WAVES
`
`David H. Staelin « Ann W. Morgenthaler « Jin Au Kong
`
`
`
`Momentum Dynamics Corporation
`Exhibit 1017
`Page 001
`
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`Page 001
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`

`

`
`
`ELECTROMAGNETIC WAVES
`
`
`
`David H.Staelin
`
`Ann W. Morgenthaler
`
`Jin Au Kong
`
`Massachusetts Institute of Technology
`
`An Alan R. Apt Book
`
`PRENTICE HALL, Upper Saddle River, New Jersey 07458
`
`Momentum Dynamics Corporation
`Exhibit 1017
`Page 002
`
`Momentum Dynamics Corporation
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`
`

`

`Library of Congress Cataloging-in-Publication Daw
`Staelin, David H
`Electromagnetic waves / David H. Staelin, Ann W. Morgenithaler, Jin
`Au Kong.
`p.
`<m.
`Includes index.
`ISBN 0-13-225871-4
`2. Electrodynamics.
`1, Electromagnetc waves
`Ann W. O. Kong,Jin Au. OF Title.
`QOCb6L.ST4
`1994
`439,2—de20
`
`I, Morgenthaler,
`
`93-15705
`CrP
`
`Publisher; Alan Apt
`Production Editor: Bayani Mendoza de Leon
`Cover designer: Bruce Kenselaar
`Cover concept: Katharine E. Staelin
`Copy editor: Shirley Michaels
`Prepress buyer: Linda Behrens
`Manufacturing buyer: Dave Dickey
`Supplements editor, Alice Dworkin
`Editorial assistant: Shirley McGuire
`
`©1998 by Prentice-Hall, Inc.
`A Pearson Education Company
`UpperSaddle River, NJ 07458
`
`The author and publisher of this book have used their best efforts in preparing this book. These efforts
`include the development, research, and lesting of the theones and formulas to determine their effecveness.
`The author and publisher shall not be liable in any event for incidental or consequential damages in
`connection with, or arising out of, the furnishing, performance, or use of these formulas.
`
`All tights reserved. No part of this book may be
`reproduced, in any form or by any means,
`without permission in wnting from the publisher.
`
`ISBN 0-13-2258
`
`—$$$—$—_—__
`=~ IAN
`
`ISBN O-13-2258?1-4
`
`Prentice-Hall [international (UK) Limited,London
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`
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`

`

`2
`
`
`
`RADIATION BY CURRENTS
`AND CHARGESIN
`FREE SPACE
`
`
`
`2.1 STATIC SOLUTIONS TO MAXWELL’S EQUATIONS
`
`In this chapter, we consider how electromagnetic disturbances are created and how
`they propagate in free space.
`In Chapter 1, we saw that Maxwell's equations pre-
`dicted the existence of electromagnetic waves, even in vacuum where no currents
`or charges were present (though sources outside the region of interest were of
`course necessary to generate these waves). We shall now consider the case where
`the sources p and J are not zero, resulting in an inhomogeneous wave equation.
`Knowledge of the exact current and charge distributions on an object(antenna) lo-
`cated in free space is then sufficient to predict the field distributions at every point
`in space.
`Before developing the inhomogeneous wave equation, wefirst discuss the
`simpler static field solutions to Maxwell's equations. Static fields do not change
`in time and therefore cannot produce waves. We shall see that the static field solu-
`tions to a particular problem are quite similar to the dynamicfield solutions to the
`same problem with a time-dependent source, but the dynamic solutions must take
`into account the finite propagation velocity of electomagnetic waves. Because in-
`formation does not propagate instantaneously from one point in space to another,
`this retardation effect must be included in the descriptionofradiation. Section 2.2
`describes the dynamicsofradiating fields, but for now we consider only the simpler
`static fields.
`
`[f we let 0/dr — 0 in the time-varying Maxwell's equations (1.2.1) 1.2.4),
`or let @ — 0 in the ime-harmonic Maxwell's equations (1.4.4)(1.4,7), we see that
`the four basic equations of electromagnetism immediately decouple into two pairs.
`
`46
`
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`Exhibit 1017
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`
`

