is 36, find a complete expression for p(x, t).The velocity of sound in air is 330 m/s.

1.2 For the pressure wave described in Example 1-1,plot the following:

(a) p(x, t) versus x at t = 0

(b) p(x, t) versus t at x = 0

Be sure to use appropriate scales for x and t so that eachof your plots covers at least two cycles.

1.3∗ A harmonic wave traveling along a string isgenerated by an oscillator that completes 180 vibrationsper minute. If it is observed that a given crest, ormaximum, travels 300 cm in 10 s, what is the wavelength?

C

DROM

1.4 Two waves, y1(t) and y2(t), have identicalamplitudes and oscillate at the same frequency, but y2(t)

leads y1(t) by a phase angle of 60. If

y1(t) = 4 cos(2π × 103t)

write the expression appropriate for y2(t) and plot bothfunctions over the time span from 0 to 2 ms.

1.5∗ The height of an ocean wave is described by thefunction

y(x, t) = 1.5 sin(0.5t − 0.6x) (m)

Determine the phase velocity and wavelength, and thensketch y(x, t) at t = 2s over the range from x = 0 tox = 2λ.

∗Answer(s) available in Appendix D.

C

DROM

Solution available in CD-ROM.

C

DROM

1.6 A wave traveling along a string in the +x-directionis given by

y1(x, t) = A cos(ωt − βx)

where x = 0 is the end of the string, which is tiedrigidly to a wall, as shown in Fig. 1-21. When wavey1(x, t) arrives at the wall, a reflected wave y2(x, t)

is generated. Hence, at any location on the string, thevertical displacement ys is the sum of the incident andreflected waves:

ys(x, t) = y1(x, t)+ y2(x, t)

(a) Write an expression for y2(x, t), keeping in mindits direction of travel and the fact that the end of thestring cannot move.

(b) Generate plots of y1(x, t), y2(x, t) and ys(x, t)

versus x over the range −2λ ≤ x ≤ 0 at ωt = π/4and at ωt = π/2.

x

x = 0

Incident Wave

y

Figure 1-21: Wave on a string tied to a wall at x = 0(Problem 1.6).

1.7∗ Two waves on a string are given by the followingfunctions:

y1(x, t)= 4 cos(20t − 30x) (cm)

y2(x, t)= −4 cos(20t + 30x) (cm)




PROBLEMS 37

where x is in centimeters. The waves are said to interfereconstructively when their superposition |ys| = |y1 + y2|is a maximum, and they interfere destructively when |ys|is a minimum.

(a) What are the directions of propagation of wavesy1(x, t) and y2(x, t)?

(b) At t = (π/50) s, at what location x do thetwo waves interfere constructively, and what is thecorresponding value of |ys|?

(c) At t = (π/50) s, at what location x do thetwo waves interfere destructively, and what is thecorresponding value of |ys|?

1.8 Give expressions for y(x, t) for a sinusoidal wavetraveling along a string in the negative x-direction, giventhat ymax = 40 cm, λ = 30 cm, f = 10 Hz, and

(a) y(x, 0) = 0 at x = 0

(b) y(x, 0) = 0 at x = 7.5 cm

1.9∗ An oscillator that generates a sinusoidal wave ona string completes 20 vibrations in 50 s. The wave peakis observed to travel a distance of 2.8 m along the stringin 5 s. What is the wavelength?

1.10 The vertical displacement of a string is given bythe harmonic function:

y(x, t) = 6 cos(16πt − 20πx) (m)

where x is the horizontal distance along the string inmeters. Suppose a tiny particle were attached to the stringatx = 5 cm. Obtain an expression for the vertical velocityof the particle as a function of time.

1.11∗ Given two waves characterized by the following:

y1(t)= 3 cosωt

y2(t)= 3 sin(ωt + 60)

does y2(t) lead or lag y1(t) and by what phase angle?

1.12 The voltage of an electromagnetic wave travelingon a transmission line is given by

v(z, t) = 5e−αz sin(4π × 109t − 20πz) (V)

where z is the distance in meters from the generator.

(a) Find the frequency, wavelength, and phase velocityof the wave.

(b) At z = 2 m, the amplitude of the wave was measuredto be 1 V. Find α.

1.13∗ A certain electromagnetic wave traveling inseawater was observed to have an amplitude of 98.02(V/m) at a depth of 10 m, and an amplitude of 81.87 (V/m)at a depth of 100 m. What is the attenuation constant ofseawater?

Section 1-5: Complex Numbers

1.14 Evaluate each of the following complex numbersand express the result in rectangular form:

(a) z1 = 4ejπ/3

(b) z2 = √3 ej3π/4

(c) z3 = 6e−jπ/2

(d) z4 = j 3

(e) z5 = j−4

(f) z6 = (1 − j)3

(g) z7 = (1 − j)1/2

1.15∗ Complex numbers z1 and z2 are given by thefollowing:

z1 = 3 − j2

z2 = −4 + j3





(a) Express z1 and z2 in polar form.

(b) Find |z1| by first applying Eq. (1.41) and then byapplying Eq. (1.43).

(c) Determine the product z1z2 in polar form.

(d) Determine the ratio z1/z2 in polar form.

(e) Determine z31 in polar form.

1.16 If z = −2+j4,determine the following quantitiesin polar form:

(a) 1/z

(b) z3

(c) |z|2

(d) Imz(e) Imz∗

1.17∗ Find complex numbers t = z1 + z2 ands = z1 − z2, both in polar form, for each of the followingpairs:

(a) z1 = 2 + j3 and z2 = 1 − j2

(b) z1 = 3 and z2 = −j3

(c) z1 = 3∠30 and z2 = 3∠−30

C

DROM (d) z1 = 3∠30 and z2 = 3∠−150

1.18 Complex numbers z1 and z2 are given by thefollowing:

z1 = 5∠−60

z2 = 2∠45 .

(a) Determine the product z1z2 in polar form.

(b) Determine the product z1z∗2 in polar form.

C

DROM (c) Determine the ratio z1/z2 in polar form.

(d) Determine the ratio z∗1/z∗2 in polar form.

(e) Determine√z1 in polar form.

1.19∗ If z = 3 − j5, find the value of ln(z).

1.20 If z = 3 − j4, find the value of ez.

Section 1-6: Phasors

1.21∗ A voltage source given by

vs(t) = 25 cos(2π × 103t − 30) (V)

is connected to a series RC load as shown in Fig. 1-19.If R = 1 M and C = 200 pF, obtain an expression forvc(t), the voltage across the capacitor.

1.22 Find the phasors of the following time functions:

(a) v(t) = 3 cos(ωt − π/3) (V)

(b) v(t) = 12 sin(ωt + π/4) (V)

(c) i(x, t) = 2e−3x sin(ωt + π/6) (A)

C

DROM (d) i(t) = −2 cos(ωt + 3π/4) (A)

(e) i(t) = 4 sin(ωt + π/3)+ 3 cos(ωt − π/6) (A)

1.23∗ Find the instantaneous time sinusoidal functionscorresponding to the following phasors:

(a) V = −5ejπ/3 (V)

(b) V = j6e−jπ/4 (V)

(c) I = (6 + j8) (A)

C

DROM (d) I = −3 + j2 (A)

(e) I = j (A)

(f) I = 2ejπ/6 (A)




PROBLEMS 39

1.24 A series RLC circuit is connected to a generatorwith a voltage vs(t) = V0 cos(ωt + π/3) (V).

(a) Write the voltage loop equation in terms of thecurrent i(t), R, L, C, and vs(t).

(b) Obtain the corresponding phasor-domain equation.

(c) Solve the equation to obtain an expression for thephasor current I .

1.25–1.29 Additional Solved Problems — completesolutions on C

DROM

.




2-1 General Considerations

In most electrical engineering curricula, the study ofelectromagnetics is preceded by one or more courses onelectrical circuits. In this book, we use this background tobuild a bridge between circuit theory and electromagnetictheory. The bridge is provided by transmission lines,the topic of this chapter. By modeling the transmissionline in the form of an equivalent circuit, we can useKirchhoff’s voltage and current laws to develop waveequations whose solutions provide an understandingof wave propagation, standing waves, and powertransfer. Familiarity with these concepts facilitates thepresentation of material in later chapters.

Although the family of transmission lines mayencompass all structures and media that serve to transferenergy or information between two points, includingnerve fibers in the human body, acoustic waves in fluids,and mechanical pressure waves in solids, we shall focusour treatment in this chapter on transmission lines used

Sending-endport

A

A'

B

B'

Transmission line

Load circuitGenerator circuit

Receiving-endport

+

–

Vg

Rg

RL

Figure 2-1: A transmission line is a two-port network connecting a generator circuit at the sending end to a load at thereceiving end.

for guiding electromagnetic signals. Such transmissionlines include telephone wires, coaxial cables carryingaudio and video information to TV sets or digital datato computer monitors, and optical fibers carrying lightwaves for the transmission of data at very high rates.Fundamentally, a transmission line is a two-port network,with each port consisting of two terminals, as illustratedin Fig. 2-1. One of the ports is the sending end andthe other is the receiving end. The source connectedto its sending end may be any circuit with an outputvoltage, such as a radar transmitter, an amplifier, or acomputer terminal operating in the transmission mode.From circuit theory, any such source can be representedby a Thevenin-equivalent generator circuit consistingof a generator voltage Vg in series with a generatorresistanceRg, as shown in Fig. 2-1. The generator voltagemay consist of digital pulses, a modulated time-varyingsinusoidal signal, or any other signal waveform. In thecase of a-c signals, the generator circuit is represented bya voltage phasor Vg and an impedance Zg.

41




42 CHAPTER 2 TRANSMISSION LINES

The circuit connected to the receiving end of thetransmission line is called the load circuit, or simply theload. This may be an antenna in the case of a radar, a com-puter terminal operating in the receiving mode, the inputterminals of an amplifier, or any output circuit whoseinput terminals can be represented by an equivalent loadresistance RL, or a load impedance ZL in the a-c case.

2-1.1 The Role of Wavelength

In low-frequency electrical circuits, we usually use wiresto connect the elements of the circuit in the desiredconfiguration. In the circuit shown in Fig. 2-2, forexample, the generator is connected to a simple RCload via a pair of wires. In view of our definition in thepreceding paragraphs of what constitutes a transmissionline, we pose the following question: Is the pair of wiresbetween terminalsAA′ and terminalsBB ′ a transmissionline? If so, why is it important? After all, we usuallysolve for the current in the circuit and the voltage acrossits elements without regard for the wires connecting theelements. The answer to this question is yes; indeedthe pair of wires constitutes a transmission line, but theimpact of the line on the current and voltages in the circuitdepends on the length of the line l and the frequency f

C

R

i

Vg VAA'

l

A

A'

B

B'

+

+

–

–

VBB'

+

–

Transmission line

Figure 2-2: Generator connected to anRC circuit througha transmission line of length l.

of the signal pagebreak provided by the generator. (Aswe will see later, the determining factor is the ratio ofthe length l to the wavelength λ of the wave propagatingon the transmission line between AA′ and BB ′.) If thegenerator voltage is cosinusoidal in time, then the voltageacross the input terminals AA′ is

VAA′ = Vg(t) = V0 cosωt (V), (2.1)

where ω = 2πf is the angular frequency, and if weassume that the current flowing through the wires travelsat the speed of light, c = 3 × 108 m/s, then the voltageacross the output terminals BB ′ will have to be delayedin time relative to that across AA′ by the travel delaytime l/c. Thus, assuming no significant ohmic losses inthe transmission line,

VBB ′(t)= VAA′(t − l/c)

= V0 cos [ω(t − l/c)] (V). (2.2)

