engineering electromagnetic fields and waves 2nd edition.pdf

_________________________________________ CHAPTER 1

Vector Analysis and Electromagnetic Fields in Free Space

The introduction of vector analysis as an important branch of mathematics dates back to the midnineteenth century. Since then, it has developed into an essential tool for the physical scientist and engineer. The object of the treatment of vector analysis as given in the first two chapters is to serve the needs of the remainder of this book. In this chapter, attention is confined to the scalar and vector products as well as to certain integrals involving vectors. This provides a groundwork for the Lorentz force effects defining the electric and magnetic fields and for the Maxwell integral relationships among these fields and their chargc and current sources. The coordinate systems employed are confined to the common rectangular, circular cylindrical, and spherical systems. To unifY their treatment, the generalized coordinate system is used. This timesaving approach permits developing the general rules for vcctor manipulations, to enable writing the desired vector operation in a given coordinate system by inspection. This avoids the rederivation of the desired operation for each new coordinate system employed. Next arc postulated the Maxwell integral relations for the electric and magnetic fields produced by charge and current sources in free space. Applying the vector rules developed earlier, their solutions corresponding to simple classes of symmetric static charge and current distributions are considered. The chapter concludes with a discussion of transformations among the three common coordinate systems.

1·1 SCALAR AND VECTOR FIELDS A field is taken to mean a mathematical function of space and time. Fields can be classified as scalar or vector fields. A scalar field is a function having, at each instant in

1

lJ

i'

2

VECTOR ANALYSIS AND ELECTROMAGNETIC FIELDS IN FREE SPACE

time, an assignable magnitude at every point of a region in space. Thus, the temperature field t) inside the block of material of Figure 1-1 (a) is a scalar field. To each point there exists a corresponding temperature T(x,]!, z, t) at any instant t in time. The velocity of a fluid moving inside the pipe shown in Figure 1-1 (b) illustrates a vector field. A variable direction, as well as magnitude, of the fluid velocity occurs in the pipe where the cross-sectional area is changing. Other examples of scalar fields are mass, density, pressure, and gravitational potential. A force field, a velocity field, and an acceleration field are examples of vector fields. The mathematical symbol for a scalar quantity is taken to be any letter: for example, A, T, Il, f. The symbol for a vector quantity is any letter set in boldface roman type, ff)!' A, H, a, g. Vector quantities are represented graphically by

(z)

6 200· 6

(x)

Temperature field at x 4 em

=

Heat source (a)

F

FIGURE 1-1. Examples of material. (b) Fluid velocity field ill,ide

fidd inside a block of

m-

Id. ny (b) ity lar ity for lce

3--

I

~

----...

1~!

~\ ~~\~I("'"

Unit vector a

B=C

y

FIGURE 1-2. Graphic representations of a vector, equal vectors, a uni t vector, and the representation of magnitude or length of a vector.

by means of arrows, or directed line segments, as shown in Figure 1-2. The magnitude or length of a vector A is written \A\ or simply A, a positive real scalar. The negative of a vector is tbat vector taken in an opposing direction, with its arrowhead on the opposite end. A unit vector is any vector having a magnitude of unity. The symbol a is used to denote a unit vector, with a subscript employed to specify a special direction. For example, ax means a unit vector having the positive-x direction. Two vectors are said to be equal if they have the same direction and the same magnitude. (They need not be collinear, but only parallel to each other.)

1·2 VECTOR SUMS The vector sum of A and B is defined in relation to the graphic sketch of the vectors, as in Figure 1-3. A physical illustration of the vector sum occurs in combining displacements in space. Thus, if a particle were displaced consecutively by the vector distance A and then by B, its final position would be denoted by the vector sum A + B = C shown in Figure 1-3 (a). Reversing the order of these displacements provides the same vector sum C, so that

A+B=B+A

( 1-1)

the commutative law of the addition of vectors. If several vectors are to be added, an associative law

(A + B)

+ D = A + (B + D)

follows £I'om the definition of vector sum and from Figure 1-3(b).

I

B

I I I

:A

+ B =C

I

I I I

(a)

(b)

FIGURE 1-3. (a) The graphic definition of the sum of two vectors. (b) The associative law of addition.

(1-2)

1·3 PRODUCT OF A VECTOR AND A SCALAR u and if B denotes a vector quantity, their produc a magnitude u times the magnitude of B, and having tht a positive scalar, or the opposite direction if u is negative same direction The f()lIowing laws hold IiII' the products of vectors and scalars. If a scalar

uB

(u

Bu

Commutative law

( 1-3)

(uv)A

Associative law

( 1-4)

o)A = uA

+ vA

Distributive law

(1-5)

uA

+ uB

Distributive law

( 1-6)

u(A +B)

1·4 COORDINATE SYSTEMS

The solution of physical problems often requires that the framework of a coordinate system be introduced, particularly if explicit solutions are being sought. The system most familiar to engineers and scientists is the cartesian, or rectangular coordinate system, although two other ii'ames of reference often used are the circular cylindrical and the spherical coordinate systems, The symbols employed for the independent coordinate variables of these orthogonal systems are listed as follows. 1. Rectangular coordinates: (x,y, z) 2. Circular cylindrical coordinates: (p, cj>, z) 3. Spherical coordinates: (r, 8, cj»

In Figure 1-4(a), the point P in space, relative to the origin 0, is depicted in terms of the coordinate variables of the three common orthogonal coordinate systems: as P(x,y, z) in the rectangular system, as P(p, cj>, z) in the circular cylindrical (or just "cylindrical") system, and as P(r, 8, cj» in the spherical coordinate system. In the cylindrical and spherical systems, it is seen that the rectangular coordinate axes, labeled (x), and ,are retained to establish proper angular references. You should observr that. the coordinate variable cj> (the azimuth angle) is common to both

: (zi I

I

P(x, y, z)

z

-(x)

y Rectangular

Circular cylindrical

Spherical

(a)

FIGURE 1-4. Notational convcnlions (a) Location of a point P in space, (Ii) The of a vector A into its orthogonal COmpOllt'nts.

in the three nnnmoll coordinate systems. p"im P Ie) The resolution

s·

~ ~ ~~ ~!a ~

~ &~ ~ ~ ("l)

("l)

-

,

,

i:

S '""

,

(.n

~

-

~

M- "'''I

0..

("l) •

...

:;:."'" :::r- (").: ("l)

(z) (z)

(z)

r= Constant

(sphere)

d> = Constant . (plane)

z= Constant (plane)

,

z = Constant

\

.-,1

d> = Constant (plane)

I

I I I

0

,

--(y) (x) - -

(y)

(yl

(x)

x=Constant

p=Constant (circular cylinder) (b)

_--®~ZA:---..,

,r::::"- ... __A _. ...P"",,-- I" t

I

,

'

I'

:(z)

:(z)

:(z)

I

I

'

~a;~;

: I

axAx~:-::=L------ ayAy

~ z

_-'-_

_- ----- 0

(x)

_-

' ,,/, .. A f-

I

_---

y

x

Rectangular

'

n!I r;

arA~P: a A

i

"

, --"" 4 ' '

I

I

a

:::,-_......

I

i : / q,A1 aA-~'biO p p z

----_ Jy)

_- -

- _---(x)

--p

Circular cylindrical (c)

I

I I

...

I

O/, aeAo': "-t': r~O

A

-----~y)

q,

,,"

"

::'>w'

-'--~;'

r--'<1>/

c:

§

-

- _---- : (x)

Spherical

(y)

?:

I:

:2

...:>-to' u u ~

~

FIGURE 1-4 (continued)

to'

::

" ~

the cylindrical and tbe spherical systems, with the x-axis taken as the = 0 reference, generated in the positive sense from (x) toward (y). (By the "right-hand rule," if the thum b of the right hand points in the positive z-dircction, the fingers will indicate the sense.) The radial distance in the cylindrical system is p, measured perpendicularly from the to the desired point P; in the spherical system, the radial distance is 1, measured from the origin 0 to the point P, with denoting the desired declination angle measured positively from the reference z-axis to 1, as shown 1-4( a). The th ree coordinate systems shown are so-called "right-handed" properly definable after first discussing the unit vectors at P.

°

A. Unit Vectors and Coordinate Surfaces To enable expressing any vector A at the point P in a desired eoordinate system, three orthogonal unit vectors, denoted by a and suitably subscripted, are defined at P in the positive-increasing sense of each of the coordinate variables of that system. Thus, as noted in Figure 1-4(b), ax, a y, a z are the mutually perpendicular unit vectors of the rectangular coordinate system, shown at P(x,y, z) as dimensionless arrows of unit length originating at P and directed in the positive X,], and;;; senses respectively. Note that the disposition of these unit vectors at the point P corresponds to a right-handed coordinate system, so-called because a rotation from the unit vector ax through thc smaller angle toward a y and denoted by the fingers of the right hand, corresponds to the thumb pointing in the direction of a z . Similarly, in the cylindrical coordinate system of that figure, the unit vectors at P(p, , z) are a p ' aq,' a z as shown, pointing in the positive p, , and;;; senses; at P(r, 0, rotates P to a new location. Thus, in certain differentiation or integration processes involving unit vectors, most unit vectors should not be treated as constants (see Example I-I in Section 1-6). I n Figure I , it is instructive to notice how the point P, in any of the coordinale systems, can be looked on as the intersection of three coordinate suifaces. A coordinate surf;tcC necessarily planar) is defmed as that surface formed by simply Ihe desired coordinate variable equal to a constant. Thus, the point P(x,], z) in the is the intersection of the three coordinate surfaces x = constant, y = constallt, constant (in this case planes), thosc constants depending on the desired location fe)r P. two such coordinate surfaces intersect orthogonally to define a line;whiIe the perpt'IHlicular intersection of the line with the third surface pinpoints P.) The unit vectors at z) are thus perpendicular to their corresponding coordinate surfaces .g., ax is perpendicular to the surface x = constant). Because the coordinate surfaces are mutually perpendicular, so are the unit vectors. Similar observations at in the cylindrical coordinate system are applicable. P is the intersection the three orthogonal coordinate surfaces p = constant (a right circular cylindrical constant (a semi-infinite plane), and ,(; constant (a plane), to each of which thee corresponding unit vectors are perpendicular, thus making a p ' aq" a z welL comments apply to the unit vectors an ao, aq, at P(r,O, coordinate system of Figure 1-4(b),

or

lce,

le, " ;ate red the the 'wn ~d"

~m,

at

:m. ors nit ote ded the to ate ng Drs reo

he .he He

m, ew ,rs, 5). :0A

)Iy

z) == ed a ~.)

lte Ite

lint

wherein the coordinate suriltces defining the intersection P in this instance become r = constant (a spherical surface), () = constant (a conical and 4> = constant (a semi-infinite plane). j

B. Representations in Terms of Vector Components A use[ill application of the product of a vector and a scalar as described in Section 1-3 occurs in the representation, at any poin t P in space, of the vector A in terms of its coordinalf components. In the rectangular system of Figure 1-4(c) is shown the typical vector A at the point P(x,y, z) in space. The perpendicular projections of A along the unit vectors ax, a y and a z yield the three vector components of A in rectangular coordinates, seen from the geometry to be the vectors axAx, ayAy, and azA z in that figure. Their vector sum, axAx + ayAy + azA z = A, thus provides the desired representatioIl of A in the rectangular coordinate system. Similar manipulations into circular cylindrical and spherical coordinate components yield the other two corresponding diagrams depicted in Figure 1-4(c), whence the representations of A in terms of its components: 1

+ ayAy + azA z A = apAp + a.pA.p + azA z A = arAr + aoAo + a.pA.p A = axAx

Rectangular Circular cylindrical Spherical

(1-7)

Because of the mutual perpendicularity of the components of any of these representations, it is clear that the geometrical figure denoted by each dashed-line representation of Figure is a parallelepiped (or box), with A appearing as a principal diagonal within each. The magnitude (or length) of each A in (1 thus becomes

+ A;) 1/2 A = [A; + A~ + A; 11/2 A = [A~ + A;

A

= [A; + A~ + A~]1/2

Rectangular Circular cylindrical Spherical

(1-8)

C. Representation in Terms of Generalized Orthogonal Coordinates Noting the several similarities in the charaeterizations of the unit vectors and the vector A in the three common coordinate systems just described, and to permit unifying and shortening many discussions later on relative to scalar and vector fields, the system or generalized orthogonal coordinates is introduced. In this system, u I , u2 , U3 denote the generalized coordinate variables, as suggested by Figure i-5(a). The generalized approach to developing properties of fields in terms of (UI' 112, 113) has the advantage of making it unnecessary to rederive certain desired expressions each time a new coordinate system is encountered. Just as I(x the three common coordinate systems already described relative to Fignre 1-4, the point P(uj, 112) 113) in generalized coordinates, as seen in Fignre 1-5(a),

If,

lit

I) ,

lThus, the components of A in the rectangular coordinate system are the vectors axA" ayAy, and azA z ' Another usage is to rekr to only the scalar multipliers (lengths) AX' and Az as the components of A, althongh these are more properly the of A onto the unit vectors.

8


/llncreaSing u:l I I

Increasing

112

Increasing u 1

Generalized orthogonal coordinates (a)

I

I, (z) y== x= Constant Constant

az

Rectangular (b)

Circular cylindrical (e)

Spherical (d)

FIGURE 1-5. The coordinate surfaces defining the typical point P and the unit vectors at P.

is the intersection of three perpendicular coordinate surfaces, Ul = constant, U2 = constant, U3 = constant. The intersections of pairs of such surfaces moreover define coordinate lines. The unit vectors, denoted aI, a2, a3, are mutually perpendicular, tangent to the coordinate lines, and intersect the coordinate surfaces perpendicularly. The one-to-one correspondence of the:;e generalized coordinate variables Ill' U2, U 3 to their coordinate surfaces, and the generalized unit vectors aI, a 2 , a3 to the equivalent vectors of the three common coordinate systems, can be better appreciated on making a direct visual comparison of the generalized sketch of Figure 1-5(a) with (b), (e), and (d) of that figure. If the vector A were the point P(uI' U2, in Figure 1-5(a), with the in the directions of the unit vectors shown, the components alAI, and expression for A would he

A

( 1-9)

construction for (1-9).

I ts magnitude is

( 1-10) The scalars AI, A2l and A specialized to the three COllllllOII and (1-8).

'"''I'IJI''''"''''' lIf A.

Kxarnples of these expressions already been given in (1-7)

1-5 DIFFERENTIAL ELEMENTS OF SPACE

9

1·5 DIFFERENTIAL ELEMENTS OF SPACE In the processes of integration in space to be considered shortly, the differential elements of volume, surface, and line are frequently needed. A differential element of volume dv is generated in the vicinity of a point P(Ub U2, U3) in space by means of the displacements dtb dt2 , and dt3 on the coordinate surfaces, through the differential changes dUll duz, and dU3 in the coordinate variables. This situation is represented geometrical! y in Figure 1-6 (a). Thus, a volume-elemen t dv is represented in generalized orthogonal coordinates by means of the product of the differential length-elements as follows

(1-11 ) The relation or the length-elements to differential changes in the coordinate variables Ul' U2' and U3 is provided by the relations ( 1-12)

(z)

z= fine anfhe 1eir

113 113

+ clu3 =Constant

=Constant

Generalized (curvilinear) coordinates

Rectangular

(al

lec-

(6)

(z)

19 a (d)

(z)

the the

,

[-9) I

I I/

-9). ·10) Ions 1-7)

(x)

(;j-

_

--"':: -

------Spheric;,i (
FIGURE 1-6. The generation of a volume-element dv = dt 1dt2dl'} at orthogonal coordinate systems.

10

VECTOR ANALVSIS AND ELECTROMAGNETIC FIELDS IN FREE SPACE

so that (I-II) is written (1-13~

The coefllcients hi' ! and h3 are called metric codfieients, needed to give the element~ dt of (1-12) their dimension of length (meter). From a consideration of tht geometry of dv in each of Figure 1-6(b), (e), and (d), it is evident that tht following and metric coefficients are applicable to the three commor: systems. dx

= dy

dt z

= h3 =

hi dp

=

=

I

p, h3

hi

Rectangular

I

dtz

pdp

= rdO

I, h2

hi

r, h3

dt3 = dz

dt3

(1-14)

= dz ( I IS)

Circular cylindrical dt3

= r sin 0

=

r sin Odp ( 1-16)

Spherical

The substitution of these results into (1-13) therefore provides the volume-element dL in each system as follows. do

dx

Rectangular

dv

II


dv

sin OdrdOdp

Spherical

(1-17)

S in space may be left in its scalar f(nm ds, although for some purposes it a vector characterization, ds, if desired. Suppose ds coincides with a cOIlrdillatt' surface Ul = constant, as shown in Figure I-7(a).

I (z) I

i

r

= Constant

ds = at

I (0)

FIGURE 1-7. Typical as a vector element through 011 the coordina te surf~~,r{' ds on the coordinate

Iht, characterization of ds (a) A surface element ds (b) A surface element

1-6 POSITION VECTOR

-13)

Lents fthe • the mon

-14)

-15)

Expressed as a scalar element, ds = dt2 dt3 = h2h3 du z dU3 for that example. An illustration in spherical coordinates is shown in Figure 1-7 (b); on the r = constant coordinate surface, ds = r2 sin 0 dO d¢. A vector quality is given dol' through multiplying it with either the positive or the negative of the unit vector normal to ds. Thus, in Figure 1-7 (b), the vector surface-element ds = a r ds is illustrated; ds = a r ds is the other possible choice on the coordinate surface r = constant exemplified. These concepts are particularly useful in the flux-integration techniques discussed in Section 1-9. Differential line-elements are frequently of interest in applications to vector integration. This subject is introduced in terms of the position vector r of spatial points treated in the next section.

*1·6 POSITION VECTOR 2 In field theory, reference may be made to a point P(Ub Ul, U3) in space by use of the position vector, denoted by the symbol r. The position vector of the point P in Figure 1-4, for example, is the vector r drawn from the origin 0 to the point P. Thus in rectangular coordinates, r is written (1-18)

-16) it dv

11

and in circular cylindrical coordinates ( 1-19)

while in spherical coordinates r = arr

( 1-20)

-17) ugh [lose '(a).

A further application of the position vector r occurs in the symbolic designation of points in space. Instead of using the symbol P(Ul' Uz, U3) or P(x,y, z), you may employ the abbreviated notation P(r). By the same token, a scalar fielel F(ub U2, U3 , t) can be more compactly represented by the equivalent symbol F(r, I), if desired. The differential element of length separating the points P(r) and P(r + dr) in space is denoted by the vector differential displacement dr. The differential change dr does not in general occur in the same direction as the position vector r; this is exemplified in Figure 1-8 (a). (The vector symbol de is sometimes used interchangeably with dr, particularly in line-integration applications.) The difierential displacement dr (or de) is written in terms of its generalized orthogonal components as follows.

dr == de

= al dt l + a 2 dt 2 + a3dt3 alh l dU l

+ a2h2 dU2 + a3h3 dU 3

(1-21 )

(1-22)

It is illustrated graphically in Figure I-8(b) by means of the usual rectangular parallelepiped construction for a vector in terms of its components. Furthermore, the magnitude dt of the vector dt is given by the diagonal of the rectangular parallelepiped; thus,

(1-23) 2Throughout the text, sections marked with an asterisk (*) may be omitted to conserve time if desired.

12


\

,

Pathf

o -~-- (y) (b)

(a)

FIGURE 1-8. The position vector r used in defining points of space and its differential dr. (aJ The position vector r and a difrerenlial position change dr along an arbitrary path. (b) Showing the components of dr in generalized orthogonal coordinates.

For example, in spherical coordinates hl and (1-23) are written

= 1, h2 = r, and h3

r sin 8, so that (1-2:

with ( 1-2. The simplest expression for a differential vector displacement dt occurs in tl rectangular coordinate system, for which, from (1-14), with hi = 112 = h3 = 1 and wi at = ax, a2 = a y and a 3 = a z , the general form (1-22) becomes

(1-2 while its magnitude dt is written, from the generalized (1-23), as

dt

( 1-2

Similarly, in the circular ~ylindrical coordinate system, the substitution of (1-1 into (1-22) and (I and with at a p ' a2 = a4> and a 3 = a z , the vector displa, ment dt and its magIlitude hecoltw

(1-:

ell 2

d
+

( 1-:

The position vector r has usdi.ll applications in the dynamics of particles sud electrons and ions, fiJI' A of Figure 1-8 reveals that if the vector t placement dr of a particle occurs in the time interval dt, then the ratio dr/dt dell( the vector velocity of the at Per). This particle velocity v is defined by

1-6 POSITION VECTOR

13

derivative of the position vector r(t) v

dr dt

. r(t + lit) - r(t) hm --'------:--'----'At-+O

lit

( 1-30)

A second such derivative of r(t) provides the vector acceleration
For example, in a rectangular coordinate system, the notations Vb 1}2, and V3 mean v Y ' and V z respectively. In all orthogonal coordinate systems except the rectangular system, some of or all the unit vectors may change direction as their location P moves in space. A graphical approach to obtaining the spatial derivatives of the unit vectors in an explicit coordinate system is described in the following example.

tf",

1-22)

1-24)

EXAMPLE 1·1. Find the following partial derivatives of the unit vector a r : (a) Oar/Br; (b) Bar/BO; (c) oa,/ocp. :1-25)

(a) The partial derivative oar/Br equals zero, since the unit vector a r does not vary in direction with r (nor does it vary in magnitnde, by the definition of a unit vector).

in the 1 with

(b) The partial derivative oa,/oa can be found graphically from the accompanying figure. If a r is allowed only the differential change dar in the a sense, then dar has

(1-26)

( 1-27) (1-15) splace-

(1-28) ( 1-29)

such as tor dis::lenotes by the

(a)

(b)

EXAMPLE 1-1. (a) Differential dar generated by rotating a r 8-wise. (b) Differential dar generated by rotating a r 8-wise.

I 14


the direction of the unit vector aoo The length of dar is given precisely by the dO, irom the dc1inition of angle divided by radius, and the radius is unit make day become

dar] r ~ constant = ae dB $

= constant

whence the desired result is dar]

dO

r ~ constant ¢=constant

The partial derivative 8aJikp is found sim.ilarly from (b) of the figmeo All only the changoe d4> in the position ofay generates the diHcrential vector dan r a direction specified by the unit vector a.p and a magnitude given by dq, sin (; makes day (for r = constant, () = constant) become a,,> sin 8 dq, as shown, '"

By means of graphic techniques simila,o to those used in Example 1-1, 011 show for spherical coordinates that all the spatial partial derivatives of the unit v in that system are zero except for Ja,o

°

J¢ = aq, 3m

[)

tJ

a y sin () while in the circular cylindrical system, all are zero except for

1·7 SCALAR AND VECTOR PRODUCTS OF VECTORS Besides the simple product of a vector with a scalar quantity discussed in Secli( two other kinds of products involving only vector quantities are now discussel lirst of these, called the scalar product (or dot product), is defined as followso

A· B == AB cos () in which () signifies the angle between the vectors A and B. Noting from (1-3 A· B may be written either (A cos 8)B or A(B cos 0) makes it evident that th, product A . B denotes the product of the scalar projection of either vector 0' other, times the magnitude of the other vector. The definition of A . B makes th,

1-7 SCALAR AND VECTOR PRODUCTS OF VECTORS

15

the angle unity), to

useful, tor example, in computing the work done by a constant force acting a distance expressed as a vector. A generalization of this idea extended to the expression for work is taken up in the next section. Definition (1-34) permits the conclusion that if A and B are perpendicular, cos () zero, making their scalar product zero. Again, if A and B happen to lie in the same then A • B denotes the product of their lengths. These observations lead to results involving the scalar products of the orthogonal unit vectors a l , a2, and 83 coordinate systems illustrated in Figure 1-5. For example, a l • a2 a2 • a3 = 8:\ • a l = 0, while at • at = a z . a2 = a3 • a3 = l. From the definition (1-34), and since B· A means BA cos 0, the commutative fhr the dot product follows.

. Allowing having sin O. This Il, whence

A·B=B·A

(1-35)

Ian

distributive law for the dot product of the sum of two vectors with a third vector

A . (B

+ C) = A . B + A . C

( 1-36)

also be proved. , one can lit vectors

IXAMPLE 1·2. Vector analysis can be used to shorten a number of proo[~ of g-eometry. Suppose one is to show that the diagonals of a rhombus arc perpendicular. Represent its sides and diagonals by means of the veetors shown in the diagram. The diagonals are A + B C and A B D. Form the dot product of C and D.

(A ( 1-32)

+ B)

. (A - B)

which must equal zero because A

=

A· A - B . B

B for a rhombus.

=

A2 - B2

Thercf()[(~

C and D are perpendicular.

1f rhe vectors A and B are expressed in terms of their generalized orthogonal c:omponents in the manner of (1-9), their scalar product can be written ( 1-33)

expanding this expression by means of the distributive law (1-36) and applying results obtained earlier fix the dot products of the unit vectors, one obtains

eclioll 1-3, ussed. The

(1-37a)

(1-34)

(1-34) that the scalar )r onto the s the scalar

t

EXAMPLE 1-2

if

16


For example, the expansion of the dot product of two vectors in rectangular nates is

COOl

( 1-3' and in circular cylindrical coordinates (1-3

EXAMPLE 1-3. (a) At the point P(3, 5, 6), shown in (a) of the figure, are given the two veet D = - 50a x + 60a y + 100az and E = 12ax - 24ay- Find the vector magnitudes and dot product D . E. Use these to determine the projection D cos of D onto E, and angk between the vectors. (b) In (b) of the figure, at point P(5, 60°, 9) are given two vectors F = IOa p + Ba", 4a z and G = - 20ap + BOa z in cylindrical coord ina Find the vector magnitudes and F . G as well as the angle 0 between the vectors.

e

e

(2)

D=

50a.,

+603.,+ 1003,

(xl

~-

5

'~

(yl

(al

G

F F II.

(xl

EXAMPLE I

17


coordi-

(a) By usc of (I

the vector magnitudes are

1-37b) while the dot product is found from expansion (1-37b) 50(12)

+ 60( -24)

2040

The latter, by (1-34), also means DE cos 0, whence the projection D cos 8 becomes vectors and the md the ven the :Iinates.

J)

D·E E

cos (J

-2040 26.833

76.03

This nCl-iative result shows that the projection D cos 0 alonl-i E is in the negative-E sense (meaninl-i that 0 exceeds 90°). The value of 0 is found from the definition (1-34), yieldinl-i

0= cos

.. 1

D .E

~- =

DE

cos

.. 1

2040

-~-..- - - - - =

126.886(26.883)

(b) The mal-iniludcs and dot product, from (1-7) and

, . ,0 126.82 in circular cylindrical

coordinates, arc

G= F' G

+ F~ + 1';]112 [10 2 + 8 2 + 4 2 ] [20 2 + 80 2FI 2 = 86.462 [F~

F

=

= 13.416

10(-20) - 4(80» = -520

The anl-ik () between F and G is found from definition (1-34), obtaining

0= cos

.. 1

F.G .. 1 520 , = cos --~"'-'-- - = 117.93 FG 13.416(82.462)

From this result you may determine that the projection of F ncgativc-G sellSe.

0

G is in the

The second kind or product of one vector with another is called the vector product cross product), defined as l()Uows

A x B = a"AB sin 0

( 1-38)

e

is the angle measured between A and B, and a" is a unit vector taken to be perpendicular to both A and B and having a direction determined {i-om the righthand rule provided that the rotation is taken {i'om A to B through the angle O. The vector product A x B is illustrated graphically in Figure 1-9. One may show from the diagram that in which

A

X

B

-B

X

A

(I

18


AxB

AxB

,f.

--",

A~--~

A

()

B

Positiv€~""

(J sense

from A to B

FIGURE 1-9. Illustrating the cross product.

which means that the vector product does not obey a commutative law. In forming t1: cross product, the ordering of the vectors, therefore, is an important consideratiOl If A and B are parallel vectors, sin is zero to make their cross product zen If A and B happen to be perpendicular vectors, then A X B is a vector having a lengt AB and a direction perpendicular to both A and B, with the ambiguity in the directio resolved by means of the right-hand rule. These observations applied to the crm products of the orthogonal unit vectors of Figure 1-5, for example, lead to the sped" results: al X a l = az X az = a3 X a3 = 0; a l X az = a 3 , az X a 3 = al, and a3 > a l = az. However, note that a i X a3 az. A distributive law can be shown to hold for the cross product

e

A

X

(B + C)

=

A

X

B +A

X

C

(1-40

Because of the noncommutativity of the cross product as expressed by (1-39), the orde of the factors in (1-40) is important. If the vectors A and B are given in terms of their orthogonal components il the manner of (1-9), then their vector product is written

The use of the distributive law (1-40) and the special results obtained for the cros products of the orthogonal unit vectors provides the following expansion.

which can alternatively be put into the compact determinentaI form

A

X

B

=

al

az

a3

Ai

Az B2

A3

Bl

B3

(I -41)

19


Pivot

P EXAMPLE 1-4

EXAMPLE 1·4. The definition of the cross product can be used to express the moment of a force F about a point P in space. Suppose R is a vector connecting the point P with the point of application Qofthe force vector F, as shown in the diagram. Then the vector moment M has the magnitude M = RF sin (} = X Fl. The turning direction of the moment, as well as its magnitude, are thus expressed by the vector product

IR

M

Ig the

Hion.

RxF

(1-42)

EXAMPLE 1·5. A force F = !Oay N is applied at a point Q(O, 3, 2) in space. Find the moment ofF about the point P(2, 0, 0). The vector distance R between P and Q)s

zero. ~ngth

'ction cross )ecial

The vector moment at P is found by means of (l-42) and the determinant (1-41).

a3 X

ax

1-40)

ay az

M=RxF= -2

3

2

o

10

0

=

-20a x -20a z N-m

M, shown at P in the sketch, is a vector perpendicular to the plane formed by F and R.

)rder Its

in

EXAMPLE 1·6. Given the two vectors F and G in (b) of the figure in Example 1-3, determine their vector cross product F x G, as well as the magnitude of the latter. Find the unit veetor an in the direction of the vector F X G. Verify that an is perpendicular to F and to G.

cross 1 1(z)

21

/1d-,. ---_9(0,3,2) I

/

P(2 0 0) I , ,

/

I

I

R

---0 --_

_(-)---M

-41 )

I

I

x EXAMPLE 1-5

1

F = lOay

I

---3'--_ "1

(y)

20

VECTOR ANALYSIS ANI) ELECTROMAGNETIC FIELDS IN FREE SPACE

From (I

in circular cylindrical coordinates, F x G becomes

I)

FxG

ap

a",

az

IO

8

-4

20

0

80

+ a",[ -4( -20)

- 10(80)]

+ a z [IO(O)

8(

20)]

160az The F G is IF X GI = [640 2 + 720 2 + 160 2 ] 1/2 vector an in the directioll of the vector F x G is given by

FxG

a n

The dot

= 'iF x

0.655ap

Gi an'

= 976.5,

0.737a",

while the uni

+ 0.1638a z

F [wcnmes, from (l-37b), the zero result 10

0.737(8)

+ 0.1638( -4)

= 0

verifying frorn til!' definition (I that an and F are perpendicular vectors. You ma) similarly show that an and G are perpendicular.

1·8 VECTOR INTEGRATION Vector integration, f()f the purposes of field theory, encompasses integrals in space along lines, over surfaces, or throughout volume regions, as well as integrals in the time domain and the domain. The subject of the present discussion concerns only integrations in space. Tne vector notation embodies compactness as an important feature, so it is always worthwhile to examine the integrand ofa vector integral carefully. The integrand may be either a scalar or a vector Thus, the integrals

[ A' Bdt

Line integral

~I'

J, (C J: F'· possess scalar hand, the

Surface integral Volume integTal

Gdll

produce scalar results on integration. On the other

amI

G

Line integral

Hx

Surfilce in tegral

J and

D) • ds

X

X

K

Volume vector results. In the last three examples, acroullt the different directions assumed by the on the surhce ,,)', or in the volume V defined.

t1H'IT/ill'C

1-8 VECTOR INTEGRATION Typical di (Vector displacement) ' "

dt (Scalar displacement) P2

Patht

21

""

P2

~~-;;J

.l-----R Pl

(b) ~:XAMPLE

\-7. (a) Integration of the scalar dt over a path t. (b) Integration the vector dt over the path t.

, unit EXAMPLE 1·7. The difterent results provided by scalar and vector integrands is exemplified by simple integrals of scalar and vector displacements dt or dt along some prescribed path in space. The integral

may

summed over the path t shown in (a) of the figure, provides its true scalar length d. On the other hand, the integral of the vector displacement dt on the same path

R=

)ace the erns vays nay

Jtr dt

produces quite a diftercnt answer, a vector result R determined only by the endpoints P l and P2 of that path rather than by the form of the path between the endpoints. This vector R is illustrated in (b) of the accompanying figure. So the line integral of dt about a closed path is zero, whereas if dt is the integrand, the perimeter of the closed pa th is the result.

An integral flllding extensive utility in work or energy calculations is the scalar line integral

LF . dt == LF dt

cos

e

(1-43)

This integral sums the scalar product F . dt over the path t, as suggested by Figure 1-10. Only the projection ofF along de at each point on the path contributes to the integral result. The line integral (1-43) can be expressed in terms of the generalized orthogonal components of F and of de in the following way, making use of (l-9), (1-21), and her

1-37 a)

(1-44)

es, he ~d.

In the rectangular coordinate system, in which hi = h2 = h3

= 1,

(1-44) is written (1-45)

22


- --

11----""'- F

(a)

(b)

(c)

FIGURE J-lO. A palh and the field F in space. (0) Division of t into vector elements dt. (c) product F· dt (to be summed over the path) shown at the typical point P on the path.

assuming (Xi,_Vl' of the path t.

.::tl

are the coordinates of the endpoints P 1 and P

and

EXAMPLE 1·8. Evaluate the line integral (1-43) between the points PI(O, 0,1) and P 2 (2, 4,1 ovcr a path t defined the intersection of the two surfaces y = x 2 and z = 1, if F is thl v(,ctor fidd

(1 The path t is illustrated ill the 2 Inserting = lOx, .')x y, and f~ it {()llows that dz = 0 from the definitiD!!

£

F . dt

the desired resnlt.

j'

(2,0,

( EXAMPLE I

fx2~O

lOx dx -

20

106.7 =

into (1-45) and since x 2 = y all(

fy4=O 5y2 dy + 0 86.7

1-9 ELECTRIC CHARGES, CURRENTS, AND THEIR DENSITIES

23

This answer can also be obtained by expressing the dificrential displacement dx along the path in terms of From the definition of l, dy = 2x dx and dz O. Thus 2 Jtr F . dt: = Jor

lOx dx

'4

j o 5y2 dy

4y =

-36.7

IXAMPlE 1·9, A line integral such as (1-4·3) in gcncral has a value depending on the shape of the path connecting the endpoints PI and P2 . Evaluate the integral of Example 1-3 for the same function F and the same endpoints PI(O, 0,1) and P2(2, 4,1), but deform t: into the straight-line path given by the intersection of the surElCes y = 2x and z = I. Integral (1-43) now becomes dy

+0

60

obviously dilll'rent from the result obtained over the parabolic path in the last example. F is f()r tbis reason called a nonconservalive field. A vector field fell' which the line integral (1-43) is independent of the shape of the path connecting a fixed pair of emlpoints is said to be conservative. More is said later of such fields in connection with static electric charge distributions in Chapter 4.

1·9 ELECTRIC CHARGES, CURRENTS, AND THEIR DENSITIES 4, I) is the

(I)

, and

The physical and the chemical properties of matter are known to be governed by the eitcctric and magnetic forces that act among the particles comprising all material sub!ltalH~es, whether inorganic or living cells. The fundamental electric panicles of matter of two varieties, commonly called positive and negative electric charges. Many experiments have provided the following conclusions concerning electric charges. 1. The algebraic sum oCthe positive and negative electric charges in a closed system never changes; that is, the total electric charge of a defined aggregate of matter is consewed. 2. Electric charge exists only in positive or negative integral multiples of the magnitude of the elect mnic charge, e = 1.60 X 10 - 19 C; this implies that electric charge is quantized. From the viewpoint of classical electromagnetic theory, an electric charge aggregate will be treated as though it were capable of being indefinitely divisible, such that a volume electric-charge density, denoted by the symbol Pv is defined as follows 3

Pv

Aq ,

=--

Ali

elm

3

( 1-46a)

This limit of this ratio is taken such that the volume-element in space does not become so small that it contains so few charged particles that the relatively smooth property of the density quantity p" is lost, although Ali is kept small enough thal thl' integration the quantities containing Av becomes a meaningful process. I-II (a) illustrates the meaning of these quantities relative to a volume eiemellt

or

3It is dear thaI Ihe symbol p, for volume ('haq;;" density should not be confused with the lIn.l11/" , the radial variahle of the circular cylindrical coordinales (p, 4>, 'c).

:!i

I 24

VECTOR

ANn ELECTROMAGNETIC FIELDS IN FREE SPACE

de

.. ex.· '~

dq = p{ dt on dt

(a)

FIGURE I-IL

(c)

(b)

used in ddining volume, surface, and line charge densities in space. Qualltit;t·, defining Ps' (el Quantities defining pt.

Aq residing within any element Av may vary from pOil point in a it is evident from (1-46a) that charge density function of space as possibly of time. Thus Pv is a .field, written in ger Pv(ur, U2' U 3 , t) or p,,(r, In some physical the charge I1q is identified with an element of su or line instead of a volume. The limiting ratio (1-46a) should then be defined as foil Aq

pS=AC/m

2

( 1-'

L.l.S

Aq

Pc

= 111' C/m

(1-

The quantltles associated with these definitions of volume, surface, and line ch. densities are illustrated in 1-11. In some systems aggregates, two species of positive and negative ch: densities may be simultaneously. A net charge density p" (volume, sud such an instance defined or line density) is

p"

p,;

+ pv

C/m 3

(1

P:

in which and denote limiting ratios defined due to the positive negative charges + ami Aq respectively in Av. occurrence of both pos: metallic ions and mobile electrons in a conductor is an example to which (1-47) rna applied. The ill this being of eqnal magnitudes but opposite s 41n some physical ent simultaneously characterized by

discharge, electrons and several kinds of ions maybe Their net density at any point in the region may th,

(I if a total of

to be (lUnd there.

25

1-9 ELECTRIC CHARGES, CURRENTS, AND THEIR DENSITIES

P;; = - p;;), cancel, providing the net density Pv 0 in such a compensated charge system. The total amount of charge contained by a volume, surface, or line region is . obtained from the integral of the appropriate density function (1-46a), (1-46b), or 1-46c). Thus in some volume region, each element dv contains the charge dq = Pvdv, making the total charge in 1) the integral

Iv dq Iv Pv du C

q=

=

Similar integral expressions may be constructed to yield the total charge on a given surface or a line in space.

EXAMPLE 1·10. (a) The radially dependent volume charge density Pv = 50r2 C/m 3 exists within a sphere of radius r 5 cnL Find tlfe total charge if contained by that sphere. (b) The same sphere of is now covered with the angularly dependent surface charge density Ps 2 x 1O~ 3 0 C/m 2 Find the total charge on the spherical surface.

point to lsity is a general

(a) Making usc of( 1-47) and dv of (1-17) obtains

q=

Sv p" dv SSS (50r2)r2 sin 0 dr dO dcjJ = 50 s:n d(p S: sin 0 dO S:·os r dr 4

,5 JO.os =

)f surface follows.

= 50(2n)2 -5

(1-46b)

(1-46e) ~

charge

°

3.927 x 10 - 5 = 39.27 j1C.

Attention is called to the "product separability" of the integrand in this example, enabling the expression of the triple integrand as the product of three separate integrals in r, 0, and cjJ.

Is

(b) Using q = Ps lis in this case, along with the scalar surhrce clement ds = r2 sin 0 dO dcjJ on this sphere or radius r = OJ)5 m, as suggested by ds shown in Figure 1-7(b), yields on the complete sphere If

=

e charge surface,

f p"d.1 = ff(2 S

X

2 x 1O~ 3(0.05)2 =

10- 3

cos 20)r2 sin OdO r/cjJ] ~

r.

In (OS2 0 sin 0 dO = Jo

2n

.0

dcjJ

r-O.OS

5 x 10-

[

_cos~Jn 3

0

20.9 /lC

( 1-47) rive and positive maybe te signs ly be presy then be

(1-47a)

A vector field F(Ul' U2, U,' t) at some given instant t, can be represented graphiby use of a myriad of vectors of appropriate lengths and directions at many in a region of space. A vector field plotted in this way is shown in Figure 1-12 (a). is, however, a cumbersome way to graph a vector field; usually a much more representation is by use of a/lux plot, a method replacing the vectors with lines (called jlux lines) drawn in accordance with the i()llowing rules. 1. The directions of the flux lines agree with the directions of the field vectors. The transverse densities of the flux lines are the same as the magnitudes of the fidd vectors. The flux plot of the vector field of Figure 1-12 (a), sketched in accordance with these is noted in (b) of that figure. If a surhlce S is, moreover, drawn in the region

26

VECTOR

(a)

ELECTROMAGNETIC FIELDS IN FREE SPACE

(b)

(c)

FIGURE 1-12. A veCWr field F, its flux and the flux through typical surfaces (a) A vector field F, denoted by "farrows. The flux map of the vector field F, showing an open surface S through a net flux passes. (e) A closed surface S, showing zero net flux emergent from it.

of space embracing that flux, then the net lines of flux r/J passing through S can be a measure of some physical quantity (such as charge,current, or power flow), depending on the physical meaning ofF. The differential amount of flux dr/J passing through any surface-element ds in space is defined by the scalar dr/J = F ds cos = F • ds, a positive or negative l-esuit, depending on the angle between F and ds. The net (positive or negative) flux of F through S is therefore the integral of dr/J over S

e

Is F' ds

(1-48)

in which ds is taken to emerge from that side of S assumed positive, as shown in Figure 1-12 (b). If S is a dosed surface, the net flux through it is given by

(1-49) as noted in Figure \-12 The la Her will integrate to zero (an indication that just as many flux Jines leave S' as enter it) unless the interior volume of S contains sources or sinks offlux lines. This view will be amplified later in the discussion of the divergence of a vector field. The current flow through a surb.ce embodies a good illustration of the flux COIlcept. Supr:iose that there are electric charges of density Pv(Ul, U2 , U3, t) in a region, and imagine that the cllarges have velocities averaging to the function v(ul' U2, U3, t) within the elements dv with which the densities Pv are identified. A current density function J may then be defined at any point P in the region by or

C/sec/m 2

( 1-50a)

This function is a measure, in the vicinity of any point P in space, of the instantaneous rate of flow of charge per unit cross-sectional area. If two species of charge density

l-9 ELECTRIC CHARGES, CURRENTS, AND THEIR DENSITIES

27

of opposite kinds, designated by P;; and Pv , exist simultaneously in a region of space, then their total current density J at each point is written

( 1-50b) In general, for n species with densities Pi and velocities Vi (e.g., electrons plus a mixture of ions)

( 1-50c) The differential current flux di flowing through a surface element ds at which the current density J exists, is di J . ds amperes, to make the net current i (current through S

i

be a :ling any ltive e or

48)

= S~ J . ds C/sec

or

(I-51 )

A

IXAMPLE 1·11. An electron bcam of circular cross-section 1 mm in diameter in a cathode ray tube (CRT) has a measured current of I itA, and a known average electron speed of 106 m/sec. Calculate the average current density, charge density, and rate of mass transport in the beam. Assuming a constant current density J = azJz in the cross-section (I-51), yields the following current through any cross-section.

in which A denotes the cross-sectional area of the beam. Thus the average current density is

Jz=

-ure

i A

=

10- 6 4 2 n(1O-3)2 = n Aim ----~

4

49)

The charge density in the beam, from (1-50a) in which becomes

ust ces Ice

J

a z 4/n imd v - - az I0 6 ,

Jz

)0-

)n, t) ds = azds

IC-

1 mm

a) us ty

Cross-section A I':XAMPLE I-II

i = l/1A

-----------l>-'(z)

28


The rate of mass transport in the heam is the current times the electronic mass-to-charge ratio; this yields 5.7 x 10- 18 kg/sec, assuming an electron mass of9.! x 10 31 kg.

1·10 ELECTRIC AND MAGNETIC FIELDS IN TERMS OF THEIR FORCES Electric and magnetic fields are fundamentally fields of force that ongmate from electric charges. Whether a force field may be termed electric, magnetic, or electromagnetic hinges on the motional state of the electric charges relative to the point at which the field observations arc. made. Electric charges at rest relative to an observation point give rise to an electrostatic (time-independent) field there. The relative motion of the charges provides an additional force field called magnetic. That added field is magnetostatic if the charges are moving at constant velocities relative to the observation point. Accelerated motiolls, on the other hand, produce both time-varying electric and magnetic fields termed electromagnetic fields. The connection of the electric and magnetic fields to their charge and current sources is provided by an elegant set of relations known as Maxwell's equations, attributed historically to the work of many scientists and mathematicians well before Maxwell's time,5 but to which he made significant contributiohs. They are introduced in the next section. Suppose that electric and magnetic fields have been established in some region of space. The symbol for the electric field intensity (or just electric intensity) is the vector E; its units are force per unit charge (newtons per coulomb). The magnetic field is represented by means of the vector B called magnetic flux density; it has the unit weber per square meter. If the fields E and B exist at a point P in space, their presence may be detected physically by means of a charge q placed at that point. The force F acting on that charge is given by the Lorentz force law F = q(E

+v

x B)

(1-52a) (1-52b)

=FE+FBN in which

q is the charge (coulomb) at the point P v is the velocity (meter per second) of the charge

q

E is the electric intensity (newton per coulomb) at P B is thc magnetic flux density (weber per square meter or tesla) at P

FE

=

qE, the electric field force acting on q

F B = qv x B, the magnetic field force acting on q In Figure 1-13, these quantities are illustrated typically in space. The force FE has the same direction as the applied field E, whereas the magnetic field force F B is at right angles to both the applied field B and the velocity v of the charged particle. The Lorentz force expression (1-52) may be used for discussing the ballistics of charged particles traveling in a region of space on which the electric and magnetic fields E and B are imposed. The deflection or the focusing of an electron beam in a cathode ray tube are common examples. 5James Clerk Maxwell (1331-1379).

I

il

29

I-II MAXWELL'S INTEGRAL RELATIONS FOR FREE SPACE

FE

B flux ---

--".,-I

I I

,I

I

I

,~ :

B

I I

I

t

FB (b)

fa)

(c)

:FIGURE 1-13. Lorentz forces acting on a moving charge q in the presence of (a) only an E field, (b) only a B Geld, and (e) both electromagnetic Gelds,

EXAMPLE 1·12. An electron at a given instant has the velocity v (3) I05 ay + 105 az m/see at some position in empty space. At that point, the electric and magnetic fields are known to be E = 400a z V/m and B = O.005ay WbJm 2 , Find the total force acting on the electron. The total force is found from the Lorentz reaction (1-52a)

F = q[E

+v

X

B) = -1.6(10-19)[a z400

+ (a y 3' 105 + a z4'

105)

X

a y O.005)

= (a x 32 - a z 6.4)IO-17 N

Although this is quite a small force, the very small mass of the electron charge provides a tremendous acceleration to the partiele, namely a F 1m = (a x 3.51 - azO. 7) 10 14 m/sec 2 .

1·11 MAXWELL'S INTEGRAL RELATIONS FOR FREE SPACE The relationships among the electric and magnetic force fields and their associated charge and current distributions in space are provided by Maxwell's equations, postulated here in integral form for the fields E and B in free space.

~s (EoE) . ds

Iv pv dv C

J. B· ds = 0 Wb :Vs J, E. dt = -~

'Ji

dt

( 1-53) (1-54)

r B . ds V

Js

(1-55)

( 1-56)

30


in which E B

=

E(Ul'

U2 , U3 ,

t) is the electric intensity field

B(ul'

U2, U3,

t) is the magnetic flux density field

Iv p" do = q(t) is the net charge inside any dosed surface S itt) is the net current flowing through any open surface S bounded by the closed line t Eo

is the permittivity offi-ee space (~10 9/36n F/m)

110

is the permeability of free space (= 4n x 10

7

Him)

The Maxwell equatious 6 (I-53) through (1-56) must be simultaneously sati~fied by the field solutions E and B for all possible closed paths t and surfaces S in the region of space occupied by these fields. This strict requirement might appear to limit severely the number of practical problems that can be solved by means of these integrals. Indeed, their application to the discovery offield solutions E(u l , U 2 , U 3 , t) and B(ul' 11 2 , 11 3 , t) is restricted, in the present treatment, to problems in which the charge or current distributions have particular symmetries that serve t~ simplify the solutions. The equivalent differential forms of Maxwell's equations, developed in the next chapter, have a somewhat wider range of application in problem solving at the introductory level. The reader is to be assured that only a low-level introduction to methods for obtaining electric and magnetic field solutions of Maxwell's integral relations (1-53) through (I-56) is attempted here. For the purposes of this introductory treatment, the Maxwell relations are simplified by considering only the field solutions of a few simple, symmetrical geometries of static charge or current distributions. In Examples 1-13 through 1-17 that follow, these simplifications are shown to enable, in one or two steps, solving for the electric or magnetic field of a given charge or current distribution. The symmetry of the distribution will be seen to be the key to providing quick solutions for the desired field. Symmetries about a point, a line, or a plane are considered.

A. Gauss's Law for Electric Fields in Free Space Maxwell's integral law (1-53) [I-53 J is also known as Gauss's law for electric fields in free space. The meanings of the quantities are illustrated in Figure 1-14. Thus, suppose that there is in free space an electric field E(Ub U2, U3, t) (denoted by the E-field flux line distribution in that figure), plus some related electric charge distribution of density Pv(Ul> U2, U3, t) as shown. Construct in this region a closed surface S, with S having any desired shape and enclosing 6 Although given the collective name Maxwell's equations, historically they were in a gradual process of evolution over many years before Maxwell's time. For an enjoyable and first-rate account of the details, you are encouraged to read the historical surveys at the beginning of each chapter in R. S. Elliott, Electromagnetics. New York: McGraw-Hill. 1966.

1-11 MAXWELL'S INTEGRAL RELATIONS FOR FREE SPACE

31

FIGURE 1-14. Typical closed surface S in a region containing an electric field and a related electric charge. Gauss's law must hold [or all closed surfaces constructed in the region, whether charges are contained or not.

all or any part of the electric charge in the region, or no charge at all, as desired. Then the Maxwell-Gauss law (I-53) means that the integral of the quantity (EoE) . ds over that closed surface S (the net, outward flux of EoE emanating from S) is a measure of the amount of electric charge Pv dv = q that is contained only within the volume V bounded by that surface S. The dosed-surface integral of (EoE) . ds thus automatically excludes any charge that happens to lie outside S. (The surface element ds on S is by convention taken as positively outward from S, as shown in Figure 1-14, or away from its interior volume v.) The constant Eo in this Maxwell-Gauss law, called the permittivity oIfree space, is approximately 1O-9/36n F/m in the mks system of units. 7 To evaluate the amount of electric charge q within some volume V surrounded by the dosed surface S, Gauss's law (1-53) can be employed to do this two ways: (1) from the right side of (1-53), by use of the volume integral of the charge density Pv contained within the volume V; or (2) from the left side of (1-53), by integrating (EoE) • ds over the closed surhtee S that bounds the volume V of interest. If a known charge distribution is static (motionless) and happens to possess a particular symmetry in free space, then Gauss's integral law (1-53) can even be used to evaluate the electric field E produced by that charge. The small class of symmetric, static charge problems that can easily be solved by use of Gauss's law are illustrated in the following example.

Iv

EXAMPLE 1·13. Find the electric field intensity E of the following static charge distributions in free space: (a) a point charge Q; (b) a spherical cloud of radius 10 containing a uniform volume density Pu; (c) a very long line charge of uniform linear density PI; (d) a very large planar (surface) charge of density Ps'

These charge distributions arc illustrated in Figure 1-15. Closed surfaces S arc shown, appropriately chosen to permit solving for E by the usc of Gauss's law (I-53). (a) Field of a charge (symmetry about a point). To evaluate the field E of the static point charge Q, choose S in Gauss's law (I-53) to be the sphere with Qat its center, as in Figure l-15(a). To show that E has only a radial component about 7The significance or the units of Eo is clarified in Chapter 4 in the discussion of capacitance. A correct interpretation of the factor Eo in (1-53) is that it is a proportionality factor accounting for the proper units (mks) of the equation.

32

i i

VECTOR ANALYSIS AND ELECTROMAGNETIC FIELDS IN FREE SPACE Sphere 8 1 ,\

/

"

Sphere 82

-

---~ ,

/ Point / charge

l\

\

"- "-

~E = arEr

P,

Qr'JJI

ds

/,- \ ~ r \

Spherical charged cloud

-,, -

"' , , ___ -,' ~Spherical

'\

~E(r
:;~It:sr ':;

\

\

'''"- closed surface S

-t

"

...:.:.,,'. "~ 3 ds Pv Clm ,,/

',-------/ E (r> ro)

(a)

(b)

Uniform surface charge density Ps

-~--

I

,I

Circular cylindrical closed surface S (e)

(d)

FIGURE 1-15. Static charge distributions having symmetries such that Gauss's law applied to appropriate closed surfaces will lead to solutions lor E. (a) Static point charge; spherical surface S constructed to evaluate E(r). (b) Charged cloud of uniform density, showing SI and 8 2 used to evaluate E(r). (t) Uniform line charge. (d) Uniform suriace charge.

the charge, obsfTve that for this time-static problem (did! 0, for all fields), (I-55) reduces to E . de' = 0 for all closed Jines e'. Then integrating E • de' about any circumferential path of radius r over the sphere in Figure 1-15(a) yields the conclusion that Eo and E are zero. Furthermore, assuming Q positive, E must be directed radially outward if the integral of E" oE over S is to yield a positive answer. Thus (I-53) yields

f

Since a r • a r and from the symmetry Er is constant on S, Er may be extracted from the integral to obtain (1-57a) or, in vector form E

(1-57b)

\-\\ MAXWELL'S INTEGRAL RELATIONS FOR FREE SPACE

33

Coulomb's law {Cll' the force acting on another point charge Q: in the presence of Qis deduced by combining (1-57b) with the Lorentz force relation (1-52a). In the absence of a B field, the force on Q: when immersed in the E field (1-57b) of the charge Qis ,

FE = Q:E = a r

Q:Q

(I-58)

z 4nEor

--

(b) Field of a charged cloud (symmetry about a point). For the spherical cloud containing a uniform charge density p" C/m 3 , two cases arise. The field outside the cloud (r > TO) can be obtained from Gauss's law (I-53) applied to a concentric sphere S1 of radius r, as shown in Figure 1-15(b). That E bas only an Er component is shown as in part . Then the charge q enclosed by Sl is obtained by integrating p" dv throughout sphere, so (I-53) becomes

Solving lor Er (constant

011

S 1) yields (I-59)

an inverse-square result. It is ofthe form of the point-charge result (1-57a), assuming the field point outside the charge cloud (r> ro). inside the cloud (r < ro), applying (I-53) to the closed surface S2 of Figure 1-15 (b) yields

in which the volume integration is carried out only throughout the interior of S2, obtaining . With l!,~ constant on S2, E = p"r r 3Eo

r<

(1-60)

TO

E inside the uniformly charged cloud is theref(Jre zero at its center and varies linearly to the samc valuc as (1-59) at the sllrface r roo (c) Field o/a long line (symmetry about a line). Construct a closed right circular cylinder of length t and radius p concentric about the line charge as in Figure 1-15(c). From symmetry, E is radially directed (apEp) and of constant magnitude over the peripheral surface So. The left side of Gauss's law (1-53) is zero over the endeaps of S because E . ds is zero on them. Thus (I-53) becomes

in which the right side reduces to a line integral over the linear charge distribution. Solving for Ep on Sp yields, with Pt dl = ptt and ds = 2npt

Js

ft

Ep

Pc 2nEop

. (1-61)

Thus, the electric intensity of an inflflitely long, uniform line charge varies inversely with p. .

34


(d) Field an planar charge (symmetry about a plane). A closed surface 8 is constructed in the form of a rectangular parallelepiped extending equally on both sides of the planar charge, as in Figure 1-15(d). The symmetry of the infinitely extensive charge requin:s that E be directed normally away from both sides of the charge as shown (E = ±a,h'x)' Flux emanates only from the ends 8 1 and 8 2 of the parallelepiped, whence Gauss's law becomes

A denoting the area of the ends of the parallelepiped. The two integrals over 8 1 and 8 2 provide exactly the same amount of outward electric flux, whence E = x

o'--------------~r

2Eo

~----~----~~r

ro

(b)

(a)

(c)

fd)

FIGURE 1-16. Flux plots of the fields of Example I-II. (a) Point charge. An inverse r2 field. (b) Uniformly charged spherical volume. The graph depicts variations with r. (c) Unilormly charged infinite line. An inverse p field. (d) Uniformly charged infinite plane. E is uniform everywhere.


35

Writing this in vector form to include the fleldB on both sides of the planar charge distribution gives

-~

E=a x

2Eo

Ps E =-a x

2Eo

x>o x
(1-62)

It is evident that the electric field to either side of a uniform, infinite planar charge is everywhere constant. Flux plots of the electric fields of the four charge distributions covered in this example are shown in Figure 1-16.

B. Ampere's Circuital Law in Free Space Maxwell's integral law (1-56)

J. B . dt 'it flo

SJ . s

ds

+ -dtd

S(EoE) •

ds

S \

= i + -difte dt

[ 1-56)

often called Ampere's circuital law for free space. Figure 1-17 illustrates the meanings the field quantities relative to any closed line t that bounds a two-sided surface S. positive direction of the typical element ds may be taken to either side of S, but positive integration sense about t must agree with the right-hand rule relative to The relation (1-56) means that the line integral of the B field (modified by 1) aw'und any arbitrary closed path t must, at any time t, equal the sum of the net electric current i plus the time rate of change of the net electric flux ift e passing through the lurface S bounded by t.

fl,o

-~-t ~

~ /

/~--

~- __ ~}~j>-\\ J Jlines~+I_+_1 - .}/\ j..'. r

..[\

ds --""' - -

n!~~;Jn1T ~ -

---

\

0rV~-r--

foE

~

~"-~.~ ---------~- --

(b)

(a)

FIGURE 1-17. Induced magnetic flclds ilnd Ampere's law. Any dosed line t such as that of (a) may be superposed anywhere on the example of (b); Ampere's law must be true for it. (a) Typical closed line t boundiug a surface S, relative \0 the fields in Ampere's law. (b) A symmetric example showing the B field induced by electric currents and displacement currents.

36


The two terms on the right side of (I-56) denote the two kinds of electric currents that occur physically in free space. The first, i, has already been discussed in relation to (I-50) and is given the name convection current when it is comprised of one or more species of moving charges in free space; it is also called conduction current if it pertains to electric charges drifting or transported within a solid, liquid, or gas. The second term, dt/! e/dt, is called displacement current and denotes the time rate of change of the net, instantaneous electric fiux t/! e that passes through the surface S. The displacement current term is the history-making contribution of Maxwell, who provided that missing link to unify the theories of electricity and magnetism and predicted the propagation of electromagnetic waves in empty space in the absence of charges and currents. The quantity fto is calJed the permeability oJJree space; it has the value 4n x 10- 7 H/m in the mks system of units. 8 A comparison of (I-56) with Gauss's law (I-53) shows that Ampere's circuital law is more comprehensive; it involves both the magnetic field B and the time-varying electric field E, as well as electric currents that might be fiowing in a region. Indeed, it specifies that either electric currents or time-varying electric fields in a region, or both, will give rise to a magnetic field B such that (I-56) must be satisfied for all possible closed lines constructed in the region. The direct application of Ampere's circuital law (1-56) to obtaining time-varying field solutions E(r, t) and B(r, t) whenever, for instance, a current distribution J(r, t) is somehow specified is not, in general, feasible. The difficulty lies in part in not knowing how to specify the current distribution without more information about the accompanying fields; the intricacies may be appreciated more fully on recognizing that the field solutions must satisfy simultaneously all four of Maxwell's integral relations, (I-53) through (I-56). No field solutions of the complete A~npere's law (1-56) are attempted at this time. Instead, consider for a moment only the magnetic B field of a static (direct) current distribution i, in which event the Maxwell-Ampere law (1-56) reduces to the form

A: ~. dt ':Yt flo

=

{'

Js

J' ds

==

i

Ampere's law for static fields

( 1-63)

To evaluate tbe amount of static electric current i passing through some surface S, tbe static Ampere law (1-63) can be employed to do this two ways: (1) from the right side of (1-63), using the surface integral ofJ' ds over any desired surface S (a fiux integral); or (2) from the left side of (1-63) by integrating (B/flo) • dt about the closed path t that bounds the surface S (a closed line integral). If a known static current distribution also happens to possess a simple symmetry in free space, then the static Ampere integral law (1-63) can even be employed to evaluate the magnetic B field produced by that current. Simple illustrations of these uses of Ampere's law (1-63) are described in the following examples.

EXAMPLE 1-14. Find the net sta-tic electric current i that flows through each of the surfaces S bounded by the paths

t

chosen for the three direct current systems of Figure 1-18.

8The significance of the units of flo is clarified in Chapter 5, in the discussion of inductance. In (I-56), flo is the factor that properly adjusts the units of the term in which it appears, to yield an equality that is dimensionally correct.

ds

J FIt sur cir


37

IA

(b)

(a)

FIGURE 1-18. Three examples of direct current systems showing closed paths t' that bound surfaces S through whir h net currents i flow. (a) Infinitely Ion£" wire carrying I A. (b) A two-mesh de circuit. (e) A four-turn coil carrying I A.

For the long, straight wire of Figure 1-18(a), the path t( shown yields a net current i = 0 through S\; while t2 embraces i = I A, a positive result if ds on S2 is taken to be positive in the upward direction. (b) Assume a positive ds in the upward direction on S. Then, by inspection of Figure 1-18(h), the net conduction current i through S becomes

i

= Js r J. ds = 21 + 1 - 31 = 0

(el For the path t constructed about the coil in Figure l-lS(e) such that the coil pierces S four times, the net current becomes

if ds is assumed positive in the direction shown.

EXAMPLE 1·15. Field DIa long, round wire (symmetry about a line). Use Ampere's circuital law (1-63) to find the B field of the static current 1 in an infmite, straight, round wire of radius 11, shown in Figurc l-19(a). Find B both inside and outside the wire. As in Figure 1-19(11), assume a symmetric, closed integration path t\ having the radius p sbown. From (1-63), the B field must be ¢-directed if the line integration counterclockwise (looking ii'om above) is to yicld the positive current 1 emerging from St. With B = aB and dt = apd¢ on the closed contour, (1-63) yields

but B", is of constant magnitude on t}, obtaining exterior to the wire 1101 2np

p>a

If P < a, the closed line t2 shown in the inset of Figure 1-19(a) bounds a surface S2 intercepting only a fraction of 1, as determined by the ratio of the area of S2 to the

38

VECTOR ANALYSIS AND ELECTROMAGNETIC FIELDS IN F'REE SPACE

Ii

(z)

!it{W I

L..L _ _-=-"'-""_~P

o

a (b)

(a)

FIGURE 1-19, A long, straight wire carrying a static current of 1 A, and the associated magJ1("tic held. (a) Portion of long, straight currcnt-carrying wirc showing symmetric dosed paths for used with Ampere's law to find B. (b) External magnetic flux field of the long, straight wire. The graph depicts the flux .density variations with p.

cross-sectional area

2 7W .

Then (1-63) becomes

yielding

Thus the B field of the infinitely long, round wire carrying a static current 1 is ¢ directed, varying inversely with p outside the conductor aud directly with p inside it as follows.

Pol B=a - "'2np

p>a

B

p
(1-64)

EXAMPLE 1-16. Held of a flat current sheet (symmetry about a plane), Use Ampere's circuital law to find B on both sides of a thin, infinite current sheet in the x = 0 plane and carrying the constant, static surface current density J8 = aJsz A/m, The infinite sheet can be viewed as pairs of thin current filaments located aty, - y, canying the diHerential current di = Jsz4Y as in Figure 1-20(a), On recalling the exterior

I-ll MAXWELL'S TNTEGRAL RELATlONS FOR FREE SPACE

39

(z)

di

,-<,

= Jsz dy

i: 2 (b)

(a)

FIGURE 1-2()' An inlinite, thill current sheet carrying a comtant surface current of cknsity a,],z A/m. Showing [,ain·d current f,laments, and rcsultalll dB-held. (b) Symmetric closed path about which circuital law is taken.

J,

fleld result (1-64) of Example 1-15, one may conclude that the paired cnrrents produce a net,}-dirccted, difkrential fidd dB at any point on the x-axis as shown. The superposed d!(:ct orthe whole current sheet is therefor\" a]-directed B field on the positive x side of the sheet, and a negative }-directed field on the other side. Then Ampere's law (1-63) becomes, fiJl' the symmetric, rectangular path shown ill Figure 1-20(h)

with the surface intq:.;ral on the right side reducing to a linr integral over any}o width of the sheet. Because both Bv and Jsz arc constants over the indicated paths, this becomes '2By}o = to yield B~ = It oJ,)2. In vector form, therefore

B i~

B

JioJsz a -y

2

JioJsz a y -'2 -

x>O

x
(1-65)

EXAMPLE 1·17. Find the magnetic ficlds of the l()llowing coil configurations, each carrying a static current J: (a) an n turn, closely wound toroid of circular cross-section; (b) an infinitely long, closely wound solenoid having n turns in every length d. The coils are illnstrated in Figure 1-21. (a) Thc magnetic flux developed by I in the toroid is q'>-directed as in Figure 1-21 (a), a result t()llowing {i'om symmetry and the application of the right-hand rnle to the positive current sense shown. Thus, inside the toroid, B = a",B"" exact if the winding is idealized into a current sheet. The application of the time-static Ampere's circuital law to the symmetric closed line t of radius p shown therefore yields §t (a",8",) • alp dt = Jion1, in which B,p, from the symmetry, is constant around t, and nl is the nct current passing through S bounded by t. Thus

Jion1 B",= ._-2np

(1-66)

40


Closed path f

IndLl( abou I

/ /

1

(b)

(a)

FIGURE 1-21. Two coil configurations, the magnetic fields of which can be found using Ampere's circuital law. (a) Toroidal winding ofT! turns, showing symmetric path t. (b) Infinitely long solenoid, showing a typical rectangular closed path t.

Fl m,

(a e~

an inverse p-dependent field inside the region bounded by the current sheet. If the radius p of t in Figure 1-21 (a) were chosen to cause t to 'fall outside the torus (with p < a, for example), then S would no longer intereept any net current i. Then from the symmetry, the B field outside the idealized toroid must be zero.

(b) The infinitely long solenoid of Figure 1-21 (b) may be regarded as a toroid !if infinite radius; its magnetic field is thus also completely contained within the coil if the winding is idealized into an uninterrupted current sheet. The symmetry requires a z-directed field, B azB" independent of z. Ampere's circuital law (1-63) is applied to the rectangular elosed path shown in Figure 1-21 (b), two sides lying parallel to the z-axis. A nonzero contribution to the line integral is obtained only over the interior path parallel to the z-axis, whence

B z is constant over the path, whence n J1 a1d

( 1-67)

the ratio nld denoting the turns per meter length. B is thus constant everywhere inside the infinitely long solenoid.

c.

Faraday's Law

Maxwell's integral law (1-55)

!!..-

r B· ds =

dtJs

dl/l m dt

[1-55 ]

is attributable to tbe work of Faraday, and is called the induced electromotive force (emf) Law. The essence of this law of electro magnetics is expressed in the symbolism of Figure

a t

s


41

1/Jm enclosed by

e

by integration sense about f (b)

(a)

t such as that of (a) may be superposed anywhere on the example of (b), such that .Faraday's law must be trne for it. (a) Typical dosed line bounding a snrface S, relative to the Gelds in Faraday's law. (h) A symmetric example showing tbe E field indun:d by a time-varying magnetic field.

FlGURE 1-22. Indnced electric fields and Faraday's law. Any closed line

1-22(a). The relationship of the positive line-integration sense to the positive direction assumed for ds is the same as for Ampere's circuital law. Faraday's law (1-55) states that the time rate of decrease of the net magnetic flux t/I m passing any arbitrary surface S equals the integral of the E field around the dosed line bounding S. This is tantamount to saying that an E field is generated by a time-varying magnetic flux. The E field, in general, must also be time-varying if ( 1-55) is to be satisfied at every instant. Faraday's law tor strictly time-static fields is (I-55) with its right side reduced to zero

f.

E . dt

= 0 Faraday's law for time-static fields

(1-68)

which states that the line integral of a static E field about any closed path is always zero. A field obeying (1-68) is called a conservative field; all static electric fields are conservative. If the electric charges that produce an electric field are fixed in space, that electric field must obey Faraday's law in time-static form, (1-68). Several examples of the electric fields of charges at rest have been treated in Example 1-13. All static distributions of ekctric charges in space may be regarded as superpositions of point-charge concentrations = p" dv in the volume-elements dv in space. The electric field of a point charge Q., on the other hand, has been shown to be (1-57b)

[1-57b] It is easily shown that this electric field obeys Faraday'S law (1-68) for a time-static field. If any dosed path, such as t = ta + tb shown in Figure 1-23, is chosen in the

42


B sat Closed path f

Direction of integration

j

can 1-17 rise imp law tha1 iter qU
rap

Q (b)

(a)

FIGURE 1-23. Closed paths constructed about a point charge an.d a charge distribution, relative to Faraday's law for static charges. (a) Point charge Q.. (b) Charge distribution PV'

ElU

space about a point charge, the integral ofE-dt from any point P l to any other point P 2 along the path ta is

=f.''=', 2 ~dr 4nEor 2

=

4~J:l -

:J

( 1-69)

This resul t 9 is seen to be independent of the choice of the path connecting P land P 2; it is a function only of the radial distances r 1 and r2 to the respective endpoints PI and P 2 . Therefore, if the integration is taken around the complete path t = ta + tb shown in Figure 1-23, the two integrals from P l to P 2 via ta and thence from P 2 back to P 1 via t b will cancel, and (1-68) follows. Static charge distributions like those depicted in Figure 1-23 (b) are, in general, just collections of differential charge-elements dq Pvdv; whereas their static electric fields are just superpositions (vector sums) of the conservative differential electric fields dE produced by each of those static chargeelements. One may thereby agree that Faraday's law (1-68) for static electric fields is true in generaL Valid field solutions E(r, t) and B(r, t) satisfying Faraday's law (1-55) must also satisfy the remaining Maxwell's integral relations of (I-52) through (I-56); however, if the time variations of the fields arc not too fast, in some cases a static solution for 9The physical interpretation of the result (1-69) is of interest. I t that the net work done in moving a unit test charge around a dosed path is zero; such an electric has already been termed conservative. Thus (1-69) forms the basis of the theory oflhe scalar potential field of static electric charges to be discussed in Chapter 4.


43

, satisfying the static form of Ampere's circuital law (1-63)

J. ~ . dt = ft flo

r J' ds = i

(1-63]

Js

:an be assumed to be known field (e.g., the field solutions of Examples 1-15 through 1-172If the current densities J are slowly time-varying, one can assume they will give rise to a slowly time-varying B field. Such a static field on which time variations are imposed is called quasi-static. On inserting the quasi-static field B(r, t) into Faraday's law (1-55), a first-order approximation to the E field can then be obtained; assuming that the field symmetry permits the extraction of the solution for E from (1-55). An iterative process can sometimes then be employed to improve the accuracy of the quasi-static solution lO although if the time variations of the fields are not excessively rapid, the first-order solution will often suffice.

EXAMPlE 1·18. The long solenoid of Figure 1-21 (b) carries a suitably slowly time-varying current i = 10 sin wt. Determine from Ampere's law the quasi-static magnetic flux density developed inside the coil of radius a, and then use Faraday's law to find the induced electric intensity field both insidc and outside the coil. From Example 1-17(b), the magnetic flux density inside the long solenoid carrying a static current I was found to be (1-67). Thus, the solenoid current l~ sin wt will to a first-order approximation provide the quasi-static magnetic flux density (i-70)

in which Bo = !lon1old, the amplitude ofB. This assumption is reasonably accurate for an angular frequency (}) which is not too iarge. The electric field E induced by this timevarying B held is found by means of Faraday's law (i-55), the line integral of which is first taken around the symmetric path t of radius p inside the coil, as shown in Figure i-24. Faraday's law becomes

in which, from the circular symmetry, E,p must be a constant on

E,p but

f

dt = -wBo cos wi

t.

Thus

Is ds

tt til = 2np and S5 tis = np2, so that , wBoP E,p = - - - cos wi 2

p
(1-7i)

is the first-order solution for the electric fieid intensity generated by the time-varying magnetic flux of the solenoid. Observe that E,p varies in direct proportion to p, as shown in Figure 1-24. The negative sign in the result implies that as the net flux 1/1 .. through lOAn iterative process applied to the differential forms of Maxwell's equations (which are developed in Chapter 2) is described in R. M., Fano, L. J Chl!, and R. B. Adler. Electromagnetic Fields, Energy and Forces. New York: Wiley, 1960, Chapter 6.

44


~

)

,"'''1 ... ,

;{

I I

.

: ',Bz = Boslnwl I

'

t_".~.~"~.~.~

"

\

\

l

,

FIGURE [-24. Showing the assumed integration path t' used for finding the induced E field of a solenoid, and the resulting E field.

,,,' is increasing (in the positive z-direction), the sense of the induced E field is negative
UJt

p>a

( 1-72)

The electric field generated outside the long solenoid by the time-varying magnetic J1nx, thcrt'i()fe, varies inversely with respect to p. Both answers arc directly proportional to (j) because they are governed by the time rate ofehange of the net magnetic flux intercepted by the surface, as noted from (I-55).

D. Gauss's Law for MagnetiC Fields Maxwell's integral law (1-54)

~sB' ds = 0

[ 1-54]

is also known as Gauss's law for magnetic fields. It specifies that the net magnetic flux (positive or negative) emanating from any closed surface S in space is always zero. This statement is illustrated in Figure 1-25; in (a) of that figure is an arbitrary closed surface S constructed in the region containing a generalized magnetic flux configuration having a density B(r, t) in space. Maxwell's integral law requires that a total of 11S01TIctiIncs

referred to as Lenz's law.

1-12 COORDINATE TRANSFORMATIONS

45

Closed surface S

(a)

(b)

FIGURE 1-25. Gaussian (closed) surCrcc relative to magnetic fields. (a) Typical closed surface S constructed in a region containing a magnetic field. (b) A symmetric example: the straight, long current-carrying wire.

zero net magnetic lines emanate from every such closed surface S. This means that magnetic flux lines always form closed lines. Equivalently, it states that magnetic fields cannot terminate on magnetic charge sources for the reason that free magnetic charges do not exist physically. This is in contrast to the conclusion drawn from Gauss's law (I-53) f()f electric fields; the presence of a nonzero right side involving the electric charge density function Pi' in that relation attests to the physical existence of free electric charges. It is easy to find physical examples that illustrate the closed nature of magnetic flux lines. The magnetic field of a long, straight current-carrying wire of Figure 1-19 is shown once more in Figure 1-25(b); observe how the uninterrupted flux lines account for' precisely as many magnetic flux lines entering the typical closed surface S as are emerging from it. Such is the case for all closed surfaces that might be constructed in space for that field.