`

`Sec. 2.1
`
`Static Solutions to Maxwell's Equations
`
`Faraday’s law and Gauss's law in free space become
`Vx E=0
`V-E=p/e
`
`47
`
`(2.1.1)
`(2.1.2)
`
`and this pair of electrostatic equations, along with appropriate boundary conditions,
`uniquely determines the electric field. The charge density p is assumed to be
`specified.
`Likewise, Ampere’s law and Gauss’s magnetic law in free space are
`VxA=Jl
`V-uoH =0
`
`(2.1.3)
`(2.1.4)
`
`where again this pair of magnetostatic equations with boundary conditions specifies
`H completely if J is known. Because E and Af are decoupled, it is impossible
`to have wave propagation, since it is the interaction between the time derivatives of
`E and the space derivatives of H and vice versa which leads to electromagnetic
`radiation. We may solve (2.1.1) and (2.1.2) for the electric field by noticing that
`the vector identity V x (V) = 0 holds for any scalar field ®. Since V x E =0,
`we can write
`
`E=-vVo
`
`(2.1.5)
`
`where ® is called the scalar electric potential, and the negative sign is chosen
`so that electric field lines point from regions of high potential to regions of low
`potential (i.¢., in the direction that a posiuve charge would moveif it were placed
`in the field). The electric potential is a particularly useful quantity because it is
`a scalar which contains complete information aboutthe three components of the
`vector electric field. Substitution of (2.1.5) into (2.1.2) yields the scalar Poisson
`equation in vacuum:
`
`V7 = —p/ey
`
`(2.1.6)
`
`It is Poisson's equation, an inhomogeneous second-order partial differential equa-
`tion, which we should like to solve, and since it is a linear equation, we shall use
`the method of superposition.
`Consider, first,
`the simplest charge distribution, namely a point charge g
`located at the origin, There are two ways to find the scalar electnc potential;
`the first makes use of Gauss's law directly. The differential form of Gauss’s law
`(2.1.2) may be converted to an integral representation by using Gauss's divergence
`theorem (1.6.7) discussed in Chapter 1:
`
`[eG v= $F ido
`
`A
`
`¥
`
`The quantities V, A, A, da, and dv have been defined in Section 1.6.
`
`(2.1.7)
`
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`

`

`48
`
`Radiation by Currents and Charges in Free Space
`
`Chap. 2
`
`Taking the volumeintegral of Gauss's law (2.1.2) and applying the divergence
`theorem (2.1.7) yields
`
`§ Efe =
`
`P dv
`
`(2.1.8)
`
`A
`v €o
`But /,, edu is simply q, the total amountof charge enclosed by the volume V. If
`there is only a single point charge in space,then by symmetry the electric field must
`be radially directed away from the charge. Therefore, we choose the volume of
`integration to be a sphere of radius r centered at the origin. The electric field thus
`is normal to the surface ofthe sphere enclosing q,so E is parallel to A, Hence, we
`just muluply &,, constant at a given radius, by the surface area of a sphere, giving
`
`f E-fda = 4nr°E,
`
`A
`
`(2.1,9)
`
`Combining (2.1.8) and (2.1.9) gives the static electric field solution for a point
`charge:
`
`4reqr?
`Because E is radially directed and dependent only on r, we expect © to be a
`function of r alone (no angular dependence). Therefore, the equation E =—-V®
`reduces to
`
` q
`
`E(r)=F
`
`ae
`g
`E,=->- Hap
`
`(2.1.10)
`
`(21.11)
`
`Integration over r yields
`
`d(r)
`
`q
`sh
`= —
`(2.1.12)
`axer + Go
`(r)
`where po is an integration constant. We usually set Pp to Zero so that the potennai
`will be zero infinitely far from the point charge.
`The second method for finding ®(r) from a point charge distribution is more
`mathematical, as it solves the Poisson equation (2.1.6) directly by wnting down
`the Laplacian in spherical coordinates. But first we need a mathematical way to
`express the fact that we have a charge singularity at the origin. We write the charge
`distribution p(F) as
`
`(2.1.13)
`oM=4qe7)
`where 57(F) is the three-dimensional Dirac delta function. This generalized func-
`uion has the following properties:
`(1) &) =0 forr #0,
`(2) &F) + co for r =0,
`+00
`
`of PF)dv=1.
`
`—to
`
`Property (3) implies that the delta function has dimensions of m~?,
`
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`Exhibit 1017
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`
`