Let us compare VBB ′ to VAA′ at t = 0 for anultralow-frequency electronic circuit operating at afrequency f = 1 kHz. For a typical wire lengthl = 5 cm, Eqs. (2.1) and (2.2) give VAA′ = V0 andVBB ′ = V0 cos(2πf l/c) = 0.999999999998V0. Thus,for all practical purposes, the length of the transmissionline may be ignored and terminal AA′ may be treatedas identical with BB ′. On the other hand, had the linebeen a 20-km long telephone cable carrying a 1-kHzvoice signal, then the same calculation would haveled to VBB ′ = 0.91V0. The determining factor is themagnitude of ωl/c. From Eq. (1.27), the velocity ofpropagation up of a traveling wave is related to theoscillation frequency f and the wavelength λ by

up = f λ (m/s). (2.3)

In the present case, up = c. Hence, the phase factor

ωl

c= 2πf l

c= 2π

l

λradians. (2.4)

When l/λ is very small, transmission-line effects may beignored, but when l/λ 0.01, it may be necessary to




2-1 GENERAL CONSIDERATIONS 43

Dispersionless line

Short dispersive line

Long dispersive line

Figure 2-3: A dispersionless line does not distort signalspassing through it regardless of its length, whereas adispersive line distorts the shape of the input pulsesbecause the different frequency components propagateat different velocities. The degree of distortion isproportional to the length of the dispersive line.

account not only for the phase shift associated with thetime delay, but also for the presence of reflected signalsthat may have been bounced back by the load towardthe generator. Power loss on the line and dispersiveeffects may need to be considered as well. A dispersivetransmission line is one on which the wave velocity is notconstant as a function of the frequency f . This meansthat the shape of a rectangular pulse, which throughFourier analysis is composed of many waves of differentfrequencies, will be distorted as it travels down theline because its different frequency components will notpropagate at the same velocity (Fig. 2-3). Preservationof pulse shape is very important in high-speed datatransmission, both between terminals as well as inhigh-speed integrated circuits in which transmission-linedesign and fabrication processes are an integral part ofthe IC design process. At 10 GHz, for example, thewavelengthλ = 3 cm in air and is on the order of 1 cm in asemiconductor material. Hence, even connection lengths

between devices on the order of millimeters becomesignificant, and their presence has to be incorporated inthe overall design of the circuit.

2-1.2 Propagation Modes

A few examples of common types of transmissionlines are shown in Fig. 2-4. Transmission lines may beclassified into two basic types:

• Transverse electromagnetic (TEM) transmissionlines: Waves propagating along these lines arecharacterized by electric and magnetic fields that areentirely transverse to the direction of propagation.This is called a TEM mode. A good example isthe coaxial line shown in Fig. 2-5; the electricfield lines are in the radial direction between theinner and outer conductors, the magnetic field formscircles around the inner conductor, and hence neitherhas any components along the length of the line(the direction of wave propagation). Other TEMtransmission lines include the two-wire line andthe parallel-plate line, both shown in Fig. 2-4.Although the fields present on a microstrip line donot adhere to the exact definition of a TEM mode,the nontransverse field components are sufficientlysmall in comparison to the transverse componentsto be ignored, thereby allowing the inclusion ofmicrostrip lines in the TEM class. A commonfeature among TEM lines is that they consist of twoparallel conducting surfaces.

• Higher-order transmission lines: Waves propagat-ing along these lines have at least one significantfield component in the direction of propagation.Hollow conducting waveguides, dielectric rods, andoptical fibers belong to this class of lines.

Only TEM-mode transmission lines will be treated inthis chapter. This is because less mathematical rigor isrequired for treating this class of lines than that requiredfor treating waves characterized by higher-order modes





2a

d

metal

dielectric spacing

w

d

metal

metal

dielectric spacing

metal

2a

2b

dielectric spacing

metal strip conductor

dielectric spacing

w

d

metal ground plane

dielectric spacingmetal ground plane

Concentricdielectriclayers

metal

(a) Coaxial line

(d) Strip line

(f) Rectangular waveguide (g) Optical fiber (h) Coplanar waveguide

(e) Microstrip line

(b) Two-wire line (c) Parallel-plate line

TEM Transmission Lines

Higher Order Transmission Lines

Figure 2-4: A few examples of transverse electromagnetic (TEM) and higher-order transmission lines.

and, in addition, TEM lines are more commonly usedin practice. We start our treatment by representing thetransmission line in terms of a lumped-element circuitmodel, and then we apply Kirchhoff’s voltage and currentlaws to derive a set of two governing equations knownas the telegrapher’s equations. By combining these

equations, we obtain wave equations for the voltage andcurrent at any point on the line. Solution of the waveequations for the sinusoidal steady-state case leads toa set of formulas that can be used for solving a widerange of practical problems. In the latter part of thischapter we introduce a graphical technique known as




2-2 LUMPED-ELEMENT MODEL 45

Vg

Rg

RL

LoadCross section

Magnetic field lines

Electric field lines

Generator

Coaxial line

+

–

Figure 2-5: In a coaxial line, the electric field lines are in the radial direction between the inner andouter conductors, and the magnetic field forms circles around the inner conductor.

the Smith chart, which facilitates the solution of manytransmission-line problems without having to performlaborious calculations involving complex numbers.

2-2 Lumped-Element Model

When we draw a schematic of an electronic circuit, weuse specific symbols to represent resistors, capacitors,inductors, diodes, and the like. In each case, the symbolrepresents the functionality of the device, rather than itsshape, size or other attributes. We shall do the same withregard to transmission lines; we shall represent a trans-mission line by a parallel-wire configuration, as shownin Fig. 2-6(a), regardless of the specific shape of the lineunder consideration. Thus, Fig. 2-6(a) may represent acoaxial line, a two-wire line, or any other TEM line.

Drawing again on our familiarity with electroniccircuits, when we analyze a circuit containing a transistor,we represent the functionality of the transistor byan equivalent circuit composed of sources, resistors,and capacitors. We will apply the same approachto the transmission line by orienting the line alongthe z-direction, subdividing it into differential sections

each of length z [Fig. 2-6(b)] and then representingeach section by an equivalent circuit, as illustratedin Fig. 2-6(c). This representation, which is calledthe lumped-element circuit model, consists of fourbasic elements, which henceforth will be called thetransmission line parameters. These are

R′: The combined resistance of both conductors perunit length, in /m,

L′: The combined inductance of both conductors perunit length, in H/m,

G′: The conductance of the insulation medium per unitlength, in S/m, and

C ′: The capacitance of the two conductors per unitlength, in F/m.

Whereas the four line parameters have differentexpressions for different types and dimensions oftransmission lines, the equivalent model represented byFig. 2-6(c) is equally applicable to all transmissionlines characterized by TEM-mode wave propagation. Theprime superscript is used as a reminder that the lineparameters are differential quantities whose units are perunit length.





R'∆z L'∆z R'∆z

G'∆z C'∆z

L'∆z R'∆z

G'∆z C'∆z

L'∆z R'∆z

G'∆z C'∆z

L'∆z

∆z

∆z ∆z ∆z ∆z

∆z ∆z ∆z

(a) Parallel-wire representation

(b) Differential sections each ∆z long

(c) Each section is represented by an equivalent circuit

G'∆z C'∆z

Figure 2-6: Regardless of its actual shape, a TEM transmission line is represented by the parallel-wire configuration shownin (a). To analyze the voltage and current relations, the line is subdivided into small differential sections (b), each of which isthen represented by an equivalent circuit (c).

Expressions for the line parameters R′, L′,G′, and C ′are given in Table 2-1 for the three types of TEMtransmission lines diagrammed in parts (a) through (c)of Fig. 2-4. For each of these lines, the expressionsare functions of two sets of parameters: (1) geometricparameters defining the cross-sectional dimensions ofthe given line and (2) electromagnetic constitutiveparameters characteristic of the materials of which theconductors and the insulating material between them aremade. The pertinent geometric parameters are as follows:

Coaxial line [Fig. 2-4(a)]:

a = outer radius of inner conductor, mb = inner radius of outer conductor, m

Two-wire line [Fig. 2-4(b)]:

a = radius of each wire, md = spacing between wires’ centers, m

Parallel-plate line [Fig. 2-4(c)]:

w = width of each plate, md = thickness of insulation between plates, m




2-2 LUMPED-ELEMENT MODEL 47

Table 2-1: Transmission-line parameters R′, L′, G′, and C ′ for three types of lines.

Parameter Coaxial Two Wire Parallel Plate Unit

R′ Rs

2π

(1

a+ 1

b

)Rs

πa

2Rs

w/m

L′ µ

2πln(b/a)

µ

πln

[(d/2a)+

√(d/2a)2 − 1

]µd

wH/m

G′ 2πσ

ln(b/a)

πσ

ln[(d/2a)+√(d/2a)2 − 1

] σw

dS/m

C ′ 2πε

ln(b/a)

πε

ln[(d/2a)+√(d/2a)2 − 1

] εw

dF/m

Notes: (1) Refer to Fig. 2-4 for definitions of dimensions. (2) µ, ε, and σ pertain to theinsulating material between the conductors. (3) Rs = √

πfµc/σc. (4) µc and σc pertainto the conductors. (5) If (d/2a)2 1, then ln

[(d/2a)+√(d/2a)2 − 1

] ln(d/a).

The constitutive parameters apply to all three lines andconsist of two groups: µc and σc are the magnetic per-meability and electrical conductivity of the conductors,and ε, µ, and σ are the electrical permittivity, magneticpermeability, and electrical conductivity of the insulationmaterial separating the conductors. Appendix B containstabulated values for these constitutive parameters forvarious types of materials. For the purposes of thepresent chapter, we need not concern ourselves with thederivations responsible for the expressions given in Table2-1. The formulations necessary for computing R′, L′,G′, and C ′ will be made available in later chapters forthe general case of any two-conductor configuration.

The lumped-element model shown in Fig. 2-6(c)represents the physical processes associated with thecurrents and voltages on any TEM transmission line.Other equivalent models are available also and are equally

applicable as well. All these models, however, lead toexactly the same set of telegrapher’s equations, fromwhich all our future results will be derived. Hence, onlythe model described in Fig. 2-6(c) will be examined inthe present treatment. It consists of two series elements,R′ andL′, and two shunt elements,G′ andC ′. By way ofproviding a physical explanation for the lumped-elementmodel, let us consider a small section of a coaxial line, asshown in Fig. 2-7. The line consists of an inner conductorof radiusa separated from an outer conducting cylinder ofradius b by a material with permittivity ε, permeabilityµ,and conductivity σ . The two metal conductors are madeof a material with conductivity σc and permeability µc.When a voltage source is connected across the twoconductors at the sending end of the line, currents willflow through the conductors, primarily along the outersurface of the inner conductor and the inner surface of the





(µc, σc)

b

a

(ε, µ, σ)

Conductors

Insulating material

Figure 2-7: Cross section of a coaxial line with innerconductor of radius a and outer conductor of radius b.The conductors have magnetic permeability µc, andconductivity σc, and the spacing material between theconductors has permittivity ε, permeability µ, andconductivity σ .

outer conductor. The line resistance R′ accounts for thecombined resistance per unit length of the inner and outerconductors. The expression forR′ is derived in Chapter 7and is given by Eq. (7.96) as

R′ = Rs

2π

(1

a+ 1

b

)(/m), (2.5)

where Rs, which represents the surface resistance of theconductors, is called the intrinsic resistance and is givenby Eq. (7.92a) as

Rs =√πfµc

σc(). (2.6)

The intrinsic resistance depends not only on the materialproperties of the conductors (σc and µc), but on thefrequency f of the wave traveling on the line as well. Fora perfect conductor with σc = ∞ or a high-conductivity

material such that (fµc/σc) 1, Rs approaches zero,and so does R′.

Next, let us examine the inductance per unit length L′.Application of Ampere’s law in Chapter 5 to thedefinition of inductance leads to the following expression[Eq. (5.99)] for the inductance per unit length of a coaxialline:

L′ = µ

2πln

(b

a

)(H/m). (2.7)

The shunt conductance per unit length G′ accountsfor current flow between the outer and inner conductors,made possible by the material conductivity σ of theinsulator. It is precisely because the current flow is fromone conductor to the other that G′ is a shunt elementin the lumped-element model. Its expression is given byEq. (4.76) as

G′ = 2πσ

ln(b/a)(S/m). (2.8)

If the material separating the inner and outer conductorsis a perfect dielectric with σ = 0, then G′ = 0.