1·12 COORDINATE TRANSFORMATIONS Occasionally, one finds the need for transforming a vector or a vector field from one coordinate system, in which it is given, to some other coordinate system. An illustration of the implications of this can be {()Und in Figure 1-4(c), which shows, at the point P, the same vector A resolved into its components in the three common coordinate systems. In the first diagram of that figure, the vector field A at P is taken to be a function of the rectangular coordinates (x,y, z) such that A(x,y, z) = axAAx,y, z) + ayAy(x,y, z) + azAz(x,y, z). To transform the latter to the circular cylindrical system, for example, a representation in the form of A( p, cP, z) = apAp(p, 4>, z) + a1>A1>(p, 4>, z) + azAz(p, 4>, z) as depicted in the middle diagram is required. To accomplish such transformations, two steps are required: (l) the vector components Ax, A y, A z must be geometrically transformed to the components Ap , A1>' Az ; and (2) the scalar coordinates variables x,y, z at the position P, appearing in analytical expressions for

46

VECTOR ANALYSIS AND ELECTROMAGNETIC FIEL[)S IN FREE SPACE

must be transfc)rmed, from geometrical considerations, to the coorin terms of which A( p, 4>, z) is being expressed. The details of transforming a vector field A from the rectangular to the circular cylindrical coordinate system considered in the following. Step I consists of transforming the components Ax, A y, A z to AI" Aq" A z . Since AI" for example, can interpreted as simply the projection of A onto the unit vector aI" the dot a p 'A, meaning a p ' (axAx + ayAy + azA z )' is seen to yield the desired AI" It thus suflident to deduce, from the geometry if desired, the projections of the unit the circular cylindrical system onto the unit vectors of the rectangular system coordinates. Thus, in Figure 1-26(a) is shown superposed the rectangular and systems, with the unit vector a p displayed for the purpose of projections onto the ullit vectors ax, a y , and a z . From the geom(~try it that ( 1-73a) vector aq, at P on Figure a¢ onto ax, a y , a z as follows.

Similarly geometry, a¢'

With the unit

4>,

aq, • a y

would reveal, from the

= cos 4>,

(1-73b)

to both coordinate systems, obviously

(1-73c) Now with the

of any vector A being simply the projections of AI' = at> • A, Aq, = aq, • A and A z = a z • A, (1-74a) (1-74b)

(I-He)

(xl

(6) F[GUR~:

Geomelli", circular cylindrical cmmli",,'," 1-2(j,

a" a, of rectangular coordinate, (a) to a p of coordinates. II"

1-12 COORDINATE TRANSFORMATIONS

coorletails al coSince rector d the mjee)1' the d the d for From

-73a)

47

These are the desired relations that permit finding the cmnponents Ap> A
x

p cos

cP,

y

= p sin cP,

(1-75)

z=z

Thus the expressions (1-74) and (1-75) provide the ingredients for transforming the vector field A(x,y, '<:), given in rectangular coordinate form, to its corresponding circylindrical coordinate I(Jrm. Conversely, if A were given in circular cylindrical coordinate fc)rm and its transformation to rectangular coordinates were desired, the reverse of the foregoing proCE?dure would be needed. The results (1-74a,b,c), as linear algebraie equations, may solved simultaneously to yield

n the

(1-76a) -73b)

(1-76b) (1-76c)

-73e) ms of z' A,

ii'om ( l o r from Figure 1-26(a), the coordinates p, functions of x,)" z, become

·74b) -74e)

The expression fIX

and

z,

expressed as

z

Ii

-74a)

( 1-77)

cP in (1-77) also means equivalently that cos

cP

x

(1-78)

needed in (1-76a, b) to complete the transformation of A to the rectangular coordinate fonn. A compilation of the f(m~going transformations is f(lUnd in Table 1-1. A similar geometrical procedure can be used to transfc)rm some vector field A between the rectangular and spherical coordinate systems. I t is left. to you to prove, with the aid of the geometry suggested by Figure 1-26(h), that the transformations of components AI) A z , A} of the vector A, as well as its coordinate position-variables l u z , u3 ), from rectangular to spherical coordinates or vice versa, will yield the results in Table 1 t.

IXAMPLE 1·19. Transl()rm the given vector lield to circular cylindrical coordinates. (1) and evaluate F at the rectangular coordinate point P( 1, 1, I) in both coordinate systems.

~ TABLE 1·1 Coordinate Transformations

~

RECTANGULAR TO CIRCULAR CYLINDRICAL

rp + Ay sin rp -Ax sin rp + Ay cos rp

Ap = Ax cos A.p =

~

CIRCULAR CYLINDRICAL TO RECTANGULAR

(l-74a)

Ax

=

rp - A.p sin rp sin rp + A.p cos rp

Ap cos

1-74b)

(1-76a) (1-76b)

~

~

tJ>

= Az

(I-He)

A z ='A z in which

(1-76c)

~

tl

in which

t'l

l""'

x = p cos

rp,

y

p sin

rp,

z= z

(1-75)

p= cos

=

Ax sin () cos

Ae = Ax cos () cos

rp + Ay sin () sin rp + rp + Ay cos () sin rp -

sin

rp + Ay cos rp

A z sin

e

rp y = r sin e sin rp z = r cos e

x

sin

rp =-===

(1-78)

( 1-79c)

t'l

~

o a::

~ ~

>-3

rp + Ae cos e cos rp - Aq, sin rp Ar sin () sin rp + Ae cos () sin rp + Aq, cos rp

= Ar cos () - Ae sin

e

o (1-8Ia)

""

t:;; l""'

(1-8Ib)

(I-8le)

S;

Z

"" t'l ~

in which

t'l tJ>

;;

r = ';x 2 +y2 + Z2

in which

x

=

AT sin () cos

=

cos () (1-79a) (1-79b)

A.p

rp

(1-77)

SPHERICAL TO RECTANGULAR

RECTANGULAR TO SPHERICAL

AT

z=z

(")

t'l

r sin () cos

(1-80)

cos

e=

cos

rp

=

sin

x

sinrp

e=

=-===

(1-82)

(1-82)

1-13 UNITS AND DIMENSIONS

With Fx = 3z, l'~ = (1-74a,b,c) yields

F

and

F~ =

49

5x, the use of the component transformations

+ al'~ + aJ:'~ + F~ sin l + a ( - f',; sin + Fy cos + 4) sin ( -3z sin + 4) cos
= ai~ =

ap(Fx cos

obtains the desired result.

Inserting the coordinate transforma tions F(p, , z) = a p (3,: cos

(P + 4p sin 2

+ a( -3z sin q> + 4p sin cos

(2) In cartesian form, (l) yields at the point P( 1, I, 1) F(I, I, I)

3ax

+ 4ay + 5az

The circular cylindrical coordinates at this point, fi'om (

q) =

cos- 1 (1/~'2) = 45'" and

z=

I. These values inserted into

F(J2' 45°,1)

= 4.95a p

(3)

and (1-73), arc p = J2, obtains

+ 0.707a + 5az

As a check, observe that

F=

= 7.07

1·13 UNITS AND DIMENSIONS The mb system of units, introduced by Giorgi in 1901, is now employed almost universally in electromagnetics. In this system, length is expressed in meters, mass in kilograms, and time in seconds. A fi)urth unit, that of either eleetric charge (coulomb) or electric current (coulomb per second, or ampere), is needed in the dimensional description of electromagnetic phenomena. The rationalized mks system, which eliminates a Ll,ctor 4n from the Maxwell equations, has been almost universally adopted, and it is used in this text. The Giorgi mks system is especially noteworthy in that it deals with the primary electromagnetic quantities directly in the practical units in which they arc measured: in coulombs, amperes, volts, watts, and ohms. The choice of the dimension of the fourth unit (charge) adopted for the mks system is seen to depend on the values chosen for the constants Eo and Ito that appear in the Maxwell equations, (I and (1-56). Only one of these constants is arbitrary, though, in view ofthe relationship 125b) {or the speed oflight, developed in Chapter 2 for uniform plane waves in a vacuum c=

= 2.99792

X

10 8 ~ 3

X

10 8

( 1-83)

an experimentally determined value. In the mks system, the unit of eharge is the coulomb, defined by setting the constant 110 equal to 4n x 10- 7 • The value of the constant EO is then obtained from (I 1

Eo = - - 2

1l0 C

( 1-84a)

50


tABI

TABLE 1-2 Physical Quantities in the mks System

PHYSICAL QUANTITY

SYMBOL

UNIT

Length Mass Time Charge Current Frequency Force Energy Power Potential, emf Electric flux Capacitance Resistance Conductance Magnetic flux Magnetic flux density Inductance Free-space permeability Free-space permittivity Conductivity

t,

meter kilogram second coulomb ampere hertz newton joule watt volt coulomb farad ohm mho weber tesla henry henry/meter fhrad/meter mho/meter

d, ...

m

t, T q, Q i,l

f F U P «1>, V

t/le C R G

t/lm B L fJo Eo (J

which, if the approximation c ::;::: 3 good approximation

X

ABBREVIATION

J W V C

F (l U Wb T H Him F/m U/m

Clsec sec- 1 kg·m/sec 2 N'm J/sec W/A = N'm/C C/V = A· sec/V VIA A/V V'see Wb/m 2 = V· sec/m 2 V . sec/A Wb/A (l·sec/m U·scc/m

10 8 m/s for the speed oflight is made, yields the

10- 9 Eo::;::: --::;:::

36n

m kg sec C A Hz N

MIJ

IN TERMS OF BASIC UNITS

8.85

X

10- 12 F/m

(1-84b)

This value for Eo substituted into the Coulomb force expression (1-58) then provides the correct scale factor to obtain the force between the charges in newtons, the charges q and q' being given in coulombs and separated a distance r given in meters. One newton of force, that required to accelerate a I-kg mass at the rate one meter per second per second (1 m/sec 2 ), is thus the product of mass (kilogram) and acceleration (meter/second 2 ), making, 1 N = 1 km/sec 2 (= 10 5 dyn). The unit of energy or work is the newton-meter, orjoule (= 10 7 erg). In Table 1-2 are listed units of the mks system by name, unit, and symbol. The symbolisms largely agree with the recommendations of the International Organization f()r Standardization (ISO) .12 The numerical designation of field quantities is facilitated through the use of appropriate powers of ten. Thus 10 6 hertz = 10 6 Hz is written I MHz, in which the 12See IEklo' Spectrum, March 1971, p. 77 for a digest of the recommendations of the IEEE Standards Commillee adopted December 3, 1970.

51

PROBLEMS

tABLE 1·3 Symbols for Multiplying Factors MULTIPLYING FACTOR

PREFIX

SYMBOL

MULTIPLYING FACTOR

PREFIX

SYMBOl

10 12 109 106 10 3 10 2 10 10- 1

tera giga mega kilo hecto deka deci

T

10- 2 10- 3 10- 6 10- 9 10- 12 10- 15 10- 18

centi milli mlCro nano pico femto atto

c m

G M k h da d

}J.

n

P f a

prefix M (mega) denotes the 1.0 6 factor by which the unit is multiplied. Similarly, 3 X 10- 12 F is abbreviated 3 pF, with p (pica) denoting the factor 10- 12 . Other literal prefixes to be used in this way are listed in Table 1-3.

REFERENCES ABRAHAM, M., and R. BECKER. The Classical Theory of Electricity and Magnetism. Glasgow: Blackie, 1943. PLEMENT, P. R., and W. C. JOHNSON. Electrical Engineering Science. New York: McGraw-Hill, 1960. FANO, R. M., L. J. CHU, and R. B. ADLER. Electromagnetic Fields, Energy and Forces. New York: Wiley, 1960. HAYT, W. H. Engineerin,g Electromagnetin, 4th ed. New York: McGraw-Hili, 198!. LORRAIN, P., and D. R. CORSON. Electromagnetic Fields and Waves. San Francisco: W. H. Freeman,

1970 . .PuILLlPS, H. B. Vector Ana(ysis. New York: Wiley, 1944. REITZ, R., and F. J. MILFORD. Foundations of Electromagnetic Theory. Reading, Mass.: AddisonWesley, 1960.

PROBLEMS

SECTION 1-2 1-1.

Use a vector sketch in a plane to show graphically that A - B = - (B - A).

SECTION 1-4 1-2. Given are the vector constants: A = 5a x + 3ay + 4a., B = 2ay + az> C = -6a•. Sketch them at the origin in the rectangular coordinate system and evaluate the following. (a) A (b)

+B +C =D

IAI,IDI

(c) 2A - C =

[Answer: 5a x

+ 5ay -

a z)

[Answer: 7.07, 7.14]

E, lEI

[Answer: lOax

+ 6ay + 14a z ,

18.22]

52


1-3. A particular vector electric Held intensity at the point PI 3,6) in rectangular coordinates has the value E = 120ax + 200a y + 100az V 1m. (a) Carefully sketch a labeled vector diagram, as suggested by Figure 1-4((;), showing E and its components E',., E y , E z along with the unit vectors ax, a y , a z at Pl' What is the magnitude ofE? Assuming that E = aEE, express the unit vector a E in rectangular coordinate form, adding it to your diagram. If the same E were given to exist at another point, say P 2 (3, 0, I) in this region, explain why these E vectors are considered equal at these different points. (b) As an exercise in identifying coordinate surfaces, sketch and label, on a diagram as suggested by Figure 1-5 (b), the\rectangular (planar) coordinate surfaces x = 2,y = 3, and z = 6 that define the intersection Pl' Identify the components of E that arc perpendicular to the coordinate surfaces at Pl' [Answer: (a) E = 253.8 Vim, a E = 0.47a x + 0.79a y + 0.39a zl . 1-4. In the circular cylindrical coordinate system, a particular vector B = 30ap + 10a + 40a z is located at PI (5,30°,6), as shown in the figure. Also shown arc the coordinate surfaces p 5, ¢ 30°, z 6, the intersection ofwhieh defines Pl' (a) Find the magnitude ofB. Determine the expression for the unit vector a B directed along B in circular eylindrical coordinates. Label aB on your reproduction of this sketch. (b) Suppose the vector defined by the given B also exists at the point P 2 (5, 90°, 6). Sketch and label the additional coordinate surface needed to identify the P z location. Sketch B and its components at Pz' What is the expression for the unit vector aB at P z? Explain why these B vectors at the different points PI and P z arc in fact not equal vectors, despite the identical expressions for B and its unit vector in this coordinate system. Identify the components ofB that are perpendicular to the three coordinai:e surfaces at P 2 . Label the coordinates of the points P 3 and P 4 shown. [Answer: (a) B 51.0, aB = 0.588a" + 0.196a.p + 0.784a z ] 1-5. A particular vector field is given to be G(x,y, z) = x 2yza x + Cy - I lay - xzza z in some region of space. Identify the components Gx , Gy , Gz • Defining the line t in this region as the intersection of the coordinate surfaces ~y = 3, z = 2, sketch t in three-dimensional rectangular coordinate space. What is the vector field expression for G(x, 3, 2) applieable over this t? Evaluate the vector G along t at the specific points x = 2, 1,0, 1,2,3, as well as the magnitude of G at these points. Sketch curves showing the variation of Gx and of versus x over this range. Sketch the vector G at the points x = - 2, 0, 2 along t on your three-dimensional diagram.

IGI

(z)

B PI (5, 30°, 6) -+,.f-''W'~J-'--

-~ (xl

~;,v'

/ /

PROBLEM

J~4

4>=30° semi~ir1finite plane //

(y)

(xl

Pl

S 1 If

PROBLEMS

53

(z)

-_ I ---_yl~--.L.(x)

,Y2

-

I

I I

--J..... ..... -

(y)

PROBLEM 1-6

SECTION 1-6 1-6. Shown is the "distance vector," R, a vector directed from the point PdXl,Yl' zd to P Z (x 2 ,)'z, zz) in space, the position vectors of the latter being r 1 and rz. Observing graphically that R = r z - rj, write the expression for Rand IRI in rectangular coordinate {arm. Also write the expression in rectangular coordinates for the unit vector a R directed along R, making use Ahc definition, aR = Rj R.

\.!.:z}'

From the geometry, it is readily seen that the relation among the radial unit vector a r of spherical coordinates and unit vectors of the circular cylindrical coordinate system is a r = a p sin 0 + a z cos O. Use this relation to show that (oa,/iJ
SECTION 1-7 1-8. Sketch two vectors A and B in the same plane, showing from the definition (1-34) and the geometry that A . B means the magnitude A times the length of the "projection of B onto A," dcf1ncd by B cos O. Sketch the two vectors F = 30ax + 40ay and G = 20ay + 50a z at the point 1'(2, 1,3) in the rectangular coordinate system, as suggested by Figure 1-4(c). Use the ddinitionof the dot product to find the projection F cos 0 ofF onto G. Sketch this projection. Find the smaller angle between F and G. [Answer: F cos 1J = 14.8,0 = 72.8°] 1-9. Two constant vectors, C = 30ax + 50ay + 80a z and D = - 20ay + 40aZ' are located at the point P(3, 5, 4) in rectangular coordinates. (a) Sketch these vectors at 1', after the manner suggested by Fif!:ure 1-4. Find their magnitudes, and find the smaller angle between them (in their common plane) by use of (1) their dot product, and (2) their cross product. (b) Find the so-called "direction angles," !Xc, {3e, and ')Ie between the vector C and the unit vectors ax, a y , a z , respectively. Label these angles on your diagram. [Hint: Employ the concept of "projection" from Problem 1-8; e.g., note that the dot product of C with the unit vector ax is C times cos !Xc (the direction cosine).] [Answer: (a) 99.0,44.7,60.2° (b) 72.36°, 59.66°, 36.08°] 1-10. (a) Give the possible conditions that two vectors must satisfy if their dot product, A . B, is zero. (b) Hit is given that A' B A· C, show from the definition of scalar product that this does not necessarily mean that B C. What does it mean? Use a graphical construction to reinforce your remarks.

1-11. Sketch a triangle of arbitrary shape using the vectors A, B, C to denote the sides such that A + B = C, with the angle () between A and B. Prove the law of cosines by use of the dot product C· C = (A + B) . (A + B) for this triangle. [Answer: CZ A2 + B2 2AB cos OJ

54


1-12.

Given the vectors A, B, and C of Problem 1-2, find (a) A' B, B' C

[Answer: 10, -6]

(b) IO(A' C)

[Answer: -240]

(c) A x B =

F, IFI

(d) A' (A x B) (A x B) . C

[Answer: -Sax - Say

+

lOaz , 12.25]

[Why is the answer zero expected?] [Answer: -60]

(f) (A x B) x C

[Answer: 30ax

(g) A x (B x C)

[Answer: -48a y + 36azJ

-

30ay ]

1-13. Shown on the figure in spherical coordinates at the point P are the two vectors F = 40a o + 30a.p and G I SOar - 100ao + 250a.p. (a) What are the spherical coordinates (r, 8, 4» of the point P in the figure? (b) Find the vector magnitudes and their dot product F . G. (c) Find the projection G cos IX of G onto F, and determine the angle IX between the vectors in their common plane. (d) Find the expansion for the unit vector a F directed along F. [Answer: (b) 50, 308.22, 3500 (c) IX = 76.87"]

G-"14:

~uld

In the spherical system of Problem 1-13, find the value to which the 4> component of need to be adjusted so as to make F and G exactly perpendicular.

1-15.

(a) Sketch the unit vectors of the rectangular coordinate system at some common point as depicted in Figure 1-4(b) or ] . Applying the definition (1-38) of the cross produ.rt and the right-hand rule of Figure 1-9, show by inspection of the sketch why one expects that ax x a y = a z and that a y x ax = -a z . Silnilarly, show why a y x a z ax, a z x a y -ax, a z x ax a y and ax x a z = -avo Why is ax x ax = O? (Avoid employing the determinant (1-41) in your arguments.) (b) Use the approach suggested in part (a) to verify that, in eylindrical coordinates, a p x a.p = a z , a.p x a p = -a., a.p x a z = a p ' a z x a.p = -ap , a z X a p = a.p and a p X az -a.p.

1-16. A parallelepiped has edges given by ax, 2a p and a z • Sketch this "box." Show that one major diagonal can be denoted by A = ax + 2ay + a z . Label it on the sketch. Another major G= 1503r

-100ao+ 2503"

(z)

,./ /

/

/ / /

I /

I I

/ I

I I I

y-"" -"" ,,;///

'"

....

-

(xl

PROBLEM 1-13

-----(y)

PROBLEMS

55

diagonal is written B = -ax + 2ay + at. Label it also (noting that as a free vector, B can be translated parallel to itself without altering its magnitude and direction). (a) l'iud the lengths of these diagonals, as well as their dot and cross products. (b) Find the smaller angle between the diagonals, first making use of the dot product and then using the cross product. [Answer: (a) 6, 4, 2a y - 4a z (b) 48.19°] 17.. Relative to the pivot point given, find the vector moment (torque) associated with the owing vector force and distance (in meters). (a) F = 20a y N applied at the point 4 m up the zaxis, with the pivot point at the origin. (b) G = :'lOaz N applied at 1\ (-1,3,0) with the pivot point at the origin. (b) G = :'lOaz N applied at PI ( - I, 3,0) with the pivot point at P z( 1,0, I). Sketch the applicable vector diagrams, indicating from the right-hand rule the rotation associated with the moment M. [Answer: (a) -80a x (b) 150ax + lOOa, N ·m]

SECTION 1-8 1-18. A non conservative force field, F a x 12xy2 + ay 15yz + az 9z 2 N, is applied along the straight line t shown, the intersection of the planes y = 3x and z = -fy + 2. Find the work done by F in traversing e from PI(O, 0, 2) to P2 ( 1,3,0) (in meter's). \ Answer: 48 N . m] 1-19.

x2

(a) An elliptical path t J is defined in space as the intersection of the right circular cylinder 1) 2 = I and the tilted plane z = 1/2 shown. Find the value of the line integralofH . dt

+ Cy -

(2)

I \

I \

Plane y= 3x

(xl

PROBLEM 1-13

(z)

I I

1_---_ . . .

/r

x2+(y-l)2=1~"-l

I I I I Pj(O,O,O)

"-

/)

I -----.-" I I I I

I

Plane

I

(xl (y)

PROBLEM 1-19

56


,

(z) I

------

(y) (x)

PROBLEM \-20

between the points PI and Pi shown, if H = a y 3(1 x 2 ) + az~v2. (b) Find the line integral between the same points, but over the straight-line path t z defined by the intersection of the x = 0 plane and the tilted plane z = y/2. Is H a conservative field? Explain [Answer: 7.33, 11.33J

1-20. The semicircular path t shown is the inter-section of the right circular cylinder P 3 cos cp and the z = 2 plane. Find the line integral ofE . dt from P l to P 2 ifE = a p l50p cos cp + a.p200 sin cp + a z 100 cos cp. Determine the line integral over the straight-line path from PI to P 2 (intersection of the cp = 0 and z = 2 planes). [Answer: -150, -675]

SECTION 1-9 1-21. Using standard scalar volume-integration methods, make use of the triple integral of p"dv given by (1-47) to find the charge q inside the following volume regions. (Sketch each configuration with dimensions, labeling a volume element at the typical point P inside .. appropriate to the coordinate system used.) (a) The charged volume region is a cube with sides located at x =.0, x = 0.1 m,] = 0,] = 0.1 m, Z = 0, Z = 0.1 m, with the nonuniform charge density inside given by PI' = 20xyz C/m 3 . (b) The volume region is a right circular cylinder bounded by the surfaces p = 0.1 m, z = 0, and Z = 0.1 m, with PI' = 20pz C/m 3 inside. (c) The volume region is a sphere ofO.I-m radius, containing the charge density Pv = 20r cos 2 0 C/m 3 . l Answer: 2.5 pC, 209 pC, 2.09 mC] . 1-22. Employing standard scalar surface-integration methods, make use of the double integral of Ps ds to find the total charge on the following surfaces. (Sketch the dimensional layout, labeling a scalar surface element at the typical point P(Ul, U2, u3) on S, appropriate to the coordinate system required.) (a) The charged surface is square, centered at the origin, bounded by the sides at x = ± 0.1 m,] = ± G.I m, and assumed covered with the nonuniform surface charge density Ps = lOx 2]2 C/m2. (b) The surface is a right circular cylinder (no endcaps) of radius p = a = 0.1 m, extending z = ± 0.1 m and possessing the surface charge density Ps = IOz 2 C/m 2 . (c) The surface is a hemisphere ofr = a = O.I-m radius, extending from 0 = 0 to n/2, with the nonuniform surface charge pf density Ps = 10 cos 2 0 C/m 2 thereon. [Answer: 4.44 pC, 4.19 mC, 209 mC] 1-23. Given is the E-field solution (1-5 7b) for the point charge Qlocated at the origin. Imagine the spherical surface of radius r = a to surround Q. (a) Use the definition (1-48) to evaluate the flux of the vector EoE passing through the surface of the spherical cap bounded by 0 = 8 1 as shown. (Add to the sketch the details of the vector surface element ds suggested by Figure 1-7(b).) (b) Use the result of (a) to find the flux of EoE through the cap S, if8 1 = 30°, 60°, 90°, 120°, 150 0 and 180°. Comment on the (}1 = 1800 case relative to the Maxwell/Gauss law (I-53). [Answer: (a) Q(l cos (1)/2J

PROBLEMS

57

(2)

PROBLE:\!! 1-2:'

(z)

I

1-24. With the same point charge Qat the origin as in Problem 1-23, use the definition (1-48) to evaluate the flux of ~he vector (EoE) emerging from the bandlike surface S of the sphere, 60° and O2 = 120°, what percent of the total flux of extending ii'om 0 1 to O2 as shown, IfO I EoE is passing through the band? [Answer: Q(cos 8 1 - cos fJ2 )/2, 50%]

SECTION 1-10 1-25. The electronic charge q = -e is shot with the initial velocity v

5 = a x lO mlsec into an evacuated region containing the uniform magnetic field B = a)O-4- Wb/m 2 Sketch these vectors (paper in the x:y plane), along with the force F acting on q, Give arguments as to why the electron should take a circular path, Add this dimensioned circle to your sketch, Find what electric field E wiH just overcome the force effect of the magnetic field, to cause the electron to travel in a stra;ght line along the x-axis.

SECTION l-llA 1-26. Show a labeled sketch and give the ddails of the proof of (1-60), the expression for the electricallield inside the static, unif()rmly charged spherical cloud (r < ro). 1-27. Suppose that inside a spherical cloud of static charge is contained the linearly varying charge density, Pv po(rlro) e/m 3 , Po denoting the charge density at the surface r ro of the

58


sphere. (a) Use the volume integral (1-47) to determine the total charge within this sphere. (b) With the charge density seen to be symmetric about the center of the cloud, employ Gauss's law (I-53) to determine in detail theE field both inside the sphere (r < ro) and outside it (r > ro). (Sketch a diagram patterned after Figure 1-15(b), showing the labeled Gaussian sur/aces employed.) Show from your solution that the exterior E field (r> ro) is identical with that expected if the total charge in the sphere were concentrated entirely at the origin. (c) If Po = 10- 3 C/m 3 and ro = 10 em, find q in the sphere and sketch E, versus r. [Answer: (a) q npo?o (b) E, = por2/4Eoro for r < ro, E, = por~/4Eor2 for r> TO]

1-28. A spherical shell of charge possesses the constant volume charge density Pv between its inner and outer radii a and b. Use Gauss's law to prove that the E field for r < a is zero; that for a < r < b, Er is pv(r3 - a3)/3Eor2; whereas outside the shell (r > b) it is Pv(b 3 - a3)/3Eor2. (Show an appropriately labeled sketch along with the details of your proof.) 1-29. Let the volume charge density within a spherical region of radius r a be given by Pv = Po(l + kr), in which Po denotes the density at the origin. Determine the k that will make the total charge in the sphere zero. For this k, why is the E field external to the sphere zero? Find E. as a function of r within the sphere, making use of Gauss's law. [Answer: k = -4/3a] 1-30. An infinitely long, cylindrical clond of radius p a in free space contains the static, uniform volume charge density p". With a suitably labeled sketch, make use of the symmetry and Gauss's law (1-53) to obtain the following. (a) The electric field outside the cloud (p> a). (b) The interior electric field (p < a). (c) Show that the exterior E field is the'same as that expected if the same total charge per length t were concentrated as a line charge along the z axis, as in Figure 1-15(c). [Answer: E = appva2/2Eop (b) appvp/2Eo] 1-31. Let an infinitely long, cylindrical charged cloud of radius a contain the static charge density Pv = po(p/a)2, varying parabolically to the density Po at the cloud surface. (a) Make use of (1-17) to determine the total charge q in any length t of this cloud. (b) Sketch a diagram as suggested by Figure 1-15(c), making use of the symmetry and Gauss's law (I-53) to find the E field outside the cloud (p > a), and then inside it (p < a). Label the Gaussian surfaces used. Show from your solution that the field oLltside the cloud is the same as that expected if the total clurge were concentrated along the z-axis. [Answer: (a) p onta 2 /2 (h) poa 2/4E op for p > a, pop 3 /4E oa2 for p < a]

1-32.

Two parallel, planar charges of the kind shown in Figure l-15(d) are located at x = d and - d, possessing the nniform, opposite surface charge densities - Ps and p" respectively. (a) Use the vector superposition (summing) of the fields of these two planar charges, as given by (1-62), to prove that the total E field between the planes ( - d < x < d) is Ps!E V /m, whereas that outside the planes > d) is zero. (Do not usc Gauss's law.) (b) Repeat (a) if both surface charge densities are positive.

(Ixl

SECTION I-llB 1-33. A hollow, circular cylindrical conductor in free space, assumed infinitely long to avoid end efiects, and having the inner and outer radii band c, respectively, carries the direct current 1. (a) Assuming a constant, z-directed current density in the conductor cross sectiou, show that the vector current density at any point therein is J = a z 1/n(c2 - b2 ). (b) Usc Ampere's law to show that the exterior magnetic field is the same as that of the solid conductor of Figure 1-19 carrying the same total current f. Show that B inside the hollow interior (p < b) is zero, whereas that within the conductor (b < p < c) is a",Jlo1(p2 - b2 )/2np(c 2 - b2). (c) Sketch a graph showing how B", varies with p. 1-34. A coaxial pair of circular cylindrical conductors, infinitely long in frec space, have the dimensions shown and carry the equal and opposite total currents f. (a) Show tha.t the current density in the inner conductor is a//na 2 , whereas in the outer conductor it is the negative of that iCJUnd in Problem 1-33 (a). (b) Show that the B fields within the inner conductor (p < a) and betweell the C'Onductors (a < p < b) are identical to those of the isolated wire of Example 1-15. Use Ampere's law, together with an appropriately labeled diagram showing the closed

PROBLEMS

59

s sphere. y Gauss's (r> ro). sur/aces '\lith that n. (c) If ;wee (a)

twecn its ; that for 2. (Show l,riven by fill make ~rc zero?

-4/3a] lC static, fmmetry (p> a). ~ as that ng the <

: chalge 1) Make diagram find the :es used. ~d if the )r p > a, at x = d ectively. as given whereas if both

PROBLEM \-34

assumed, to prove in detail that the B field within the outer conductor (b < p < c) is IIIP/loJ(C 2 p2)/2np(c 2 b2 ) and that it is zero for p > c. (e) Sketch a graph of BIP versus p over the (0, c) range, assuming a = 3 mm, b = 6 mm, c = 3 mm, 1 = 100 A. Find the current in each conductor, expressed in A/Cln 2 • .

1..35. Show that the static B fields of the coaxiallinc of Problem 1-34 arc the superposition of the fields of the hollow conductor of Problem 1-33' and those of the isolated conductor of Example 1-15.

1..36.

Two parallel, indefinitely thin eurrent sheets ofinfinite extent in free space are located at = +d, possessing the unii()rm but oppositely directed surface-current densities ±Jsv respectively. (The currents are assumed charge-compensated, making electric fields absent in this problem.) (a) Employ resulls ofExamplc 1-16 and superposition (not Amp(~rc's law) to show that B between the sheets ( -d d) is zero. (b) Sketch a flux plot of the net By field. Show that ifJ,z = 100 Aim, then By = 1.257 mWb/m 2 • Comment on the changes in the fields if both current densities wcre assumed in the same ( direction.

1..37.

,An infinite, planc conducting slab of thickness d in free space has its sides coincident with the x = - d/2 and d/2 plancs. Assume the constant volume current dcnsity J = azJz A/m 2 within In the manner of Example 1-16, use Ampi~rc's law to show that the B field thc conductor. outside the conducting slab > d/2) is ±ay1/loJzd. (h) Make use of Ampere's law to find B inside the slab. [Hint: Choose a rectangular dosed path t with one side parallel to the known field of (a), and its other side aligned with the unknown field.]

(Ixl

o avoid current ow that , law to Irc 1-19 Nhereas I graph ave the current : of that a) and "ample closed

1.38.

Two parallel, round conductors, infinitely long and carrying the currents 1, 1, are 2d m apart. Assume them parallel to the <-axis and centered about the origin on the x-axis. Sketch a top view of the conductors in the x:y plane, with the current in conductor 1 at x = d assumed + z-directed. Show its vector field contribution Bl at the normal distance P1from 1 to the typical location P(x,]), making use of (1-64). Showing Bl decomposed into its Bxl and Byl components, use the geometry to develop the expression lor Bl solely in terms ofx and]. (b) Doing the same for conductor 2, 0xpress the total B at P, due to both conductors, entirely in rectangular coordinate form. (c) Il'l = 10 A and 2d = 5 em, find B at the origin. Find also the vector B at the following (x,.y) locations expressed in centimeters: (1.25,0), (3.75,0), (0,1.25), (1.25,1.25), 1.25), (3.75,1.25), (0,2.5), . [Answer (b):

60


PROBLE,\1 1-39

in which i, 'i

1-39. Find, by superposition, the magnetic fields of a pair of coaxial, ideally closely wound toroids of circular cross section as shown, assuming the same number of turns and the identical currents /, Assume the currents first in the same direction; then, in opposing directions.

SECTION 1-12 1-40. Introducing the unit vectoraql at Pon Figure 1-26(a), from the geometry verify (1-73b). Similarly verify (1-73c). 1-41. (a) From the geometry of Figure 1-26(b), verify that the projections of the unit vector a r onto ax, ay' and a z yield aT' ax = sin 0 cos cP, aT' a y = sin 0 sin cP, aT' a z = cos O. (b) Modify Figure 1-26(b) to enable deducing the following projections: ae' ax = cos () cos cP, ae' a, = cos 8 sin cP, ao' a z = -sin 8. Show similarly from the geometry that a",' ax = -sin cP, a",' a,y = cos cP, a", . a z = O. (c) Expressing A in rectangular coordinate form, A = axAx + ayAy + a~z' use the fOl'egoing results and methods discussed in Section 1-12 to deduce the expressions for the spherical coordinate components of A in terms of its rectangular components, that is,

e cos IjJ + Ay sin fJ sin IjJ + A Ax cos e cos 1> + Ay cos () sin IjJ - A z sin e

Ar = a r ' A = Ax sin Ae =

AqI = - Ax sin IjJ

z

cos 0

+ Ay cos IjJ

[I -79a,b,c J

1-42. (a) A sphere of radius a and centered at the origin is expressed in rectangular coordinates by x 2 + y2 + Z2 = a 2 . Use the appropri
1-43. Transform the following veetor fields to the circular cylindrical coordinate system. aplO cos IjJ c- a",10 (a) A = lOax , (b) B = IQyax , (c) D = 3(1 - x 2 )a y + a z1Y2. [Answer: A sin 1jJ, B apIOp sin IjJ cos IjJ - aqllOp sin 2 1jJ, D = a,,3(1 - p2 cos 2 1jJ) sin IjJ + a",3(1 2 2 p2 cos 1jJ) cos IjJ + a z 4p2 5in 1jJ] 1-44. Transform the given vector fields to the spherical coordinate system. (a) A = lOa x , (b) = a},lOOx. [Answer: A = arlO sin cos IjJ + aolO cos eos IjJ aqllO sin 1jJ, E = a,IOOr sin 2 () sin IjJ cos IjJ + ao 1001' sin cos sin ¢ eos ¢ + aql 100r sin cos 2 1jJ]

E

e

e

e

e

e

. ."----------------------------------------CHAPTER2

Vector Differential Relations and Maxwell's Differential Relations in Free Space

lind

leal

.he ~n-

cJ es 1C )IJ

n.

o

In this chapter is considered the development, in generalized orthogonal coordinates, of the gradient, divergence, and curl operators of vector analysis, with forms in the common coordinate systems taken up in detail. The divergence theorem and the theorem of Stokes are used to derive the differential forms of Maxwell's divergence and curl equations in free space fi'om their integral versions postulated in Chapter L The appropriate manipulations of Maxwell's time-varying differential equations are seen to lead to the wavc equations in terms of the Band E fields, and the wavelike nature their solutions is exemplified by considering in detail the field solutions of uniform waves in free space. A pursuit of these ideas requires some background in the differentiation of vector fields, to be discussed in the following section.

2·1 DIFFERENTIATION OF VECTOR FIELDS In many physical problems involving vector fields, a knowledge of their rates of change with respect to space, time, or perhaps some parameter is often of interest. This notion has already been introduced in Section 1-6 in connection with the position vector r. I t is now considered in general for any differentiable vector field. IfF(u) is a vector function ofa single scalar variable u, iL<; ordinary vector derivatiye with respect to u is defined by the limit dF

l)

du

. L1F . F(u lim - = lIm

IlU'~O

L1u

Ilu->O

+ L1u) L1u

F(u)

(2-1)

61

62

VECTOR DIFFERENTIAL REI"ATIONS AND MAXWELL'S DIFFERENTIAL RELATIONS

}'IGURE 2-1. A vector fimctiol1 F in space, and its variation with respect to some variable u.

l~F

provided that the limit exists (i.e., the limit is single-valued and finite). As in tl stance of the derivatives of the position vector r considered earlier, the vector i ment L\F is not necessarily aligned with the vector F, implying that the direction t vector F may change with the variable u. This circumstance is exemplified in I" 2-1, in which the conventional triangle construction is used to define AF, the diffe between F(u + Au) and F(u). The derivative (IFjdu defines a function, the deriv of which in turn defines a second-order derivative fimction dZF /du 2 , and so on. The derivatives of the sum or product comhinations of scalar and vector I tions are often of interest. For example, iff and F are respectively scalar and VI functions of the variable u, the derivative of their product is, from (2-1) "\

'

dUF)

.

U

+ AI) (F + L\F)

-.IF

- - - = hm - - - - - . - . - - - - - - -

du

Au

au"" 0

= f . dF + F .

du

df du

Note that this result resembles in form a similar rule of the scalar calculus (in wi both fu nction8 are scalars). IfF is a function of more than one variable, say OfUl' U2, u3 , t, its partial del' tive with respect to one of the variables (U1) is defined lim F(u l aUI-O

+ L\u 1 , 112, u3,

t)

F(Ul>

UZ, u 3 ,

t)

(~

AUI

with similar expressions for the partial derivatives with respect to the remain variables. Successive partial differentiations yield functions such as jJ2FIe 2 F/oul eJu z . If F has continuous partial derivatives of at least the second order, i permissible to differentiate it in either order; thus

a

(2·

The partial derivative of the sum or product combinations of scalar 'and vect functions sometimes is useful. In particular, one can use (2-3) to prove tbat t following expansions are valid

)Ns

2-2 GRADIENT OF A SCALAR FUNCTION

63

of + F at at (2-6) o(F x G)

-~-=Fx

at

DG

DF

+~xG

(2-7)

at

if f is any scalar function and F and G are vector functions of several variables, among which t denotes a typical variable.

in the intor increion of the in Figure liffercncc erivative on. tor funcld vector

(2-2) n which

2·2 GRADIENT OF A SCALAR FUNCTION The space rate ofehange ofa sealar fieldf(ull Uz, U3, t) is frequently of physical interest. For example, in the scalar temperature field T(ull Uz, U3, t) depicted in Figure I-I(a), one can surmise from graphical considerations that the maximum space rates of temperature change OCCllr in di rections normal to the constan t temperatu re surfaces shown. Generally, the maximum space rate of change of a scalar function, induding the vector direction in which the rate of change takes place, can be characterized by means of a vector di!li:?rential operator known as the gradient of that scalar function. It is developed here. If, al allY fixed time t, a single-valued, well-behaved scalar field f(u l , U2, U3, t) is set equal to any cons/ant fo so that f(UlJ uz, U3, t) = Jo, a surface in space is described, as depicted by .)1 ill Figurc 2-2. A physical example of such a surface is any of the constant temperature surfaces of Figure l-l(a). Another SllrfaCe, 8 2 , an infinitesimal distance from 81> is described by letting f(u 1 + dUI' Uz + dU2, U3 + dU3) = Jo + df, in which dl is taken to mean a very small, constant, scalar amount. Suppose that two nearby points, P and P', are located a vector distance dt apart on these two surfaces

deriva-

(2-3)

82 (defined

by f =

to + df) 82 (f= (o + dt)

laining 2F/ouI, ~r, it is

(2-4) -1 I

vector at the

(a)

(b)

FIGURE 2-2. Two nearby surfaccs 1=10 and 1 .f~ + dl" rdatiVt" [0 a discussion or grad (a) Points P and P' separated by dt and on snr/aces defincd by 1 =.f~ and 1 = 10 + til Points P and P' on the same slIJ'face 1 = 10' to show that grad 1 and dt are perpendicular.

64

VECTOR DIFFERENTIAL RELATIONS AND MAXWELL'S DIFFERENTIAL RELATIONS

as in Figure 2-2(a), recalling from (1-21) that one may express dt = aldt as (2-8)

df is

the amount by which f changes in going from P to P' from the first surface to the second, written as the total differential

(2-9)

curvil

from in S(

The presence of the components of dt in (2-9) permits expressing df as the dot product

in tl Calling the bracketed quantity the gradient of the function f, or simply grad f, as follows

and

or

(2-10)

tha one may write the total differential df of (2-9) in the abbreviated form

df = (grad I) . dt

(2-11 )

Two properties of grad I are deducible from (2-11); 1. That the vector function grad I defined by (2-10) is a vector perpendicular to any I = Io surface is appreciated if the points P and P', separated by a distance dt, are placed on the same surface as in Figure 2-2(b). Then the amount by which I changes in going from P to P' is zero, but from (2-11), (gradf) . dt 0, implying that grad I and dt are perpendicular vee tors. Grad I is therefore a vector everywhere perpendicular to any surface on which I constant. 2. If a displacement dt from the point P is assigned a constant magnitude and a variable direction, then from (2-11) and the definition (1-34) of the dot product it is seen that dI = Igrad dt cos 0, 0 denoting the direction between the grad f and dt. The magnitude of gradf is therefore df/(dt cos 8), but from Figure 2-2(a), dt cos 0 = dn, the shortest (perpendicular) distance from the point P on the surface SI to the adjacent surface S2 on which I Io + dj, whence

hoI Co

Po

Fr be

II

df IgradII = dn

OJ

p: 0'

(2-12)

d (~

2-2 GRADIENT OF A SCALAR FUNCTION

65

The vector grad j therefore denotes both the mag"nitude and direction of the maximal space rate of change of j, at any point in a region. Note that the magnitude ofgradj can also be expressed in terms of its orthogonal lfvilinear components, given in the definition (2-10) by \gradji=

/ OJ)2 - + (or)2 -"- + (OJ - -)2J1 2 [(hi OUI h2 AU h3 OU3

(2-13)

2

The expressions for grad j in a specific orthogonal coordi nate system are obtained rom (2-10) on substituting into it the appropriate symbols {CH" U; and hi as discussed n Section 1-5. Thus, in the rectangular system

grad j = ax III

the circular

oj

oj

ox + a oy + a z - -

(2-14a)

y

system (2-14b)

and in the .>jJherical coordinate system

_ grad)

oj

loj

1

oj

= a,-or + a o -r oe + a.p -'-e r 5111 (3'"' 'V

(2-14c)

An integral property of grail j, of considerable importance in field theory, is that its line integral over any dosed path t in space is zero. Symbolically

~ (gradj)

. dt = 0

(2-15)

holding fell' all well-behaved scalar functions j, and proved in the riJlIowing manner. Consider (2-15) integrated over an open path between the distinct endpoints po(u7, u~) and P(u 1 , 112'

ug,

C'P Jpo

(grad j) . dt

(2-16)

From (2-11) it is seen that (grad j) . dt denotes the lotal differential df, so that (2-16) becomes

fP

Jpo

(gradj) . dt =

fP df = IJP

Jpo

Po

(2-17)

or the difference of the values of the function I at the endpoints P and Po. Thus, any path connecting Po and P will provide the same result, (2-17). Carrying out (2-16) over some path A from Po to the point P and then back to Po once more over a different path B, the contributions of the two integrals would cancel exactly, making (2-15) the result. The integral property (2-15) of any vector field grad f is sometimes

66


called the conservative property of that field, from the applications of integrals of that type to problems involving certain kinds of energy, Any field gradf is a conservative field,

2·3 THE OPERATOR V (Del) Recall that the gradient of a scalar field (2-14a)

f

is expressed in rectangular coordinates by

[2-14a] The presence of the common function f in each term permits separating from this expression a vector partial differential operator represented by the symbol V (pronounced del) as follows (2-18) to permit writing gradf in an alternative symbolism, Vf

8f 8f of gradf=Vf=ax-+a - + a z -

ax

Yay

(2-19)

oz

The notations grad f and Vf will henceforth be considered interchangeable, It may be noted that the operator V defined by (2-18) in the rectangular coordinate system can be defined in other coordinate systems as well, including the generalized orthogonal curvilinear system. This is not done here because of its lengthy form and because it serves no particular need in connection with the objectives of this text. You may wish to consult other sources relative to extending (2-18) to other coordinate systems.}

EXAMPLE 2·1. Suppose a scalar, time-independent temperature field in some region of a space is given by

T(x,y) = 200x

+

100y deg

with x andy expressed in meters. Sketch a few isotherms (constant temperature surfaces) of this static thermal field and determine the gradient of T. The isotherms arc obtained by setting T equal to specific constant temperature values. Thn5, letting T = 100° yields 100 = 200x + 1O(!y, the equation of the tilted plane y = 2x + I. Th.is and other isothermie surfaces are shown in the accompanying figure. The temperature gradient of T(x,y) is given by (2-14a)

vT

== grad T

aT + a -aT + a -aT- = ox y oy z a.::;

= ax -

200ax

+

I OOay0 1m

example, sec M.,Javid, and P. M. Brown, Field AnalYsis and Elfictromagnetics. New York: McGraw-Hill, 1963, p.477.

1 For

:I... TI

:r

Ih

-

i

, II!

2-4 DIVERGENCE OF A VECTOR FUNCTION

\

,

67

(y)

(b)

(a)

EXAMPLE 2-1. (a) Graph of T

constant. (b) Side view of (a).

a vector everywhere perpendicular to the isotherms, as noted in (b) of the figure. The x andy components of the temperature gradient denote space rate of change of temperature along these coordinate axes. From (2-13), the magnitude is

denoting the maximal space rate of change of temperature at any pomt. One may observe that heat will flow in the direction of maximal temperature decrease; that is, along lines perpendicular to the isotherms and thus in a dir'cction opposite to that of the vector grad T at any poin t.

1-4 DNERGENCE OF A VECTOR FUNCTION

The flux representation of vector fields was described in Section 1-9. If a vector field r is representable by a continuous system of unbroken flux lines in a volume region as for example, in Figure 2-3(a), the region is said to be sourcefree; or equivalently, field F is said to be divergenceless. (The divergence ofF is zero.) On the other hand,

,

111l1li;;;;

fjl!"! /%

"~ 1/

l;r '- " (a)

Ijffk/"/.,

\-y / / / (b)

(e)

FIGURE 2-3. Concerning the divergence of flux fields. (a) A vector field F in a source-free As many flux lincs enter S as leave it. (b) A vector Geld F in a region containing sources posse,;sirlg net outgoing flux). (c) The meaning of div F: net outward flux per unit volume as -0.

68


if the flux plot of F consists of flux lines that are broken or discontinuous, as depicted in Figure 2-3(b), the region contains sources of the field flux; the field F is then said to have a nonzero divergence in that region. The characterization of the divergence of a vector field on a mathematical basis is described here. The divergence of a vectorfield F, abbreviated div F, is defined as the limit of the net outward flux ofF, F • ds, per unit volume, as the volume !lv enclosed by the surface S tends toward zero. Symbolically,

fs

div F

==

F· ds

lim

!lv

fl ux lines/m

3

(2-20)

Av-+ 0

Thus, as the closed surface S is made very small, as depicted in Figure 2-3(c), the limiting, net outward flux pCI' unit volume in the neighborhood of the point P defines the divergence of the vector field F there. The shape of S is immaterial in this limit, as long as the dimensions of !lv tend toward zero together. The definition (2-20) leads to partial diHerential expressions for div F in the various coordinate systems. For example, in generalized orthogonal coordinates, div F is shown to become

FIC the (:on

oPI

ter (2-21 )

The derivation of the differential expression (2-21) for div F in generalized orthogonal coordinates proceeds from the definition (2-20). Express the function F in terms of its generalized components as follows F(Ul' U2, U3,

I) = a1Fdul,

U2, U3,

t)

+ a 2 F 2 (u l , li2,

li 3 ,

t)

+ a 3 F 3 (u l ,

li2, U3,

I) (2-22)

The definition (2-20) requires that the net effiux ofF be found over the closed surface S bounding any limiting volume !lv, which from (1 II) or (1-13) is expressed

It in di 01

(2-23) The net, outward Hux ofF is that emanating from the six sides of !lv, designated by Llsi> !lS'I' and so forth, in Figure 2-4(a). The contribution !It/ll entering element !lSl is just F· Lls i = (alF I )' !lSI' or

T tJ

(2-24) (2-25 ) the negative sign being the consequence ofassuming a positively direeted Fl component the outward !lSI = - al !lt2 Llt3; that is, the flux Llt/ll enters !lSI' In the limit, as the separation Lltl between !lSl and Lls'! becomes sufficiently small, the flux Llt/l'l leaving !ls'! in Figure 2-4(b) differs from !It/ll entering !lSI by an amount given by the second

(


69

FIGURE 2-4. A volume-clement L'iu in the generalized orthogonal coordinate system nsed in the development of the partial diflcrcntial expression for div F. (a) A volume-element !lv and I:omponents ofF in the neighborhood of 1'(u 1 , 112' U3)' (b) Flux contribntions entering and leaving opposite surfaces of !lv. The remaining four sides are similarly treated.

term of the Taylor's expansion of L\I/;~ about the point P; that is, A ./,

lJ.'f'1

=

+ 0(L\1/; tl '" UUl

F'1L\t2L\t3

L\ Ul

+ [~(F'lM2L\t3)]L\Ul OUI

Fl L\t2 L\t 3 +

[a~l (Flh2 h3) ] L\Ul L\uz L\U3

(2-26)

It is permissible to remove L\U2 and L\u3 from the quantity affected by the O/OUI operator in the foregoing because each is independent of Ul, in view of the orthogonal coordinate system being used. The net outgoing flux emerging from the sides As[ and L\S'l of Figure 2-4(b) is thus the sum of (2-24) and (2-26) (2-27a) The two remaining pairs of surface elements tribute outgoing flux in the amounts

L\s~, As~,

and L\S3',

As~

similarly con-

(2-27b)

(2-27c) seen to be obtainable from the symmetry and the cyclic permutation of the subscripts of (2-27a). Finally, putting (2-27a, b, c) into the numerator of (2-20) obtains the result

70

VECTOR DIFFERENTIAL RELATIONS AND MAXWELL'S DIFFERENT1AL RELATIONS

anticipated in (2-21)

whence

I n rectangular coordinates, div F is found from (2-28) by setting hI and

U1

=

X, U 2

= y,

U3

=

hz

h3

=

1

Z

.

oFx

oFy

i3F

z ox + -oy + --

dlV F = - -

Rectangular

(2-29a)

whereas in the circular cylindrical and the spherical coordinate systems, the expressions become

. (hv F

div F

I

=-

a

P i3p

, + 1 -of,,, + -p i3

(pF )


(2-2% )

Spherical

(2-29c)

P

13 2 (r p".) Or

1

+. r 8m 8

a (Flj sin 8)

138

1

+ r SIn . 8

"",

U
The form (2-29a) ofdiv F in the rectangular system is the basis for another notation using the del operator (V) defined by (2-18). On taking the dot product of V with F in the rectangular system of coordinates, one finds

(2-30)

or precisely (2-29a). This is the basis for the equivalent symbolisms divF

== V' F

(2-31 )

The notations div F and V' F will be considered interchangeable regardless of whiCh coordinate system is used, even though the symbol V has for our purposes been defined only in the rectangular system.


(6)

(aJ

71

( c)

L=

(d)

ap

K

p

(eJ

EXAMPLE 2-2

IXAMPLE 2·2. Sketch nux plots [or each o[the following vector fields, and find the divergence of each: (a) F = axh', G axKy, H axKx; (b) J = apK, L ap(Kjp). (II) Applying (2-29a) to the fimctions F, G, and H in the rectangular system obtains

div F =

ox

=

0

divG

ox

=0

divH

ox

K

Their llnx plots an: shown in (a) through (c). Tnspection reveals a zero value or divergence is obtained for the fields F and G; a tcst closed surface placed anywhere in the region will have zero net flux emanating from it. The nonzero div H, on the other hand, is evident from its flux plot because of the discontinuous flux lines, here required to possess an increasing density with x, yielding a net nonzero outgoing flux emerging from the typical dosed surf~1Ce S shown. (b) From (2-29b)

div J

I i!

K

P P

p

-0 (pK)

divL

1 i! pop

(p~) o p

the flux plots of which are illustrated looking along the z-axis of the circular cylindrical system in (d). The divergencclcss character of L is evidcnt from its lip dependence, which, in this cylindrical system, provides an uninterrupted system of

72


outgoing flux lines. The radially directed field J, having a constant flux density of magnitude K, on the other hand, e1early must pick up additional flux lines with an increase in p. It is therefore required to possess a divergence.

EXAMPLE 2·3. Find the diverge nee of the E field produced by the uniformly charged cloud of Figure 1-15(b) at any location both inside and exterior to the cloud. The field E(r) outside the cloud (r> (0) is given by (I-59). Its divergence III spherical eoordinates is

/ /

divE

=

/

(2-32)

Or

/

/

/ /

This null result signifies a flux plot in the region' > ro consisting of unbroken lines, as noted in Figure 1-16(b). All inverse r2 radial ficlds behave this way. Inside the charged cloud (r < r0), the E field (1-60) being proportional to , has the divergence

p" Eo

divE

r

<

TO

(2-33)

a nonzero, eonstant result, proportiollal to the density p" of the e1oud. Note that bringing Eo inside the divergence operator puts (2-33) into the form div (EoE) = Pv C/m

3

,<

;' /

FIG!. them insid, V.

clos'

'0

making the divergence of (EoE) the same as the charge density Pv inside the cloud. It is shown in Section 2-4B that this result is true in general, even for nonuniform charge distributions in free space.

A. Divergence Theorem If F(uj, U Z , U3, t) is well-behaved m some regIOn of space, then the integral identity Sv(divF)du

~sF'ds

(2-34)

is true for the dosed surface S bounding any volume V. Equation (2-34) implies that the volume integral of (div F) dv taken throughout any V equals the net flux of F emerging from the dosed surface S bounding V. A heuristic proof of (2-34) proceeds as follows. Suppose that V is subdivided into a large number n of volume-elements, any of which is designated AUi with each endosed by bounding surfaces ,S; as in Figure 2-5(a). The net flux emanating from AUi is the surface integral ofF· ds over S;, but from (2-20), this is also (div F) Av; for Au; sufficiently small, that is,

~Si F· ds

(div F)

AVi

(2-35)

The fluxes contributed by every Si will sum up to yield the net flux through the exterior surface S bounding the volume V. Thus the left side of (2-35) summed over the

ge


73

,, (b)

(a)

fIGURE 2-5. Geometry of a .typical closed surface S, used in relation to the divergence theorem. (a) A volume V bounded by 8, with a lypical volume-dement t.v, bounded by S, inside. (b) Surfaces 8 2 and S, constructed to eliminate discontinuities or singularities from

V. do~cd

surfaces .1s i inside S yields

i

;= 1

[rh 1s,

F.dsJ = 1srh F.ds

(2-36) to the right side of (2-35) summed over the n volume elements Llv; as the number n tends toward infinity (and as .1vi --+ dv)

rh

Ie;;

F . ds

=

lim

Avc-~O

f

i=l

(div F) dv

=

r

Jv

(div F) do

(2-37)

just (2-34), known as the divergence theorem. If the limiting process yielding (2-37) is to be valid, it is necessary that F, together its first derivatives, be continuous in and on V. IfF and its divergence V . Fare not continuous, then the regions in Vor on S possessing such discontinuities or possible must be excluded by constructing closed surfaces about them, as typified 2-5(b). Note that the volume V of that figure is bounded by the multiple surface S = SI + S2 + S3, with S2 and S3 constructed to exclude discontinuities or singularities inside them. The normal unit vectors an, identified with each vee tor surface element ds = an ds on Sl' S2, and S3, are assumed outward unit vectors pointing away from the interi'or volume V. The following examples illustrate the foregoing remarks concerning the diver. gence theorem.

EXAMPLE 2·4. Supposc the one-dimensional field H(x) = axKx of Examplc 2-2(a) exists in a region. Illustrate the validity of the divergence theorem (2-34) by evaluating its volume and surface integrals inside and on the rcctangular parallelepiped bounded by the coordinate surfaccs x = 1, x = 4, Y = 2, y = - 2, z = 0, and z = 3, for the given H.

ail 74


ds = - ax dy dz on 82

2 (y) (x)

EXAMPLE 2-4

Since div H = K, the volume integral of (2-34) becomes (Jl

Evaluating the surface integral requires summing the integrals ofH . ds over the six sides of the parallelepiped. Because H is x-directed, however, H . ds is zero over four of these sides, the surface integral reducing to the same result as (I)

J. H 1s

. ds = ('3

JFO

('2 Jy~

= 48K - 12K = 36K

(2)

EXAMPLE 2·5. Given the p-dependent field: E = R"K/pl I 2, with K a constant, illustrate the validity of the divergence theorem by evaluating both integrals of (2-34) within and on a right circular cylinder of length L, radius R, and centered about the <-axis as shown.

(Detail of thin tube used to exclude singularity)

:~

EXAMPLE 2-5

a p ' I'

_ _111_

75

2-4 DlVERGRNCE OF A VECTOR FUNCTION

Since E has it singularity atp = lJ, a thin, tubular surface S2 of radius a is constructed as shown, to exclude the singularity from the integration region, making S = SI + S2 + S3 + 84 , The divergence orE, by use of (2-29b), is

v.E =

1

a

K

-;-- (p~~) = ~

pup

2p

yiclding the following volume integral

With E P directed, the surface integral of (2-34) reduces to contributions from only 81 and S2 (the end caps yielding zero outward flux), whence

~sE'dS

Sz.

f,L (a Z~O

2nKL(Rl/2

-K-) . (a PRd"'dz) 'I'

p R112

+

S2.

f,L (a Z~O

K ) . (-a adA-.dz) --P 'I'

P a112

a1/ 2 )

(2)

agreeing with the result (I). [Note: Each answer has the limit 2nKLRl/2 as a

->

0.]

The usefulness of the divergence theorem embraces more generally the interchange of volume for closed-surface integrals required for establishing several theorems of electromagnetic theory. An example occurs in Poynting's theorem of electromagnetic power considered later in Chapter 7.

B. Maxwell's Divergence Relations for Electric and Magnetic Fields in Free Space The definition (2-20) of the divergence of a vector field serves as a basis for deriving the dilferential, or point, forms of two of Maxwell's equations from their corresponding integral forms (1-53) and (1-54) for free space

rh fs (EoE) -

• ds

= Jv r Pvdv C

rh B. ds = 0 Wb fs

[1-53] [1-54]

These laws apply to closed surfaces S of arbitrary shape and size. If S is the surface bounding any small volume element Av, dividing (1-53) by Av yields

fs (EoE) • ds Av

= Iv pv dv Av

(2-38)

The Iimil of the left side, as A1I becomes sufficiently small, is div (EoE) from the definition (2-20). The right side denotes the ratio of the free charge Aq inside Av to Av itself; its limit is PV' As A1I -t 0, therefore, (2-38) becomes

div (EoE)

= Pv C/m 3

(2-39)

76


the differential form of Maxwell's integral expression (1-53). Note that expressing (2-39) in rectangular coordinates using (2-29a) yields the partial differential equation

aEx+ __ oE y+ aE z ax

~y

oz

Pv Eo

(2-40)

It is evident that the divergence of (EoE) at any point in a region is precisely Pv, the volume density of electric charge there, implying that the flux sources of E fields are electric charges. Equivalently, if electric field lines terminate abruptly, their termini must be electric charges. By a similar procedure applying (1-54), one obtains the following partial differential eq uation in terms of B div B

= 0 Wb/m 3

Pa

FI th

(2-41 )

implying that B fields are always divergenceless and therefore source free. The flux plot of any B field must, therefore, invariably consist of elosed lines; free magnetic charges are thus nonexistent in the physical world. A divergenceless field is also called a solenoidal field; magnetic fidds are always solenoidal.

EXAMPLE 2·6. Suppose that Maxwell's diHcrential equation (2-39), instead of its integral form (I-53), had been postulated. Execute the reverse of the process just described, deriving (1-53) from by the latter over an arbitrary volume Vand applying the divergence theorem. Integrating over an arhitrary volume V yields

01

fi

CI

b

VI

tl fi t f

v

c

Assume that E is well-behaved in the region in question. From a use of (2-34), the left: side can be replaced by the equivalent closed-surface integral ts (EoE) . ds, and (I-53) follows

[ 1-531

2·5 CURL OF A VECTOR FIELD From (2-15) it is established that the line integral of (grad f) • dt around any closed path is always zero. Many vector functions do not exhibit this conservative property; a physical example is the magnetic B field obeying Ampere's circuital law (1-56). For example, in the steady current system of Figure 1-19, the line integral of'B· dt taken about a circular path enclosing all or part of the wire, a nonzero current result is anticipated. Nonconservative fields such as these are said to possess a circulation about closed paths of integration. Whenever thc elosed-line integral of a field is taken about a small (vanishing) closed path and the result is expressed as a ratio to the small area enclosed, that circulation per unit area can be expressed as a vector known as the curl of the field in the neighborhood of a point. It follows that a conservative field has a zero value of curl everywhere; it is also called an irrotational field.

2-5 CURL OF A VECTOR FIELD

B Paddle wheel

77

-(xj

FIGURE 2-6. A vc\ocity field in a'fluid, with an interpretation "fits curllrom the rotation of a small paddle wheel.

Historically, the concept of curl comes from a mathematical model of effects in hydrodynamics. The early work of Helmholtz in the vortex motion of fluid fields led ultimately to the mathematical postulates by Maxwell of Faraday'S conceptiollS of the electric fields induced by time-varyin~ magnetic fields. A connection between curl and fluid phenomena can be established by supposing a small paddle wheel to be immersed in a stream of water, its velocity field being represented by the flux map shown in :Figure 2-6. Let the paddle wheel be oriented as at A in the figure. Th(' eH<';ct of the greater fluid velocity on one side than on the other will cause the wheel to fotate- clockwise, in the example shown. In this example, the velocity field l ' is said to have a vector curl directed into the paper along the axis of the paddle wheel, a s('nse determined by the thumb of (he right hand if the fingers point in the direction of the rotation; the vector curl of v has a negative z direction at A. Similarly, physically rotating the paddle wheel axis at right angles as at B in the figure provides a way 10 determine the x component of the vector curl of v, symbolized [curl v]x. In rectangular coordinates, the total vector curl of v is the vector sum occurrin~

Generally, the curl 2 of a vector field F(ub U2, U3, t), denoted curl F, is expressed as the vector sum of three orthogonal components, as follows (2-42)

Each component is defined as a line integral ofF, dt about a shrinking closed line on a per-unit-area basis with the al component defined (2-43)

The vanishing suriace bounded by the closed line t shown in Figure 2-7 is As l , with the direction of integration around t assumed to be governed by the right-hand rule. 3 2In

f~uropean

texts, curl F is written rot F, and is read rotation of F.

3The integration sense coincides with the direction in which the fingers of the right hand point if the thumb points in the direction of a l '

78


sn

tl ec

FIGURE 2-7. A closed line l bouuding the vanishing area As lo used in defining the a l component of curl Fat P.

v

Similar definitions apply to the other two components, so the total value of curl F at a point is expressed

curlF A difierential expression for curl F in generalized coordinates is found from (2-44) by a procedure resembling that used in finding the differential expression for div F in Section 2.4. The shape of each closed line l used in the limits of (2-44) is of no consequence, as long as the dimensions of Lls inside l tend toward zero together. Thus, in finding the a l component of curl F, t is deformed into the curvilinear rectangle of Figure 2-8(b) with edges Lllz and Lll3 . The surface bounded by t is Lls l = a l Lltz Llt3 = alhzlz3 LlU2 LlU3, the only components ofF contributing to the line integral in the numerator of (2-43) being F2 and F 3 . Thus, along the bottom edge Llt z , the contribution to ~t F • dt becomes (2-45) in which LlW2 denotes that contribution. Along the top edge, F2 changes an incremental amount, but in general so does the length increment, Lltz , because of the curvilinear coordinate system. The line-integral contribution along the top edge is found from a Taylor's expansion of ~W2 about P. The first two terms are sufficient if ~U3 is suitably

\ (U3) \

\ \ \

\ I

(a)

(b)

FIGURE 2-8. Relative to curl F in generalized orthogonal coordinates. (a) The components ofF at a typical point P. (6) Construction of a path l rdative to the a l component or curl F.


79

small; thus

LlW~(U1' U2, U3 + Llu

3)

[LlW2

=

+ -,---'- LlU 3 ] (2-46)

the negative sign resulting from integrating in the sense of decreasing U 2 along the top edge. Similarly, the contribution along the left edge Llt3 in Figure 2-8(b) is (2-47)

whereas along the right edge, it is I '

LlW3=F3Llt3+

0(F3 M 3)

Lluz

(2-48)

OU2

The substitution of (2-45) through (2-48) into the definition (2-43) obtains for the a 1 component of curl F

ad curl FIt

. t F ' dt

.

=a 1 hm - - =a 1 hm ASI---+O

AS l

AS1---+0

(2-49)

A similar procedure yields the two remaining vector components of curl Fin (2-44); although from symmetry, a simple cydic illterchange of the subscripts in (2-49) leads directly 10 them. The expression /<)r curl F in generalized orthogonal coordinates is thus

(2-50) which is identical to the determinemal form

curlF

a1

a2

a3

h2 h3

h3 hj

h1hz

0

0

0

oU I

OU2

hlF j

h2F2 h3 F3

(2-51 )

80

VECTOR DIFFERENTIAL RELATIONS AND MAXWELL'S nfFFERENTIAL RI'~LATIONS

a result simplifying in the rectangular coordinate system to ax

curlF

=

(J

iJx Fx

ay

az

a a

(2-52)

ay Fy

On comparing (2-52) with the cross product A x B of (I definition (2-18) f()l' V, one is led to the equivalent

curl F

== V x

and recalling the

F

(2-53)

Although V has been defined only in the rectangular and V x F are customarily considered interchangeable system used. It is seen that I) also leads to the following pvt.t'pq", cylindrical system

symbolisms curl F of the coordinate

curl F in the circular

curlF (2-54)

and in the spherical coordinate system

h tl tl t a

curlF

C (2-55)

EXAMPLE 2·7. Find the curl of G = aJCy, a flux plot of which is sketched in Examplt' 2-2. Because G has only ay-dependcnt x componcnt, from (2-52) one obtains

curlG

ax

ay

az

()

D ily

0

hy

()

0

a z [-

D~)J

-azK

a negative ,c-directed result for K> O. So if G were a fluid velocity field with a paddle wheel immersed in it as in Figure 2-6, a clockwise rotation looking along the negative z direction would result, agreeing with the direction of curl G.

EXAMPLE 2·8. Find the curl of the B fields both inside and outside the long, straight wire carrying the steady current J shown in Figure 1-19.

81

2-5 CURL OF /I. VECTOR FIELD

from

The B fidd is (1-6+), a 4>-directed function of p. The curl ofB, obtained I), inside the wire (p < a) ap

p curl B =

az

aq,

J

8

ill'

84>

0

Ii [fi.OJP

P

8 ilz

J

-~.

2na 2

a

z

~ [p fi.olp2

I' elP

2na

J=

azfi.o

-!na

2

0

a result proportional to the current density ]z = flna 2 in the wire. This special case demollstrates the validity of a Maxwell's diflerential relation to be developed in Section 2-SB. You may fi.nthcr show from (2-5+) that curl B outside the wire is zero, in view of the inverse p dependence of B there.

A. Theorem of Stokes If F(ub u2, U3, t) is well-behaved in some region, then the integral identity

1

(V x F)' ds

rf:

'Yt F

· dt

(2-56)

holds [i)r every closed line t in the region, if S is a surELee bounded by t. This is ealled the theorem (Jf Stokes. J\ heuristic proof follows along lines resembling the proof of the divergence theorem. Suppose the arbitrary S is subdivided into a large number n of surface-elements, typical ofwhieh is ~Si bounded by til as in Figure 2-9(a). The line integral ofF' de around Ii is ini<:rred hom lhe definition (2-43) of the componenl of the curl F in the directioll oC ~Si to be

J, F· de ~ti

=

[curl F] . ~Si

S ~Positive side)

'-,,---

Integration'"'" sense of

ifF>dt

(bl

FIGURE 2-9. Relative to Stokes's theorem. (a) Showing a typical interior surfaceelement ~Sl bounded by t't· (b) Closed lines t'2 and l3 comtructcd to eliminate discontinuities from S.

(2-57)

82


for ~Si sufficiently small. If the left side of (2-57) is surnmed over all closed contours t; on the surface S of Figure 2-9(a), the common edges of adjacent elements are traversed twice and in opposite directions to cause the integrations about t; to cancel everywhere on S except on its outer boundary t. Summing the left side of (2-57) over the n interior elements ~s; therefore obtains

f [rf-j{, F . dt]

;=1

=

rfF • dt j{

(2-58)

and equating to the right side of (2-57) summed over the same elements yields the result, as n approaches infinity

rf- F . dt ::Yt

=

lim

f

[(curl F) • ~s;]

As,->O t=1

=

r (curl F) . ds

Js

(2-59)

which is Stokes's theorem (2-56). As with the divergence theorem, It IS necessary in (2-56) that F together with its first derivatives be continuous. Ifnot, the discorHlnuities or singularities are excluded by constructing closed lines about them as in 2-9(b), causing S to be bounded by the closed line t = tl + t2 + t3' The connective strips, of vanishing widths as shown, are however, traversed twice so their integral contributions cancel. The positive sense of ds should as usual agree with the integration sense around t according to the right-hand rule.

w

E EXAMPLE 2·9. Given the vector field (1)

illustrate the validity of Stokes's theorem by evaluating (2-56) over the open surface S defined by the five sides of a cube measuring 1 m on a side and about the closed line t bounding S as shown.

(z)

(z)

Positive side of S

P4

____ Positive integration

s~: x= 0 ds =- axdydz

(x)

(x)

(b)

(a)

EXAMPLE 2-9. (a) l.ine elements on

(y)

t.

(b) Surface elements on S.


The line integral is evaluated first. The right side of (2-56) applied to making usc of figure (a)

= 0

+ Jz--o rl_ zdz + Jx-l ro_

5xdx

+ Jz-l ro_ zdz =

t:

83

becomes,

(2)

The surface integral of (2-56) is found next. From (2-52). ax

curl F =

ay

az

a

a a

ox

oy oz yZ yz

=

axz

+ a y5xy -

a z 5xz

(3)

whence the surface integral of (2-56) evaluate over S\, ... , '')5 yields, using figure (b),

r (curl F) • ds Js \

=

rl Jx~o r1

Jy=o

(4)

which agrees with (2).

EXAMPLE 2-10. Given the veetor field

F(ti) = a",K cot ti

(I)

in which K is a constant, illustrate the validity of Stokes's theorem evaluating (2-55) for the hemispherical surface S with a radius a, bounded by the closed jine t: ti = 90°, r = a as shown. There is a singularity in F on Sat () = 0°; it must be excluded to assure the validity of Stokes's theorem on the given surface. To accomplish this, a small circle t:3 at. (1 = til and r = a is constructed as in (b). Ifds is assumed positive outward on S, then the sense of the line integration is as noted, the integrals cancelling along (z and (4 oftlw connective strip

jr----Integration sense (a)

EXAMPLE 2-10. (a) Open hemispheric surface S. (h) Exclusion

(b)

or the singular point.

84


as its width vanishes, The line integral around t

r2~ F

J~-o

0

a",r

sin 8 d]

r=a

O~,,/2

= tl + t3

+ J
thus hecomes 0

a",r

sin 8 d]

at

r=a

O~O,

converging to -2naK as {)l --> O. The surface integral is evaluated using

Mal

ae

a4>

r sin

r

ar

a ao

a a

0

0

(r sin 8) K cot ()

curlF

-ar

K K - ao - cot 0 r r

whence

Jsr(curIF)ods=

r2:

l~

unit vect'

r1t~2 (-It'asinOdOd)

J",-o JO-Ol

=

~2naKeos81

(3)

which agrees with (2). You might consider how the results would have compared had one ignored the singularity.

B. Maxwell's Curl Relations for Electric and Magnetic Fields in Free Space In Section 2-4B, the divergence of a vector fUllction was put to use in deriving the differential Maxwell equations (2-39) and (2-41) from their integral versions (1-53) and (1-54). The definition of the curl may similarly be used to obtain the differential forms of the remaining equations (I-55) and (i-56), Because the latter are correct for closed lines of arbitrary shapes and sizes, one may choose t in the form of any small closed path bounding a j Lls 1 in the vicinity of any point, as in Figure 2-7. Taking the ratio of (I-55) to Lls 1 yidds, with the assignment of the vector sense a l to each side, d dt

r

Jl1s, Lls 1

the the Th sib] sati

reI;

It cu

aF fo]

Bods

(2-60)

(2-43), the left side, as AS l -40, becomes a1rcurlEl 1 . The right side denotes time rate of decrease of the ratio of the magnetic flux I1ljJm to I1S1> but this is just compon(~nt Bl at the point P. The limit of (2-60) therefore reduces to

£,

(~

(2-61 )

te

''''UIII: the at component of curl E to the time rate of decrease of the at component tnagneticJlux densilY B at any point. 4 The choice of the direction assigned by

2

al

[curl EJI

differentiation symbol alDt the lild that the field B is a a function of t only, I,)r a fixed

in (2-61) replaces the total differentiation did! in (2-60), of space as well as of time, whereas the volume integral

t.\,

C fl C

2-6 SUMMARY OF MAXWELL'S EQUATIONS: COMPLEX, TIME-HARMONIC FORMS

85

is arbitrary, implying that two similar results aligned with the directions of the unit vectors az and a 3 and independent of (2-61) are also valid. Combining these Ilectorially thus obtains the total curl of E at the point 1.1

Making use of the notation of (2-42) yields the more compact form

aB

VxE=--Vjm

2

at

(2-62)

the differential form of Faraday's law (I-55). Equation (2-62) states that the curl of the field E at any position is precisely the time rate of decrease of the field B there. This implies that the presence of a time-varying magnetic field B in a region is responsible for an induccd time-varying E in that region, such that (2-62) is cverywhere satisfied. A procedure similar to that used to derive (2-62) is applicable to the Maxwell relation (I-56), yielding the differential equation

V

B = Ilo

X -

a(EoE) J +- - A/m z

at

(2-63)

It states that the curl of B/llo at any point in a region is the sum of the electric current density J and the displacement current density a(EoE)/ot at that point. If the electric and magnetic fields in free space are static, the operator Ojat appearing in (2-62) and (2-63) should be set to zero. This restriction provides the following curl relations for time-static fields

VxE=O B V x-=J Ilo

(2-64 ) Curl relations for static E, B fields (2-65)

Equation (2-64) stales that any static E field is irrotational (conservative), whereas (2-65) specifies that the curl of a static B field at every point in space is proportional to the current density J there. 2·6 SUMMARY OF MAXWELL'S EQUATIONS: COMPLEX, TIME·HARMONIC FORMS

One may recall that in Sections 2-4B and 2-5 the differential Maxwell equations for free space were obtained Irom their integral forms, (1-53) through (I-56). These are collected for reference in Table 2-1, columns I and III. The integral Maxwell equations

~

TABLE 2-1 Time-dependent and complex time-harmonic forms of Maxwell's equations in free space

Differential forms

Integral forms TIME-DEPENDENT

~s EOE' ds

j:

~s EoE' ds

p,du

¢. B· ds = 0 v

~sB' ds

S

~

E. dt =

d

dl

v {

A: ~ . dt 'fr flo

i' B· ds

i' J' ds + r!.-, i' EoE' ds

Js

dt Js

J,B . dt ::rc fto

V' (EoE)

vx

Is B' ds

V' (EoE) =

V'B

0

E

cB

v x

fJt

i' j . ds + jw i' EoE' ds

Js

p,

V' B

0

A: E' dt = :Yt

[I

Js

Iv p,d"

IV, COMPLEX, TIME-HARMONIC

TIME-DEPENDENT

COMPLEX, TIME-HARMONIC

Js

V x

a

B

J + _ (EoE)

flo

Vx

ct

P" 0 r2-71]

E -jwB [2-72J ~

13 flo

~

J + jWEoE

[I-56]

,.,

<""'

a .....

c

~.;2. ~

1',

"'l"::h .... "

""Or.I2~::J'"'o

~ ~

;. g.