`

`Sec. 2.1
`
`Static Solutions to Maxwell's Equations
`
`49
`
`We can now rewrite (2.1.6) using the spherical form of the Laplacian operator
`and the charge distribution given by (2.1.13). Since p is spherically symmetric,the
`angular derivatives of ® in the Laplacian are zero, and we find that only the radial
`derivative contributes:
`
`1d (6) =-29H)
`r dr?
`€o
`
`(2.1.14)
`
`For r # 0, the delta function is zero by property (1):
`
`ag
`
`(2.1.15)
`rit?) = 9 r¥#0
`By inspection, (2.1.15) has the solution ® = K/r + 09, where again Do may be
`set to zero so that @(r — 00)=0, The constant A is determined by integrating
`both sides of (2.1.14) over a sphere of radius r > 0
`
`[af rdo | rsinddp sre) =
`r
`2x
`1
`ad?
`Fr
`Bi
`ely ve ues
`-| ar [ rao | rsinodd +8) =—4
`
`€o
`
`&q
`
`0
`
`0
`
`0
`
`(2.1.16)
`
`where the last equality follows from properties (1) and (3) of the delta function.
`Performing the 6 and ¢ integrations on the left side of (2.1.16) is trivial since the
`integrand is independent of both coordinates. Thus, (2.1.16) becomes
`r
`‘2
`Le ey dear?
`o rdr?
`€0
`
`(2.1.17)
`
`Integratingthe left side of (2.1.17) by parts and dividing by 47 gives
`
`d
`
`oo
`
`ia d
`
`
`rs (r®) ft -f[ arear Mae
`
`q
`
`(2.1.18)
`
`where the surface term [the first term on the left side of (2,1.18)] is zero because
`d(r&’)/dr = d(K)/dr =0. The second term isan integral of a total derivative, so
`q
`Pere 4n€
`We conclude, then, that for a point charge located at the origin, the scalar potential
`is
`
`
`om=—
`4m eqr
`
`(2.1.19)
`
`is the distance from the charge (at the origin) to an observer at Fr.
`where r = |7|
`Both approaches therefore give the same result for the potential of a point charge
`located at the origin.
`
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`

`

`i]
`
`Radiation by Currents and Charges in Free Space
`
`Chap, 2
`
`If the point chargeis not located at the origin, but is instead found at 7", then
`(2.1.19) becomes
`
`oF) =
`
`=,
`4req|F —F!
`where |F —F'| measures the distance between the observerat 7 and the point charge
`at 7’. Andif, instead of a point charge, we have an incremental! volume év’filled
`with charge 6g‘ = p(F)dv’, then this incremental amount of charge will give rise
`to the incremental] potential
`
`pr)dv'
`——
`)
`4req|F -F'|
`We now integrate the potential created by each infinitesimal amount of charge
`(2.1.20) over each ofthe infinitesimal volume elements Sv’ where p(F') #0to find
`the total electric potential:
`
`5O(7F) =
`
`2.1,20
`
`oF) =[E45eae
`
`(2.1.21)
`
`4ne\F —F|
`This integral is designated as the scalar Poisson integral or the superposition in-
`tegral. The latter name comes from the fact that since Maxwell's equations are
`linear, we can simply sum the potentials contributed by each incremental amount
`of charge acting alone to get the total potential. Notice that all of the fF depen-
`dence disappears after integration over dv’ in (2.1.21), so that ® is only a function
`of the observer coordinate r. (In performingthis integral, r is kept constant.) The
`scalar potential may be computed by using the superposition integral if p is known,
`and then E can be determined by taking the negative gradient of ©. Figure 2.1 is
`a pictorial representation of the superposition integral.
`The static magnetic field equations (2.1.3) and (2.1.4) give rise to a similar
`vector Poisson equation. Because V . 8 =0and the identity V' (V x A) =0 is
`obeyedby any vector A, we can express B = oH as the curl of an unknown
`vector A
`
`(2.1.22)
`B=wH=VxA
`where A is called the vector potential. As usual, it is necessary ultimately to
`check that (2.1.22) does not conflict with any of the other Maxwell’s equations.
`Substitution of (2.1.22) into (2.1.3) gives
`Vx (Vx A) spol
`
`(2.1.23)
`
`which may be simplified by the vector identity (1.3.6)
`(2.1.24)
`Vx(VxA=V(V-AD—- VA
`Because only the curl of A is specified, we are free to choose its divergence to be
`anything we wish; this choice is known as setting the gauge. To seethat (2.1.22)
`
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`

`Sec. 2.1
`
`Static Solutions to Maxwell's Equations
`
`51
`
`Source p (F") He
`
`Figure 2.1 Charge distribution and observer coordinates.
`
`is satisfied even if the divergence ofAis arbitrary, we let A =A+Vw where
`is any scalar. Taking the curl of both sides gives V x A =¥V xA since the curl
`of the gradient of any scalar is zero. Thus A is not unique. For static fields, we
`simplify (2.1.24) by letting
`
`V-A=0
`
`(2.1.25)
`
`which is called the Coulomb gauge. Combining (2.1.23), (2.1.24) and (2.1.25)
`gives the vector Poisson equation for vacuum
`
`V7A=—poJ
`
`(2.1.26)
`
`which mayalso be written as the three scalar equations
`
`V?*Aj = —pod;
`
`where i =x, y,z.' Because we can transform the previously derived scalar Pois-
`son equation (2.1.6) into each of the Cartesian components of the vector Poisson
`equation (2.1.26) by making the substitutions @ — A;, p — J;, and €9 + 1/19,
`where i can be x, y, or z, We can immediately find the vector solution to (2.1.26)
`by similarly transforming the superposition integral (2.1.21), yielding a superposi-
`tion integral for each Cartesian componentof the static vector potential:
`Modi’) dv’
`v 4afF—F|
`
`Ai?) =
`
`1. Note that this separation of a vector differential equarion into three scalar equations is straight-
`forward only in Cartesian coordinates; for example, V7A, # —jzo/, in spherical coordinates,
`
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`