The last line parameter on our list is the capacitanceper unit length C ′. When equal and opposite charges areplaced on any two noncontacting conductors, a voltagedifference will be induced between them. Capacitance isdefined as the ratio of charge to voltage difference. Forthe coaxial line, C ′ is given by Eq. (4.117) as

C ′ = 2πε

ln(b/a)(F/m). (2.9)

All TEM transmission lines share the following usefulrelations:

L′C ′ = µε, (2.10)

and

G′

C ′ = σ

ε. (2.11)




2-3 TRANSMISSION-LINE EQUATIONS 49

If the insulating medium between the conductors is air,the transmission line is called an air line (e.g., coaxialair line or two-wire air line). For an air line, ε = ε0 =8.854 × 10−12 F/m, µ = µ0 = 4π × 10−7 H/m, σ = 0,and G′ = 0.

REVIEW QUESTIONS

Q2.1 What is a transmission line? When should trans-mission-line effects be considered?

Q2.2 What is the difference between dispersive andnondispersive transmission lines? What is the practicalsignificance?

Q2.3 What constitutes a TEM transmission line?

Q2.4 What purpose does the lumped-element circuitmodel serve? How are the line parameters R′, L′, G′,and C ′ related to the physical and electromagneticconstitutive properties of the transmission line?

EXERCISE 2.1 Use Table 2-1 to compute the lineparameters of a two-wire air line whose wires areseparated by a distance of 2 cm, and each is 1 mm inradius. The wires may be treated as perfect conductorswith σc = ∞.

Ans. R′ = 0, L′ = 1.20 (µH/m), G′ = 0,C ′ = 9.29 (pF/m). (See C

DROM )

EXERCISE 2.2 Calculate the transmission line parame-ters at 1 MHz for a rigid coaxial air line with an innerconductor diameter of 0.6 cm and an outer conductordiameter of 1.2 cm. The conductors are made of copper[see Appendix B for µc and σc of copper].

Ans. R′ = 2.08 × 10−2 (/m), L′ = 0.14 (µH/m),G′ = 0, C ′ = 80.3 (pF/m). (See C

DROM )

2-3 Transmission-Line Equations

A transmission line usually connects a source on oneend to a load on the other end. Before we consider thecomplete circuit, however, we need to develop equationsthat describe the voltage across the transmission line andthe current carried by the line as a function of time tand spatial position z. Using the lumped-element modeldescribed in Fig. 2-6(c), we begin by considering adifferential length z as shown in Fig. 2-8. The quantitiesv(z, t) and i(z, t) denote the instantaneous voltage andcurrent at the left end of the differential section (nodeN ),and similarly v(z + z, t) and i(z + z, t) denote thesame quantities at the right end (nodeN+1). Applicationof Kirchhoff’s voltage law accounts for the voltage dropacross the series resistance R′N

v ( z + z, t)


Upon dividing all terms by z and rearranging terms, weobtain

−[v(z+ z, t)− v(z, t)

z

]= R′ i(z, t)+ L′ ∂ i(z, t)

∂t. (2.13)

In the limit as z → 0, Eq. (2.13) becomes a differentialequation:

− ∂v(z, t)

∂z= R′ i(z, t)+ L′ ∂i(z, t)

∂t. (2.14)

Similarly, application of Kirchhoff’s current law at nodeN + 1 in Fig. 2-8 leads to

i(z, t)−G′ z v(z+ z, t)

− C ′ z∂v(z+ z, t)

∂t− i(z+ z, t)= 0. (2.15)

Upon dividing all terms by z and taking the limitas z → 0, Eq. (2.15) provides a second differentialequation,

−∂i(z, t)∂z

= G′ v(z, t)+ C ′ ∂v(z, t)∂t

. (2.16)

The first-order differential equations given by Eqs. (2.14)and (2.16) are the time-domain form of the transmissionline equations, otherwise called the telegrapher’sequations.

Except for the last section, our primary interest inthis chapter is in sinusoidal steady-state conditions. Tothis end, we shall make use of phasors with the cosinereference notation as outlined in Section 1-6. Thus, wedefine

v(z, t)= Re[V (z) ejωt ], (2.17a)

i(z, t)= Re[I (z) ejωt ], (2.17b)

where V (z) and I (z) are phasor quantities, each of whichmay be real or complex. Upon substituting Eqs. (2.17a)and (2.17b) into Eqs. (2.14) and (2.16) and utilizing the

property given by Eq. (1.62) that ∂/∂t in the time domainbecomes equivalent to multiplication by jω in the phasordomain, we obtain the following pair of equations:

−dV (z)dz

= (R′ + jωL′) I (z), (2.18a)

−dI (z)dz

= (G′ + jωC ′) V (z). (2.18b)

These are the telegrapher’s equations in phasor form.

2-4 Wave Propagation on a TransmissionLine

The two first-order coupled equations given byEqs. (2.18a) and (2.18b) can be combined to give twosecond-order uncoupled wave equations, one for V (z)and another for I (z). The wave equation for V (z) isderived by differentiating both sides of Eq. (2.18a) withrespect to z, giving

−d2V (z)

dz2= (R′ + jωL′)

dI (z)

dz, (2.19)

and upon substituting Eq. (2.18b) into Eq. (2.19) fordI (z)/dz, Eq. (2.19) becomes

d2V (z)

dz2− (R′ + jωL′)(G′ + jωC ′) V (z) = 0, (2.20)

or

d2V (z)

dz2− γ 2 V (z) = 0, (2.21)

where

γ = √(R′ + jωL′)(G′ + jωC ′) . (2.22)




2-4 WAVE PROPAGATION ON A TRANSMISSION LINE 51

Application of the same steps to Eqs. (2.18a) and (2.18b)but in reverse order, leads to

d2I (z)

dz2− γ 2 I (z) = 0. (2.23)

Equations (2.21) and (2.23) are called wave equations forV (z) and I (z), respectively, and γ is called the complexpropagation constant of the transmission line. As such,γ consists of a real partα, called the attenuation constantof the line with units of Np/m, and an imaginary part β,called the phase constant of the line with units of rad/m.Thus,

γ = α + jβ (2.24)

with

α = Re(γ )

= Re(√(R′ + jωL′)(G′ + jωC ′)

)(Np/m),

(2.25a)

β = Im(γ )

= Im(√(R′ + jωL′)(G′ + jωC ′)

)(rad/m).

(2.25b)

In Eqs. (2.25a) and (2.25b), we choose the square-rootvalues that give positive values for α and β. For passivetransmission lines, α is either zero or positive. Mosttransmission lines, and all those considered in thischapter, are of the passive type. The active region of alaser is an example of an active transmission line with anegative α.

The wave equations given by Eqs. (2.21) and (2.23)have traveling wave solutions of the following form:

V (z)= V +0 e

−γ z + V −0 e

γ z (V), (2.26a)

I (z)= I+0 e

−γ z + I−0 e

γ z (A), (2.26b)

where, as will be shown later, the e−γ z term representswave propagation in the +z-direction and the eγ z

term represents wave propagation in the −z-direction.Verification that these are indeed valid solutions iseasily accomplished by substituting the expressions givenby Eqs. (2.26a) and (2.26b), as well as their secondderivatives, into Eqs. (2.21) and (2.23). In their presentform, the solutions given by Eqs. (2.26a) and (2.26b)contain four unknowns, the wave amplitudes (V +

0 , I+0 )

of the +z propagating wave and (V −0 , I

−0 ) of the −z

propagating wave. We can easily relate the current waveamplitudes, I+

0 and I−0 , to the voltage wave amplitudes,

V +0 and V −

0 , respectively, by using Eq. (2.26a) inEq. (2.18a) and then solving for the current I (z) to getthe result

I (z) = γ

R′ + jωL′[V +

0 e−γ z − V −

0 eγ z]. (2.27)

Comparison of each term with the corresponding termin the expression given by Eq. (2.26b) leads to theconclusion that

V +0

I+0

= Z0 = −V −0

I−0

, (2.28)

where

Z0 = R′ + jωL′

γ=√R′ + jωL′

G′ + jωC ′ (), (2.29)

is defined as the characteristic impedance of the line.It should be noted that Z0 is equal to the ratio of thevoltage amplitude to the current amplitude for each of thetraveling waves individually (with an additional minussign in the case of the −z propagating wave), but it is notequal to the ratio of the total voltage V (z) to the totalcurrent I (z), unless one of the two waves is absent. Interms of Z0, Eq. (2.27) can be rewritten in the form

I (z) = V +0

Z0e−γ z − V −

0

Z0eγ z. (2.30)





In later sections, we will apply boundary conditions atthe load and at the sending end of the transmission lineto obtain expressions for the remaining wave amplitudesV +

0 and V −0 . In general, each will be a complex quantity

composed of a magnitude and a phase angle. Thus,

V +0 = |V +

0 |ejφ+, (2.31a)

V −0 = |V −

0 |ejφ−. (2.31b)

Upon substituting these definitions in Eq. (2.26a) andreplacing γ with Eq. (2.24), we can convert back tothe time domain to obtain an expression for v(z, t), theinstantaneous voltage on the line:

v(z, t)= Re(V (z)ejωt )

= Re[(V +

0 e−γ z + V −

0 eγ z)ejωt]

= Re[|V +0 |ejφ+

ejωte−(α+jβ)z

+ |V −0 |ejφ−

ejωte(α+jβ)z]= |V +

0 |e−αz cos(ωt − βz+ φ+)

+ |V −0 |eαz cos(ωt + βz+ φ−). (2.32)

From our review of traveling waves in Section 1-3, werecognize the first term in Eq. (2.32) as a wave travelingin the +z-direction (the coefficients of t and z haveopposite signs) and the second term as a wave travelingin the −z-direction (the coefficients of t and z are bothpositive), both propagating with a phase velocityup givenby Eq. (1.30):

up = f λ = ω

β. (2.33)

The factor e−αz accounts for the attenuation of the+z propagating wave, and the eαz accounts for theattenuation of the −z propagating wave. The presenceof two waves on the line propagating in oppositedirections produces a standing wave. To gain a physicalunderstanding of what this means, we shall first examinethe relatively simple but important case of a lossless line

(α = 0) and then extend the results to the more generalcase of lossy transmission lines (α = 0). In fact, we shalldevote the next several sections to the study of losslesstransmission lines because in practice many lines can bedesigned to exhibit very low-loss characteristics.

Example 2-1 Air Line

An air line is a transmission line for which air is thedielectric material present between the two conductors,which renders G′ = 0. In addition, the conductors aremade of a material with high conductivity so thatR′ 0.For an air line with characteristic impedance of 50 and phase constant of 20 rad/m at 700 MHz, find theinductance per meter and the capacitance per meter ofthe line.

Solution: The following quantities are given:

Z0 = 50 , β = 20 rad/m,

f = 700 MHz = 7 × 108 Hz.

With R′ = G′ = 0, Eqs. (2.25b) and (2.29) reduce to

β = Im(√(jωL′)(jωC ′)

)= Im(jω

√L′C ′)

= ω√L′C ′ ,

Z0 =√jωL′

jωC ′ =√L′

C ′ .

The ratio is given by

β

Z0= ωC ′,

or

C ′ = β

ωZ0

= 20

2π × 7 × 108 × 50

= 9.09 × 10−11 (F/m) = 90.9 (pF/m).




2-5 THE LOSSLESS TRANSMISSION LINE 53

From Z0 = √L′/C ′,

L′ = Z20C

′

= (50)2 × 90.9 × 10−12

= 2.27 × 10−7 (H/m) = 227 (nH/m).