~

9.., 7~ 9"::; . . ?"" r"';

2-6 SUMMARY OF MAXWELL'S EQUATIONS: COMPLEX, TIME-HARMONIC FORMS

87

were s(~en in Section 1-11 to be well suited for finding the field solutions of static charge or current distributions possessing simple symmetries, though methods relying 011 symmetry are UnflJI'tunately limited to a few isolated problems. The difTerential Maxwell equations usually oH;:,r a much hroader class of solutions; obtaining a number of these solutions will constitute the task of rnueh of the remaining text. Also of importance are the sinusoidal steady state, or time-harmonic solutions of Maxwell's equations. Time-barmonie fields E and B are generated whenever their charge and current sources have densities varying sinusoidally in time. Assuming the sinusoidal sourees to have been active long enough that the transient field components have decayed to negligible levels permits the further assumption that E and B have reached a sinusoidal steady state. Then E and B will vary according to the factors cos (wt + f)e) and cos (0)[ + 0b), in which Oe and Ob denote respective phases and w is the angular frequency. The alternative and equivalent t(wmulation is achieved if the fields are assumed to vary according to the complex exponential factor . This assumption leads to a reduction of the field fimetions of space and time to fimctions of space only, as ohserved in the following. The held quantities in the real-time forms of Maxwell's equations presented in columns I and III of Table 2-1 are symbolized E

E(l1b 112, 11 3, t)

J ~J(111. 112,

113.0

(2-66)

The linearity of the Maxwell relations guarantees that sinusoidal time variations of charge and current sources produce E and B fields that in the steady state are also sinusoidal. Then one may replace the functions of space and time with products of only, multiplied by the as follows

E(111'

U2, 11 3 ,

t) is replaced with E(u[,

B(u J ,

112, 113'

t) is replaced witb B(ul' u2 ,

J(11 1, 112,

Ii},

U2 ,

t) is replaced withj(l1[) 11 2, (2-67)

If the complex vectors E, B, and j are written in terms of the generalized coordinate system as f()llows, that is, (2-68)

then on illserting obtains

into tlte Maxwell equations

o

or column

Ill, Table 2-1, one

(2-69)

The parti'll-derivative oper'ltors V . and V x of (2-69) affect only the space-dependent functions E(l1lo 112, and B(U1' 112, whereas a/at operates only on the tl mt factors

88

VECTOR DIFFERENTIAL RELATIONS AND MAXWELL'S DIFFERENTIAL REI,ATIONS

common to all the fields. Equations (2-69) therefore yield, after cancelling the ~wt factors,

Pv

V- (EoE)

V-B =0

V

X

Vx

E = -jwB B

j + jWEoE

110

C/m 3

(2-70)

Wb/m 3

(2-71 )

V/m

2

(2-72)

A/m

2

(2-73)

These are the desired complex, time-harmonic Maxwell equations for free space. They represent a simplification of the real-time fOIIns in that the tipe variable t has been eliminated. On finding the complex solutions E(u l , U2, U3) and B(ul' U2, U3) that satisfy (2-70) through (2-73), the sinusoidal time;.depen~ence can be restored by multiplying each space-dependent complex solution E and B by eJwt and taking the real part of the result as follows E(Ul'

U2) U3,

t)

= Re [E(ul>

U2,

u3 )eirot ]

B(Ul'

U2, 113,

t)

= Re

U2,

U3)~wt]

[B(Ul'

(2-74)

Considerable use is to be made of (2-70) through (2-74) in subsequent discussions of the time-harmonic solutions. One can show that a similar procedure using the replacements (2-67) leads to a complex, time-harmonic set of the integral forms of Maxwell's equations in free space. A comparison wi th their time-dependent versions is provided in Table 2-1. Applications of the complex time-harmonic forms (2-70) through (2-73) to elementary wave solutions in free space are considered in Section 2-10. A preliminary discussion or the Laplacian operator and a development of the so-called wave equations are desirable prerequisites to finding such solutions. These are discussed next.

2·7 LAPLACIAN AND CURL CURL OPERATORS The gradient of a scalar field was seen in Section 2-2 to yield a vector field. Moreover, the divergence of the vector function grad], denoted symbolically by V - (V]), is by the definition (2-20) a scalar measure of the flux source-per-unit-volurne condition of V] at every point in a region. The expansions (2-10) and (2-28) for V] and its divergence can be combined to obtain V - (Vf) in a desired coordinate system, a result to be found useful for obtaining both time-varying and time-static field solutions. Thus, in generalized coordinates, the gradient of] is expressed by (2-10)

(2-75)

2-7 LAPLACIAN AND CURL OPERATORS

To find the divergence of VI, the components of and F3 of (2-28), obtaining

89

become the elements Fb Fll

(2-76)

This scalar result has a particularly simple form in rectangular coordinates, becoming

V- (Vf)

(2-77)

The definitiolls of the dot product and of V are seen to permit the following operator notations

V-V == ;j2

+

a2 + a2

-=

V2 .

(2-78)

in which the notation V2 , called the Laplacian operator, is equivalent to V - (V ) V- V( ) div (grad ). From (2-76), the Laplacian operator in generalized coordinates is, therefore

V2 == V - V

yielding in the circular

r1J/'lnrlr'}NI/

system

V-VI

(2-80)

while in spherical coordinates

2'

Vj

=

J

1) ara (2iJ ra~ +

(2-81 )

90

VECTOR DIFFERENTIAL RELATIONS AND MAXWEl,L'S DIFFERENTIAL RELATIONS

The Laplacian operator (2-79) is also applicable to a vector field F(Ul, U z , U3, t), the result of which is shown to be useful in the expansion of curl (curl F), Apply (2-79) to define VZF, the Laplacian of a vector field, as follows

(2-82)

The term-by-term expansion of the latter can be tedious, since in general a l , aZ and a 3 are not constant unit vectors in a region; that is, their directions depend on Uil U z , and U3' In rectangular coordinates, however, (2-82) yields the relatively simple result (since ax, a y , a z are constant unit vectors) l

(2-83) in which the components V 2 F x , and so on, are specified by (2-77). No corresponding simplicity occurs in other coordinate systems because of the spatial dependence of the unit vectors already noted. For example, if the space partial derivatives of the nnit vectors are properly accounted for, as in Example 1-1 of Section 1-6, one can show £l'om (2-82) that V 2 F in the circular I]lindrical system becomes 2 of

pi

01>

a result decidedly not of the form of (2-83) with }<~" F, F z merely taking the places of Fx, Fy , Fz· St ill another vector result, the curl of the vector curl F, designated V X (V X F), is of importance. The function V X F provides the three components given in (2-50); then performing another curl operation yields

(2-85)

I

2-7 LAPLACIAN AND CURL OPERATORS

91

Because of its complexity, this result is examined only in rectangular coordinates, becommg

v

X

(V x F)

= a {~(OFy

oy ox

x

OFx) oy

0

(OFozx_ O}~)} ox

_ OFy) _ ~ (OIi~ _ OFx)} OZ oy oz ox ox oy y {() (OFx_ OFz) _ () (OFz_ OF )}

+ a {~(OFz y

+a z

ox oz

ox

oy

(?Y

(2-86)

OZ

A comparison of the latter with the vector V(V' F) is now made. In rectangular co6rdinates, using (2-10) and (2-28) obtains

V(V' F) (2-87) and adding and subtracting six properly chosen terms puts (2-87) into the following form

On comparing the terms of the latter with (2-83) and (2-86), it is seen that one has precisely V(V' F)

V2F

+V V

X

X

(V

(V

X

X

Fl. This is a vector identity, usually written

F) = V (V . F) - V 2F

Equation (2-88a) provides a useful equivalence for V F is divergenceless (V, F = 0). Then

X

(V

(2-88a)

X

F), especially if the field

if V . F = 0

(2-88b)

92


as

TABLE 2-2 Summary of vector identities

v(

Differential (II) V(f+ g) = Vf + Vg (12) V'(F+G)=V'F+V'G

Algebraic

(1) F·G=G·F F x G -G x F

(3) (4) (5) (6)

(13) V x (F + G) = V x F + V x G (14) VUg) =fVg + gVf (15) V'(jF)=F'Vf+f(V'F) (16) V' (F x G) = G· (V x F) - F' (V x G) (17) V x UF) = (Vf) x F + f(V x F) (13) V'Vf=V 2f (19) V' (V x F) = 0 (20) V x (Vf) = 0 (21) V x (V x F) = V (V • F) - V2F' (22) V x UVg) = Vfx Vg

F' (G+H) =F'G+F'H F x (G + H) = F x G + F x H F x (G x H) = G(H' F) - H(F' G) F' (G x H) = G . (H x F) = H • (F x G) Integral

(7)

(3)

Ps F . ds =

Iv V . F dv

rf

Js

';ft

F • dt = f (V x F) • ds

Iv [f V g + (v/) . (Vg)] dv gV!]' ds Iv UV2g gV2/) 2

(9) PJ(Vg) . ds = (10)

Ps rfV,<:

=

Cl

do

Although the proof of (2-88a) was carried out in the rectangular system, such differential'results are independent of the coordinate system, meaning that (2-88a) and (2-88b) are true for any system. It is worth wile to observe that one can more easily expand V 2 F by use of the vector identity (2-88a) than by ddinition 2-82). Thus

V 2F = V (V . F) - V

X

(V

X

F)

(2-89)

is useful f(:>r expanding V 2 F in a coordinate system other than the cartesian .. Several vector identities involving the difterential operators grad, div, and curl are listed in Table 2-2 along with vector algebraic and integral identities. Proo[~ of the algebraic and the diHcrential identities are achieved in the manner used to prove (2-88a), that is, expanding both sides in rectangular coordinates leads to an identity. The integral identities (7) and (8) are recognized as those of diver'gence and Stokes's theorem, respectively. Extensions of the divergence theorem lead 'to Green's integral identities (9) and (10), proved in the next section.

2·8 GREEN'S INTEGRAl THEOREMS: UNIQUENESS One can specialize the divergence theorem (2-34) to a particular class of vector functions and ohtain the integral identities known as Green's theorems. Suppose F to be a scalar fieldJmultiplied by a conservative vector field Vg; let F = JVg. Then (2-34) takes on the special form

~~ UVg)

. ds

=

Iv V· UVg)

dv

(2-90)

Ie ft

s

2-9 WAVE EQUATIONS FOR ELECTRIC AND MAGNETIC FIELDS IN FREE SPACE

93

assuming the functions well-behaved in and on the volume V. The integrand in the volume integral may be expanded by use of (15) in Table 2-2, whence (2-90) becomes Green's first integral identity

~~ (fVg) . ds =

Iv [JV2g + (VJ) • (Vg)] dv

(2-91 )

If one chooses to define a vector function G g VJ instead, the same procedure leads to a result like (2-91) except for the interchange of the roles of the scalar functions I and g

Ps (g VI) . ds = Iv [gV J + (Vg) . CVllJ dv 2

Subtracting the latter from (2-91) obtains Green's second integral identity (2-92)

also knowll as Green's symmetric theorem. Green's theorems (2-91) and (2-92) are important in applications to theorems ofbound,lIy-value problems of field theory, as well as to special theorems concerning integral properties of scalar and vector functions. One such. theorem concerns those dil~ ferential properties of a vector field F that must be specified in a region to make F unique. This theorem, not proved here,5 shows that the specification of both the divergence and the curl of a vector function F in a region V, plus a particular boundary condition on the surface S that bounds V, are sufficient to make F unique. Maxwell's equations (:2-39), (2-4: 1), (2-62), and (2-63) specify the divergence and the curl of both the E and the B fields in a region (in terms of charge and current densities as well as the B or E field), so that these relationships, together with appropriately specified boundary conditions, can similarly be expected to provide unique field solutions. Finding solutions of Maxwell's differential equations is facilitated for some problems hy first manipulating them simultaneously to obtain differential equations in terms of only B or E, as is discussed next.

*2·9 WAVE EQUATIONS FOR ELECTRIC AND MAGNETIC FIELDS IN FREE SPACE Electromagnetic field solutions Band E in free space must, by the uniqueness discussion of the previous section, satisfy the Maxwell divergence and curl relations (2-39), (2-4:1), (2-62), and (2-63). In a time-varying electromagnetic field problem, one is generally interested in obtaining E and B field solutions of the tour Maxwell relations, a process that can often be facilitated by combining Maxwell's equations such that one of the fields (B or E) is eliminated, yielding a partial differential equation known as the wave equation. This is accomplished as follows. 5For a pro()f~ sec S. Ramo, J. R. Whinnery, and T. Van Duzer. Fields and Waves in Communication Electronics, 2nd cd. New York: Wiley, 1985, p. 130. *For the purposes of the next section, Section 2-9 may be omitted if desired.

94

VEcrOR DIFFERENTIAL RELATIONS AND MAXWELL'S DIFFERENTIAL RELATIONS

The Maxwell differential equations for free space are

here for conven-

lence.

V· (EoE)

[2-39 J

Pv

V ·B= 0 VxE=

[2-41]

oB ot

[2-62]

B

o(EoE)

Jlo

ot

Vx---=j+

[2-63 J

To eliminate B, taking the curl of both sides of (2-62) obtains

v

X

(V

X

E)

(2-93)

Substituting (2-63) into the right side of (2-93) yields, after transposing terms containing E to the left side

(2-94)

a vector partial differential equation known as the inhomogeneous vector wave equation for free space. A wave equation similar to (2-94) can be obtained in terms of B. Thus, taking the curl of (2-63) arid substituting (2-62) into the result yields the inhomogeneous vector wave equation

V X (V X B)

iJ2B

+ JloEo ot 2

(2-95 )

= Jlo V X j

From (2-41), B is always divergenceless, and with V· E written

= pjE,

(2-94) and (2-95) are

Inhomogeneous vector wave equations for charge-free region

(2-96)

(2-97) A further simplification is possible if the region is empty space; that is, it is both charge free and current free (Pv j = 0). Then the simpler homogeneous vector wave

2-9 WAVE EQUATIONS FOR ELECTRIC AND MAGNETIC FIELDS IN FREE SPACE

95

equations hold

o

(2-98) Homogeneous vector wave equations for empty space (2-99)

If in a problem the rectangular coordinate system is appropriate to the E and B fields

governed by (2-98) and (2-99), making use of (2-83) provides the following scalar wave equations in terms of field components

a Ex 2

V2Ex

JLoE0aT =

V2Ey

JLoEo

V 2 Ez

JloE o

tPEY

ap

a j'.,'z

0

(2-100a)

= 0

(2-100b)

=0

(2-100c)

2

with an analogous trio of equations in Bx, By, and Bz yielded by (2-99). The complex time-harmonic forms of the wave equations may be obtained by replacing Band E with their complex exponential forms, (2-67). If this is done for (2-98) and (2-99), one obtains after cancelling eiwt

2~

VE

+ (t)

2

~

JLoEoE

=0

(2-101)

Homogeneous vector wave equations in complex time-harmonic 2~ 2 ~ i: V B + (t) JLoEoB = 0 lorm, for empty space

aiy aJ;z

(2-102)

aii

Since E = aJix + + and j = a}ix + aliz + z , (2-101) and (2-102) expand to obtain the following homogeneous, scalar wave equations in complex timeharmonic form (2-103) (2-104) (2-105)

96


with a similar triplet of equations in Ex, By, and Bz yielded by (2-102). The simplest solutions of these scalar wave equations are uniform plane waves, involving as few as two fIeld components. They are considered in the next section.

(

a t s

*2·10 UNIFORM PLANE WAVES IN EMPTY SPACE The simplest wave solutions of Maxwell's equations are uniform plane waves, characterized by uniform fIelds over infinite plane surfaces at fixed instants. Simplifying features are that the solutions are amenable to the rectangular coordinate systeln, and the number offIeld components reduces to as few as two. These simplifications provide a background for the more complex wave structures discussed in later chapters. Uniform plane waves have the property that, at any fixed instant, the E and B fields are uniform over plane surfaces. These planes are arbitrarily chosen; for present purposes, assume that they are defined by the surfaces z = constant. This is equivalent to stating that space variations of E and B are zero over Z = constant planes; thus assume 1. The fields have neither x nor y dependence; that is, a/ax = %y = 0 for all field components. It will be shown that waves propagating in the z direction result from this restriction. If the waves propagate in empty space, one requires an additional assumption. 2. Charge and current densities are everywhere zero III the region; that IS,

Pv

=

J

=

O.

The complex time-harmonic forms of the Maxwell differential equations determining the wave solutions are (2-70) through (2-73). With assumption (2) they become (2-106)

o

(2-107)

E = -jwB

(2-108)

V-B v

X

(2-109)

Combining these equations has been shown to produce the wave equations (2-101) and (2-102) 2~

V E

+w

2

~

floEoE

=

0

[2-101]

o

[2-102]

*This section on plane waves, pins Section 3-6 in Chapter 3, may optionally be omitted at this time, if desired, and taken up immediately heR) .. e beginning Chapter 6. Plane wave concepts are included here becanse of their universal relevance to all dynamic field phenomena, and because they are essential to a more complete understanding of conduction and polarization eflects in materials under other than purely static comtitions.

2-10 UNIFORM PLANE WAVES IN EMPTY SPACE

97

One should bear in mind that no new information is contained in the latter that is not already expressed by the preceding Maxwell's equations. Before atlempting to extract solutions from the wave equations, one may note that the curl relations, (~-l 08) and (2-109), furnish some interesting properties of the solutions, restricted by assumptions (1) and (2). Assuming that all six field components are present, 08) becomes, with a/ax a/ay = 0 of assumption (l),

VxE=

ax

ay az

0

0

a

-jOJ(a)3 x

+ a/3 y + a)jz)

Ex Ey Ez expanding into the triplet of diflerential equations

(2-110a)

of'x

(2-110b)

(2-110c)

Similarly,

109) provides

(2-1 11 a)

(2-111b)

0=

(2-I11c)

From these ditrerential expressions, the following properties apply to the solutions about to be /clUnd

1. No z component of either E or :B is obtained, thus making the field directions entirely transverse to the .~ axis. 2. Two indetJendent pairs offieldl', (Ex> By) and (Ey,~Bx), are yielded under the assumptions. This is seen to be the case on setting Ex 0 in (2-11 Ob), for example, forcing By to vanish while yet leaving the field pair (Ey, Bx) intact, the lall!'!' being governed only by (2-11 Oa) and (2-111 b). When field pairs are of each other, they are said to be uncoupled.

,

2&

98


tiuppoJe one desires wave solutions involving only the field pair ('Ex, By). Then put Ey = Ex = 0, reducing the pertinent differential equations tojust (2-1 lOb ) and (2-11Ia)

DEx

(Co

[2-110b]

[2-11lal The field solutions are obtained on combining (2-110b) and (2-11la) to eliminate E:x or By, yielding a scalar wave equation from which solutions can be found. Alterllatively, one can make use of either vect.?r wav~ equation (2-102) or (2-103), subjecting it to the same assumptions. (Only Ex and By are present and olox ~= olDy = 0.) Either approach obtains the following wave equation in terms of Ex:

res

ca In

(2-112) This is a partial differential equation in one variable (z); thus it can be written as the ordinary differential equation (2-113) It~ solution is the familiar 6 superposition of two exponential solutions

(2-114) wherein Dl and (;2 are arbitrary (complex) constants and the coefficient Po, called the constant, is given by Po = W.J/toEo. It is to be shown that the exponential solutions and DzeifJ oz arc representations of constant amplitude waves travelytg in tl;e positive z and negative z directions, respectively. The complex coefficients C't and C2 must have the units of volts per meter, denoting arbitrary complex amplitudes of t;.he positive z neg';!:tive z traveling waves. Employing amplitude symbols E:' and E;;' instead of I and C z puts (2-114) into the form

EAz)

= E,~e-jfloz =

+ E;;'e jfJoz Vim

E; (z) + E:.~ (z)

~+

(2-115)

~

The complex amplitudes Em and E;;' may be represented by points in the complex plane using the Argand diagram of Figure 2-10, so from their polar rqJresentations and

E;;'

(2-116)

4> + and 4> dt'noting arbitrary phase angles. hi! assumed that the reader is familiar with the details of this solution, found in any text on ordinary dill<'rential equations.

l'

q 1


99

(Complex plane)

FIGURE 2-10. Complex amplitudes represented in the complex plane.

Once a solution of one wave equation has been obtained, the remaining ~field can be l()und by use of Maxwell's equations. Thus, the solution (2-115) for Ex(:::.) inserted into the Maxwell relation (2-11 Ob) yields A

B (,z;)

I DEx

= --;-

}w D<-

Y

130

= --

w

= ;;;;~E~e-j/loz

r

A

n

E~~e-jpoZ

+ E,;;~/loZ] A

- J/toEoE;;,~(Joz Wb/m 2

(2-117)

in which flo once more denotes the space phase factor Po

== w~J1.oEo rad/m

(2-118)

The real-time, sinusoidal steady state expression for the electric field component is found from (2-74). Taking the real part of (2-115) after multiplying by eiwt obtains

Ex\<-, t) =

Re [(E,!ei1>+e- j /l oz +

= E~

cos (wt

Po<-

E,;;~1>-~fJoZ)ejwt]

+ ¢+) + E';;

cos (wt

+ Po<- + ¢-)

(2-119)

Note that E~ and E';; denote the traveling wave real amplitudes, whereas ¢ + and ¢- are arbitrary phase1 relative to the instant t = 0 and the location <- 0 in space. The real-time f()rm of By of (2-117) is similarly found to be

By«., t)

r.JJ1.oEoE,~ J cos (Wi

Po<-

+ ¢ +)

- [~J1.oEoE';;] cos (wt

+ Po<- + ¢ (2-120)

The traveling wave nature of unii()rm plane waves can be grasped fi'om a graphic interpretation of (2-119) and (2-120). Consider only the first terms of each: the positive <- traveling wave. The following symbols are chosen to denote them. (2-121a)

B+ y

(2-121b)

Their positive <- traveling nature may be observed if (2-121) is plotted as a of cosine waves versus <-, at successive instants of time t. (When observing the

100


Thu:

will the a co --""

Wave motion, with increasing t (a)

(x)

Bee of a V

/

0

r\

V /

1\

"-

I \

/

\

\

I

\

\

I

\

"-

\

/

z

I \

(or {3 oz)

the

I

I - - " " Motion (b)

t"IGURE 2-11. Electric field sketches ofa positive traveling unil()rm plane wave. Vector plot along at success; vc inst ants, (b) Flux plot of the electric field at t = 0,

time or space variations of a field, it is usually best to hold space or time fixed, while the other is allowed to vary,) At t=O, (2-12Ia) becomes £';(:::,0) = cos ( #0:: + cjJ+) = E,! cos (#0':: cjJ+), shown plotted against the z variable as solid line in Figure 2-1] (a), With the period T defined by

T=

sec

J'

an.

(2-122)

one-eighth period later, fl.)r example, (2-121) becomes E; T/8) = cos (#oz 2n/8 cjJ +), The cosine function is thus shifted in the positive z direction the time lapse of the eighth period as shown, yielding a positive motion of the wave with increasing time. The vector field plot of Figure 2-11 (a) shows only a~;; (z, t) a typical z-axis ill the region. To display the field throughout a cross section any x-z plane, the flux plot of Figure 2-11 (b) is more suitable. The lllotion of the wave with increasing t is related to the phase factor flo = W~Eo appearing in the wave expressions, with #oz having the units of radians (dimensionless), implying that #0 is given in radians per meter. The z distance that the wave must travel such that 2n rad of phase shift (one complete cycle) occurs is called wavelenl',th, designated by the symbol A and defined by

In

ne ra

to

ec

WI

POA

= 2n rad

(2-123)

a'


rhus, the wavelength in iree space is related to the phase factor

2n

2n

c

Po

I

101

Po by (2-124)

m

An observer moving with the wave such that he experiences no phase change will move at the phase velocity of the wave, denoted by Vp' The equiphase surfaces of the positive z traveling wave are defined by setting the argument of (2-121) equal to a constant; that is, wi Poz + 4> + = constant', whereupon differentiating it to evaluate dz/dt yields the phase velocity

dz vp

= dt =

w

Po m/sec

(2-125a)

Because {jo = wji;~~, and with flo = 4n x 10 7, Eo ~ 1O- 9/36n, the phase velocity of a uniform plane wave in empty space is I

--- = c ~ JfloEo

3

X

10 8 m/sec

(2-125b)

-

the speed of light. 7 ._ A comparison of the complex expressions, (2-115) and (2-ll7), for Ex(z) and By(z) shows that their separate traveling wave terms are paired into ratios producing the sarne constant. Thus, write (2-115) and (2-117) in the forms = E~ e -

j{Joz

+ E;;' ei{loz

E; (z) + E; (z)

(2-126)

and -+

- Eme - j/1oZ B- (Z ) y

C

--

Em

_ilJoz

--I:""

C

(2-127) in which E;(Z), E;(z) and B;(x), fJ;(z) symbolically denote the positive z and negative Z traveling wave terms directly above them. Then the following complex ratios hold at any point in the region

c~ 3

X

10 8 m/sec

(2-128)

to provide j means for finding one of the fields whenever the other is known_ A more common variation of this technique is achieved by modifying the B field in empty 7Experiments have shown that the speed oflight, c, is more nearly 2.99792 x 10 8 m/sec. This value, together with the assumed permeability for free space fJ-o = 4n x 10- 7 H/m, inserted into (2-125b), is seen to a value for Eo that departs slightly from the approximate value 10 - 9/36n given.

DIFFERENTIAL REI"ATIONS AND MAXWELL'S DIFFERENTIAL RELATIONS (x)

t;:

--Z B/

-~

t) ""

Ei' (z, t) = Ei;,

cos (wi - f30z

+ 1>+) Equiphase surface

Motion

cE;:;

cos(wt - f30z

+ .... .,.. +)

(at t = 0) (z)

[nOURE 2-12. Vector plot ofthc fields ofa uniform plane wave along the z axis. Note the typical equiphase surface, depicting fluxes of

E;

and 13;.

through a division by /lo, defining a magnetic intensity field denoted by the symbol lor empty space as follows.

B

= H A/m

For empty space

(2-129)

/lo

Thus, denoting B: (Z)//lo by if: (z), and B; (Z)//lo by if; (z) the following ratios the traveling wave terms are valid for plane waves in empty space

E: (z) =~: (z)

/lo

(z)

fly

(z)

=

/loc

=~= J/loEo

r;;;, == 110 ~ 120n Q

V~

r;;;, == '10 ~ 120n Q

V~

(2-130a)

(2-130b)

2·

J

The real ratio, /lo/Eo the units volts per meter per ampere per meter, or ohms), ill called the intrinsic wave !11l;(JI!llran.ce empty space, and is denoted by the symbol 110' The advantage of (2-130) over ill that the ratio 110 is a usefully smaller number. The real impedal1ce ratio of 30) shows that the electric and magnetic fields of uniform plane waves in are in phase with one another, a condition evident on comparing tht~ the negative Z traveling solutions of (2-126) and (2-127). Each contains argument in the exponential factors, ample evidence of their 2-12 depicts the real-time electric and magnetic fields of in space at t = O.

T dl Ul

m E

li sl 81

p p

IXAMPLE 2·11. Suppose

p

empty space has the electric field (I)

a a

2-11 WAVE POLARIZATION

103

its frequency being 20 MHz. (a) What 1~ it§. direction of travel? Its amplitude? Its vector direction in space? (b) Find the associated B field and the equivalent H field. (e) Express E, :8, and H in real-time form. (d) Find the phase factor Po, the phase velocity, and the wavelength of this electromagnetic wave. (a) A comparison of (1) wit!: (2-115) 2r (2-126) reveals a positive z}raveling wave, whence the symbolism: E(z) = axE; (z). The real amplitude is E~ = 1000 Vim, with the vector field x directed in space. (b) Using either (2-127) or the ratio (2-128)

The use of (2-130a) obtains the magnetic intensity ~+

- Ex-(z) _1000 _ 2 65 -jPoz A/ z) -- e -jPoz . e m H~+( y '10 120n

(c) The real-time fields are obtained from (2-74) by taking the real part after multiplication by Jillt

E; (z, t) = Re [1000e- iPozJwt] =

B;

t) = 3.33

H;

t) = 2.65 cos (wt - Poz) A/m

X

Poz) V/m

1000 cos (wt 2

10- 6 cos (wt - Poz) Wb/m (or T)

(d) Using (2-118), (2-125), (2-124), and (2-122) yields

r--:-

150 =0 OJ,>! f.loEo vp = c = 3 ~

X

2n(20 x 10 6 )

c

3 x 10

.f

8

= 0.42

rad/m

10 8 m/sec 3 x 10 8

2n

Po

OJ

= - =

20

x 106

=15m

2·11 WAVE POLARIZATION

The vector orientation, or polarization, of an electromagnetic wave in space is usually described with reference to its electric field direction. Thus in Figure 2-12, the z traveling uniform plane wave shown with the field components Ex) Hy is said to be polarized in the x direction (or simply x-polarized). Similarly, the plane wave with the components Ey , Hx described in Problem 2-43 is polarized in the y direction. Both these waves are linearly polarized, because the electric field vector in any fixed z plane describes a straight-line path as time passes. Because Maxwell's equations are linear equations, a vector superposition, or summing, of the two linearly polarized uniform plane waves just introduced will also provide a valid field solution. The resultant vector sum will not necessarily be linearly polarized, however, depending on the phase condition between the x- and the ypolarized electric field components. For example, with Ex(::., t) Emx cos (wt Poz) and Ey(z) t) = Emy cos (wt - Poz) propagating in phase and at the same frequency along the z-axis, their sum, E = axEx + ayEy, would appear as depicted in Figure

104


l'

(x) I

Locus of E in z=O plane

I

.E)O.
',I

I(y)

I

t

oJ

I

(f,"(O<) --

t

(y)

)

tht In

an It

--

-----

(z)

Motion (a)

(x)

I

I

En Locus of E in z=O plane (x)

~

W

E)O,t~ / I

(y)\

I

t

'-'

1:,,(00 /

--

t

(y)

-

n w 01

.

tl o

--

Locus of tota I E

(z)

fr

Motion

'V"

(b)

FIGURE 2-13. Uniform plane waves shown in space at t = 0 and added to produce difl'ercnt wave polarizations. (a) Cophasa,1 Ex and E y, their sum yielding- a linearly polarized result. (b) Elliptical polarization of E prodn'ced with 90° phasing. (Related H field components arc omitted for clarity.)

St

o 11 1]

c c

2-13(a), c/J = arc may be forming

producing a linearly polarized field E, tilted in any fixed z plane by the angle tan (Emy/Emx) from the x-axis, as shown. The equation of the straight line found by conveniently inserting z = 0 into the Ex and Ey expressions and their ratio to eliminate OJt, yielding (2-131)

This is evidently the equation of the straight line (form:)! = mx) as shown in the inset diagram of Figure 2-13(a), regarding Ex and Ey as the variables in lieu of x andy.

I

REFERENCES

105

On the other hand, if the two component fields were 90° out of phase, such that Ex = Emx cos (OJt - Poz) and Ey = Emy cos (OJt Poz + 90°) as in Figure 2-l3(b), the sum E = axEx + ayEy would produce the spiraling locus of the E vector about the z-axis as noted. In the fixed Z = 0 plane, the component fields are written Ex = Emx cos OJI and Ey = Emy cos (OJI + 90°) = -Emy sin OJt = -Emy -Jl - C05 2 OJI. Inserting the Ex expression into Ey to eliminate OJl yields the locus ofE in the Z = 0 plane. (2-132) the equation of an ellipse with principal axes of half lengths Emx and E my , as seen in Figure 2-13(b). Thus, the tip of the total E vector describes an elliptical locus in any fixed Z plane as the wave moves by, indicating the elliptical polarization of the wave. It is also evident that a circular polarization of the E vector would occur if Emx = Emy in (2-132). You may show that if the 90° phase condition between Ex and Ey were replaced by the general angle 0, making Ey = Emy cos (OJ! + 0) in the z = 0 plane, then the polarization locus would acquire the form of sin 2 0

=0

(2-133)

an ellipse with its major axis tilted, depending on the choice of O. Wave polarization is of practical importance in radio communication transmitreceive links because the power extracted by a receiving antenna from the arriving wave is usually dependent on the orientation of the antenna relative to the polarization of that wave. The common half-wave, thin wire dipole antenna, for example, picks up the maximum power from a linearly polarized oncoming wave when the electric field of the arriving wave is aligned with the antenna wire, while accepting zero power from the wave if the electric field and the wire are at right angles. If the arriving wave were circularly or elliptically polarized, a component of the arriving E-field vector is made available to the receiving dipole regardless of its tilt in the plane ofE, so that the orientation of the receiving antenna, in any fixed z-plane, would have little or no effect on the amount of signal received. This could be of considerable importance in satellite communications, in which the receiving antenna on the satellite is tumbling in space and therefore changing its attitude relative to the oncoming wave. Antennas capable of transmitting circularly polarized waves, such as helical antennas or phased crossed dipoles, are readily constructed to accommodate this need.

REFERENCES ABRAHAM,

M., and

R. BECKER.

The Classical Theory of Electricity and Magnetism. Glasgow: Blackie,

1943.

R. M., L. T. CIlD, and R. P. ADLER. Electromagnetic Fields, Energy and Forces. New York: Wiley, 1960. KRAUS, J. D. Electromagnetics, 2nd ed. New York: McGraw-Hill, 1984. PHILLIPS, H. B. Vector Analysis. New York: Wiley, 1944.

FANO,

S., J. R. WHINNERY, and T. Electronics. New York: Wiley, 1984.

RAMO,

VAN DUZER,

2nd ed. Held" and Waves in Communication

106

VECTOR DIFfERENTIAL RELATIONS AND MAXWELL'S DIFFERENTIAL RELATIONS

PROBLEMS

SECTION 2-2 2-1.

From the substitution of the appropriate coordinate variables and metric coefficients into the gradient expression (2-10), show that (2-14a,b,c) follow in the three common coordinate systems. Also convert the magnitude expression (2-13) to correct forms in those systems.

2-2.

Express as a vector function the gradient (maximum directional derivative) of the following scalar fields (a) f(x) = 20x 2 ; (b) g(x,y, z) = 20x 2 + 30xy2 + 40xyz; (c) F(r) = 100!T; (d) G(p, , z) = 5p sin - 6p 2 Z cos ; (e) h(T, . [Answer: (b) a x (40x + 30y2 + 40yz) + a y (60xy + 40xz) + a z 40xy (d) a p (5 sin - 12pz cos (5 cos 1> + 6pz sin 1» - az 6p2 cos 1>]

e,

2-3.

Prove, by expression in rectangular coordinates, that V(f + g) tity (11) in Table 2-2).

= Vf

+ Vg (vector iden-

With f and g given to be scalar, differentiable fields, by expansion in rectangular coordinates prove the identity (14) in Table 2-2, that V(fg) = fVg + gVJ.

2-4.

2-5.

In Problem 1-6 is depicted the "distance-vector," R, defined as the difference r 2 r 1 of the position vectors to the endpoints ofR. Relabel the point P2 now as P(x,y, z) with arbitrary coordinates (taken to be differentiation variables), so that now R = r rl' (a) Write the expression for R as well as its magnitude R in rectangular coordinates. (b) Show that V R is also the unit vector in the direction of R.

SECTION 2-4 2-6.

Carry out a direct proof resembling that leading to the expression (2-28) fi)r div F, but carried out in the rectangular coordinate system. Begin with (2-22), expressed in rectangular coordinates with reference to a diagram like Figure 2-4 but adapted to the rectangular system.

2-7.

By the substitution of the appropriate coordinate variables and metric coefficients into (2-28), show that the expressions (2-29a,b,c) follow, in the three common coordinate systems.

2-8.

Determine for each of the following vector fields whether or not it has Hux sources; that is, find its divergence. (a) A = 3ax + 4ay (constant vee tor field in a region) (b) F(x,y, z) = 3xzax + 4xya y + (5x 2 + y)a z (c) G(x,y, z) = 3yax + 4zay + (5x 2 + y)a z (d) H(x,y, z) = 6xa x + 6ya y + 6za z = 6a,r (determine it 111 both rectangular and spherical eoordinates) (e) J (p, , z) = a p 5pz sin + a4>lOpz cos (f) K(r, 1» = a,100/r 2 + a820/r + a4>10r cos 1> [Answer: div F = 3z + 4x, fields A, G, and J are sourceless]

e,

2-9. Prove, by expansion in rectangular coordinates, that V . (F identity (12) in Table 2-2.

+ G) =

V • F + V . G, the

2-10. By expansion in the rectangular coordinate system, prove the identity (15) in Table 2-2, V' (fF) =F·Vf+f(V·F).

2-11. Show that the following fields are, divergen~eless (source-free). (a) The p-directed, inverse-p dcpendent field F = a,,/p, for p > 0; and (b) the r-directed, inverse-r 2 dependent field, G = aJ r2, for r > O. (By comparison with results found in Example 1-13, with what kinds of static-charge sources are these field-types identified?) (a) Given the class of electric fields E(p) = apK/ pn with K a constant and n a parameter, find div E. What choice of n yields a divergenceless (charge-free) field everywhere (excluding p = OJ? Comment on this conclusion relative to (1-61), applicable to the uniform line charge. (b) Given the class of electric fields E = a,K/rn , find div E for r > O. Which choice of the parameter n provides a divergenceless field? Comment on this conclusion with respect to (1-5 7b), the electric field of the point charge.

2-12.

PROBLEMS

107

SECTION 2-4A 2-13. Assuming the same six-sided closed surface S to bound the box-shaped interior volume as in Example 2-4, assume the field G(x,y, z) = a z lOxy 2 z3 exists in the region. Illustrate the validity of the divergence theorem (2-34) by evaluating its volume and surface integrals in and on the given parallelepiped. [Answer: 10,800] 2-14. Assuming the same right circular cylindrical region of radius p = a and length t as for Example 2-5, illustrate the correctness of the divergence theorem for this region, given the electric field E = appop3/4Eoa2, that corresponds to the nonuniform charge density of Problem 1-43. [Note for this case that no singularity exists within the given V or on S, thereby obviating any need for the exclusion surface S2 used in Example 2-5.} [Answer: nLpoa 2/2E o] 2-15. The first octant of a sphere centered at the origin is bounded by the four coordinate surfaces: r = a, 4> = 0, 4> = n/2, and on the bottom by the plane = n/2. Sketch it. Given that the field F(r, 4» arlO - a.,,30r sin () cos 4> exists in this region, illustrate the truth of the divergence theorem (2-34) by evaluating the volume and surface integrals within and on the defined region for the given field. [Answer: lOa 2 (a + n/2)]

e

e,

SECTION 2-48 2-16. In Problem 1-28, the electric field within the uniformly charged spherical shell (a < r < b) was found to consist of only thl! component Er = pv(r3 - a3 ) /3Eor2. Show that inserting this field into the Maxwell divergence relation (2-39) yields the charge density originally assumed. 2-17. It was found by use of Gauss's law in Problem 1-29 that the choice of the nonuniform charge density Pv = Po(l - 4r/3a) within a sphere of radius a yields the electric field therein given by Po EO

(r3

r2) 3a

Show that div (EoE) for this field yields the charge density originally assumed, thereby satisfying Maxwell's equation (2-39).

2-18. (a) In Problem 1-30(b) it was found, using Gauss's law, that the static electric field within the uniformly charged cylindrical cloud is E = a p pvp/2E o ' Determine div (EoE), to prove that Maxwell's divergence relation (2-39) is satisfied. (b) Show similarly, from the E-field solution of Problem 1-31 (b), that E inside the nonuniformly charged cylindrical cloud of that problem satisfies the Maxwell divergence relation (2-39). 2-19. By the application of (2-28) in the appropriate coordinate system, show that the Maxwell relation (2-41), div B = 0, is satisfied for each of the B fields given by (1-64) for the long, straight wire, by (1-65) for the current sheet, and by (1-66) and (1-67) lor the toroid and solenoid. What is the physical interpretation of the zero value of the divergence expected of each and every B field?

SECTION 2-5 2-20. With reference to a diagram resembling Figure 2-8 but adapted to the rectangular coordinate system, give the details of a proof of the curl expression (2-50) carried out in rectangular coordinate form. 2-21. (a) By the substitution of the appropriate coordinate variables and metric coefficients into the determinant (2-51) for the curl of a vector field, show to what result it expands in the rectangular coordinate system. (b) Similarly show that (2-54) and (2-55) are the results of expanding (2-51) in the circular cylindrical and the spherical coordinate systems. 2-22. Find the curl of each of the vector fields given in Problem 2-8. Which of those fields are irrotational (conservative)? [Answer: (c) -3ax - IOxa y - 3a., (e) -aplOp cos 4> + a.,,5p sin 4> + a z l5Z cos 4>]

108


2-23.

Find in detail the curl of the vcctors Vg, VG, and Vh generated in Problem 2-2, [These results exemplify the validity of the vector identity (20) in Table 2-2.]

2-24.

By use of expansions in rectangular coordinates, prove the vcctor identity (17) in Table 2-2, that V x (iF) = (Vf) x F + f(V x F).

2-25. Given the vector field F(x,y, z) 2xz + 5YZ 2, find the following.

=

3xy 3 ax

+ 4y 2 z 2 ay

and the scalar field f(x,y, z) =

(a) Vf (b) V' F (c) V x F (d) V· (iF) (e) V' (Vj) =- V 2f (f) V x (V x F) (g) V· (V X F) (h) V X (Vf) [Answcr: (a) 2zax + 5z 2 a y + + lOyz)a z (c) -3y 2zax - 9xy 2 a z (e) lOy (g) 0: also by identity (19) in Table 2-2]

2-26. Given are the fields G(p, , z) = aq,5p sin - a z 6p 2 z 2 and g(p, , z) = 3pz sin . Find the functions (a) Vg (b) V· G (c) V x G (d) V· (gG) (el V· (Vg) =- V 2 g (f) V X (V x G) (g) V(V·G) [Answer: (b) 5 cos 12p2z, (e) zero (f) ap(lOp-I cos - 24pz) + a z 24zZ]

2-27.

Given functions (a) Vh (b) (d) V· (Vil) [ Answer: (a) (f) 0]

thc fields H(r, 0) = arlOr cos 0

+ aq,20r 2

and

her,

0, , find thc

V· H (c) V x H V 2h (e) V x (V x H) (1') V· (V x H) a r 8 sin 0 cos + a o3 cos 0 cos - aq,8 sin (c) a r 20r cot 11 - a o60r + aq, 10 sin 0

SECTION 2-5A Illustrate the validity of Stokes's theorem using the same closed line t and vector ficld of Example 2-9, but this time employ the surface S, consisting simply of the square located at y I. (What is the required expression for ds on S, ifit is to contorm to the line integration sense chosen about t?)

2-28.

Given is the vector field E(p, , z) = a p 5p,c - aq,8z 2 + a z I OZ2 sin . (a) Find curl E. Is E conservative? (b) Evaluate thc line integral of E . dt about the closed path t = tl + t z + 1'3 + t4 on the portion of the circular eylinder of radius 2 and height 3 located in the first octant as shown. (c) Obtain the answer to (b) by usc of an appropriate surface integral via Stokes's theorem. (One such surface S is shown.) [Answer: (b) 316.2]

2-29.

2-30.

A G-directed field is defined by F(r, 0, in a region of space. (a) Find curl F at any point. (b) Evaluate the integral of (curl F) . ds over the surface S of a sphere of radius r = R appearing within the first octant as shown, bounded by the closed line t = ta + tb + tco (c) Find the answer to (b) another way by usc of Stokes's theorem, from the line integral of F . dt taken in the correct: sense about t. [Answer: (b) 5R2]

PROBLEM 2-29

PROBLEMS

109

I (z) ) r=R

fa :1<1> = 0

I

r=R

tC:=7f/2

(x)

tb

(y)

.\r= R '10 =

7f/2

PROBLEM 2-30

SECTION 2-5B 2-31.

(a) In Problem 1-37 it was shown that the field within the conducting slab carrying the constant current density J = a;:;}z is B = ayJ.lo}zx. Show that this B field satisfies the time-static Maxwell curl relation (2-65). (b) In Problem 1-33 was derived the expression for the B field within the hollow conductor, B = a",J.loI(p2 - b2 )/2np(c 2 - b2 ). Show that this magnetic field satisfies (2-65).

2-32.

Show that the B fields, found for the coaxial conductor pair of Problem 1-34, all satisfy the Maxwell curl equation (2-65) in the three regions p < a, a < p < band b < P < c.

SECTION 2-7 2-33. From the substitution of the appropriate coordinate variables and metric coefficients into (2-76), show that the Laplacian operator ofa scalar field, v, (VI) == V21, becomes (2-77), (2-80), and (2-81) in the rectangular, cylindrical, and spherical systems, respectively.

2-34.

Substituting the correct coordinate variables and metric coefficients, show that the definition (2-82) of the Laplacian operator of a vector field becomes (2-83) in rectangular coordinates. 2-35. Repeat Problem 2-34, except this time show that the definition (2-82), in the cylindrical coordinate system, yields (2-84). [Hint: Make use of (1-33) in accounting (elr the space derivatives of the unit vectors a p and a",.] 2-36. Show that the use of the vector identity (2-89) expanded in the cylindrical coordinate system yields the result (2-84).

SECTION 2-9 2-37. In a manner similar to that employed in obtaining the wave equation (2-94) in terms ofE, derive the vector wave eqnation (2-95) in terms ofB. Show how it may be reduced to (2-99) for empty space. 2-38. Using the replacements (2-67) for real-time with complex time-harmonic fields, convert the vector, inhomogenous wave equations (2-94) and (2-95) to their corresponding complex time-harmonic forms. With the proper assumptions, show how these reduce to (2-10 I) and (2-102), appropriate to source-free empty space.

SECTION 2-10 2-39. (a) Show that combining the time-harmonic Maxwell differential equations (2-ll0b) and (2-111a) yields the scalar wave equation (2-112). (b) The wave equation (2-ll2) is to be

110


obtained via a different rout~: Beginning wi~h the wave equation I) and the electric field component Ex (along with By) to exist, show that it reduces to (2-112).

2-40. Suppose that you are told that the complex wave fLlIlction Ex(;;:.) = E:' ejw"/;'o
IS

a

2-41. A particular ~niiorm plane wave in empty space has the electric field given, in timeharmonic form, by E:(z) = 1885e- j /i oz Vim, at the frequency f 100 MHz. (a) Describe this electric field wave: What is its amplitude? Its direction of travel? Its vector direction in space (polarization)? (b) Express it in its real-time form, (z, I). What is the value of the ftee-space phase factor {Jo'? What is the waveleng~h A? (c) Find the associated magnetic field (z) well as the equivalent magnetic field (z)) in time-harmonic form. Express the H field in its real-time form, H;(z, t). (d) Sketch a labeled wave diagram in the manner of Figure 2-12, showing the real time fields E-; and H; at t = O.

E-;

B;

H;

2-42. The unit.orm plane wave electric field, E(,:) aJ;,~,(z) = axl50S,!lIoz Vim, is given in some region of space. Let the frequency of the source producing this wave be f = 150 MHz. Answer the questions asked in Problem 2-4·1 concerning this given traveling wave. 2-43. Begin with the other indepepdellt pair of Maxwell dillcrential equations (2-110a) and (2-111 b), involving the field-pair E y , Bx. Defining this plane wave to be '>-po/arized" in view of they-direction of its electric field, obtain the following. (a) Manipulate these equatiops to obtain a wave equation resembling (2-112) but in terms of the component E y • Express the solution of this wave equation in the mauner of (2-115). (b) Show that the corresponding magnetic-field traveling-wave solution can be expressed

Using 129), determine the equivalent expression lor Hx(z). (e) Use the results of (a) and (b) to establish the wave impedance ratios il'~ IH: and Ey- Ifr;. Compare them with the ratios applicable to the x-polarized case. (d) Sketch a labeled wave diagram suggested by Figure 2-12, showing only the forward-traveling real-time sinusoidal waves (Z, I) and H~ I) at the instant t O. Compare the results with the x-polarized case depicted in Figure 2-12, looking for similarities. (e) Sketch a wave diagram, this time showing only the real-time (Z, t) and H; t) at t = O.

E;

E;

SECTION 2-11 2-44. (a) Prove (2-131) for the linear polarization trace formed by cophasal Ex and Ey plane-wave components. (b) Prove (2-132) for the elliptical polarization trace of Figure 2-13(b). Show that it becomes circular polarization when the component amplitudes are equal. (c) Repeat (b), but let Ey lag Ex in time by 90°. Use a polarization diagram in the z 0 plane as suggested by Figure 2-13(b) to prove which or these two polarization cases has the electric field vector rotating cloekwise in time, and which counterclockwise (looking in the positivc-z direction. ) 2-45. Prove (2-133) for the polarization ellipse obtained whcnver Ex and Ey differ in phase by the general angl<' fl. Sketch a labcled polarization diagram in the Z = 0 plane fc)r this case as suggested by Figure 2-13(b). Comment on the analogy between this diagram and thc "Lissajou figures" observable with an oscilloscope on exciting its vertical and horizontal amplifiers with sinusoidal signals differing in phase. (As an added option, modify the three-dimt'llsional diagram in Figure 2-13 to illustrate the details of this polarization problem.)

2-46. A uniform plane wave has the time-harmouic electric field: E(z) = 500e - jiJoZ(a x jay) V 1m. (a) Write the real-time exprt'ssions in thez 0 plane. What kind of polarization exists here? Is it clock,:vise or counterclockwise (looking along + (b) Find the expression for the accompanying H(z) field.

_--------------------CHAPTER 3

Maxwell's Equations and Boundary Conditions for Material Regions at Rest

Materials in nature are invariably composed of atoms or arrangements of atoms into ions or molecules, each made up of positively and negatively charged particles having various configurations in empty space and varying states ofrelative motion. An electric or magnetic field impressed on a material exerts Lorentz forces on the particles, which undergo displacements or rearrangements to modify the impressed fields accordingly. The Maxwell equations that describe the electric and magnetic field behavior in a material are thus expected to require modifications from their free-space versions to account for whatever additional fields the material particles produce. It is the task in this chapter to diseuss these extensions of the free-space Maxwell equations. The topic of conduction is discussed from the viewpoint of a collision model. The chapter continues with a consideration orthe added effects of electric polarization within a material, providing a Maxwell divergence relation valid for materials as well as free space. Next is treated the added effect of magnetic polarization, yielding a suitably altered Maxwell curl expression lor the magnetic field. The field vectors D and H are thereby defined. Boundary conditions prevailing at interfaces separating difterently polarized regions are developed fi'om the integral forms of the Maxwell equations, to compare the normal components ofD and the tangential components ofH at adjacent points in the regions. The discussion continues with related treatments of the Maxwell div B and curl E equations for material regions, their integral forms, and corresponding boundary conditions. The chapter concludes with a discussion of uniform plane waves in a material possessing the parameters (J, E, and j1, exemplifYing the use of the Maxwell equations for a linear, homogeneous, and isotropic material. 3·1 ELECTRICAL CONDUCTMTY OF METALS

The electric and magnetic field behavior of material regions, solid, liquid, or gaseous, may he characterized in terms of threc effects.

111

112

MAXWELL'S EQUATIONS AND BOUNDARY CONDITIONS

Applied E ~

\'

\

Direction of acceleration

\

e

e

(Start) ~

(~

FIGURE 3-1. A representation of the production of a drift component of the velocity of free electrons in a metal. (a) A typical sequence of electron free paths resulting from collisions with the ion lattice. (b) Exaggerated view of the effect of drift in the direction of the acceleration due to an applied E field.

1. Electric charge conduction

2. Electric polarization 3. Magnetic polarization For large classes of materials, these effects are often adequately described through use of three parameters: (7, the electric conductivity; E, the electric permittivity; and /.1, the magnetic permeability of the materiaL These parameters will be defined in the course of the ensuing discussions. In terms of their charge-conduction property, materials may for some purposes be classified as insulators (dielectrics), which possess essentially no free electrons to provide currents under an impressed electric field; and conductors, in which free, outer orbit electrons are readily available to produce a conduction current when an electric field is impressed. An electrically conductive solid, commonly known as a conductor, is visualized in the submicroscopic world as a latticework of positive ions in which outerorbit electrons are free to wander as free electrons1-negative charges not attached to any particular atoms. On this structure are superposed thermal agitations associated with the temperature of the conductor --the light, agile conduction electrons moving about the more massive ion lattice, imparting some of their momentum to that lattice in exchange for new random directions of flight until more interactioIls occur. This circumstance is depicted in Figure 3-1 (a) for a typical conduction electroIl. The velocities of the free electrons are randomly distributed so that a mean velocity, averaged at any instant over a large number N of particles in the volume element,2 is given by Vd

I = -

N

N

I

Vi

m/sec

(3-1)

i=l

This quantity, called the drift velocity of the electrons, averages to zero in the absence of any externally applied electric field. lIn the atomic view, the free (condnction) electrons are those associated with the unfilled outer orbit, or valence band, of particular elements known as metals. 2 The volume-element used in characterizing the average velocity (3-1) is chosen sufficiently large that it contains enough ions and associated conduction electrons to yield a meaningful average, and yet it is taken small enough that the averaged velocity may be characterized at a point in the region. That a very large number of particles are present in a small volume increment is appreciated on noting that a typical conductor, sodium, possesses about 2.5 x 10 19 atoms/mm 3 at room temperature.

3-1 ELECTRICAL CONDUCTIVITY OF METALS

113

A mean free time, represented by the symbol To denotes the average interval between collisions in a volume element. When free electrons collide (interact) with the ion lattice, they give up, on the average, a momentum rrtvd in the mean free time Tc between collisions, ifm is the electron mass. Thus the averaged rate of momentum transfer to the ion lattice, per electron, is rrttfd/Te N of force. On equating this to the Lorentz electric field force applied within the conductor, one obtains (3-2) and solving tor

Vd

yields the steady drift velocity (3-3)

The expression (3-3), linearly relating the drift velocity to the applied E field, is of the form (3-4)

in which the proportionality constant Pe' taken to be a positive number, is termed the electron mobility, which from (3-3) is evidently eTc

2

m m 'IV-sec .

(3-5)

A high value of electron mobility is thus associated with a long mean free time T e • Making use of (1-50a) and multiplying Vd by the volume density Pv = -ne of the conduction electrons obtains the volume current density

J

(3-6)

with n denoting the free electron densi ty in electrons/m 3 . Eq uation (3-6) is an expression exhibiting a linear dependence ofJ on the applied E field in the conductor. Experiments show that this is an exceedingly accurate model for a wide selection of physical conductors. Equation (3-6) has the form of

J = O"E

(3-7)

sometimes given the name point fornl of Ohm's in which the factor 0" is called the conductivity of the region, having the units ampere per meter squared per volt per meter, or mho per meter. For the present model to which (3-6) applies, the conductivity is expressible as the positive number (3-8)

114


It is thus seen that both the electron mobility and the conductivity are proportional to the mean free time,
(3-9)

EXAMPLE 3·1. Find the mean free time and the electron mobility for sodium, having the measured dc conductivity 2.1 x 107 Dim at room temperature. Sodium has an atomic density of 2.3 x 10 28 atom/m 3 at room temperature, and with one outer-orbit electron available, n has the same value. Thus from (3-8), the mean free time becomes rna

1:

= -ne2 = c

(9.1 x 10- 31 ) (2.1 x 10 7 ) (2.3 x 10 28 )(1.6 x 10 19)2

=

33

X

.

10- 14 sec

Its electron mobility is found from either of the relations 2.1

X

10 7

---~~----~

=

5.7

X

10- 3 m 2 fV -sec

This implies from (3-4) the very slow drift velocity Vd = 5.7 mm/sec for an applied field of I V /m, emphasizing the sluggish, viscous nature of electron drift in a conductor.

The foregoing picture of direct current in a conductor is readily extended to the time-varying case, assuming that E varies slowly in comparison" to the mean free time,
(3-10) This differential equation has the complementary solution, assuming the initial condition Vd = VdO at t = 0, as follows: (3-11)

a transient solution denoting a decay or relaxation in the drift velocity on suddenly turning offthe applied field E. Thus the mean free time, <0 introduced into force relation (3-2), has acquired the interpretation of a relaxation time in the event of applying or removing an electric field from a conductor. The relaxation phenomenon furthermore occurs in an exceedingly short time for typical good conductors; thus, from Example 3-1 it was shown to be of the order of 10- 14 sec for a metal having a conductivity of about 10 7 O/m. The current density (3-7) is proportional to the drift velocity Vd, implying from (3-11) that current decays with time at the same rate on removing the E field. The differential equation (3-10) can be simplified ifE is assumed sinusoidal. Replacing E and Vd with the time-harmonic forms Eeirot and Vdeirot obtains, after cancelling the factor eiillt , the complex algebraic relation

3-1 ELECTRlCAL CONDUCTIVITY OF METALS

yielding the time-harmonic solution for

vd e

~

E

The complex current density due to this drift velocity is therefore, from

j

115

=_m __

EA/m2

J = Pvvd

(3-13)

The coefficient of E denotes the conductivity of the metal as in the de result (3-7), though now a complex quantity is obtained

8= __ m_V/m 1

.

(3-14 )

- + Jill Tc

However, for typical good conductors having a mean free time, To of the order of 10- 14 sec (Example 3-1), (3-14) reduces to the real, dc conductance result (3-8) [3-8] provided the angular frequency ill of the electromagnetic field is of the order of 1013 rad/sec or less (below the optical frequencies). Additional confidence is gained for this rather heuristic model of metallic conduction by experimental measurements made in the microwave range of frequencies, showing that the E and J fields in good conductors are in phase, implying that (f is real in the relationship J = (fE, even up to very high frequencies. The model of electrical conductivity just described is essentially that proposed by Karl Drude in 1900. The advent of quantum mechanics since that time has provided comprehensive techniques for describing, among other things, why the conductivities of various materials behave differently with temperature and how the vast range of conductivities of physical materials comes about-· -of the order of 10 8 V/m for the best conductors at room temperature to 10- 16 V/m for the best insulators-a range of some 24 orders of magnitude. The so-called band theory of solids, an outgrowth of quantum mechanics, is useful for describing the intrinsic differences among the conductors, semiconductors, and insulators. 3 3See T. S., Hutchison, and D. C. Baird. The Physics ~f Engineering Solids. New York: Wiley, 1968, for details.

116


3-2 ELECTRIC POLARIZATION AND DIY D FOR MATERIALS Insulators, or so-called dielectrics, incapable of carrying appreciable conduction currents under impressed electric fields of moderate magnitudes, are the subject of this discussion. The mechanism of the dielectric polarization effects resulting from applied electric fields may be explained in terms of the microscopic displacements of the bound positive and negative charge constituents from their average equilibrium positions, produced by the Lorentz electric field forces on the charges. Such displacements are usually only a fraction of a molecular diametcr in the material, but the sheer numbers of particles involved may cause a significant change in the electric field from its value in the absence of the dielectric substance. Dielectric polarization may arise from the following causes. 1. Electronic poLarization, in which the bound, negative electron cloud, subject to an impressed E field, is displaced from the equilibrium position relative to the positive nucleus. 2. Ionic polarization, in which the positive and negative ions of a molecule are displaced in the presence of an applied E field. 3. Orientational polarization, occurring .in materials possessing permanent electric dipoles randomly oriented in the absence of an external field, but undergoing an orientation toward the applied electric field vector by amounts depending on the strength of E. The tendency for the so-called polar molecules of such a material to align parallel with the applied field is opposed by the thermal agitation effects and the mutual interaction forces among the particles. Water is a common example of a substance exhibiting orientational polarization effects,

In each type of dielectric polarization, particle displacements are inhibited by powerful restoring forces between the positive and negative charge centers. In Figure 3-2 is illustrated the polarization mechanism in a material involving two species of charge. One should imagine thermal agitations superimposed on the average positions of the particles shown. If an external field E is impressed on the material, Lorentz forces EE = qE will be exerted on the positively charged nucleus and the negative electron cloud to produce displacements of both systems of particles. Displacement equilibrium is attained when the applied forces are balanced by the internal attractive Coulomb forces of the couplets. The moment Pi of the ith displaced charge pair in a collection of polarized dipoles as in Figure 3-2(a) is defined by Pi = qd i C'm

15 )

in which q denotes the positive charge of the couplet (q, q), and d i the vector separation of the couplet, directed from the negative to the positive charge, The average electric dipole morneHt per unit volume, called the electric polarization field and denoted by P, is defined by

(3-16) for a volume element ~v containing N electric dipoles. If no E field were applied to the material, no dipoles would be induced in the case of electronic or ionic polarization; even if the material were polar (containing permanent dipoles), their orientations

3-2 ELECTRIC POLARIZATION AND

6

~/

(8) ~/

(0, 6) (0'1 '-.:"" (0 to) (0) (0) "'''''

"'''''

(0) '-.-/

(0) '-0:/

{)l (0'1 (",,/, 'C:::/

'-0/

'-C/

(0) 'C'l

(OJ

'C/

(

,,~

'C"

(~0

(0)

()\ I

(0' "

No E field applied

D FOR MATERlALS

117

(0 (0 CD (0 E0 (0 (0 (0 CO

"''l

'-0"

mv

(~B

I

CD CD (0 (0 E field applied

Electron 'r;-.. Positive cloud I--,:!;; nucleus

-q

CO CD CD

'-Ud, +q (a)

\

\

p ,/

I

I

/ p . --.,..

/-:...

/'

~ /_~ \,

/" Random orientations of a small sample of electric dipoles, no E field applied

"'"

__ lPj = Pdv

Orientations influenced by applied E field, to produce P

(b)

FIGURE 3-2. Electric polarization eflects in simple models of nonpolar and polar dielectric materials. (a) A nonpolar substance. (b) A polar substance (H 2 0).

would under usual circumstances be random as illustrated in Figure 3-2(b), in which case the numerator of (3-16) would sum to zero to make P = O. IfE were applied in the x direction as shown, a net component ofP would be induced. If p + and p _ denote the densities of the positive and the negative charges that constitute the dielectric material, (3-16) can be written N

.2:

P

[=1

N

Pi

.2:

[=1

N

qidi

Nq.2: d i ,=1

--=---=---=p+

I1v

I1v

I1v

N

d

(3-17)

in which p + = Nq/l1v is the density of only the positive charges comprising the dipolefilled region, and d denotes ('LdiljN, the dipole displacement averaged over the N dipoles in Av. An examination orthe polarization field P = p+d of (3-17), characterized as the vector in Figure 3-3(a), reveals the establishment of a bound charge excess within I1v, giving rise to a so-called polarization charge density, wherever P has a divergence. Since, from (3-17), P is the positive charge density times the vector displacement d of the positive charge cloud with respect to the negative charge cloud, the definition (2-20) reveals that the divergence of P amounts to the limit of a net, positive charge

118


Py

(a)

(b)

FIGURE 3-3. Relative to (3-21), {Iv --div P. (a) Polarization field P at a volume-element in a dielectric. (iI) Efkcl of nonnniformity of Px , leaving an excess of negative polarization charge within Av.

out of the volume ~ V, divided by ~ V. In other words, div P becomes a negative, polarization-created, effective charge density, given the symbol pp. A formal proof of the observation that div P = - PP' relative to Figure 3-3, follows. Consider a typical volume element ~v = ~x ~y ~z in a region containing, in general, a nonuniform polarization field P as shown in Figure 3-3.(b). The x component, 1\ p accounts for a net, positive bound charge passing through the left-hand face 8 1 into ~v, amounting to

(3-18a) while through the opposite side S'1> the positive bound charge commg out of expressed

~v

is

(3-18b) A net, negative, polarization bound charge therefore remains inside the difference of (3-18a) and 18b), or

~v,

amounting to

19)

With similar contributions over the other two pairs of sides, one obtains the total, negative hound charge remaining inside ~v (3-20) a measure of the net, bound charge excess Pp~v within ~v, if.p p denotes the volume

ELECTRIC POLARIZATION AND IllV ]) FOR MATERIALS

density of the polarization view of (2-29a)

excess. Equating

119

to Pp liZ! thus obtains, in

(3-21 )

Pp is thus a "negative, effective charge density" created by the dielectric polarization process, whenever the polarization density field P has a divergence. Thc divergence of EoE, inIree space, has been given by (2-39) to express the density Pv of fi'ee charge. By (3-21), the polarization-charge-excess developed in a material, by the nonuniformity ofthe polarization density field P, is seen from its divergence property (3-21) to contribute the added, negative, effective charge density Pi' = div P, a bound-charge excess over whatcver free charge of density Pv may exist in the polarized material. The divergence of EoE in a material in general, then, becomes (3-21) with the effective polarization charge density Pp added in, whence div (EoE) Pv + Pp = Pv div P, to yield

v . (EoE + P)

=

(3-22)

p"

a divergence expression t()[ E in a material region. A more compact version is obtained using the abbreviation D· for (EoE + P) as follows

to permit writing (3-22) in the preferred form

V'D

= Pv C/m 3

(3-24)

Experiments reveal that many dielectric substances are essentially linear, meaning that P is proportional to the E field applied. For such materials 4

PocE (3-25) in which the parameter Xc is called the electric Eo is retained in (3-25) to make Xe dimensionless. Then

V' [(1 Comparing

+ Xe)EOE] = Pv

becomes (3-26)

with (3-24) shows that the bracketed quantity denotes D, that is,

4S trictly speaking, the expressions arc static (or forms. These results are more usefully written in phasor {()nns, with field quantities replaced with the phasors P, E. XC' Pv, D, and E,. Thus. with E complex in (3-30c), {t)r cxampk, D = EE shows D and E are in general out of phase.

120


I t is usual to denote I

+ Xe

by the dimensionless symbol (3-28)

Er is called the relative permittivity of the region. Finally, choosing the symbol E, called the permittivity of the material, to denote (1 + XelEo as follows (3-29a) (3-29b) permits writing (3-27) in the following successively more compact forms

D = (1

+

(3-30a)

XelEoE

(3-30b) (3-30c) In free space, Xe = 0, to reduce (3-30) properly to D in the form

= EoE.

E \

Also, expressing (3-29b)

(3-31 )

\

emphasizes that Er denotes a material permittivity relative to that of empty space. To summarize, note that Maxwell's relation (3-22) or (3-24 1 is expressible in any of the equivalent forms

V· [(1

+ XelEoE]

Pv

(3-32a) (3-32bl

V· (EEl

=

Pv

V· D = Pv Cjm 3

(3-32c) (3-32d)

No dielectric material is strictly linear in its electric polarization behavior, though many are very nearly so over wide ranges of applied E fields. If E is made strong enough, a material may experience polarization displacements that result in permanent dislocations of the molecular structure, or a dielectric breakdown, for which case (3-25) does not hold. In a nonlinear material the magnitude of D is not proportional to the applied E field, (though the E and the P vectors may have the same directions). Then (3-25) is written more generally (3-33) in which the dependence of Xe on E is noted.

3-2 ELECTRIC POLARIZATION AND DlV D FOR MATERIALS

121

A. Dielectric Polarization Current Density If the electric field giving rise to dielectric polarization effects is time-varying, the resulting polarization field is also time-varying. Then the displacements of the positive charge constituents in one direction, together with the negative charges moving oppositely, give rise to charge displacements through cross sections of the material identifiable as currents through those cross sections. Applying a time-derivative operator to the Pi terms of (3-16) thus yields a current density interpretation as follows

I ;=1

api

at

(3-34)

The resulting time derivative of the polarization field, current density, is given the symbol jp as follows

j

aplat, having the units of volume

ap

p

=-A/m2

at

(3-35 )

and is called the electric polarization current density. The field jp, along with the polarization charge density field Pp described by (3-21), acts as an additional source of electric and rnagnetic fields. In particular, the special role played by jp in relation to magnetic fields in a material region is discussed later in Section 3-4.

B. Integral Form of Gauss's Law of Materials The dielectric polarization effects attributed to material regions have been seen to lead to the divergence expressions (3-21) and (3-24-), relating the field quantities p and D to the polarization charge and free charge sources. The divergence theorem can be used to transform these differential equations into corresponding integral forms. The most important of these is (3-24) for the D field; that is, V . D = Pv' Multiplying both sides of by dv and integrating throughout an arbitrary volume region V yields

fv V • D dv = fv Pv dv

(3-36)

By the divergence theorem (2-34), the left side can be replaced by a closed-surface integral to yield

~s D . ds =

fv P dv C v

(3-37)

in which S bounds V. Equation (3-37) is the integral form of Maxwell's equation (3-24) for a material region, sometimes called Gauss's law for material regions. It states that the net outward flux of D over any closed surface is a measure of the total free charge contained by the volume V bounded by S, at any instant of time. As expected, it becomes the free-space Gauss's law (1-53) if Xe = 0, reducing D to EoE.

122


Pp of (3-21), has the equivalent integral

Another divergence relation, V . P = form

(3-38) obtained by the method analogous to that used in converting (3-24) to Gauss's integral (3-37). Equation (3-38) states that the net outward flux ofP emanating from the surface of V is a measure of the net polarization charge summed throughout V.

C. Spatial Boundary Conditions for Normal D and P In many electromagnetic field problems ofphysical interest, it becomes necessary to discuss how the fields behave as one traverses the boundary surfaces, or interfaces, separating the various material regions that comprise the system. In such problems, a matching or fitting of the field solutions is required so that the boundary conditions at the interfaces may be satisfied. The proper boundary conditions for the fields are determined, as will be shown, from the integral forms of Maxwell's equations for material regions. The Maxwell integral relation (3-37), D· ds = pvdv, can be used, through an appropriately constructed closed surface, for comparing the normal components ofD that appear just to either side of an interface separating two materials of different permittivities. Denoting the materials as region 1 and region 2 with permittivities E1 and E 2 , define a pillbox-shaped closed surface of small height (5h and end areas 1\s so that both regions to either side of the interface are penetrated as in Figure 3-4. Calling the fields Dl and D2 at points just inside regions 1 and 2, respectively, the application of the left-hand integral of (3-37) to the closed pillbox yields the net outward flux from the top and bottom surfaces 1\s. At the same time, the right side is the charge enclosed by tQe pillbox; this is Pv1\s(5h, so (3-37) becomes

fs

Iv

(3-39) The right side of (3-39) vanishes as (5h - 0, assuming Pv denotes a volume free charge density in the region. If, however, a surface charge density denoted by Ps and defined by the limit

P.

=

lim PvJh

(3-40)

~h-+O

Region 1: (tt) As

Region 2: (€2)

D2 (a)

(b)

FIGURE 3-4. Gaussian pillbox surface constrncted for deriviug the boundary condition on the normal component ofD. (a) Pillbox-shaped closed surface showing total fields at points adjacent to interface. (b) Edge view of (a), showing fields resolved into components.

3-2 ELECTRIC POLARIZATION AND DIV D FOR MATERIALS

123

is present on the interface, (3-39) reduces to the general boundary condition

(3-41 )

Equation (3-41) means that the normal component oj D is discontinuous to the extent oj the jree surface charge density present on the interface. Since Dnl = n . Dl and Do2 = n· D z , with n denoting a normal unit vector directed from region 2 toward region 1 as in Figure 3-4(b), (3-41) is written optionally in vector notation as follows.

(3-42)

The boundary condition (3-41) is true in general, but for some physical problems a free surface charge densi ty Ps may be absent. Two special cases of (3-41) of physical interest are mentioned in the following, while a more general result is left for discussion in Section 3-11.

A. Both regions perfect dielectrics. A perfect dielectric, for which the conductivity (J is zero, cannot furnish free charges, so that if no excess charge is supplied to the interlace by an external agent (rubbing it with eat's fur, for example), then Ps = a on the interface. Then (3-41) reduces to

CASE

(3-43)

The normal component oj D is continuous at an interface separating two perfect dielectrics, as illustrated in Figure 3-5(a).

Region 1: (0-1

= 0; fl)

Region 1: (El)

Region 2: (0-2 = 0; <2) (a)

(b)

FIGURE 3-5. Two cases of the boundalY condition lor normal components of D. (a) Continuous D. at an interface separating perfect dielectrics. (b) r;quality of normal D. to a surface charge dcnsi ty on a perlect conductor.

124 CASE


B.

One region is a perfect dielectric; the other is a perfect conductor.

Electric currents are limited to finite densities in the physical world. Thus from (3-7), J = O"E, the assumption of a pe~lect conductor in region 2 of Figure 3-4 (0" z -4 CIJ), implies that E z in that region must be zero if the current densities are to have, at most, finite values. Moreover, with electromagnetic fields satisfying (2-108), V x :E = -jroB, one can see that if E z is zero in region 2, then :8 2 must be zero there also. Thus for time-varying fields, implies

E=B=O

(3-44)

in a perfect conductor. The boundary condition (3-41) or (3-42) then must reduce to Dnl = p" or in vector form

n' D = Ps C/m 2

(3-45)

The su~lace charge density residing on a pe~fect conductor equals the normal component of D there, as illustrated in Figure 3-5(b). In a static field problem involving only fixed electric charges and no static currents, the boundary condition (3-45) holds true even though region 2 may be only finitely conducting, for the assumption of no static currents in the finitely conducting region 2 implies from (3-7) that E z = 0 there, making D2 = 0 as well. Thus (3-42) reduces to (3-4,5). A boundary condition similar to (3-41) can be derived comparing the normal components of the dielectric polarization vector P. Noting the similarity of Maxwell's integral law (3-37) and the polarization field integral (3-38) and using another pillbox construction, one can show that Pn1 - P n2 = Psp' or in vector form

(3-46) !

in wKich Ps p denotes the net surface bound charge density lying within the pillbox. The net density includes the eHect of both species of surface polarization charge (positive and negative) accumulated just to either side of the interface. A simpler picture is obtained if'region 1 is free space, for which Xe1 0 (or E 1 = Eo). Then P 1 = 0, reducing to the special case (3-47) The sUijace polarization charge density residing at a free-space-to-dielectric interface equals the normal component of the P field there. EXAMPLE 3-2. Two parallel conducting plates of great extent and d m apart are statically charged with ±q C on every area A of the lower and upper plates, respectively, as noted in (a), The conductors are separated by air except for a homogeneous dielectric slab of thickness c and permittivity E, spaced a distance b from the lower plate. (a) Use Gauss's law to establish D in the three regions. Sketch the flux ofD. (b) Find E and P in the three regions and show their flux plots. (c) Determine Ps on the conductor surfaces, Pp in the dielectric, and Pps at y band y b + c.