`

`52
`
`Radiation by Currents and Charges in Free Space
`
`Chap. 2
`
`Again, the subscript i stands for a Cartesian componentof the field, and we may
`make the notation more compact using vector notation:
`
`dor |r — "|
`
`2.1.27)
`A®) = [eee
`This means that if we know the current distribution in a problem, we can
`calculate A directly and then take its curl
`to get oH 2 Therefore,
`in theory
`we can find the electric or magnetic fields in a statics problem if we know the
`charge or current distribution simply by performing an integration. Although this
`integra] may be difficult to evaluate, a more serious problem also exists, as the fields
`themselves may alter the source distributions, Fortunately, for most of the simple
`problems solved in this text, the charge and/or current distributions are known
`exactly or may be approximated quite accurately.
`
`2.2 RADIATION BY DYNAMIC CURRENTS AND CHARGES
`
`If we now consider the fully dynamic form of Maxwell's equationsin free space,
`we see that we can no longer Jet E be the¢gradient of a scalar potential because
`Vx E=-dB/at 40. But since V- uo =0, we may again express oH as
`the curl of a vector potential; (2.1.22) is still a valid equation.
`Substitution of(2.1.22) into Faraday’s law (1.2.1) gives
`
`(2.2.1)
`vx E+ )=0
`which means that E + @A/dr (instead of E alone) may be expressedas the gradi-
`ent of a scalar
`
`— 0A
`«2.
`—=-+-V
`(2.2.2)
`fe)
`E+ a
`since VY x V@=0 for any scalar®, This equation has the correctstatic limit:
`as d/dr — O, (2.2.2) reduces to E =-—V¢®. Ampere’s law (1.2.2) can now be
`rewritten with substitutions for H and £ using (2.1.22) and (2.2.2):
`
`=
`
`ay
`
`Vx (Vx A)= Hed — Motors ~ woe($)
`
`A
`
`a®
`
`(2.2.3)
`
`__ 2. Using this procedure, we arnve ar the useful Biof-Savart aw, which relates the magnetic field
`H directly to the current J without the vector potential as an intermediary:
`
`Fone “40 - [ I) xF-F) 4,
`
`Snir —FPP
`
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`

`Sec. 2.2
`
`Radiation by Dynamic Currents and Charges
`
`53
`
`After use of the identity (2.1.24) and rearrangement of the terms in (2.2.3),
`we find:
`
`=—pol +V¥ (v A+ poco) (2.2.4)
`
`am
`
`5
`
`aA —
`—
`A — 1L0€0 a2
`Since we are free to choose the gauge (divergence of A), welet
`(2.2.5)
`Pid Supkjer
`i
`= eNOS
`which is called the Lorentz gauge. Combining (2.2.4) and (2.2.5) now gives the
`inhomogeneous vector Helmholtz equation:
`vA
`z
`(2.2.6)
`V7A — Loeo——at = —poJ
`If J =0, we have a wave equation (the homogeneous Helmholtz equation dis-
`cussed in Section 1.3) which is consistent with the fact that both E and A obey
`wave equations in source-free vacuum. If E and H obey wave equations, then A
`and ® will also.
`
`a
`
`2V
`
`In a similarfashion, we may= Gauss's law (2.1.2) with(2.2.2) toyield
`VO + —av. A)= -£
`(2.2.7)
`
`0
`
`Again using the Lorentz gauge(2.2.5) to eliminate V - A from (2.2.7), we obtain
`the inhomogeneous scalar Helmholtz equation for the scalar potential ©:
`am
`(2.2.8)
`Vb — poeo—> = —P/€o
`at
`If p and J are time-harmonic sources, then the scalar and vector Helmholtz equa-
`tions can also be expressed as
`(2.2.9)
`VA + wrLeA = —Hod
`(2.2.10)
`VW+ w*Uperd = —0/€o
`where the phasor nature of A, ©, J, and p is made explicit. Notice that both
`inhomogeneous Helmholtz equations (2.2.6)and (2.2.8) or (2.2.9)}-(2.2.10) reduce
`to the static Poisson equations (2.1.26) and (2.1.6) respectively in the limit 0/dr +
`0 (w— 0), The dynamic solution to (2.2.9) or (2.2.10) is thus simply the static
`solution modified by a factor which accounts for the finite propagation time of the
`electromagnetic waves through space;this time delay or retardation is |F — F'|/c.
`We would now like to find solutions to the inhomogeneous Helmholtz equa-
`tions analogous to the superposition integrals (2.1.21) and (2.1.27). For simplic-
`ity, we shall consider time-harmonic fields since they form the basis for more
`arbitrary solutions. We first demonstrate that if the static point charge g at the
`origin starts to oscillate with time dependence g(f) = q coswy,
`then the static
`
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`