EXERCISE 2.3 Verify that Eq. (2.26a) is indeed a solutionof the wave equation given by Eq. (2.21). (See C

DROM )

EXERCISE 2.4 A two-wire air line has the following lineparameters: R′ = 0.404 (m/m), L′ = 2.0 (µH/m),G′ = 0, and C ′ = 5.56 (pF/m). For operation at5 kHz, determine (a) the attenuation constant α, (b) thephase constant β, (c) the phase velocity up, and (d) thecharacteristic impedance Z0. (See C

DROM )

Ans. (a) α = 3.37 × 10−7 (Np/m), (b) β = 1.05 ×10−4 (rad/m), (c) up = 3.0 × 108 (m/s), (d)Z0 = (600 − j2.0) = 600∠−0.19

.

2-5 The Lossless Transmission Line

According to the preceding section, a transmissionline is characterized by two fundamental properties, itspropagation constant γ and characteristic impedanceZ0,both of which are specified by the angular frequency ωand the line parameters R′, L′, G′, and C ′. Inmany practical situations, the transmission line canbe designed to minimize ohmic losses by selectingconductors with very high conductivities and dielectricmaterials (separating the conductors) with negligibleconductivities. As a result, R′ and G′ assume very smallvalues such that R′ ωL′ and G′ ωC ′. Theselossless-line conditions allow us to set R′ = G′ = 0in Eq. (2.22), which then gives the result

γ = α + jβ = jω√L′C ′ , (2.34)

which means that

α = 0 (lossless line),

β = ω√L′C ′ (lossless line). (2.35)

Application of the lossless-line conditions to Eq. (2.29)gives the characteristic impedance as

Z0 =√L′

C ′ (lossless line), (2.36)

which is now a real number. Using the lossless-lineexpression for β given by Eq. (2.35), we obtain thefollowing relations for the wavelength λ and the phasevelocity up:

λ= 2π

β= 2π

ω√L′C ′ , (2.37)

up = ω

β= 1√

L′C ′ . (2.38)

Upon using the relation given by Eq. (2.10), which isshared by all TEM transmission lines, Eqs. (2.35) and(2.38) may be rewritten as

β = ω√µε (rad/m), (2.39)

up = 1√µε

(m/s), (2.40)

where µ and ε are, respectively, the magnetic permeabil-ity and electrical permittivity of the insulating materialseparating the conductors. Materials used for this purposeare usually characterized by a permeability µ = µ0,where µ0 = 4π × 10−7 H/m is the permeability of freespace, and the permittivity is usually specified in termsof the relative permittivity εr defined as

εr = ε/ε0, (2.41)





where ε0 = 8.854×10−12 F/m (1/36π)×10−9 F/m isthe permittivity of free space. Hence, Eq. (2.40) becomes

up = 1√µ0εrε0

= 1√µ0ε0

· 1√εr

= c√εr, (2.42)

where c = 1/√µ0ε0 = 3 × 108 m/s is the velocity of

light in a vacuum. If the insulating material between theconductors is air, then εr = 1 and up = c. In view ofEq. (2.41) and the relationship between λ and up givenby Eq. (2.33), the wavelength is given by

λ = up

f= c

f

1√εr

= λ0√εr, (2.43)

where λ0 = c/f is the wavelength in air corresponding toa frequency f . Note that, because both up and λ dependon εr, the choice of the type of insulating material used ina transmission line is dictated not only by its mechanicalproperties, but by its electrical properties as well.

When the phase velocity of a medium is independentof frequency, the medium is called nondispersive, whichclearly is the case for a lossless TEM transmission line.This is an important feature for the transmission of digitaldata in the form of pulses. A rectangular pulse or a seriesof pulses is composed of many Fourier components withdifferent frequencies. If the phase velocity is the same forall frequency components (or at least for the dominantones), the pulse shape will remain the same as the pulsetravels on the line. In contrast, the shape of a pulsepropagating in a dispersive medium gets progressivelydistorted, and the pulse length increases (stretches out)as a function of distance in the medium, thereby imposinga limitation on the maximum data rate (which is relatedto the length of the individual pulses and the spacingbetween adjacent pulses) that can be transmitted throughthe medium without loss of information.

Table 2-2 provides a list of the expressions for γ , Z0,and up for the general case of a lossy line and for several

types of lossless lines. The expressions for the losslesslines are based on the equations for L′ and C ′ given inTable 2-1.

EXERCISE 2.5 For a lossless transmission line,λ = 20.7 cm at 1 GHz. Find εr of the insulating material.

Ans. εr = 2.1. (See C

DROM )

EXERCISE 2.6 A lossless transmission line uses adielectric insulating material with εr = 4. If its linecapacitance is C ′ = 10 (pF/m), find (a) the phasevelocity up, (b) the line inductance L′, and (c) thecharacteristic impedance Z0.

Ans. (a) up = 1.5 × 108 (m/s), (b) L′ = 4.45 (µH/m),(c) Z0 = 667.1 . (See C

DROM )

2-5.1 Voltage Reflection Coefficient

With γ = jβ for the lossless line, the expressions givenby Eqs. (2.26a) and (2.30) for the total voltage and currenton the line become

V (z)= V +0 e

−jβz + V −0 e

jβz, (2.44a)

I (z)= V +0

Z0e−jβz − V −

0

Z0ejβz. (2.44b)

These expressions contain two unknowns, V +0 and V −

0 ,the voltage amplitudes of the incident and reflectedwaves, respectively. To determine V +

0 and V −0 , we need

to consider the lossless transmission line in the contextof the complete circuit, including a generator circuit atits input terminals and a load at its output terminals, asshown in Fig. 2-9. The line, of length l, is terminatedin an arbitrary load impedance ZL.For convenience, thereference of the spatial coordinate z is chosen such thatz = 0 corresponds to the location of the load. At thesending end at z = −l, the line is connected to asinusoidal voltage source with phasor Vg and an internal





Table 2-2: Characteristic parameters of transmission lines.

Propagation Phase CharacteristicConstant Velocity Impedanceγ = α + jβ up Z0

General case γ = √(R′ + jωL′)(G′ + jωC ′) up = ω/β Z0 =√(R′ + jωL′)(G′ + jωC ′)

Lossless α = 0, β = ω√εr/c up = c/

√εr Z0 = √L′/C ′

(R′ = G′ = 0)

Lossless coaxial α = 0, β = ω√εr/c up = c/

√εr Z0 = (60/

√εr)

ln(b/a)


√εr Z0 = (120/

√εr)

two wire · ln[(d/2a)+√(d/2a)2 − 1]Z0 (120/

√εr)

ln(d/a),if d a


√εr Z0 = (120π/

√εr)(d/w)

parallel plate

Notes: (1) µ = µ0, ε = εrε0, c = 1/√µ0ε0, and

√µ0/ε0 (120π) , where εr is the relative

permittivity of insulating material. (2) For coaxial line, a and b are radii of inner and outer conductors.(3) For two-wire line, a = wire radius and d = separation between wire centers. (4) For parallel-plateline, w = width of plate and d = separation between the plates.

impedance Zg. At the load, the phasor voltage across it,VL, and the phasor current through it, IL, are related bythe load impedance ZL as follows:

ZL = VL

IL. (2.45)

The voltage VL is equal to the total voltage on the lineV (z) given by Eq. (2.44a), and IL is equal to I (z) givenby Eq. (2.44b), both evaluated at z = 0:

VL = V (z=0) = V +0 + V −

0 , (2.46a)

IL = I (z=0) = V +0

Z0− V −

0

Z0. (2.46b)

Upon using these expressions in Eq. (2.45), we obtain theresult:

ZL =(V +

0 + V −0

V +0 − V −

0

)Z0. (2.47)

Solving for V −0 gives

V −0 =(ZL − Z0

ZL + Z0

)V +

0 . (2.48)

The ratio of the amplitude of the reflected voltage waveto the amplitude of the incident voltage wave at the load





Vg

IiZg

Z0ZL

~

Vi~~

+

+

–

–

VL~

IL~+

–

Transmission line

Generator Loadz = –l z = 0

Figure 2-9: Transmission line of length l connected onone end to a generator circuit and on the other end to aload ZL. The load is located at z = 0 and the generatorterminals are at z = −l.

is known as the voltage reflection coefficient . FromEq. (2.48), this definition gives the result

= V −0

V +0

= ZL − Z0

ZL + Z0

= ZL/Z0 − 1

ZL/Z0 + 1(dimensionless), (2.49a)

and in view of Eq. (2.28), the ratio of the currentamplitudes is

I−0

I+0

= −V−

0

V +0

= −. (2.49b)

We note that is governed by a single parameter, theload impedance ZL, normalized to the characteristicimpedance of the line, Z0. As indicated by Eq. (2.36),Z0 of a lossless line is a real number. However, ZL is ingeneral a complex quantity, as in the case of a series RL

circuit, for example, for which ZL = R + jωL. Hence,in general may be complex also:

= ||ejθr, (2.50)

where || is the magnitude of and θr is its phase angle.Note that || ≤ 1.

A load is said to be matched to the line if ZL = Z0

because then there will be no reflection by the load ( = 0andV −

0 = 0). On the other hand, when the load is an opencircuit (ZL = ∞), = 1 and V −

0 = V +0 , and when it is

a short circuit (ZL = 0), = −1 and V −0 = −V +

0 .

Example 2-2 Reflection Coefficientof a Series RC Load

A 100- transmission line is connected to a loadconsisting of a 50- resistor in series with a 10-pFcapacitor. Find the reflection coefficient at the load fora 100-MHz signal.

Solution: The following quantities are given [Fig. 2-10]:

RL = 50 , CL = 10 pF = 10−11 F,

Z0 = 100 , f = 100 MHz = 108 Hz.

The load impedance is

ZL = RL − j/ωCL

= 50 − j1

2π × 108 × 10−11= (50 − j159) .

CL

RL 50 ΩZ0 = 100 Ω

10 pF

A

A'

Transmission line

Figure 2-10: RC load (Example 2-2).





From Eq. (2.49a), the voltage reflection coefficient isgiven by

= ZL/Z0 − 1

ZL/Z0 + 1

= 0.5 − j1.59 − 1

0.5 − j1.59 + 1

= −0.5 − j1.59

1.5 − j1.59= −1.67ej72.6

2.19e−j46.7 = −0.76ej119.3.

This result may be converted into a form with positivemagnitude for by replacing the minus sign with e−j180

.Thus,

= 0.76ej119.3e−j180 = 0.76e−j60.7 = 0.76∠−60.7

,

or

|| = 0.76, θr = −60.7.

Example 2-3 || for Purely Reactive Load

Show that || = 1 for a purely reactive load.

Solution: The load impedance of a purely reactive loadis given by

ZL = jXL.

From Eq. (2.49a), the reflection coefficient is

= ZL − Z0

ZL + Z0

= jXL − Z0

jXL + Z0

= −(Z0 − jXL)

(Z0 + jXL)=

−√Z2

0 +X2L e

−jθ√Z2

0 +X2L e

jθ

= −e−j2θ ,

where θ = tan−1XL/Z0. Hence

|| = | − e−j2θ | = [(e−j2θ )(e−j2θ )∗]1/2 = 1.

EXERCISE 2.7 A 50- lossless transmission line isterminated in a load impedance ZL = (30 − j200) .Calculate the voltage reflection coefficient at the load.

Ans. = 0.93∠−27.5 . (See C

DROM )

EXERCISE 2.8 A 150- lossless line is terminated ina capacitor whose impedance is ZL = −j30 .Calculate .