3-2 ELECTRIC POLARIZATION AND DIV D FOR MATERIALS

125

(y)t

I I

I

I

d

L___~h'

~ ~charged Negatively surface

Conductor

I

I I

(-q/A)

b+Cr--r I c b~-

_L~~~________~~+-

Dielectric slab

__~~__~~

~~

I

I

charged surface

I

(+q/A)

--oi---\!1~~~~:t;--c;=~~=;=:;-=-=!~l-------(:) I

I (a)

Dt Dt t+-+-+-f-+-+-1r-rt-+-t-t-h11 II

E t'r-'-r""""".,---,-rr.,---,'-r-r--r'

pt ltttft~~+t!

Et

yt + + + + + + + + + + -+

\

-P'P

+t!m~77:~~~~~?i7m

(d)

(c)

EXAMPLE 3-2. (a) Charged parallel conductor system. (b) Flux of D. (c) Flux of EoE. (d) Flux ofP.

(a) E exists only between the conductors and by symmetry is independent of x and

z.

A Gaussian closed-surface S in the form of a reetangular box is placed as in Figure 1-15(d), to contain the free charge q. With static E inside the conductor zero, aD flux of a constant density emanates from the top ofS, making the left side of Gauss's law (3-37) become

Dy

r

JS(top)

ds

DyA

I:':quating to the right side of (3-37), the free charge q = DyA, whence

D=aD =aY J.. y Y A

(I)

a result correct for all three regions between the conductofs because no free charge exists in Of on the dielectric. The flux plot of D is shown in (b). (b) E is obtained using (3-30c), so in the dielectric slab,

D E=-=a E

YEA

b
(2)

126


whereas in the air regions it is

D

q

E=~=a Eo Y EOA

o <)! <

b+ c
band

(3)

Since E > Eo for a typical dielectric, E in the air regions exceeds the value in the dielectric, as shown in (c). P in the dielectric is found by use of

P

D

q(~)

a Y

A

Er

(4)

For Er > 1, P in the slab is positive.y directed, as shown in (d). In air, P is zero. From (3-35), no polarization current density JP is established in the dielectric because the fields are time-static. The free charge densities on the conductors are obtained from (3-45), yielding Ps = ±q/A. The polarization charge density Pp from (3-21) is zero because P is a constant vector throughout the slab. The surface polarization charge density Ps p is tound by inserting (4) into (3-47), yielding (5)

These surface densities are noted in (b) and (d) of the figtire.

3-3 DIY B FOR MATERIALS: ITS INTEGRAL FORM AND A BOUNDARY CONDITION FOR NORMAL B In Section 3-2 the Maxwell relation for V • D in a material was developed by adding the effect of the electric polarization charge density Pp to the free-space Maxwell relation. Thc form of the expression for V • B ill a material can be developed analogously. N0iadditive term is required in this casc, however, because no free magnetic charges cxist physically in any known material. Thus B remains divergenceless in materials; that is,

V' B = 0 Wb/m 3

(3-48)

Equation (3-48) is converted to its integral form using a technique analogous to that employed in obtaining (3-37). Multiplying both sides of (3-48) by dv, integrating it throughout an arbitrary V, and applying the divergence theorem

#SB'

ds = OWb

(3-49)

1 c ]

3-4 MAGNETIC POLARIZATION AND CURL H FOR MATERIALS

127

the form of (3-49) states that the net outgoing Hux orB over ally closed surface is always zero, implying that B flux always forIns dosed lines. A boundary condition concerned with the normal components ofB and analogous to (3-42) can be j()Und by applying (3-49) to a vanishing Gaussian pillbox like that of Figure 3-4. The resulting boundary condition is

(3-50) that is, the normal comt)onent or the B .field is continuous at an interface separating two adjacent


The magnetic properties of a material are attributed to the tendency for the bound (un-mis, circulating on an atomic scale within the substanee, to align with an applied B field. Three types of bound currents are associated with atomic structure: those attributed to orbiting electrons, and those associated with electron spin and with nuclear spin. Each of these phenomena, represented in Figure 3-6(a), is equivalent to the circulation of a current 1 about a small closed path bounding an area ds, the positive sense of which is related by the right-hand rule to the direction of 1 as in Figure 3-6( b). The product Ids defines the magnetic moment tn contributed by those bound currents of the atomic or molecular configuration. J t is shown that applying an external magnetic fIeld B to the typical moment tn = I ds yields a torque exerted on tn, lending to align tn with the applied B field. One can in this manner explain the magnetic behavior of a malerial as though it were a collection, in empty space, of many magnetic moments tn per unit volume. The tendency to align with the applied B fidd is shown to provide an equivalent magnetization current of density Jm, serving to modify the magnetic field in a certain way. A desuiption of this process, beginning with it discussion of the torque produced by the B field on a current element, follows. A current loop of microscopic size has an external magnetic field behavior independent of its shape in a plane, so a square loop is assumed in lieu of the circular conflguration of Figure 3-6(h). It is shown in Figure 3-7 (a) in the z 0 plane, immersed in the applied field B = axBx + ayBy + azB z . The Lorentz force acting on each of the four edges of the square current loop is obtained from (I-52)

dF E

= dqv x B N

(3-51 )

Orbital motion vector ...

Electron spin

m= Ids

NU~I:~;o~Pin

vect? ( ;I 'J bor, Electron

----:ti - e

orbital motion

'

~~

Nucleus

(a)

I

LdS

~ (b)

FIGURE 3-6. The clements of bound currents that exist in atomic structure. (a) Constituents of circulating currents associated with particles of a simpk atom. (b) Magnetic moment III of a current I circulating about an area ds.

128

MAXWELL'S EQUATIONS AND BOUNDARY CONDITIONS :(Y)

.

r<,-:-; -~ _1- ------,1

I

....

-

1

,:

(

dq.,=Idf ,

By:

1-;:)

~

dY',II.;

v

:'

.'

ds

I! II 'I

II j'

C

'.

I..:--..:::.--~_~-J:

dx -"

ds = azdxdy (a)

(b)

(c)

FIGURE 3-7. Development of torque expression for a current loop immersed in a B field. (a) Current loop immersed in arbitrary B field. (b) A moving charge element, dqv, of the loop. (c) Development of torque dT produced on edge dl, .

if the charge dq moves with a velocity v along the edges dx and dy. One may cast (3-51) into the following forms, noting that dq = p"dv p"dt ds from Figure 3-7(b), and usi ng (1-50a)

[J dt ds] x B

I dt x B

(3-52)

with the direction denoted by assigning a vector property to each edge length dt. The origin of the torque arm R is for convenience taken at the center of the loop. Along t l , the differential torque dT 1 is given by Rl x dF B (Example 1-4), yielding

ax

1B dxdy y

2

with the same result obtained for edge t 3, while that acting on t z and t4 becomes + dT 4 ayIBxdxdy. Thus the torque on the complete loop becomes

dT 2

dT

(a z x B)Jds

l(a z ds) x BIds

X

B

and with 1 ds denoting the magnetic moment (3-53) one may abbreviate the result dT=tnxBN·m

(3-54)

It is clear from (3-54) that only the components of the applied B field in the plane of the current element act to produce a torque on it. Htn and B were parallel, dT would become zero; thus the torque dT is such that it tends to align the current element with the applied B field.

A very large number of current loops like those of the atomic model in Figure 3-2 comprise a magnetic material, susceptible to such magnetic alignment eHects. In the absence of an applied B field, they possess random orientations accompanied by thermal

3-4 MAGNETIC POLARIZATION AND CURL H FOR MATERlALS

129

~ N

j

::::Em;: Mdv i=l

(a)

(b)

FIGURE 3-8. Current loop constituency or a magnetizable material, alTected by an applied B field. (a) Random magnetic moments, in the absence o[ B. (b) Partial alignment of magnetic moments, B applied.

agitation effects, as depicted in Figure 3-8(a), if one may avoid the su~ject of permanent magnetism occurring in some materials. Impressing a B field develops a torque on each current loop, as specified by (3-54), such that the loops tend to align more or less in the direction ofB as depicted in Figure 3-8(b). The magnetization density M is defined in essentially the way the dielectric polarization field P is defined by (3-16), that is, by summing the magnetic moments In within a volume-element Av and expressing the sum on the per-unit-volume basis N

LIn;

M=~A/m

Av

(3-55 )

This becomes a smooth functional result if the number N of current elements within Av is quite large, while Au is yet small enough to be considered suitable for manipulation in differential or integral expressions. Thus M furnishes a characterization of the circulating atomic currents within matter from a smoothed-out, macroscopic point of VIew.

An important derivative fi.mction of the magnetization field M is its curl, shown in the following to yield a volume density 1m of uncanceled bound currents within a

130


rnagnetic material according to

(3-56)

A formal derivation of (3-56) proceeds with the aid of Figure 3-9. The examination of an incremental volume-clement of a material, depicted in Figure 3-9(a), reveals the presence of surface current contributions on bov as in (b) of that figure, assuming for the present that only the effects of the z component of M are considered. If two such volume increments are considered side by side as in Figure 3-9(c), then the bound surface currents along their common sides, with densities designated by ]sm,y, cancel partially to produce a net upward flow of current in the region given by

This current passing through the cross-sectional area box boZ is depicted by the bold arrow in the figure. They component of the bound current density 1m through box boz

(y)

(y)

Jsm,x=-M z

(z)

(z)

(x)

M

J::. trI ,y= -Mz

(b)

(a)

ll.velements (separated to show J sm , y and J;m,y)

(x)

(c)

(d)

FIGURE 3-9. Relative to Jm = V x M. (a) Bound current clements producing surface currents on 8v. (h) Bound surface currents smoothed into rectangular components, assuming M z only. (c) Net volume current All through 8x Az: the difference of bound surface current densities. (d) The other contribution to the Jm" component.

131


is thcn AldAx Az oMz/ox. Anothcr contribution, shown in Figure 3-9(d) is obtained from the x component ofM in the vicinity ofthc point; it contributes the density oMx/oz through Ax Az. The totaly component ofJm therefore becomes Jm,y = 8Mx/8zoMz/ox, which from (2-52) is evidently they component of curl M. A similar development yields the other components Jm,x and Jm,z of Jm, obtaining (3-56)

Jm

ax

ay

az

0 = ox Mx

8 oy My

0 =V xM oz Mz

[3-56]

The significance of (3-56) in revealing the presence of "olume currents inside a material whenever its interior is nonuniformly magnetized is des@1Jed inanexample to follow. A side effect is the pr~se~nceotsUrface currentdensities Jsm established by M on the surface of the material.

EXAMPLE 3-3. Suppose a B field is applied to a cube of magnetic material, h m on a side, such that M is z-direeted and. varies linearly with x according to M = a z lOx A/m, as shown in (a). Find the magnetization current density Jm in the material, as well as the surface magnetization current density. Sketch the bound current fields in and on the cube. The magnetization current density Jm is obtained from (3-56)

Jm = V X M =

ax 0

ay 0

az iJ

0

0

lOx

2 ox oy oz = -aylO A/m

(I)

negative y-directed and of constant density as in (b). The uncanceled segments of the bound currents at the surface of the block constitute a surface density of magnetization currents denoted by Jsm (A/m). On the end x = b, Jsm is y-directed and has a magnitude equal to that of M there; that is, (2)

Surface bound current flux

( z)

(x)

J m = 'i7xM

= - aylO (a)

(b)

(c)

EXAMPLE 3-3. (a) Material sample magnetized linearly with increasing x. (b) Volume magnetization currents produced by transverse variations or M. (c) Surface currents produced by uncanceled segments of bound currents,

132


while on the top and bottom of the block -a x

J sm] y=O

This is to mat!

M]

p=M

:z y=b

axMz]y=o

=

a,JOx A/rn

(4)

No bound currents exist on the end at x = 0, since M = 0 there. These surface effects are shown as flux plots in (c).

The modeling of the bound currents in Figure 3-9 reveals that, on any sampling cube ,1v, the surface bound-current densities J8m are oriented perpendicularly with respect to the local M field. It is therefore evident that J8m on any surface element of ,1v can bc found from the cross product of that M with the normal unit veCtor emerging from the surface. Thus,

Jsm =

(3-57)

-n X M

In Example 3-3, on the surface y = b of the magnetized block, with n = ax and M = azlOb there, one obtainsJsm = - 0 X M = -ax X a)Ob = aylOb, which agrees with thc result (2) obtained in that example. The curl of B/flo in free space has been expressed by (2-63) as the sum of a convection or a conduction current density J plus a displacement current density a(EoE)/at at any point. Two additional types of current densities occur generally in materials: J P = ap/at of (3-35) and Jm = V x M of (3-56), arising from dielectric and magnetic polarization effects, respectively. Adding these together accounts for the total current density at any point, yielding a revision of (2-63) for a material region.

V

a(E E)

ap

at

at

J+_o_+

X (:)

+VxM

Grouping the curl terms and the time-derivative terms together obtains

VX

(.! - M) = J + _o(_Eo_E..,...+_P_l at flo

Recalling from (3-23) that EoE + P defines D, and further abbreviating B/flo (3-57) by use of the symbol H, sOrhetimes called the magnetic intensity field

H

B

-MAjm

(3-58a) Min

(3-58b)

flo permits writing (3-58a) in the compact form

(3-59)

I Mpro gouS 1 prove! H as I

in wI (3-6(


133

This is the desired Maxwell curl expression for the field H defined by (3-58b), applicable to material regions. Note that it properly reduces to its free-space form (2-73) on setting P;::M;::O. In a linear region possessing a magnetization M, one might be inclined to express M proportional to the B field in the material (i.e., M IX B) to provide a result analogous to (3-25) for a linear dielectric (P IX E). Historically, however, this has not proved to be the assumption used; instead, it is customary to set M proportional to H as follows: MIXH (3-60) in which the dimensionless Xm is called the magnetic susceptibility of a material. Inserting (3-60) into (3-58b) therefore yields

B

H = - - M flo

B

=- flo

XmH

which, on solving for B, obtains (3-61 ) The quantity (1

+ Xm),

abbreviated fl" flr;:: I

+ Xm

(3-62)

is called the relative permeability of the material. Further choosing the symbol fl, called the permeabili~y, to denote the product (3-63a)

Ii;:: flrlio HIm

(3-63b)

permits writing (3-61) in the compact form for linear materials (3-64a) (3-64b) B

= ,uHWbjm2

(3-64c)

One should note the analogy of the steps yielding (3-64c) to those leading to (3-30), connecting D and E for linear, electrically polarized materials. It is seen from (3-64b) that the relative permeability expresses the permeability of a material relative to that of free space, ,uo, if one writes (3-65) This is evidently analogous to

(~31),

the expression for the relative permittivity Er .

134


A. Integral Form of Ampere's Law for Materials Maxwell's curl relation (3-59), V X H = J + (aD/at) can be transformed into an integral relationship by using Stokes's theorem. Forming the dot product of (3-59) with ds and integrating over any surface S bounded by the closed line t yields

f Js

Js J' ds

(V x H) 'ds = f

+!!..-dt Jsf D • ds

From Stokes's theorem (2-56), the left side can be expressed as an integral of H • dt over the closed line t bounding S, assuming H suitably well-behaved; thus

1

~

H' dE

=

1J . + !£l ds

s

dt s

D . ds A

(3-66)

the desired integralform of Maxwell's differential equation (3-59). Equation (3-66) is also known as Ampere's circuital law for materials. I t states that the net circulation ofH about any closed path t is a measure of the sum of the conduction (or convection) current plus the displacement current through the surface S bounded by t. Another curl relation, (3-56), Jm = V X M connecting the magnetization field M with a volume magnetization current density, was treated in the last section. It has an integral form analogously obtainable by use of Stokes's theorem, becoming,

(3-67) This means that the circulation of the M field about a closed path t is a measure of the net magnetization current through it. For example, a surface integration of Jm over a cross section in the x-z plane of the magnetized cube in Example 3-3 is seen to yield a bound magnetization current 10b 2 A flowing vertically through the specimen, also obtainable from a line integral of M • dt around a horizontal perimeter of the cube.

B. Boundary Conditions for TangenHal Hand M In a manner resembling the derivation of the boundary condition (3-41), one can compare the tangential components of H adjace'nt to an interface separating two materials, by applying Maxwell's integral law (3-66) to the small, rectangular closed line t shown in Figure 3-10. With the magnetic fields in the adjacent media labeled HI and H2 and resolved into normal and tangential components as in Figure 3-10, integrating the left side of (3-66) clockwise around t yields Htl L'1t - Ht2 L'1t, if the height bh is taken so small that the ends do not contribute to the line integral. The right side of (3-66) involves integrations ofJ and D over the vanishing surface S bounded by t, obtaining

(3-68) if In and Dn denote the components normal to L'1s. The last term of (3-68) vanishes as (jh -+ 0; similarly, the contribution of the In term would also vanish if J were a volume


135

Region 2 (J.Lzl FIGURE 3-10. Rectangular closed line I: constructed to compare Ht! and H,z using Ampere's law.

current density. In some physical problems, however, one can assume a free surface current flowing solely on the interface with a density Js defined by

Js =

lim

J oh

(3-69)

3h~O

(It develops that 1. is of interest only if one of the regions is a perfect conductor, a case to be discussed shortly.) Thus, the general boundary condition resulting from the substitution of (3-69) into (3-68) becomes

(3-70a)

in which the subscript (n) denotes a surface current flowing normally through the side of the rectangle, as noted in Figure 3-10. Equation (3-70a) states that the tangential component of the H field is discontinuous at an interface to the extent of the surface current density that may be present. Using n to denote a normal unit vector directed from region 2 toward region 1 as in Figure 3-11, a vector form of (3-70a) is written

(3-70b)

to include direction as well as magnitude information. The boundary condition (3-70) is true in general, though in its application to a boundary-value problem, it becomes two cases.

136


(a)

(b)

FIGURE 3-11. The two cases of the boundary condition (3-70b) on tangential H,. (a) Continuous H, at interface separating regions of finite conductivities. (b) Equality of the H, and surface current density on a perfect conductor.

A. Both regions have finite conductivities. In this case, free surface currents cannot exist on the interface, reducing (3-70a) to

CASE

(3-71 )

Thus, the tangential component ofH is continuous at an interface separating two materials halling, at most, finite conductivities. This boundary condition as illustrated in Figure 3-11 (a).

B. One region is a perfect conductor. From (3-44) it has been noted, under time-varying conditions, that no electric or magnetic field can exist inside a perfect conductor. ]f region 2 were a perfect conductor, then H2 = 0 reducing (3-70a) to Htl = Js(n); or in vector form, (3-70b) becomes

CASE

n X HI =

JsA/m

(3-72)

the boundary condition depicted in Figure 3-11 (b). At the interface separating a region from a perfect conductor, the surface current densify Js ~as a magnitude equal to that of the tangential H there, and a direction specified by the right-hand rule. It is shown later that no normal component of H or B may exist at the surface of a perfect conductor, implying that the tangential magnetic field is also the total magnetic field there. A similarity in form is noted between Ampere's circuital law (3-66) and the relationship (3-67) for M. Thus, by analogy with the boundary condition (3-70a), derived by applying (3-66) to the closed rectangle as in Figure 3-10, one may establish from (3-67) the boundary condition Mtl -

Mt2

= Jsm(n) A/m

(3-73a)

This result expresses the continuity of the tangential component of M as one traverses an interface between two adjacent, magnetized regions. The subscript (n) denotes a


137

surface magnetization current density normal to the tangential M components at the boundary. The vector sense of the surface magnetization current density J8m is included in the boundary condition (3-73a) by expressing it n X

(Ml

M 2)

a result analogous with (3-70b). Ifregion I is nonmagnetic, then Ml

= J8m A/m

(3-73b)

= 0, reducing (3-73b) to (3-57) (3-74)

An illustration of the latter has already been noted in parts (b) and (c) of the figure accompanying Example 3-3.

EXAMPlE 3·4. Suppose a very long solenoid like that of Figure 1-21(b) eontains a coaxial magnetic rod of radius a, as in figure (a), the rod having a constant permeability /1. The winding is closely spaced with n turns in every d m of axial length, carrying a steady

(a) 000 000 000 0

0

® ® 0 ® 0 ® 0 0 0 ® 0

o

0 0 0

o0

000 0 0 0

0 0

0 0

(b)

(c)

000 0

o

0

0 0

0 0

0 0

0 0 0

0 0

0 0

0 0

0

(d)

J,m=-n x M

=a
Flux of J sm (e)

EXAMPLE 3-4. (a) Solenoid with magnetic core. (b) H field flux. (e) B field flux. (d) M field flux. le) The J,m field on the iron.

138


current /. Determine Hand B in the air and iron regions, making use of Ampere's law (3-66) and symmetry. (b) Find M in the rod, and determine whether any volume magnetization current density 1m exists in it, as well as magnetization current densities on its surface. (e) Sketch the flux ofH, B, and M in the air and iron regions. (a) With dc in the wire producing time-static fields, (3-66) becomes ~tH' dt Ss 1 . ds. From the axial symmetry and the implications of Ampere's law in relation to the current sense, H is positive <: directed within the winding and essentially

zero outside it. Constructing the rectangular closed path t shown, H . dt integrated between PI and P 2 yields

nl

;tH'dt in which Hz is constant over the path PI to P z , yielding

nl

H=d

(I)

Z

This result is correct in both the air and iron regions because nl is the current enclosed by t regardless of whether P 1 and P 2 fall within the air or the iron. The turns per meter in the winding are denoted by n/d. The corresponding B field is obtained from (3-64c)

0< p < a

B

B = J1.oH = a z

J1.on/ d

-

Iron

a < p < bAil'

(b) The volume magnetization field M is zero in air; in the ferromagnetic region it is given by (3-60)

M

(3)

M is constant in the iron rod for this example, yielding 1m = 0 from (3-56). The surface magnetization current density, however, is determined from (3-74), calling the iron region 2. With n = a p on the interface

nl aXm d

(4)

(c) Sketches of the H, B, and M flux-fields are shown in (b), (c), and (d) of the accompanying figure. It is seen from (b) that H is the same in the air as in the iron for this example; thus the boundary condition (3-71) is satisfied. The consequence is that B is IIr times as strong in the iron as in the adjacent air region. Finally, 15m has a uniform surface flux density on the iron rod as shown in (e).

EXAMPLE 3-5. Obtain a refractive law for the B field at an interface separating two isotropic materials of permeabilities J1.1 and J1.2; that is, find the relation between the angular deviations from the normal made by BI and B2 at points just to either side of the interface. Assume the total B fields tilted from the normal by the angles ()I and O2 as in (a). The boundary conditions relating the tangential and the normal magnetic field compo-

7


139

Region 2: (1l2) (a)

Air: III

= IlO

Air: fJ.1 = fJ.o

-L/h Small I

74.6° ~

(b)

I

(Iron: 112 »> fJ.o)

EXAMI)LE 3-5. (a) B flux refraction. (b) Refraction at air-to-magnctic-region interfaces.

nents are (3-50) and (3-71); Bnl = Bn2 and Hll = H'2' The latter can be written (3-75) From the geometry of the figure, the tilt-angles obey tan 01 = Btl/Bnl and tan O2 = B'2/Bu2' which combine with (3-75) to yield J1.2 B

,I

Inserting the expression for tan 0 1 obtains (3-76)

140


As a numerical example, compare the tilt of the B lines at an interface separating two regions with /11 = /10 and /t2 = I O/to. Assume at some point on the interface that B in region I is tilted by 8 1 = 45°. From (3-76),8 2 = arc tan (10 tan 45°) = 84.3°. Similarly, if8 1 = 20°, then 8 2 = 74.6°, and so on. In the event of an air-to-iron interface (/12» Jltl, one may show from (3-76) that for nearly all f)2, the corresponding 8 1 values are ve~y small angles (essentially 0°); that is, the flux leaves the iron nearly perpendicularly from its surface. These examples are noted in (b).

c.

The Nature of Magnetic Materials

The classical macroscopic theory of the field phenomena associated with magnetizahle suhstances, introduced in Section 3-4, attributes their magnetic properties to the magnetic moment n:l provided by the orbiting electrons, electron spins, and nuclear spins. Moreover M denotes from (3-55) the averaged volume contributions of the magnetic moments n:l in the vicinity of any point inside the substance. The net magnetic effects are altered significantly by the temperature-the random thermal agitations that inhibit the alignment of the magnetic moments. Although noteworthy advances in the understanding of magnetic processes on the microscopic scale have been provided by applying quantum mechanics and electromagnetic theory to models of the magnetic elements, there is yet much speculation in the deduction of the magnetic properties of the many complex alloys and compounds. Magnetic effects in materials have been classified as diamagnetic, paramagnetic, ferromagnetic, antiferromagnetic, and ferrimagnetic. The following discussion is intended to provide a glimpse of some of the classical models of magnetism to help explain the origins of these magnetic properties. 5 In a diamagnetic material, the net magnetic moment n:l of each atom or molecule is zero in the ahsence of an applied magnetic field. In this state', the classical picture of the electron speeding at an angular velocity w about a positive nucleus is accompanied by a balance of the centrifugal and the attractive Coulomb forces between those opposite charges. The application of a mag~etic field provides a Lorentz force, - ev x B, on the orbiting electron snch that if a fixed orbit is to be maintained, an increase or decrease ± L'lw in the electron angular velocity must occur, depending on the direction of the applied B field relative to the orbital plane. This amounts to a change in the electronic orbital current, thereby generating a small magnetic field, the direction of which is such as to oppose the applied field. The net, opposing magnetization field M thus created in any typical volume-element L'lv of the material leads to a slightly negative susceptibility Xm for such a material. Diamagnetism is presumed to exist in all materials, though in some it may be masked by other magnetic effects to be discussed. Typical small, negative values of Xm for diamagnetic solids at room temperature are - 1.66 x 10 5 for bismuth, -0.95 x 10 5 for copper, and 0.8 x 10- 5 for germanium. It is to be expected that the less dense gases have even smaller diamagnetic susceptibilities, which is borne out by both calculation and experiment. Another weak form of magnetism is known as paramag"netism. In a paramagnetic material, the atoms or molecules possess permanent magnetic moments due primarily to electron-spin dipole moments, randomly oriented so that the net magnetization M of (3-55) is zero in the absence of an applied magnetic field. The application of a B field to gaseous, paramagnetic nitrogen, for example, produces a tendency for the moments n:l to align with the field, a process inhibited by the collisions or interactions among the particles. In a paramagnetic solid, thermal vibrations within the molecular 5 An

excellent digest of the theories of magnetic phenomena, including ample references, is found in Chapter 7 of R. S. Elliott, Electromagnetics. New York: McGraw-Hill, 1966. .


141

Transition region

m

+++ +++ + ++++ ++++++ + ++++ + +++ + ++ t + ++ (b)

(a)

No applied field

Weak applied field

Moderate Saturation field field

- -

.....B

(e)

B

B

(d)

FIGURE 3-12. Alignment of magnetic moments in a ferromagnetic material: domain phenomena. (a) Magnetic moment alignments in a ferromagnetic material. (b) A perfect single crystal, showing domains and domain walls. (c) A transition region between adjacent domains. (d) Domain changes in a crystal with increase in the applied B field.

lattice tend to lessen the alignment effect., of an applied magnetic field. The room temperature susceptibilities of typical paramagnetic salts such as FeS04, NiS0 4 , Fe203, and CrCl 3 are of the order of 10 - 3 and inversely temperature-dependent, according to a law discovered by Pierre Curie in 1895. The importantjerromagnetic materials are characterized by their strong, permanent magnetic moments, even in the absence of an applied B field. They include iron, cobalt, nickel, the rare earths gadolinium and dysprosium, plus a number of their alloys and even some compounds not containing ferromagnetic elements. It was originally postulated by Weiss in 1907, and much later confirmed experimentally in photomicrographs by Bitter,6 that a ferromagnetic material in an overall unmagnetized state in reality consists of many small, essentially totally magnetized domains, randomly oriented to cancel out the net magnetic field. Domain sizes have been found to range from a few microns to perhaps a millimeter across for many ferromagnetic materials. Weiss further postulated that strong intrinsic coupling or interaction forces exist between adjacent atoms to provide the fully magnetized state within a given domain. It was not until 1928 that Heisenberg of Germany and Frenkel of the U.S.S.R. independently verified, using quantum theory, that the extraordinarily strong forces holding the domain atoms in parallel alignment is attributable to the coupling forces between the net electron spins of the adjacent atoms. 7 The parallel orientation of the spin moments in a ferromagnetic domain is depicted in Figure 3-12 (a). An idealized, perfect crystal might have a domain structure, in the absence of an applied B field, like that shown in Figure 3-l2(b), although flaws such as lattice imperfections and impurities would modify this idealized picture somewhat. The walls between the domains (Bloch walls), having the appearance suggested by Figure 3-l2(c), are transition regions between the spin alignments of the adjacent domains, and they are of the order of 100 atoms 6F. Bitter, "A gencralization of thc theory of ferromagnetism," PIl)'s. Rev., 54, 79, 1938. 7W. Heisenberg, "On the theory of ferromagnetism,"

Zeit. I

Phys., 49, 619,1928.

142

MAXWELL'S EQUATlONS AND BOUNDARY CONDITIONS

Magnetization

M

B Irreversible magnetization rotation region

T-

I I

I I I I

j

I

I I

Irreversible wall motion region

I I

______ 1

-c-l"'-t--/--Ll'

"ilIi I I I I

Reversible wall motion region

o

i Applied

I I I

Applied

I

I

I

I I I Bias I I

i

I

I I I

Ho-dJ : J> :

value

(a)

Itt

:

; - - - - -_ _ 1

Sinusoidal applied If field

I

I

rt (b)

FIGURE 3-13. Magnetization effects due to an external magnetic field applied to a fcrromagnetic matniaL (a) Magnetization proe,"ss (solid line) ill a virgin ferromagnetic region. Irreversible behavior shown dashed. (b) B If hysteresis loops for a ferromagnetic materiaL

thick. The domain division by such wall structures occurs in such a way that a minimal external magnetic field is supported by the structure, to minimize the work done in forming (he structure. As all external B field is increasingly applied to a ferromagnetic crystal containing domains, as denoted in Figure 3-12(d), tht'B!och walls first move to hwor the growth of those domains having magnetic moment\ aligned with the applied field, a reversible condition on removing the field ifB is not too large. For higher applied fields, domainwall motion occurs, which is not reversible, as noted in the third sketch of Figure 3-12(d). For a sufficiently large applied field, the domain magnetic moments rotate until an essentially total parallel alignment with the applied field occurs, a condition called saturation. The averaged effect of such changes on the bulk magnetization M, in a sample volume element containing a sufficient number of domains, is shown in Figure 3-13 (al. The arrows denote the direction ofincreasing or decreasing the applied H field. 8 One may note, on decreasing the applied H field to zero fi:'om the values P2 or P 3 , that a permanent magnetization Mr! (or M r2 ) is retained in the ferromagnetic sample, signifying an irreversible and distinctly nonlinear, multivalued behavior. These My values are termed the remanent (remaining) magnetizations of the specimen. The applied field must be further decreased to the reverse value He! (or ffel) as shown, before BIt has become cllslOmary to denote the applied magnetic field in the material than B.

the H fidcl, rather


143

the permanent is removed from the material. The value He is rather loosely called the coercive free, the field required to reduce the magnetization to zero within a specimen. If the M-H plot of Figure 3-l3(a) is replotted in terms of the B field in the ferromagnetic material, the B-H curve of Figure 3-13(b) results, recalling from (3-58) that these quantities are related to B = !lotH + M). In (b) is depicted a complete cycle of the events of (a), such as might occur if the applied field were varied sinusoidally as noted below the B-H curve. After the virgin magnetization excursion from 0 to P 3 , obtained over the first quarter cycle of the sinusoidal H field, the subsequent decrease in H provides the sequence of values passing through the remanent value Bn the coercive force He> and thence to the maximum negative flux density in the material at P4' With the applied H field going positive once more, a reversed image of the prior events takes place. The multi valued curve obtained in this cyclic fashion is called the hysteresis (meaning lagging) loop of the ferromagnetic region. Note that for smaller amplitudes of the applied H field, correspondingly smaller hysteresis loops are obtained, whether centered about origin 0 as just described, or appearing about Po as the consequence of a bias field Ho. The incremental permeability of a ferromagnetic material is defined as the slope of the B-H curve. The slope at the origin 0 of the virgin curve is called the initial incremental permeability. If the material is used such that it possesses a fixed (dc) magnetization Ho with a small sinusoidal variation about this value as noted at the point Po in Figure 3-13(b), the minor hysteresis loop formed there has an average slope defining the incremental permeability there. These events take place in the ferromagnetic core of an inductor or trans/ormer coil carrying an alternating current superimposed on a direct current, lor example. Energy must be expended in supplying the losses incurred in the hysttTesis effects accompanying the sinusoidal variations of an applied field. For this reason, ferromagnetic materials with low coercive forces (having a thin B-H loop) arc desirable for transformer and inductor designs. On the other hand, a ferromagnetic material used for permanent magnets should have a high coercive force He and a high remanent, or residual, flux density Br (corresponding to a fat B-H loop). 'Table 3-1 lists a few representative ferromagnetic alloys along with some of their magnetic properties. An additional and usually undesirable side effect, occurring in the magnetic core of devices such as transformers, is that of the free-electron conduction currents circulating within the core material due to an electric field E generated inside it by a time-varying magnetic field. The densities of these currents are limited by the conductivity (J of the core material through (3-7), that is, J = (JE, and are given the name tJddy currents because of their vortexlike nature within the conductive core, resulting from their relationship to the time-varying B field through (2-62)

VxE

8B

[2-62]

In the next section (2-62) is shown to be valid for a material region as well as for free space. Thus, with a conductive, ferromagnetic core in the solenoid as shown in Figure 3-11·( a), a sinusoidally time-varying current in the winding produces a sinusoidal B field in the core material to generate an E field, and from (3-7) also an eddy current field therein. Its sense is thus normal to the time-varying B field. The losses may be reduced substantially by subdividing the conductive core into a fibrous or laminar structure, as suggested by Figure 3-14(b), in which the subdivided conductors are insulated from one another. Small, spherical magnetizable particles serve the same purpose.

t"'"

TABLE 3·1 Magnetic Properties of Ferromagnetic Alloys

(A) Transfortner alloys Pertneabilities MATERIAL

PERCENT COMPOSITION

INITIAL

MAXIMUM

SATURATlON B (lNb/m2)

COERCIVE FORCE He (Aim)

CONDUCTlVllY (x 10 7 U/m)

Silicon iron

4 Si, 96 Fe 3.5 Si, 96.5 Fe 78 Ni, 0.6 Mn, 21.4 Fe 79 Ni;SMo, 16 Fe

400 1,500 9,000 100,000

7,000 35,000 100,000 800,000

2 2 1.07 0.7

40 16 4 0.16 to 4.0

0.16 0.2 0.12 0.15

H ypersil (grain oriented) 78 Permalloy Supermalloy

(B) Pertnanent tnagnet tnaterials MATERIAL

Carbon steel . Alnico V

PERCENT COMPOSITION

1 Mn, 0.9 C, 98;1 Fe 8 AI, 14 Ni, 24 Co, 3 Cu, 53 Fe

COERCIVE FORCE (Aim)

REMANENT B,0Nb/m2)

4,000 44,000

1.25


145

@

aB

Solenoid winding

~-O::::O::::'O::::O::::~0::::~~"Z>' ~"IOSinwt,mJo B = a Z --

L

Conductive, magnetic core (g,o)

at

sin wt

aB

at

--,

E and J flux ) (induced eddy currents)

aB

at (6)

(a)

FIGURE :1-14. Eddy current, in con
(a) Eddy enrFibrous and laminar

constrains the eddy curren ts to much smaller volumes, limiting their densities lIubstantially if the cellular substructures are made sufficiently small or thin. In the previous discussions it was seen that paramagnetism is a characteristic materials possessing permanent magnetic spin moments that arc randomly oriented, condition depicted in Figure 3-15(a). Ferromagnetic materials, due to the effects of Ihort-range couplings between adjacent atoms, possess parallel-oriented atomic magwithin given domain boundaries that comprise the material as suggested in Figure If such a material is heated until the thermal energies exceed the coupling energies, the material becomes disorganized into a paramagnet, though on cooling it reve~ts to a ferromagnet once more. The critical temperature at which this occurs is .known as the Curie temperature. Variations of the coupling phenomena responsible for ferromagnetic materials can even produce anti parallel alignments of electron spins in materials known as anti,,,·yrtlrYJIHTri'PJ· , as depicted by Figure . In this state, an antiferromagnet is characby a zero magnetic field. Manganese fluoride, for example, is paramagnetic at

m

t t t t 1t t t (a)

( c)

(b)

t

tttttttt (d)

'XGURE 3-15. Orientations of the spin moments of various magnetic ma(a) Paramagnetic. Ferromagnetic. (c) Antifcrromagnetic. (d) Ferrietic mat~rials, or

146


room temperature, but on cooling it to 206°C: (called its Neel temperature, after the French physicist), it becomes antiferromagnetic; below this temperature it exhibits no magnetic effect. An important variation of this phenomenon is ferrimagnetism, associated with noncanceling antiparallel arrangements of the coupled spin moments as suggested by Figure 3-15(d). Thus, in magnetite, the magnetic iron oxide FeO' Fe203, two of the three adjacent spins are reversed such that a somewhat weaker form of ferromagnetism is produced. Magnetite is an example of the group of ferrimagnetic oxides XO . Fe203, in which the symbol X denotes a divalent metallic ion Cd, Co, Cu, Mg, Mn, Ni, Zn, or divalent iron. When synthesized in the laboratory, these brittle, ceramiclike compounds are particularly useful for magnetic cores in highfrequency transformers and special applications ranging into the microwave frequencies because of their low conductivities comparable to those of the semiconductors, usually from 10 - 1 to 10 - 4 U/m. They are thus desirable because they limit eddy current losses in such applications. These conductivities may be compared with the much higher values of the typical alloys for lower-frequency applications as listed in Table 3-1, in which the values of the order of 10 6 U/m appear. A general account of the theory of ferro- and ferrimagnetism, together with a number of microwave applications of the latter, is found in the book by Lax and Button. 9

3·5 MAXWELL'S CURL E RELATION: ITS INTEGRAL FORM AND BOUNDARY CONDITION FOR TANGENTIAL E In Section 3-3, the Maxwell relation (3-59) for curl H in a material region was developed by adding in those current densities contributed by the electric and magnetic polarization fields. The form of the curl E relationship for materials is obtained by analogy, but retaining its form (2-62) for free space .

VxE=

oB

at

(3-77)

That the free-space Faraday's law (2-62) remains correct for a material region is evident on observing that an additive magnetic-current-densiry term, analogous to the electric-current-density term J of (3-7), is physically impossible iffi'ee magnetic charges cannot exist. Thus (3-77) correctly applies to both materials and free space. Equation (3-77) is readily converted to ,an integral form. The scalar multiplication of (3-77) with ds, integrating the result ~ver any surface S bounded by a closed line t, and applying Stokes's theorem yields \

~E 'ft

. dt =

-~ dt

f B . ds V s

(3-78)

again unchanged from the free-space version (1-55). The determination of (3-77) and (3-78) completes the development of Maxwell's differential and integral relations applicable to material regions, and they are summarized in the first two columns of Table 3-2. 'B, Lax, and K.J. Button. Microwave Ferrites and Ferrimagnetics. New York:'McGraw-Hill, 1962.

TABLE 3-2 Summary Maxwell's Equations and the Corresponding

Spatial Boundary Conditions at an interface DIFFERENTIAL FORM

V·D

INTEGRAL FORM

[3-24]

Pv

~s D

. ds =

CORRESPONDING BOUNDARY CONDITION

Iv Pv dv

[3-37]

Dn2 =

Dnl -

Ps

or

Case A: a 1, a 2 zero Dnl

V·B=O Vx H =

[3-48]

J+

3D

[3-59]

~B'ds =

0

J, H . dt

f J . ds + dtd f D . ds

'ft

=

[3-49]

S

S

[3-66]

Bn1

H'1

=

H,2 =

]s(n)

Case A: aI' a 2 finite H'l = H,2 Vx E

aB

3t

..:a.

J

J, E . dt = -~ 'ft dt

f B' ds s

[3-78]

Et1

E'2

or

[3-45]

[3-50]

n' (B1 - B 2 ) = 0

or

[3-71]

-> CfJ

Dnl =

or

Bn2

D 2 ) = Ps

Case B: a2 Ps

[3-42]

Dn2

n' (Dl

n X (HI - H 2)

Case B: a 2 -> n x HI = Js n X (El - E2l = 0

Js CfJ

[3-72] [3-79]

148

MAXWELVS EQUATIONS AND BOUNDARY CONDITIONS

A boundary condition, comparing the tangential components of the E fields to either side of an interface, may he obtained from Faraday's integral law (3-78). The details of the derivation may be avoided if one recalls that Ampere's line-integral law (3-66) leads to the boundary condition (3-70a), lIt! - 1It2 = }s(n)' The boundary condition comparing the tangential components ofE can be analogously found by applying (3-78) to a similar thin rectangle, yielding the analog of (3-70a)

(3-79)

Thus the tangential component of the E field is alwOeYs continuous at an inte~face. The right side of (3-79) is evidently zero because no magnetic currents are physically possible. A summary of the four boundary conditions derived from Maxwell's integral laws for material regions in Sections 3-2C, 3-4B, and in the present section, is given in Table 3-2.

EXAMPLE 3·6. (a) Derive a refractive law for E at an interface separating two noneonductive regions. (b) Deduce from boundary conditions the direction ofE just outside a perfect conductor. (a) The boundary conditions for the tangential and the normal eomponents of E at an inter1i:lce separating nonconductive regions are (3-43) and (3-79); that is, EIEni = E2En2 and E'l = E'2' From the latter and the geometry of (a), one obtains

(3-80) a result analogous with (3-76) of Example 3-5 concerned with the refraction of B lines. (b) From (3-44), a perfectly conductive region 2 implies null fields inside it. Then (3-79), Etl in the adjacent region I must vanish also. The remaining normal component in region 1 is given by (3-45). ])n1 p" yielding Ps E lE"1 as shown in (b).

Region 1:

Cl.q. € 1. <11

= 0)

Region 1: (I.t), tl, (11) E! = nE n !

n +

+

+

4

+

Region 2: (<12 ---+ 00) (b)

EXAMPLE 3-6. (a) E flux refraction at an interface separating nonconductivc regions. (b) E is everywhere normal to the surface of a perfect conductor.

149

3-5 MAXWELL'S CURL E RELATION: ITS INTEGRAL FORM

Region 2:

(0"2 -,>-00)

Wave motion

---(z)

(b)

(a)

EXAMPLE 3-7. (a) Parallel-plate system supporting a uniform plane wave field. (h) Charge and wrrent distributiou on conductor inner surfaces.

EXAMPLE 3·7. A uniform plane wave is described by the electric and magnetic fields

and propagates in air between two perfectly conducting, parallel plates of great extent, as in (a). The inner surfaces of the plates are located at x = 0 and x = a. Obtain expressions for (a) the surface charge field and (h) the surface currents on the two conductors.

(a) The given E is everywhere normal to the plates at x = 0 and x = a, satisfying the boundary condition of (b) in Example 3-6. The surface charge distributions thus become

x=o P" =n'·D 1

x

a

implying that E lines emerge from positive charges and terminate on negative ones. (b) The given H, to satisfy (3-72), must be everywhere tangential to the perfect conductors at x = 0 and x = a, yielding there

x=o -a z

E~

cos (wt

Poz)

x

a

110

It is seen that, in any fixed Z plane, current flows in opposite two conductors.

z directions

in the

150

MAXWELL'S EQUATIONS AKD BOUNDARY CONDITIONS

3·6 CONSERVATION OF ELECTRIC CHARGE A relationship between eharge and current densities is obtainable from Maxwell's equations, assuming that electric charge can neither be created nor destroyed. Let a charge density Pv(u 1 , Uz, U3, t) occupy some volume region V. Then the net charge in Vat any instant is

Note that even though p" is in general a function of both space and time, the net q enclosed is a function of t only, because the definite limits on the integral dispose of the space variables. For brevity, the latter is written with the function notation understood as follows. (3-81a) The time rate of change of q within V is a measure of the current flowing into the closed surface S bounding V; hence

aq = lap" ' at v -at- dv C/sec or A

-

(3-8Ib)

With ds directed normally outward from S, the current flowing out of S becomes

aq at

1

(3-82a)

implying that the net positive charge q inside V is decreasing in time. The postulate that electric charge is neither created nor destroyed permits equating the negative (3-81b) to (3-32a), yielding

~s J

0

ds = -

ap" l v -at- dv

(3-32b)

This means that the net outflow of current from any volume region is a measure of the time rate of decrease of electric charge inside the volume. Equation (3-32h) is thus the expression of the conservation f.!f electric charge. The relation (3-82b) has an equivalent differential, or point form

ap" ,3 VoJ= --Aim at

(3-32c)

a result obtained by applying (3-82b) to any limiting volume-element and using the definition (2-20) of divergence. While (3-82e) is true for any volume-element of a current-carrying region, it is also applicable to the surface currents and charges at the interface between a perfect conductor and a perfect insulator, as in the system of the forthcoming Example 3-8. With currents and charges confined to the interface so that J -+ Js and p" -+ Pso the

3-6 CONSERVATION OF ELECTRIC CHARGE

151

becomes 10

charge-conservation relation

V1"Js=

ops Am / 2 -Tt

(3-82d)

if Vl' ' Js is taken to mean a tangential (two-dimensional) surface divergence ofJs. For example, if the interface coincides with the y-z plane, implying Js a;]" + az.Jsz' the two-dimensional divergence of Js is written

v

l'

'J = oJs y + oJsz s

oy

oz

I n a time-static field problem, steady current densities are divergenceless, so (3-82c) reduces in that case to

VoJ

=

0

(3-82e)

Direct currents are therefore always characterized by uninterrupted, closed current flux lines. EXAMPLE 3·8, Show that the surface current and surface charge fields at the conductor dielectric interfaces of Example 3-7 satisfy the two-dimensional charge-conservation relation (3-82d). At the lower interface (at x 0), the left side of (3-82d) yields

'10

E:' sin (w[

+ wEoE:'

(Jot)

sin (wt

(Jot)

r--'

i

on substituting {Jo = Wy ftoEo and 110 = V flO/EO' With a surface charge density Ps = + EoE:' cos (w! (Joz) on the lower conductor,

oPs = + WEoE'+m SIn . (w! ot whence (3-82d) is satisfied.

EXAMPLE 3·9. Determine the relaxation expression for the time deeay of a charge distribntion in a conductor, if the initial distribution at t = 0 is Pvo(u l , U2, U3, 0). The desired result is obtained by combining (3-82c) with the expression [or div D. Replacing J with O"E for the conductive region obtains, from (3-82c)

V· (O"E)

+

opv =

at

0

(1 )

The region being homogeneous makes E and 0" constants, so (3-24) is written V . E = pJE, and snbstituting it into the first term of (I) yields

0Pv

0"

-+ ot E

Pv =0

(2)

IOlt should be noted that the relationship C:J-32c), connecting current density and charge density at any point in a region, is consistent with the Maxwell curl expression (3-59). This is evident from taking the divergence of thc latter, which promptly yields (3-32c) on making nse of the identity (/9) in Table 2-2.

152


Integrating yields the desired result

(3-83a) assuming the initial charge distribution at tOto be PvO(ub U2, U3, 0). This implies that if the internal, free electric charge in a conducting region is zero, it will remain zero for all subsequent time. One may conclude that in a material having a nonzero conductivity (J, there can be no permanent volume distribution of free charge. Thus, the static state of a free charge supplied to a conducting body is that it must ultimately reside on the surface of the conducting body through the mutually repulsive (Coulomb) forces among the free charges.

The time constant 3-9 is given by

T

of the free charge density decay process (3-83a) in Example

E T = -

sec

(3-83b)

(J

a quantity called the relaxation time of the conductor. Good conductors, for which (J may be of the order of 10 7 Vim, have relaxation times around 10- 18 sec, assuming a permittivity essentially that offree space. In poor conductors, T may be of the order of microseconds, though a good insulator may have a relaxation time of hours or even days. *3-7 UNIFORM PLANE WAVES IN AN UNBOUNDED CONDUCTIVE REGION The topic of uniform plane waves propagating in empty space was discussed in Section 2-10, in which the influence of the free-space parameters Po and Eo on the various wave characteristics was observed. The study of a plane wave propagating in a material having the parameters E, p, and (J is considered in this section. It is shown that the important new effects produced by the conductivity (J is to provide wave decay in the direction of propagation, as well as a phase shift between E and H. The assumptions made for the problem of wave propagation in an unbounded, linear, conductive region are

1. The components of E and H have neither x nor y dependence; that is, a/ax = D/iJy 0 for all field components. 2. Free-charge densities Pv in the conductive region are in general nonzero if the charge-continuity relation (3-82c) is to be satisfied; while the current density J in the conductor 11 is related to the E field therein by (3-7), J = (JE. 3. The parameters of the region, assumed linear, homogeneous, and isotropic, are p, E, and (J. ~

J=

The problem will employ time-harmonic forms of the fields. With Pv = 0 and Maxwell's equations for the region are obtained from (3-24), (3-48), (3-59),

(JE,

*As an option, this section may be omitted for now, to be taken up (along with Section 2-10) bef()re beginning Chapter 6, if desired. However, its relevance to an improved understanding of the material parameters over the broad frequency spectrum makes it desirable for study in this chapter. 11 Although this assumption refers explicitly to waves in a conductive region, the extension to wave propagation in a lossy dielectric through the use of a loss tangent, E" IE', is described .in Section 3-8.

3-7 UNIFORM PLANE WAVES IN AN UNBOUNDED CONDUCTIVE REGION

153

and (3-77), becoming

= Pv

(3-84a)

V'B=O

(3-84b)

V' (EE)

v X E= V

X

(3-84c)

-jwB = -jwflH

H = j + jwD =

6E

+ jWEE

(3-84d)

in which B = flH and D = EE of (3-30) and (3-64) are applicable. These equations need not in fact be solved, since this has already been done analogously in Section 2.10 for plane waves in empty space. To obtain the solution by analogy, compare (3-84a) through (3-84d) with (2-106) through (2-109) applicable to the empty-space case V'(EoE)=O

[2-106]

V'B=O

[2-107] [2-108J [2-109]

in which B = floH and D = EoE apply. A comparison of the two cllri expressions in these two groups of Maxwell's equations reveals that the two V X E expressions are precise analogs of each other, with (3-84c) obtainable From (2-108) on simply replacing 110 in (2-108) with fl. Comparing (l-84d) with (2-109), however, reveal an additional conduction-current-density term O'E in (3-84d). On collecting terms of the right side of (3-84d) as follows V

X

H

0' E

+ jWEE =

(0'

+ jWE) E

= jw ( E

j;;)

E

(3-85 )

the analogy of the latter with (2-109) is evident on replacing EO of (2-109) with the complex permittivity, E - j6lw. Thus, each of the Maxwell's equations (2-108) and (2-109) is seen to become (3-84c) and (3-84d) on replacing in the former flo with fl

and

(3-86)

These replacements applied to the wave solutions of lOS) and (2-109) are therefore expected to yield the solutions of (3-S4c) and (3-S4d) in an unbounded conductive region. Recalling the solution (2-115) for empty space

+

E; (z) [2-115]

154


the replacements (3-86) in the latter yield analogous plane wave solutions for an unbounded conductive region

E;; (z)

+

E;(Z)

(3-87)

In (3-87), the pure phase factor jWJlloEo of (2-115) becomes a complex factor abbreviated with the symbol y, called the propagation constant (3-88) and y can be separated into real and imaginary parts

y=ot+jpm- 1

(3-89)

in which ot, the real part of y, is called the attenuation constant, and P is termed the phase constant of the uniform plane waves (3-87). Explicit expressions for ot and P are found by replacing y of (3-88) with ot + jp, squaring both sides to remove the radical, and equating the real and imaginary parts of the result. The following positive, real solutions for ot and P are obtained.

[J ((J)2 J2 w~ J2 [J ((J )2

w~ ot=--

1 12

J

1+-

Np/m

(3-90a)

+ 1J1 2rad/m

(3-90b)

1

WE

P=

I

+

WE

/

The dimension of ot and P is (m) -1, though the artificial dimensionless terms neper and radian are usually mentioned to emphasize their attenuative and phase meanings in the wave expressions. With the substitution of (3-88) into the exponent of the wave solution (3-87), one may express it

E;(z)

E:'e- YZ

+ +

E;(z)

(3-91a) (3-91b) (3-91c)

in which the complex amplitudes in (2-116)

E;; of the traveling wave terms are denoted again as (3-92)

A comparison of the conductive region wave solution (3-91) with the empty-space wavc solution (2-115) reveals the presence of two real factors, e-a.z and efl.Z, accounting for wave decay as the positive z and the negative Z traveling waves proceed in their corresponding directions of flight with increasing time. An additional view ofthc decay (attenuation) property of the waves is gained by converting (3-91) to its real-time form,


155

obtained as usual by use of (2-74)

EAz, t) = Re [EAz)i/rot]

= Re

E~e-az cos (wt

fh

[E~ i/
+ E; i/r eazi//Jzejrot]

+ cf>+) +E;eocz cos (wt +

fh

+ cf>-)

(3-93)

These positive z and negative z attenuated traveling waves are depicted in Figure 3-16(a) and (b). A comparison of (3-93) with the real-time unilorm plane wave solution (2-119) in empty space

Ex(z, t)

=

Poz + cf>+) + E;

E~ cos (WI -

cos (wt

+ Poz + cf>-)

[2-119]

shows that the important new characteristic introduced by the nonzero conductivity (1 is the wave attenuation occUlTing in the direction of the wave motion. Note that setting

(x)

E;(z,t)

(x)1 _I

=

E~,e-az cos (wI - (3z) (at t.= 0)

Wave motion

----

-(x)

0

II!..

/

/

(y)

E; flux (x)

(a)

(x)

I

I

1

(x) E;(z, t)

=

--

Wave motion

(z)

(b)

FIGURE 3-16. Attenuated wave solutions for E~(x, t) in a conductive region. Flux plots are shown, emphasizing field independence of x and y. (a) Positive z traveling, positive::. attenuated wave. (b) Negative::. traveling, negative::. attenuated wave.

156

MAXWELL'S EQUATIONS AND BOUNDARY CONDiTIONS

the region parameters equal to the empty-space values E = Eo, 11 = 110' and (J = 0 reduces the attenuation and phase constants to a = 0 and /3 = /30 in (3-90a) and (3-90b). The wave attenuation in a conductive region is governed by the size of the (J/WE term relative to unity in (3-90a). As (J becomes larger so does ex, causing the plane wave to decay more rapidly with distance. Denote only the positive z traveling wave term of (3-93) by the symbol (z, t); that is,

E;

(3-94) This wave penetrates a conductive region as shown in Figure 3-17, attenuating with distance according to the factor e- az such that at the particular depth z = b, its amplitude has decayed to e -1 of its value at the reference surface z = O. The depth
(3-95 )

ex

A current density by (3-7); that is,

J

accompanies the Ex field in the conductive region as given

] ; (z, t)

= (JE:'e- az cos (wt -

/3z

+ (V)

A/m

(3-96)

a result in phase with the electric field. For a highly conductive region (with a large a/WE), b is seen from (3-95) and (3-90a) to be correspondingly small; so in the limiting

case of a perfect conductor (a -+ CX)), the skin depth vanishes with a indefinitely large. This provides the limiting surface current phenomenon of the boundary condition (3-72), involving the tangential H field at the s'lrface of a perfect conductor. The magnetic field accompanying Ex of (3-91) is obtained by substituting (3-91) into Maxwell's equation (3-84c); or alternatively by invoking the analogy with the wave solutions in empty space, using the replacements (3-86) in (2-130). Then, for (x)

(at t = 0)

--

(2)

(M, t, a)

-

Motion

;' ;'

~FIGURE

of e-

1,

3-17. The penetration depth (j associated with an amplitude attenuation for a unil<:mn plane wave in a conductive region.


157

uniform plane waves in an unbounded conductive region, the following complex impedance ratios are found to apply.

r~

(3-97)

,(J

J-;;;

E -

with the complex ratios denoted by q, the intrinsic wave impedance. The field written in terms of the solutions E: (z) and E; (z) in (3-91) as follows.

Hy(Z)

Hy(z) is thus

H: (z) + H; (z)

=

(3-98a)

E:(Z)

q

(3-98b)

q

~+

Em -yz =-e

q

E-

m eYz

Aim

(3-98c)

The intrinsic wave impedance defined by (3-97) can be expressed in complex polar form as follows:

J ~ ~)2J1/4 ,,"~ p

';1 ",.

(J

E-j-;;;

[

1+

(

(.jo,1

n

(3-99a)

(J

WE

seen to be of the form (3-99b) with 17 and

e taken to mean

e = zarc tanWE 1

(J

Evidently letting (J = 0 reduces fj to the real result .JPji., applicable to uniform plane waves in a nonconductive (perfect dielectric) region. Moreover, the positive phase angle associated with fj means that (z) lags the accompanying (z) in time phase, as shown in the real-time sketches of Figure 3-18. The crank method for simulating the motion of the wave with increase in the time variable t is depicted in Figure 3-19. The additional characteristics of plane wave propagation in a conductive region are the wavelength defined in (2-123) (setting (U 2n rad), yielding

e

H:

E:

2n {J

A=-m

(3-100)

158

MAXWELL'S EQUATIONS AND BOUNDARY CONDrTIONS (x);

(=0

(z)

~ l17otion FIGURE 3-18. Positive shown at t = O.

z

traveling fields of a uniform plane wave in a conductive region,

1m

Crank counterclockwise, simulating time increase in e)wt

E; (z, t)

= Re [E; (z) ejwtJ

-?f> r ~, I'<£'" " ~'«

~ '.)"'/<"

7

FIGURE 3-19. Wave of f"(z, t) showing the complex phasors displayed along z at t = 0, and its real-time projection below. Do not confuse the electric-field phasors (arrows) depicting the fields changing phase along the {Jz axis in the complex plane as in the top view of this figure with the vector direction of the electric field in three-dimensional space. I t should be dear that the electric field in the present problem has only an x-directed component in space, as shown in the real-time diagram of Fignre 3-18.


the phase velocity 1J p ' obtained by putting the argument of and differentiating in time, to obtain

w

vp

159

equal to a eonstant

,

=7f m/sec

(3-101 )

T = ]sec 1

(3-102)

and the period

The applicable value oj' fJ is of wurse that of (3-90b).

EXAMPLE 3·10. Suppose a uniform plane wave with the amplitude 1000ei°- Vim propagates in the + z direction at I = 10 8 Hz in a conductive region having the constants Jl = Jlo, E 4Eo, a/WE 1. (a) Find {J, ct, and ~ for the wave. (b) Find the associated H field, and sketch the wave along the z axis at t = O. (c) Find the depth of penetration, the wavelength, and the phase velocity. Compare A and >u p with their values in a lassies.> (a = 0) region having the same Jl and E values. Assume only l';x and 11y components for the wave. (a) The attenuation and phase factors are given by (3-90a) and (3-90b)

2w

P=

2w

[0.414]11 2

1.90 Np/m

(I)

[2.414p I 2

4.58 rad/m

(2)

The propagation constant is therefore,)! = 1.9 impedance is given by (3-99a)

-:-:--'-~.,..,..,.. ei(112) arc tan 1

60n

+ j4.58 m -- 1.

The complex wave

I 59ei(n/8) Q

e;(l/2)(n I4)

1.19 whence 11 (b) The

= 159 Q

H field

(3)

and 0 = n/8 or 22.5°.

is found by use of (3-97)

iI;(z)

(4)

to yield thc real-time expressions

E;

Re[E;(z)eiwt ]

t)

= 1000e-1.

11; (z, I)

=

9z

Re[lOOOe-a"e-iPz~wt]

cos (wt - 4.58z) Vim

6.2ge -1.9z cos rwt - 4.58z

(n/8)] Aim

(5) (6)

160


(c) The depth of penetration is found using (I): 8 = a- 1 = 0.52 m, the distance the wave must travel to diminish to e- 1 (or 36.8%) of any reference value. The wavelength is obtained using the value of {I

211: {I

A=

=

211: = 1.37 m 4.58

comparing with that for a lossless region (flo, 4Eo) as follows

211:

c

3

10 8

X

1.5 m

The effect of finite conductivity is thus to foreshorten the wavelength. The phase velocity in the conductive region is

v p

OJ

=

{I

=

211:( 10 8 ) = 1.37 4.58

X

10 8 m/sec

which compares with that in the lossless region as follows: (0) _

OJ

_

c

OJ

Vp - {I(O) - OJJ;;;4~o

2

1.5

X

10 8 m/sec

Conduction thus serves to slow down 1J p ' The foregoing numerical results may be added to Figure 3-18 to provide a picture of the wave motion in the conductive region.

3-8 CLASSIFICATION OF CONDUCTIVE MEDIA Conductive materials can be classified with reference to the magnitude of the conduction current density term O'E relative to the displacement current density term jWEE appearing in Maxwell's relation (3-85)

0') ~

-

E

[3-85]

W

Denoting the complex permittivity,

E -

A

E

jO'lw, in (3-85) by the symbol

== E

-

J

.0'

W

Flm

(3-103)

one may represent E 111 the complex plane as in Figure 3-20. The angle (jd is called the dissipation angle, which vanishes for a lossless region. Its tangent, defined by (3-104)

is called the loss tangent, or dissipation factor, of the material.

3-8 CLASSIFICATION OF CONDUCTIVE MEDIA Iml

161

1m

Re

Re

(a)

(b)

FIGURE 3-20. Complex permittivity for conductive and losslcss regions. (a) Conductive region (general). (b) Lossless region (0' -> 0).

The importance of the loss tangent is recognized from its appearance in the expressions (3-90a) and (3-90b) for a and {3, and in (3-99a) for the wave impedance of uniform plane waves propagating in a conductive material or a lossy dielectric at a given frequency. Under impressed electric fields that are time-harmonic, the microscopic (atomicscale) mechanisms contributing to the electric polarization P in a dielectric material, as discussed in Section 3-2, are often modified by damping (loss) effects. The classical model, inspired by experimental measurements on dielectric materials, assumes an oscillating system of interacting atomic or moleeular particles, in which the response of the dielectric to the applied electric field involves damping mechanisms plus resonances about certain frequencies. The damping is taken as proportional to the velocity of the particles oscillating under the impressed fields, to produce results similar in some ways to the conductivity mechanism discussed in Section 3-1 for the Drude model of a conductor when a time-harmonic field is applied. The resonances in the dielectric polarization arise from the inertia of the particles, displaced by the sinusoidal applied field and interacting with the restoring Coulomb forees. The response of the dielectric to the applied field resembles that of a three-dimensional system of masses interconnected through springs and dashpots and subjected to applied distributed vibrational forces, or analogously, a network of reactive and resistive circuit elements excited by sinusoidal voltages, with maximum losses occurring at the resonant frequencies. For typical dielectric materials, the lowest resonance is usually in or above the microwave range, with higher resonances occurring in the optical range. 12 The large-scale Of macroscopic effect of these interaction phenomena is observed experimentally to make the permittivity of a dielectric become complex at a given excitation frequency, to permit writing it in terms of its real and imaginary parts

E = E' - jE"

(3-105)

Since a complex permittivity E has already been defined by (3-103) in connection with the loss mechanism in a conductive region, a comparison with (3-105) is in order. One may see that the substitution of the complex permittivity E of (3-105) into the Maxwell relation (3-85) yields

V

X

it = jWEE = jW(E'

jE")E = wE"E

+ jWE'E

(3-106)

12Details of damping and resonance phenomena in dielectrics ii-om the microscopic point of view and using classical Or quantum-theory approaches are found in A. R. von Hippel, Dielectric Materials and Ajlplications. Cambridge, Mass.: MIT Press and New York: Wiley, 1954; and R. S. Elliott. Electromagnetics. New York: McGraw-Hill, 1966, Chapter 6. A brief digest is to be found in S. Ramo, J. R. Whinnery, and T. Van Duzer. Fields and Waves in Communication Electronics. New York: Wiley, 1965, pp. 330-334.

162


The conduction current density j = aE is omitted, since comparing (3-106) with the form of (3-85) shows that an equivalent conduction loss mechanism is already accounted for by the term wE"E in (3-106). Thus WE" assumes the role of conductivity a in a lossy dielectric, while E' in the last term of (3-106) is identical with the real E in (3-85), corresponding to electric field energy storage in the dielectric. With the equalities

a=WE

E = E'

If

(3-107)

the complex permittivities expressed by (3-103) and (3-105) therefore have equivalent meanings. It is evident that Elf, the imaginary part of the complex E, is descriptive of all loss mechanisms in the dielectric at a given frequency. With the replacements (3-107), the loss tangent tan I<>dl of (3-104) is written in the equivalent forms (3-108) Denoting the loss tangent by becomes N

=

Elf/E',

WJf J.lE [ 'Y £.

2

(3-90a) for the wave attenuation in a lossy dielectric

1+

1 (E")2 7 - I J1 2Np/m

(3-109)

Corresponding expressions for the phase constant f3 and the complex wave impedance are obtained from (3-90b) and (3-99a), yielding

~

a __ f'

WJf J.lE [ 'Y £.

2

I

(Eff)2 + 1J1 /2rad/m

+ 7

(3-110)

(3-111)

From the foregoing it is concluded that the characterization of the loss tangent (3-108) by a/WE is better suited to a conductor, whereas the form Eff/E' is more desirable for a dielectric region. A conductive material or a lossy dielectric supporting electromagnetic waves at a frequency W may in general fall within one of the following three classifications: (a) it is a good conductor if the conductivity a is sufficiently great that its loss tangent, a/WE, becomes very large compared with unity (i.e., a/wE» 1); (b) it is a good insulator ifits loss tangent is sufficiently small (Eff/E'« 1); and (c) it may be called moderately conductive (semiconducting) if it falls somewhere between these extremes (i.e., if the loss tangent is roughly of the order of unity). The expressions [or the attenuation constant, phase constant, and instrinsic wave impedance associated with uniform plane wave propagation in such regions simplify to the following, for the classifications (a)

3-9 LINEARITY, HOMOGENEITY, AND ISOTROPY IN MATERIALS

163

and (b):

1. For a good conductor, assuming

(f/WE

» 1, (3-90a), (3-90b), and (3-99a) reduce to (3-112a)

(3-112b)

A

f/

2. For a lossy dielectric, if

=

(1

~ + j) ~ 2(;

(3-112c)

EU/E' «1, (3-109) through (3-111) become IX = w~ (EU) 2

(3-113a)

E'

fJ = w~ [ 1 + 81 (EU)2] -;; f/ A

_

flE[ -

(EII)2 +J.-1 (Ell)] E' 2 E'

1- 3 8

(3-113b)

(3-113c)

The (3-113) equations are obtained by including only the first two terms of the binomial expansions of the square root quantities in (3-109) through (3-111), assuming a very small loss tangent. In the limiting case of a lossless dielectric, (3-113) reduce to = 0, fJ w~, and ~ = 0L/E as expected. Note in view of (3-95) that the inverse of (3-112a) can be used to express the depth of penetration, b, in a good conductor

IX

(3-114) a result inversely dependent on the square root of the frequency, the permeability, and the conductivity of the material. Thus, for copper having a conductivity of 5.8 x 10 7 U/m with Jl = Jlo, the skin depth at 1000 Hz is about 2 mm, while at a frequency 10 6 times as large (J = 1000 MHz), b is reduced by the factor 10-3, becoming 0.002 mm.

3·9 LINEARITY, HOMOGENEITY, AND ISOTROPY IN MATERIALS Electric and magnetic polarization effects in materials have been accounted for by the polarization field P and the magnetization field M, defined by (3-16) and (3-55), respectively. Their additive effects, yielding the Maxwell relations for a material region,

164


have provided the definitions of the fields D and H by (3-23) and (3-5B)

D

= EoE + P

(3-115 ) (3-116)

If the region is also capable of transporting free charges in a conduction process, the conductivity parameter (J assigned to the region expresses the proportionality of the current density J to the applied field E by (3-7)

J

= (JE

(3-117)

Avoiding the questions of anisotropy for the moment, one may recall that the relations of the electric polarization field P and the magnetizations M to the applied fields may be expressed by (3-25) and (3-60) 13 [3-251 [3-601 With the foregoing as a background, the questions of linearity, homogeneity, and isotropy in material media are discussed. In this framework, one should assume that the temperature of the material and the sinusoidal frequency of its impressed fields are constants when defining its parameters It, E, and (J, since the dependence of the latter on temperature and frequency callnnt, in general, be ignored.

A. Linearity and Nonlinearity in Materials If the susceptibilities Xe and Xm are constants and thus independent of the applied fields, the material is said to be linear with respect to electric and magnetic polarization effects. A straight-line relationship between an applied field component Ex and the resulting polarization component P x characterizes this linearity property. With the substitution of (3-25) and (3-60) into (3-115) and (3-116), the compact results (3-30c) and (3-64c) D=EE

r3-30c] [3-64c]

have been seen to result. Nonlinearify in a material is characterized by one or more of the parameters, fl, E, and (J, being dependent on the level of the applied fields. Then one may choose to write (3-30c), (3-64c), and (3-117) in forms signifying this dependence D

E(E)E

(3-118)

B

/l(H)H

(3-119)

J = (J(E)E

(3-120)

13Attention is here drawn to footnote 4 relative to expressions (3-25) through (3-33), which by extension pennits a similar generalization of (3-60) and (3-64c) to their phasor forms, M = XmH and 13 = .aH.

3-9 LINEARITY, HOMOGENEITY, AND ISOTROPY IN MATERIALS

165

An example of (3-119) is depicted by the multivalued B-H curve of a ferromagnetic material in Figure 3-13(b). *B. Isotropy and Anisotropy in Materials In some physical materials such as crystalline substances possessing a well-ordered atomic or molecular lattice throughout a given sample, the polarizations P or M resulting from the application of an E or B field may not necessarily have the same directions as the applied fields. Such materials are termed anisotropic 14 , meaning that different values of fl, E, or (J are measurable in different directions within the substance. Differences in the polarization responses to the direction of an applied E field in crystals, for example, are due to the disparities in the interatomic spacings associated with the several symmetry axes of the crystalline lattice. In some crystals, in which three orthogonal principal axes may be identified, the cartesian coordinates can be chosen along the same axes. Then, for an applied field E = axEx + ayEy + azE z , the components of the electric polarization field P become P x = Xe11 (EoEx) Py

Xe22(EOEy)

Pz

= Xe33(E OE z )

(3-121 )

in which the susceptibility components Xell' Xe22' and Xe33 are generally different. (The static field values for gypsum, for example, are about 8.9, 4.1, and 4.0, respectively.) These circumstances are depicted in Figure 3-21 (a), showing the development of a polarization vector P having a direction different from that of the applied E field, a result of the unequal susceptibilities associated with the coordinate directions. It is evident that (3-121) reduces to the vector result (3-25), P = xeeoE, whenever Xell == Xe22 == Xe33' Again, if rectangular coordinates are selected so that the applied field has only an x component, that is, E = aj;~" applying it to an arbitrarily oriented anisotropic material yields all three components of dielectric polarization (3-122) a result exemplified in Figure 3-21 (b). In general, if E possesses all three rectangular components, applied at an arbitrary angle with respect to the crystal principle axes, one must write

Px = Xell(EOEx) P y = Xe2t(EOEx) Pz

= Xe31 (EoEx)

+ Xe12(EOE~) + Xe13 (EoEz ) + Xe22(EOEy) + Xe23(E OE z ) + Xe32(EOEy) + Xe33(€OE z )

(3-123a)

This triplet of expressions is denoted by the matrix form

Px] _ [xeu P v - Xe2l [P Xe31 z

Xe12 Xe22 Xe32

(3-123b)

14From the Greek an (not), plus iso (same), plus trope (turning); hence, not having the same (property) with different directions.

166


(x)

(a) Ey

"'--Crystal sample

(b)

FIGURE 3-21. Aspects of dielectric anisotropy in a crystal. (a) Polarization components rcsulting from an x-directed E-field component applied to an arbitrarily oriented crystal. (b) Polarization components resulting from applied E-field components, if principal axes in a crystal arc aligned witb the cartesian coordinate axes.

having the compact representation

[P]

(3-123c)

The linear relations (3-123) involve the components x.eij of a susceptibility matrix [x.e]. One may observe that if the three principle axes of a particular anisotropic material are aligned with the cartesian coordinates, the off-diagonal coefficients (i #- j) of (3-123) become zero, reducing it to (3-121). Applying D = EoE + P to (3-123), one can verify that the expressions connecting D and E in an anisotropic substance are

Dx

= EllEx + E12 E y + E13 E z

Dy =

E21Ex

+

Dz =

E31Ex

+ E32E~ + E33 E z

E22Ey

+

E 23 E z

(3-124a)

having the matrix form

[D] = [E][E]

(3-124b)

3-10 ELECl1WMAGNETlC PARAMETERS OF TYI'ICAL MATERIALS

167

should be evident that expressions analogous to 123) and (3-1 can be estabamong the cartesian components of the vector Band H for anisotropic magnetic materials.

C. Homogeneity and Inhomogeneity in Materials A material region having parameters jJ, E, and (J independent of the position it is termed homogeneous. Conversely, if one or more of the parameters is of the space-dependent form

(3-125) the material is said to be inhomogeneous. The mixture of earth and water occurring ncar the suriace after a rain is an instance of an inhomogeneous region having parameters E and (J that vary with depth. The ionosphere, a gaseous mixture of positive, negative, and neutral particles, must be regarded generally as electromagnetically inhomogeneous. Artificial inhomogeneous materials are created by the variable spacing of small metal spheres within a Styrofoam or other supporting insulating· material, to yield an electrically polarizable region having a variable E depending on the average spatial densities of the spheres. Devices constructed in this way, using metal spheres, rods, or plates, have been used in artificial lenses for microwave applications. ls The complications of nonlinearity, inhomogeneity, and anisotropy in materials are for the most part avoided in subsequent treatments in thi~ text. The emphasis is restricted essentially to discussions of electric and magnetic fields in materials that are linear, homogeneous, and isotropic.

3·10 ELECTROMAGNETIC PARAMETERS OF TYPICAL MATERIALS A tabulation of measured parameters at room temperature for typical nonmetals and nonferrous metals is given in Table 3-3. The frequency dependence of Er and the loss tangen t Elf IE' for nonmetals is evident from their values at the three widely different sinusoidal fi'equencies listed. Laboratory methods for measuring material parameters differ considerably, depending on the frequency at which the parameters are to be determined. The permittivity and loss tangent ofa nonmetal at frequencies up to several megahertz can be found using lumped-eircuit methods; thus, a capacitor making use of the test material as a dielectric and connected in a Q-meter arrangement might be employed. At higher fi·equencies, the measurement of the influence of the material on the wave transmission properties of a coaxial transmission line or a waveguide can be useful for obtaining its parameters. Several source books may be consnlted for further information on such methods. 16 15For example, sec W. E. Kock, "Metallic delay lenses," Bell Syst. Tech . .lour., 27, 58, January 1948. 16For example, see A. R. von Hippell, Dielectrics and Waves. New York: Wiley, 1964.

..