`54
`
`Radiation by Currents and Charges in Free Space
`
`Chap. 2
`
`potential &, = g/4mepr can be replaced with the time-harmonic potential phasor
`= %,¢7!*"". Using the chain rule and the spherical Laplacian,wefind
`2
`am,
`=,
`
`Vo= (ve. —2jk (>= t *)-~kKoie~kr
`
`The second term in this expression vanishes since ®, is inversely proportiona! to
`r.and we conclude that
`
`(2.2.11)
`26 = (V2, —kd,) e/*"
`If we nowsubstitute (2.2.11) into the inhomogeneous time-harmonic wave equation
`(2.2.10), we find that
`(V2@, — k2,) e+ wen be" =- 2 8H£0
`which reduces to the simpler equation
`(2.2.12)
`(¥20,) e-*" = =2 PF) = ae e#r 537)
`0
`if the dispersionrelation in free space k? = @jz9éy holds. The second equality in
`(2.2.12) follows because e~/*" is unity at 7 = 0, the only value for which the delta
`function is nonzero. Therefore, (2.2.12) simply reduces to the static scalar Poisson
`equation for a point charge and we have shown that the potential phasor
`
`= gir
`4ireor
`is a valid solution to (2.2.10) in free space when the source p is just an oscillat-
`ing point charge. Superposition of the incremental charge elements 5g’ = p(F’)dv'
`over the volume V‘, which encloses any arbitrary charge distribution, gives a dy-
`namic superposition integral similar to the static version (2.1.21):
`_
`ef)
`=jHF-F'l
`ges
`by) = [| 3
`
`0
`
`dv
`
`(2.2.13)
`
`'
`
`(2.2.14)
`
`Therefore, (2.2.14) is a general solution to the inhomogeneous time-harmonic
`scalar Helmholtz equation (2.2.10). The inhomogeneous vector Hetholtz equa-
`tion may be written down by inspection from (2.2.14) by noting u.
`(2.2.9) is
`identical to (2.2.10) when the substitutions p+ J;, ®— A;, and €&) + 1/o
`(i =x, y,z) are made. Recombiningthe three scalar equations into the more com-
`pact vector notation gives the dynamic vector Poisson integral?
`HolF’)
`e7jir-F|
`;
`Af) = [SSin?"
`
`(2.2.15)
`
`dv
`
`Notice that in the static limit (w= =0), the dynamic superposition integrals
`(2.2.14) and (2.2.15) reduce to the static Poisson integrals (2.1.21) and (2.1.27).
`
`3, Although we have chosen to derive time-harmonic superposition integrals, it should be realized
`thar time-dependent forms of these integrals also exist. We omit their lengthy derivations, but include
`
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`