Ans. = 1∠−157.4 . (See C

DROM )

2-5.2 Standing Waves

Using the relationV −0 = V +

0 in Eqs. (2.44a) and (2.44b)gives the expressions

V (z)= V +0 (e

−jβz + ejβz), (2.51a)

I (z)= V +0

Z0(e−jβz − ejβz), (2.51b)

which now contain only one, yet to be determined,unknown, V +

0 . Before we proceed toward that goal,however, let us examine the physical meaning repre-sented by these expressions. We begin by deriving anexpression for |V (z)|, the magnitude of V (z). Uponusing Eq. (2.50) in (2.51a) and applying the relation|V (z)| = [V (z) V ∗(z)]1/2, where V ∗(z) is the complexconjugate of V (z), we have

|V (z)| = [V +0 (e

−jβz + ||ejθrejβz)]

· [(V +0 )

∗(ejβz + ||e−jθre−jβz)]1/2

= |V +0 |[1 + ||2

+ ||(ej (2βz+θr) + e−j (2βz+θr))]1/2

= |V +0 | [1 + ||2 + 2|| cos(2βz+ θr)

]1/2,(2.52)

where we have used the identity

ejx + e−jx = 2 cos x (2.53)





0

0.20.40.60.81.01.21.4 V

–λ –3λ 4

–λ 2

–λ 4

|V(z)|~

z

|I(z)|~

z–λ –3λ

4–λ 2

–λ 4

0

51015202530 mA

(b) |I(z)| versus z~

lmax

|V |max~

|V |min~

|I |max~

|I |min~

(a) |V(z)| versus z~

lmin

Figure 2-11: Standing-wave pattern for (a) |V (z)| and(b) |I (z)| for a lossless transmission line of characteristicimpedance Z0 = 50 , terminated in a load with areflection coefficient = 0.3ej30

. The magnitude ofthe incident wave |V +

0 | = 1 V. The standing-wave ratiois S = |V |max/|V |min = 1.3/0.7 = 1.86.

for any real quantity x. By applying the same steps toEq. (2.51b), a similar expression can be derived for |I (z)|,the magnitude of the current I (z).

The variations of |V (z)| and |I (z)| as a function of z,the position on the line relative to the load at z = 0,are illustrated in Fig. 2-11 for a line with |V +

0 | = 1 V,|| = 0.3, θr = 30, and Z0 = 50 . The sinusoidalpattern is called a standing wave, and it is caused bythe interference of the two waves. The maximum valueof the standing-wave pattern of |V (z)| corresponds

to the position on the line at which the incident andreflected waves are in phase [2βz + θr = −2nπ inEq. (2.52)] and therefore add constructively to give avalue equal to (1 + ||)|V +

0 | = 1.3 V. The minimumvalue of |V (z)| corresponds to destructive interference,which occurs when the incident and reflected waves arein phase opposition (2βz + θr = −(2n + 1)π ). Inthis case, |V (z)| = (1 − ||)|V +

0 | = 0.7 V. Whereasthe repetition period is λ for the incident and reflectedwaves individually, the repetition period of the standing-wave pattern is λ/2. The standing-wave pattern describesthe spatial variation of the magnitude of V (z) as afunction of z. If one were to observe the variations ofthe instantaneous voltage as a function of time at anylocation z, corresponding to one of the maxima in thestanding-wave pattern, for example, that variation wouldbe as cosωt and would have an amplitude equal to 1.3 V[i.e., v(t) would oscillate between −1.3 V and +1.3 V].Similarly, the time oscillation of v(z, t) at any location zwill have an amplitude equal to |V (z)| at that z.

Close inspection of the voltage and currentstanding-wave patterns shown in Fig. 2-11 revealsthat the two patterns are in phase opposition (when oneis at a maximum, the other is at a minimum, and viceversa). This is a consequence of the fact that the secondterm in Eq. (2.51a) is preceded by a plus sign, whereasthe second term in Eq. (2.51b) is preceded by a negativesign.

The standing-wave patterns shown in Fig. 2-11 are fora typical situation with = 0.3 ej30

. The peak-to-peakvariation of the pattern depends on ||, which can varybetween 0 and 1. For the special case of a matched linewith ZL = Z0, we have || = 0 and |V (z)| = |V +

0 |for all values of z, as shown in Fig. 2-12(a). With noreflected wave present, there will be no interference andno standing waves. The other end of the || scale, at|| = 1, corresponds to when the load is a short circuit( = −1) or an open circuit ( = 1). The standing-wavepatterns for these two cases are shown in Figs. 2-12(b)





|V0+|

0–λ –3λ 4

–λ 2

–λ 4

z

(a) ZL = Z0

(b) ZL = 0 (short circuit)

(c) ZL = ∞ (open circuit)

|V(z)|~

2|V0+|

0–λ –3λ 4

–λ 2

–λ 4

z

|V(z)|~

2|V0+|

0–λ –3λ 4

–λ 2

–λ 4

z

|V(z)|~

Matched line

2λ

2λ

Figure 2-12: Voltage standing-wave patterns for (a) amatched load, (b) a short-circuited line, and (c) anopen-circuited line.

and (c), both of which have maxima equal to 2|V +0 | and

minima equal to zero, but the two patterns are shifted in zrelative to each other by a distance of λ/4.

Now let us examine the maximum and minimumvalues of the voltage magnitude. From Eq. (2.52), |V (z)|is a maximum when the argument of the cosine function isequal to zero or multiples of 2π . Noting that the locationon the line always corresponds to negative values of z

(since the load is at z = 0), if we denote lmax = −z asthe distance from the load at which |V (z)| is a maximum,then

|V (z)| = |V |max = |V +0 |[1 + ||], (2.54)

and this occurs when

2βz+ θr = −2βlmax + θr = −2nπ, (2.55)

with n = 0 or a positive integer. Solving Eq. (2.55) forlmax, we have

−z= lmax = θr + 2nπ

2β= θrλ

4π+ nλ

2,

n = 1, 2, . . . if θr < 0,n = 0, 1, 2, . . . if θr ≥ 0,

(2.56)

where we have used the relation β = 2π/λ. Thephase angle of the voltage reflection coefficient, θr,is bounded between −π and π radians. If θr ≥ 0,the first voltage maximum occurs at lmax = θrλ/4π ,corresponding to n = 0, but if θr < 0, the first physicallymeaningful maximum occurs at lmax = (θrλ/4π)+ λ/2,corresponding to n = 1. Negative values of lmax

correspond to locations “beyond” the load at the endof the line and therefore have no physical significance.As was mentioned earlier, the locations on the linecorresponding to voltage maxima also correspond tocurrent minima, and vice versa.

Similarly, the minimum values of |V (z)| occur atdistances lmin = −z corresponding to when the argumentof the cosine function in Eq. (2.52) is equal to −(2n+1)π ,which gives the result

|V |min = |V +0 |[1 − ||],

when (θr − 2βlmin) = −(2n+ 1)π, (2.57)

with −π ≤ θr ≤ π . The first minimum corresponds ton = 0. The spacing between a maximum lmax and the





adjacent minimum lmin is λ/4. Hence, the first minimumoccurs at

lmin =lmax + λ/4, if lmax < λ/4,lmax − λ/4, if lmax ≥ λ/4.

(2.58)

The ratio of |V |max to |V |min is called the voltagestanding-wave ratioS, which from Eqs. (2.54) and (2.57)is given by

S = |V |max

|V |min= 1 + ||

1 − || (dimensionless). (2.59)

This quantity, which often is referred to by its acronym,VSWR, or the shorter acronym SWR, provides a measureof the mismatch between the load and the transmissionline; for a matched load with = 0, we get S = 1, andfor a line with || = 1, S = ∞.

M2.1

REVIEW QUESTIONS

Q2.5 The attenuation constant α represents ohmiclosses. In view of the model given in Fig. 2-6(c), whatshould R′ and G′ be in order to have no losses? Verifyyour expectation through the expression for α given byEq. (2.25a).

Q2.6 How is the wavelength λ of the wave travelingon the transmission line related to the free-spacewavelength λ0?

Q2.7 When is a load matched to the line? Why is itimportant?

Q2.8 What is a standing-wave pattern? Why is itsperiod λ/2 and not λ?

Q2.9 What is the separation between the location of avoltage maximum and the adjacent current maximum onthe line?

Example 2-4 Standing-wave Ratio

A 50- transmission line is terminated in a loadwith ZL = (100 + j50) . Find the voltage reflectioncoefficient and the voltage standing-wave ratio (SWR).

Solution: From Eq. (2.49a), is given by

= ZL − Z0

ZL + Z0= (100 + j50)− 50

(100 + j50)+ 50= 50 + j50

150 + j50.

Converting the numerator and denominator to polar formand then simplifying yields

= 70.7ej45

158.1ej18.4 = 0.45ej26.6.

Using the definition for S given by Eq. (2.59), we have

S = 1 + ||1 − || = 1 + 0.45

1 − 0.45= 2.6.

Example 2-5 Measuring ZL

A slotted-line probe is an instrument used to measurethe unknown impedance of a load, ZL. A coaxial slottedline contains a narrow longitudinal slit in the outerconductor. A small probe inserted in the slit can be usedto sample the magnitude of the electric field and, hence,the magnitude |V | of the voltage on the line (Fig. 2-13).By moving the probe along the length of the slotted

Vg~

40 cm 30 cm 20 cm 10 cm

Probe tip Slit

Sliding probeTo detector

ZL

+

–

Zg

Figure 2-13: Slotted coaxial line (Example 2-5).




2-6 INPUT IMPEDANCE OF THE LOSSLESS LINE 61

line, it is possible to measure |V |max and |V |min andthe distances from the load at which they occur. Useof Eq. (2.59) then provides the voltage standing-waveratio S. Measurements with a 50- slotted line connectedto an unknown load impedance determined that S = 3.The distance between successive voltage minima wasfound to be 30 cm, and the first voltage minimum waslocated at 12 cm from the load. Determine the loadimpedance ZL.

Solution: The following quantities are given:

Z0 = 50 , S = 3, lmin = 12 cm.

Since the distance between successive voltage minima isequal to λ/2,

λ = 2 × 0.3 = 0.6 m,

and

β = 2π

λ= 2π

0.6= 10π

3(rad/m).

From Eq. (2.59), solving for || in terms of S gives

|| = S − 1

S + 1= 3 − 1

3 + 1= 0.5.

Next, we use the condition given by Eq. (2.57) for thelocation of a voltage minimum to find θr:

θr − 2βlmin = −π, for n = 0 (first minimum),

which gives

θr = 2βlmin − π

= 2 × 10π

3× 0.12 − π = −0.2π (rad) = −36.

Hence,

= ||ejθr = 0.5e−j36 = 0.405 − j0.294.

Solving Eq. (2.49a) for ZL, we have

ZL = Z0

[1 +

1 −

]= 50

[1 + 0.405 − j0.294

1 − 0.405 + j0.294

]= (85 − j67) .

EXERCISE 2.9 If = 0.5∠−60 and λ = 24 cm, find thelocations of the voltage maximum and minimum nearestto the load.

Ans. lmax = 10 cm, lmin = 4 cm. (See C

DROM )

EXERCISE 2.10 A 140- lossless line is terminated in aload impedance ZL = (280 + j182) . If λ = 72 cm,find (a) the reflection coefficient , (b) the voltagestanding-wave ratio S, (c) the locations of voltagemaxima, and (d) the locations of voltage minima.

Ans. (a) = 0.5∠29 , (b) S = 3.0, (c) lmax = 2.9 cm+nλ/2, (d) lmin = 20.9 cm+nλ/2, where n = 0, 1, 2, . . . .(See C

DROM )

2-6 Input Impedance of the Lossless Line

The standing-wave patterns indicate that for a mis-matched line the voltage and current magnitudes are os-cillatory with position on the line and in phase oppositionwith each other. Hence, the voltage to current ratio, calledthe input impedance Zin, must vary with position also.Using Eqs. (2.51a) and (2.51b), Zin is given by

Zin(z)= V (z)

I (z)

= V +0 [e−jβz + ejβz]V +

0 [e−jβz − ejβz] Z0

= Z0

[1 + ej2βz

1 − ej2βz

](). (2.60)




2-6 INPUT IMPEDANCE OF THE LOSSLESS LINE 63

applying the boundary condition at the load, we were ableto relate V −

0 to V +0 through , and, finally, by applying

the boundary condition at the sending end of the line, weobtained an expression for V +

0 .