TABLE 3-3 Material Parameters at 20°C (Unless Otherwise Stated)

8;l

A. Nonmetals Elf

Er ,

MATERIAL

jl,

BM 120 Douglas fir Miearta 254 Nylon (Dupont) Plexiglas Polyethylene Polystyrene (Dow) Silicone fluid SC 200 Soil, sandy, dry Soil, sandy, 2.2% H 2 O Styrofoam 103.7 Tam Ticon B (barium titanate) Teflon (n°G) Teflon (lOODG) Water, distilled

- , At Frequency

At Frequency

60

10 6

10 '0

4.87 2.05 5.45 3.60 3.45 2.26 2.55 2.78 3.45 3.25 1.03 1240 2.10 2.04 81

4.74 1.93 4.51 3.14 2.76 2.26 2.55 2.78 2.60 2.50 1.03 1140 2.10 2.04 78.2

3.68 1.80 3.30 2.80 2.50 2.26 2.54 2.74 2.50 2.50 1.03 150 2.10 2.04 50

60

10 6

0.080 0.028 0.004 0.026 0.098 0.036 0.018 0.022 0.064 0.014 ( <0.0002) ( <0.003) 0.0001 0.0003 0.020 0.200 0.700 0.025 ( <0.0002) 0.056 0.010 ( <0.005) «0.001) 0.040

·B. Metals MATERIAL

Silver Copper Aluminum Sodium Brass Tin

/1,

(J

(Ujm)

6.17x107 5.8 x 10 7 3.72 x 10 7 2.1 x 10 7 1.6 X 10 7 0.87 x 10 7

DIELECTRIC STRENGTH

E'

DEPTH OF PENETRATION

ij

FOR PLANE WAVES (m)

0.064/J] 0.066/J] 0.083/J] O.lljJ] 0.13jJ] 0.1 7/[r

10 '0

(Vjmil)

0.Q410 0.030 0.0400 0.0110 0.0050 0.0005 0.0003 0.010 0.0040 0.0650 0.0001 0.60 0.0004 0.0005 0.200

300 1,020 400 990 1,200 600

1,500

3-11 GENERAL BOUNDARY CONDITIONS FOR NORMAL D AND J

169

"'3·11 GENERAL BOUNDARY CONDITIONS FOR NORMAL D AND J

In Section 3-2, the boundary condition (3-41) (3-126) comparing the normal components of D to either side of an interface was derived. Special cases were cited concerning (a) two perfect dielectrics and (b) a perfect dielectric and a perfect conductor. In this section are treated the remaining cases involving regions with finite conductivities, in which E gives rise to current densities specified by (3-7): J = (jE. It is shown in general that a free charge density Ps accumulates at an interface, in an amount determined by the ratios of the conductivities and permittivities of the adjacent regions. To this end, the boundary condition (3-126) cannot in itself reveal the proportions of Dnl and Dn2 yielding Ps there. Another boundary condition is required, obtained from the current continuity relation (3-82b) [3-82b J Equation (3-82b) is applied to a pillbox region of vanishing height, as used in deriving (3-41). The surface integral applied to the upper and lower surfaces of the pillbox in Figure 3-22 (a) yields contributions ]n1 fJ.s and - Jn2 fJ.s to the net outward current flux. The tangential components Jtl and ]t2 contribute only a vanishing amount of current from the sides of the pillbox, as (jh -+ O. However, if a surface density Js exists on the interface (permissible if region 2 is a perfect conductor), then a nonvanishing current outflow from those sides is possible, occurring if Js exhibits longitudinal changes, that is, if J., has a surface divergence as shown in Figure 3-22(b). Then the current outflow

Edge view

+7----- J1 J

I

J n 1,

~'

__

Region 1:

,( ) ,J.l.1,tj,O'j

.,JJ'l

••

(a)

(b)

FIGURE 3-22. Gaussian pillbox constructed for comparing the normal components of J at an interlace. (a) Components OfJ1 and J2 to either side of the interface. (b) Showing the variation of the x component of J3'

170


through the four sides of the pillbox becomes

to which is added the current flow from the upper and lower surfaces, yielding

Upon canceling terms and eliminating Lls

= Llx Lly factors, one obtains the

boundary

condition

(3-127) This can be written

oPs

Jn1 - Jn2 + VT • Js = -7it Aim

2

(3-128)

Tbe general boundary condition involving the continuity of the normal components of the volume current density at an interface is (3-128). It states that the normal component ofJ is discontinuous at an interface to the extent of (a) the time rate of decrease of the surface charge density, -ops/ot and (b) the tangential divergence possessed by the surface current J •. An alternate form of (3-128) is, with J = O"E (3-129) The general boundary condition (3-128) or (3-129) simplifies depending on the adjacent regions, three cases of which are discussed in the following. I. One region nonconductive; the other a perfect conductor. Assuming region 1 lossless (0"1 = 0) implies that J 1 = 0, and with region 2 a perfect conductor and containing no fields, J2 = 0 also. Then (3-128) becomes

ops

ot

0"1

=

0,

(0"2

---+

CI))

(3-130)

Thus at the surface of a perfect conductor adjoining a perfect dielectric, the time rate of decrease of Ps equals the surface divergence of J., but (3-130) is just a restatement of the charge conservation relation (3-82d).

171

3-\1 GENERAL BOUNDARY CONDITIONS FOR NORMAL D AND J

2. One region hasfimte conductivity; the other is a perfect conductor. With a 2 reducing (3-128) to

a 1 finite,

~

00, J2

= 0,

(3-131)

The normal outflow Of}nl from a perfect conductor into an adjacent conductive region is dependent on the time rate of decrease of Ps and on the surface divergence of Js. 3. Both fI?gions have finite conductivities. In the absence of a perfect conductor, Js O. Then (3-128) yields (3-132) lfthis is combined with (3-126), Dnl - Dn2 = p., one can develop a relationship between the normal components of 0 (or E) at an interface, besides an expression for Ps' To avoid the use of a/at in the result, it is desirable to replace the fields with time-harmonic forms according to (2-67). Thus, after canceling the eiro ! factors and replacing D with d~ and j with aE, (3-132) and (3-126) become (3-133) (3-134) These must be simultaneously satisfied at the interface. Eliminating {is obtains

whereupon factoringjw yields the boundary condition

Using the complex permittivity notation of (3-103) obtains (3-135) The boundary condition for the normal component ofE is, therefore, that EE~ is continuous at an interface separating finitely conductive regions. An expression for ~he fre£ charge density {i accumulated at the interface is obtained by eliminating Enl or B"2 fi'om (3-133) and (3-134), yielding the equivalent results (3-136)

172


in which Eland E2 are given by (3-103). One conchldes that a surface charge induced on the interface by the normal components of E if at least one region is conductive. On the other hand, no free surface charge exists at the interface if (a) both regions are nonconductive ((Jl = (J2 = 0) o[ (b) the special proportion EdEz = (Jl/(J2 is true among the region parameters, presumably a rare event and oflittle importance. For both regions nonconductive, putting Ps = 0 into (3-134) yields the special case (i1E,,) EzE"z = 0, or just (3-1 a result agreeing with (3-43) for the I1ot1conductive case.

EXAMPLE 3·11. Determine the refractive law for direct currents at an interface separating two isotropic conductive regions. Specialize the result for one conductivity much larger than the other. Assume the J vectors tilted by amounts 0 1 and O2 as shown in (a). The boundary condition (3-132) for dc becomes

(I)

(a)

Region 1: (
Region 1: (
Region 1: (
./

Region 2: (
-'---~-"---

Region 2: (
»

Region 2: (
(d)

EXAMPLE. 3·11. (a) Refraction of currents. (b) Examples of current flux refraction if
PROBLEMS

173

while the boundary condition involving tangential components is obtained from (3-79), with J = O"E

(2) From the geometry, the tilt angles obey

The latter combines with (1) and (2), whereupon inserting the expression for tan (J 1 obtains the refl'active law (3-138) The analogy with the refraetive laws (3-76) and (3-80) for Band E might be noted. For an example in which 0"2 = 10(T 1, the refractive effects of direct current streamlines at an interface are shown typically in (b) of the accompanying figure. For (T2» (Tl' the ncar perpendicularity of the current flux occurs in regions 1, as noted in (c). If 0"1 were reduced to zero, thert J 1 = 0, constraining the current flow in region 2 to paths tangentialto the conductor-insulator boundary as in (d), a result evident from the insertion ofl. l = lt1 = 0 into the boundaIY conditions (4-133) and (4-134).

REFERENCES ELLIOTT, R. S. Electromagnetics. New York: McGraw-Hill, 1966. JAVID, M., and P. M. BROWN. Field Analysis and Electromagnetics. New York: McGraw-Hill, 1963. JORDAN, E. C., and K. G. BALMAIN. Electromagnetic Waves and Radiating Systems, 2nd ed., Englewood Cliffs, N.J.: Prentice-Hall, 1968. LORRAIN, P., and D. R. CORSON. Electromagnetic .Fields and Waves. San Francisco: Freeman, 1970. REITZ, R., and F. J. MILFORD. Foundations of Electromagnetic Theory. Reading, Mass.: AddisonWesley, New York: 1960.

PROBLEMS SECTION 3-1 3-1. Pure copper, with a free (outer orbit) electron concentration of about 10 29 electrons/m 3 , has the conductivity (T 5.8 x 10 7 mho/m at room temperature (Table 3-3). (a) Find the mobility orthe free electrons in copper. (b) Express the free electron charge density in coulombs per cubic millimeter for this material. (c) Find the drift velocity of the electrons for the unit applied electric field E = ax V 1m. What is the corresponding volume current density in this specimen? Sketch the vectors depicting Vd, J, and E in the sample. (Explain from physical reasoning why Vd and E are in opposite directions, although J and E are in the same sense.) 3.6ax mm/sec] [Answer: (c)

3-2. Find the current density (expressed in A/cm l ) in the following conductors, possessing only negative electronic charge carriers under the given conditions. (a) The average drift velocity is -a z4.5 mm/sec and the charge carriers have the density 2 x 10 28 electrons/m 3 . (b) The volume density of electronic charge carriers is 3.5 x 10 8 C/m 3 and the carrier average drift lOa x V /m within the conductor. What is the conductivity of velocity is 4.2 mIll/sec, with E the region in the latter case? [Answer: (a) a z 1440 A/cm 2 (b) a x 147 A/cm 2 , 0.147 MU/m]

174


SECTION 3-2 3-3. At some particular temperature, helium gas has 10 25 atoms/m 3 and is measured to have the dielectric susceptibility of 1.5 x 10- 4 . What is its electric polarization field P for the applied field E 10 3 V /m? What is the charge density p +, and the average displacement d of the nucleus relative to the electron cloud for the given E? What is E/ [Answer: p + 3.2 X 10 6 C/m 3 J 3-4. At low frequencies, the measured relative permittivity of water is 81 Crable 3-3). What is then its electric susceptibility? What electric field E must be applied to produce, at the sinusoidal frequency w, the polarization field P = a x lO sin Wi JiC/m 2 in a water sample? (Express E in kV/m.) Find the corresponding electric displacement density D, expressed in JiCjm z . Without using field values, (orm appropriate ratios to determine by what factor the magnitude of D is larger than that ofP; similarly, compare P with EoE. 3-5. (a) To make the electric polarization density P and the applied field EoE exactly the same in a material, what must its relative permittivity be? (b) What is the relative permittivity of a material if P has 10 times the value of EoE therein? (c) If D has 10 times the strength of EoE in a material, what is its relative permittivity? (d) IfD is lOP in some region, what is its E/ [Answer: (b) II (d) l.IlI] 3-6. The same electric field, E 10 3 a z Vim, is applied to the following regions having the dielectric susceptibilities: (a) zero (what sort of region is this?); (b) 10-\ (c) I; (d) 10 3 . Determine the relative permittivity, the applied field EoE, the electric polarization field P, and the eleetric displacement field D for each region.

SECTION 3-2A 3-7. The following E fields are given to exist in some block of polyethylene, for which E, = 2.26 (from Table 3-3): (a) a x l0 3 x 2 sin wtV/m; (b) a p 10 3 p sin wi Vjm; (c) a r (10 3 /r2) sin Wi VIm. Find the fields EoE, P, D, the polarization (bound) charge density PP' and the volume polarization (bound) current density JP for each applied E field. [Answer: (c) Pp = 0, . JP = ar(1 Ll4/r 2 )w cos wI nA/m 2 ] 3-8. Corresponding to the electric polarization field P = a x lO sin wi JiC/m 2 of Problem 3-4, find the polarization (bound) current density J P at the frequencies: (a) I kHz; (b) I MHz.

SECTION 3-2B 3-9. Apply the Gauss-Maxwell integral law (3-36) to a vanishing volume element tl.v in dielectric region, to rederive its differential form (3-24). [Hint: Divide (3-36) by tl.v and consider the meaning of each ratio as tl.v -> 0.]

3-10. Making use of the divergence theorem, show how the differential expression (3-21) can be manipulated to yield the integral form (3-38). Explain the physical meaning of this result.

SECTION 3-2C 3-11. The coaxial, circular cylindrical conductor pair (coaxial line) of great length and with the dimensions shown contains a homogeneous dielectric sleeve with the permittivity E.' Assume the static surface charges totaling ± Qon every axial length t of the inner and outer conductors respectively. (a) Making use of the symmetry and Gauss's law (3-37), determine for each region between the conductors the D and the E fields. (b) Determine P in the dielectric region. By use of the criterion (3-21), determine whether there is any volume density of excess polarization (bound) charge, of density Pp' within the dielectric. (c) Making use of the appropriate boundary conditions, find the free charge densities on the conductor surfaces at p = a and d, as well as the surface polarization (bound) densities at p = band c. (d) Letting a = 2 mm, b = 4 mm, C 8 mm, d = I cm, Qlt = 10- 2 JiC/m, and Er = 2.26 (polyethylt;:ne), find the values ofE and P. at the conductor surfaces at p a and d. Find also D, P, and E at the surface p = b + (just within the dielectric), comparing their values with those at p = b- (just outside the dielectric). [Answer: (d) E(a) 90 kV/m, ps(a) 0.796 JiC/ml, E(b+) = 19.9 kV/m]

PROBLEMS

175

PROBLEM 3-11

Assume that the region a < p < d between the coaxial conductors of Problem 3-11 is filled with a single, inhomogeneous dielectric material for which the permittivity is E(p), a hmetion of only p, (a) Make usc of the symmetry and Gauss's law (3-37) to establish the limctional dependence of E on p required to make E between the conductors independent of p. Express the answer such that E(p) lIas the value Er at the outcr radius p = d. What is then E? (b) Find both the polarization density field P and the volume density Pp of polarization (bound) charge Note that this nonuniform design of the dielectric region provides a for this choice of fields in a coaxial configuration, thus reducing the possibility of diway to avoid high electric breakdown. Suggest how the nonuniform permittivity conditions of this problem might be met say, three or four diflerent but homogeneous dielectric materials. [Answer: Pp = QEo/2nE r dtpJ The concentric, spherical conductor pair is separated by two dielectric shells of permittivities El and E z as shown, the interface between them appearing at r = b. Assuming the static surface charges totaling ± Q on the inner (r = a) and outer (r = c) conductor surfaces, respecand the symmetry to deduce D and E tivcly, answn the following. (a) Use Gauss's law within the two regions. Both these fields arc normal to the interiace at r = h. Which boundary condition in Table 3-2 is applicahle at this interlace? (b) Find the expression for P in each region. From I), deduce whether there is a polarization (hound) charge density Pp within either dielectric region. (c) Employ the proper boundary conditions to find the free surface charge density Ps on the conductor surfaces, as well as the surface polarization charge density Psp at r = b. (d) With a \ m, b 1.02 1Il, c 1.05 m, Erl = 2.26, Er2 = I and Q = 0.\ flC, find the values of E at the radii b- and b+, as well as on the conductor surfaces r = a and c. Sketch E, versus r from a to c. [Answer: (a) Ed = QI4nElr2 (d) Erl(a) = 398 k V 1m1

." 1111

PROBLEM 3-13

176


SECTION 3-3 3-14.

Based on a pillbox construction suggested by Figure 3-4, prove the boundary condition (3-50) concerning the continuity of the normal components of B to either side of an interface.

SECTION 3-4 3-15.

Beginning with the force (3-51) acting on each edge of the current loop of Figure 3-7 (a), fill in the remaining details to prove (3-54).

3-16. Prove that the net magnetic force F B, acting on an arbitrary, closed, thin (filamentary) circuit carrying the uniform current f and immersed in a uniform B field, is zero. [Hint: Integrate (3-52) about the circuit path t, noting that' B x dt can be written B x ,dt. See also Example 1-6.]

3-17. Expand the dimensions of the square current loop, located in the z 0 plane as in Figure 3-7(a), to a large scale by assuming each side to be 2a meters long (side t1 located at y = a in the z = 0 plane, etc.). The current flows clockwise when looking in the +z direction. (a) Assume the current loop to be immersed in a magnetic field having only the component B z . Sketch the system. Determine the force dF 1 due to the magnetic field acting on any element -laxdx of the side ( I ' as well as the total force on (1' showing that F I = ay2aIB z • By symmetry, show that the net magnetic force on the loop is zero. Show also that the differential torque acting on a current element of ( I ' relative to the moment arm R axX + aya measured from the origin, is dT = aJBz 2xdx, and that the total torque on (1 is zero. (b) Repeat (a), assuming this time that only Bx is present. Show why the forces on the sides t1 and t4 are zero, and that the total torque on the loop, due to B x , is ay4a2fBx- Find the total torque, due to By only, by analogy. \\lith all three components ofB present, what is the total torque on the loop? (c) Defining the magnetic moment of this finite sized current loop as Dl = Is = 4a 2 fa" show that the total torque expression obtained in (b) is equivalent to (3-54), T = Dl X B. (d) What is the magnetic moment of a thin, sqnare current loop of sides 2a 10 em and carrying 10 A, immersed in the field B = 0.3a x + OAa y + O.6a z Wb/m 2 ? 3-18. A particular magnetizable material has 3.3 x 10 28 atoms/m\ and with the steady magnetic field H = 4a z A/m applied to its interior, there results the average dipole moment Dl 1.2 X 10- 24 A_m 2 (a) Find the density of magnetization M, the magnetic susceptibility, the relative permea bility, and the permeability of this materiaL (b) Find B in this materiaL [Answer: M 39,600 A/m, p 12.4 mH/mJ 3-19. The magnetic susceptibility of a particular specimen of magnetic material is measured to be 59. \\lhat is the magnetic polarization M and the magnetic intensity H, if the field B in the material is O.Ola x Wb/m 2 ? 3-20. Given the following volume magnetization fields M within certain regions of magnetic materials, find the volume densities 1m produced by the bound currents therein. a 4,200/p (a) 150xay (c) a4>320 (cylindrical) (d) aolOOrcos 0 (e) a4>160/r 2 [Answer: (b) 0 (d) a4>200 cos OJ 3-21. Employ Stokes's theorem (2-56) to show how the curl relationship (3-56), relating the magnetic polarization density M to the magnetization current density 1m, is transformed to the integral relation (3-67).

SECTION 3-4A Show that the magnetization current density 1m = aylO A/m 2 associated with the bound currents in the sample of Example 3-3 yields, from an appropriate surface integral, the total (bound) current flow of lOP A through any fixed z cross section of that sample. Obtain the samc answer by use of the line integral of (3-67). Sketch the system, appropriately labeled.

3-22.

PROBLEMS

177

SECTION 304B 3-23. Employ a suitable sketch, showing how the quantity n x Hb used in the magnetic-field boundary condition (3-72), specifies the lanJ!,ential component of the surface current density Js in both magni tude and direction. 3-24. Apply the appropriate boundary condition in answering the following. (a) An air-toperfect-conductor intcrhce is at z = 0, the region z > 0 being air. With H = 150ax Aim in the air region, what is the snrface current density on the perfect conductor? How much total current I flows in a 20-cm-wide x-directed strip of this conductor surlace? Sketch this system showing H, J" and a [(OW current flux lines. (b) Find the current density on the conductor surface of (a), this time assuming H = 30ax + 40a y A/m. Sketeh this system. (c) Suppose in the geometry of Figure 1-19(a) that the long, straight wire shown is a perfect conductor, and that surface eurrents totaling I flow on the conductor surface p = a. The B field for p > a is still given correctly (1-64). Use this field to deduce the surface current density Js on the wire. Formulate a vector integral relationship between J and J" showing a related sketch. 3-25. What two simultaneous boundary conditions arc being satisfied by the magnetic field refraction expression (3-76)? Establish that, if region I is air and region 2 is iron with 11,2 = 104 (a case of high contrast in permeabilities), the tilt angle 01 of Br from the normal in region I is very slllall for most values or0 2 . For example, find 1 if0 2 = 0, 45°, 89°, and then 89.9°. How lin from the normal must O2 be if 8 1 is to become as large as 10°? Sketch this example.

°

3-26. The toroidal iron core ofrectangular cross section partly fills the closely wound toroidal coil of!l turns and carrying the direct current 1 as shown. (a) Usc the right-hand rule (thumb in the sense or 1) to establish the direction of H inside the winding. (b) Use the static form of Ampt're's law C)-66) to deduce H at any radius p within the winding, and determine B for is being satisfied the two regions. Which boundary condition for magnetic fields Cfable at the air-iron interface? (c) From H deduce expressions for the magnetization density field M in the two regions. Sketch flux plots showing (in side views) the relative densities of H, B/l1o, and M in the two regions, assuming 11, » 1 for the iron. (d) Find Jm within the iron as well as J8m on the timr sides of the iron core. Sketch representative vectors or fluxes depicting these quantities. (c) If a = 1 em, b = 1.5 cm, C= 2 em, d = I cm, fl, = 1000, n = 100 turns, 1= 100 rnA, find t he values of Hand B at p = a + and b (just within the iron), at p b + and p = c.

3-27. As a simple exercise iu applying boundary conditions, an air space (region 1) defined for all > 0 and a magnetic substrate with 11, = 4 (region 2) occurring for all z < 0 are separated by the inllnitc pbmc interlace at z O. The constant, static magnetic field in region I is given to be BI = O.3a x + O.4ay + 0.5a z Wb/m 2 . Sketch BI (shown for convenience at the origin) and the Ilormal unit vcctor n at the interface (its direction taken as going from region 2 to region I). (a) Make use of the boundary conditions (Table 3-2), concerning the continuity of appropriate tangential or llormal field components at the interface, to deduce the vector fields HI> B 2 , and H2 in the as well as the field magnitudes. (Leave H expressions in terms of

PROBLEM 3-26

178


If"\!, ~----------

, f

\

~+-..,...-,

\-+4---:/ -11-+"""-,

(

----t--- (z)

1: (1-'0)

'-:1-+.......- ;

' -2;-(p) 3;

(1-'0)

PROBLEM 3-28

the symbolic 110.) (b) By use of the definition of n • B, find the angles (), and (}2 between n and B (or H) in the two regions. (Label (J, on the sketch.) Check your answers by use of (3-76). [Answer: (a) B2 = 1.2a x + 1.6ay + 0.5a z Wb/m 2 (b) (}2 = 76°]

3-28. A very long, nonmagnetic conductor (fl. I) of radius a carries the static current I as shown. The conductor is surrounded by a cylindrical sleeve of nonconducting magnetic material with a thickness extending from p = a to p = b and the permeability fl. The surrounding region is air. (a) Make use of symmetry and Ampere's law (3-66) to find Hand B in the three regions. (Label the closed lines employed in the proof, depicting H in the proper sense on each line.) (b) Find the M field in the magnetic region. If 1= 628 A, a = I em, b = 1.5 cm, fl. = 6 for the magnetic sleeve, sketch H"" B"" and M", versus p for this system. Comment on the continuity (or otherwise) of these tangential fields at the interfaces. (c) By use of (3-56) and (3-73b), find the volume magnetization current density 1m and the bound surface current densities J,m within and on the magnetic sleeve.

SECTION 3-5 3-29. Two semi-infinite regions, air (region I) for z > 0, and a dielectric (region 2, in which E = 4Eo) for z < 0, are separated by the interface at z = O. In the air region, the constant electric field El = l5a x + 20ay + 30az Vim is given. Sketch El for convenience at the origin. (a) Find D and E for both regions, making use of boundary conditions (Table 3-2). (Leave Eo explicitly in the D expressions.) (b) Find the refraction angles ()l and ()2 from the normal in both regions, making use of the definition ofn . E ifn is directed from region 2 to region 1. Use the refraction law (3-80) as a check. Answer: E2 = - 15ax + 20a y + 7.5a z V /m, ()2 = 73.30°]

r

SECTION 3-7 3-30. Prove the expressions (3-90a) and (3-9Gb) for the attenuation constant IX and the phase constant fJ associated with uniform plane waves in an unbounded, lossy region. 3-31. Assume uniform plane waves to be traveling at the frequency f= 100 MHz in a lossy region having the constitutive parameters fl = flo, E = 6Eo, (J = 10- 2 mho/m. (a) By direct substitution into (3-88), determine the value of the complex propagation constant associated with the waves, expressing y in its complex rectangular 101'm denoted by (3-89). From this result infer the values of the wave attenuation constant and phase constant. (b) Find the attenuation constant and the phase constant by use of (3-90a) and (3-90b). [Answer: IX 0.761 Np/m, f3 = 5.187 rad/m] 3-32. Repeat Problem 3-31, this time assuming the parameters of the lossy region to be fl = flo, E I.B Eo, (J = 10 mho/m, and in which uniform plane waves are traveling at the frequency f= 10 GHz. [Answer: y = 597.7 + j660.5 m- i ] 3-33. M.aking use of the free-space parameters fl flo, E Eo and (J 0, show that the expressions (3-90a), (3-90b) and (3-99a) reduce to the free-space results IX = 0, fJ = fJo of(2-11B), and I] = 1]0 of (2-13Gb). 3-34. Prove that the penetration of three skin depths by a plane wave into a conductive region produces an amplitude reduction to 5% of the reference value. Show that six skin depths yields 0.25'Yo'

PROBLEMS

179

3-35. Given the electric-field plane wave solution (3-91 b) in which the propagation constant is defined by (3-88), show by substitution into the Maxwell curl equation (3-83) that the corresponding magnetic field solution becomes (3-98c), if the intrinsic wave impedance q is defined by (3-99a). [Hint: Show that the coefficient y/jWfl reduces to q-I.] 3-36.

Show that the expression for intrinsic wave impedance

q,

defined by (3-97) as

can be re-expressed in complex polar form by the last expression given in (3-99a).

E;

H;,

3-37. A positive z-traveling, uniform plane wave has the field components and with the electric-field amplitude E; = 200 Vim, and operates at the frequency f = 100 MHz. It travels in a lossy region with the parameters given in Problem 3-31 (flo, 6Eo, (J = 10-2 ). The propagation constant in this region at 100 MHz was found to be y = 0.761 + jS.IS7 m- I ). (a) Determine the wavelength of this wave. Find its depth of penetration, b. What is the phase velocity of this wave? (b) Determine the intrinsic wave impedance q fo!, this region, at the (z) accompanying given frequency. Use this to obtain the expression for the magnetic field the given electric field. (c) Show a labeled sketch, patterned after Figure 3-18, showing the real-time E; t) and H; (z, t) fields of this uniform plane wave, at t = O. Label the depth of penetration as well as the wavelength on your diagram. [Answer: (a) b = 1.314 m (b) ~ = 150.6e iS . 4 ' il]

H;

3-38. A vehicle located far above the surface of the sea transmits an electromagnetic signal at the frequency j. Upon striking the air-sea interface, a transmitted wave penetrates the sea. The waves at the surface are presumed to be sufficiently far from the source that they,may, locally at least, be considered to be uniform plane waves. Supposing the net transmitted electric field amplitude is f-; = I Vim, how far will the wave penetrate before reaching of its surface value? Perform this calculation at two very low radio frequencies: 10 kHz (in the VLF range) and 1000 Hz (ELF), assuming sea water has the constants E, = 81 and (j = 4 U/m at these frequencies. Comment on the effectiveness of undersea radio communication, bascd on your results.

SECTION 3-8 3·39. Use the general expressions (3-90a, 90b, 99a) to derive (3-112), applicable to a good conductor (for which a/wE> 1).

3-40. for ex,

p,

Use the general exprcssions (3-109) through (3-111) to prove (3-113), approximations and q applicable to waves traveling in good dielectrics, for which EN/E' « 1.

3-41. An electromagnetic uniform plane wave is specified in some lossy region by the fields E; (z) = 3142e- Yz Vim, fI; (z) = fI;e- YZ A/m at the frequcncy f = 1000 MHz. The region has the parameters 11 = /-10, E = 24Eo, and (j = 48 mho/m. (a) Show that the loss tangent of this region at the givcn frequency is 36. Is the region classified as a "good conductor" or not? Explain. (b) Find the attenuation and phase constants of the region at this frequency. (Reasonable approximations are allowed.) Show that the "depth of penetration" of the wave into the region at this frequency is about 2 mm. Use a sketch of the real-time electric field (in the vicinity of the z-origin) to explain the meaning of "depth of penctration." (c) Find the wavelength of this wave, labeling it on the sketch of part (b). Compare this wavelength with that occurring in this region assuming now that it is completely lossless (samc fl and E, but now with a = 0). Comment. (d) Evaluate the complex amplitudc fI; of the magnetic field of this plane wave, making use of the intrinsic wave impedance of the region. (To what fact do you attribute the angle of q being close to 45°?)

CHAPTER 4 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __

Static and Quasi-Static Electric Fields

Electric fields of stationary charge distributions in space are considered in this chapter. Maxwell's equations, subjected to the time-static assumption, tJ/Dt = 0, provide an uncoupling of the static electric fields from the static magnetic fields. Gauss's law is applied to symmetrical systems; and the scalar potential field (]) is derived to supply an intermediate, often simplifying, step useful f(lr finding the static E field. Expressions fix the stored energy of an electrostatic system are derived and applied to twoconductor capacitance systems. Boundary-value problems of electrostatics are treated by means of Laplace's equation and extended to finite difIerence methods of solution for arbitrary, two-dimensional boundary shapes. Image and fIux-mapping techniques are discussed as alternative approaches to capaci lance problems, all(i. a capacitanceconductance analog is developed. The chapter is concluded with a consideration of the fc)rces of electric charge systems.

4·1 MAXWELL'S EQUATIONS FOR STATIC ELECTRIC FIELDS In Chapter 3, the Maxwell equations and boundary conditions f(lr time-varying electromagnetic fields in material media at rest were developed. Time-varying B (or H) and E (or D) fields are produced in a region whenever the charge and current sources of the fields are time-varying. For certain generic classes of field problems, it is advantageous to consider the sources to be non-lime-varying, that is, time-static (or just static). Then the charges and possibly currents responsible for the fields are stationary. The governing Maxwell equations for time-static fields are (3-24), (3-48), (3-59), and with the operator a/at set to zero, yielding V . D PI!' V x E 0, V' B = 0, and V x H = J. The static fields are designated D(u j , 11 2 , Pv(Ub U2, and so on.

180

4-2 STATIC ELECTRIC FIELDS OF FIXED-CHARGE ENSEMBLES iN FREE SPACE

181

An inspection of the static Maxwell relations reveals a new property not valid for their more general, time-varying forms. Thus, the static electric fields D and E are governed solely by the divergence and curl properties V' D

= Pv

VxE

(4-1 )

0

(4-2)

whereas the behavior of the static magnetic ,fields Band H is dictated by

V·B = 0

(4-3)

v x H=J

(4-4)

The coupling between the electric and magnetic field quantities, generally provided under time-varying conditions by the terms aB/at and aD/at appearing in (3-77) and (3-59), is seen to be missing in these pairs of equations. The sources of electrostatic fields are, from the divergence expression (4-1), static charges of density Pv' Magnetostatic fields, on the other hand, have static (direct) currents for sources, as noted in (4-4). In the present chapter, solutions of the electrostatic field equations (4-1) and (4-2) are considered from several points of view, whereas a detailed discussion of magneto statics by use of (4-3) and (4-4) is deferred until Chapter 5. The differential equations of electrostatics are, together with their integral forms and boundary conditions, given TABLE 4·1 Maxwell's Equations of Electrostatics DIFFERENTIAL FORM

V- D = Pv

vx

E

=

0

INTEGRAL FORM

[4-1]

~s D

[4-2]

1:

~

. ds = q

(4-5)

E - dt =O

(4-6)

BOUNDARY CONDITION

Etl

Et2 = 0

(4-3)

in Table 4-1. For a linear, homogeneous, and isotropic material, moreover, (3-30c) is applicable

D=EE

(4-9)

4·2 STATIC ELECTRIC FIELDS OF FIXED-CHARGE ENSEMBLES IN FyE SPACE

••

The Maxwell equations in Table 4-1 apply to fixed charges in free space, as well as to systems of rlielectrics and conductors into (or onto) which charges have been introduced such that static equilibrium of the charge distribution has been reached. Examples of the applications of the Gauss law (4-5) are given in Section 1-9. One of the results, (1-58), is Coulomb's force law (4-10a)

182

STATIC AND QUASI-STATIC ELECTRIC FlELDS

FIGURE 4-1. Illustrating quantities appearing in Coulomb's [()rcc law,

glvmg the force acting on q m the presence of the field E produced by a second charge q as shown in Figure 4-1, The symbol R is used instead of the spherical coordinate variable r because the source q is not necessarily located at the origin 0, The field of q was deduced from Gauss's law in Section 1-9 to be

E = aR--'---;;:-N/C

or

Vim

lOb)

Thus (4-10a) is a special case of the Lorentz [(lITe law (1-52) in the absence of a magnetic field; that is, F = q'E, Maxwell's equations (4-1) and are linear equations; therefore, any sum of their solutions in free space constitutes a solution, Suppose an aggregate of point charge of arbitrary posi tive or negative strengths is located at fixed points P' as in Figure 4-2, The total electrostatic field at the field point P is the sum of n terms like (4-lOb)

E Moreover, if a charge

Ij

10e)

is placed at P, the [()rce on it, from (1-52), becomes (4-10d)

If a system contains a large number of fixed charges, it is undesirable to use a surnmation like (4-lOc) or (4-10d), It is pre/tTable to replace the charge ensemble

(z)

FIGURE 4-2. Electrostatic fIeld of n discrete charges.

4-2 STATIC ELECTRIC FlELnS OF FIXED-CIIARGE ENSEMBLES IN FREE SPACE

183

with a jime/ion representing the average charge density in every volume-, surface-, or line-clement of the region. The symbols Pv, p" or Pc have been used to denote these density functions as discussed in Section 1-9 relative to Figure 1-1]. A continuum of charges distributed throughout some region with a density Pv thus possesses the charge dq = Pv dll in every dv element. Generally, Pv is a function of position and time, though for static Helds, the variable I is missing. With dq p"dv' located at the source point P' (x',y', the fidd dE at P due to dq is obta.ined from (4-10h), written

dE(x,y, <:) = a R

pv(x',y', z') 4nEoR

2

dv'

(4-11 )

The unit vector directed from the source point P' to the field point P to give the proper direction to dE is denoted by aI{, whereas R is the scalar distance Irom P' to P, as in Figure 4-3. In rectangnlar coordinates

R=

(4-12)

The total static E field at P in Figure 4-3(a) is thus the volume integral of (4-11) (4-13) Iflhe static charges are distributed over a surface S or a line l as in Figure 4-3(b) and (e), the following in1egrals apply (4-14)

waf (4-15 )

y

Field POint P(r) P (x,y,z)

dE (x,Y,z)

}f

P (x,y,z)--.....

~

R /

Source point P' (r') P' (x',y',z')

=

" .\ / \

I I

~/ /<~

I _ I/ ~

0

/

R

. dE

IdE

/ //r

/

dq = Pv dc'

/

/

)

dq

= Psds'

~ Volume charge

P' (x',y', z') Surface charge

----_________---.. (a)

(b)

(c)

FIGURE 4-3, Geometries rdative to the electrostatic field integrals in terms of volume, suri~,ce, and line charge distributions. (al Volume charge distlibution. (b) Surface charge distribution. (e) Line charge distribution.

'fIlII'

184

STATIC AND QUASI.STATIC ELECTRIC FIELDS

The foregoing integrals are not always readily evaluated for charge distributions in space, mainly because the unit vector a R changes in direction as the source point P ranges throughout the charge region. In some cases, the symmetry disposes of this problem, as in the following example.

EXAMPLE 4-1. The linear charge of length 2L, centered in free space about the ongm as shown, possesses the static uniform charge density Pc C/m. By direct integration of (4-15), find E at any distance p perpendicularly away from thc ccnter of thc linc charge. Find E as the line charge becomes infinitely long. In evaluating (4-15), only the two circular cylindrical coordinates (p, z) are required here, because of the axial symmetry. Comparison with Figure 4-3 shows that the position vector r of the field point P (from the origin to P) is r a"p, while that of the typical source point P' is r' = a.z'. This makes the vector distance R from P' to P becomes R r - r' app azZ' with magnitude R .Jp2 + (<:,)2. The unit vector aR in (4-15), directed from P' to P, thus becomes from definition

(4-16) Then (4-15) yields at P

E

(4-17)

The vector field E at P is thus seen from (4-17) to consist of two components, E apEp + azEz . The axial component azE z , given by the second term of (4-17), can be shown to integrate to zero. However, this integration is not even necessary hel-e, since an inspection of the symmetry in the diagram shows that canceling dE. components are produced at P by paired, symmetrically located source-elements on the line charge. Thus, using only the a p term of (4-17) (while integrating over only the range (0, L) (z)

I

IL

+ +

-

~

+-----+

p,dz'

+ -L + I I

EXAMPLE 4-1

____R

p

P(p,O)

-----_\ --~ --~--------(p)

- --;; --

..,...----

~

dE=aRdE

4-2 STATIC ELECTRIC FIELDS OF FIXED-CHARGE ENSEMBLES IN FREE SPACE

185

aml doubling the result) yields

L

=a p

(4-18)

2nEo

If the field point location P were quite close to the finite line charge compared to its length (p« L), then a good approximation to (4-\8) is seen to become E(p)

PI a-I' 2nEop

(4-\9)

It is evident that the latter also applies to the intinitely long line charge case (L which checks with (1-6\) obtained by use or Gauss's law.

->

(f)),

EXAMPLE 4·2. Consider the same linear charge system of Example 4-\, except move the field (observation) point to the more general location P(p, z) as shown. Find E at P by direct integration. In this case, the vector distance R from P' to P becomes R = r r' = a"p + ao(Z yielding the unit vector determininf!; the sense of dE at P: app

+ az(Z-

J+ (,~Then (4-15) obtains

E =

JL-

L aR

PtdZ' 4nEoR2

= -Pt -- [ a p

4nEo

o -L

f!

I I

EXAMPLE 4-2

P

JL

-I.

l)z

(1-20)

186

STATIC AND QUASI-STATIC ELECTRIC FIELDS

which, with the help of integration tables, provides the result

E(p,

z)

=

4:: 0

[apC~p2 ~ (Zz_ L)2 + p~p2 ~~z\ L)2)

+az

-~p2+;Z+L)2)]

I

(4-22)

As the line-charge length is made infinitely long (L -> ex:», it is seen from (4-22) that the a z component ofE vanishes, leaving only the p component.

. E hm

L~ 00

(I + -I) =

= Pr a 4nEo

P

_. P

P

a -Ptp 21tE oP

(4-23)

which agrees with (4-16).

EXAMPLE 4-3. Assume the circular disk region, of radius a as shown, to possess a static charge of uniform surface density P. C/m 2 • By direct iutegration, find the field E at any point P(O, z) on the z-axis. Use the applicable integral (4-14). The vector distance from P' to P is given by R = r - r' = + azz. Making use of a R = R/R, (4-14) in circular cylindrical coordinates becomes

-app'

(I) wherein note the positive sign of the a p component in the bracketed quantity, conforming to the positive sense of the radial component of dE at the point P (0 +, z) located very close to the z-axis on the diagram.! The axial symmetry of the charge distribution here provides only the E z component on the z-axis at P, allowing the a p component in (1) to be discarded.

p,ds' = p,(p' dp' d.p')

P'(p',.p',O)

EXAMPLE 4-3 'The unit vector of a p is not defined on the z-axis; it requires an infinitesimal radial displacement to the location P(O +, z) as noted. However, this observation is here largely academic, for the symmetry causes the Ep component to integrate to zero.

4-3 GAUSS'S LAW REVISITED

187

Thus, E at P becomes (2)

Ps

= a z

2Eo

l

I

(4-24)

As the disk radius a is made infinitely large, the result (4-24) should agree with the field (1-62) obtained by use of Gauss's law. With a -> CfJ substituted into (4-24), one obtains,

as expected,

E

=

a!!!z 2Eo

(infinite charged plane)

(4-25)

Note in the foregoing problem that instead of writing (I) for the totallleld dE, one might instead have discarded the Ep component initially (based on the symmetry argument) and then proceeded to write the integral expression lor just the component E z at P. Thus, E =

P ds' cos IX r dEz = Jsr dE cos IX = f,s -:~--4'1l:EOR2

z Js

(4-26)

in whieh IX denotes the angle between dEz at P and the total dillerential field aligned with R there. Putting cos IX = z/R into (4-26) is seen to lead directly to the integral (2) as obtained before.

4-3 GAUSS'S LAW REVISITED In Section 1-11, it was shown that Gauss's law (I-53) for free space may be used to obtain the electrostatic field of charge distributions having particular symmetries. Parallel-plane, concentric-circular-cylindrical, and concentric-spherical charge distributions in free space are particularly amenable to analysis by means of Gauss's law. If charge distributions are combined with conducting and dieltxtric materials having shapes that yet preserve the symmetry, then Gauss's law in the form of (3-37) or (4-5)

91D'ds =q C

[3-37]

is useful fiJI' finding D in the various regions, taking into account the possible polarization effects in the dielectric regions.

EXAMPLE 4·4. A pair of long, coaxial, circular conducting cylinders are separated by concentric air and dielectric regions as shown, the inner ring bcing a dielectric with = 4(a < P < b). A static chargc q is assumed distributed over each length t of the inner conductor, with - q on the other conductor. Use Gauss's law to find D and E. from the symmetry, the D lines between the conductors are radial and independent of 4>. The field is found by use of a symmetric, closed cylinder S enclosing the inner conductor as in (c). The procedure may be compared with Examplc I-II (fl. With S

"r

188

STATIC AND QUASI-STATIC ELECTRIC FIELDS Gaussian surface S

(a)

(c)

(b)

EXAMPLE 4-4. (a) Showing the dimensions of the coaxial pair. (h) Depicting the continuity ofD lines. (e) Symmetric closed S for finding the fields.

having a length Gauss's law

SO)

t

and radius p, and the p directed D piercing the peripheral surfacc becomes

r (aP D P ) . a P ds = q Jso Dp is constant over So) whence D = p

q 2npt

(4-27)

a result applying to both the dielectric and air regions. Using (4-9), E in the respective regions becomes

=--_1:£-

E p

With D

2n( 4Eo)pt

a
(4-28)

b
(4-29)

0 inside the conductors, the boundary condition (4-7) reduees to (4-30)

which, applied to (4-27) at p = a and p = (, yields the free charge densities on the ductor surfaces

PsJ - = p-a

COIl-

2navI'

4-4 ELECTROSTATIC SCALAR POTENTIAL Any electrostatic field E(ul' U 2 , u3) must satisfy the curl relation (4-2), V X E = 0, which states that any static E field is irrotational, and therefore conservative. In view of the identity (20) in Table 2-2, that V X (V<])) = 0 for any difIerentiabl.e scalar function, (4-2) means that E is derivable from an auxiliary scalar function <])(Ul' U2, U3)

4-4 ELECTROSTATIC SCALAR POTENTIAL

189

by means of the gradient relation

= -V Vim

E

(4-31 )

The nature of the function having this correspondence to some E field is not evident from (4-31), but it is clarified by two related methods described in the following. The first obtains the potential from the known charge distribution of density Pv, and once has been found, the E field is obtained using (4-31). The second method presumes E known at the outset of the problem; is found from an appropriate line integral of E over a path beginning at a designated potential reference.

A. Potential $ Obtained from a Known Charge Density in Free Space The relation of the electrostatic E to its charge sources injree space is (4-13) E(x,y,z) =

pv(x',y', z') , aR----2--dv Vim 4nEoR

1 v

[4-I3J

A dependence on the variables z') is evident in this integral because, r-----~.-----~~-~----~~.c~ by (4-12), R x') + (y . One can show by direct expansion that

V(~) = - ~~

(4-32)

assuming that V is defined in terms of derivatives with respect to the field point variz) as follows ables (4-33) This permits writing (4-13) E(x,y, z)

(4-34)

in which an interchange between the integration and the gradient operations is permissible because the only quantity affected by V is R, while the integration is to be carried out with respect to the source point variables (x' ,y', . Comparing (4-34) with (4-31) shows that the integral in (4-34) is the desired scalar function <1>; hence

(x,y, z)

'-'--'--"--_-'.. dv' V

is called the scalar potential field of the static E field.

(4-35a)

190


It is further evident that if the charge density in (4-35a) takes the form of a surface or line charge density Ps or Ph then the integral becomes

"'(

"I'

)_fsPs- 4( - -, RZ')- dJ '

X,], Z

X,],

-

(4-35b)

j

·1[E o

(4-35c)

(x,], z)

B. Potential

(J)

Obtained from a Line Integral of E

The potential field
~ E· dt =

[4-6J

0

true ft)r all closed lines t in space. Physically, (4-6) states that the work done on a test charge q, in moving it around any closed path t in the presence of a static field E, is precisely zero. This is equivalent to saying that the work done on q in moving it between two fixed points 1\ and P 2 in the field is independent of the shape of the open path connecting the points. This is evident if two dilkrent paths tl and t2 are used to connect PI and P 2' If the closed contour t of IS taken to be tl + t 2, then (4-6) yields (4-36) correct fix aU paths connecting PI to P2' The property (1-36) makes it possible to derive a single-valued potential field equivalent to (4-31) as follows. Suppose Po(u?, ug, ug) is fixed ill space, called the jlOtentiat and defined such that <1>0 there. The line integral ofE, over any path t connecting P () and any arbitrary P( u b u2 , as in Figure 4-1, is written in the fbrms, making use of I),

5,

p

Po

From

E· dt = -

5,P. (V
• dt

= -

5,1' (3<1> -- dx + -3<1> d] + -3<1» dz Po

ax

~y

oz

11), the integrand of the right side denotes the total difrerential d, whence

(1' E. dt

Jpo

- (P d

Jpo

the latter of which can be integrated to yield (1-37) Thus, the line integral of the static E field over any path connecting two points in space is just the difference of the potentials at those points. For most purposes.it is desirable to call <1>0 = 0 at the potential reference; then (4-37) becomes

(P) = -

(p(U\,U 2 ,U3)

JPo(uY,u~~u~)

E . dt V

(4-38a)


/

191

p (X,y,z)

/ /

/'

./

./ ./ ,/

/' ,/

---- --FIGURE 4-4, Development of the
This potential field <1>(P) can be made to agree exactly with results obtained from (4-35), ifone observes that (4-35) provides a zero potential at an infinite distance from a charge distribution grouped within a finite distance from the origin, Thns, with the reference Po at infjnity (4-38a) provides the absolute potential

<1>(1')

= - S~ E· dl V

(4-38b)

yielding the same results as (4-35), Sometimes, as in problems of academic interest involving charges that extend to infinity (e.g" the uniform line charge of infinite extent), the integrals and (4-38b) yield infinite potentials. (The integrals then do not exist.) In such cases, one should make use of (4-38a), using a zero reference value at a finite distance from the origin. Surfaces defined by setting becomes more positive as one approaches positive charges; the opposite is true lor negative charges.

EXAMPLE 4·5. (a) Employ the charge integral to determine first the potential at the field point P(p, 0) of the line charge offinitc length '2L centered on the z-axis, shown in Example 4-1. From the result, determine Eat P by use of (4-31). (b) Verify that the same potential result <1> as {()Hnd in part (a) is obtained by use of the line integral

(4-33b),

192


(a) In (4-35c), only the scalar distance R from the source-point P'(O, z') on the line charge to the fixed observation point pep, 0) is needed; this is R .) p2 + «;,)2. Then

from (4-35c), the potential <1J at P becomes

<1J

(p, 0)

=

fL

-L

= 47CE

dz' 47CEoR

tn(z' o

=

fL

P? dz'

-L-----;=O;=====:;c

+ #+'-iZ')2j]L L

.)p2+L2+L

Pr = -47CE o

t n -'-7=======-----

(I)

- L

which may also be written

<1J(p,O)

L + .)p2 +IJ· tn ----------

Pc

= -.-

[volt]

(2)

+L ==2tn-----p

(3)

p

27CE o

in view of the identity tn

.)p2

+ L2 + L -L

Then E at pep, 0) is found by inserting (2) into the gradient relation (4-31).

o<1J -a - -

-V<1J

E(p, 0)

pap

(4)

the desired result. (b) The potential field (2) can also be obtained by inserting the known E field (4) into the line integral (4-38b) and integrating the latter from the potential reference to the desired field point. The integration path chosen, between the zero potential reference at infinity and the field point pep, 0), is completely arbitrary for this conservative E field; but since the available E field expression (4) is limited to the Z = 0 plane, the integration path must be restrictcd to this plane. Thus, with dt = a p dp OIl the radial path from Cf) to p in the z 0 plane,

f~ E • dt

<1J(p, 0) =

= -

PeL 27CE o

27CE o

=

fP --;::=;===;c 00

tn-'-"-----p

which agrees with the result (2) as expected.

(5)


193

(z) --< , ' ' ' "

--' "\"P{ elm

(fO)

(a)

(b)

EXAMPLE 4-6. (a) Geometry ofinfmite line charge. (b) Equipotential surfaces.

EXAMPLE 4·6. Find the electrostatic potential at any field point located a normal distance p ii"om an infmitc line charge in free space having the constant density Pc elm as shown. Sketch a few equipoAssume the zero potential reference at the position Po(Po, 1>0, tential surfaces. The potential (lJ at any location P rdative to a fixed reference Po is (4-38a). The field [rom Example 4-1 was f(lUnd to be E = appt/2nEoP. Inserting this into (4-38a) and integrating over any path connecting Po to P as in (a) obtains (lJ(P)

-

f, .

p [ Po

yielding the result independent of (I> and

J

PI a -._. (a dp p

2nEoP

P

+ a 4> p d1> + a z

z (4-39)

1t is evident that pntting the zero potential reference at infinity in this result (Po --> C1J) is not desirable, for (lJ(P) then becomes infinite; a finitely located reference position is necessary. Equipotential surfaces are obtained by setting (lJ(P) of (4-39) equal to the constant values (lJl) (lJ2' (lJ3,· .. ; yielding P Pi' P2) P3' ... ) the circular cylindrical

sur/aces shown in (b).

EXAMPLE 4-7. (a) Find the absolute potential of a point charge q located at the origin r = 0 ill Figure , making usc of the field (4-]Ob). Describe its equipotential surfaces. (b) Show that the potential lield can also be obtained directly from the volume integral (4-35a) applied to concentrated charge q. Determine the potential at P if q is located at a general source point P' as in Figure 4-5( b).

(11) The E field in Figure 4-5(a) is (4-1 Ob). Integrating E ovcr any path between Po(to) and the arbitrary P(1") yields, from (4-38a)

(4-40a)

194


! tr-Y" Direction of integration

Potential reference \ Po \

E

=ar

q

Field

point~

'

\ \

R

\

\

\

\

r

\

\ q~\

(r)

/

//

P(Fieldpoint)

rol

q

./'E

;/X\PP «xx"Y, z) (R)

"/ p ' (x' ,y , ,Z ') /

/ X

r'/ I

//

/r

//

(b)

(a)

50V

-20V

(d)

(e)

FIGURE 4-5. Point charge q: geometry and equipotential plots. (a) Geometry ofa point charge at 0, showing t over which E is integrated to find tl>(P) relative t.o Po. (b) Geometry of a point charge located at P'(r'). P is the field point at which tl> is obtained. (e) Equipotential surfaces of point charge, potential reference at '0 assumed. (d) Equipotelltial surfaces of point charge: potential reference at infinity.

=

q 4nEor

(4-40b)

The latter is plotted in Figure 4-5(d). (b) The absolute potential ofa static charge can also be found from the volume integral (4-35a). Here the point charge is concentrated at P', so let Po dv -+ q and no integration is required. Then (4-35a) becomes
(4-40c)


195

The result (4'-40c) is useful for constructing the absolute potential of an aggregate of n charges in free space like that of Figure 4-2, yielding the sum of the potential contributions of each charge

fL"

<1>(P)

k=1

_q_k-V

4nE oRk

Ab . I - soI ute potentIa

(4-41 )

The absolute potential of the most general configurations of static charges in free space is one accounting for discrete charges plus line, surface, and volume charge density distributions, the sum of (4-35a), (4-35b), (4<~5c), and (4-41)

(4-42)

EXAMPLE 4·8. Find the static potential, and hom this, the E field of the fixed dipole charges (q, q) located at the positions (d/2, on the z-axis as in (a). Express the answer in spherical coordinates, assuming r » d. The absolute potential at P is given by (4-41), if n = 2

(P)

Assuming r »d, one can approximate R J R 2 ~ (a). Then (P) reduces to qd cos 8

(P)~--

4nEor2

r2

and R z - R J ~ d cos 0, as noted from

d«r

(4-43)

Note that (P) of a static dipole is an inverse r2 function (lor r» d), as contrasted with the inverse r potential (4-40b) of a single static charge.

P (field pOint) (z)

Rl (J

dJJ~ (x)

//

R2

r

T-q~ (y)

(a)

(b)

EXAMPLE 4-3. (a) Geometry of the static charge dipole. (b) E field plot of the static charge di pole.

196


One obtains E trom (4-31) in spherical coordinates

E=

V = - [ a r

J

8

8 + a,p - -8- ar + ao -.80 r sin 0 81>

(4-44) an mverse in (b).

r3

function with both rand 0 directed components. Tts flux plot is shown

4-5 CAPACITANCE Of considerable practical use in electrical circuits is the capacitor, commonly used to store or release electric field energy. Basically, a capacitor consists of two conductors separated by free space or suitable dielectric materials of arbitrary permittivities. Its form is generalized in Figure 4-6(a). A capacitor with the charges q and -q can be brought to this charge state by means of a source of electric charge such as the battery shown, although it is perhaps more common to connect it to a source of sinus0idal or pulsed voltages. In this event, the charges become functions of time, q(t) and - q(t). The viewpoint of the present discussion is that if the time variations are sufficiently slow, a static field analysis of the system will provide results of sufficient accuracy to serve the purposes of many time-varying applications of practical interest. A capacitor, brought to the charge state of 4-6, has the properties 1. The free charges q and - q reside entirely on the conductor surfaces, accounting for a charge density Ps on each such that on their surfac.es Sl and S2 reside the charges q=

f

s,

-q

p,ds

r

J

p, ds

(4-45)

S2

2. From the boundary com:li tion (4- 7), the E field originates normally from the positively charged conductor and terminates normally the negative one, with the total D flux equaling q (Gauss's law).

C) +

+

+ + +

\":>-'

+q

+

"

1

(+

+

+

~

--Of~ --q 2

-=-v

~

~ -I 1 - _

'"\\

L __________ ..J

\

(a)

(b)

(e)

FIGURE 4-6. The two-conductor capacitor. (a) Generalized two-conductor capacitor. (0) Electric field about (a). (c) Variation of (b): one conductor surrounds the other.

4-5 CAPACITANCE

197

3. A consequence of the perpendicularity orE at the conductor surfaces is that they are equipotential lurJclees (<1> <1>1 and <1> = <1>z). Thus a single-valued potential difference <1> 1 - <1>2 = V exists bctween the conductors, obtainable from (4-38b) as follows:

v = <1>1

- <1>2

=[-

S~l E . dtJ -

[

S E· dt P2

00

]

=

(4-46)

in which 1\ and P2 can he located anywhere on the respective conductors, and the latter integral is over any path connecting PI and Pz. To make V positive, the potential reference P 2 (the lower limit) is assumed on the negative conductor.

If a linear dielectric medium is used in a capacitor, the effect of doubling the charges q and q on the conductors results in a doubling of the E field everywhere. From (4-46), then, the potential difference is also doubled. Thus in a linear capacitor, V is proportional to the charge q so that q oc V, or equivalently,

q = CV

(4-47)

The proportionality constant C, having the units coulomb per volt or farad, is called the capacitance of the two-conductor system. It is positive whenever an increase in V (the potential of the positive conductor relative to the negative one) results in an increase in the charge q on the positive conductor (accompanied hy a negative increase of q on the other conductor). For a passive element, C is always positive, its value depending on physical dimension and the dielectric properties of the system. An expression useful for evaluating C is obtained from substituting (4-46) into (4-47)

C

q V

-----"--~-

- Jp2 rp , E· dt

F

(4-48)

EXAMPLE 4·9. Determine the capacitance of a coaxial conductor pair of length t with the dimensions shown in (a) of the accompanying figure. Assume that a dielectric of permittivity E separates the conductors. Avoid the effects offield-fringing in (b) by assum-

ing that the system of length t is part of the infinite system in (c). To find the capacitance of a length t, assume the conductors charged, for every length t, with + q and - q C on the inner and outer surfaces. The D field is found using Gauss's law as in Example 4-4, yielding the E field E

q

= a p

2nEpt

(4-49)

The potential difference V between a reference P2 on the negative condnctor and PIon the positive condnctor is

v

(4-50)

198

STATIC AND QUASl-STATIC ELECTRIC FIELDS

(a)

---------~~~~~----------

:JJ~[[t-'-'-'-'+-'--'-'--'---'----'--1--.I...----'-I]J]]~L~, ~III~[-~IrlrI~ ~----t--------~ (c)

EXAMPLE 4-9. (a) Circular cylindrical coaxial capacitor and dc sourcc. (b) Fringing of electric field at ends of finite length system. (e) Showing field independent of 1> and z in a section of an infinite system.

Thus, thc capacitance of a length

c

t, q V

neglecting end effects, is ohtaincd ti'Oll1 (4-48)

2nd

--'--=--F q b b tilto1l-

2nEt

a

I)

a

Note that thc rcsult is independcnt of 1(, as of a linear function only of the dimensions and E. If {l rnm, b = 6 mm, and E = Cjt becomes 31 x 10 .. 12 F/m (or 31 pF/m). Using a dielectric with E = !(JUr times as large.

(a)

Hence C is a (air dielectric), 4Eo yields it result

Eo

(b)

FIGURE 4-7. Common two-conductor capacitance devices. (a) A parallel-plate capacitor. Fringing effects are neglected. (b) A spherical capacitor.

1-6 ENERGY OF THE ELECTROSTATIC FIELD

199

By the techniques of the last example, one can show that the capacitance of the parallel-plate system of Figure , neglecting field-fringing, is

c

(4-52)

whereas that of the concentric spheres in (b)

C

IS

4nE --F 1 1

a

(4-53)

b

4·6 ENERGY OF THE ELECTROSTATIC FIELD The concept of stored energy in the electrostatic field has important physical interpretations and applications. As in mechanics, many problems ofelectrostatics can often be simplified if an energy viewpoint is adopted. Although generally systems of electric charges possess both potential and kinetic energies due to their positions and motional states, in the electrostatic case only the charge positions determining the potential energy of the system need be of concern. To establish the n charge aggregate of Figure 4-2, mechanical work must be done by some external agent in bringing the charges to their final positions. Whenever two char~cs q 17' are brought within a distance R of each other, work is done against the Coulomb fixer lOa) in consummating this process. Once the charges are in place, the persistence of the Coulomb t(HTe makes the stored energy potentially available whenever demanded. The discharge of a capacitor bank through a resistor exem plifies this reverse process. The electrostatic energy stored in a system of discrete, or point, charges is found by building up the assemblage one charge at a time until all are in their intended locations. It is assumed that if they are moved slowly enough that their kinetic energies may be ignored and that~:adiation effects, significant if rapid charge accelerations occur, can be neglected. Assume initially that all n charges, Ql' Q2' Q3.' . . . , are located at infinity in their zero potential state. On bringing only Ql from infinity to its final location f\, no work is done because only ql is present; at least two charges are required if Conlomb forces are to exist. 2 On next bringing Q2 limn infinity to P z as in Figure 4-8, the work done against the field of fI 1 is U 2 = q2~1), in which ~1) denotes the electrostatic potential at P 2 and due to (11. Thus one obtains, using the absolute potential expression (4-40c),

(4-54a) (4-54b) in which an interchange of fli and

fl2

is seen to yield equivalent work expressions.

2The selFenerf!..Y of each discrete charge, that is, the energy required to create each diminutive electron cloud, is Ileglected in this development.

200

STATIC AND QUASI.STATIC ELECTRIC FIELDS

~-F;om= P2

/..~"~h' ,_ <.~_ I

q2 (Brought

. in from =)

I

I

I I

/R12

/

PI

q3 (Brought

/

in from 00)

""'\5 ql (Fixed) (a)

(b)

FIGURE 4-8. Two steps in the construction of an n charge aggregate. (a) From infinity q2 is brought in in the presence of Q1' (b) From infinity q3 is brought in in the presence of ql and Q2'

Again, if a third charge 13 is brought in to P3 as in Figure 4-3(b), the work done against the fields of q1 and q2 is expressible two ways q3 = q1<1>~3J

4nEo

R

13

+ 13 4nEoR23

+ q2<1>~3J

(4-SSa) (4-SSb)

and so all, for Q4, q5, ... , qu· Continuing the preceding development shows that the total energy, U e V 1 + V 2 + ... + Vn, can be written two ways 1. On adding (4-S4a), (4-SSa), etc.,

+ q3<1>~1) + q3<1>~2) + q4<1>:t) + 14<1>~J + q4<1>ZJ + ... + qn<1>~11 + qn<1>~21 + ... + qn~1-1)

U e = Q2<1>~1)

(4-S6)

2. Adding (4-S4b), (4-SSb), etc. yields

Ve

=

+ ql<1>~31 + q2<1>~31 + ql<1>~4) + 12<1>~4) + 13<1>~4) + ... + q1<1>\n1 + q2<1>1f1 + ... + qn_l<1>~n) 1

ql<1>~21

which can be regrouped

+ <1>~31 + ... + <1>\"») + q2(<1>~31 + <1>~4) + ... + <1>1f1) + q3 (<1>~4) + ~5) + ... + <1>~n») + ... + qn - 1~I~ 1

Ve = q1 (<1>\2)

(4-S 7)

Adding (4-S6) and (4-S7) and dividing by two obtains

+ <1>\3) + <1>~4) + ... + <1>\"1] + 12[<1>~1) + \i) + <1>\4) + ... + 1f)J + q3[~1) + ~2) + <1>~4) + ... + <1>~n)] + ... + qn [<1>(1) + <1>(2) + <1>(3) + ... + <1>(nn 1)]} n n n

U e = Hql

[<1>~2)

201

4-6 ENERGY OF THE ELECTROSTATIC FIELD

The meaning of each bracketed sum in the latter is now assessed. In the first term, the sum [<1>\2) + <1>\3) + ... + <1>~')J, abbreviated <1>1, is the total potential of P1 (position of qd due to all the charges except q1 itself. Thus the bracketed factors signify the potential at the location of the typical charge qk> a potential due to all the charges except qk' Denoting the bracketed factors by <1\, <1>2, ... ,<1>,. respectively, the desired [q1<1> 1 + q2<1>2 + ... + qn<1>nL or result becomes U e =

m

n

/

Ue

=! k=L1 qk<1>d

(4-58a)

in which qk is the charge of the typical (kth) particle <1>k is at Pk> the absolute potential due to all the charges except the kth

If the assemblage of charges is not discrete, but rather a continuum of density Pv distributed throughout some volume region V, then (4-58a) becomes an integral on replacing qk with dq = Pvdv, obtaining

(4-58b)

wherein <1> is the absolute potential at the position of Pv- For charge continua comprised ofswlace or line distributions as discussed in Sections 4-2 and 4-5, the following expressions are used in lieu of the preceding ones.

ull (4-58c)

(4-58d)

«II, In computing Ue £i-om one or a combinatioll of these four expressions, only the charge distributions and the potentials at the charge locations need to be known. The energy integrals (4-58), expressed in terms of the potential distribution <1> accompanying static charge distributions in space, can also be written in terms of only the D and E fields that occupy the whole of space. The result becomes

(4-58e)

202


To prove the latter, suppose that surface charges of density Ps exist on the closed conducting surface S, where S may consist of n individual conductors such that S = S1 + S2 + ... + Sn, with the additional possibility of a volume charge density Pv occupying the region V enclosed by S.3 The two-conductor capacitor of Figure 4-6(b) or (c) represents such a system. The electrostatic energy of the system is the sum of (4-58b) and (4-58c) (4-59) in which S denotes the simply connected closed surface of the charged conductors, and V is the region between the conductors. Using the boundary condition (3-45) but with the unit vector n directed away from the volume V so that Ps = - n ' D, (4-59) becomes

Iv p/Pdv Iv V • (D) dv + ! Iv pv dll

Ue = -! ~s (D) • nds +!

(4-60)

in which the transformation of the dosed-surface integral to the volume integral4 is accomplished by use of the divergence theorem (2-34). The use of the vector identity (15) of Table 2-2, V- (fG) =IV'G+G-Vf, yields

Iv D • (VV - D dv + 1 Iv p,, dll Since Pv = V - D, the last two integrals cancel, and with V
one conductor does not enclose the other, as in the capacitor shown in Figure 4-6(b), the dielectric volume region V extends to infinity. The surface S enclosing V must then include a sphere at infinity, hut because of the manner in which the and D fields vanish at remote distances, it develops that the surface integral contribution over this sphere is zero.

4J[

4-6 ENERGY OF THE ELECTROSTATlC FIELD

203

EXAMPLE 4·10. Find the energy stored in the electric field of the coaxial line of Example 4-9, making use of (4-58e). In a coaxial line, D and E were found in Example 4-4 to be

D=a p

2npt

E=a - q p 2nEpt

These substituted into (4-58e) and integrated throughout the volume of the dielectric yield ~

(4-62) A useful application of the energy integral (4-58c) is to the capacitor of Figure 4-6. The fact that the two conductors, carrying q and q, are at the equipotentials <1> = <1>1 and <1> = <1>2 permits simplifying (4-58c) as follows

in which the surface integrals, from (4-45), denote q and -q on the conductors. Thus (4-63a) wherein V for <1>1 - <1>2 has been substituted from (4-46). Putting (4-47) into (4-63a) yields alternatively (4-63b) (4-63c) which show that the stored electric field energy is proportional to the square of either Vor q. The equivalence of (4-63) to (4-61) enables finding the capacitance of a twoconductor device in terms of energy. Thus, solving for C in (4-63b) or (4-63c) and substituting for Ue with (4-6Ia) yields the equivalent results (4-64a)

(4-64b)

EXAMPLE 4·11. Determine C of the coaxial capacitor of Example 4-9 trom its stored energy. From Example 4-10, the energy of the coaxial pair of length t is U

e

=t 4nE

b tJ't

a

204


This is in terms of q, so putting it into (4-64b) yields

which agrees with Example 4-9.

4·7 POISSON'S AND LAPLACE'S EQUATIONS In the previous sections, the solutions of electrostatic field problems were obtained by the methods 1. Integrating (4-13) throughout the given static charge distribution in free space to find E.

2. Integrating Gauss's law (4-5) with respect to certain symmetric charge and dielectric configurations to find D, and thus E. 3. Integrating (4-35) throughout a static charge distribution in free space to find the potential (Ul' 112, U3) can be derived by combining the Maxwell relations I) and (4-2). With DEE, (4-1) is written

V' (EE)

(4-65)

Pv

and E is conservative so that (4·31) applies; thus (4-65) becomes

Pv

(4-66)

a partial differential equation known as Poisson's equation. In this form it is correct even though the dielectric region is inhomogeneous (E a function of position). If E is a constant, (4-66) takes the more usual form: V' V = pJE or with the notation V . V V2
Pv E

(4-67)

4-7 POISSON'S AND LAPLAOE'S EQUATIONS

205

Sometimes V 2 <1> is called the Laplacian oj <1>, expansions of which arc given by (2-77), (2-80), and (2-81) in the common coordinate systems. If no free charge exists in the region (Pv 0), the generalized Poisson equation (4-66) reduces to V . (EV
The common form of Laplace's equation (4-68), together with the particular space boundary conditions that is required to satisfy, constitute a boundary-value problem in a charge-fi'ee region. EXAMPLE 4·12. A pair of long, coaxial, circular conductors is statically charged with its inner conductor at the potential = V relative to the outer conductor, assumed at zero potentiaL The intervening region is a homogeneous dielectric with a permittivity E. Solve Laplace's equation, subject to the boundary conditions, for the potential anywhere between the conductors. Obtain also E in the dielectric, q on the conductors, and the capacitance (of a length t) of the system. From symmetry, the fields are independent of l/i and z, assuming fringing effects are nq:;lected. Then Laplace's equation (4-68), by use of (2-80), reduces to the ordinary ditl'erential equation (4-69) Integrating once obtains p ajap = C1 , and a second integration yields the solution

(I) The boundary conditions are applied to evaluate C 1 and C2 . At () = b, = 0 so that (I) yields 0 C1 tn b + C 2 , to permit expressing C 2 in terms of (;1 as C2 = -C 1 tn b. Substituting this back into (1) yields

(p)

EXAMPLE 4-12

(2)

206


The second boundary condition, <1l(a) whence (2) becomes

V, applied to (2) produces G\ = - Vltn (bla),

<1l(p)

V

b

b tn-

P

= -tn

(4-70)

a

the desired solution for <1l anywhere between the conductors, written in terms of V. As a check, note that setting p a and p b yields, respectively, the boundary values <1l V and <1l = O. One finds E from (4-70) by llse of (4-31). The expansion of (2-14b) yields D<1l -a -

E

V

a

PDp

P

1

bP tna

(4-71 )

To find the total charge on either conductor, the charge density Ps is required, obtained from the boundary condition (4-30). At the inner surface {I = a

EV

(Is = Dn = Dp],,=a = EEp]p=a = --ba tn a obtaining the charge in a length

(4-72)

t 2nEtV

q = ps(2nai')

b tna

( 4-73)

The definition (4-47) of capacitance thus yields

q

2nd'

c=-=-V b

(3)

t?
a

which checks with (4-51) in Example 4-9.

Although this example does not exhibit a great economy of effort when compared with the previous methods used to solve this one-dimensional problem, the chief merit of boundary-value methods for solving electrostatic field problems lies in their applicability to two- and three-dimensional systems lacking useful symmetries and not possessing known charge distributions. The latter is taken up in Section 4-9.

*4-8 UNIQUENESS OF ELECTROSTATIC FIELD SOlUTIONS I t is of importance to know, once one has obtained (by whatever means) a solution to an electrostatic field problem that it is the only solution possible; that is, it is a unique solution. The mathematical model furnished by potential theory would be oflittle use if it furnished several solutions to a given problem, among which the correct solution

4-8 UNIQUENESS OF ELECTROSTATIC FIELD SOLUTIONS

207

of the physical problem might have to be verified by experiment or in some other manner. 1t can be shown that potential solutions of the f()llowing classes of boundaryvalue problems are unique solutions. 1. The Dirichlet Problem. A poten tial solution <1>( U1, U2, unique if $ satisfies a specified boundary condition

of Laplaee's equation is

(4·-74 ) on the clos~d boundary S of the region. 2. The Neumann Problem. A potential solution $(Ul' U2, U3) of Laplace's equation is unique within a constant value if the normal derivative of $ satisfies a specified boundary condition

o(D

Dn =

O<1>J

a;;

(4-75)

s

on the closed boundary 8' of the region. 3. The Mixed Boundary- Value Problem. A potential solution of Laplace's equation is unique if it satisfies (4-74) on a part of S, and (4-75) on the remainder. A proof of (I) is established by supposing that there are two solutions, <1> and $', each of which satisfies Laplace's equation (V2$ = 0 and V 2 $' 0) everywhere within the volume V bounded by the dosed surbce S shown in Figure 4-9(a), and both of which satisfy the same boundary condition <1>" as f()lIows. (4-76) TI](~ specified boundary conditioll (D s (u 1 , U2, U3) is, in general, it fUllction of posi tion on S. For some problems, S' may consist of several (n) conductors as suggested

ds = n ds

(a)

(b)

FIGURE 4·-9. Closed surface configuratiuns relative to boundary-value problems of electrostatics. (a) Volume region V bounded by closed surface S on which the boundary condition is specified. (b) Variation ()f (a): V bounded by " I interior surfaces and exterior surface S,. A special case occurs if S, Soo.

208

STATIC AND QUASI-STATlC ELECTRIC FIELDS

by Figure 4-9(b), in which the boundary condition (4-76) is a sequence of potentials

(4-77)

0 on S

The uniqueness of
Iv [JV2g + (Vf) • (Vg)] dv

=

~f(Vg)

• ds =

0 in V. To this

~f~! ds

(4-78)

true for any pair whatsoever of well-behaved functions f and g. It must therefore hold iff = g, and equally well for f =

-O
an

With
Iv [V(
-O
on

(4-79)

0,

(4-80)

Because of (4-77), the surface integral of (4-80) is zero, obtaining

Iv [V(

constant in V

(4-81)

but even the value of this constant is zero in the Dirichlet problem, in view of the boundary condition (4-77). Thus
4-9 LAPLACE'S EQUATION AND BOUNDARY-VALUE PROBLEMS

209

for a dielectric partitioned into several homogeneous regions with difterent E values. The prooJ'follows on subdividing V by means of surfaces lying just to either side of the interfaces, but it is not given here. 5

*4·9 LAPLACE'S EQUATION AND BOUNDARY·VALUE PROBLEMS In Example 4-12 of Section 4-7, an instance of the direct integration of Laplace's equation (4-68b) in one dimension was described. In lhe present section, a method for extending the dir~ct-integration procedure to two-dimensional conductor systems is given. The separation of variables method is used, which, via the assumption of a product-type solution, permits a conversion of the Laplace equation in two or three space variables into the same number of ordinary differential equations, solutions of which are obtained by standard methods. J .aplace's equation has been found separable by this method in some 11 orthogonal coordinate systems. 6 The present discussion is confined to the cartesian system. Consider the solution of Laplace's equation in the two-dimensional cartesian system. In a charge-free, homogeneous, linear and isotropic region, (4-68) is written, by use of (2-77)

o

(1-82)

The separation of variables method begins by assuming a product solution of the form

= X(x) 'fly)

Q>(x,y) ill which into

(4-83)

and r(y) respectively denote functions of x and ofy only. Substitution yields

X"y+ XY"

0

in which the double primes denote differentiation with respect to x or y, whichever applies. Dividing by XY Xfl

Y"

X

Y

-+--=0

(4-84)

stating that the sum of a function of x only plus a function ofy only eq uals a constant . This is possible fCJf all values of x and y in an assigned region only if each term of (4-84) equals a constant. Denoting them by and yields

f;

X"

__ =

X

__ k2

-k;

Y"

(4-85a)

x

5A proof or this extension of the lmlOLH:l1eSS theorem is found in W, R. Smythe, Static and IJ.ynamic Electricity, New York: McGraw-Hill, 1950, p,

6S ce L P. Eisenhart, "Separable systems of Stach:l," Annals

oI Math., 35,

1934, p. 284.

210


and if (4-84) is to be satisfied, one obtains (4-85b) This means

k; (4-85c)

implying that ifone constant (k x or ky) is real, the other must be imaginary. Thus (4-85a) are ordinary differential equations, being functions of one independent variable (x or y), and so they are written (4-86a)

(4-86b)

If kx is taken to be real, (4-86a) has exponential solutions expressible in either imaginary exponential or trigonometric form as follows. (4-87a)

X(x)

or (4-87b) From (4-85c), real kx requires ky to take on the imaginary values ± jkx, to make the solutions of (4-86b) become the real exponential or equivalent hyperbolic forms 7 (4-87 c) or Y(y)

=

C~

cosh kxY

+ C4 sinh kxY

(4-87d)

Static potential field solutions of physical problems are real solutions, making the real trigonometric solutions (4-87a) preferable to (4-87b). Moreover, choosing the real exponential solutions (4-87c) in lieu of their hyperbolic form:'> yields for the product solution (4-83) (4-88a) 7The hyperbolic functions in (4-87d) are defined as the linear sums of exponential functions

e" + e- a cosha=---

2

ea _ e- a sinha = - - - -

2


211

in which G\, ... ,(;4 are real constants. If the preceding development had begun with the assumption in (4-85c) ofa real ky instead ofkx, making kx imaginary in that event, then (4-88a) would become (f)(x,y)

(4-88b)

The choice of (4-88a) or (4-88b) depends on the boundary conditions of the given problem. Indeed, almost all boundary conditions of practical interest are such that a single solution of the form of (4-38) is insufficient to satisfy the potential conditions at the boundaries; becat.!se Laplace's equation is linear, an infinite sum of solutions like (4-88), containing differeht but proper values of kx or ky, constitutes a valid representation of(f)(x,y). It is shown in the following example that the methods of Fourier series are important in the evaluation of the coefficients of such series representations.