`Sec. 2.3
`
`Radiation from a Hertzian Dipole
`
`55
`
`Once ®(F) and A(F) are found from the superposition integrals (2.2.14) and
`(2.2.15), the electnc and magnetic fields may be calculated by applying (2.1.22)
`and (2.2.2),
`
`2.3 RADIATION FROM A HERTZIAN DIPOLE
`
`The simplest radiating source is called a Hertzian dipole and is illustrated in Fig-
`ure 2,2(a).
`It consists of two reservoirs separated by a distance ¢ which contain
`equal and opposite amounts of charge; the charge oscillates back and forth between
`the reservoirs at frequency w. The charge reservoirs are located on the 2-axis at
`z= +d/2, constituting an electric dipole with dipole moment
`
`p=igd
`
`(2.3.1)
`
`In the limit that d + 0 and g — o© but the gd product remains finite, we have a
`Hertzian dipole.
`Conservation of charge dictates that a current J must flow between the two
`charge reservoirs as the charge oscillates, where J = 0q/0rt or J = jwg in time-
`harmonic form.
`(We shall find the electric and magnetic fields for the Hertzian
`dipole at a single frequency w, where we understand that for an arbitrary frequency
`spectrum we can superpose single-frequency solutions.) Therefore, the current
`density J is
`
`J=21d3@)
`
`(2.3.2)
`
`where we check the dimensions to make sure that (2,3,2) makes sense. Because
`the delta function has the dimensions of m~*, Jd 5°(7) has dimensions of A/m?*
`which is also the proper dimension for J. We find (2.3.2) by considering the
`dipole to have finite (though small) width w, length /, and depth d as shown
`in Figure 2.2(b). Because the current flows in the 2 direction, J=i1/lw=
`zIid/lwd. But lwd is the volumeofthe dipole, and as /, w,d— 0, 1//wd ~
`5°(F), giving the result (2.3.2).
`If we substitute the current distribution for a Hertzian dipole (2.3.2) into the
`
`_[e@t-F-Fi/e)
`
`the time-dependent superposition integrals for @(F,1) and A(F,t) for completeness:
`er) = [.
`4neqiF —7|
`du
`4e,1)=[ Modst = F=File) ay
`
`we
`
`4n|F-—F'|
`
`If the charge density oscillates sinusoidally as p(r’, 1) = Re{p(F') e/“], then the retarded charge den-
`sity o@'.1 —|F —F l/c) = pF) e/=-/**-7) is observed in the integrand of (2.2.14) (with the e/“
`factor implicit) since k = a/c. Therefore, the time-dependent equations above yield the correct time-
`harmonic equations when a sinusoidally varying source is chosen.
`
`Momentum Dynamics Corporation
`Exhibit 1017
`Page 013
`
`Momentum Dynamics Corporation
`Exhibit 1017
`Page 013
`
`

`

`56
`
`Radiation by Currents and Chargesin Free Space
`
`Chap. 2
`
`Figure 2.2 Hertzian dipole.
`
`expression for the vector potential (2.2.15), we find that the vector potential for the
`dipole is
`
`
`
` = Id
`
`(2.3.3)
`AP =2 a elke
`because all of the current is concentrated at the origin.* Since A varies with r, it
`is natural to use spherical coordinates r, 9, and > as defined in Figure 2.3. The
`z-unit vector may be written as
`
`% =Fcosé —Asind
`
`which mayalso be seen from Figure 2.3, and therefore
`A(r, 0) = (F. cos @ — 6 sin@) Hold et
`4orr
`
`(2.3.4)
`
`(2.3.5)
`
`4. Mathematically, the integration
`
`is carried out to obtain (2.3.3).
`
`£O =F)\8F)du'= fF)
`
`Momentum Dynamics Corporation
`Exhibit 1017
`Page 014
`
`Momentum Dynamics Corporation
`Exhibit 1017
`Page 014
`
`

`

`Sec. 2.3
`
`Radiation from a Henzian Dipole
`
`57
`
`Ne (b)
`
`Figure 2.3 Spherical coordinates.
`
`directly by taking the curl of A in spherical coordinates
`Wecan find
`_ 7
`Fr
`ré
`sr sinod
`(2.3.6)
`H=—VxaA= 7 a/dr
`93/3é
`d/dg
`
`
`Ho Hor’ sin@|4 rAg orsindA,
`
`where 0/0 and Ay are zero for the Hertzian dipole.
`a wi
`a
`a
`
`
`
`
`
`H=¢o— Fea 7 (4)
`ats
`'
`¢ anr *
`re az|
`
`
`a J
`
`oss
`
`-jkr
`
`+
`1+— 0
`
`(2.3.7)
`
`We can find E from H using Ampere's law (1.2.2) where J =O for r #0:
`=
`1
`-
`E=—VxdH
`J @€o
`\"
`L
`1
`Ho jkId i, }.|
`
`
`oes ee ——4(=) |se088
`
`maarAA Jee Naked es (2.3.8)
`
`
`+6/1+— +(3
`in@
`jkr a) ng
`
`Momentum Dynamics Corporation
`Exhibit 1017
`Page 015
`
`Momentum Dynamics Corporation
`Exhibit 1017
`Page 015
`
`