M2.2-2.3

Example 2-6 Complete Solution for v(z, t)and i(z, t)

A 1.05-GHz generator circuit with series impedanceZg = 10 and voltage source given by

vg(t) = 10 sin(ωt + 30) (V)

is connected to a load ZL = (100 + j50) through a50-, 67-cm-long lossless transmission line. The phasevelocity of the line is 0.7c, where c is the velocity of lightin a vacuum. Find v(z, t) and i(z, t) on the line.

Solution: From the relationship up = λf , we find thewavelength:

λ = up

f= 0.7 × 3 × 108

1.05 × 109= 0.2 m,

and

tan(βl)= tan

(2π

λl

)= tan

(2π

0.2× 0.67

)= tan 6.7π = tan 0.7π = tan 126,

where we have subtracted multiples of 2π . The voltagereflection coefficient at the load is

= ZL − Z0

ZL + Z0= (100 + j50)− 50

(100 + j50)+ 50= 0.45ej26.6

.

With reference to Fig. 2-14, the input impedance of theline, given by Eq. (2.63), is

Zin = Z0

[ZL + jZ0 tan βl

Z0 + jZL tan βl

]= Z0

[ZL/Z0 + j tan βl

1 + j (ZL/Z0) tan βl

]= 50

[(2 + j1)+ j tan 126

1 + j (2 + j1) tan 126

]= (21.9 + j17.4) .

Rewriting the expression for the generator voltage withthe cosine reference, we have

vg(t)= 10 sin(ωt + 30)

= 10 cos(π/2 − ωt − 30)

= 10 cos(ωt − 60)

= Re[10e−j60ejωt ] = Re[Vge

jωt ] (V).

Hence, the phasor voltage Vg is given by

Vg = 10 e−j60(V) = 10∠−60 (V).

Application of Eq. (2.66) gives

V +0 =(

VgZin

Zg + Zin

)(1

ejβl + e−jβl

)

=[

10e−j60(21.9 + j17.4)

10 + 21.9 + j17.4

]· [ej126 + 0.45ej26.6

e−j126]−1

= 10.2ej159(V) = 10.2∠159 (V).

The phasor voltage on the line is then

V (z)= V +0 (e

−jβz + ejβz)

= 10.2ej159(e−jβz + 0.45ej26.6

ejβz),





and the instantaneous voltage v(z, t) is

v(z, t)= Re[V (z)ejωt ]= 10.2 cos(ωt − βz+ 159)

+ 4.55 cos(ωt + βz+ 185.6) (V).

Similarly, use of V +0 in Eq. (2.51b) leads to

I (z)= 0.20ej159(e−jβz − 0.45ej26.6

ejβz),

i(z, t)= 0.20 cos(ωt − βz+ 159)

+ 0.091 cos(ωt + βz+ 5.6) (A).

M2.4

2-7 Special Cases of the Lossless Line

We often encounter situations involving lossless trans-mission lines with particular terminations or lines whoselengths exhibit particularly useful properties. We shallnow consider some of these special cases.

2-7.1 Short-Circuited Line M2.1D

The transmission line shown in Fig. 2-15(a) isterminated in a short circuit, ZL = 0. Consequently, thevoltage reflection coefficient defined by Eq. (2.49a) is = −1, and the voltage standing-wave ratio given byEq. (2.59) is S = ∞. From Eqs. (2.51a) and (2.51b),the voltage and current on a short-circuited losslesstransmission line are given by

Vsc(z)= V +0 [e−jβz − ejβz] = −2jV +

0 sin βz, (2.67a)

Isc(z)= V +0

Z0[e−jβz + ejβz] = 2V +

0

Z0cosβz. (2.67b)

The voltage Vsc(z) is zero at the load (z = 0), as it shouldbe for a short circuit, and its amplitude varies as sin βz,whereas the current Isc(z) is a maximum at the load and itvaries as cosβz. Both quantities are displayed in Fig. 2-15as a function of negative z.

–1

1

–λ –3λ 4

–λ 2

z

Vsc(z)

2jV0+

~

(b)

(a)

–1

1

–λ –λ 2

z

Isc(z)Z02V0

+

~

(c)

2-7 SPECIAL CASES OF THE LOSSLESS LINE 65

The input impedance of the line at z = −l is given bythe ratio of Vsc(−l) to Isc(−l). Denoting Zsc

in as the inputimpedance for a short-circuited line, we have

Zscin = Vsc(−l)

Isc(−l)= jZ0 tan βl. (2.68)

A plot of Zscin/jZ0 versus negative z is shown in

Fig. 2-15(d).In general, the input impedance Zin may consist of a

real part, or input resistance Rin, and an imaginary part,or input reactance Xin:

Zin = Rin + jXin. (2.69)

In the case of the short-circuited lossless line, the inputimpedance is purely reactive (Rin = 0). If tan βl ≥ 0, theline appears inductive, acting like an equivalent inductorLeq whose impedance is equal to Zsc

in . Thus,

jωLeq = jZ0 tan βl, if tan βl ≥ 0, (2.70a)

or

Leq = Z0 tan βl

ω(H). (2.70b)

The minimum line length l that would result in aninput impedance Zsc

in equivalent to that of an inductorof inductance Leq is

l = 1

βtan−1

(ωLeq

Z0

)(m). (2.70c)

Similarly, if tan βl ≤ 0, the input impedance iscapacitive, in which case the line acts like an equivalentcapacitor Ceq such that

1

jωCeq= jZ0 tan βl, if tan βl ≤ 0, (2.71a)

or

Ceq = − 1

Z0ω tan βl(F). (2.71b)

Since l is a positive number, the shortest length l for whichtan βl ≤ 0 corresponds to the range π/2 ≤ βl ≤ π .Hence, the minimum line length l that would result in aninput impedance Zsc

in equivalent to that of a capacitor ofcapacitance Ceq is

l = 1

β

[π − tan−1

(1

ωCeqZ0

)](m). (2.71c)

These results mean that, through proper choice of thelength of a short-circuited line, we can make substitutesfor capacitors and inductors with any desired reactance.Such a practice is indeed common in the design ofmicrowave circuits and high-speed integrated circuits,because making an actual capacitor or inductor is oftenmore difficult than making a shorted transmission line.

Example 2-7 Equivalent Reactive Elements

Choose the length of a shorted 50- losslesstransmission line (Fig. 2-16) such that its inputimpedance at 2.25 GHz is equivalent to the reactanceof a capacitor with capacitance Ceq = 4 pF. The wavevelocity on the line is 0.75c.

Z0Zinsc

Zinsc

shortcircuit

Zc =

⇒

l

1jωCeq

Figure 2-16: Shorted line as equivalent capacitor(Example 2-7).





Solution: We are given

up = 0.75c = 0.75 × 3 × 108 = 2.25 × 108 m/s,

Z0 = 50 ,

f = 2.25 GHz = 2.25 × 109 Hz,

Ceq = 4 pF = 4 × 10−12 F.

The phase constant is

β = 2π

λ= 2πf

up= 2π × 2.25 × 109

2.25 × 108= 62.8 (rad/m).

From Eq. (2.71a),

tan βl = − 1

Z0ωCeq

= − 1

50×2π×2.25×109×4×10−12= −0.354.

The tangent function is negative when its argument isin the second or fourth quadrants. The solution for thesecond quadrant is

βl1 = 2.8 rad or l1 = 2.8

β= 2.8

62.8= 4.46 cm,

and the solution for the fourth quadrant is

βl2 = 5.94 rad or l2 = 5.94

62.8= 9.46 cm.

We also could have obtained the value of l1 by applyingEq. (2.71c). The length l2 is greater than l1 by exactlyλ/2. In fact, any length l = 4.46 cm + nλ/2, where n isa positive integer, is also a solution.

2-7.2 Open-Circuited Line M2.1E

With ZL = ∞, as illustrated in Fig. 2-17(a), we have = 1, S = ∞, and the voltage, current, and inputimpedance are given by

Voc(z)= V +0 [e−jβz + ejβz] = 2V +

0 cosβz, (2.72a)

Ioc(z)= V +0

Z0[e−jβz − ejβz] = −2jV +

0

Z0sin βz, (2.72b)

Zocin = Voc(−l)

Ioc(−l)= −jZ0 cot βl. (2.73)

Plots of these quantities are displayed in Fig. 2-17 as afunction of negative z.

2-7.3 Application of Short-Circuit andOpen-Circuit Measurements

A network analyzer is a radio-frequency (RF) instrumentcapable of measuring the impedance of any loadconnected to its input terminal. When used to measureZsc

in , the input impedance of a lossless line terminated ina short circuit, and again Zoc

in , the input impedance of theline when terminated in an open circuit, the combinationof the two measurements can be used to determine thecharacteristic impedance of the line Z0 and its phaseconstant β. The product of Eqs. (2.68) and (2.73) givesthe result

Z0 = +√Zsc

in Zocin , (2.74)

and the ratio of the same equations leads to

tan βl =√

−Zscin

Zocin

. (2.75)




2-7 SPECIAL CASES OF THE LOSSLESS LINE 67

–1

1

–3λ 4

–λ 2

–λ 4

z

Voc(z)

2V0+

~

(b)

(a)

–1

1

–3λ 4

–λ 4

z

Ioc(z)Z02 jV0

+

~

(c)

–3λ 4

–λ 4

z

ZinjZ0

oc

(d)

Z0Zinoc

–λ 2

–λ

–λ

–λ –λ 2

Voltage

Current

Impedance

Figure 2-17: Transmission line terminated in an opencircuit: (a) schematic representation, (b) normalizedvoltage on the line, (c) normalized current, and(d) normalized input impedance.

Because of the π phase ambiguity associated with thetangent function, the length l should be less than or equalto λ/2 to provide an unambiguous result.

Example 2-8 Measuring Z0 and β

Find Z0 and β of a 57-cm-long lossless transmis-sion line whose input impedance was measured asZsc

in = j40.42 when terminated in a short circuit andas Zoc

in = −j121.24 when terminated in an opencircuit. From other measurements, we know that the lineis between 3 and 3.25 wavelengths long.

Solution: From Eqs. (2.74) and (2.75),

Z0 = +√Zsc

in Zocin = √(j40.42)(−j121.24) = 70 ,

tan βl =√

−Zscin

Zocin

=√

1

3.

Since l is between 3λ and 3.25λ, βl = (2πl/λ) isbetween 6π radians and (13π/2) radians. This placesβl in the first quadrant (0 to π/2) in a polar coordinatesystem. Hence, the only acceptable solution for the aboveequation is βl = π/6 radians. This value, however, doesnot include the 2π multiples associated with the integerλ multiples of l. Hence, the true value of βl is

βl = 6π + π

6= 19.4 (rad),

in which case

β = 19.4

0.57= 34 (rad/m).





Q2.11 What is a quarter-wave transformer? How is itused?

Q2.12 A lossless transmission line of length l isterminated in a short circuit. If l < λ/4, is the inputimpedance inductive or capacitive?

Q2.13 What is the input impedance of an infinitely longline?

Q2.14 If the input impedance of a lossless line isinductive when terminated in a short circuit, will it beinductive or capacitive when the line is terminated in anopen circuit?

EXERCISE 2.11 A 50- lossless transmission line usesan insulating material with εr = 2.25. When terminatedin an open circuit, how long should the line be for itsinput impedance to be equivalent to a 10-pF capacitor at50 MHz?

Ans. l = 5.68 cm. (See C

DROM )

EXERCISE 2.12 A 300- feedline is to be connected to a3-m long, 150- line terminated in a 150- resistor. Bothlines are lossless and use air as the insulating material,and the operating frequency is 50 MHz. Determine(a) the input impedance of the 3-m long line, (b) thevoltage standing-wave ratio on the feedline, and (c) thecharacteristic impedance of a quarter-wave transformerwere it to be used between the two lines in order to achieveS = 1 on the feedline. (See C

DROM )

Ans. (a) Zin = 150 , (b) S = 2, (c) Z0 = 212.1 .