EXAMPLE 4·13. A two-dimensional, air-filled, infinitely long channel of semi-infinite depth in the y dimension as shown, is formed of conducting planes on three sides, insulated at the corners. The bottom plate is at V V relative to the sides at .~ = 0 and x = a. Find the potential anywhere inside the channel region. A solution <1>(x,y) of Laplace's equation (4-82) (1)

is to be found, subject to the boundary conditions

<1>(0,))

0

(2)

<1>(a,y) = 0

(3)

<1>(x, OCJ) = 0

(4)

<1> (x, 0) = V

(5)

=

The solution of (1) was shown to be (4-88). In view of the boundary condition (4) at y -> OCJ, choose the form (4-88a) (6)

The unknowns C 1 through C4 and kx are evaluated by use of the boundary conditions. Applying (2) to (6) yields

to obtain C 1 = O. Then (6) becomes

(7) Applying the boundary eondition (3) to the latter obtains

211

212


(x)

(a)

4> (x, 0) = V V I--L------,

o

a

1>

=V

o

(x)

4>

a

(x)

=- V

(b)

( c)

I

I

I Constant 100 V

I I

'on y= 0 plane

(y)

OV

OV

x=a Q

2

(d) EXAMPLE 4-13, (a) Potential well of infinite height. (b) Boundary condition on Iht' physical half range (0, a). (c) Odd function

satisfied only ifsinkxa Then (7) becomes

0, whence kxa

213

(mn/a) (m= 1,2,3,.,.).

mn, making kx

(8)

The third boundary condition (t}) yields

having a zero limit as y --> 00 only if C3 = 0, since any nonzero C3 would prodnce an infinite at the remote boundary,y --> 00, a nonphysical result. 'Tbus (8) becomes m = 1,2, ...

(9)

a function exponentially decreasing iny. An attempt to apply the last boundary condition to (9) yields (x, 0) = V = C2 C4 sin

mn

x

a

an equality impossible to satisfy for all x within the (0, a) x range at they = 0 boundary, hut since m can assume any positive integer value m = 1,2,3, ... , the linearity of the differential equation (I) permits forming an infinite sum of solutions like (9) ranging over all the m integers, that is,

I

Ii)

(x,y)

Ame-(m"la)y

m=l

sin

mn

x

(10)

11

Equation (10) is a Fourier series (trigonometric series) representation tell' (x,y) with by respect to the variable x. The unknown coefficient Am are to be determined at y = applying the boundary condition (5) to the series (10), yielding

°

aJ

(x, 0) = V =

I

m=l

mn

Am sin- x a

(II)

a Fourier representation of the boundary condition (5). Standard Fourier techniques yield the unknown coefficients Am. The spatial period of the Fourier representation must first be defined, however. Note that the boundary condition (x, 0) = V is specified over the physical x range (0, a) between the channel walls, as in (b) of the accompanying figure. By defining (0, al as one-half of a total spatiallJeriod (-a, a), the rest of the range ( -a, 0) may be filled in with an arbitrary function <1>, as long as the Fourier expansion of (x, 0) converges to = 0 V at the endpoints of the physical half-range (0, al as required by the boundary conditions (2) and (3). The latter is accomplished nicely by assuming (x, 0) of (II) to be an odd function defined over the period ( - a, a) as in (c 1of the figure. Thus, represent on this interval the potential

(x 0\ = { ,I

V -V

O
(12)

291

214


by the Fourier sine series representation

(()

nrrx'

I An sin n=l

(13)

a

The eodIieients An are found by the standard Fourier procedure of multiplying (13) by sin in which rn 1, 2, ... , and integrating the result over the orthogonality interval ( a, a) on x; that is

f

rnnx

a

0) sin

-a
a

A

dx

11

f

a ' ••

rnnx

nnx

a

(l

sin - - sin (4

dx

(14)

The right-hand integral of (14) reduces to just the single termS Ana, in view of the orthogonality of the trigonometric functions on the interval (-a, a) (i.e., all terms for which 111 n to zero). The left side or with (12) inserted for
*

f

a -a

nnx
2Va mt

2 Va

nnx

V sin

a

a

dx = - nn

[1-- (-I )"1

(1

cos nn)

1,2, ...

n

(15)

Thus, (14) yields An

2V

= -

nn

fl --

(-I)"J

(16)

implying that Al 4V/n, A z = 0, A3 4V/3n. A4 0, .... Inserting (IG) into (10) thus obtains the desired Fourier representation of
2V

(J)(x,y)

n =

f n~

_1_____ ( _-~l__

4 V ,- e n

L

e - ("nla)y

11

1

(nla)y

sin nx a

+

sin nnx a

}e - (3n/a)y

sin

~nx + ... a

J

(17)

A sketch of
8'rhat Sa_a over the

S111

dx = a for rn = n follows from the orthogonality propnty of the sinc functiollS n, n): sin mO sin nO dO =

{n, 0,

rn

n

rn ¥= n

tn,n

1,2, ...

=

Letting 0 = nx/Il, whence dO -= (n/a) dx, the latter integral becomes

f as stated relative to (14).

a -a

sin

mnx unx sin --- dx a a

m

n

rn ¥= n

4-10 FINITE-DIFFERENCE SOLUTION METHODS

215

E(x,y) = - V
a I )"]e -

(nnla)y

rnnx sin ----a

(18)

The flux orE, orthogonal to the equipotential surfaces, is also depicted in (d). To ill~trate the use of (17) and (18), suppose V = 100 V and one desires

= 400

n

[e -

nl2

sin

n2 + .le-

= 127.3[0.2079

3nl2

3

sin 32'. 2

~ (0.00898)

+ ...

+ ... J =

J

2G.08 V

The E ficld there is [(lUnd using (18), a result seen to depend on t he a dimension. Choosing a= 1 m,

E(;,D

= -axO - a y400[O.2079 - .00898

+ .00039 -

.. J

a y79.72 Vim From (18) it is seen that E is invtrscly depl'lldcnt on a. Decreasing II to I cm thus increases E by the factor 100 to yield E(a/2, a/'l) = -a y 7972 V 1m, a consequence of compressing the eqnipotcntial contours more closely together.

4·10 FINITE·DIFFERENCE SOLUTION METHODS

Because many physical problems of electrostatics may involve boundaries that do not coincide with tbe coordinate surfaces of standard orthogonal coordinate systems, or may involve mixed systems, the analytical methods described in the previous sections may not be very useful ill such applications. A powerful technique {clr solving Laplace's or Poisson's equation, subject to conditions on boundary surfaces of arbitrary shapes, makes use or finite-difference approximations. This method replaces the partial differential equation, correct for the potential at all points in the region, with expressions for <]) in terms of the average of the potentials at nearby, incrementally distant points, these being located at the finite intersections formed by a grid system laid over the region in question. The grid may be either two- or three-dimensional. With the potential <]) being given or known at points lying Oil the region boundaries (usually conductors), the potentials at all grid points in the region are to be determined. Consider the Poisson eq ua tion (4-67) in two dimensions, expressed in rectangular coordinate f()rm.

Pv E

(4-89)

216

STATIC AND QUASI-STATIC ELECTRIC FIELDS AtP l : $1-1[¢2+0+ V+O]

(y)

4

"'-I ----r-i---t I

I

I

(I)=V

_~_+~_-:11

(I>(x+h,y)

I

L_~...J

(!>(x-h,y)

¢=V

Gap

QZ?1\vzzFlTil/?/A/

(x,y+h) (t)(x,y) '-C~-,

y

\

~-----,,"""-

__ +_

2' 3 -

I ¢(x,y-h) I I

!

-+ -

;6

'4 I

__ ¢=O

I o L - - - - - - ' - - ' - x - - L - - - - (x)

(6)

(a)

FIGURE 4-10. (a) Geometry relative to the finite-difference method. (b) Triangular-trough example of a two-dimensional electrostatic problem, showing a squaregridded overlay with a labeling of points at which the potentials
To permit expressing (4-89) in an equivalent finite-difference form, it is convenient to use the Taylor's series expansion of (x,y) in the neighborhood of P(x,y), including terms through only the second derivative. From Figure 4-1 O(a), the Taylor'S expansions of (x + Llx,y) and (x - Llx,y), expressed in terms of (x,y) , are expressed with reasonable accuracy by the three-term approximations (with Llx denoted by h for simplicity, andy held fixed). . (x

+ h,y)

~ = (x,y)

(x - h,y) ;:; (x,y)

o(x,y}

+ h - ,ox ,-- + h (J(x,y) h -'--ox

2

+h

(J2(x,y) (J 2

(4-90a)

(J2(X,y) - - - -2. (Jx

(4-90b)

x .

2

Adding these obtains an expression for the second x partial derivative in terms of at neighboring points. (4-91a) A similar procedure leads to an approximation for the secondy partial derivative, with x assumed fixed and letting Llx = Lly = h as in Figure 4-10. (4-9Ib) Substituting these results into Poisson's partial differential equation (4-89) yields its finite-difference equivalent. (x

+ h,y) + (x -

h,y)

+ (x,y + h) + (x,y -

h) - 4(x,y) ;:;

Pv h2

(4-9lc)

E

Because of the approximations incurred by the truncation of the Taylor's series (4-90) to just three terms, it is evident that the accuracy of (4-9lc) is improved by keeping the interval h = Llx = Lly in Figure 4-10 suitably small.

4-10 FINITE-DIFFERENCE SOLUTION METHODS

217

Uthe region described by (4-91c) is charge-free (p" = 0), then (4-89) becomes Laplace's equation, and its finite-difference form (4-91 c) reduces, on solving for <1>(x,y) , to

<1>(x,y) ~ H<1>(x

+ Il,y) + <1>(x -

h,y)

+ (x,y + h) + <1>(x,y -

h)]

(4-91d)

Thus, <1>(x,y) is simply the average of the potentials at the lour points adjacent to P(x,y) in the x:y plane, as shown in Figure 4-10 (a). 9 The finite-difference solution of an electrostatic boundary-value problem governed by Laplace's eq uation can be managed by dividing the two-dimensional region into a square-gridded system defining unknown potentials <1>(x,y) at the grid intersections and inch/ding known potentials <1> = Vb V2 , • . . on different segments of the boundary, as sug@ested by the example of Figure 4-1 O(b). The interior dielectric region of that example, encompassed by the indicated conducted boundary, is shown covered by a grid defining six intersection points at which the unknown potentials <1>1' <1>2' ••• , <1>6 are to be found. In general, there are n unknown potentiaL~. The potential expression (4-91d) is written for each potential <1>\, <1>2, •.• ,<1>n at the n interior grid points, each being written as the indicated average of the four adjacent potentials (some of which may be known boundary potentials). Two methods for finding the n unknown potentials are evident:

1. The n equations are simultaneous linear algebraic equations in n unknowns. If n is quite large, the inversion of the n x n matrix could become prohibitive, even if aided by a large computer. For a sufficiently coarse grid (making n reasonably small), matrix inversion is feasible. 2. The n simultaneous equations can be solved by the process of iteration, consisting of successively improving the estimate or the potentials at the grid points. This approach submits readily to computer programming. Examples of these methods are described in the f()llowing.

EXAMPLE 4·14. The very long rectangular conducting channd d~scrjbed in Problem 4-32 has the dimensions shown, with (a/b) = ~. Assume the conducting cover plate at = V =

100 V, with the remaining three sides at = 0 V. The system is divided into the gridded configuration shown (with a = 4h and b = 3h). Use finite dilTcrcnce methods to find the potentials, by writing the expression (4-91d) for each interior point. Rearrange and solve for the unknown potentials. By inspection of the figure geometry, symmetry requires that <1'>1 = <1'>5 and <1'>2 <1'>6, yielding only four unknown potentials. Thus, at each interior point labeled I, 2, 3 and 4, (4-91d) is written to obtain (1)

(2) (3) (4) By an extension of (4-91d), it evident that in a three-dimensional charge-free region, the potential (x, y, z) becomes simply Ii times the sum of the potentials (x + h, y, z), (x - h. y, z), (x, y + h, z), and so on at the six face-centers ofa cube with edges 2h surrounding P(x, y, z).

9

218


Small gap

100V

-~ I

1>= 100V

I

I

hi

./

I I I e-~~~1_ +4 I 1>2 ~-2-t--4t 6t(2;I 1 I '!J=oj i i

l

41.6

50.9

41.6

15.5

20.5

15.5

2

4

•

OV

•

• 3

5

•

(1)

•

6 (2)

r, I (a)

(b)

EXAMPLE 4-14

These lineal' equations are rearranged in terms of <1> I through <1>4 as follows 4<1>1 -<1>1

<1>2 -

<1>3

=

100

+ 4<1>2 (5)

and can be put in matrix fi)rm if desired.

l

-~

-1

-I

4

o

-2

o

4

o

-2

-1

(6)

By determinantal solution or matrix inversion, the potential solutions become <1>3 = 50.9;

These values are labeled in (b) of the figure. They must be regarded as only approximations of the exact potentials at these points, to the extent that the precision of the three-term Taylor's approximation (4-90) depends on the grid size It chosen for the grid overlay. The exact solution for this problem also happens to be available from the Fourier result (4-164), to provide a convenient comparison. For example, at point 3 of the figure, (4-164) can be shown to yield the exact potential <1>(a/2, a/2) = 52.462 V. The value <1>3 = 50.9 V found by the present finite-difference method is about 3'10 lower, not an unreasonable estimate when cOllsidering the coarseness of the grid used. To reduce the errors, a finer grid structure must be employed with a resulting increased solution complexity.

EXAMPLE 4·15. Rework Example 4-14, this time using iteration. Assume, as needed, initial potentials of 0 V for the unknowns. Write the average-potential relationships (I) through (4) of Example 4-14 once again for the potentials at the grid intersections 1, 2, 3, and 4, except substitute the initial "guess" of 0 V (unless a more recent value has becn obtained). This yields it)r the first iteration

4-11 IMAGE METHODS

219 (1) (2) (3) (4)

Next, these new potential values (or more rcccnt ones) are substituted back into (1) through (4), obtaining from this second iteration the improved values

$1 =1(100 + 6.25 + 37.5) = 35.94 \

$2

1(35.94+12.5)

12.11

$3 = 1(2 x 35.94 + 100 + 12.15) = 46.10 $4=1(2 x 12.11 +46.10) = 17.58 This iteration process is repeated until a reasonable convergence to thc desired values is obtained. A tabulated of values for five iterations is given here, along with thc values obtained from the matrix solution of Example 4-14 to indicate the actual values toward which the iterations are converging. Finally, the exact potential values, calculated by use of the Fourier expansion (4-164) of Problem 4-34, are listed (partly) in the last column.

GRID NUMBER (k)

I

2 3 4

1ST

2ND

3RD

4TH

5TH

MATRIX, METHOD (EX. 4-13)

25 6.25 37.5 12.5

35.94 12.11 46.10 17.58

39.55 14.28 49.17 18.98

40.86 14.96 50.17 20.02

41.28 15.32 50.64 20.32

41.68 15.53 50.93 20.50

k FROM ITERATION:

EXACT

(4-164)

52.462 20.788

4·11 IMAGE METHODS The method of images about to be described takes advantage ofthe uniqueness property of potential solutions. It consists of replacing a problem, involving one or more statically charged conductors, with an equivalent problem of suitably located point or line charges (so-called image charges) that yield precisely the same electrostatic field as the original problem. The well-known fields of point or line charges can then be used to obtain a solution of the original boundary-value problem. The number of chargedconductor configurations that can be solved in this manner is relatively small, but included are enough examples of physical importance to make the method worthy of treatment. The image method is illustrated by an example in Figure 4-10. Suppose two point charges, q and -q, are spaced 2d m in free space as in (a) of that figure. The combined potential II> at any position P is given by two terms of (4-41)

lI>(x,y, z) = -----;=:::-::=:~~==::?

(4-92)

291

220

STATIC AND QUASI-STATIC ELECTRIC FIELDS I(y)

(y)

~1111~1~01;:-'"--1--_

(x)

Equipotential surface

I

["'-1>=0

(a)

1>

1>0 f--1>=0 (b)

(y) I(y)

Point charge +q

(x)

(c)

(d)

FIGURE 4-10. Three examples of charged conductor systems, the exact fields of which are obtained from image system (a). (a) Two electrostatic point charges and their E and q, fields. (b) Replacing interior of surfaces (q, = ±(Jlo) with conductors. (e) A variation of (b). (d) Replacing region to left of(Jl = 0 in (b) with a conductor.

The equipotential surfaces are found by equating (4-92) to constant potentials; a family of equipotential surfaces obtained in this way is shown dashed in Figure 4-10(a). Recalling that a conductor immersed in an electrostatic field has its surface at a constant potcntial, replacing the interior of the equipotcntial surfaces <1> = <1>0 and <1> <1>0 with conductors as in }"igure 4-10(b)cannot alter the E field exterior to the conductors. The original image charges ± q of Figure 4-1O(a) moreover appear as conductor surface charges totaling ± q, a conclusion reached from Gauss's law (4-5) integrated over the conductor surfaces_ The image charge system of Figure 4-1O(a) therefore yields the desired fields of the two-conductor system of Figure 4-1O(b), obtaining the same <1> and E solutions outside the conductors in the latter. A complementary system (one conductor within another) is shown in Figure 4-1O(c); its fields are also obtainable from the image system of Figure 4-1O(a)_

221

4-11 lMAGE METHODS

One can see that the symmetry plane x = 0 of Figure 4-l0(a) is the equipotential surface = 0, evident from setting the potential expression (4-92) to zero. Thus, if a conductor having the shape of one of the equipotential surfaces is located to the right of the conducting plane at x = 0 as in Figure 4-1O(d), the field between the conductors is once more specified by the image problem of Figure 4-10 (a). The field to the left of the plane is nullified, in terms of boundary condition (4-30), by the presence on its surface of the charge density

Ps

= Dn] x-o _ = EoEx] _ = -Eo aa] x-o x x=o

Hence, the x derivative of (4-92), with x

(4-93)

= 0 in the result, yields

\ (4-94) Extensions of the image system of Figure 4-IO(a) can be deduced from superposition as depicted in Figure 4-11. For example, a system of fixed point charges Ql, Qz, ... , placed near a .large conducting plane as in Figure 4-11 (a) has a static field in the right-hand space given by the sum of the fields of the original charges and their images shown. The zero potential on the median plane is maintained by that image

(fO)

Conductor charge system


(a)

(b)

,

l(y)

,I

(y)

a I a

a

--er q o

jb

-q'l--r--"1 q

'" =0

I i

Ib

I

Ib

mm~~% (x) --t-oi--i~) I

qb--+---!-q Conductor charge system

Image system (c)

I Image system


(d)

FIGURE 4-11. Image equivalent~ of static charge near infinite conducting planes. (a) Discrete charges ncar a conducting plane. (b) Arbitrary line charge near a conducting plane. (e) Line charge parallel to a conducting plane. (d) Point charge near intersection of two condncting planes.

291

222


system. A line of arbitrary shape pla~ced near a plane conductor provides another image equivalence as in Figure 4-11 (b), a special case of which is the straightline charge of Figure 4-11 (c). These schemes can be extended further with the image equivalent of a charge q near the perpendicular intersection of two conducting planes as in Figure 4-11 (d); three image charges are needed to establish zero potential on both planes. The parallel-line charge system of Figure 4-11 (e), duplicated in Figure 4-12(a), is an important image system that enables finding the electrostatic fields of parallel, round conductors as developed in the following. Assume two infinitely long, parallel line charges separated 2d and possessing the uniform charge densities PI' and PI'. The latter are denoted by the ratios q/t and - q/t, signifying the charges per length t of each line. Because of the infinite extent of the system, the analysis is confined to the z = 0 plane, restricting it to two dimensions (x,y) as in the section view of Figure 4-12(b). The equipotential surfaces ofthis parallel line charge system are right circular cylinders. To show this, note that the potential (x,y) at P in Figure 4-12(b) is found from the superposition of the potentials <1>(+) and <1>(-) due to each line. Each produces the potential field (4-39); so with 0 chosen as the potential reference, the potentials at P due to q/t and - q/t become <1>( +)

=

q

t:Ft

d

(4-95)

Rl

2nd Their sum is the total potential at P (x,y) =

<1>(+)

+ <1>(-)

=

q

2nd

tn R2

(4-96)

Rl

in which (4-97)

(E)

(-d,O,O) q

f

.++ ++ + q

/

fl{={ (a)

(b)

FIGU RE 4-12. Geometry of the parallel-line charge image system. (a) Parallel-line charges of uniform densities. (b) End view of (a) showing the two-dimensional geometry in the z 0 plane.

223

4-11 IMAGE METHODS

Observe Ii-om (4-96) that <[> ranges over all the real numbers, {()r as P approaches - q/t(R2 --+ 0), there <[> --+ - 00; whereas <[> --+ 00 at the positive line charge_ Equipotential surfaces are obtained by equating (4-96) to any desired constant potential <[> <[>0 (4-98a) This means that any fixed, real ratio (4-98b) defines an equipote\ltial surface on which <[> <1)0 prevails_ Thus, K R 2 /R J = 1 defines the plane x = 0 bisecting the system_ (Substituting K = 1 into (4-98a) reveals that <[>0 = 0 on it.) Other equipotential surfaces given by other K values are, in general, circles in the sectional view of Figure 1-l2(b); if the z-axis is included, they become circuLar ~ylindrical surfaces_ This is proved by substituting (4-97) into (4-98b) as f()llows

which expands into (4-99) This reduces to the equation ofa circle, (x - h)2 + f = R2, ifd 2 [(K2 is added to each side of (4-99) to complete the square, obtaining

+ 1)/(K2 -

1)]2

(1-100)

This result shows that the equipotential surfaces are a family of circular cylinders with centers displaced from the origin by

h

K2

+I

= d-;:--

1

(4-101 )

and having the radii

R

2Kd

(4-102)

Typical equipotential circular cylinders defined by (4-100) are illustrated in Figure 4-13. K-values less than 1 correspond to equipotential cylinders to the left of the origin, whereas K> 1 yields the cylinders on the right.

291

224

STATIC AND QUASI-STATIC ELECTRIC FIELDS : (y) I I I

I

I

: R2.. I /'

~:;--1--

:0 I

I I

':"~d I :

h

'------_. I

FIGURE 4-13. Equipotential surfaces ofa parallel-line charge system, that is, circular cylindrical surfaces.

Taking the difference of the squares of (4-101) and (4-102) eliminates .K to obtain

h2

R2 = d2 , whence

-

d=

(4-103)

This gives the locations ± d of the image charges in Figure 4-13 in terms of Rand h. On now replacing the interior (or exterior) of any pair of equipotential cylinders of Figure 4-13 with conductors (carrying the su dace charges q and q in every length t), the electrostatic field problems such as those of Figure 4-14 can be considered to

./

2h (b)

(a)

/"

--- --/

./

./

--- . / --- ---

./

--./

./

./

./

. /

--- ---(c)

./

/"

/"

(d)

FIGURE 4-14. Two-dimensional conductor systems. Solutions obtainable /i'om image system of Figure 4-12. (a) Circular conductor paralic I to a plane conductor. (b) Parallel circular conductors of equal size. (c) Parallel cylinders ofuncqual size. (d) Cylindcrs ecccntrically locatcd onc inside the other.

4-12 AN APPROXIMATION METHOD FOR STATICALLY CHARCED CONDUCTORS

225

have been solved. The capacitance C of a length t of the systems of Figure 4-14(a) and (b), for example, are found as follows. Dividing (4-101) by (4-102) to eliminate d obtains a quadratic expression in K, yielding _ h K= + R -

j(h)-2 -1

R

h±d R

(4-104)

with d given by (4-103). The positive and negative signs correspond to the positive and negative equipotential surfaces to the right and left of x 0, respectively, in Figure 4-13. The potential <1>0 of any equipotential cylinder in the right-half region is thus found from substituting (4-104) into (4-98a), but with the plane x 0 at <1> 0 V, the potential difference V between a circular cylindrical conductor and the conducting plane of Figure 4-14(a) becomes <1>0 0 V, yielding

V = <1>0 - 0 The

capacitan~e

= -q tn 2nd

-

[~+ n(~)2 ~J R Y\R

(4-105)

of that system, Irom (4-48), is theref()re

Wire above plane conductor

(4-106)

In the parallel-wire line system of Figure 4-14(b), the potential (4-104) of an identical conductor in the left-half plane is just the negative of that of other conductor, yielding a potential difference between the conductors just twice that of (4-105) for the cylinder plane system. Its capacitance is therefore Parallel-wire line

(4-107)

*4·12 AN APPROXIMATION METHOD FOR STATICALLY CHARGED CONDUCTORS Occasionally, approximation methods can be used for rapidly assessing the potentials and the capacitance of conductor systems. The technique described here depends on conductor dimensions being small compared to their separations, assuring that their SurfilCC charge distributions are not altered signiticantly by the proximity of the conductors. Then the electrostatic potential in the region can be obtained by simply superposing the potentials of the conductors taken separately. An illustration of this concept is given in Figure 4-15. Suppose a long circular conductor, isolated as in (a) of the figure, possesses for every length t, a total'charge q distributed uniformly over its surface. Its potential field is given by (4-39) <1>(P)

q Po =---tn2nd

P

(4-108)

291

226


(a)

(e)

(b)

f'JGURE 4-15. Circular cylindrical conductors, showing the effect of proximity on charge distributions and the superposed potential fields. (a) A round conductor and its potential field. (b) Wide spacing: Dla large. (c) Close spacing: Dla smalL

-qh q~al 12hl

-q~al

~a2 12h2 I I

q-G

a2

(b)

f'IGURE 4-16. Examples of charged conductor systems amenable to approximate analysis. (a) Parallel round cylinders, and spheres. (b) Conductors (cylinders or spheres) above ground (lift), and image equivalent (right).

4-12 AN APPROXIMATION METHOD FOR STATICALLY CHARGED CONDUCTORS

227

Two such conductors, possessing glt and g'lt as in Figure 4-15(b) and kept reasonably apart as shown, produce a potential at P that is the sum of the potentials due to each conductor, yielding very nearly

(4-109) This result is subject to an increasing error as the conductors are brought closer together as in Figure 4-15(c), in view of the charge redistribution taking place due to the attractive forces acting between the charges. These arguments provide a basis for finding the approximate capacitance between a pair of conductors having known potential fields when taken separately. Conductors of practical interest in this class of problems are spheres and round wires. Figure 4-16 shows a few examples.

EXAMPLE 4·16. Find the approximate eapacitance of the parallel-wire system of Figure 4-l6(a), two cOllductors of unequal radii al and az separated by the center-to-center distance D. I The potential diflcrence V between the conductors is obtained by superposing the potentials of each isolated conductor. Let the static charges on the conductors be qjt and - qjt Glm as shown, and the potential reference be at Po on the negative conductor. In the presence of only the conductor of radius at, the potential at P relative to Po in Figure 4-l7(a) is obtained from (4-103), yielding

---1> = Constant

(a)

(b)

FIGURE 4-17. Relative to the superposition of potentials for fmding approximate capacitance. (a) Field of positive conductor taken alone. (b) Field of the negative conductor only.

291

228

STATIC AND QUASI-STATIC ELECTRIC F'IELDS

with the distance from the source to the reference Po observed to be D. Similarly, for the negative conductor in Figure 4-17(b), the potential at P relative to the same rcfi::rence Po is

The sum is the total V between the conductors (neglecting charge redistribution effects); that is,

From (4-48) the approximate capacitance becomes (4-110)

For conductors of equal radii at

a2

a, (4-1 ]0) becomes

c:::;;

nEt D

(4-111)

tn-a

a result comparable to the exact expression (4-107) deduced from the image approach.

4-13 CAPACITANCE OF TWO-DIMENSIONAL SYSTEMS BY FIELD MAPPING Methods are now examined for the graphic sketching of electrostatic flux line; and equipotential surfaces of two-dimensional conductor systems. For any two-dimensional system possessing a uniform charge distribution along the z-axis, the same electrostatic field sketch is seen to apply to every cross section. Examples of electrostatic fields between conductor pairs ofarbitrary cross sections and possessing the charges q, -q, in every length t are shown in Figure 4-18. The sketches of the electric field flux and equipotentials of two-dimensional systems are executed in accordance with the following rules.

1. The conductors comprise equipotential surfaces between which additional equipotential surfaces may be constructed, their shapes varying gradually fi'om that of one conductor to that of the other. Equipotential surfaces must intersect the electric flux lines orthogonally. 2. Electric flux lines form the boundaries of so-called flux tubes, as in Figure 4-18(b). In a charge-fi.·ee region, a flux tube contains a fixed amount of flux /iiI"" over any cross section. The capacitance per meter depth of a two-dimensional system can be found with good accuracy from a carefully executed field sketch. Given the system of Figure 4-19(a), flux lines originate from an assumed charge q distributed over a length t of the inner conductor, terminating on q in the same length-of the outer conductor.

4-13 CAPACITANCE OF TWO-DIMENSIONAL SYSTEMS BY FIELD MAPPING

229

291

(a)

(b)

FIGURE 4-18. Typical two-dimensional condnctor systems and flux tube interpretations. (a) Examples of two-dimensional condnctor system, electric flux and equipotential plots. (b) Flux tubes in a two-dimensional conductor system.

On replacing the equipotential surfaces with a very thin conducting loil, that system can be regarded as the series combination of three capacitors G\, C2 , C3 between the conductors. Furthermore, if the equipotentials are located such that V between the conductors is divided into three equal amounts Vi V 2 = V3 = Vo , then the series capacitances are the same, that is, C 1 = Cz = C3 = Co, in view of the identical charges ± q on each. The total capacitance is therefore C = Co/3 for that example. Generally, if ns denotes the number of elements Co in series, the total capacitance is (4-112)

Each of the series capacitors of Figure 4-19(a) can further be subdivided into a parallel capacitance increment L\C associated with each field cell of the system, as in Figure 4-19(b). With Up parallel clement'>, Co = up(L\C), yielding the total capacitance

c = up I1C ns

(4-113)

It remains to determine the field cell capacitance I1C. Assuming charges I1q, - L\q induced on the conducting-foil walls at the top and bottom of each cell as in Figure 4-I9(c), one obtains, fi'Orn (4-48), I1C = I1qlVo , in which the potential difference beE . dt, also expressed in terms of an average electween the boundaries is Vo tric field hy Vo = Eav L\hav> wherein L\hav is the median height of the typical cell; but

H;

230


Equipotential surfaces

Electric field flux

(a)

(b) ~

,

-

Constant ~

\

't

_\'---f_---P~

~-i~Wav

__}!I

- - - -; _-----.. ++ .. + ....

(d;

ac aWav -=ft ah av

-f1----~--P-I ~hav\ I

• :>:.I /

~ = Constant

/\-;1 .;-

Flux \ hnes

Vo

~

=

~

constan~"i

I

\ \

+~q

I

~Wav

.c

-= t

E

--i- __ I(~= Constant

(e)

I

I

~hav-+ I

(d)

FIGURE 4-19. Capacitance determination from a two-dimensional electric field map. (il) Insertion of conducting foil at equipotential surfaces, yielding scries capacitance equivalence. (b) Subdivision of region between equipotentials into parallel field cells. (c) Enlargement of field cell ~C of (b). (d) End view of field cells. A curvilinear rectangle and square.

Eav DaviE and Day equals Aq/Asav , As. v denoting the average area of the cell cross section: its length t times the average cell width Aw av as in Figure 4-19(c). Thus AC becomes

EAq

yielding the capacitance per meter depth of a field cell

(4-114) If the cells are sketched as cur1lilinear squares defined by Aw av = l1hav as shown in Figure

4-13 CAPACITANCE OF TWO-DIMENSIONAL SYSTEMS BY FIELD MAPPING

4-19(d), then

231

14) simplifies to

I1C

(4-115)

E

t

The incremental capacitance per meter depth of a eurvilinear square (tux cell thus equals the permittivity E of the dielectric filling the cell. In air, for example, eaeh square cell contributes Eo = 8.84 pF1m. The total capaeitance between the conductors, obtained from the series parallel comhination of all cells, is found from the substitution of (4-115) into (4-113)

C

-

t

ltp

= -E

Flm

(4-116)

lts

From the development of 113), it is evident that be integers, as noted in the following example.

ltp

and

lts

in (4-116) need not even

EXAMPLE 4-17. Sketch the electrostatic tlux plot of the coaxial capacitor of Figure 4--20, obtaining its capacitance per meter depth. Assume air dielectric and b/a = 2. Because of tile symmetry, a flux plot for only onc quadrant suffices. If the interval between the conduc'tors is subdivided, by trial, into two equal potential difference intervals as in Figure 4--20(b), a tlux map consisting of the curvilinear squares plus two leftover rectangks as shown is obtained. Then ns 2 and ltp = so (4--116) yields

c

-- (8.84 x 10- 12 ) = 79.5 pF/m

t

2

(1)

Another flux plot, dividing the quadrant into five flux tubes as shown in (e), yields ltp = and sketching in the equipotential surfaces to obtain the curvilinear squares as shown, lis = 2.3 to yield

c~ p

,

(a)

2.3

(8.84 x 10-

( b)

12

)

=

(2)

77 pFjrn

( c)

FIGURE 4-20. A coaxial capacitor and typical flux plots. (a) Coaxial capacitor: b!a = 2. (b) A flux plot using equal potential intervals. (c) A flux plot using five flux tubes per quadrant.

291

232 np


= 4 (3.33)

Ins = 2

I

(b)

(a)

(e)

FIGURE 4-21. Examples of flux plots for two-dimensional conductor systems. (a) Elliptical cylinder inside a pipe. (b) Rectangular cylinder inside a pipe. (c) Toothed structure above a plane.

The discrepancy between (I) and (2) is due to the unavoidable errors of estimation. It happens that this example can be checked by use of the exact (4-51), yielding

C

t

2rrEo

b

tn-

2rr(8.84 x 10-

0.093

= 80.3

pF/m

(3)

a

The chief merit of the flux-plotting method for two-dimensional electrostatic systems lies in its applicability to systems for which no analytical approach is feasible. In Figure 4-21 are shown such examples. Note that care must be exercised to assure the perpendicularity everywhere of the equipotential and flux lines; observe the tendency toward the compression of the flux lines at convex curves and corners because of the higher surface charge concentrations there. Advantage should always be taken of the symmetry, with no more equipotentials being employed than necessary to obtain satisfaetory curvilinear squares. A suitable procedure in Figure 4-21 (a), for example, is to begin at section A-A' by placing a trial equipotential surface at point C, inserting appropriate orthogonal flux lines while progressing toward the right, and checking continuously for the squareness of the flux cells that develop. Needless to say, an eraser is a valuable adjunct to these trial-and-error procedures. Further suggestions and examples are found in a number of sources. lO

4-14 CONDUCTANCE ANALOG OF CAPACITANCE A system is said to be analogous to another if a quantity in one system varies in the same way as some quantity in the other. An analogy may even exist between two quantities in the same system. If the quantities are vector fields, to be analogous they must satisfy comparable divergence and curl relationships as well as similar boundary conditions. I t is to he shown that the capacitance system of F'igure 4-6 in Section 4-6 leads to a conductance analog. In the capacitance system of Figure 4-22(a), applying a IOFor example, see S. S. Atwood, Electric and Magnetic Fields, 3,.d ed. New York: Wiley, 1949; S. Ramo,

J. Whinnery, and T. Van Duzcr. Fields and Waves in Communication Electronics. New York: Wiley, 1965, p. 159.

233

4-14 CONDUCTANCE ANALOG 01' CAPACITANCE

\

\

V

'----+-11\1--------' (b)

(a)

FlGURE 4-22. Analogous capacitance and conductance systems. (a) Capacitance system: conductors at potential difference V, separated by dielectric. (b) Conductance system: a small conductivity (J supplied to the dielectric.

voltage difference V between the conductors separated by a dielectric results in static charges +q and -q being deposited on the conductors. In the charge free dielectric, D = EE, obeying V' D = 0 and V X E = 0 of (4-1) and (4-2). These properties state that D between the conductors consists of uninterrupted flux lines, with the conservative E field implying a related potential field such that E = - V
-f

D'ds

P1 P2

E'dt

f'

(4-117)

In obtaining this ratio, D· ds is integrated over the posltlve conductor of Figure 4-22(a), while P 1 assumed on that conductor makes V positive. A dc conductance analog of (4-117) can be established for the system if the dielectric possesses a small conductivity (f. The dielectric then carries a current of density j = (fE, from (3-7); j is the analog of D in the dielectric, since from (3-82e), V' j = O. Thus j consists of uninterrupted current flux lines, supplied by V. Assuming A and B good conductors and with the dielectric a relatively poor conductor, from the refraction Example 3-11 one conc\udf's that the current enters or leaves A and B essentially perpendicularly. The boundary condition (3-136), moreover, reveals what charge density Ps exists on each conductor surface. With OJ 0 and (fj «(f2' one obtains

Ps

(4-118)

if the good conductor is denoted by the subscript 2 and the lossy dielectric by 1. Thus the boundary conditions of Figures 4-22(a) and (b) are essentially the same. It is thus seen that adding a small amount of conductivity to the dielectric produces virtually no

291

234

STATIC AND QUASI-STATIC ELECTRIC

FlEI~DS

change in the E field configuration in the dielectric. Thus besides C, an analogous positive parameter G, the conductance of the system, is defined by the ratio of the total current J through the dielectric to the voltage difference V between the conductors

Js J. ds

J

G

= V = ---"-,J""Pl'---E-.-d-ct

(4-119)

P2

The surface S of conductor A excludes the cross section of the connecting wire so that in (4-119) only the outflow of J into the dielectric is taken into account. The analog of C is G because J in (4-119) is the analog of D in (4-117). With (f and E constants for the homogeneous, linear, and isotropic dielectric, (4-117) and (4-119) become E Js E· ds - _JPl E· dt

C_

(f

G

P2

Js E· ds

= -_-;;JFP1-=E:c-.-dC""C"t

(4-120)

P2

yielding the ratio

G C

(f

(4-121 )

E

This is also written (RC) -1 = (fIE if I/G = R, the resistance between the conductors. Equation (4-121) implies that if C is known, the analogous G can be found from the applicable ratio (fIE. In view of the relaxation result (3-83a) of Example 3-9, (4-121) has further implications. If V applied to the conductive capacitor system of Figure 4-22(b) were suddenly removed, the surface charges on each conductor would decay in time according to (3-83a) (4-122) if (f and E are the dielectric parameters. Integrating (4-122) over the positive conductor surface S yields the charge q on it at any instant e-«T/E)t

q(t)

rp

Js

ds sO

(4-123a)

which, by use of (4-121) obtains a form familiar in circuit theory (4-123b)

q(t)

The charge on the positive conductor thus decays exponentially with the time constant 1:

=

RC

E

= (f

sec

(4-124)

Thus 1: is expressible either in terms of the derived lumped constants Rand C, or the parameters E and (f of the dielectric. The time-decay behavior of q on the positive conductor is depicted in Figure 4-23(a), with the equivalent circuit shown in (b).

4-14 CONDUCTANCE ANALOG OF CAPACITANCE

235

~<>--;=SWl - ,

it

v-=-

C

+

R

R V(t)

=

Vm cos wt (b)

(c)

FIGURE 4·23. Behavior of a capacitor with dielectric losses. (a) Time decay of charge from initial value qo. (b) Voltage removed from capacitor equivalent circuit. (c) Equivalent circuit, ac voltage applied.

The so-called quality factor, Q., of the capacitor is shown to be the reciprocal of the loss tangent of its dielectric. Define its Q. under time-harmonic applied-voltage conditions as

w energy stored Q. = ----'-------"'-''-----'''---'---'Average power loss over a cycle

(4-125)

in which w is the radian frequency. Assuming V(t) = Vm cos wt applied as in Figure 4-23(c), the maximum energy is stored when the voltage is Vm , to yield Umax = CV;,/2 {l'om (4-63b). Also, V(t) is impressed on the loss resistance R, yielding the timeaverage power loss V;'j2R. Thus, (4-125) becomes (4-126)

which from (4-124) is also WEj(J, the reciprocal of the loss tangent (3-104), which was to have been proved. By use of (4-121) and (4-124), the Q.ofthe capacitor can be written in the various forms WE

Q. = wRC =

(J =

E' E"

= wr

wC =

R

G = Xc

(4-127)

wherein Xc == (wC) -1 denotes the reactance of C at the frequency w. Thus, a material with a loss tangent E" jE' = 0.001 at some frequency will yield a capacitor with a Q. of WOO. If; moreover, its reactance Xc were 100 Q at the giveu fi'equency, the equivalent circuit of Figure 4-23 would need to incorporate a parallel resistance R = Q.Xc = 10 5 Q to represent the dielectric losses.

EXAMPLE 4·18. Assume that the spherical capacitor of Figure 4-7 (b) contains a dielectric with the constants E = 3E Q and (J' = 10- 5 U/m at some frequency. Letting a = I em and b = 2 em, find C and G, and sketch the equivalent circuit. Using (4-53) , C=

4n(3 x 1O-9/36n) ., I 1 = 6.67 pi'

0.01

0.02

291

236

STATIC AND QUASI·STATIC ELECTRIC FIELDS

c=

R=

6.67 pF

0.40 Mfl

EXAMPLE 4·18

while G, from (4-140), is merely C with

E

replaced by

4n(10

,(J

G=G-=

=2.5/1U

I

E

0.01

(J

0.02

yielding the resistance between the spheres, R = G- 1 0.40 MQ. The equivalent circuit diagram consists of C in parallel with R as in the sketch.

A. Capacitance-Conductance Analog and Field Mapping For two-dimensional capacitors, the infinite length makes it desirable to express (4-121) as the ratio Ca

G

t

t

E

(4-128)

Vim

Thus, from the Cjt ratio (4-51) for the coaxial line G

2nE a

2na

t

tn

tn-

b a

b

E

( 4-129)

a

Such results are also applicable to the two-dimensional field-mapping techniques of Section 4-12, assuming the dielectric to possess a small conductivity (J. Then each field cell as in Figure 4-19(c) contributes a conductivity per meter depth obtained by putting (1-114) into (4-128) !J.G

(4-130a)

t For any curvilinear square cell, letting !J.wav !J.G T =

(J

=

!J.hav yields

Sq uare cell

(4-130b)


(a)

237

(b)

FIGURE 4-24. Typical two-dimensional system showing analogous quantities used in analysis of Cit and cit. (a) Capacitive system with perfect dielectric and flux cell. (b) Capacitive and conductive system and analogous current flux cell.

the analog of (4-115)
G

C(J

t

t

n,

E

(J

Ujm

(4-131 )

The latter can be viewed as the conductance produced by the mesh of cell conductances connected between the equipotential conductor surfaces.

EXAMPLE 4·19. Use the flux plot of Figure 4-20(b) to deduce Cjt and Cjt for that coaxial system, assuming a dielectric with Er = 2.5 and (j = 10 - 8 U jm. What resistance is seen by a de voltage connected to the input of 1000 m of this line, assuming an open-circuit termination? The flux plot of Figure 4-20(b) yields from (4-116)

C

t

4(4.5)

np

- = -

E

ns

_

= - - - (2.:>

2

x 8.84 x lO

_

= 198

pFlm

Then, !i'om (4-131) C

t

(j

rts

4(45) = - - '- IO- s = 9 x IO- S Ujm 2

For t = 1000 m of open-circuited line, C = (9 x 10- 8 ) 10 3 = 9 R = C- 1 = 11.1 kfl, the resistance seen by the applied de voltage.

X

10- 5 U, making

2tJl

238


Resistive paper

(b)

(a)

FIGURE 4-25. Models of two-dimensional conductive or capacitance systems. (a) Model using resistive paper and silver paint electrodes. (b) Electrolytic tank with immersed metal electrodes. [Figure (a) depicts a resistive paper model of a spatial period of the repetitive tooth structure shown in Figure 4-21 In view of the entirely tangential current flow occurring at the two sides of this model, boundary condition there is iJ/iJn 0, agreeing with that of the actual system of Figure 4-21(c).1

Current-conduction models of two-dirnellsional systems as described can be constructed using commercial resistive paper, or by use of a shallow tank of electrolyte to simulate the conduction region, with electrodes of desired shapes placed in contact with the conductive medium as suggested in Figure 4-25. In (a) of that figure, intimate contact of the electrodes with the resistive paper is assured by using a good-conducting silver paint to produce the desired electrode shapes on the parler. A battery serves as a source of current, with a high-impedance voltmeter and pointed probe used to map equipotential contours onto the resistive paper. 1\ low-frequency source (up to 1000 Hz or so) can be used if ac detection methods are prefhred, and they are especially usefu.! for eliminating polarization effects occurring; when direct currents pass through an electrolytic liquid. The latter yields ion accumulation ncar one or both electrodes, causing; distortions in the equipotential distributions obtained from electrolytic tank models. 11 1\n advantage of the current model of a flux map is that il obviates the errors of estimation incurred in the hand-plotting methods described in Section 4-12, yielding highly accurate equipotential maps when careful measurements are taken. Moreover, an ohmmeter or bridge measurement between the electrodes of a conduction model leads directly to the capacitance and conductance per meter of the system being studied, without a need for the rtp and rts values required by hand-plotting techniques. The electrolytic tank can be extended to axially symmetric g'eometr'ies associated with electrostatic beam-focusing electrodes such as used in cathode ray tubes and electron microscopcs.12 Such maps in circular cylindrical coordinates are often very difficult to obtain analytically or by hand-plotting schemes. A halt:cylindrical tank containing semicylindrical electrodes and revealing their sectional views at the surface of the electrolyte permits probing the equipotential surhces in the vicinity of the z-axis. Indeed, the axial symmetry permits using merely a thin, wedge-shaped trough III which correspondingly small sectors of the cylindrical electrodes are immersed. liThe electrolytic tank was first used by C. L. Fortescue. See Transactions 12See E. Weber, Electromagnetic Fields, Vol. 1 ~JtI(Jpping

rif Fields.

rif the

A.I.E.E., 32, 1913, p. 893.

New Y Ol-k: ·Wiley, 1950, pp. I87~ 193.


239

EXAMPlE 4·20. A sheet of resistive paper measuring 1000 Q per square (with 1000 Q between opposite equipotential sides of a square sheet, regardless of size) 13 is used to model a two-conductor cable of rather unusual, though uniform, cross-sectional shape. The conductor shapes are painted on the paper with silver paint, and a measurement yields 160 Q between those conductors. Find Cjt and Glt of the actual cable if the dielectric has the constants Er = 2.5 and (f = 10"-8 U/m. The conductancc betwecn electrodes of the resistive sheet model is specified by (4-150)

(1) if Gr denotes the measured l~O U and ( f i r is the product of the conductivity of the resis1 tive sheet and its thickness. For the resistance paper used, (J,t, 10 00 U, the conductance of a curvilinear square of any flux plot applicable to this model. The usual ratio for such a plot is denoted by and irom (1), this is

6.25 Applying the latter to (4-116) and (4-131) obtains Itn' the cable C

(b.25)(2.5 x 3.34 x 10- 12 )

t t

np

133 pF/m

(J = 6.25 x 10 - 2 flU/m

ns

B. Dc or Lo~.Frequency Resistance of Thin Conductors Thin conductors (of small diameter compared to length) are of common occurrence in electric circuits. It is of interest to determine the resistance offered by a thin conductive circuit to a driving source, as depicted iu Figure 4-26. The circuit is t : median path

FIGURE 4-26, A thin de electric circnit. "From this it is inferred that any curvilinear of a nux map on this paper has 1000 Q resistance between opposite equipotential sides, or 0.001 U CUI1ULLll
291

240


immersed in a nonconductor (e.g., air). The direct current in the conductor has a density given by (3-7), J = aE. Steady currents are, by (3-32e), divergenceless, so the current consists of uninterrupted flux lines totaling I A through any cross section. The static E field in the conductor, obeying (4-6), E . dt 0, is thus conservative, so equipotential surfaces exist in the conductor, normal to the E and J field as denoted in Figure 4-26. Equation (4-6) is equivalent to Kirchhoff's voltage law for the circuit shown as follows. Taking Eg and E as the fields in the battery and the conductor respectively, integrating (4-6) clockwise over any closed path t about the circuit of Figure 4-26 obtains

ft

{'pz E . dt JPI

+

{,PI E . dt = () Jpz 9

(4-132)

But the second integral denotes -V, the negative of the battery voltage;14 with E in the conductor, (4-132) is written

=

J/a

(4-133) That (4-133) expresses the Kirchhoff law V = IR is seen by noting that the current through every cross-section A is the constant value 1=

LJ'

(4-134)

ds

in which J is not, in general, constant at each point of the cross section. The need for knowing J at every point is disposed of, if I is expressed in terms of an average density Jav; that is, (4-135) wherein lav is tangential to a properly chosen median line t as denoted in Figure 4-26. For a thin wire, t in (4-133) may be taken as the wire axis, and with Jav = atllA into (4-135), (4-133) becomes

yielding

V 1=---{'pz dt Jp, aA(t)

(4-136)

This is of the form I VIR, the Kirchhoff voltage law for the circuit, in which the de resistance of the conducting path is

R= 14Thc negative sign of the the source, which go from

+

1

PZ

P,

dt

--VA aA(t) I

or

0

(4-137)

V term in (4-132) is justified from the direction of the generated E. fields in to -.

4-15 ELECTROSTATIC FORCES AND TORQUES

241

Its reciprocal, K 1 = G, is its conductance. The notation A(t) emphasizes that the conductor cross-sectional area might not be uniform, depending generally on the position along t. For a conductor of constant cross section, (4-137) reduces to t

R=-Q

(4-138)

o-A

if t denotes the conductor length. These resistance expressions, correct for direct currents, are reasonable approximations at suHlciently low frequencies for which the skin effect, associated with reduced field penetration into a conductor with increasing frequency, is neglected. As an example, the dc resistance of 10 m of 0.1 in. (0.00254 m) diameter copper wire (obtaining (f fl-om Table 3-3) is 10

R = ~~"-" ---"-.--~ (5.8 x 10 7 )(0.00254 2 n/4) A wire this size made of aluminum, for which (f tance about 56(Y<) greater than the copper one.

=

0.034 Q

3.72 x 10 7 (jIm, will have a resis-

*4·15 ELECTROSTATIC FORCES AND TORQUES

Section 4-6 developed expressions for the work done by an external source in establishing a system of electrostatic charges in a region. Such charges reside physically on the conducting bodies of the system, which may also include dielectric regions. The force on any of the conductors or dielectric bodies can be deduced from an assumed diflerential displacement dt of that body, nn computing the change in energy dUe accompanying the displacement. It is shown that the electrostatic force can be found from the gradient of the electrostatic energy of the system, if the energy is expressed in terms of the coordinate location of the body being displaced. l<'orces obtained in this way are said to be fcilind by the method of virtual work. This method is developed for two cases.

A. System. of conductors with fixed charges. Suppose one is concerned with the system of dielectric and conducting bodies of Figure 4-27 (a), the conduetors being assumed isolated from one another so that they possess fixed amounts of free charge. (Batteries or other sources used to bring them to their charge states have been

CASE

Displacement de

(a)

FIGURE 4-27. Two electrostatic systems of conducting and dielectric bodies. A virtual displacement of dt ofa body is assumed for the purpose of calculating the l(Jn.,e on it. (a) System with fixed charges. (b) Conductors at fixed potentials.

291

242


removed.) Let one element (conductor or dielectric) be displaced by a differential distance dt, due to electric field forces acting on it. The mechanical work done by the system is (4-139) Since no additional energy is being supplied (sources are disconnected), the work (4-139) is done at the expense of the stored electrostatic energy of the system, energy being conserved, such that

+ Electrostatic energy change'

dU

=0

(4-140)

'Mechanical work done

implying an energy decrease in the amount (4-141) but dUe can be written also in terms of the moves dt = axdx + aydy + azdz

X,],

and z variations in U e as the body

(4-142) the latter being evident from the vector representation (2-11) for a total differential. A comparison of (4-141) and (4-142) reveals that

F = -VUeN

(4-143a)

implying the cartesian components of F given by

aUe F =-x

ax

, 1" y

aUe

= --~y

Fz

=

(4-143b)

I t is seen from (4-143) that knowing how the total electrostatic field energy U e of the system changes with dx, d], and dz displacements of one of its elements is sufficient to determine the force on that element. This is called the virtual work method for finding the lorce, since no actual physical displacements are required. If, instead of being subjected to a translation, the desired body is rotated about an axis, assuming constant charges on the conductors, then (4-139) is written

dU

T' dO

(4-144)

in which T a 1 Tl + a Z T 2 + a3T3 is the torque developed, and dO is the vector differential angular displacement. One can analogously show that the components ofthe vector torque T become (4-145)

4-15 ELECTROSTATIC FORCES AND TORQUES

243

B. System conductors atfixed potentials. The system consists ofn charged conductors held at the fixed potentials <1>1, <1>2) ... ,<1>. by charge sources (such as batteries). Dielectric bodies may also be included, as in Figure 4-27(b). The displacement dt of an element is in this case accompanied by changes in the charges on each conductor. For example, if two parallel conducting plates connected to a battery were moved apart, the positive and negative charges on the plates would both decrease to maintain the constant, impressed voltage difference. This means that the total electrostatic energy on the system changes with the displacement, but also it means that work is done by the sources in producing the changes in the charge states of the conductors, to maintain their fixed potentials. The work done by the sources (batteries) during the displacement of the desired element is

CASE

n

dUs =

L <1>k dqd

(4-146)

k= 1

in which the potentials <1>k on the n conductors are constants. The energy conservation relation now becomes

dUe

+

dU

dUs

~~

Electrostatic energy change

(4-147)

-----~---~

Mechamcal work done

Work done by sources to maintain fixed potentials

Since each conductor charge undergoes a change dqk while being maintained at the potential <1>k, fi·om (4-58a) the total electrostatic energy changes by n

dUe

=1L

<1>k dqk

(4-148)

k= 1

or just one-half the work (4-146) done by the sources. Thus (4-149) stating that the work done by the sources is twice the change in the total electrostatic energy; the remaind~ is the mechanical work dU done in moving the element in question by the distance dt. Putting (4-149) into (4-147) therefore yields (4-150) which means

F

= VUe N

(4-151)

EXAMPLE 4-21. Find the force between two point charges ± q separated a distance x in free space, using the concept of virtual displacement. The electrostatic energy is obtained using (4-53a), with n 2. From (4-40), the potential
29]

244


-------:;.0-

(x)

EXAMPLE 4-21

In this isolated system, the force of

+ q is

found from (4-162)

to the left (attractive) as noted in the accompanying figure. This answer agrees with that obtained from (1-52), making use orE (due to q) at the location of +q

=

EXAMPLE 4·22. Two parallel conducting plates an area A, separated a distance x as shown. f(lrce on either plate fi-om the field energy, the plates and (b) constant charges ±Qon

ax

arc separated by an air dielectric. Each has Neglecting fi'inging at the edges, obtain the assuming (a) a constant voltage V between the plates.

(a) Assuming a constant voltage V between the plates, and with the plate at x = 0 held fixed, a virtual displacement dx orthe other yields fi, = oUe/ox from (4-151). With C = one obtains

EA/x,

__ oUe _ 0 I . 2 _ V,2 D ('EoA) ----(zCV)----

j<

x

ox

ox

2

ax

x

The negative result denotes au al tract ivc f(lrCe, since the stored energy increases with a decrease in the plate separation x.

EXAMPLE 4-22

PROBLEMS

245

(b) With V disconnected, fixed charges ± Qreside on the plates. Then a constant Ex = Vlx exists between the plates regardless of their separation (neglecting fringing). The electrostatic energy iB conveniently expressed by Ue = (t)QV = (t)QExx, and with Q and Ex both independent of x, (4-143) obtains

REFERENCES ELLIOTT,

R. S. Electromagnetics. New Y ork:

MeGraw~

Hill, 1966.

LORRAIN, P., and D. R. CORSON. Electromagnetic Fields and Waves. San Franciseo: Freeman, 1970. REITZ, R., and F. J. MILFORD. FOllndat£ons of Electromagnet£c Theory. Reading, Mass.: Wesley, 1960.

Addison~

PROBLEMS

SECTION 4-2 4-1. A static point charge ill tree space is located at the arbitrary position P'(x',y', z'). Use (4-lOb) to develop the vector expression for the E field at the observation point P(x,y, showing that

Sketch the system, labeling the pertinent details. Show to what result this expression reduces if (a) q is located at prO, 0, z') on the z.-axis; if (b) q is located at the origin.

4-2. Use the expression developed in Problem 4-1 to find the vector E field produced at the (b) pro, 3,0) and (c) pro, 0, 0), by the charge q = I JlC located following points: (a) prO, 3, at prO, 0, I), with distances given in meters. Find the magnitude ofE at prO, 3, 5), as well as the vector force exerted on a second charge q' = I JlC located there. 4-3.

M.

In Example 4-1, prove that (4-18) becomes (4-19) as the line length L

-> 00.

J-2,

(a) In Example prove by use of integration tables that (4-21) yields the result (4-22). (b) Show, for the observation point P(p, z) located in the z = 0 plane, that (4-22) reduces to (4-18) of Example 4-1. (c) Show that the limiting expression (4-23), as L -> 00, is correct.

• 4-5. (a) A thin conductor in free space is bent into a circle of radius a, charged with the uniform linear charge density Pt, and centered at the origin in the z = 0 plane. Use a direct integration for E to show that on the z-axis

With Pt = 10-- 6 C/m and a = 10 cm, graph E z versus z from 0 to ±50 em. (b) Show that E z at great distances (z» a) converges to the form of (1-57b) for the point charge, if the answer is expressed in terms of the total charge q on the charged circle.

4-6. In Example 4-3, make use of integration tables to show in detail that (4-24) is obtained. With Ps = 10- 8 C/m 2 , a = 10 cm, graph E z versus z from 0 to ±50 em.

291

246


°

• 4-7. An annular disk lies in the = plane, centered at the origin, with the unitt1rm charge density Ps between its inner and outer radii a and b. Use direct integration and symmetry to show that the electric tield on the z-axis is (4-152)

SECTION 4-3 4-8.

By usc of Gauss's law (3-37), deduce all approximate expression for the static E field vcry elose to the ring charge of Problem 4-5 .

• 4-9. A very long circular coaxial line has two concentric dielectric insulating sleeves of permittivities E[ and E2 filling the space between the inner and outer conductors as shown. Assume the interface at p = b to be midway between the conductor radii a and c, on which reside the charges (Land - (~, respectively, for every length t of the system. (a) Deduce from Gauss's law the expressions II)r D and E in the two dielectric regions. (b) By usc of graphs sketched to relative scales, show the variations with p exhibited by E" wilhin the dielectric materials (()r two cases: (1) E1 = 2E 2 , and (2) E1 = E2 /2. Assuming that the aim is to maintain approximately tbe same average electric field within the two dielectrics, which case provides this? 4-10. In Problem 4-9, let Erl = 4 and b = 2a. \Nhat value of Ep fields of both regions have the same value?

Er2

will make the maximum

SECTION 4-4 4-11.

By expansion in rectangular coordinates, prove the relation (1·-32) concerning V(I/R).

4-12. (a) Rework Problem 4-5 for the thin ring of charge, this lime by integrating i()r the potential $ at the position P(O, "c) on the z-axis. From the result, determine E at P by use of (4-:-11). (b) Using the fidd E(O, z) found in (a), verily by means of the line integral (4-38b) that the identical potential result $(0, z) is obtained. " 4-13. Rework Example 4-3 lor the charged disk, by first finding the potential $ at the axial position P(O, z), and then making use of (4-:11) to find E(O, z). (b) Inscrtjng the E field found in (a) into the line integral (4-38b), verily that the identical potential result of (a) is obtained at P(O, z). 4-14. Make usc of the pOlen tial expression (4--43), obtained ji)l' the static dipole charge of Example 4-8, to derive in detail its E field expression (4-44) _ (b) Beginning with the E field found in (a), employ the line integral (4-38b) to verify that the same potential $(r, 0) as found in (a) is obtained.

PROBLEM 4-9

PROBLEMS

247

291

SECTION 4-5 .. 4-15. (a) For the coaxial line of Example 4-9, instead of determining the full potential difference V between the two conductors as given by (4-50), determine the potential

2nd

tn b p

Note that this result hecomes the voltage difference V =

V

h

--tn-b p ('II

a Usc (4-l53b) to derive a corresponding expression f()r E(p) in the coaxial line. (c) Let the applied voltag-c across a particular coaxial line be V = 100 V, with a = 2 em, b = 10 ern, and Er = 2 till' the dielectric. What is the capacitance per meter length of this line? Show that (4-153b) can be written
4-16. The large commercial coaxial line, type RG-213/U, has a polyethylene dielectric with an inner conductor of 0.195 in. (= 0.495 em) diameter and is specified to have a distributed capacitance of 29.5 pF/ft ( 96.3 pF/m). What is the inner diameter of its outer cOllductor?

,4-17.

(a) Use Gauss's Jaw and symmetry to obtain the static D Held exprcssion It)r the parallel-pia te capacitor of Fig-ure proving 1hat the E field given in that fig-nrc is correct. Employ the lil1e integration (4-:i3a) to obtain the potential

If

EA

(d - x)

(b) Usc (4-154a) to infer the total voltage V between the conductors, hom whieh verify the capacitance of this parallel-plate system. Make use of (4-52) to convert to the alternative j()rm expressed in terms of V:

V

x)

(4-1 54 b)

Use (4-154b) to derivc a corresponding expressioll I()r E(x) between the plates. (d) The voltage applied across a particular parallel-plate capacitor is 100 V. The plates arc square, 1 m on a side, and separated by use ofa l-ll1ln thick polyethylene sheet. What is the capacitance? Show that 54b) here becomes
4-18. A parallel-plate capacitor used in a microcircuit is made of a I mil (=0.001 in. = 25.4 tim) thick sheet of polyethylene (Er sandwiched between two square conducting plates 3 mm ou a side. Find the capacitance.

, lot

248


4-19.

(a) Usc Gauss's law and symmetry to derive the expression for D(r) of the spherical capacitor of Figure 4-7(b), showing that the E field denoted on the figure is correct. Use line integral (4-38a) to express the potential (»(r) at any location between the conductors, using the negative conductor (r = b) as the potential reference; that is, prove that (»(r) = - q 41tE

(1 i) - - -

b

r

(4-155a)

(b) Infer from (4-155a) the total voltage V between the conducting spheres, whence verify the capacitance (4-53) of this system. (cl Use (4-53) to rc-exprcss (4-155a) in the alternative form (»(r)

~(! 1 J r a

(4-155b)

b

(d) Assume V = 100 V aeross a spherical capacitor with a = 5 em, b = 10 em, and Er = 2.26 for the dielectric. What is its capacitance? Show lor this example that (4-155b) is written (»(r) = I OOO(r - 1 - b - 1) if distance r is expressed in cm. Graph the potential (» versus r between the conductors (in D.5-cm steps). Plot Er versus r on the same graph, noting the location and value of Emax. What is the potential at the exact midpoint between the conductors?

4-20.

(a) With the coaxial line E field given by (4-49), make use of its capacitance (4-51) to obtain the relationship between E and the voltage V impressed between the conductors.

V E=a - P b p tn

(4-156b)

a

At which p location between the coaxial conductors docs the E field have its maximum value? (b) Given that the maximum allowable static E tleld in polyethylene should not exceed 1200 V imil ( = 1.20 MVlin. = 472 k V jcm) if dielectric breakdown is not to occur, calculate the maximum dc voltage permitted between the conductors of the RG-2l3/U coaxial line ofProblem 4-16.

SECTION 4-6 4-21.

(a) Make use of (4-58e) to find the expression for the energy stored in the electrostatic field of the parallel-plate capacitor of Figure 4-7 (a). (b) Make use of the energy result of (a) to determine the capacitance C of the parallel-plate system.

4-22.

Repeat Problem 4-21, this time for the spherical capacitor of Figure 4-7 (b).

• 4-23.

A long circular coaxial line uses two concentric dielectric layers as illustrated for Problem 4-9. With Q, -Qresiding on each length t of the conductor surfaces a, c, respectively, the Dp field between the conductors has been found to be Q/2npt. (a) Find, by usc of the line integral (4-46), the voltage difference V between the conductors, assuming the zero potential reference at p c. (Explain why a sum of two integrals is required.) Use the V result to find the capacitance C of any length t of the system. Identify the answer as the series combination of the capacitance contributions associated with the two dielectric regions between the conductors. (b) Use (4-58e) to determine the electrostatic energy of a length t of the system. (Explain why the sum of two integrals is needed.) Use the U e result to obtain the capacitance C of the system.

SECTION 4-7

J

The parallel-plate capacitor of Figure 4-7 has the upper plate (x = d) at zero potential while that of the lower plate is at the potential V. Ignoring fringing, solve Laplace's equation (4-68) for (»(x) subject to the proper boundary conditions. Find E from (4-31).

4-24.

PROBLEMS

,

249

4-25. For the spherical capacitor of Figure 4-7, assume <1>(a) V and <1>(b) O. Integrate Laplace's equation (4-68) for <1>(r) , subject to the given boundary conditions. Find E using (4-31). 4-26. A pair of conducting cones, ideally of infinite extent, are located at (il and (iz as shown, separated by a charge-Irec dielectric of permittivity E. A voltage difference V is impressed at the infinitestimal gap between their apices at the origin to provide the potentials <1> 0 and Von the conductors as indicated. (a) Argue from the points of view of symmetry and the boundary conditions for <1> as to why the Laplace's equation (4-68) must here reduce to

d ( d<1» I ~ sin ( i - = 0 sin 0 dO df) (b) Solve the cones

(4-157)

157) by means of two integrations, obtaining the solution [or the potential between

in which C1 , Cz are arbitrary COllstants. (cl Apply the given boundary conditions at 0 and O2 to evaluate the arbitrary constants, showing that

01

tan (0/2)

t71-'-'-'~

<1>(0) = V

tan (0 2 /2) tan (Od2)

(4-158)

tn---tan (0 2 /2)

(d) Find the electrostatic field between the cones, obtaining

V

E

(4-159)

4-27. For a particular biconical conductor system as analyzed in Problem 4-26, let 01 = 30°, O2 = 150 0 , E = Eo, V = 100 V. (a) Sketch this system. Show from (4-158) that on its symmetry

(z)

~~-----r----

____'/

/

/

P(r.H)

'['=0 0

PROBLEM 4-26

291

250


7r - -

--

=0

PROBLEM 4-28

plane 0

= 90°,

o=

2 are tan

tan ((11/2»)IV 02J -----tan [( tan (0 /2) 2 2

(4-160)

Use this to graph 11 versus

• 4-28.

The circular coaxial line shown here in sectional view uses two dieleetrics of perm itt ivities E1 and E2 extending over the 4> intervals (0, n) and (n, 2n), respectively. Assume the conductors statically charged with q, - If on the inner and outer conductors to make = 0 and 4> = n?) What are thus thc solutions for the fields 0 1 and O 2 in the regions? (b) Use the boundary condition (3-45) at p a to dctcrmine the expression for the total charge q on any length t of the inner conductor. Obtain the capacitance by usc of (4-48), showing that

C = nt(El

+ (2)

tn

(4-161 )

b a

Identify the answer as cquivalent to the parallel combination of thc capacitances contributed by the top and bottom halves of this coaxial systcm. Furthcr show that the eapacitance of this two-dielectric system is the same as that of a coaxial capacitor with a single dielectric having an E that is the average OfEl and E2 • (e) If this system consisted of half polyethylene and half air with a = 2 mm and b = 7 mm, find its Cit.

4-29.

In the two-dielectric coaxial system of Problem 4-23, make use of (4-53e) to determine the electrostatic energy of any length t. Use the result to deduce its capacitance .

.. 4-30.

The coaxial line illustrated is similar to that of Problem 4-28 except that the first region (E1) extends over the arbitrary angular interval (0,4>1) as shown. (a) Confirm by the methods of Example 4-12 that the 4> and E solutions in the two regions are unchanged. Determined 0 1 and O 2 in the two rcgions. (b) Make use of the boundary condition (3-45) on the inner conductor (p = a) to determine the total surface charge If on any length t. By use of (4-43), find the capacitance C, obtaining C = t[El4>l

+ E2(2n tn

b a

4>1)]

(4-162)

PROBLEMS

251

29]

PROBLEM 4-30

(c) With a = 2 mm, b = 7 mm, region 1 a polyethylene wedge 5° wide and region 2 air, find

CIt for this system. Sketch it. Compare its Cjt with that obtained for a completely air dielectric.

SECTION 4-9 4-31. Employ the definitions of the hyperbolic functions to show that the solution (4-88a) of the two-dimensional LapJace equatiolJ (4-82) can be expressed equivalently as (4-163a) in which C1 , C2 , C~, C~ are arbitrary constants. The equivalent form of the alternative solution (4-88b) is seen to be (4-l63b)

4-32. The very long rectangular conducting channel, with interior dimensions a, b as shown, is insulated at its top corners [rom a conducting cover plate that is at = Va V. Use an appropriate solution of thc two-dimensional Laplace equation, subject to the appropriate boundary

iL -==b

1

Vo

\" <1>=0

(x)

a

PROBLEM 4-32

y(

252

STATIC AND QUASI-STATIC EI,ECTR1C FIELDS

conditions, to obtain the potential (x,y) at any interior point, showing that

(x,y) = 2 Vo n

f: _1_-_(,--_1,-- sinh nny sin nnx n= 1

1mb a

.

n smh

l

sinh ny 4 Vo a nx - - - - - - sin n nb a sin

a

a

+

I 3

a

3ny sinha sinh

3nx sin 3nb a

+ ...

j

164)

a

[Hint: Noting that both terms lxY and e -kxY of solution (4-BBa) arc needed to satisfy the boundary condition (x, 0) = 0, it is more convenient to usc the equivalent hyperbolic solution developed in Problem 4-31. Observe that sinh u -> 0 and cosh u -> 1 as u -> 0.]

4-33.

For the long, covered conducting channel of Problem 4-32, assume V = 100 V and a = b (square cross section) and usc (4-164) to calculate the potential (x,y) in the cross section at the nine points detemdned by the intersections ofx = a/2, a/2 and 3a/4 withy = a/4, a/2, and 3a/4. Sketeh the cross section, labeling the potentials found at the indicated points. Usc the sketch as a basis for estimating the shapes of the = 25 V and 50 V equipotential contours in the cross sections.

4-34.

(al Make use of 164) to find the series expression for the E field at any P(x,y) in the conductive channel described in Problem 4-32. (b) Assmning V = 100 V and a = b I m, evaluate E(x,y) at the following points: (0, , (a/2, and (a, a/2). (c) Assuming air dielectric, use results of (b) to determine the surface charge density at the conductor locations (0, and a/2).

SECTION 4-10 4-35.

The very long rectangular conducting channel viewed sectionally in the figure has dimensions as indicated by the square-grid overlay. Thl' cover plate is at 100 V with the remaining sides at 0 V. Write the expression (4-9Id) for the potential at the three indicated interior points, taking advantage of the symmetry about the plane A. Solve the simultaneous linear equations for the potentials. Show a labeled sketch denoting the potential values obtained.

4-36.

Use the exact Fourier expression (4-164) to verify the potentials in the last column of the table at the end of Example 4-15; also calculate the values of <1>1, <1>2'

4-37.

Repeat Problem 4·35 for the square conducting channel shown, making use of the symmetry about the plane B. Find the six potentials at the indicated points (a) by matrix methods; A

B Insulated corners

II

(1)= 100V

q,= 100V \ I

I I

I

,.._1+_+_ I

2'

I

I

I

--..1-- .. - I

31 --.,.-

1-:

\

<1'-0

I

I

I

I

I I

21

I

61

31

I

:

i

I

4' f--t 51

1+- 1--

--t--.,.--~

,.....-+---t-- -+-I

(!)=O

I I

I

PROBLEM 4-35

I

PROBLEM 4-37

PROBLEM 4-33

PROBLEMS

253 29]

C I

I I

I

I

-.--+ "--+---1 21 4 I 51 61 --+-.-.I 11 31 -+-+I

-+I +I

I

I

I

I

I

-+- +-+ I

+--1- -1I

I

I

4>=0

PROBLEM 4-40

(b) by iteration. (c) Use the Fourier expression (4-164) to check for the correct potential at point 2, the center of the channel.

.. 4-38. The very long conducting channel, of triangular cross section as shown, has dimensions given by the square-grid overlay. The upper cover is at 100 V relative to the other two sides. Solve for the six unknown potentials by (a) matrix methods and (b) iteration. 4-39. Double the linear grid density of the potential points in both directions in Problem 4-38 to provide 28 unknowns, making use of symmetry). Solve for the unknown potentials by iteration (use of a computer is advised). Showing labeled potential values on a reasonably sized reproduction of this system (using gridded papcr), sketch in equipotential contours at
.. 4-40. Inside a square, hollow conductor is coaxially placed a square conductor canted by 45° with respect to it, as shown in the figure. Relative dimensions are suggested by the square grid overlay, and the inner conductor is at the potential of 100 V relative to the outer one. Noting the symmetries about planes Band C, 'write the expression (4-9Id) f()r the potentials at the six indicated points. Solve for the unknown potentials using (a) matrix methods; (b) iteration. On a sketch of the system, label the potentials obtained.

tI

SECTION 4-11 • 4-41. Begin with the expression (4-96) for the potential at P(x,.Y) of the two-line charge system to derive, by completing any omitted details, the equation (4-100) for the equipotential circles defined by the parameter K. Show the details leading to determining (4-104), the expression for K as a function of hjR, as a quadratic solution from (4-101) and (4-102). ClarifY the meanings of the ± sign~ (4-104). Finally, assume two conductors, spaced 2h center-to-center and of radius R each, to fill in two corresponding equipotential cylinders as depicted in Figure 4-13. Sketch this system of Figure 4-14( b) in a sectional view. Establish the conductor potentials, given now by (4-96), as being q h+d <1>0 = - - t n - -

2nd

respectively, with d tance (4-107).

R

and

-
2nd

tn

h-d R

From this information deduce the parallel-wire line capaci-

Iltll

254


4-42.

(a) For the parallel-wire line of Fii.(ure 4-H(b), convert its potential field (x,]) , i.(iven by (4-96), to a form dependent on the vol tai.(e V between the conductors by use of the capaci tanee result (4-107), showini.( that (x,y)

V

R2

=----tnIi + d RI 2 tiN

(4-165)

R

in which R J and R2 arc defined by (4-97), and the quantity (h other equivalent ti:lrms, for example,

h+d ~~--

=

R

d+h-R --~"---"~

d

=

h+R

+ d)jR

can be shown to have

h+R+d h+R

(4-166)

d

JizZ -

provided that d Ri. (h) As a check, show that (4-165) reduces to the expected potential values at the f()llowing conductor-surface points: PI (Iz - R, 0), P2(h + R, 0) on the positive conductor; also a\ P 3 ( -Iz - R, 0), P 4 ( Iz + R, 0). Sketch the cross-sectional view of the line, laheling the point locations. (c) Apply (4-:)1) to 65) to t1nd the E-field expression at any P(x,y), showing that

v

E(x,y) = 2 tn~

{(x d x+ d) (] aX-if - R~ + ay l~

(4-167)

R in which RI and R2 are defined by (4-97).