`

`55
`
`Radiation by Currents and Chargesin Free Space
`
`Chap, 2
`
`We can also find the complex Poynting vector for the Hertzian dipole from these
`electric and magnetic fields:
`S=ExH"*
`kid\*
`
`|,
`
`DAE og
`
`(2.3.9)
`
`anr {eli (a) Jove
`= M0
`6(ae) - Ge) [=
`
`I
`
`itis
`
`From the complex Poynting vector, we may find the time-averaged powerin the
`usual way:
`
`(S) =5Ree(S}=
`
`
`z No
`m2
`
` uel
`
`sin’@
`
`(2.3.10)
`
`In the Hertzian dipole limit (d — 0), these fields are valid for all r. For
`a short dipole where we can approximate the current as constant over thedipole
`length, (2.3.7)(2.3.9) are valid for r>>d. The electric field E(r,@, 1)
`is dis-
`played in Figure 2.4 at four instants of time, Were we to watch the field pattems
`evolve in time,
`it would appear that new dipole pattems were constantly being
`formed at r = 0 and that each pattern would begin to propagate outward once it
`was about a wavelength from the origin, until far from the source the fields would
`resemble plane waves.
`It is evident that for r <A/2m (kr < 1), the field resem-
`bles the static field of an electric dipole, and that for r > A/2z it resembles a
`uniform plane wave with gradually diminishing amplitude,
`In the region where
`r * 4/20, , the field lines appear to detach from the dipole field structure and begin
`propagating. We examine the exact electric and magnetic field solutions (2.3.8)
`and (2.3.7) in the limit where kr < | (the nearfield zone) and also where kr > 1
`(the farfield zone).
`First, we consider the nearfield case where kr <1. Using the dispersion
`relation k = w.//ip€p and (2.3.1), we write
`(2.3.11)
`Id = jwqd = jwp = jkp//uoeo
`Substituting (2.3.11) into the expressions for E (2.3.8) and H (2.3.7) and keeping
`only the leading terms (highest powers of 1/kr) gives the following expressions
`for the fields:
`
` E=p (?2cosO+Osind)
`(2.3.12)
`(2.3.13)
`H=o i? sin@ (kr <1)
`(Note that e~/* — |
`in the nearfield zone.) Equation (2.3.12) is the nearfield
`expression for the electric field, and (2.3.13) is the induction field. We note that H,
`
`(kr <1)
`
`Momentum Dynamics Corporation
`Exhibit 1017
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`
`Momentum Dynamics Corporation
`Exhibit 1017
`Page 016
`
`

`

`Sec. 7.3
`
`Radiation from a Hertzian Dipole
`
`59
`
`Figure 2.4 Electric fields for a Hertzian dipole as a function of time.
`
`which is proportional to 1/r?, is much smaller than E (proportional to 1/r>) when
`+ €A/2,
`so that the near field is dominated by electric field (capacitive) effects.
`The terms that dominate the complex Poynting vector (2.3.9) in the near field are
`proportional to 1/r° and may be seen to be purely imaginary. This means that
`although electric energy is stored in the near field, there is no net power flow out of
`the nearfield region, consistent with the fact that there is no wave propagation near
`the dipole. All of these observations are completely consistent with the quasistatic
`solutions to the electric dipole, and we make the additional observation that in the
`completely static limit (w = & = 0), the magnetic field vanishes identically, and the
`dipole is completely capacitive.
`By contrast, in the farfield zone (kr > 1), we neglect terms of order 1/(kr)*
`(or smaller) in (2.3.7) and (2.3.8), resulting in electric and magnetic fields which
`both have 1/r dependence:
`E=0 inka e/* sind
`4nr
`
`(kr > 1)
`
`(2.3.14)
`
`Momentum Dynamics Corporation
`Exhibit 1017
`Page 017
`
`Momentum Dynamics Corporation
`Exhibit 1017
`Page 017
`
`