2-8 Power Flow on a LosslessTransmission Line

Our discussion thus far has focused on the voltage andcurrent aspects of wave propagation on a transmissionline. Now we shall examine the flow of power carried

by the incident and reflected waves. We begin byreintroducing Eqs. (2.51a) and (2.51b), the generalexpressions for the voltage and current phasors on alossless transmission line:

V (z)= V +0 (e

−jβz + ejβz), (2.78a)

I (z)= V +0

Z0(e−jβz − ejβz). (2.78b)

In these expressions, the first terms represent theincident-wave voltage and current, and the termsinvolving represent the reflected-wave voltage andcurrent. At the load (z = 0), the incident and reflectedvoltages and currents are

V i = V +0 , I i = V +

0

Z0, (at z = 0), (2.79)

V r = V +0 , I r = − V

+0

Z0, (at z = 0). (2.80)

2-8.1 Instantaneous Power

The instantaneous power carried by the incident wave,as it arrives at the load, is equal to the product of theinstantaneous voltage vi(t) and the instantaneous currentii(t),

P i(t)= vi(t) · ii(t)= Re[V iejωt ] · Re

[I iejωt]

= Re[|V +0 |ejφ+

ejωt ] · Re

[ |V +0 |Z0

ejφ+ejωt]

= |V +0 | cos(ωt + φ+) · |V +

0 |Z0

cos(ωt + φ+)

= |V +0 |2Z0

cos2(ωt + φ+) (W), (2.81)

where use was made of Eq. (2.31a) to expressV +0 in terms

of its magnitude |V +0 | and phase angle φ+.

Similarly, upon replacing in Eq. (2.80) with||ejθr and then following the same steps, we obtain





= −||2 |V +0 |2

2Z0, (2.89)

which are, respectively, identical to the expressions givenby Eqs. (2.84) and (2.85).

EXERCISE 2.13 For a 50- lossless transmission lineterminated in a load impedance ZL = (100 + j50) ,determine the fraction of the average incident powerreflected by the load.

Ans. 20%. (See C

DROM )

EXERCISE 2.14 For the line of Exercise 2.13, what is themagnitude of the average reflected power if |V +

0 | = 1 V?

Ans. P rav = 2 (mW). (See C

DROM )

REVIEW QUESTIONS

Q2.15 According to Eq. (2.82), the instantaneous valueof the reflected power depends on the phase of thereflection coefficient θr, but the average reflected powergiven by Eq. (2.85) does not. Explain.

Q2.16 What is the average power delivered by alossless transmission line to a reactive load?

Q2.17 What fraction of the incident power is deliveredto a matched load?

Q2.18 Verify that

1

T

∫ T

0cos2

(2πt

T+ φ

)dt = 1

2,

regardless of the value of φ.

2-9 The Smith Chart

Prior to the age of computers and programmablecalculators, several types of charts were developed toassist in the solution of transmission-line problems. TheSmith chart, which was developed by P. H. Smith in1939, has been and continues to be the most widelyused graphical technique for analyzing and designingtransmission-line circuits. Even though the original intentof its inventor was to provide a useful graphical tool forperforming calculations involving complex impedances,the Smith chart has become a principal presentationmedium in computer-aided design (CAD) software fordisplaying the performance of microwave circuits. As thematerial in this and the next section will demonstrate, useof the Smith chart not only avoids tedious manipulationsof complex numbers, but it also allows an engineer todesign impedance-matching circuits with relative ease.The Smith chart can be used for both lossy and losslesstransmission lines. In the present treatment, however, wewill confine our discussion to the lossless case.

2-9.1 Parametric Equations

The reflection coefficient is, in general, a complexquantity composed of a magnitude || and a phaseangle θr or, equivalently, a real part r and an imaginarypart i,

= ||ejθr = r + ji , (2.90)

where

r = || cos θr, (2.91a)

i = || sin θr. (2.91b)

The Smith chart lies in the complex plane of . InFig. 2-20, point A represents a reflection coefficientA = 0.3 + j0.4 or, equivalently,

|A| = [(0.3)2 + (0.4)2]1/2 = 0.5




2-9 THE SMITH CHART 73

Γr

Γi

θr = 180°

θr = 90°

θr = 0°

θr = 270° or –90°

–1

–1

–0.9 –0.7 –0.5 –0.3 0.1

0.1

–0.1–0.2–0.3–0.4–0.5–0.6–0.7–0.8–0.9

0.30.40.50.60.7

0.80.9

1

0.2

0.3 0.5 0.7 0.9 1

|ΓB| = 0.54B

|ΓA| = 0.5

|Γ| = 1

A

θ r

= 202°

θr =53°

D C

Unit circle

Open-circuitload

Short-circuitload

Figure 2-20: The complex plane. Point A is at A = 0.3 + j0.4 = 0.5ej53, and point B is at B = −0.5 − j0.2 =

|0.54|ej202. The unit circle corresponds to || = 1. At pointC, = 1, corresponding to an open-circuit load, and at pointD,

= −1, corresponding to a short circuit.

and

θr = tan−1(0.4/0.3) = 53.

Similarly, point B represents B = −0.5 − j0.2,or |B | = 0.54 and θr = 202 [or, equivalently,θr = (360 − 202) = −158]. Note that when both r

and i are negative numbers θr is in the third quadrant inthe r–i plane. Thus, when using θ = tan−1(i/r) to

compute θr, it may be necessary to add or subtract 180to obtain the correct value of θr.

The unit circle shown in Fig. 2-20 corresponds to|| = 1. Because || ≤ 1 for a transmission line, onlythat part of the r–i plane that lies within the unit circlehas physical meaning; hence, future drawings will belimited to the domain contained within the unit circle.

Impedances on a Smith chart are represented bynormalized values, withZ0, the characteristic impedance





of the line, serving as the normalization constant.Normalized impedances are denoted by lowercase letters,as in z = Z/Z0. The normalized load impedance is thengiven by

zL = ZL/Z0 (dimensionless), (2.92)

and the reflection coefficient , defined by Eq. (2.49a),can be written as

= ZL/Z0 − 1

ZL/Z0 + 1= zL − 1

zL + 1. (2.93)

The inverse relation of Eq. (2.93) is

zL = 1 +

1 − . (2.94)

The normalized load impedance zL is, in general,a complex quantity composed of a normalized loadresistance rL and a normalized load reactance xL:

zL = rL + jxL. (2.95)

Using Eqs. (2.90) and (2.95) in Eq. (2.94), we have

rL + jxL = (1 + r)+ ji

(1 − r)− ji, (2.96)

which can be solved to obtain explicit expressions forrL and xL in terms of r and i. This is accomplishedby multiplying the numerator and denominator of theright-hand side of Eq. (2.96) by the complex conjugateof the denominator and then separating the result into realand imaginary parts. These steps lead to

rL = 1 − 2r − 2

i

(1 − r)2 + 2i

, (2.97a)

xL = 2i

(1 − r)2 + 2i

. (2.97b)

These expressions state that for a given set of values forr and i there corresponds a unique set of values for rL

and xL. However, if we fix the value of rL, say at 2, manypossible combinations of values can be assigned to r

and i, each of which can give the same value of rL. Forexample, (r, i) = (0.33, 0) gives rL = 2, as does thecombination (r, i) = (0.5, 0.29), as well as an infinitenumber of other combinations. In fact, if we were to plotin ther–i plane all the possible combinations ofr andi corresponding to rL = 2, we would obtain the circledenoted by rL = 2 in Fig. 2-21. Similar circles applyto other values of rL, and within the || = 1 domainall these circles pass through the point (r, i) = (1, 0).After some algebraic manipulations, Eq. (2.97a) can berearranged to give the following parametric equation forthe circle in the r–i plane corresponding to a givenvalue of rL:(

r − rL

1 + rL

)2

+ 2i =(

1

1 + rL

)2

. (2.98)

The standard equation for a circle in the x–y plane withcenter at (x0, y0) and radius a is given by

(x − x0)2 + (y − y0)

2 = a2. (2.99)

Comparison of Eq. (2.98) with Eq. (2.99) shows that therL circle is centered at r = rL/(1 + rL) and i = 0,and its radius is 1/(1 + rL). The largest circle shown inFig. 2-21 corresponds to rL = 0, which is also the unitcircle corresponding to || = 1. This is to be expected,because when rL = 0, || = 1 regardless of themagnitude of xL.

A similar examination of the expression forxL given byEq. (2.97b) also leads to an equation for a circle given by

(r − 1)2 +(i − 1

xL

)2

=(

1

xL

)2

, (2.100)

but the xL circles in the r–i plane exhibit adifferent character from that for rL. To start with, thenormalized reactance xL may assume both positiveand negative values, whereas the normalized resistancecannot be negative (negative resistances are physically





Γr

Γi

rL=0

rL=0.5

rL=1

rL=2

xL=0.5

xL=0

xL=1

xL=2

xL= –0.5 xL= –1

xL= –2

Figure 2-21: Families of rL and xL circles within the domain || ≤ 1.

meaningless). Hence, Eq. (2.100) can generate twofamilies of circles, one family corresponding to positivevalues of xL and another corresponding to negativevalues of xL. Furthermore, as shown in Fig. 2-21, onlypart of a given circle falls within the bounds of the unitcircle. The families of circles of the two parametricequations given by Eqs. (2.98) and (2.100) plotted for

selected values of rL and xL constitute the Smith chartshown in Fig. 2-22. A given point on the Smith chart,such as pointP in Fig. 2-22, represents a normalized loadimpedance zL = 2 − j1, with a corresponding voltagereflection coefficient = 0.45 exp(−j26.6). Themagnitude || = 0.45 is obtained by dividing the lengthof the line between the center of the Smith chart and the





EXERCISE 2.15 Use the Smith chart to find the valuesof corresponding to the following normalized loadimpedances: (a) zL = 2 + j0, (b) zL = 1 − j1,(c) zL = 0.5 − j2, (d) zL = −j3, (e) zL = 0 (shortcircuit), (f) zL = ∞ (open circuit), (g) zL = 1 (matchedload).

Ans. (a) = 0.33, (b) = 0.45∠−63.4 ,(c) = 0.83∠−50.9 , (d) = 1∠−36.9 , (e) = −1,(f) = 1, (g) = 0. (See C

DROM )

2-9.2 Input Impedance

From Eq. (2.61), the input impedance looking toward theload at a distance l from the load is given by

Zin = Z0

[1 + e−j2βl

1 − e−j2βl

](). (2.101)

To use the Smith chart, we always normalize impedancesto the characteristic impedance Z0. Hence, the normal-ized input impedance is

zin = Zin

Z0= 1 + e−j2βl

1 − e−j2βl(dimensionless).

(2.102)The quantity = ||ejθr is the voltage reflectioncoefficient at the load. Let us define

l = e−j2βl = ||ejθre−j2βl = ||ej (θr−2βl) (2.103)

as the phase-shifted voltage reflection coefficient,meaning that l has the same magnitude as , but thephase of l is shifted by 2βl relative to that of . Interms of l , Eq. (2.102) can be rewritten as

zin = 1 + l

1 − l. (2.104)

The form of Eq. (2.104) is identical with that for zL givenby Eq. (2.94):

zL = 1 +

1 − . (2.105)

The similarity in form suggests that, if is transformedinto l , zL gets transformed into zin. On the Smith chart,transforming into l means maintaining || constantand decreasing the phase θr by 2βl, which correspondsto rotation in a clockwise direction on the Smith chart.Noting that a complete rotation around the Smith chart isequal to a phase change of 2π , the length l correspondingto such a change is obtained from

2βl = 22π

λl = 2π, (2.106)

or l = λ/2. The outermost scale around the perimeterof the Smith chart (Fig. 2-22), called the wavelengthstoward generator (WTG) scale, has been constructedto denote movement on the transmission line towardthe generator, in units of the wavelength λ. That is, lis measured in wavelengths, and one complete rotationcorresponds to l = λ/2. In some transmission-lineproblems, it may be necessary to move from some pointon the transmission line toward another point closer tothe load, in which case the phase is increased, whichcorresponds to rotation in a counterclockwise direction.For convenience, the Smith chart contains a third scalearound its perimeter (in between the θr scale and the WTGscale) for accommodating such a need. It is called thewavelengths toward load (WTL) scale.