4-43. A parallel-wire line consists of two conductors, each of 5-cm radius, separated in air by 20 em center-to-eenter. (a) Sketch this system (cross-sectional view), labeling R, It, and d. Show a few representative E-flux lines connecting the virtual image charges at d, -d. Use physical reasoning to explain why you expect the maximum E field of this system to be at the points on the x-axis whCl'e the conductors are nearest to each other. (b) Assuming 1000 V between the conductors, make use of (4-167) in Problem 4-42 to calculate the electric field at the surface point P(11 - R, 0) on the positive conductor. Show this V(:ctO[ on your sketch. (c) Use the appropriate boundary condition to determine the surfzlce charge density at P(h R,O). (el) What must the voltage between the conductors be, to produce the E-field magnitude of 1 MV!m at P(h - R, OJ?

4-44.

Repeat Problem 4-43, except assume the wire radii to be R

2 mm.

4-45. (a) A parallel-wire telephone line uses 165-mil (OA19-cm) diameter bare copper conductors with 12-in. (30.5-cm) center-to-eenler spacing. Find the capacit,ince between the conductors in pF/m; in J1FJmile. (b) Repeat (a), assuming il-in. spacing.

SECTION 4-12 '"" 4-46.

(a) Use superposition combined with the method of images to derive an approximate expression for the capacitance between two long, parallel wires of radii al and a 2 , each spaced Iz above an infinite conducting ground plane as in Figure 4-16( b), assuming the center-lo-center separation D. Sketch this labeled system. Show that

c t (b) Show, in the limit as It result (4-110).

-> 00

2nE

(4-168)

(the ground plane is removed), that (4-168) becomes the

PROBLEMS

255

291

PROBLEM 4-50

4-47. A parallel-wire line in air, with conductors of unequal radii and located above a ground plane as in Figure 4-16(b), has the dimensions at 3 mm, a 2 = 1 mm, D = 20 em, and It[ = h z = It = 10 em. (a) Sketch the system, and find its distributed capacitance by usc of (4-168). (b) Obtain the capacitance of this system with the ground plane removed, finding it by usc of (4-110). 4-48. Repeat Problem 4-47, this time assuming the parallel-wire system spacings It[ and hz above the ground plane, as shown in Figure 4-16(b).

to

have different

SECTION 4-13 4-49. For each of the three Hux plots given in Figure 4-21, find the Cit, assuming air dielectric. In (e) of that figure, find the value of Cit corresponding to the width between tooth-centers. 4-50. Shown is the sectional view ora conducting elliptic cylinder within a round conducting pipe, dimensioned as noted. (a) Use Iield-mapping techniques to lind the capacitance per unit length (Cjt) of this system. Assume air dielectric. (Employ at least one equipotential surface, at I]) = V12, sketched between the conductors, using curvilinear squares as a basis as exernplilied in Figure 4-21. Take advantage of symmetries to reduce the amount of sketching needed, observing that straight flux lines coincide with the symmetry planes.) (b) Denote the location(s) in the system where the maximum electric field is to be found, explaining your reasoning. (c) If the dielectric permittivity were 4E o , what new value of Cjt would be obtained·? 4-51. Repeat Proqlem 'i-50, in this case for the round cylindrical conductor eccentrically located within the rotind pipe as shown. Compare this Cit result with that obtained on moving

PROBLEM 4-51

lIat r oj

tht

01

256


PROBLEM 4-52

the inner conductor to its coaxial location, obtaining the latter answer analytically. Comment on this comparison.

4-52.

Repeat Problem 4-50, this time for the square conducting bar located inside the round pipe as shown. (Note that symmetry planes can be drawn through opposite corners of the square conductor.)

4-53.

Repeat Problem 4-50, in this instance lor the round cylindrical rod above the infinite ground plane shown. Compare your graphical solution with that obtained using the exact expression (4-106), assuming air dielectric.

4-54.

Repeat Problem 4-50, here for the infinitely wide, equispaced gridded system of rods as shown. The grid-rods are assumed electrically neutral ("floating"), with no net charge on each. Compare the Cit result obtained, for every width 4a as shown, with that obtained in the absence of the grid rods, assuming air dielectric.

Tl d=2"

«()=o

W////X1//////-//ffi///7//mM PROBLEM 4-53

-l-

f8-

o o =V

-j---jL--j---

(1)='''

_ t-;_% _ _ :/;;:/~/~_%/;:;:/~/~~~:0~;:'ij~~~:;;:;?:~~~:?2~~~~~~%/CP:7;_/~=;:;;o~~:/~/,?// PROBLEM 4-54

PROBLEMS

SECTION 4-14 4-55. Let the system of Figure 4-21 (a) employ a lossy dielectric with

Er =

4 and

257 (J

29]

=

10-- 6 mho/m. Find the capacitance and the conductance per meter depth. What resistance is

seen by de voltage source connected between the conductors of a 5 m long sample of this twoeonductor system? (a) A coaxial line with a = 1.5 m, b = 4.8 mm has a dielectric with Er = 2.60 and loss tangent E" /E' = 10 - 4 at the frequency I = 10 6 Hz. Find its capacitance and conductance per meter. (b) If a 4-em-long section of this coaxial cable is used as a capacitor at 10 6 Hz, find its equivalent parallel RC circuit, and its Q. How arc its Q ancl loss tangent related?

4-56.

4-57.

Two circular conductive rods 2.5 em in diameter are driven into wet earth 15 em between centers, to a depth of 1 m. 1000-1-1z bridge measurements between the conductors show the system to be equivalent to a 67.3 pF capacitor in parallel with a 985.9 Q resistor. Determine the Er and (J of this soil, neglecting field fringing at the bottom of the rods. [Answer: Er = 6.0, (J = 0.0008 mho/m]

4-58. A rectangular box has the inside dimensions 12 x 8 x 2.5 cm, the opposite 12 x 8 em sides being conducting plates. A sample of the wet earth of Problem 4-57 is packed inside. Using the answers to that problem, deduce the resistance and capacitance values expected to be measured between the plates. Which of these two measuring schemes is the more precise? Explain. 4-59.

Deposited conductive films, like conductive paper, can be used in the modeling of two dimensional systems in Figure 4-25), or in the evaporation or beam deposition of thin resistive dements on a suitable Honconductive surface. (a) Determine, by use of (4-131), the resistance R between the ends of a thin conductive film of uniform thickness. If the film is subdivided into a number of curvilinear cells, with rls in series between the constant-potential ends and rip in parallel in the transverse direction, show that (4-1,11) can be wri tten in the f(Jl'm (where R = C- 1) (1)

in which Rsq = l/ad, called the "resistance per (curvilinear) square" of the film. (Note that the symbol d replaces t in (4-131) to denote the film thickness.) (b) Find the resistance per square of a metal film 0.15 Jim thick, if the metal conductivity is (J = 104 mho/m. If (J = 107 mho/m. (c) A film of aluminum (a 3.6 x 10 7 mho/m) is deposited 0.1 Jim thick on an SiO z substrate. What is its per-square resistance? This film is deposited as rectangular stripe of width w = I mil = 25.4 Jim and length t = 12 mil (an interconnection on a VLSI layout). Use (1) to determine quickly the resistance between the ends of the stripe. [Answer: (b) 667 Q pcr squarel

4-60. Using a resistive paper model as suggested by Figure 4-25(a), the toothed structure of Figure 4-21 (c) is modek,d by applying silver paint electrodes of the shapes shown onto resistive paper measuring 1000 Q per square. From the curvilinear square field sketch of Figure 4-21(c), what G and R values would be measured between the electrodes?

SECTION 4-15 4-61. A variable air capacitor, using a rotating, multiplate rotor, provides a linear capacitance variation from 30 to 500 pF as the rotor rotates from 0 to 260 0 • Determine the electrostatic torque on this rotor at any arbitrary angle setting, when 5000 V are applied.

tl

CHAPTER5 ______________________________________________

Static and Quasi-Static Magnetic Fields

The static magnetic fields of steady currents and the electromagn~tic fields of relatively slowly time-varying currents are considered in this chapter. Ampere's law is applied to symmetrical current configurations and to magnetic circuits containing high permeability cores, for the purpose of obtaining their magnetic fields. The static magnetic potential, a vector function, is inferred, and from this, the Biot-Savart law. Faraday's law then leads into the concepts of self· and mutual inductance and the energy and forces of the magnetic field.

5-1 MAXWELL'S EQUATIONS AND BOUNDARY CONDITIONS FOR STATIC MAGNETIC FIELDS In Section 4-1 it was pointed out that static magnetic fields are required to satisfy only the Maxwell equations (4-3) and (4-4)

v

V·B=O

(5-1)

H=J

(5-2)

X

The divergenceless property (5-1) specifies that B flux lines are always closed, whereas (5-2) states that the sources of static magnetic fields are steady currents of density J. The divergenceless property of any direct current distribution in space is moreover assured by (3-82c)

VoJ=O

258

(5-3)

:;-2 AMPERE'S CIRCUITAL LAW

259

291

although this property of direct current is not independent of Maxwell's equations, in view of the fact that (5-3) is a consequence of taking the divergence of (5-2). The three foregoing differential equations have integral counterparts given by the static versions of (3-49), (3-66), and (3-82b) as follows.

~sB' ds = 0

(5-4)

i

(5-5)

H ' dt

~sJ'ds=O

(5-6)

whereas the constitutive relationship between Band H at any point, for the linear, homogeneous, and isotropic materials considered in this chapter, is given by (3-64c)

B= tlH

(5-7)

The boundary conditions for magnetic fields have already been derived in Chapter 3 under the general assumption of time variations for the fields, though they remain unaltered under static conditions. These are given by (3-50), (3-71), and (3-132) as follows. (5-8) (5-9)

Jnt - Jn2 = 0

(5-10)

assuring the continuity of the normal components of the static Band J fields at any interface, as well as the tangential components of H. The presence of a current in a finitely conductive region implies the presence of an E field, in view of relation (3-7) that J = O"E, yielding the possibility of coupling the static magnetic field with an electrostatic field. 5-2 AMPERE'S CIRCUITAL LAW

Ampere's circuitalbJ:w for the static magnetic field in free space was initially discussed in Section 1-11. The presence of a magnetic material with a permeability Jl in the region of interest was taken into account by the definition (3-58) of the field H, the law in this event becoming (5-5)

rna

[5-5] Figure 5-1 illustrates (5-5) relative to a conductor compelled to carry a steady current 1. Thus, the line integral ofH, around the dosed path tl shown, yields the value zero because the current i enclosed by that particular choice of path is zero. On the other hand, the current piercing S2 is precisely the current I carried by the conductor, whereas i = 0 for the assumed path t3 because the current I flows both into and out of 8 3 to provide canceling contributions to i.

tot

260

STATIC AND QUASI-STATIC MAGNETIC FIELDS

ds

Positive integration sense

~ ~-----"'-m-:~--=----r-;l

ds FIGURE 5-1. Showing typical closed paths t" t z , and the interpretation of the current i.

t3

chosen to illustrate Ampere's law and

Two important interpretations of Ampere's circuital law are the following 1. Sleady current sources possess magnetic flux line distributions that, at positions in space near the sources, are directed in accordance with the right-hand rule. 2. Ampere's circuital law may be used as the basis for finding the H field (and thus B) of a steady current if the physical symmetry of the problem permits extricating the desired field from the integral. Applications of2 to finding the static magnetic fields of systems exhibiting simple symmetries have been given in Examples 1-13, 1-15, and 3-4. Additional examples involving conductors wound about symmetrically shaped magnetic materials are given here.

EXAMPLE 5·1. Two long, coaxial, circular conductors carry the steady current I as shown in Figure 5-2. Assume constant current densities over each conductor cross section. The region a < p < b is filled with a magnetic material of constant permeability /1; the region b < p < c is air. Find Band H in the two regions. Sketch their graphs versus p, assuming /1 = 100/10' From symmetry and the application of the right-hand rule, the magnetic field is everywhere

and because

f dt = 2np, solving for H", obtains (5-11 )

5-2 AMPERE'S CIRCUITAL LAW

261 291

Magnetic sleeve (region 1)

\

Air (region 2)

o'--a.1..--==,"-c--_" -- P b (b)

(a)

FIGURE 5-2. Coaxial line partly filled with magnetic materiaL (a) Cutaway view of the line. (b) Fields produced by I.

This result is independent of Jl, which mcans that it applies to both thc magnetic region 1 and the air region 2. Thus the B field in each region is found by inserting (5-11) into (5-7) B=a.p-

JlI 2np

a
Jlol a.p2np

b
B

(5-12)

fll lOr ;

These results show that if region I had a permeability Jl 100110, B.p just inside the magnetic region (at p = b -) would be 100 times as dense as on the air side. This is illustrated by the solid curve of Figure 5-2(b). Thus nearly all of the flux of B resides within the magnetic material, if Jl » Jlo·

EXAMPLE 5·2. Suppose an n turn, closely wound toroidal coil with a rectangular cross section is filled with a magnetic material of constant permeability Jl from a to b as in Figure 5-3(a). With a current Fin-the coil, find Band H in the two regions; sketch their relative magnitudes if 11 = 100Jlo for the magnetic materiaL Compare the total magnetic flux t/lm in the core ifit is all air with that obtained ifit is all magnetic material, assuming Jl = lOOllo. From the symmetry it is evident that Ampere's law is useful for finding Hi choose t as a circle with the radius p shown in Figure 5-3. From the symmetry and the righthand rule, H must be 11 directed and of constant magnitude on t. Equation (5-5) then yields

Ct

:d led

urH

m;

11th tol

1-5 whence

H -

nI

.p - 2np

(5-13)

262


Magnetic material (iJ.)

~'fa <,<

Air (iJ.o)

:

I

I I

I I

"'

I

:

: ''1

b)

: . . . '-i . . ..Hcp I

Bcp =

o

0:

I I

Bcp (b

< p < c)

abc

-?<-p

(b)

(a)

FIGURE 5-3. A toroid of rectangular cross sectiou, partially filled with a magnetic materiaL (a) Dimensions of toroid. (b) Interior fields.

Using (5-7), B in the magnetic and air regions of the core becomes

a",

/1 n1 2np

/10 111 B = aq)5", = a", 2np

a
b
The graphs of thcsc quantities are shown in Figure 5-3(b). If the core is all air, the total B flux in it becomes

r

JS(core)

B· ds

f. iC d

z=O

p=a

/10 711 ·--dpdz

2np

/1onJd

2n

tn

c

a

For a completely magnetic core, /1 = 100/10 would appear in the foregoing answer in lieu of /10, demonstrating the considerable increase in magnetic flux possible if an iron core is used.

5-3 MAGNETIC CIRCUITS It has been noted that a magnetic material oflarge permeability can aid in producing large magnetic flux densities compared to what would exist without its use. From (5-1) it is evident that physical magnetic fields must always consist of dosed flux lines. By constraining the B flux to occupy the interior of closed (or nearly closed) paths of magnetic material, one may speak of magnetic circuits with reference to those closed paths. Figure 5-4(a) shows an idealized magnetic circuit: a closely spaced toroidal winding establishing a magnetic field within it, with essentially no magnetic flux outside the core, whether or not the core material is magnetic. If the winding is localized on the core as in (b), the effect of a high-permeability core material (fl »flo) is such that the magnetic flux t/lm generated by the current I in the coil still appears almost wholly within the boundaries of the core. The magnetic flux must consist of closed

5-3 MAGNETIC CIRCUITS

263

291

Median line

I Magnetic

flux

(b)

(a)

\

(c)

FIGlJRE 5-4, Development of magnetic circuit concepts, Toroidal core with closely (e) A generalized magspaced winding. (h) With a localized winding, showing leakage netic circuit: leakage Hnx neglected,

lines as required by the divergence property (5-1), and because of the constraint supplied by the refractive law (3-76) (requiring that B flux leave the surface of the highpermeability magnetic core very nearly perpendicularly), one concludes that very little can appear outside the core as leakageJlux if the permeability of the core is sufficiently large. In f(~rromagnetic cores having relative permeabilities of 10 2 to 104 or more, the leakage flux developed external to a core may thereic)fe ordinarily be neglected. The analytical determination of the leakage flux usually requires a rigorous solution of the boundary-value problem of the magnetic system; in general, this is a difficult process. For present purposes concerned with magnetic circuits as in Figure 5-4, the magnetic core is assumed linear, homogeneous, and isotropic; furthermore, the leakage flux is ignored, implying a constant flux t}! m through any cross section of a single-mesh core. This flux is (5-15a)

. fh Jr (

til

d

cd if S is any cross section. The need for knowing B at each point in the cross section i S T C I obviated if t/lm is expressed in terms of an average flux density Bav over S; that is, 1m ./

(5-15b)

cm assuming Bav lies tangent to a median line t, as in Figure 5-4(c). Even for a toroid of constant cross section, the median line will not lie precisely at the core center, for one may recall from the solutions obtained in Example 5-2 the inverse p dependence of B 1> in the core, In the following, the Bav is assumed at the center of the core cross section; that is, the median line t is taken as the core center line. Then (5-15b) becomes a good approximation if the core is thin. To find the flux t/I m developed by the current I in the core of the single magnetic circuit of Figure 5-4(c), apply Ampere's law to the median path t; that is,

J. jt H' dt =

rtl A

(5-16)

tot i-5

)-5 Yc

264


=

in which dt = atdt, and H

RIp,

=

atBav/p,. Making use of (5-15b)

In the generalized case, the cross-sectional core area A can be a variable depending on the location along the median path t, and it is designated A(t). The core flux tf1m through any cross section along t is constant if the leakage flux is neglected, obtaining tf1m =

nI dt

~t p,A(t)

(5-17a)

Wb

= VIR and applicable to the thin, de circuit of Figure 4-26, reproduced here in Figure 5-5 (b). Thus (5-17a) can be written in the analogous form

It is seen that (5-17a) is the analog of Ohm's law (4-136), of the form I

(5-17b) in which ;g; = nI is called the applied magnetomotive force (mmf), the analog of the applied voltage V (or emf; electromotive force) in the analogous electric circuit of Figure 5-5(b). The denominator q{ of (5-17b) is called the reluctance of this series magnetic circuit, defined by q{=

~ --A/Wb p,A(t) dt

t

or

(5-18)

which from Figure 5-5 is seen to be the analog of the resistance R, given by (4-137)

Median line {

Median line {

if/m

(
(JL) (a)

= 0)

(0-) (b)

FIGURE 5-5. Dc magnetic and electric circuit analogs. Leakage flux is neglected in the magnetic circuit. (a) Magnetic circuit. Magnetic flux is generated by the source ni. (b) Electric circuit. Current flux is generated by the source V.


265

291

for the electric circuit of Figure 5-5 (b). Its reciprocal (analogous to conductance) is called the permeance of the magnetic circuit. If the magnetic core has a constant cross-section A and a constant permeability j1, the reluctance (5-18) reduces to.o/I = t/j1A, whence (5-17) yields the special result for the core flux

./, __ nI __ ~ 'I'm

~

t

Single-mesh; constant A, j1

(5-19)

This result applies to the magnetic circuit of Figure 5-4(b) neglecting leakage flux and assuming a reasonably thin core. More general magnetic circuits might consist of series arrangements of magnetic materials as in Figure 5-6. A narrow air gap (oflength t g ) can also be included, of interest in the design of relays and in the linearization of iron core inductors, or a gap might be a mechanical necessity as in a motor or generator. For the series system of Figure 5-6(a), applying the reluctance integral (5-18) to the successive portions tb t 2 , t3 and t 9 over which the permeabilities and cross sections are constants, obtains (5-20a) analogous to the resistance of the series electric circuit shown. The field-fringing effects near the edges of a small gap are neglected in Ihe air gap reluctance term ~g. Substituting the total, series reluctance ~ given by (5-20a) into 7) obtains (5-20b)

t\a or 0 th,

t

1se 0

eel i-

v

~d

~ -R,+R.+R3+Rg

led rren

V~R2

urm

ar

Ra

-1 r-

(

fg (air gap) ,f,

nf

,/'f'm=

9i'J+i!l2

Ai

rna thin tot:

-- t2

5-5: (a)

(b)

FIGURE 5-6. Examples of series magnetic circuits and their electric circuit analogs. (a) A series magnetic circuit and its electric circuit analog. (b) A rectangnlar configuration of high permeability materials.

Yo

266


which is analogous to the Kirchhoff voltage expression R1i + R2i + R3i + Rgi = Vfor the analogous, four-resistor electric circuit of Figure 5-6(a). Thus, the similarity of the "voltage drop" terms Ri in the latter to the analogous terms of (5-20b) suggests that the source-term nI should be called an "mmf rise," and the fJtt/lm terms be viewed as "mmf drops" (also called "nI-drops" or "ampere-turn drops") in this characteristic magnetic-circuit expression. Extending (5-20b) to the general case of any number n of magnetic reluctances connected in a series magnetic circuit, you have (5-20c) The series magnetic circuit result (5-20b) can alternatively be expressed in terms of the Hav fields appearing within the series reluctance elements. Applying Ampere's law to the series magnetic circuit of Figure 5-6(a), for example, yields (5-20d) which has the general form n

L

k=l

Hav,kt k

= nI

(5-20e)

for the n-element series magnetic circuit. A comparison of (5-20e) with (5-20c) shows that these governing relations for magnetic circuits are identical term-for-term; that is, the "mmf drop" associated with any reluctance element of a series magnetic circuit can be expressed either as the product fJtt/lm or as Havt. The identity (5-20f) is evident from the definitions of the quantities.

EXAMPLE 5·3. A toroidal iron core of square cross section, with a 2-mm air gap and wound with 100 turns, has the dimensions shown. Assume the iron has the constant p. = 1000p.o. Find (a) the reluctances of the iron path and the air gap and (b) the total flux in the circuit if I = 100 mA. (c) Find Bav and Hav in the iron core and in the air gap. (el) Show, irom an integration ofR' dt about the median path, that Ampere's law (5-16) is satisfied. (a) The reluctance of the iron path, having a median length tl ~ 2n(0.05) and cross-sectional area A 1 = 4 x 10 - 4 In 2 , is

tl

:J£l~--= - IllAl

0.314-0.002 6 4=0.621 x 10 H 103(4n x 10 7)4 x 10

The air gap reluctance, assuming no fringing, becomes

:J£ = ~ = 9

p.oAl

0.002 4n x 10-

1

= 0.314 m


267

291

--l2cm~ EXAMPLE 5-3

(b) The magnetic flux is given by (5-17), that is, the magnetomotive force nl of the coil divided by the reluctance of the series circuit

10 2 (0.1 ) 4.6 x 106

2.18

X

10- 6 Wb

With the air gap absent, V! mis limited only by the reluctance 9f I of the iron path, becoming V!m = 15.97 X 10- 6 Wb. (c) Since only the total magnetic flux in the iron and air-gap cross section is available, no detailed p-dependence of the corresponding o/-directed Band H fields is obtainable; only average values can be found. With the same V!m in both the iron core and air-gap cross section, the same Bau is expected in each, becoming

V!m

Bau = Al

=

, 2.18 X 10- 6 = 5.45 ml 4 x

th use ( e ce

It

The continuity of this Bau at the iron-air interface satisfies the boundary condition Bnt = B.2 of (3-50), while producing an abrupt discontinuity in the average H fields there. Thus, in the iron,

5.45

X

4.34 A/m

unr

les (

while in the air gap

1m

Bav Havy = = 4340 A/m , lio

d.til tot

just Ii, = 1000 times as large as HaD,Fe' (d) With the substitution of Hau,Fe and Hau.g into the line integral (5-16) Hav,Fetl

:~d

lied Irn:1

10- 3

--,;----;;- =

~l H· dt =

e f1a lOr (

+ Hau,gty

4.34(0.312)

which agrees with the right side of (5-16), i = nl

+ 4340(0.002)

=

10.0 A

10 A. Yo

268

STATIC AND QUASI-STATIC MAGNETIC .FIELDS

In (5-20a) the air-gap permeability ~o ordinarily is much smaller than ~1' ~2' and of the magnetic materials in a bonafide magnetic circuit. This means that for even a small air gap, the gap reluctance term can often be orders of magnitude larger than the reluctance of the rest of the circuit. A good approximation in such cases is that the core flux is determined essentially by the air-gap reluctance only; that is, ~ ~ ~g. For practical reasons concerned with fabrication problems, magnetic cores of rectangular shape, like that of Figure 5-6(b), are in common use in devices such as relays, inductors, and transformers. The approximations of the magnetic circuit concept become greater in such configurations because of the difficulty in assigning correct median lengths to the various legs of the rectangle, particularly if the cross sections are large compared with the overall core dimensions. An extension of the theory of magnetic circuits to systems having more than one magnetic path is possible again through the use of the electric circuit analogy, as illustrated in Figure 5-7. Because the fluxes divide among the branches of the magnetic circuit in just the way the currents do in a dc electric circuit, it is seen that writing Ampere's law around the two magnetic meshes of Figure 5-7(a), for example, yields the following equations

~3

nI

o=

~lt/tml

+ .oJl Zt/tm2

~3(t/tml - t/tm2)

+ .iJl 2t/tm2

(5-21 )

in which ~1' ~l' ~3 are found from the mcdian paths t l , t l , t3 in Figure 5-7(a). For linear core materials, (5-21) can be solved simultaneously to find the magnetic fluxes t/tml and t/tm2' The accuracy of the analysis of magnetic circuits through reluctance methods is affected not only by the leakage flux problem and the assignment of median paths, but also by the nonlinear B-H curves of ferromagnetic materials. Nonlinearity, as exhibited in Figure 3-13, requires that the permeability be expressed as afunction of the H field in the core, or ~(H). One cannot find H, on the other hand, until a value of ~ has been assigned to the circuit (or values of ~ to its branches). Iterative processes are frequently successful in such problems. Thus, if a trial value of magnetic flux is assumed for the

(Choice of median paths corresponding to Jl'h Jl'2, ~3)

(a)

V-=-

(b)

FIGURE 5-7. Two-mesh magnetic circuits and their electric circuit analogs. (a) A !w{Hnesh magnetic circuit and its electric circuit analog. (b) A variation of (a).

5-4 VECTOR MAGNETIC POTENTIAL

269

:91

circuit, the value of Il may be obtained; this result can then be used to determine a new value of magnetic flux. This process is repeated until the desired accuracy of the answer is obtained.

5·4 VECTOR MAGNETIC POTENTIAL Section 4-5 showed how the irrotational property (4-2) of the static E field permits expressing E as the gradient of some auxiliary scalar potential function
B=VxA

(5-22)

in view of the vector identity (19) in Table 2-2, V' (V x A) O. The function A defined by (5-22) is called the veclor magnetic potential field. The vector magnetic potential A is related to steady current density sources J responsible for the field B as follows. In a static magnetic field problem, the relation (5-2), V x H J, is satisfied by the H field. It is also written

VxB

IlJ

(5-23)

fla r (

(b; th

for a region in which Il is constant; substituting (5-22) for B into (5-23) yields V

x

(V X A)

= IlJ

(5-24)

e(

ce This vector differential equation is simplified by use of the vector identity (2-88a) I

V x (V x A)

2

V(V' A) - V A

(5-25)

To assure the uniqueness of the potential A, both its curl and divergence must be specified. The curl is given by (5-22), and div A appearing in (5-25) has not yet been assigned. Assuming V . A = 0 does not conflict with any prior assumption, permitting V x (V X A) in (5-24) to be replaced with -V2A to yield

~d

,er nn al

tit

(5-26)

This result, sometimes called the vector Poisson equation because of its similarity to (4-67), is an inhomogeneous, linear diflerential equation relating A to its sources J, with Il a constant in the region in question. The virtue of (5-26) lies in the availability of several methods for finding its solutions, among which are the method of separation of variables, and an integration approach described in the next section.

-5

270


5·5 AN INTEGRAL SOLUTION FOR A IN FREE SPACE: BIOT-sAVART LAW

An integral solution of (5-26) can be inferred as follows, assuming an unbounded region of free space (It = Ito). In cartesian coordinates, the left side of (5-26) is written, using (2-83),

whence (5-26) becomes the three scalar differential equations ( 5-27) Each of the latter is analogous to the Poisson equation (4-67)

Pv E

the integral solution of which, in unbounded free space (E charge of density Pv, has been shown to be (4-35a)

[4-6 7 1 EO) containing a static

[4-35a]

There/ore, the analogous solutions of the three scalar differential equations (5-27) in free space must he .

Adding these three integrals vectorially yields the desired integral solution of (5-26)

(5-28a)

The meaning of R in (5-28a) is the same as in (4-35a); it denotes the distance from the source point P' to the field point P at which A is to be found. Once the A has been obtained by means of (5-28a), the corresponding B field is ohtained from the curl of A, using (5-22). The geometry of a system with current sources of density J producing the vector magnetic potential A given by (5-28a) is shown in Figure 5-8. Note that the integrand

5-5 AN INTEGRAL SOLUTION FOR A IN FREE SPACE: BIOT-SAVART LAW

291

dA =!1Q~d,:,'

dA

47TR

Field

po~P I

271

=~J$ds'

./

47TR

pr \

\

\R \

\ \ \

\

~

~

FIGURE 5-8. Three types of steady current distributions in space. (a) Volume distribution of elements Jdv'. (b) Surface distribution of elements Jsds'. (e) Line distribution of elements

J dv'

-t

I dt".

or (5-28a) is a differential dA given by

tI-om which it is seen that the current source J dv' at the typical source point P' (U'l' u~, u~) produces, at any fixed field point P, a vector contribution dA parallel to the element J dv'. Moreover, the magnitude of its influence at P is inversely proportional to the distance R. These relationships are depicted in Figure 5-8(a). In case of either a surface current (a current sheet) or a line current,! as noted in Figures 5-8(b) and (el, (5-28a) reduces to the following surface and line integral

A(u

u 1,

2,

u) 3

-

r floJs( U'l' u~, 113) ds' 4nR

Js'

r flol dt' Jt

4nR

(5-28b)

Hal r 0'

(b) tht

eo cel rg) b) d

(5-28c)

I n practice, steady surface and line currents are approximated by physical currents flowing in thin sheet conductors or thin wires. The vector magnetic potential results (5-28a,b,c) deserve comparison with the analogous results (4-35a,b,c) for the scalar electric potential fields of static charge distributions. a: Ita

EXAMPLE 5·4. Find the vector magnetic potential in the plane bisecting a straight piece ofthin wire of tinite length 2L in free space, assuming a direct current J as in Figure 5-9. Find B from A. lThe line element of current is shown in Fignre 5-8( c), enlarged into the volume element J dll' atJ dC' tis' (] ds') (at dt') , which becomes just I dt' if the product] ds' denotes the finite current I in the line source.

)0

272

STATIC AND QUASi-STATIC MAGNETIC FIELDS : (z') z' =L

I

~Idt' =:

azldz'

Source point P'(O.O.z')

(p)

I z'=: -L,

, I

FIGURE 5-9. Geometry of a thin wire carrying a steady current I.

The fixed field point is on the plane z = 0 at pep, 0, 0). The typical current source element at 0, z') is I dt' = azI dz', and R from P' to Pis R = ~ p2 + (Z')2, putting the line integral (5-28c) in the form

reo,

A ( p0 , ,0 )-

i

L

z'=-L

_. floaJdz' Tz 4nvp

The unit vector a z has the same direction at all

+

(z')

2

r, yielding at P

(5-29) One finds B at P using (5-22) in circular cylindrical coordinates ap p

B=VxA= 8 8p

0

a",

az

p

0

0

0

Az

flo I

-a

=a - -

'" 8p

L

(5-30)

'" 2np

For p «L, (5-30) simplifies to flo I 2np

B=--a

1>

(5-31)

a result very nearly correct when near a finite-length wire, or correct at any p distance for an infinitely long wire. In Example 1-13 (5-31) agrees with (1-64).

EXAMPLE 5·5. Find the A and B fields of a thin wire loop of radius a and carrying a steady current I, as in Figure 5-10(a). Make approximations to provide valid answers at large distances from the loop (assume a « r).


273 291

I dtz'

I (y)

dA= dA l+ dA z

Field point P

dA

R

t Source point

P' (y)

(b)

(a)

FIGURE 5-10. Circular loop, finding the static magnetic field at P. use of symmetry to obtain fields at P.

the spherical coordinate geometry adopted for Circular loop carrying a current I. (b) Making

Without detracting from the generality, the field point P can be located directly above the y-axis as shown in Figure 5-10(b). The A field at P is given by (5-28c), in which f dt' = aq,fad(f/. The variable direction of aq, in the integrand is handled by pairing the eflcets of the current clements 1 dt l and f dt~ at the symmetrical locations about the y-axis in Figure 5-IO(b). From the geometry, 1dt'l = aq,fa d4>'

ax sin

f dt~ = (- ax sin

4>' -

a y cos

flat r 01

4>' + a y cos 4>') fa d4>'

(b) ,

4>') 1a d4>'

(1)

the' e oj

to provide a cancellation of the y components of the potential contributions of the pair of clements at P, leaving a net dA at P that is - x directed. Thus (5-28c) becomes (2) From the law of cosines applied to the triangle POP' in the figure, R2 = a2 + r2 2ar cos (l = a2 + (2 2ar sin 0 sin (//. If r »a, one can approximate, making use of the binomial theorem,2 R

~

r [ 1 - 2 ;a sin 0 sin

4>' J1

1

2~ r [ I -

;a sin 0 sin

4>' + ...

J

1

~

1

a.

.

--( + -sm esm 4> r2

2From the binomial theorem one may see that, in the expansion of (I

rg)

b) d en' Ill~

an OJ

la) :ll!

a ta

The reciprocal, for small a, is similarly approximated

R

eel,

iO ,

± b)", ifb« 1 then (I ± b)";?; 1 ± nh.

274


This puts (2) into the form

2/-lola .1:"/2 4n q,'=

A~--

q, -

n/2

[~r + a sin esin ¢'] sin ¢' d¢'

(3)

The integral of the (sin ¢')jr term is zero, so integrating the second term yields the answer

(5-32) Taking B = V

X

A in spherical coordinates therefore yields

(5-33) if a « r. The duality between the B field (5-33) of a small current-carrying loop and the electric field (4-44) of a small electrostatic dipole is noted. This gives rise to the name magnetic dipole, when reference is made to the field of a small loop earrying a steady current.

Taking the curl of (5-28a) leads to an alternative free-space integral expression for the B field of a static current distribution as follows.

B = V

X

A = V

X

r JicJ(U'l, u~, U3) dv'

Jv

4nR

(5-34)

One may note that the differentiations imposed by the V operator in this expression are with respect to the field point variables (u 1, Uz, U3), whereas the integration is performed within V with respect to the source point variables (U'l' . Thus R is a function of both the source point and field point variables, since R = so (5-34) becomes B

One can write V X

[J/R]

=

r

Jio V x

JV4n

[!J R

dv'

from the vector identity (17) in Table 2-2

The last term is zero because J is a function of only the source point variables; flll'thermore, the factor V (1/R) can be expressed

if aR is a unit vector pointing from P' to P. Thus


275

n

Field point P

R

(z)

iI

Source point P'

I I

------0----(x) (y) FIGURE 5-11. A volume distribution ofcurreuts, showing the dB contribution ofa typical current element J dv' from the Biot-Savart law.

obtaining (5-35a) This integral for B, expressed directly in terms of the static current distribution J in free space, is known as the Biot-Savart law. It provides an alternative approach for obtaining the magnetic fields of static current distributions in free space. Figure 5-11 shows the geometry relative to (5-35a), depicting a system of steady currents with densities J, and a typical field point P at which B is found by means of (5-35a). The differential contribution dB is given by the integrand of (5-35a)

hi o

b)

tht

~o

~el

meaning that dB contributed at P by J dv' is mutually perpendicular to both the current element vector J and the unit vector aR, as depicted in Figure 5-11. Specializations of the Biot-Savart law to surface or to line currents are readily obtained. Thus, if the volume current of Figure 5-11 is contracted to a thin filament of negligible cross section, putting J dv' --+ I dt' into (5-35a) obtains

rg~

ly 3; en in ar

:c (5-35b)

la

11l

)t;

EXAMPLE 5·6. Usc the Biot-Savart law to find the B field of the thin wire of length 2L and carrying a steady current, as given in Example 5-4. The form (5-35b) of the law is applicable. In the circular cylindrical system as shown in Figure 5-12, Idt' = azIdz', while a R is resolved into components as follows: azz'). With R = .}p2 + (zy, (5-35b) becomes a R = a p sin Ct: - a z cos tt. = Ir 1 (app

5(

'0

276


:(z')

£'

TIdf' R

Source point

P'

z'

z'

o ------------:.-:..----P

o

Field point

a -

",,~

I

--p----~~

---------

P

. . . --.,,'\

"',

P

-£ ,

-£ FIGURE 5-12. Geometry of'the straight wire of length 2L, using the Biot-Savart law to find B.

and integrating obtains

(5-36) Close to a wire of finite length (p« L), or for an infinitely long wire, (5-36) becomes B

(5-37)

results that agree with those of Example 5-4.

5-6 QUASI-STATIC ELECTROMAGNETIC FIELDS In previous sections of this chapter, only purely stalic magnetic fields, associated with steady current distributions, were considered. Such fields are required to satisfy the Maxwell integral laws (5-4) and (5-5) for all closed surfaGes or lines in the regions in question, or equivalently the differential laws (5-1) and (5-2) for all points in the regions. The boundary conditions, also to be satisfied at all interfaces, are (5-8) and (5-9). If the current sources are generalized to the time-varying case, their fields are then no longer purely magnetic but become electromagnetic, governed by all four Maxwell equations, (3-24), (3-48), (3-59), and (3-77), with the boundary conditions embracing the relations (3-42), (3-50), (3-70), and (3-79). For current sources that vary slowly in time, however, approximate methods, termed quasi-static, may sometimes be employed to advantage. An instance has already been given in Example 1-16. Quasi-static field solutions can be termed first-order solutions, because they do not satisfy Maxwell's equations exactly except in the zero frequency limit. Another view, bctter appreciated in Chapter 11 on radiation and antennas, is that the dimcnsions of the current-carrying system must be small compared with the wavelength AO in free space 3 if the system is to be amenable to a quasi-static method of attack. This 3Suppose one assumes that a device such as a coil or capacitor should not exceed 0.01,1,0 in its maximum dimension, adopted as a criterion for sufficient smallness to enable employing quasi-static analysis in the description of its fields. Operation of the device at a frequency of 100 MHz implies that its size should then not exceed 0.03 ill (3 em), since ,1,0 = 3 m at this frequency.

5-7 OPEN-CIRCUIT INDUCED VOLTAGE

277

91

constraint is equivalent to ignoring the finite velocity of propagation of the field from the sources to the nearby field points of interest, amounting to ignoring field radiation effects. A more sophisticated approach to quasi-static field solutions, using an appropriate power series representation of the fields, is described elsewhere. 4 The quasistatic approach to field problems is sometimes the only method that provides ready solutions to an otherwise difficult boundary-value problem. It has applications in the discussion of the voltages induced in stationary or moving coils immersed in magnetic fields that mayor may not be varying in time, as well as in the development of circuit theory, particularly regarding concepts of self- and mu tual inductance, to be discussed in subsequent sections.

EXAMPLE 5·7. Demonstrate that the approximate quasi-static fields of the long solenoid of Example 1-18 obey the Maxwell's equations (3-59) and (3-77) exactly only in the static field limit (}J -,> O. The quasi-static Band E fields inside the solenoid were found to be

wpBo E(p, t) = -a.,--cos wt

B(l) = azB o sin wt

2

Testing whether these fields satisfy (3-77), V ~ p

VxE=

a ap

0

X

a.,

az

0

0

pE.,

0

E =

(1)

-aB/at, one finds

p =

-azwB o cos wt

(2)

'!at oj

revealing that Band E of (I) do indeed satisfy (3-77). This is to be expected, because E was originally obtained using the integral form of (3-77), but Maxwell's equation (3-59), reducing to V x H = aD/at within the solenoid, is not satisfied by (I). This is evident on obtaining V X H = V x (B/Ilo) = 0, since B of (I) is independent of position inside the solenoid; whereas aD/at becomes

b), the ~

oj

:ell rg) b) j "

en'

a vanishing result only if w -> O. Thus (3-59) is satisfied only in the static field limit, though an approximate equality prevails if w is sufficiently small.

in~

an o la'

5·7 OPEN·CIRCUIT INDUCED VOLTAGE The transformer makes use of Faraday's law (3-77) to couple electromagnetic energy from one electric circuit to another through the time-varying magnetic field. Typical physical arrangements are diagramed in Figure 5-13. In (a) is shown the configuration of Figure 1-25(b): a primary coil consisting of a long solenoid, encircled by a secondary coil. Single-turn secondary coils are shown for simplicity; many turns are commonly

53

4See R. M. Fano, L. J. Chu, and R. B. Adler. Electroma.gnetic Fields, Energy and Forces. New York: Wiley, 1960, p. 221 If.

or

278

STATIC AND QUASI-STATIC MAGNETIC FIELDS B flux

B flux

B flux

Primary Secondary

(!Lo)

8;" Bj :' ':3 :3

~C-~;~

: ttttttt: I I ; I ! 11

1111111

(c)

(b)

(a)

FIGURE 5-13. Typical transformer configurations. (al Primary coil a long solenoid. (b) Short solenoid primary, secondary laterally displaced. (c) Gonfiguration of (b) with ferromagnetic core.

used to enhance the induced voltage V(t). A ferromagnetic or a ferrite core can also be used in a magnetic circuit arrangement as in f'igure 5-l3(c), to augmentsubstantially the magnetic flux intercepted by the secondary coil. The voltage V(t) developed at an open-circuit gap in the secondary coil 5 of a transformer is shown to be V(t)

dl/l m

--V dt

(5-38)

in which l/I m denotes the magnetic flux intercepted by the surface S bounded by the secondary winding. Suppose the coil shown in Figure 5-14(a) carries a time-varying current 1(/). In the surrounding region, the accompanying magnetic field B(Ut, U2, U3, t) induces an azimuthally directed, time-varying E field as described in Example 1-18 and depicted in the cross-sectional view of Figure 5-14(b). The secondary coil is shaped such that Hux orB passes through the surface S bounded by the coiL This assures the alignment of the conductor with the induced E field such that the free electrons in the conductor are urged by the E field forces to move along the conductor as noted in Figure 5-14(c). Thus an excess of electron charge accumulates at one end of the wire, while a dearth of electrons (a positive charge) is established at the other, producing about the gap another electric field denoted by Eo. Then the total E field about the system becomes E = El + Eo. Faraday's law (3-78) written about the closed path including the secondary coil and its gap thus becomes

J. E. dt == Jrconductor Yt

(E t

+ Eo)

. dt

+ Jrgap

(EI

+ Eo)

. dt =

dl/l m dt

(5-39)

The relationship between the total electric field El + Eo along the conductor and the current density J within it is given by (3-7), bccoming J = a(El + Eo) along the coil. SIt should be borne in mind that the designation secondary coil is arhitrary; either coil ora transformer may be designated as the primary coil, with the other coil taking the name secondary.