`

`60
`
`Radiation by Currents and Charges in Free Space
`
`Chap.
`
`2?
`
`© nt
`
`SEPGH=¢2 ew! sin@ (kr > 1)
`
`(2.3.15)
`tatic electric field decays as 1/r? away from the
`Wenotefirst that since tl.
`origin and the far fields decay as 1/r, the near field rapidly becomes unimportant
`in comparison with the far field for large r. The electric and magnetic far fields
`have identical spatial dependencies, are orthogonal, and have an amplitude ratio
`Ex / Ag = no, the impedance of free space. These were just the properties of uni-
`form plane waves discussed in Section 1.3, with the exception that the plane waves
`in Chapter 1 had constant amplitudes in space. Forthe farfield solutions (2.3.14)
`and (2.3.15), the amplitude gradually decreases as ]/r away from the origin, bulif
`we are observing the waveal very large distances (r >> 4/270 ), we see whatlocally
`resembles a uniform plane wave propagating in the +r direction.
`(If we are far
`enough away, the 1/r factor doesn’t change significantly as the wave propagates
`outward. We also look at the plane wave segmentovera sufficiently small range of
`@ so that the sin @ factoris relatively constant, especially near @ = 2/2.)
`We can calculate the time-averaged power (W/m?) found in this wave from
`the complex Poynting vector:
`
`
`a
`ees
`ae OP IP
`S=Ex H*=F 19 | sin? @
`
`It follows that
`
`(5)
`
`|kld/?
`JEP
`1m
`
`e{s1= =#7 4r
`—-
`=5R S|} =F —
`
`
`
`in
`
`*g
`
`2.3.16
`
`f
`
`)
`
`which is identical with the exact expression (2.3.10) calculated from the complete
`Hertzian dipole fields. Thus (2.3.16) is valid everywhere in space, which is not
`surprising since power must be conserved.
`Integration of (2.3.16) over a spherical
`shell of radius r must be a constant independentof r, since this integration yields
`the total power emitted by the dipole. The sin?@ dependence of the Poynting
`yector meansthat most of the poweris radiated to the sides of the dipole; there is
`no radiation along the dipole axis. Figure 2.5 illustrates the E, Hi, and § vectors
`for the far field of a Hertzian dipole, as well as the electric field and Poynting vector
`magnitudes as functions of @.
`The total power P radiated by the dipole may be calculated by integrating
`(S), the average powerradiated per unit area, over the surface of a sphere enclosing
`the dipole, as described above:
`
`cj
`
`an
`
`p=(3)-Ada=[ aersindd¢(S,)
`tiet[d@sin?b= —— jkId?
`
`= To
`
`(2.3.17)
`
`Momentum Dynamics Corporation
`Exhibit 1017
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`
`Momentum Dynamics Corporation
`Exhibit 1017
`Page 018
`
`

`

`Sec. 2.3
`
`Radiation from a Hartzian Dipole
`
`61
`
`Figure 2.5 Radiation from a Hertzian dipole.
`
`The total power radiated thus increases as the square of the frequency and as the
`square of the dipole moment.
`Sometimesit is useful to think of the dipole element as part of an equivalent
`circuit; this concept is pursued in much greater depth in Chapter 9. We define the
`radiation resistance of any radiating element as the total power radiated by that
`element divided by one half the magnitude of the current squared; for the Hertzian
`dipole, wefind
`
`P W
`
`y?
`
`Raa =
`
`i
`if we approximate np = 377 2 = 120m 2. Physically, this resistor R.4 would
`dissipate the same total power P that the dipole with current / radiates.
`
`= © Kay? = 20(kd?
`67
`
`(2.3.18)
`
`Momentum Dynamics Corporation
`Exhibit 1017
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`
`Momentum Dynamics Corporation
`Exhibit 1017
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`
`

`

`G2
`
`Radiation by Currents and Charges in Free Space
`
`Chap. 2
`
`We also define the gain G(@,@) of a radiating element as the ratio of the
`power density (W/m?) radiated in the direction (f , @) to the power density radiated
`by an isotropic source emitting uniform power flux (Sisouopic) = P/4ir? incall
`directions. Clearly, if the radiating elementis isotropic, the gain is | for all angles.
`For the Hertzian dipole,
`
`(S(6,¢,7))
`3 23
`(2.3.19)
`Gié, @) = Paar = 3 a
`as can be seen by combining (2.3.16) and (2.3.17). The radiation pattern for
`a source is determined by the gain and is a plot of G(8,) over all angles but
`normalized so its maximum value is unity. Since G(@,@) is independent of ¢
`for the Hertzian dipole, the radiation pattern is symmetric around the Z-axis. The
`dipole radiates most strongly at @ = 7/2 (in the x—-y plane), and has a null at
`@ = 0,2 (along the z-axis). Thus, the radiation pattern looks like a doughnut with
`a vanishingly small hole, oriented along the Z-axis. Figure 2.5 plots the gain for a
`Hertzian dipole.
`If we integrate the gain over the solid angle dQ = sin@d@dé, we should
`always find that
`
`‘in
`
`0
`
`[ ao| sin6de@G(é, ¢) = 4x
`
`n
`
`0
`
`(2.3.20)
`
`no matter what type of radiating element we choose, since the integral of (S,)
`over the surface of any sphere is equal to the total power P of the source. Equa-
`tion (2.3.20) is readily derived from (2.3.17) and the definition of gain (2.3.19); it
`may be easily verified for the Hertzian dipole.
`We also define Go to be the maximum gain of the antenna. For the Hertzian
`dipole, Gy = 1.5 whereas for an isotropic radiator, Gg = 1. Thus the dipole is 50
`percent more directive than an isotropic antenna (which isn’t directive at all).
`In
`Chapter 9 we shall discuss how to make antennas that are much more directive than
`the Hertzian dipole—useful if one wants to maximize the power transmitted by the
`antenna to a given regio