To illustrate how the Smith chart is used to findZin, letus consider a 50- lossless transmission line terminatedin a load impedance ZL = (100 − j50) . Our objectiveis to find Zin at a distance l = 0.1λ from the load. Thenormalized load impedance is zL = ZL/Z0 = 2 − j1,and it is denoted by point A on the Smith chart shownin Fig. 2-23. On the WTG scale, the location of point Ais at 0.287λ. Using a compass, a circle is drawn throughpoint A, with the center of the circle being at the centerof the Smith chart. Since the center of the Smith chartis the intersection point of the r and i axes, all pointson the drawn circle have the same value of ||. This iscalled the constant-|| circle, or more commonly the





0.1

0.1

0.1

0.2

0.2

0.2

0.3

0.3

0.3

0.4

0.4

0.4

0.50.5

0.5

0.6

0.6

0.6

0.7

0.7

0.7

0.8

0.8

0.8

0.9

0.9

0.9

1.0

1.0

1.0

1.2

1.2

1.2

1.4

1.4

1.4

1.6

1.6

1.6

1.8

1.8

1.8

2.02.0

2.0

3.0

3.0

3.0

4.0

4.0

4.0

5.0

5.0

5.0

10

10

10

20

20

20

50

50

50

0.2

0.2

0.2

0.2

0.4

0.4

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0.4

0.6

0.6

0.6

0.6

0.8

0.8

0.8

0.8

1.0

1.0

1.01.0

20-20

30-30

40-40

50

-50

60

-60

70

-70

80

-80

90

-90

100

-100

110

-110

120

-120

130

-130

140

-140

150

-150

160

-160

170

-170

180

±

0.05

0.05

0.06

0.06

0.07

0.07

0.08

0.08

0.09

0.09

0.1

0.1

0.11

0.11

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0.13

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0.14

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0.15

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0.16

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0.17

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0.190.19

0.20.2

0.210.21

0.220.22

0.230.23

0.24

0.240.25

0.25

0.26

0.26

0.27

0.27

0.28

0.28

0.29

0.29

0.3

0.3

0.31

0.31

0.32

0.32

0.33

0.33

0.34

0.34

0.35

0.35

0.36

0.36

0.37

0.37

0.38

0.38

0.39

0.39

0.4

0.4

0.41

0.41

0.42

0.42

0.43

0.43

0.44

0.44

0.45

0.45

0.46

0.46

0.47

0.47

0.48

0.48

0.49

0.49

0.0

0.0

AN

GLE

OF R

EFLE

CT

ION

CO

EF

FIC

IEN

T IN D

EG

RE

ES

—>

WA

VE

LEN

GTH

S T

OW

AR

D G

EN

ERAT

OR

—>

<— W

AVEL

EN

GTH

S T

OW

AR

D L

OA

D <

—

IND

UC

TIV

E R

EA

CTA

NC

E C

OM

PONENT (+

jX/Z

o), OR C

APACITIVE SUSCEPTANCE (+jB/Yo)

CAPACITIVE REACTANCE COMPONENT (-jX

/Zo)

, OR IN

DUCTIVE

SU

SCE

PTA

NC

E (-

jB/Y

o)

RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)

A

B

0.100 λ

0.287 λ

0.387 λ

SWR Circle

Figure 2-23: Point A represents a normalized load zL = 2 − j1 at 0.287λ on the WTG scale. Point B represents the lineinput at 0.1λ from the load. At B, zin = 0.6 − j0.66.

SWR circle. The reason for this second name is that thevoltage standing-wave ratio (SWR) is related to || byEq. (2.59) as

S = 1 + ||1 − || . (2.107)

Thus, a constant value of || corresponds to a specificvalue for S. As was stated earlier, to transform zL to zin,we need to maintain || constant, which means stayingon the SWR circle, and to decrease the phase of by 2βl.This is equivalent to moving a distance l = 0.1λ towardthe generator on the WTG scale. Since the location





of point A is at 0.287λ, we need to move to location0.287λ+0.1λ = 0.387λ on the WTG scale. A radial linethrough this new position on the WTG scale intersects theSWR circle at point B. This point represents zin, and itsvalue is zin = 0.6−j0.66. Finally, we unnormalize zin bymultiplying it by Z0 = 50 to get Zin = (30 − j33) .This result can be verified analytically using Eq. (2.101).The points between points A and B on the SWR circlerepresent different points along the transmission line.

EXERCISE 2.16 Use the Smith chart to find thenormalized input impedance of a lossless line of length lterminated in a normalized load impedance zL for each ofthe following combinations: (a) l = 0.25λ, zL = 1 + j0,(b) l = 0.5λ, zL = 1 + j1, (c) l = 0.3λ, zL = 1 − j1,(d) l = 1.2λ, zL = 0.5−j0.5, (e) l = 0.1λ, zL = 0 (shortcircuit), (f) l = 0.4λ, zL = j3, (g) l = 0.2λ, zL = ∞(open circuit).

Ans. (a) zin = 1 + j0, (b) zin = 1 + j1,(c) zin = 0.76 + j0.84, (d) zin = 0.59 + j0.66,(e) zin = 0 + j0.73, (f) zin = 0 + j0.72,(g) zin = 0 − j0.32. (See C

DROM )

2-9.3 SWR, Voltage Maxima, and Minima

Consider a load with zL = 2 + j1. Figure 2-24 showsa Smith chart with a SWR circle drawn through zL

(point A). The SWR circle intersects the real axis (r)

at two points, designated Pmax and Pmin. Thus, at bothpoints i = 0 and = r. Also, on the real axis, theimaginary part of the load impedance xL = 0. From thedefinition of ,

= zL − 1

zL + 1, (2.108)

points Pmax and Pmin correspond to the special case

= r = rL − 1

rL + 1(for i = 0), (2.109)

with Pmin corresponding to the condition when rL < 1and Pmax corresponding to the condition when rL > 1.Rewriting Eq. (2.107) for || in terms of S, we have

|| = S − 1

S + 1. (2.110)

For points Pmax and Pmin, || = r; hence

r = S − 1

S + 1. (2.111)

The similarity in form of Eqs. (2.109) and (2.111)suggests thatS = rL. However, since by definitionS ≥ 1,only pointPmax (for which rL > 1) satisfies the similaritycondition. In Fig. 2-24, rL = 2.6 at Pmax; hence S = 2.6.In other words, S is numerically equal to the value of rLat Pmax, the point at which the SWR circle intersects thereal axis on the right-hand side of the chart’s center.

The points Pmin and Pmax also represent the distancesfrom the load at which the magnitude of the voltage onthe line, |V |, is a minimum and a maximum, respectively.This statement is easily demonstrated by consideringthe definition of l given by Eq. (2.103). At pointPmax, the total phase of l , that is, (θr − 2βl), is equalto zero (if θr > 0) or 2π (if θr < 0), which isthe condition corresponding to |V |max, as indicated byEq. (2.55). Similarly, at Pmin the total phase of l is

equal to π , which is the condition for |V |min. Thus,for the transmission line represented by the SWR circleshown in Fig. 2-24, the distance between the load andthe nearest voltage maximum is lmax = 0.037λ, obtainedby moving clockwise from the load at point A to pointPmax, and the distance to the nearest voltage minimum islmin = 0.287λ, corresponding to the clockwise rotationfrom A to Pmin. Since the location of |V |max is also thelocation of |I |min and the location of |V |min is also thelocation of |I |max, the Smith chart provides a convenientway to determine the distances to all maxima and minimaon the line (the standing-wave pattern has a repetitionperiod of λ/2).





5

.0

5

.

0

10101

0

20202

0

50505

0

0

.

2

0

.

2

0

.

2

0

.

2

0

0

0

.

4

0

.40

.

6

0

.

6

0 0

0

.

8

0

.

8

0

0

1

1

1

1

20-

2

03

0

-

3

0

40-4050-506

0

-

6

0

70-

7

080-809

0

-

9

0

100-100110-

1

101

2

0

-

1

2

0

130-130140-1401

5

0

-

1

5

0

160-

1

60170-1701

8

0

–

0.050.05 0-060-06 0-070-07 0.

0

80.

0

8

0-09

0

-

0

9

0

-

1

0

-

1

0

-

1

1

0

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3

0

-

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-

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0-14

0

-

1

5

0

-

1

5

0-160-16

0.

1

70

-

1

7

0-180-18 0-190-19 0-20-2 0-210-21 0

-

2

2

0

-

2

2

0

-

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3

0

-

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0

-

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40-24

0

-

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5

0

-

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0-260-26

0

-

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7

0

-

2

7

0

-

2

8

0

-

2

8

0

-

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9

0

-

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9

0-30-3 0-310-31 0-

3

20-32

0

-

3

3

0-

3

3

0-340-34 0

-

3

5

0

-

3

5

0-36

0-36

0

-

3

7

0

-

3

7

0

-

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8

0

-

3

8

0-39

0-39

0

-

4 0

-

4

0-410-41 0-

4

20-

4

2

0-430-43

0

-4

4

0

-4

4

0-450-45

0

-

4

6

0

-

4

6

0

-

4

7

0

-

4

70

-

4

8

0

-

4

8

0-49

0

-

4

9

0

-

0

0

-

0

AN

G

L

E OF REF

LECTI

ON

C

O

E

F

F

I

C

I

E

N

T

I

N DEGR

E

ES

>

W

AV EL ENGTH

S

T

O

WA

R

D

G

E

N

E

R

A

T

OR

>

<

W

A

V

E

L

E

NG

T

H

S

T

OW

A

RD

L

O

A

D

<

I ND

UCT

IV

E

RE

A

C

T

A

NC

E C

O

M

P

O

N

ENT

(

+

jX

/

Z

o)

,

O

R

C

A

PA

C

ITIV E SU

SCE

PTANCE

(+j

B

/Yo

)

CAP

ACIT

IVE

R EA

C

TAN

C

E CO

MPO

NENT

(

-

jX/ Z

o )

, OR IN

D

U

C

T

I

V

E

S

U

SC

E

P

TA

N

C

E

(-

jB

/

Y

o

)

RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)

0.037

0.287

A

P

max Pmin

0.213

Figure2-24: Point A

representsanormalizedloadwithzL=2+

j1.Thestandingwaveratiois

S=2.

6(at

P

max ),the distancebetweentheloadandthe“rstvoltagemaximumis

l

max =

(.25 Š0.213

=0.037,andthedistancebetweenthe loadandthe“rstvoltageminimumis

l

min =

(.037 +0.25

=0.287

2-9.4ImpedancetoAdmittanceTransformationsI

ns

ol

vi

ng

ce

rt

ai

nt

yp

es

of

tr

an

sm

is

si

on

li

ne

pr

ob

le

ms

,

it

is

of

te

nm

or

ec

on

ve

ni

en

tt

ow

or

kw

it

ha

dm

it

ta

nc

es

th

an

wi

th

im

pe

da

nc

es

.A

ny

im

pe

da

nc

e

Z

is

in

ge

ne

ra

l

ac

om

pl

ex

qu

an

ti

ty

co

ns

is

ti

ng

of

ar

es

is

ta

nc

e

R

an

da

reactanceX:

Z=

R+

jX

((2

.1

12

)





The admittance Y corresponding to Z is the reciprocalof Z:

Y = 1

Z= 1

R + jX= R − jX

R2 +X2(S). (2.113)

The real part of Y is called the conductance G, and theimaginary part of Y is called the susceptance B. That is,

Y = G+ jB (S). (2.114)

Comparison of Eq. (2.114) with Eq. (2.113) leads to

G= R

R2 +X2(S), (2.115a)

B = −XR2 +X2

(S). (2.115b)

A normalized impedance z is defined as the ratio ofZ to Z0, the characteristic impedance of the line. Thesame concept applies to the definition of the normalizedadmittance y; that is,

y = Y

Y0= G

Y0+ j

B

–a introduction: waves and phasors - higher education · ﬁrst electric battery, by alessandro...

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