5-7 OPEN-CIRCUIT INDUCED VOLTAGE

279

11

B

",

I

~

~~~nductor ~ with gap ;I Eo field of . / displaced charges (b)

(a)

(c)

HGURE 5-14. Development of the open-circuit voltage V(t) ora transfOlmer. Transformer configuration nsed to prove (5-38). (b) Sectional view of E, induced by timevarying B of (a). (c) Showing charges displaced by El to produce Eo, canceling total E along wire.

Assuming the coil a jJe~lect conductor, E + Eo must tend toward zero if J is kept to a necessary finite value, making El + Eo = 0 along the conductor portion of the closed path t. This simplifies (5-39) to obtain

A: E' dt = r (El + Eo) . dt ~ Jgap

dt/lm dt

(5-40)

ne

tt

implying that the total E' dt generated by the time-varying magnetic flux t/lm embraced by t appears wholly at the gap. The closed-line integral of (5-40) is sometimes called the induced electromotive force (emf) abou t t, and is denoted by the voltage symbol V(t). Then (5-40) is written

V(t)

==

V ~ E· dt = --dl

dlPm

t

at of I),

(5-41)

of ell ~y

)y a nt :lg re or

ay Thus the induced emf: or equivalently the gap voltage V(l), depends only on the time rate of change of magnetic flux through the surface S bounded by the closed line t described by the wire. The explicit values ofE 1 and Eo are not required to be known on/he path. Furthermore, the wire path t may be distorted, if desired, into any arbitrary shape; for example, a square or a helix, in which case (5-4,1) is still valid. A h~lix shaped (many turn) conductor is useful for increasing the induced voltage across the gap, and it is commonly used in practical transfi)fmer and inductor designs. If in the foregoing discussions a finitely conducting wire had been assumed, the result (5-41) would have been modified only trivially if the conductivity (J were sufficiently large (of the order of 10 7 U/m, as for most good conductors).

llg a~

:a'

0: 3'

280

STATIC AND QUASI.STATIC MAGNETIC FIELDS

Long solenoid

(J turn/m)

FIGURE 5-15. Showing open-circuit coils ( and (' and the induced voltages Vet) obtained from the time-varying t/lm.

EXAMPLE 5·8. A thin wire is bent into a cirele of radius b and placed with its axis concentric with that of the solenoid in Example 1-16. Find Vet) induced across a small gap left in the conductor, for the two cases of Figure 5-15: (al b > a and (b) b < a. Include the polarity of Vet) in the answers. (a) If b > a, (5-41) combined with (1-67) yields, for the solenoid current 10 sin

Vet) =

-~ r B . ds = dt Js

d dt

r[

Js a z

llon10 sin d

wt,

wt] • (azds)

b>a

( 5-42)

since Ssds = na 2 • The polarity of Vet) is found by use of a right-hand-rule interpretation of the induced voltage law (5-41). Assuming, at a given t, that t/Jm through t is increasing in the positive z sense in Figure 5-15, aligning the thumb of the right hand in that direction points the fingers toward the terminal P2 at the gap, which at that moment is the positive terminal. The presence of the negative sign in the answer (5-42), however, requires that the true polarity of Vet) becomes the opposite of the indicated polarity in Figure 5-15, at that instant. (b) If b < a, the surface S' bounded by the wire t' is smaller than the solenoid cross section; (5-41) then becomes

Vet) =

-~ r B· ds = dt Js'

b
(5-43)

5-8 MOTIONAL ELECTROMOTIVE FORCE AND VOLTAGE The Faraday law (3-78) provides the connection between the time-varying magnetic flux t/lm passing through a surface S and its induced E field. It states that the closed-line integral ofE over the closed path t bounding S exactly equals the time rate of change of the magnetic flux through S, or

~E' dt =

dt/lm dt

[3-78]

5-8 MOTIONAL ELECTROMOTIVE FORCE AND VOLTAGE

BI~

281

1

dt

,~~ (b)

(a)

FIGURE 5-16. A closed line t moving in spacc with a velocity'/} in the presence ofa time-varying B field. (a) A moving contour t in the presence ofa time-varying B field. (b) The contour t shown at the time t and t + dt.

In (3-78), the closed-line integral ofE' dt about t, termed the circulation ofE about the closed path, has also come to be known as the induced electromotive force, or just emf, produced by the time-varying magnetic field through S6. In the present section, the emf induced by a motion of the path t relative to the frame of reference of the magnetic field is discussed. Faraday's law (3-78) can conveniently be resolved into two terms on its right side, accounting for the induced emf's about t produced (a) by the time variation of the B field over the surface bounded by t and (b) by the relative motion of the closed path t with respect to the coordinate frame of reference of the B field, as shown in Figure 5-16(a). This form of the law, useful in the analysis of moving-coil devices such as generators, motors, and d' Arsonval type instruments, is developed in the following. If, as in the previous section, the closed path t in Faraday's law (3-78) is chosen to coincide with a conductive wire path immersed in a steady magnetic field, but the conductor is now moving with the velocity v (not necessarily constant about t), then the free charges q available within the conductor would be subjected to the Lorentz forces F B = qv X B given by (1-52a) in Chapter 1. The quantity v X B, having from (1-52a) the units of force per charge, is evidently a motional electricjield, Em, defined by

at of

I) , he of ell gy

by a nt ng

vxB The free conduction charges urged in the conductor by this electric field v X B will establish the voltage difference V(t) at a gap left in the conductive path t and produce current flow on closing the circuit; it is given by

V(t)

=

1, Em . df = 1, (v X B) . df

1

Je

(5-44a)

If the moving conductor were immersed in a time-varying magnetic field, then the added emf due to the time-varying magnetic field through the surface S bounded by the circuit t would be required. This emf, accountable only to the time-varying B(t)

f

6Elcctromotive force (emf), E· dt, obviously does not have the units of force, but rather volts, or joules per coulomb (work per unit charge about t). For this reason, the word eleclrornotance has been suggested as an improved term for emf. Becanse it is widely used, however, emf is the term employed in this text.

,rt' OJ

a) n~

a:

ta 10

282


field, has been given by (5-41)

dljlm dt

-~ dt

• ds r B· ds _Jsr aB at

Js

Adding the latter to (5-44a) thus obtains the desired total emf V(t) =

1"

E' de = -

r aB. ds + 1" x B) . de (v

Js at

(5-44b)

accounting {or two eontributions to the induced emf as follows

"" for the induced emf about 1. The first term of the right side of (5-44b) accounts t provided by the time rate of change of the B field integrated over the surface S bounded by t. 2. The second term yields the additional induced emf arising from the motion of the path t relative to the coordinate fi-ame of reference in which B is specified. If t is

stationa~y

in space (v

0), (5-44b) reduces to

rtJ. E . dt = _ Jsr aBat • ds V

t

stationary

(5-44c)

Suppose that a wire loop t is immersed in a steady magnetic field, but t is moving in space. Then (5-44b) becomes

x B)' dtV

Static B

(5-44d)

The correct polarities of the contributions to the voltage Vet) appearing at the gap of a wire contour can be grasped from Figure 5-17, involving two cases

A. In (a) of Figure 5-17, the polarity ofthe induced gap voltage Vet) obtained from the surface integral (5-44c) of the time-varying magnetic field is desired. The positively directed ds is chosen on that side of the open surface from which the positive B field emerges. On aligning the thumb of the right hand with the positive ds (or B) sense, the fingers will point toward the positive terminal of the gap voltage V(t).

CASE

B. In Figure 5-17 (b), the polarity of the gap voltage obtained from the motionalline integral (5-44d) is desired. For illustrative purposes, the wire contour t in that figure is assumed to be shrinking in size while immersed in a steady B-field (directed toward the observer as shown). If the line integration sense is chosen for (5-44d) in the same sense as v x B, the polarity is designated as positive at that gap terminal toward which the integration is being taken (this is the direction in which positive charges will be urged by Em = V X B).

CASE

283


-:1 r.

V(t) = -

B_~ .'

-

Is ~~ ods

---\

B

~--,...

~ (8)

l.(t

~--

ds

+ At)

aB (in time at)

)

--;----

-------r--~

Line integration sense

I I

Line integration sense

I

J vlCB

(b)

(a)

FIGURE 5-17. Conventions relative to Faraday's law. (a) Polarity of V(t) induced by time-varying B, with t stationary. (b) A shrinking wirc contonr, showing scnse of induced field" X B, with B stalic.

A rigorous calculus proof of the motional-emf result (5-44b) can also be obtained directly from the Faraday law (3-78)

r£ E· dt Jt

d dt

r B· ds

Js

(3-78)

fla r (

With the circuit t in motion, both Band ds in the right-hand integral of (3-78) are, in general, functions of time. The time derivative of that integral can be found from a three-dimensional vector extension of the rule of Leibnitz. Assuming the closed path t, as well as the surface S which it bounds, to be moving with velocities v thereon, the time derivative of (3-78) can be expanded into the following general form 7

r

d B. ds dt Js

= ~ (v

X

B) • dt

+

Is .ds + Is (V . B)v . ds 'oB

in which the last integral disappears, in view of the Maxwell relation (3-48), V • B = O. Substituting the resulting expression into (3-78) thus obtains the desired motional form (5-44b) of the Faraday law.

(b tb

e( ce :r~

t

l

~d

rei

rll a :s m:

ti EXAMPLE 5-9. A rigid, rectangular conducting loop with the dimensions a and b is located between the poles of a permanent magnet as shown. Let B = azB o , constant as shown over the left portion of the loop, and assume the loop is pulled to the right at a constant velocity v a x1-'o. Find (a) the emf induced around the loop, (h) the direction of the current caused to flow in the loop, (c) the force on the wire resulting from the current flow, and (d) the magnitude and polarity of the open-circuit voltage V(t) appearing at a gap in the wire at P shown.

1,

to

y,

7See S. W. Maley, "Differentiation of Line, Surface and Volume Integrals," Scientific Report 60, Electromagnetics Laboratory, University of Colorado at Bonlder, March, 1981.

284


(y) I I I

Region of a constant magnetic field /--i--.t' B = azBo l ; / 0 0 ir~0~"",,===='i1 (y)

I

i

/

{0

0

0

0

10

0

0

....

\0

0

0

0

0

0

'\.

I

P1lb

Positive integration sense of fE-de

~

0 \

J

01

P

~==;z!/===i:I=i=:==r:!J- --- - 7;)

' ..... --.a...-El/

Gap

Conductor

(a)

(b)

EXAMPLE 5-9. (a) Moving wire loop in a constant magnetic fielfl'. (b) Geometry showing assumed line-integration sense.

(a) The sense of the line integration is assumed counterclockwise looking from the front, as in (b) of the accompanying figure. The emf induced about the loop is found by use of (5-44d),

A:.

;Yt(t)

On P 1 (v

X

D) . dt

[(axvol

E' dt X

A:. jj(t)

(v x B) . dt

(azB o)]' aydy = -voBody, obtaining

A:. E· dt = Soy~1J (-'!JoB o ) dy = voBob

Ye(t)

(I)

(b) The positive sign of (1) denotes that the induced emf about t is in the same sense (counterclockwise) as the direction of integration. The result (1) therefore causes a current to flow in the same direction. (e) The force acting on the wire carrying the current I immersed in the field D is obtainable from (1-52a), FB = q(v x D). The force on a differential charge dq = Pv dv being dF B = dq(v X D), and with pvv of (1-50a) the current density J in the wire, one obtains dF B =

The product becomes

J dv

J

x Ddv

(5-45a)

defines a volume current element I dt in a thin wire, so (5-45a)

dF B

Idt x B

(5-45b)

Integrating (5-45 b) over the length OP 1 of the wire obtains the total force (2)

in which the integration in the direction of the current produces the proper vector sense of the force. F B is a force to the left in the figure, opposing the motion of the wire. (d) A small gap at P in the wire renders the loop open-circuited, reducing I to zero and yielding Vet) = E . dt = voBob across the gap. The polarity is determined by the direction of v X D, directed around the loop toward the positive terminal of the gap as in (b) of the figure.

ft


285

1

EXAMPLE 5-10. A small open wire loop of radius a in air is located in the y-z plane as shown in (a), immersed in a plane wave composed of the fields E:

t) = - E~ sin (rot - Poz)

(I)

E+

H; (z, t) =

m

sin (rot - (1oz)

(2)

tlo

Assume a « Ao. Find V(t) induccd at the loop gap. The gap voltage is obtained from (5-44c) because the loop is stationary. Using B = flo" and (2) obtains

aB = -a [axfloE~ - - - sin (rot at at tlo

-

Paz)

J

floroE~

= ax - - - cos (rot - PoZ)

tlo

With a « Ao, the coil occupies essentially the position past, so (5-Hc) yields fl roE+ fs [ ax -"~ cos (rot -

=-

V(t)

PoZ)

1

tlo

0 on the plane wave moving

Z =

• ax ds

-

2

roE+ na = ___ m_ _ cos rot

(3)

=0

IfE~ = 100 flVjm,f = I MHz, and a = I m (satisfying the criterion a« A), the induced voltage becomes V(t) = 396 cos rot flV. The five-turn coil shown in (h) provides a gap voltage five times that of (a), in view of the structure of the surface S bounded by the

B or H hnes\

ds = 8 xds

'¥

Gap voltage

V (t)

~~t'-"" ~ ~F'I-t-I. S)I

Loop t

I~

~

/Integratlon sense of

ftE.df

(z)

(a)

B lines

t

\I~'\~~ ~-~\~_-I -J--!::::-~ ±l-¢il~ /I

--B lines

j

;t-:----

Ga~ (b)

'\

\

"0

~~--l ____ I

o-

I

/

I

s

\~-~I

\I

~S X-~Ga~

0

19 lS

al

}) \)

(c)

EXAMPLE 5-10. (a) Loop immersed in a field. (b) Five-turn coil immersed in a time-varying field. (e) Effect of spreading the turns; fewer B lines intercept S.

·k

286


wire contour t, while from (e), the effect of opening the structure is to reduce cepting 51, reducing Vet) accordingly.

l/Im

inter-

5·9 INDUCED emf FROM TIME-VARYING VECTOR MAGNETIC POTENTIAL The emf induced about a closed path t linking a time-varying magnetic flux I/Im can also be expressed in terms of the vector magnetic potential A developed in Section 5-4. This is accomplished by use of (5-22), B = V x A, to permit writing Faraday's law (3-78) as follows

rh

:Yt

f

E. dt

f

d (V x A)' ds dt s

B· ds d dt s

_~rh A. dt

:Yt

dtV

(5-46a)

on applying Stokes's theorem (2-56) to obtain the last result. [f t is stationary (motional emf is absent), (5-46a) is written simply

rh

~t

E . dt = -

rh aA . dt ':Yt at

V

Stationary path t

One can see from (5-46a) that the flux through the path

./,

'I'm

=

f B . ds S

=

'rh :Yt A

t

(5-46b)

is expressed two ways

. dt Wb

(5-47)

Thus it follows that a knowledge of the vector magnetic potential A on the closed t determines the magnetic flux I/Im passing through S bounded by t. EXAMPLE 5-11. Find the emf induced about a rectaugular stationary path t in free space, in the plane of two long, parallel wlres carrying the currents 1(t) and -l(t) as shown. Find the emf two ways using the Faraday law expressed (al in terms of the B field and (b) in terms of A. (a) Fronl (1-64), the quasi-static B field exterior to a long, single wire carrying l(t) is B = a",p ol(t)/2np, if p is the normal distance from the wire to the field point. In the present example involving two wires, a cartesian coordinate system is adopted as in (a). At any x on 51 bounded by t, the quasi-static B field due to both wires is the vector sum • B - a

-

J

Po1(t) - - Ilo1(t) -------y [ 2n(x d) 2n(x + d)

The latter into (5-44c) obtains the induced emf about

t (1)

the desired result.

5-9 INDUCED emf FROM TIME-VARYING VECTOR MAGNETIC POTENTlAL

I(z)

i(t)

Closed

r __ L:1Path t

-i (t)

-i (t)

--=b=~{!)dz: b dx

= 1 coswt A

~

cm +.Vet)

I

1(8) I.: ____ ...JI _ _ __ (x)

~

, -d

291

~~E'df

l{X-~P(X.Z) I

287

=4 em

, 2d = 4 I

(a)

(x)

Wire loop

h

d

2cm

em

(b)

EXAMPLE 5-11. (a) Geometry ora parallel-wire system and a rectangnlar dosed path t. Showing polarity of gap voltage V(l), corresponding to sense of dt integrated about loop.

JE .

(b) To find the induced emf using A, note from Example 5-4 for a 2L that /1o I (t)

A= a

~.. -. z

411:

tn

..}L2

wire of length

+ p2 + L

-'-;==c=c.==c---·-

+-L

The latter is improved by noting that A for p « L is desired. The first two terms of the binomial expansion l')r the square root quantities obtains

flat r of

(b) , the e oj

cell rg) b) d ;;

valid for p « L. For parallel wires, A is the vector sum

'en'

2L A = a)'.o!. [tn _ - tn -'!:.~-J = 211: X- d x+d

a_lloI(t~ tn x + d "211:

X -

m~

an

d

; 0'

With A z-directed, the induced emf by use of (5-46b) becomes

~

jt

E • dt =

-~ ~ . dt

= -

\

jt

Ill! a Ita

at

~~ /10 {[t :-~!!.J jb dz + ot 211: x d x=h Jz~o 'It

/1o~tn 211:

_+__c.~___._.~ (h - d)(h

+ a + d)

[t

n

~~J fO dZ} x - d x=h+a JZ=b

50 (2)

which agrees with (1). Note that the integration has been taken clockwise about to conform to dle assumption in (a) of a positive y-directed ds on S.

')3

t, Dri

288


h/2

If the current J(t) = 1 sin wt A flows in the wires, withf= 1000 Hz and d = 2 em, the induced emf (2) becomes

= a= b=

~~ E . dt

- (2n x 10 3 ) cos wt

-17.4 cos wt flY This is also the gap voltage Vet) developed if an open-circuited wire loop replaces with a polarity as in (b).

t,

From (5-46b) for the induced emf about a fixed closed path,

~E'dt

r£ aA. dt

at

~

[5-46b]

one might be inclined to argue, because (5-46b) is true f(Jr all closed paths t, that the electric field can be expressed at any point in the region fi'om equating the integrands; that is,

aA

E=

at

J t is, however, noted that adding an arbitrary function - V to the latter, obtaining E

V

aA

(5-48)

provides an E field that still satisfies (5-46b), in view of the property (2-15), ft(V. Thus (5-48) is in general the correct expression for E in terms of its potential fields A and <1>. The physical meanings of each contribution to E in (5-48) is appreciated on noting, in the time-static limit, that (5-48) reduces to E =

V

a/at

°

(5-49)

Comparison of (5-49) with (4-31) identifies as the scalar potential field established by the free-charges of the system, whether they be volume, surface, or line charges. The potential integral (4-35a) provides this relationship, extended in (5-48) to timevarying charge distributions. Secondly, the A field in (5-48) is connected with the current distributions of the system through the integral (5-28). Summarizing, the total electric field (5-48) is written 8 E

Eo

~--.,~---'

Due to charges ~

E

V

+

E]

-----~------

Due to time-varying currents

aA

(5-50)

8From Section 5-7, one may observe that the notations Eo and E t with the meanings defined in (5-50) were used in that discussion.

289

290

5-10 VOLTAGE GENERATORS AND KIRCHHOFF'S LAWS

291

R

SWitCh\Q,i g

-

Rg

+ (a)

(b)

FIGURE 5·20. The electrochemical generator connected to an external resistive circuit. (a) Actual circuit and equivalent symbolism. (b) Magnetic flux by I.

I/tm generated

no current is delivered, the electrolyte is an equipotential region noted by the flat central plateau in the potential diagram, with no E field inside it. The behavior of the electrochemical system is thus equivalent to the lower diagram of Figure 5.19(b), a series pair of charge double layers maintained by the chemical reactions at the electrode-electrolyte interfaces. 1o To maintain Vg , energy is supplied at the expense of one or more of the materials comprising the celL When they are used up, the cell might be restored by replacing the materials, or in some instances by applying energi externally to reconvert them to their original forms. A cell that must be restored by' adding new materials is called a primary cell; it is not rechargeable, A cell is called a secondary or a storage cell if it can be rejuvenated by externally driving the current backward through the cell to reverse the electrolytic action that took place during discharge. The reactants as well as the products of the electrochemical reactions are in general gaseous, liquid, or solid. They can be stored in one or both electrodes or in the electrolyte as the reaction proceeds, or, as in the case of fuel cells, they may be removed continuously. When a cell is connected to a resistive loop as in Figure 5-l9(c), the resulting current is predictable fi'om a Kirchhoff voltage law, derived from field theory, as described in Figure 5-20( a) to emphasize the role of the external conductor. The total E field, at any field point either inside or outside the conductor or cell, obeys (5-50) V _

oA

at

10I,'or details of the chemical reactions, see Encyclopedia of Chemical Technology, 2nd ed., Vol. 3. New York: Intersciencc, 1964.

292


field

The

system (5-54)

in which is the electric field associated with the current flow in both the conwhereas Eg is the electric field within the cell, present dueting wirc and tlw interfaces whether or not current flows, as depicted in at the and (5-54)

aA the closed path of Figure

Now

+ electrodes)

obtains

aA . dt

J

i at t

(J

fielr! Eg exists only inside the the first term of open-circuit voltage, - V g • with the last integral of

in which the

, to

' aA J a . tit t

I

d dt

f A.dt

Jt

d f B. ds dt Js

dt

(5-57)

Then

voltage expression {()r the

whieh can be rearranged illto the Kirchhoff circui t of Figure 5-20

(R +R)I+ d!/lm dt

9

The term d!/l",/dt is the time-varying ~I m linking t region is magnetically thus, ifJ m OC I. l-'he nr,o)""W' designated L as

voltage generated about the circuit t by the called the self-flux of the circuit). If the surrounding the self-flux is proportional to the current producing it; constant, called the self-inductance of the circuit, is

!/1m = to

(5-58)

(5-59)

LIWb

a definition irw the seH:inductance

I~ in which

!/1m denotes

!/1m

the flux linked

I

Wh/A

or

the circuit.

H

(5-60)

S-IO VOLTAGE GENERATORS AND KIRCHHOFF'S LAWS

293

RI

sr~ R 1-I + Rg

+ -=-Vg

M

~

~

FIGURE 5-21. Series electric circuit and models. (a) The physical de circuit. (Ii) Circuit model depicting voltage terms of (5-61). (c) Circuit model using inductance symbol L

With (5-59) inserted into (5-58), the Kirchhoff voltage expression for the circuit of Figure 5-20 is written Vg = (R + Rg)I + d(LI)jdt, and if L is a constant (independent of time), one obtains the Kirchhoff voltage law

dI

Vg = (R + Rg)I + L - V dt

(5-61 )

The transient and de (steady state) solutions of this circuit differential equation are well-known and are omitted here. A further discussion of the self-inductance parameter L, from the energy point of view, is discussed in the next section. The Kirchhoff equation (5-61) leads to the circuit model shown in Figure 5-21. The effects of a time-changing magnetic flux linking the circuit, as noted in (a) of the figure, is to produce a back voltage term, dt/l"jdt, seen from its polarity markings in (b) to oppose any tendency for the current to change. The circuit convention representing this phenomenon is the lumped-inductance element L of Figure 5-21(c), across which the back voltage L dljdt is imagined to be generated.

B. The Electromechanical Generator Another example of a generator is the electromechanical energy converter, or rotating machine. Its emf is derived from a magnetic flux linked by the machine windings, a flux that, in one version of such machines, becomes time-varying by virtue of the motion of the conductor windings relative to a static magnetic field. A generic model is diagramed in Figure 5-22(a). The magnetic flux is obtained from a permanent magnet or a field winding as shown. A cylindrical iron armature forming part of the magnetic circuit carries a winding that intercepts magnetic flux when the armature is rotated. The purpose of the armature is to provide physical support for the winding and to decrease the reluctance of the magnetic circuit by leaving only a small air gap, thereby enhancing the magnetic flux intercepted by the armature winding. A singleloop winding is illustrated for simplicity, although practical machines use many turns

-

294

STATIC AND QUASI-STATIC MAGNETIC FlELDS

Winding contour t (one turn)

(a)

(bJ

(e)

FIGURE 5-22. The simple electromechanical energy convertor (generator). (a) Simplc generator showing field and armature windings. (b) Enhancement of air gap using armature slots. (c) Voltage-inducing effect of armature-winding motion.

distributed about the armature to increase the induced emf. The wires are usually in slots, as noted in Figure 5-22 (b), to lessen the eftective air gap even more and reduce the mechanical forces on the armature conductors through a transference of the forces to the core materiaL The armature iron is also laminated to reduce eddy current losses . (see Figure 3-14). If the armature of the generator is left open-circuited and rotated with an angular speed w rad/sec, the gap voltage V(t) is obtained from (5-44d) V(t)

[5-44dl

as seen hom details in Figure 5-22(c). Thus, a radially directed magnetic field of constant value Bo imposed on a single-turn coil of radius a and length d produces an open-circuit voltage V 2Bo daw V, as long as the rotating coil is immersed in a constant magnetic field. The polarity is shown in Figure 5-22(c), determining the direction of the current in an externally connected load. If the voltage were taken oft' slip rings, the waveform of V would approximate a square wave as the sides of the coil are moved from the B field of one pole of the stator into the reversed magnetic flux lines of the other pole, A proper shaping of the poles, to make the air-gap width variable with the angular position of the armature winding, could produce an essentially sinusoidal voltage V(t), making a sinusoidal alternator of the machine. Finally, the use of an interrupted contactor (commutator) instead of the slip-ring arrangement produces a rectified or unilateral output voltage polarity, to yield a direct current machine. An analysis of the induced emf of such machines is left to appropriate books on the subject. 11 If the output terminals of an electromechanical energy convertor are connected to an external circuit, the resulting current is influenced not only by the external IlFor example, see G, J. Thaler, and M. L. Wilcox. Electric Machines: Dynamics and Stead} State. New York: Wiley, 1966.

5-10 VOLTAGE GENERATORS AND KlRCHHOFF'S LAWS

295

circuit, but also by the reactions of the armature winding itself. One of these reaetions is the back torque that must be supplied by the motor driving the generator to keep the latter at the desired speed. Because of the presence of iron in both the field and the armature structures, the forees and torques developed between the armature and the stator are best expressed in terms ofthe changes taking place in the system magnetic energies with rotation. An interpretation is developed in Section 5-13 dealing with virtual forces. Another important reaction to current flow in the generator is the effect of the armature-winding inductance. The linking of the winding current with the self-flux produced by that current yields an opposition to changes in current with time resulting from the self-voltage generated by the changing self·flux, a phenomenon already observed relative to the circuit of Figure 5-20. In this way, an armature self-inductance can be defined as in (5-60) (5-62)

,

expressed as the ratio of the self-flux produced by the armature current, to the current itself. An equivalent circuit of the armature winding with a connected load is depicted in Figure 5-23, showing the generated voltage V(t) resulting from the rotation of the armature winding in the impressed static B field, the self-inductance La of the armature winding, and a series resistance Ra representing the ohmic winding losses. (Other losses such as iron hysteresis and eddy current losses, as well as rotational wind resistance and bearing friction losses, may be represented in more elaborate equivalent diagrams; they are omitted here.) If the externally connected load has the resistance Rand self-inductance L as developed in relation to the external circuit of Figures 5-20 and 5-21, one can deduce the equivalent circuit of the loaded rotating machine as in Figure 5-23. The Kirchhoff voltage differential equation for this system is evidently

V(t)

= (R + Ra)! + (L + La)

dI dt

(5-63)

with V(t) denoting the machine-generated voltage deduced from the basic expression

(5-44d).

( Armature self-flux y~

I

I

t :::

Load Machine

Impressed B flux (changing linkage with rotation)

Equivalent circuit

FIGURE 5-23. Development 0[' an equivalent circuit of a rotating machine connected to a load.

296

STATIC AND QUASI-STATIC MAGNETIC FIELDS E=_V_iJA iJt

J and ds elements at source V(t)

FIGURE 5-24. Electric and magnetic field qnantities associated with a cnrrent-carrying circuit.

5-11 MAGNETIC ENERGY AND SELF-INDUCTANCE 12 In this section, the glib assertions of the last section concerning the inductance of a current-carrying circuit are examined from the viewpoint of the energy required by the circuit to supply its heat losses and to build up the magnetic field. The generalized definitions of the self-inductance of a single circuit, and in the next section, the mutual inductance between pairs of circuits, are established in this way. This point of view regards the inductance parameter as the basic criterion of the magnetic field energy, or work done in establishing the magnetic field.

A. Self-Inductance in Terms of A and J Consider the series circuit of Figure 5-24. An external energy source of terminal voltage V(t) is connected to a conductive circuit of arbitrary shape, carrying a current I. It is assumed that the currents form closed paths, that is, the current-continuity relation is (4-22), V' J = O. Strictly speaking, the latter requires that the current be dc, although it is very nearly satisfied up to fairly high frequencies as long as the overall circuit dimensions are not an appreciable fraction of a free-space wavelength. At the higher frequencies, however, the current penetration into the conductor is severely limited by the skin effect, with negligible electromagnetic field penetration occurring at very high frequencies. 13 The work done by the source V(t) in bringing the current up to the value J, expressed in terms of the electric and magnetic fields developed in and around a conductive circuit, leads to the circuit parameters (resistance and inductance) as shown in the following. Observe in Figure 5-24 that the conductive circuit, the interior denoted by v;, , is bounded by S (conductor surface), with endcaps at the gap where the voltage V is impressed. At the gap V is specified by the quasi-static equipotentials = <1>1 and <1>2 at the endcaps such that V <1>1 <1>2. With the current 1 = J . ds delivered by V into the endcap at the positive terminal, and J . ds coming out rif the

f

f

12 If you desire a shorter treatment of self-inductance, studying selected portions of Parts A, C, and D in this section should provide a reasonable background, with emphasis on the important energy and flux-linkage methods for finding L.

13See Appendix B, part A for a discussion of the skin effect.

297

5-11 MAGNETIC ENERGY AND SELF-INDUCTANCE

negative side, the energy supplied by V(t) in the amount V dq = VI dt is written (5-64) The electric field anywhere in the conductor is (5-48)

E

aA

= -V--

(5-65)

at

The latter is given an energy rate interpretation by dotting (5-65) with integrating the result throughout the volume v" of the conductor; thus

J dv

and

f E· J dv = - f (V
Jvc

Jvc

Jvc

at

By the identity (15) in Table 2-2, J' (VJ), since div J = o. With this into the second volume integral and applying the divergence theorem (2-34), one obtains

f E· J dv + f J' aA dv :rs (J) • ds = Jvc Jvc at

_J.

(5-66)

From the continuity of the current flux, only tangential currents appear at the conductor walls in Figure 5-24, except at the gap endcaps. There, <1>1 on one end cap and = <1>2 on the other, reducing the surface integral of (5-66) to just

_J. fs(gap)

(J) • ds == (<1>1 - <1>2) f

JS(gap)

J' ds

= VI

the power delivered by V to the circuit at any instant. The second term of (5-66) is a measure of the irreversible heat energy expended in the volume; its value is 12 R, defining the low-frequency conductor resistance 14 R by (5-67) Inserting the last two expressions into (5-66) obtains

VI = RP

+

i

Vc

aA

J . :;dv vt

but the energy expended by V in the time dt is

VI dt

RI2 dt

+ Ivc J'

(dA) dv

(5-68a)

symbolized (5-68b) l4The question of conductor resistance, defined in terms of the heat generated by it, is examined in ample 7-1.

298


By integrating (5-68a) with respect to time, the result (5-69) is obtained, yielding the work done by V in bringing the circuit to its final state. The last term is interpreted as the energy Urn expended in establishing the magnetic field (the energy stored in the field)

(5-70)

The interpretation of (5-70) is straightforward. The current density at any point in the conductor is J, with A the vector magnetic potential there. Both J and A are fields, so they are generally dependent on position in v;,. Equation (5-70) states that the energy stored in the magnetic field is the integral of [J~ J . dA] dv throughout the conductor volume, in which S~ J . dA denotes, at any dv, the integral of] dA cos 0 as the potential A there is built up from zero to its final magnitude A. Note that the integrand has the units of joules per cubic meter. For a linear circuit (a linear magnetic environment), A anywhere in the conductor is proportional to the current density J (hence, to the total current J). If the circuit were nonlinear, the relationship between A and the value of J at each volumeelement in the conductor would not be a straight line, but for a linear circuit, the energy expression (5-70) simplifies as in the following. The integration within the brackets of (5-70) entails a buildup in time of the vector magnetic potential from zero to its final value A. For a linear magnetic environment, the vector potential anywhere in Vc is proportional to the densities J therein. Suppose J is built up in a straight-line fashion from zero to its final (quasi-static) value J
Urn =

t Jvr

c

A· J dv J

Linear circuit

FIGURE 5-25. Simultaneous buildup of J and A at a typical volumcelement dv in a conductor, as current is brought from zero to final value I ..

(5-71 )


299

if the final value (f) superscript notation is dropped. Note that (5-71) is applicable to a linear system only. Like its more general version, (5-70), it expresses the energy expended in establishing the magnetic field, through an integration required to be taken only throughout the conductor volume region possessing the current densities J. The self·inductance of a linear circuit can be defined in terms of the energy (5-71). Tt contains a product of J and A and is thus proportional to /2, whence

(5-72) in which the proportionality constant L is termed the self-inductance of the circuit, expressed in joule per square ampere, or henry. Solving for L thus permits expressing the self-inductance in terms of the magnetic energy as follows

L

2U

I

= ---i'= 2I Jvc r A· J dv H I

(5-73)

assuming the circuit is linear (i.e., immersed in a linear magnetic environment).

*B. Self·lnductance of a Circuit in Free Space For a linear circuit devoid of magnetic materials (e.g., an air core coil or a parallel-wire line), (5-72) and (5-73) can be simplified by 'use of the free-space integral (5-28a) for A [5-28a] The circuit in Figure 5-26 depicts the quantities needed in the evaluation of A at a typical field point P by use ,of (5-28a). Substituting it into (5-72), the magnetic energy

FIGURE 5-26, Circuit in iree space, showing source point P' and field point P relative to energy and self-inductance integrals.

300


integral (5-71) becomes

which can also be written

U

=.1 m

2

r r

JioJ"

Jvc Jv c

J du' du J

4nR

'

Free space

(5-74)

The result (5-74) is independent of the order of integration, but note the use of the primed current density J' at the source point P' to avoid confusion with J at the field point P. The corresponding self-inductance expression becomes, using (5-72)

(5-75)

:Free space

No explicit use is made here of (5-75) in self~inductance calculations. If you are interested in applications of (5-75), consult other sources on this subject. 1s

*C. Self-Inductance from an Integration throughout All Space Another expression for the magnetic energy of a circuit can be obtained from

(5-70) in terms of the Band H fields of the system. The current densities J in the conductor are related to H therein by (5-2) for quasi-static fields: J = V X H. Making use of the vector identity (16) in Table 2-2, V . (F X G) = G· (V X F) F· (V X G), J' dA in (5-70) can be written

J'

(dA)

= (dA) • (V

X

H)

= V' [H

X

(dA)]

V' [H

X

(dA)]

=

+ H· V X + H . dB

(dA)

Inserting this into (5-70) and applying the divergence theorem (2-34) to the first volume integral yields

Urn =

Iv [IOA V, (H dA)JdU + Iv [I: H· dBJdV ~s [I: dA) J. + Iv [I: J X

(H

X

ds

H . dB dv

but the surface integral in the latter vanishes as S is expanded to include all of space, because H and A decrease at least as r- 2 and r- 1 respectively in remote regions, 15For example, see R. S. Elliott, Electrvmagnetics. New York: McGraw-Hill, 1966, p. 309.


301

whereas surface area is picked up only as r2. Thus the magnetic energy expended in establishing the fields of a quasi-static circuit becomes

(5-76)

As with (5-70), the energy (5-76) is correct whether or not the circuit is linear, although (5-70) requires integration only throughout the conductor volume, whereas (5-76) must be integrated throughout all space to obtain the same result. One can simplify (5-76) if the system is linear, by making use of the fact that (5-76) is analogous in form to (5-70). Since the latter becomes (5-71) for a linear system, one should thus expect (5-76) to yield I

um = 1.2 Jv f B· HdvJ

Linear circuit

(5-77)

The integrand B • H/2, seen to have the units of joules per cubic meter, is called the magnetic energy density in the volume region V. Another expression for the self-inductance of the circuit of Figure 5-24 is obtained by equating (5-77) to the definition (5-72) for L, whence

I

Iv B· Hdv

(5-78)

One can separate (5-77), if desired, into two volume integrations as follows

um = 1.2 JVi f B· H dv + 1.2 JV f

B· H dv

(5-79)

e

attributing the total energy Um to two contributions: one associated with the volume i'i in the conductor, plus another outside it. With (5-79) substituted into (5-78), the total inductance is expressed

L

m = -2U 2 - = ?1 1 1-

1 B . H dv + 11 1 B . H dv = L; + Le Vi

-2

V.

(5-80)

The first term, L;, is called the internal se(f-inductance; the remaining integration taken outside the conductor yields the external self-inductance, Le.

EXAMPLE 5·12. Find only the internal self-inductance associated with every length t of a very long straight wire carrying a low-frequency current I.

302

STATIC AND QUASI-STATlC MAGNETIC FIELDS

j

EXAMPLE 5-12

For any length of the single infinitely long wire shown, the energy in the external magnetic field is infinite, a [act revealed on integrating (5-77) for the energy associated with the exterior fields Band H; however, the energy stored within a length t of the conductor is finite. The associated internal inductance is obtained from (5-80) ! I

fVi

B • Hdv

I)

By use of (1-64) for B1> inside the wire (the factor p used in the event of a magnetic conductor), one obtains from (5-81)

Li

1 = -2

I

~ Vi

pH~dll =

I

il' i 2" _

__

z-o 1>-0

J,"_

p-O

p(lp)2 pdpdc/)dz =pt ~ -

4n a

8n

(5-82)

a result independent of the wire radius. A nonmagnetic wire therefore has the internal inductance per unit length, Ldt po/Sn 0.05 pH/m.

t of a long coaxial line with the dimensions shown. Assume uniform current densities in the conductors. The total seH~illductance is obtained using (5-78). The magnetic fidds within and between the conductors, obtained by the methods of Example 5-1, are

EXAMPLE 5·13. Find the total self·inductance of every length

Ip 2na 2 H1>=

H1> =

1

2np

-;Tir(~/-b2) (~ p)

O
5-ll MAGNETIC ENERGY AND SELF-INDUCTANCE

303

V ,(2)

I 1: V

(1) L

I

I

~

I I

I : '

I

:,--t---J 1, ... __ 1__ ... ./

I

EXAMPLE 5- i 3

with H1> = 0 outside the system. The internal selt:inductance of the inner conductor (1) is

2U(l\

I

P

P

L\l)=~=_ f

,

Jv!')

BoHdv=Jl

t

(\)

8n

a result the sam(' as (5-82) for the isolated wire. An external self-inductance, attributed to the field hetween the conductors, is

~

rt f2rr fb dp d=o Jp=a p

=

/lot t n ~ 2n a

(2)

Another internal contribution by the field in conductor (2) yields LF)

2U(2). m.tn

]2-/It

s: s:rr s:

l

4

c

t9'bc

(~ b2 )

2

p) P dp d
+ hc4

-

b4 )]

(3)

The total self-inductance L of the coaxial line is thcrefore the sum of (1), (2), and (3). If at high frequencies the two conductors are assumed perfectly conducting to prevent field penetration into them, the self-inductance reduces to (2) (5-83)


IJIAMPLE 5·14. Determine the low-frequency self-inductance oflength t of the long parallelwire line in free space shown, by use of (5-73). The integration of (5-73) is performed inside the conductors where J exists, so A need be found only in the conductors. One might employ (5-28a) to evaluate A, but for a single wire, applying symmetry to (5-22) obtains the answer more quickly. Thus, with 0/0;:; = tJ/tJ¢ = 0 and only a Bq,component, B = aq,Bq, = V X A = aq,( -oAz/op) , implying that A has only a ;:; component as noted in part (b) of the accompanying figure. From Bq, = - oAzltJp, integrating yields Az

= -

f Bq,dp + C

(1)

with C an arbitrary constant, depending on the potential reference chosen [or A z • Thus A z values inside and outside the wire become, from (1)

=

A z

_fl-loIP d 2naz p

+ C = _'t olp22 + C 4na

1

1

p
(2)

p>a

(3)

In the presence of both conductors, carrying I and - I as in (c) of the figure, the total A z is obtained by adding the contribution of each conductor, called A~l) and A~2). For conductor 2, choosing zero potential at p = flZ such that from (2), A~Z) = 0 = l-l oI/4n + Cl' From (3), with A~) = 0 = - (l-loI/2n) tn az + C 2 , one obtains C l = l-l oI/4n and

A z = 0 on 2

lCiQl---

2

I

k--(b)

d -----'0'>1 (c)

EXAMPLE 5-14. (a) Parallel-wire line. (b) Vector potential A z associated with a single wire. (el Sum of vector potentials at typical field point in one of the wires.

/


305

(flol (n az )/4n. Thus the contribution of wire 2, satisfying A~2) = 0 on its surface, is

A(Z)

= _flo! (n az

2n

z

P2

pz

>

(4)

a2

Similarly, the potential of wire I satisfying A~I) = Ao (an arbitrary value) at its surface PI = al becomes flol 4n

(1 (5)

The total potential A z from part (c), is

A~I)

+ A~2) inside conductor 2, using PI

a2

+ p~ + 2P Z il cos e/>

fli [ A = 1 - -p~ - 2 {n r=~==;;:====== z 4n a~

In conductor 2, Jz IlnaL and with dv = pz dpz de/> dz, the inductance contribution of a length (of conductor 2 only, from (5-73), is

IP) = .~1 12

j'

V

A . J dv - -Ao} 2

na2I

pzdpzde/>

The integral contribution of the third term in the integrand can be written

in which the second term integrates to zero (Peirce's integral 523 16 ), obtaining L(2)

= ito{ - 1

[ 8n

+

dJ

I (n-

2n

a2

A similar consideration of conductor I yields by analogy

making the total inductance L(l)

+ IPl

of the two-wire system

L= 16B.

/

O. Peirce, A Short 'fable oj Integrals. Boston: Ginn, 1910.

(5-84)

306


Comparison with (5-82) shows that the leading term of (5-84) is the internal inductance, making the last term the external inductance.

D. SeH-lnductance by the Method of Flux Linkages The resolution of the self·inductance of a circuit into the sum of internal and external self-inductances provided by (5-80) is closely related to another technique known as the method if flux linkages. This approach is based on the use of the energy definition, (5-78), but with the integration in all space replaced by a surface integral intercepting all the magnetic flux of the system, the self-inductance being thereby characterized by the linkage of that flux with the circuit current. The method is described here. For most circuits, the total magnetic flux generated by the current can be partitioned (exactly or approximately) into two amounts: that lying entirely outside the conductor, plus that flux wholly internal to the conductor. Such a flux division occurs precisely for the single round wire noted in Figure 5-27 (a), and very nearly so for the parallel two-wire line shown in the same figure,17 especially for wires with diameters small compared to their separation. Another example is the loop shown in Figure 5-27(e); for thin wire, the flux tubes can be separated into those wholly inside or outside the wire as shown. The volume occupied by the magnetic field (all space) is thus divisible into closed flux tubes that surround or are embedded in the current. The magnetic energy contained in all space has been given by (5-77)

u

m

=.1 2

r

Jv B· Hdv

[5-77]

Suppose the volume of the typical flux tube in Figure 5-27(b) is subdivided into elements dv = dsdt', in which dt' is aligned with the tube wall (and therefore with the B field) and ds denotes the cross-sectional area of the tube. Then B • H dv = (a,tB)· Hdsdt' = (a,tdt')· HBds = H· dt' dt/lm. Thus, if the integration (5-77) of the latter is to include all elements dv where Band H prevail, H • dt' should be integrated about the closed median line (' of the flux tube shown, with the remaining surface integration taken over an open surface S chosen to intercept all the flux tubes of the circuit. For the single-turn circuit of Figure 5-27 (b) or (e), the appropriate S intercepting all flux tubes is that bounded by a closed line { essentially coincident with the wire axis. Thus the energy integral (5-77) can be written

(5-85)

with S bounded by the circuit t. [Note: The last integral is the consequence off H . d{', integrated about any closed flux tube (I, being just the current i({I) enclosed by (I.] 170wing to the proximity effects of the low-frequency currents, the magnetic flux in the interior of parallel wires is not concentric about the centers of the wires, but about points moved slightly apart from the centers. This efleet is responsible for some of the magnetic flux being partly inside and partly outside the wire, as noted in Figure 5-27(a).


307

, ,,

,,/ ,,

\

\

""\ \

,\

,, \ \ ,,

/

:

\

\ I

\

,

,// \ \

-----_ ..

\

\

--

/

,

,

, ,,

/

,,

\ \

,,

Parallel-wire line

/

Single wire (a)

dv = dsd/' Typical flux tube carrying dl/;m

I

dl/;m

dl/;m

Circuit

t

t' over which J,[H·dt=I ,

r

(b)

Flux tube {'

Conductor

, , '"

(c)

FIGURE 5-27. Concerning the method of flux linkages. (a) Examples of internal and external flux-partitioning. (b) Single-turn circuit (left) showing external flux tube linking lance, and a two-turn circuit (right) with a flux tube linking I twice (passing through Sex twice). (c) Wire loop, showing internal and external flux (lef!), and a typical internal flux tube (right) linking i(t'), a fraction of 1.

For all exterior flux tubes, passing through Sex as shown in Figure S-27(b), the total current I is linked by t', whereas a variable fraction itt') of I is linked by flux tubes t' located inside the conductor and passing through Sin, as shown in Figure S-27(c). In the event ofa circuit t having more than one turn as in Figure S-27(b), an exterior flux tube t' may even encompass I more than once (in general, as many as n times for

308


an n turn coil). It is thus evident in such cases that the same flux tuBe t' can contribute to (5-85) over the surface Sex several times, thereby increasing the magnetic energy and the self-inductance correspondingly. Whenever the magnetic flux of a circuit is separable into internal and external linkages passing through Sin and Sex as depicted in Figure 5-27 (e), it is convenient to separate (5-85) into the contributions (5-86)

In the latter, one is cautioned to observe that the quantity t/lrn.ex JSex B . ds appearing in the external energy term denotes a total flux through Sex, which can be the result of some or all the flux tubes passing through that surface more than once, for example, as in Figure 5-27 (b), or for the many turn coils illustrated in Figure 5-28. By use of (5-86), the self:'inductance of the circuit t is expressed in terms of internal and external contributions as follows

L

2Um

= J2 =

r . ,

I

Js z(t) dt/lm

I =

r

JSin itt') dt/lm +

(5-87)

I

such that the external inductance is given by

t/lrn.ex

L =

I

e

~

=

t

r

JSex

B· ds

(5-88a)

or just the total magnetic flux penetrating Sex divided by the current

t. The internal

n-turn coil (circuit f)

Circuit

t

BfIUX

t

-'-~' ~s~ ~ /

~ .---

l'::::;;: I

(a)

.1-.

,

,.exl

I' f"

B flux-...

I I~Y.

'~

I~el " ~

"'Y).!b~'< ~, , (b)

Core flux (c)

FIGURE 5-28. Examples of many-turn coils having negligible interual self-inductance. (a) Air core solenoid. (b) Toroidal winding. (c) Coil and iron core.

I


309

inductance is

(5-88b)

a consequence of i(t') dljlm integrated over the appropriate internal strip Sin connecting with the wire axis, as depicted in Figure 5-27 (c). An illustration of the use of the latter for a long straight wire is taken up in Example 5-15. Although the internal conductor volume of a circuit may be small, the magnetic fields may be relatively large there; individual circumstances will dictate whether or not the internal inductance is negligible. For circuits having large external fluxes, such as those with iron cores, the total self-inductance is generally well approximated by (5-88a), the external selfinductance. II

EXAMPLE 5-15. Determine the internal self-inductance of every length t of the infinitely long wire shown, carrying the low-frequency current I. Use (5-38b), employing the method of flux linkages. A typical flux tube t' carrying dl/l m = B • ds through Sin is shown in the accompanying figure. With the internal B obtained from (1-64), the flux in the tube is

dl/l m = B· ds = Bq,dpdz =

/lIp --2

2na

dpdz

The current i(t') intercepted by dl/l m is the fraction I(np2jna 2 ) = I(p2ja 2), obtaining from (5-83b) (5-89) which agrees with (5-32).

~

iU')

=I ~ A, encompassed by dl/l m a

EXAMPLE 5-15

310

STATIC AND QUASI-STATIC MAGNt:TIC FIELDS

EXAMPLE 5-16. Determine the approximate self-inductance of a length t' of a long parallelwire line shown in (a), using the flnx linkage method. Assume the radii small compared to the spacing d. For well-separated conductors as in (b), the internal field is essentially that of an isolated conductor, making the internal self-inductance for both conductors just twice (5-89), that is, Li = JIt'/4n. The external inductance is found by using (5-88a), the ratio of the magnetic flux through Sex of (a), divided by I, but the total flux is just twice that through Sex due to one wire, given by

r

JSex

B' ds

i

t

z=O

J,d-a (JIOI -- a ) 2np 4> p=a

. a dp dz 4>

~

JIoIt' t' n d2n a

(1)

yielding for both wires

(2) The total self-inductance is the sum

L

=

Li

+ Le =

pt' 4n

+

JIot' n

d a

t'n - H

(5-90)

a result seen to agree with the exaet expression (5-84) on putting a 1 = a z into the latter and assuming nonmagnetic wire. For a nonmagnetic parallel-wire line with d = 12 in. and a = 0.1 in., one obtains

L/t'

=

lO-7

+ (4

x 10- 7 ) t'n 120

=

2.02 JIH/m

Neglecting the internal inductance would incur about 5% error' in this example, not a negligible amount.

In calculating self-inductance at low frequencies, the internal-inductance contribution is in some cases quite small; in others it should not be neglected. The internal inductance of a single-layer air core coil of several turns, .illustrated in Figure 5-28(a), contributes little to the self:inductance if the volume of the wire is small compared to

(a)

(b)

(c)

EXAMPLE 5-16. (a) Parallel-wire line and surfaces linked by internal and external magnetic fluxes. (b) Division of internal and external Huxes for thin wires. (c) Proximity effect for thick wires.

I 11.1


311

EXAMPLE 5-17

,

the region where the significant fields are located. In the closely spaced toroid as in Figure 5-23(b), with every turn intercepting all the core flux, the self·inductance is proportional to the square of the turns, as seen in Example 5-17. The addi tion of an iron core in the form of the low-reluctance magnetic circuit of Figure 5-23(c) increases the selfinductance substantially more. In these cases, the added effect of the internal inductance is insignificant.

EXAMPlE 5·17. Find the self-inductance of an n turn toroid with a rectangular cross section as shown, for two cases: (a) with an air core, assuming closely spaced turns, and (b) with the core a linear ferromagnetic material (constant /l). (a) The magnetic flux in the air core, from Example S-2, is /lonld

b

I/Imcore=--tn , 2n a

An inspection of the sUlface Sex bounded by the circuit t, as given in Figure S-28(b), reveals that Sex intercepts the core flux n times,18 yielding t/lm,ex = nt/lm,core through Sex. Thus the self-inductance from (S-88a) becomes n!/lm, core ell

L~L

!/Im,ex

2

=--=-_._-=

/lon d 2n

b a

tn~

(S-9Ia)

with the internal inductance neglected. Thus, a 100-turn air core toroid with dimensions a = 1 em, b = 3 em, d = O.S cm has the inductance L = (4n x 10- 7 x 100 2 x O.OOS

tn 3)/2n = 11.0 /lH

Doubling the turns to 200 is seen to quadruple the inductance. (b) Inserting an iron core with the permeability /l, (S-91a) becomes L

(5-91b)

Using a linear ferromagnetic material with /lr = 1000 makes the inductance of the 100-tnrn toroid just 1000 times as large, yielding L = 11.0 mHo ISOr equivalently, every flux tube

dJ/!m encompasses the culTent

In times in this example.

I

I

312


t~eumann's Formula for External Inductance in Free Space An extension of the flux linkage expression (5-87) leads to Neumann's formula, applicable to circuits in free space. Equation (5-87) consists of internal and external self-inductance terms as follows

*E.

[5-87] Consider first only the external inductance term (5-88a) of (5-87), involving the flux t/lm,ex linked by the external surface Sex bounded by the circuit t. From (5-47), this is expressed

t/l m~ex =

r

JSex

B· ds =

r

(V

JSex

X

A) . ds =

~ A· dt Wb Yt

(5-92)

With (5-92), (5-88a) becomes L

e

t/lmex = = -'I

II

-

I

Sex

B • ds = -l~ A· dt I t

(5-93)

In free space, the vector magnetic potential A can be found by use of (5-28a) [5-28a] Applied to the circuit of Figure 5-29(a), (5-28a) obtains A at the typical field point P located on t bounding Sex. Another integration of A . dt about t in accordance with (5-93) then obtains the external self-inductance of the circuit. These steps are combined by inserting (5-28a) into (5-93), yielding

Le

= -1

I

f [i t

Vo

dV']

JloJ --R. dt H 4n

t (a)

(b)

FIGURE 5-29. A closed circuit in free space, relative to external self-inductance calculations. (a) Wire circuit, showing source and field points P' and P. (b) Simplification of (a), with sources I dt' concentrated on the wire axis.

(5-94a)


313

This quadruple integration is simplified for a thin-wire circuit if I is considered concentrated on the wire axis as in Figure 5-29(b). Then J dv' becomes I dt', reducing (5-94a) to

floIdt'] L - -1 ~ [~ - -dte I { C' 4nR -

~~ t

('

flodt' - dt 4nR

(5-94b)

a result known as Neumann's formula for the external inductance of a thin circuit in free space. The order of the integrations relative to dt' and dt, and hence, relative to the source point and field point coordinates, is immaterial. From (5-87), the total self-inductance L is obtained by adding (5-94b) to the internal inductance term L i • Since the latter is a measure of the internal stored magnetic energy, Li is expressible using either (5-S1) or (5-8Sb); thus L = Li + Le becomes, in Fee space

,

i

L= 1

V;n

r JSin

~~

flo dt' - dt 4nR

+ J. J.

Il odt' - dt 4nR

B-Hdv + i(t') dl/tm

t

t'

'ft 'ft'

(5-95)

EXAMPLE 5·18. Find the self~inductance of a thin circular loop of wire in free space, with dimensions as in (a). Use the Neumann formula (5-94b). The current assumed concentrated on the wire axis as in (b) allows the use of (5-94b). In cylindrical coordinates, dt' = a",bdf/J' at the souree point P'(b, f/J', 0) on the

axis t'. From the circular symmetry, the location of the field point P on t is immaterial, so put P at f/J 0; that is, at P(b - a, 0, 0). The distance from 1" to P is given by the law of cosines

R=

(a)

(1)

(b)

EXAMPLE 5-UI. (a) Circular loop of round wire. (b) Axial, line current approximation of (a).

314


while dt' • dE in (5-94b), from part (b), means dE' dE cos cp' (implying that only the component of A parallel to dE at P is required in the integration.) Then (5-94b) becomes L e -

J.2l< J.2l< 1>= 0 1>' =0 ---;~================== (5-96)

This result is not integrable in closed form, though with numerical values of a and b it yields to computer solution. An alternative makes use of tabulated values of the complete elliptic integrals K(k) and E(k). A conversion of (5-96) in terms of such integrals is accomplished as follows. Change the variable cpt to 21X, making dcpt = 2 dlX and cos cpt = cos 21X = 2 cos 2 IX - 1, with the limits on IX going from 0 to n. Then R in (5-96) becomes

=

R

.jb 2

+ (b

- a)2 - 2b(b - a)(2 cos 2 IX - 1)

.j(2b - a)2 - 4b(b - a)

COSzlX

(2) if k 2

=

4b(b - a)/(2b - a)2. The complete elliptic integrals, defined by

de

J,fo"!2 -r===::;===

K(k)

(5-97)

are incorporated into (5-96) as follows. The integral in (5-96), making use of (2), becomes

21t fIt J.1>'=0 = Jo

2 cos 21X dlX (2b - a)

---;:===

S:

rk=co="s~2:=IX=d=lX=.::==

(3)

but the numerator of (3) is written

k cos 21X = k(2 cos 2 IX - I) = 2k cos 2 IX

2

2

k+k

k

to yield a further conversion of (3)

I

"

.jb(b - a) So

=

( -2 k ) dlX "k 2 - "7 { .jb(b - a) SO.jl - k 2 cos 2 IX k 1

" .j 1 - k 2 cos 2 IX dlX}

So

(4)

An inspection of the last integral shows that

which from (5-97) is just 2E( k). A similar consideration of the preceding integral in (4)

315

5-11 MAGNETIC ENERGY AND SELF-INDUG'TANCE

reveals that it is just 2K(k), so (5-96) becomes (5-98) The tabulated values 19 of K(k) and E(k) can be used in (5-97) to evaluate Le ofa circular loop with desired dimensions. For thin wires (a« b), the elliptic integrals are approx· imated by

E(k)

~

1

a«b

yielding the simplification

a«b

(5-100)

I

For example, a 2-mm diameter wire bent into a circle of 10-cm radius has the external inductance Le = (4n x 10- 7) (0. l)(tn 800 2) = 0.588/lH. The internal magnetic field of the loop is virtually that of a straight, isolated wire, making their internal inductances nearly the same. Applying the results of Example 5-12, the approximate internal inductance of the loop becomes

L. ~ /l(2nh) ,- 8n

ftb 4

(5-101)

With b = 10 cm and assuming nonmagnetic wire, L; = 0.031 /lH. Thus the self-inductance expressed by (5-93) becomes L = Le + L; = 0.619/lH, in which Le is seen to be the predominant term.

A summary of expressions for magnetic energy described in the foregoing discussion, together with expressions for the circuit inductance when the system is linear, is given in Table 5-1. *F. Kirchhoff VoHage Relation from Energy Considerations In concluding the remarks about the circuit of Figure 5-30(a), a Kirchhoff-type voltage equation resembling (5-63) can be obtained for it from the energy expression (5-68a)

VI dt = RI2 dt

+ Jvc r J' dA dv

[5-68a]

abbreviated [5-68b] Dividing by dt obtains

VI =

RP + dUm dt

19For example, sec E. Jahnke, and F. Emde. Tables oj Functions, 4th cd. New York: Dover, 1945.

(5-102)

316

STATIC AND QUASI.STATIC MAGNETIC FIELDS

TABLE 5-1 Summary of Magnetic Energy and Self-inductance Relations SELF-INDUCTANCE

MAGNETIC ENERGY

In terms of A and

J

integrated throughout conductor volume

In general (5-70)

Linear circuit (5-71 )

(5-73)

(5-74)

(5-75)

In free space

U =

m

11 1

fioJ" J du' dv

Jvc Jvc 4nR

In terms of Band H integrated throughout all space In general (5-76)

Linear circuit Um

= ~2 Jv 1 B· Hdv =1 1

JVin

(5-77)

B'Hdv+1 1

JVex

L

m 2U== -121 ~V B . H dv 12

(5-78)

B'Hdv(5-79)

Extension to method of flux linkages

Urn

=

1 Is itt') dt/lm

=

~2

I

Stn

itt') d,I,

'I'm

+ It/lm.ex 2

(5-85 )

L=

(5-86)

=

)2 Is itt') dt/lm ~ 12

I

Stn

itt') dt/l

m

+ t/lm.ex 1

(5-87)

In free space _

L= I

Is

in

i(t') dt/lm

+ rC, rC, j-t 'fe'

de" dt 4nR

(5-95)


317

v

(a)

(b)

( c)

FIGURIfS-30. Development of circuit models of the circuit oLFigure 5-24. (a) Physical circuit driven by V(t). (Ii) Circuit model depicting terms of (5-106). (c) Circuit model using lumped elements.

signifying the instantaneous power delivered by V: the sum of the instantaneous heat loss plus dUm/dt, the power delivered to the magnetic field (rate of magnetic energy storage or release). Dividing by I produces a voltage relation 1 dUm V=RI+-I dt

(5-103)

in which Urn, the im;tantaneous magnetic stored energy, is specified by any of the expressions listed in Table 5-1, depending on whether the system is magnetically linear. For a linear circuit, a self~inductance L is attributable to the circuit energy by (5-78) (5-104 ) With L constant, the last term of (5-103) becomes (5-105) making (5-103) a voltage relation comparable to the Kirchhoff expression (5-61); that is,

V= RI + d(LJ) dt

(5-106)

Equation (5-106) states that the applied voltage V(t) supports two effects: (a) a voltage drop RI associated with the circuit resistance Rand (b) a back voltage d(LI)/dt or LdI/dt produced by the time-varying magnetic flux linking the circuit, a flux produced by 1. Because of the separation of these effects into two terms, one may properly lump the resistive voltage and the self-induced voltage to yield the series circuit model shown in Figure 5-30.

318

STATIC AND QUASI-STATIC MAGNETIC FIEI,DS

5-12 COUPLED CIRCUITS AND MUTUAL INDUCTANCE Besides the single circuit of Figure 5-24, also of physical interest is a pair of such circuits, coupled electromagnetically by the time-varying fields generated by their currents. Examples are the iron core and air core transformers of Figure 5-31 (a), which may have active sources in one or both windings. A generalization is illustrated in (b). The analysis of coupled circuits from the magnetic energy point of view closely parallels that for the single circuit. Consider the circuit pair of Figure 5-31 (b) with one driving source V(t) in circuit 1, producing the primary current 11 (t). The latter generates a field B 1 , the flux of which links not only circuit 1 but some fraction of that flux (governed by the geometry and the presence of ferromagnetic bodies) also links circuit 2, generating an emf about each circuit in accordance with the Faraday law, (3-78). The ensuing current 12 produces a field B2 reacting similarly on circuit 2 while also partly linking circuit 1, therehy establishing an additional back emf in each to modify 12 and 11 accordingly. The influence of these mutual coupling effects on current flow can conveniently be treated by use of Kirchhotf voltage equations, developed later in this section. The mutual magnetic coupling between the circuits leads to their mutual inductance parameters, developed in the following. A simple extension of the power integral (5-66) to the pair of circuits of Figure 5-31(b) yields

- J, (J) . ds j-'Sl

=

r

JVl

E· J dv

+

r

JV2

E· J dv

+

r aA. r. aA. at J dv + JV2 at J dv

JVl

(5-107) in which Vj , V2 denote the volumes inside the two conductors, with S] taken as the surbce enclosing VI exclusive of the driving source V(t). The left side of (5-107) denotes the instantaneous power Vi l delivered, whereas the two-volume integrals ofE· J are the ohmic losses Rili and R2I~ within the conductors. Multiplying (5-107) by dt

V«)~L._._(_fJ.)_~_.-...J I2

h

BI flux (of

h only)

RL

With iron core

V(t)

Vj (Conductor volume) With air core (a)

(b)

FIGURE 5-31. Magnetically coupled circuits. (a) Typical coupled circuits. (b) Generalized coupled circuits.

5-12 COUPLED CIRCUITS AND MUTUAL INDUCTANCE

319

yields (5-108a) abbreviated (5-108b) Integrating (5-1 08b) obtains

denotil)g the work done by V(t) in bringing the system up to the levels II and 12 at the instant t. The volume integrals in (5-109) represent the energy expended by V in establishing the magnetic fields of the coupled circuits; that is, the energy stored in the magnetic fields in the amount

(5-110)

The integrations are required only within the conductors, since no densities J exist outside them. Equation (5-110) is correct whether or not the system is linear. If the system of Figure 5-31 (b) is linear, one can assert that the contributions to the total A at any point in the region are proportional to the current densities J in the circuits. Then

Urn = 1 Iv! A" J dv

+ 1 IV2 A" J dv J

Linear system

(5-111 )

obtained analogously from (5-110) in the manner that (5-70) led to (5-71). I t is advantageous to rcexpress (5-111) in terms of the vector potential contributions of each current. Let the total vector potential at any field point P in either conductor be written (5-112) with Al and A2 denoting the potentials at P due to the currents in circuits 1 and 2, respectively. Then (5-111) splinters into the four contributions Um

=1'Jv!

A 1 "Jdv+1' JV2 A 2

·Jdv+1' JV2 A 1 ·Jdv+1'JVl A 2 "Jdv

(5-113a)

abbreviated as follows (5-113b)

320


Note that Urn 1 , for example, denotes the magnetic self~energy of circuit 1 taken alone (with circuit 2 open-circuited), with (5-71) revealing that Urn1 is the energy associated with the self-inductance of circuit 1, called L 1 . A similar remark applies to U m2 , leading to the self-inductance L2 of circ·uit 2. With J in conductor 1 as well as At both proportional to 11 , U rnl becomes proportional to Ii. Similarly, U m2 , U m12 , and U m21 are proportional to 1~, 1112, and 11 12 , respectively, yielding from (5-113a)

(5-1 14a) (5-1l4b) (5-114c) (5-114d)

The constants M12 and M21 appearing in (5-114c) and (5-114d) are known as the mutual inductances of the pair of circuits, related to the additional mutual magnetic energies associated with the magnetic coupling of the circuits. It is now shown that the mu tual inductances M 12 and M 21 are identical for linear systems, namely (5-115)

with the symbol M chosen to denote either parameter. That (5-115) is true for a linear system is demonstrated on reexpressing (5-113a) in terms of the volume integral of B . H by use of (5-77)

Um

= ~2 Jv f B· Hdv

[5-77]

This result, derived for the single circuit of Figure 5-24, is equally valid for the coupled circuits of Figure 5-31. Suppose Band H of the coupled system are expressed as the sums B = Bl

+ B2

H = HI

+ H2

(5-116)

in which Bl V X Al ,uH1' B2 V X A2 = ,uH 2 , whence B 1 , Hi are taken to be due to II in circuit I, while B 2 , H2 arc proportional to 12 in circuit 2. Then (5-77) expands into the four terms

in which the integrations are to be taken throughout all the space where the fields B and H exist. A comparison of the four integrals in (5-117) with those of (5-113a) reveals


321

a one-to-one energy correspondence, implying that the self and mutual inductances defined in (5-114) can also be written

(5-118a)

(5-118b)

(5-118c)

, (5-118d)

but in the latter, the product Bl . H2 equals B2 • HI because (5-119) Thus (5-1 18c) and (5-118d) are identical, proving (5-115), that M12 = M 21 . Combining (5-114) and (5-115) into (5-1 13b) permits writing the total magnetic energy in the form

(5-120)

Hence, a knowledge of the inductance parameters and instantaneous currents determines the magnetic energy state of coupled circuits at any instant. Since the selfinductance expressions (5-118a, b) have already been considered in detail, the expressions (5-118c,d) concerning the mutual inductance M will occupy the attention of the remainder of this section. For coupled circuits in free space, M can be expressed by a volume integral in terms of the current sources, yielding a result resembling (5-75) for self-inductanee. Hence, substituting (5-28a) for A into (5-118c) or (5-118d) obtains

M21

= M=

I

1112

11 Vi

V2

lloJ" J v, dvH --d 4rcR

Free space

(5-121)

with primes again used to distinguish the souree point current element J'dv' from the unprimed field point element as in (5-75). In Figure 5-32(a) is shown the geometry

322


(a)

Current fiiamentt'2 linking 1{12

\

S2

(Cross- section) (b)

Circuit

(1

t~ (c)

FIGURE 5-32. Generalized coupled-circuit configurations pertaining to mutual energy and inductance calculations. (a) Linear coupled circuits in free space. (b) Linear coupled circuits in general (iron present or not), showing the portion tfr12(t'~) of the flux of I, liuking current filament t~. (e) Special case of (b): thin circuits. Depicting portions tfr12 (If:ft) and tfr21 (r(lflll) of the fluxes of I, and 12 ,

relative to the integrations. The Neumann integral (5-121) is not discussed further here; refer to other sources for applications. 20 More general expressions for M can be derived from magnetic flux and current linkage interpretations of (5-114c) and (5-114d), to include the effects of magnetic materials. Subdivide circuit 2 into closed current filaments t~ carrying the differential current di as in Figure 5-32(b), each linking a portion t/J12(t~) of the flux of circuit 1. 20See R. S. Elliott, Electromagnetics. New York: McGraw-Hill, 1966, p. 309.


323

Equation (S-114c) for the mutual energy Um12 then becomes

a result that follows on noting that Ai . J dv = Ai . (atJl dt'ds Ai' dt" di, andobserving from (S-47) that Ai . dt" denotes t{! 12(t2), the portion of the flux of Ii linking t 2. Thus Um12 is found by integrating t{!12(t'2) di over the cross section S2 of wire 2, as depicted in Figure S-32(b). Similarly, (S-114d) becomes

f

I

(S-122b)

The use of the flux linkage expressions (S-122a, b) is facilitated by assuming Ii and 12 to be concentrated along the wire axes. Then t{!12(t2) and t{!21(t'1) in (S-122a) and (S-122b) become constants, yielding the simpler results

(S-122c) (S-122d)

in which

t{! 12 = the portion of the flux of Ii linked by circuit 2 t{! 21 = the portion of the flux of 12 linked by circui t 1 The simplifications (S-122c) and (S-I22d) are excellent approximations if the circuits are thin, as depicted in Figure S-32(c). The mutual inductance M is finally obtained by substituting the energies (S-122) into the definitions (S-llSc) and (S-11Sd), making use of M12 = M21 = M of (S-IIS); thus

M

Exact

(S-123a)

For thin circuits

(S-123b)

324


EXAMPLE 5-19

The latter approximations (5-123b) are usually acceptable in practical mutual inductance calculations.

EXAMPLE 5-19. Find M for the iron core toroidal transformer illustrated, the windings having n 1 and n2 turns and assuming no leakage flux. Compare M2 with the product LIL z . M for thin coils is conveniently found by use of (5-123b). For II in t l , the core flux obtained in Example 5-2 is

but t/I 12 linked by t2 (i.e., passing through obtaining from (5-123b)

Sex.2

bounded by

t 2) is

n2

times

t/lm,core,

(2)

The same answer is obtained using M = t/lzl/I2 , The se1f:'inductances of the coils, from Example 5-17, are (3)

Thus the product LIL2 equals the square of M given by (2). This is expected for coupled circuits whenever all the magnetic flux links each turn of the windings.

The idealization that all the magnetic flux produced by one circuit completely links the other, as in Example 5-19, is never quite attained in practice, even when high-permeability cores are used to minimize flux leakage. There is invariably some leakage, as depicted in Figure 5-33(a), causing M2 to be less than L I L 2 . This circumstance is expressed by the so-called coefficient of coupling between circuits, symbolized


(a)

325

(b)

FIGURE 5-33. Magnetic coupling between circuits yielding high and low coupling coefficients.

(a) Iron core transformer with small leakage (k .... I). (h) Circuits coupled in air, Jar high-frequency applications.

by k and defined

M k=-.JL 1 L 2

(5-124)

The latter permits expressing M as a function of the self-inductance of each circuit whenever k is known; that is, (5-125) The maximum value attainable by k is unity, while for circuits totally uncoupled, k = O. If coils are coupled using high-permeability cores, k may have a value as high as 0.99 or better, though with air as the coupling medium as in Figure 5-33(b), a much smaller k is usual, in view of one circuit linking a correspondingly smaller fraction of the total self-flux of the other. The circuit model of coupled circuits can be deduced in the same manner as for single circuits. Since a pair of circuits is involved, two Kirchhoff voltage relations are desired. Three interrelated methods can be employed to obtain the Kirchhoff voltage equations: (a) a method based on the scalar and vector potentials I]) and A of the electromagnetic fields, described in Section 5-10; (b) a technique based on energy considerations, treated in Section 5-11, part F; and (c) an approach making use of the Faraday law, (3-78). The Kirchhoff voltage equations of coupled circuits are derived from application of the Faraday law, (3-78)

J: E. dt = _ dt/Jm ~

dt

[3-78]

to the closed paths tl and t2 defining the circuits. In (3-78), E denotes the total field existing at the elements dt of the paths tl and t 2 , with o/m the total flux intercepted by each circuit-flux generated by both 11 and 12 , To help visualize this process, in

326


(b)

(a)

FIGURE 5-34. Se!f- and mutual fluxes produced by 11 and 12 in coupled circuits. and emfs induced. (a) Flux of 11 only. The self-flux l/i I links t p inducing

~ ~tJ The mutual flux

1/112

=

IS

2

El 'dt

(1)

BI . ds links t 2 , inducing (2)

(b) Flux of 12 only. The self-flux l/i2links /:2, inducing

~ E . d/:

:Yt2 The mutual flux

t/121

=

= _dl/i2

2

Is! B2 . ds links t

1,

(3)

dt

inducing (4)

Figure 5-34 are shown the separate fluxes of 11 and 12 , Only one independent voltage source V(t) is used. The senses of 11 and 12 are arbitrary, being assumed as shown. Figure 5-34 shows that the total E generated along the closed path of circuit II and appearing in the left side of (3-78) consists of three contributions: a field El induced along II by -dl/1t/dt, in which 1/11 is the self~flux linking the circuit II and due to the current 11; another field E21 induced along II by -dl/121/dt, in which 1/121 is the "mutual flux" linking II and produced by 12 ; plus the generated field Eg produced only within the independent voltage source V(t). Thus, the total E· dl contribution to the integrand of the left side of the Faraday law (3-78) becomes E . dl = (E1

+ E21 + E 9 ) . dt = ! . dt + E 9 . dt (J

(5-126)

in which the current density J is that induced in the conductor via (3-7) and by the continuity of the tangential portion of the total electric field Econd = El + E21 appearing at the conductor surface.

5-l2 COUPLED CIRCUITS AND MUTUAL INDUCTANCE

(e)

(b)

(0)

327

FIGURE 5-35. Magnetically coupled circuits and circuit models. (a) The physical coupled circuits, with assumed current directions. (b) Circuit model showing elements corresponding to terms of (5-128). (e. Circuit model using symbolic convention to denote circuit self-indnctances.

The right side of the Faraday law (3-78) concerns the two magnetic flux contributions t/lm = t/ll + t/l21 linking the surface SI bounded by the circuit tl as shown in Figure 5-34(a) and (b). With this and (5-126), (3-78) finally becomes

~

:Yt,

l. dt + (J

j( +) E . dt J(-)

9

= _ dt/ll _ dt/l21 dt

dt

(5-127)

At low frequencies, the leftmost integral of (5-127) becomes Rllb Rl being the resistance of the conductive path by the arguments of Section 4-14B. Thus (5-127) may be written

The fluxes t/ll = Is, BI . ds and t/lzl = Is, B z . ds linked by tl are the positive quantities t/ll = LIIl and t/lZl = M12 , since those fluxes emerge from the positive side of S1 bounded by tl in Figure 5-34. With these substitutions one obtains (5-128a) the desired Kirchhoff voltage relation for the circuit t l . Applying a similar line of reasoning to the other circuit, one obtains the desired Kirchhoff voltage relation for t z

o

(5-128b)

These coupled diflcrential equations correspond to the circuit model in Figure 5-35. The use of this model makes it evident, without recourse to field theory, that on removing Rv tor example, the open-circuit voltage obtained across gap terminals at cod is just M dlddt. Other features of coupled circuits from the point of view of this model are treated in standard texts on circuit theory.21 liSee, for example, S. 1. Pearson, and G. J. Maler. introductory Circuit Analysis. New York: Wiley, 1965, pp. 54-63.

328


5·13 MAGNETIC FORCES AND TORQUES Although the force acting on a current-carrying circuit in the presence of an external magnetic field can often be obtained by use of the Ampere force law, (5-45a), frequently it is more expedient to obtain it from the stored magnetic field energy. It is shown how the force or torque acting on a current-carrying circuit or a nearby magnetic material region is deduced from an application of the conservation of energy principle to a virtual displacement or rotation of the desired body. This process is analogous to the determination of forces or torques exerted on charged conductors or dielectrics in the presence of an electrostatic field, discussed in Section 4-15. Suppose the magnetic circuit of Figure 5-36(a), having an air gap of variable width x, derives its energy from the source V supplying a direct current to the winding. If the armature were displaced a distance dt at the gap due to the magnetic field force F acting on it, the mechanical work done would be (5-129) This work is done by V at the expense of the energy in the magnetic field such that the following energy balance is maintained

+ dU dUs dUm ~~ '-M-a-g-n~etostatic\ Mechanical Work done work done by source V energy change

(5-130)

The change in the magnetic energy, on changing the air gap in Figure 5-36(a), produces a corresponding inductance change. The magnetostatjc energy Urn is 1/2LP from (5-72), so the energy change occurring with 1 help constant becomes (5-131) Omitting the 12 R heat losses associated with the coil resistance in the equivalent circuit of this system depicted in Figure 5-30(c), the work dUs exerted by V to maintain (5-130) is done against the voltage induced by the flux change dl/l m in the time dt such

I

v-=-

(a)

(b)

FIGURE 5-36. Single circuits using magnetic cores subject to relative translation or rotation. (a) Armature translates. (b) Armature rotates.

5·13 MAGNETIC FORCES AND TORQUES

329

that V = -dljim/dt. With ljim = LI from (5-88a), and with I maintained at a constant value, the induced voltage becomes V = -dljimfdt = -ldL/dt. The work dUs done by the source in the time dt to overcome this voltage is therefore

dUs = - VI dt = 12 dL

(5-132)

which is just twice (5-131), the change in the stored energy. Combining (5-129), (5-131), and (5-132) into the energy balance, (5-130) thus yields (i)PdL+F'dt= [2 dL, reducing to F' dt = (1)P dL, or (5-133) The latter shows that the mechanical work just equals the change in the magnetostatic field energy. Thus, of thc electrical energy supplied by V, one-half goes to increasing the magnetic ener[jY of the system, whereas the other half is used up as mechanical work done by the magnetic force. The differential magnetostatic energy change dUm can be written in terms of the coordinate variations of Urn as the armature moves the distance dt = axdx + a y 4Y + a. dz if desired; that is, (5-134) gradient form allowable in view of (2-11). A comparison of (5-134) with (5-129), making use of (5-133), leads to the cartesian components ofF

(5-135a) Since Urn = (i)LP from (5-72), the force components with I constant can also be written in terms of the derivations of the self-inductance L as follows

I 2 0L

F=-y 2 oy

(5-135b)

To evaluate F, the magneLostatic energy Urn (or the self~inductance L) should be given in terms of the coordinates of the displaced element of the system. In Figure 5-36(a), for example, U m would be expressed in terms of the single coordinate x denoting the air-gap width. Suppose a portion of the iron core, instead of being translated, is constrained to rotation about an axis as in Figure 5-36(b). Then the differential work (with dU = dUm) done by the magnetic force in the angular displacement dO a1 dOl + a2 d0 2 + a3 d0 3 becomes (5-136) wherein T = a 1 T1 + a2 T2 + a 3 T3 denotes the vector torque due to the magnetic force. Then results analogous with (5-135a, b), in terms ofthe variations ofthe magnetic

330

STATIC AND QUASI-STATIC MAGN.ETIC FIELDS

energy with respect to angular changes, obtain as follows

(5-137a) and in terms of the variations in the circuit self-inductance with respect to the angular motions, one obtains

(5-137b)

EXAMPLE 5-20. A magnetic relay has a movable armature with two air gaps of width x as shown in the accompanying figure. The n turn coil carries a current 1 derived from the source V. The core and armature, both of permeability f-t, have the median lengths and cross-sectional areas t 1 , AI; t 2 , A 2 , respectively. (a) Find the expression for the magnetic flux, the magnetic energy stored, and the self-inductance of the system, expressed as functions of the gap width x. (b) Determine the force acting on the armature. Express this force in terms of magnetic flux in the air gap, and in terms of the air-gap Bav field.

(a) The core flux is obtained by use of the magnetic circuit methods in Section 5-3. The reluctances are 91 1 = tdf-tAl' 91 2 = tzlf-tAz, and that of the two air gaps in series is 2x/f-toA 1; whence

nl

t/lm,core =

------2-x91 1 + 91 2 + --_. f-toA l

(1)

L is well approximated by the extcrnal self-inductance (5-88a). The core flux passes n times through the surface Sex bounded by the coil, so that

L = nt/lm.core = 1 ----------2-x91 1 + 9l z +-f-toAl

(

I

I

1

I I L

I \

EXAMPLE 5-20

(2)

PROBLEMS

331

The magnetic energy of the system is therefore

(3)

It is evident that increasing the air gap results in a decrease in the core flux, the self-inductance, and the stored energy. (b) The force on the armature is obtained from (5-135a) or (5-135b); F has only an x component, as expected from the physical layout; thus

2x JloA [ [Jt 1 + [Jt 2 + JloA

J2

(4)

The negative sign means Fx is in the direction of deereasing gap width x, corresponding to an increase in magnetic energy. ''''ith the core flux expression (1), rewrite the air-gap force (4) as

F

x

With

!/Im,eore

=

BovA,

thi~

I = __

Jlo

A

.1, 2 'P m,core

(5-138a)

is also written (5-138b)

showing thc air-gap force to be proportional to the air-gap flux squared, as well as to the flux-density squared.

REFERENCES ELLIOTT, R. S. Electrornagnetics. New York:

McGraw~Hi1l,

1966.

LORRAIN, P., and D. R., CORSON. Electrornagnetic Fields and Waves, 2nd ed. San Francisco: Freeman, 1970. REITZ, R., and F. J. MILFORD. Foundations of Electrornagnetic Theory. Reading, Mass.: Addison~ Wesley, 1960.

PROBLEMS SECTION 5-1 5-1. From the divergence of the static diflerential Ampere law (5-2), show that the differential property of static current density (5-3) follows. Explain the physical meaning of (5-3). Show how (5-6) follows from (5-3), from an appropriate integration and by an application of the divergence theorem.

SECTION 5-2 5-2. In the figure is shown a toroid of permeability Jl

= JloJl" through which a long wire carrying the steady current I is coaxially threaded. (a) Making use of the symmetry, Ampere's and boundary conditions, argue why the same H field exists in the toroid as in the surrouuding air. Find B in the two regions. (b) With 1 = 10 A, Jl = 500Jlo, a = I cm, b = h = 2 cm,

332


I

I h

L PROBLEM 5-2

find Hand B to either side of the interface at the inner radius p in the toroid.

=

a. Determine the core flux

SECTION 5-3 5-3. A particnlar ferromagnetic core with an air gap is similar to that shown in Example 5-3. I t has a 5-cm 2 cross-sectional area, a median core length in the iron of 20 em, a 4-mm air-gap length, and is wound with a 200-turn coil carrying 0.1 A. The iron core has the constant permeability It = 5000lto. (a) Sketeh the analogous dc electric circuit and the equivalent magnetic circuit diagram, labeling the symbolic quantities that apply. (b) Calculate the reluctances of the iron path and the air gap. Find Bov and Hav values in each region. (c) At the iron-to-air-gap interface, which boundary condition (from Table 3-2) applies there? (d) Show that the Ampere integral law (5-5) is satisfied, by integrating H • dt about the closed median path. To which Maxwell law is the Ampere law analogous, but applicable to the electric-current analog? (e) If the air gap were missing and the applied mmfnI were the same, by what factor would Bov and the core flux increase? [Answer: (b.) Bav 6.22 mT]

5-4.

Suppose the toroidal magnetic circuit of Exam pie 5-3 had no air gap. With the dimensions and parameters as given, find the H field in the core along the median path (p = 5 cm) two ways: (a) using the magnetic circuit method; (b) using Ampere's law. Compare the answers. If this toroid has an air gap, explain why the Ampere law cannot be applied to find H directly.

5-5.

Given is the two-mesh magnetic circuit with an n-turn winding as shown in Figure 5-7 (b). Let the iron core /1 104 Ito and the coil wound about the middle leg carry 0.1 A with 80 tUfns. The median path length of the middle leg is t3 = 4 crn, whereas the outside legs have tl = 2 (2 = 12 em, with all cross-sectional areas fixed at 2 em . Sketch the schematic diagram of the magnetic circuit appropriately labeled, along with the analogous dc electric circuit. (a) Using the analogous circuit, employ simple circuit reduction methods borrowed from the analogous electric circuit to calculate the magnetic flux in each branch, neglecting leakage. Find Bav in each branch. (b) Find Hav in each branch. Check your solution by verifying whether Ampere's law is satisfied around one closed loop that includes the mmf source rd. [Answer: (a) ifJm3 = 0.201 mWbl

PROBLEMS

333

5-6.

Given is the same two-mesh magnetic circuit as in Problem 5-5, except that, additionally, a O.5-mm air gap is sawed through the middle branch t3' (a) What is the air-gap reluctance? Sketch the new analogous electric circuit, labeling appropriate quantities and their analogies. (b) Find the new value of current required in the n-turn coil to establish the same magnetic flux in each branch as was obtained for Problem 5-5. By what factor does the current need to be increased? Comment on the effect of the air gap. (c) If the air gap had instead been placed in the outer branch t 1 , comment qualitatively on its effects in this event. [Answer: (b) 1= 5.1 A]

5-7. Given is the two-mesh magnetic circuit of Figure 5-7(a), with the mmfsource nl wound on the outer leg t 1 • Sketch this system, along with a labeled schematic magnetic circuit. Assume the identical dimensions and parameters of Problem 5-3. (a) Repeat part (a) of Problem 5-5 for this new configuration. (b) Calculate Hap in each branch. Check your solution by verifying whether Ampere's law, of the form (5-20e), is satisfied around the closed loop defined by the branches tl and t 2. 5-8. A particular I %-silicon (Si) steel, useful in magnetic circuit applications, has the type of nonlinear B-H curve depicted in Figure 3-13(b). Only points on the virgin curve OP3 are considered lkre. (1'he hysteresis efleet is disregarded.) Tests on this steel show a curve having the (B, H) coordinates: (0.04,20), (0.13,40), (0.24,50), (0.39,60), (0.53,70), (0.63,80), (0.76,100), (0.87,125), (0.95,150), (1.06, 200), (1.19, 300), (1.25,400) in mks units. (a) Graph this B-H eurve on linear graph paper with reasonable care. (b) With fl, defined by B/floH, calculate the static fl and fl, values for each given point, and graph fl, as a function of Hover the given range. 5-9. The gapless toroidal ring shown is made of the Si steel described in Problem 5-8, with R = 10 em, r = 2 cm. Let the current in the 100-turn winding be 1.257 A. (a) Use (5-20e) to find Hav in this core. Find also Bap and the magnetic flux in the core. Employ the B-H characteristic given in Problem 5-8. [Answer: I/Im = 1.33 mWb] (b) Use answers obtained in (a) to deduce tbe values of fl and fl, of the core at its operating point. Find the reluctance of this magnetic core. Making use of the latter, check the value of the core flux obtained in (a). (c) Explain why the use of (5-20c) would have been unsuitable in part (a). 5-10. A toroidal magnetic circuit with an air gap has dimensions the same as those of Example 5-3. The core is made of the Si steel described in Problem 5-8. (a) Suppose that the maximum magnetic density Bav at which this device is to be operated is 1.06 T. Determine the corresponding core flux, the field Hav established in the steel core and in the air gap, and the mmf drops across the two regions. What mmf is required of the 100-turn coil to produce the desired Bav? What coil current? (b) If there were no air gap, what coil current would then be needed? Comment on the effect of the air gap on the required driving current to produce a desired Bap in the magnetic core. . 5-11. In the toroidal magnetic circuit with air gap of Problem 5-10, assume 1= 10 A flows in the 100-turn coil. Find Bav and the core flux. [Hint: Since neither (5-20c) nor (5-20e) is amenable to a direct solution for Bap , assume as a first approximation that the applied mmf due to nl is entirely across the air gap only, using successive approximations to find Bap from the B-H graph of Problem 5-8.]

PROBLEM 5-9

334

STATIC AND QUASI-STATIC MAGNETIC FlELDS

SECTION 5-4 5-12. Given a very long, round conductor of radius a carrying the static current I in free space and that its exterior B field is aq,1l01/2np, use (5-22) as the basis for finding the potential A outside the wire. [Hint: Expand (5-22), noting it has only a z-eomponent and that its %z operator is zero (why?). Integrate the resulting differential equation to obtain 1101

tnp+C

(p?:a)

2n

If desired, put the arbitrary potential reference (where A z

= 0)

at p

= a to eliminate C.J

5-13. Repeat Problem 5-12, but this time find A inside the wire, given that B there aq,llolp/2na2 Show that

IS

To what docs this result reduce, if the wire surface is taken as the potential reference?

SECTION 5-5 5-14. A finite length of this wire, in air, carries the static current I and lies on the z-axis as in Example 5-4, except it is displaced so that its lower end is at z = L J and its upper end is at L z . Sketch and label it. Find the vector magnetic potential A at any location P(p, 0, 0) on the p-axis by integrating (5-28c), showing that

5-15. (a) In Example 5-5 concerning the small curre'nt loop, show the details of inserting (5-32) into (5-22) to obtain B of (5-33). (b) Comment on the duality existing between the B field (5-33) of the current loop of Figure 5-10 and the E field (4-44) of the electric dipole charge of Example 4-8. How do their field sketches compare? The strength of the electric dipole moment in (4-44) is qd. Recalling the definition (3-53) of the magnetic moment of a current loop, what is the "magnetic dipole" moment inferred from (5-33)?

°

5-16. A square loop of thin wire centered in the z = plane and of sides 2a parallel to the X,] axes in air carries the current I flowing counterclockwise looking from the top. Sketch this geometry, and show details of how the Biot-Savart law (5-35b) is used to obtain the B field at P(O, 0, 0), yielding

1l0J21 B(O, 0, 0) = a z - - 11.a Make use of symmetry to show that integration along only one side of the loop is needed.

5-17.

(a) Show that B along the z-axis of the thin, square loop of Problem 5-16 is given by B(O, 0, z) = a z

11.(z

2

+

21loIa2 2 1/2 Z 2a) (z

2

+a )

\

[Hint: Make use of results of Example 5-4, if desired.] (b) To what result does this reduce at the center of the loop? (See Problem 5-16.) If 1= 10 A, a = 1 ern, find B(O, 0, 0). (c) Show, as z becomes sufficiently large, that B at great distances falls oft' as the inverse cube of the distance.

°

5-18. A thin, circular loop of thin wire centered in the z plane is of radius a and carries the current I (going couHterclockwise seen from the top) in air. Sketch it. (a) Use a direct inte-

/

PROBLEMS

335

: :

free Iltial

a/az

t

d[JU con~~~tlng b

I(t)

'~1

t

IS

I(t)

i-'- Viti I I

:e

t

(a)

I I I

I I

I I I

(b)

I'ROBLEM 5-19

ofthc Biot-Savan law (5-35b) to show that B along the

as is at the

IS

z axis

is given by

what result does this reduce at the loop center? Find B there iff = lOA, a = I em. Iff = lOA, 10 cm. (b) Show that this B field agrees, as the distance from the loop is made large, with f()!' the B field of a small loop.

SECTION 5-7 5--UI. ting 'ield e of lent

That the this j at

I

A highly conductive wire loop, of the rectangular dimensions as noted, is placed in the ('ommon plane of a nearby long wire carrying the current I(t) = 1m sin rot as shown in (a). What (quasi-static) B field is produced by the current'? (b) Use the Faraday law (5-41) to the open-circuit voltage V(t) at the loop gap. (Show on a sketch the direction orB on the bounded by the loop and the choice of a positive surface element.) What is the polarity at the gap? Explain. If 1m 10 A, f 20 kHz, d 4 mm, a b 10 em, find V(t), its polarity. (c) Repeat (b) for the parallel-wire system of figure (b), making use of ~ymmetry.

5--20.

A high-11 magnetic toroid has a rectangular cross section as shown, and is wound with n-turn coil carrying the current I(t) 1m sin rot. A one-turn secondary loop of wire embraces core as shown. (a) Use Ampere's law to deduce the quasi-static B(p, t) field in the toroidal Find the "core flux. (Sketch the flux in a side view of the system, noting its direction in relation to the positive current sense.) (b) Use Faraday's law (5-41) to deduce the open-circuit V(t) at the gap of the secondary loop t 2 . If a = 1 em, b = 3 em, d = 2 em, n1 = 150 turns, kHz, core 11 = 4000Jlo, and 1m = 2 A, find t/lm(t) and V2 (t). Label the polarity of V2 (t) thc gap, explaining your choicc.

by

-(z)

the be1ce. nes Ilte-

('ROBLEM 5-20

. E7

II

336


I

I I I 1'1

It

"I

T I l'--------'~

!

V(t)

p

1 I

I I 1 I

PROBLEM 5-21

SECTION 5-8 5-21. The long, straight, round wire shown carries the static current 1. The thin, rectangular loop shown is located with its nearest side at the distance p from the wire center. The rigid loop is moved radially away from the long wire, all points on the loop moving at the velocity v apv" relative to the wire. Use (5-44d) to determine V(t) induced at the loop gap, including its polarity and the reason for your choice. (Sketch the system, labeling typical v, B, and v X B symbols thereon, as required by the integration.)

SECTION 5-10 5-22. In Figure 5-22(c) assume, in the end view of the simple generator shown, that the radial magnetic field has the constant Eo magnitude in the gap over a ± 60° angular interval measured {i'om the vertical, and is zero outside the gap. Use (5-44d) to derive the motional voltage V(t) generated by the rotating coil across its open terminals, assuming the coil has n turns. (Show its polarity on a sketch, justifying your choice.) Show that V(t) = 2B odawn. If Bo = 0.3 T, d = 12 cm, a 4 cm, n 20 turns, and the r'otor is spinning at 50 revolutions per second, find V(t).

SECTION 5-11C 5-23.

(a) Make use of (5-77) to show that the magnetic energy stored in the toroid of Problem 5-20 is U m = (J.ldn 2 J2/4n) tn (b/a). Deduce its self inductance therefore to be L

/.ldn

2n

2

R b ,n-

a

and compare this wi th the result obtained in Exam pie 5-17 by the flux-linkage method. (b) For the toroid with dimensions as given in Problem 5-20(b), find its magnetic energy if J = 2 A, and its self-inductance. Under what condition would the self~inductance be a function of the current in the device? 5-24. (a) Find the magnetic energy stored in the toroidal inductor of Examplc 5-3, using average magnetic field values. What percentage of the total energy is stored in the air gap? What is the self-inductance? (b) Repeat the energy and inductance calculations of (a), but for no air gap in the core. Comment on the comparative results. 5-25. Determine, from results obtained in Example 1-17, the magnetic energy stored in a length d of a very long solenoid in air, with n/d closely spaced turns per meter. Show that its selfinductance per meter, L/d, is /.lonb 2 (n/d)2. For a long solenoid with b = 3 em and 10 turns per centimeter, find its inductance per meter.

PROBLEMS

337

PROBLEM 5-31

5-26.

(a) For the coaxial line of Example 5-13, verify the results (I), (2), and (3) obtained for its internal and external inductances, giving ample details. (b) The expression (5-83) is sometimes used for the inductance of a length t of the coaxial line. Under what condition(s) would this result be accurate?

SECTION 5-11D 5-27. For the toroidal inductor of Example 5-3, use the external flux linkage to lind its selfinductance. With no air gap, by what factor docs its inductance increase?

5-28.

Find, using the flux linkage method, the expression for the self-inductance of every length d of the very long solenoid in air of Example 1-17. Check the result with that given in Problem 5-25.

51.29.

For the two-mesh magnetic circuit with parameters as given in Problem 5-5, it was found that 0.1 A in its 30-turn coil produced 0.201 mWb of magnetic flux through the coil. Find its external self~inductancc, using the flux linkage method.

5-30.

In the two-mcsh magnctic circuit with an air gap, as described in Problem 5-6, it was found that the coil current of5.1 A produced the magnetic flux of 0.201 mWb through the coil. the flux linkage method to find the coil self-inductance. Neglect internal inductance.

5-31.

The toroidal magnetic core of circular cross section has a coil ofn turns as shown. Ncglectthe winding intcrnal inductance and the flux leakage and assuming the iron permeability p to be constant, use the flux-linkage expression (5-83a) to determine the approximate selfinductance. Use magnetic circuit methods to determine the core flux. Show that L = 1lT/ 2r2/2R. If Ilr 10 5 , n 50, r = .5 mm, R 3 em, find L. 5-32.

Rework Problem 5-31, this time employing Ampere's law to find the exact expression

H in the core, whence deduce the core flux from the integration ofB . ds over the core cross lIection. From this, deduce the external self~inductance by usc of (5-33a) and flux linkages. Calculate L for the values given in Problem 5-31. 5-33.

A wire circuit is threaded through a small toroidal low-loss ferrite bead of permeability shown. How much self-inductance is added to the circuit? [Hint: Reason that the H field or without the bead is essentially the same. The fields within the bead (sec enlarged figure) essentially those for the straight-wire Problem 5-2.]

I'ROBLEM .5-33

338


= 0.5 em

I

<:!;50.5cm I

Loop of (a)

0.1 mm

(b)

(a)

PROBLEM 5-36

SECTION 5-11E 5-34.

Employing the elliptic integral approximations (5-99), provc (5-100) for the external inductance of a circular loop of wire.

5-35.

Add (5-100) and (5-101) to obtain the expression for the self-inductance of a circular wire loop in air. Use the result to calculate the low-frequency and high-frequcncy inductances (explain the difference) of a loop of nonmagnetic wire 4 mm in diameter, forming a IO-cm diameter circle. Is the internal inductance negligible in the low-frequency case?

5-36.

(a) The wire loop in figure (a) has the dimensions shown. Calculate by use of (5-100) and (5-101) its self:inductance. Assuming low-frequency operation, what percentage of this is internal inductance? (b) Now wind the wire loop about the iron core as in figure (b), with the mean radius R = 1.5 cm and the cross-sectional radius r = 5 mm. Assume no leakage flux and J1. = 5000J1.0 for the core. Determine the factor by which L increases ovcr its free-space value in (a). Comment on the effect of the closed, high-permeability magnetic path. Is internal inductance of importance here?

SECTION 5-12 5-37.

Beginning with (5-65), prove the result (:>-107) for the power delivered to coupled circuits. [See thc proof of (5-66) for a single circuit.]

5-38.

From the expression (5-110) for the magnetic energy of coupled circuits, derive (5-111) for linear circuits. [Hint: Observe how the linear result (5-71) was obtained from the general expression (5-70) for a single magnetic circuit.]

5-39.

Usc (5-121) to deduce thc Neumann formula for two thin circuits in free space M=

11 {,

12

J1. o dt'· dt ...._ .4nR

Sketch a pair of circuits with labeling appropriate to the usc of this intcgral.

5-40.

Use the Neumann formula for thin circuits given in Problem 5-39 to derive the mutual inductance between two coaxial, circular loops with radii a and b, and separated by the distance

PROBLEM 5-40

PROBLEMS

339

free space as shown, obtaining

which K(k) and E(k) are the complete elliptic integrals (5-97), and k=

Proceed along lines suggested by Example 5-18, noting that the distance between a source P' and a field point P is

5-41. Given a fixed circuit tl in free space as shown, suggest, with respect to the flux-linkage definition (5-123b) of M, how the mutual inductance varies with respect to the second circuit on relocating it according to the three cases illustrated. Explain briefly, showing roughly the extent to which the flux of 11 (in t 1 ) links t z . 5-42. Suppose a second coil tz with Tt2 = 250 turns is wound on the iron core with an air gap, described in Example 5-3. Employ flux linkage methods to determine the self-inductance of each winding. Find the mutual inductance between these windings two ways: (I) by usc of the flux linkage result (5-123b); and (2) using (5-125), assuming zero leakage flux in this system.

~-43.

(a) In the coaxial coupled circuit system (in air) of Problem 5-40, assume the radius b of circuit t2 to be small compared to a, the radius of circuit t 1. Then the current 11 in tl would produce an essentially uniform B field over the smaller circuit t z . Using the solution to Problem 5-18 for Bl along the z-axis, show that the mutual inductance between these circuits is essentially J1.o1f.(ab)z/2(a 2 + d2 )3/2. (b) Find M between these circuits if a = 12 em, b 2 em for two cases: (I) if d = 20 cm (coaxial circuits), and (2) if d = 0 (coaxial and coplanar). Let the wire diameter be Imm. (c) If 11 = 10 A flows in circuit iI' find the magnetic flux 0/12 linking the second circuit. (If 12 were 10 A, then from (5-123b), how much flux 0/21 would link the first circuit?)

5-44. Make use of the inductance expressions (5-100) and (5-\ 0 1) for a circular wire loop to determine the self inductance of each of the two loops with dimensions as given in Problem 5-43(b). Usc these and the value of M to deduce the coupling coefficient k for both circuit separations d 0 and d 20 cm. 5-45. (a) For the same rectangular circuit near a long, straight wire in air as shown in figure (a) of Problem 5-19, find the expression for the mutual inductance between the two circuits. Sketch this labeled system. Find the value of M, using the dimensions given in Problem 5-19. (b) If ll(t) 10 sin Wi, make use of (5-123b), 0/12 MIll to find the amount of flux 0/12(t) linking the rectangular circuit having the given dimensions. (c) Use the Faraday law (5-41) to

(a)

(b)

(c)

PROBLEM 5-41 (a) Coaxial circuits. (b) Coplanar circuits. (e) Coaxial and coplanar circuits.

340


PROBLEM 5-46

deduce the open-circuit voltage V(t) (ineluding its polarity) appearing at the gap in the rectangular circuit at the frequency f= 20 kHz specified in Problem 5-19. (Identify the flux l/l", in (5-41) here as precisely 1/112' the flux produced by 11 and linking the circuit whcre t/1l2 = MIl' Evaluate V(t) making usc oftlle latteL) 5-46. Clamped firmly about the long, straight wire shown is a split toroidal core of permeability I' and the given dimensions, with n turns wound about it. The long wire carries the currentl j (t) 1m sill wt. (a) Based on thc flux produced in the toroidal core, obtain an expression for the mutual inductance betwecn circuits tj and t2 using the flux-linkage definition (5-123b). (Note tbat the flux t/112 linked by t z , that is, passing through the surface Sex,2 bounded by t z , is rl Limes the core flux.) (b) Find the value of Nt, if a 5 nnll, b = 1.5 cm, d = 3 cm, n = 200, 11 (t) = 50 sin wt A at the frequency f = 60 Hz, with I'r 5000. (c) For the values given in (b), use the Faraday law (5-41) to obtain the open-circuit voltage (including polarity) at the terminals of circuit l2' Do this two ways: (l) by usc of (5-41), or V -dl/l 12 /dt; and (2) making use of 123b) to express the flux t/l 12linked by l2 as t/1 12 M I], yielding V 2 (t) d(MItl/dt -Mdl l /dt.

SECTION 5-13 5-47. In Example 5-20, let n 150, 1 0.2 A, tl 10 cm, t2 5 em, 5 cm 2, A2 I cmz, gap x 1 mm, and I' = 80001'0' (a) Find the core flux, the densities Bav in the U-shaped stator and in the armature, and the force on the armature at the given gap width. (b) Repeat for the gap closed. 5-48. The hinged, movable iron armature provides a variable air gap oflength x with respect to the fixed iron U-shaped stator shown, both having the same cross-sectional area Ac- Assume that the small armature displacement x is linear translation. (a) Write the expression tor the core flux of this system, neglecting leakage. (b) Obtain an expression for the self-inductance of the coil, using the flux-linkage method. Find Ii'om this the expression for the magnetic stored

PROBLEMS

341

I I ~'-X

I I

t,.

r---------

I I I I

(Ill

I

.. I

(ILl

'- _ _ _ _ _ _ _ _

I I I I ---.I

PROBLEM 5-48

energy. (e) Determine the expression for the force on the armature, as a function of x. (d) If

t c = 12 em, Ac = 4 cm 2, x = 1.5 mm, I = 1.25 A, n = 200 turns, and J.l = 10 5J.lo (assuming linear iron), find the values of the core flux Eav and Hav in the iron and air-gap regions, the selfinductance, the stored magnetic energy, and the force on the armature. (e) [f the gap length x were reduced to 0.75 mm, by what factor would the force increase? If x were reduced to zero?

5-49. A magnetic relay has a rotating armature as in Figure 5-36(b). Label (as for the relay of Example 5-20) mean paths t 1 , t2 and cross-sectional areas A l , A z in the iron stator and armature, each of permeability J.l = J.lrPO' The air gap is produced by the small angle 0 = x/tz, x being the mean air gap length. Find expressions for the magnetic flux, self-inductance, stored energy, and torque, each in terms of the small angle 8.

,

(

CHAPTER 6 - - -_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __

Wave Reflection and Transmission at Plane Boundaries

This chapter is concerned with plane-wave boundary-value problems in one or two dimensions. The reflection from a perf(~ctly conducting' plane on which a uniform plane wave is incident is considered first. Replacing the perfect conductor with a lossy dielectric extends the problem into a two-region system, f()r which the wave transmitted into the dielectric is also of interest. The definition of wave impedance and reflection coefficient permits a systematic analysis of the multiple-layer problem, dealing with the reflected and transmitted waves excited by a normally incident wave. Next, a developmen1 of the Smith chart is discussed, with applications to the foregoing problems. Then the concept of standing waves and standing-wave ratio fiJr a lossless region is treated. The chapter concludes with a discussion of wave reflection and transmission at oblique incidence on a plane boundary.

6-1 BOUNDARY-VALUE PROBLEMS A boundary-value problem in electromagnetics is one involving two or more regions (separated by one or more intedaces) lew which solutions are desired such that (a) Maxwell's equations are satisfied by those held solutions in each of the regions, and (b) the boundary conditions discussed in Chapter 3 are satisfied at the interfaces. Examples are illust.rated in Figure 6-1. Figure 6-1 (a) shows a rudimentary boundaryvalue problem: a plane wave normally incident on a perfect conductor, yielding a reflected wave. In (b) is a two-region system separated by a plane interface. A given plane wave traveling in region I leads to the additional waves shown, such that the boundary conditions at the interf~tce are satisfied. In these problems, the given incident wave is presumed to originate hom an appropriate electromagnetic source (a generator) at the far left.

342

6-1 BOUNDARY-VALlIE PROBLEMS Region 2

Region 1

~

Reflected wave m~~n

{

- 7'

343

Region 2

Region 1 Incident

Transmitted

"-

/

-+-----,//

/

~

To sources of plane wave

/

Perfectly conducting plane boundary

-E-- To

sources of plane wave (b)

(a)

C~~

~ ~~\\ ;f\ \ \ \ \

(Region 2); Air

Monopolt;: Region 1

Voltage source

(d)

(c)

G;,,",'holo. . '

."

~z,

:vegUlde"" "'~~:'-"",Z

~ ~ 0'

-

~-

J:

"-

'"

-------tJli_____..... L

-~

Rectangular, hollow waveguide

(e)

Linear Biconical Biconical Spherical (thin) (fat)

(f)

I"IGURE 6-1. Examples of boundary-value problems in electromagnetic thCOIY, (Il) Reflection of a plane wave from a perfectly conducting plane. (b) Reflection of a plane wave from, and transmission into, a dielectric region 2. (el Monopole antenna at the earth's snrface. (d) Two types of conducting pairs, carrying waves from a generator to a load. (e) Two types of hollow waveguides, carrying waves from a generator to a load. (f) Four types of driven antennas in free spacc.

Whenever the source of electromagnetic energy is included in a boundary-value problem, you can say that you are discussing the complete boundary-value problem. If the reflected wave does not couple signiticantly with the generator, a discussion of the complete problem may not be necessary. [n Figure 6-1(c) is shown a three-region problem consisting of a driven monopole antenna source transmitting electromagnetic energy into the surrounding space (regio1l 2) and into the earth (region 3). In Figure 6-1 (d) and (e) are showIl other complete boundary-value problems involving generators (sources) driving waves down one- or two-conductor systems (waveguides or tr::msmission lines) to a load at the far end. Systems such as these are considered in Chapters 8 through 10.

(

344

WAVE REFLECTION AND TRANSMISSION AT PLANE BOUNDARIES

6-2 REFLECTION FROM A PLANE CONDUCTOR AT NORMAL INCIDENCE A fundamental boundary-value problem of electromagnetics involves the reflection of a normally incident uniform plane wave from a plane perfect conductor. Assuming a plane of infinite extent avoids edge (diffraction) effects, and with the simplification of normal incidence, the problem is reduced to two dimensions (t and z). The geometry is shown in Figure 6-2. The sources of the incident wave are assumed at the far left in lossless region 1. Assuming x polarization, the incident wave is given in the real-time domain by (2-121)

E; (z, t) = E:'

cos (wt

-/3z)

Vjm

(6-1)

letting the phase angle + = 0 for convenience, but the incident wave (6-1) alone cannot satisfy the tangential field boundary conditions (3-72) and (3-79) at the interface. One must add a reflected wave solution, its effect being such as to cancel the incident field everywhere on the perfect conductor at every instant t. This occurs only if the second solution has the same frequency and if its equiphase surfaces are parallel to the walL The only other independent solution of Maxwell's equations that meets these requirements is the negative Z traveling wave solution of (2-119)

E; (z, t) = E~ cos (wt

+ 13z + (P-)

(6-2)

The unknown amplitude E~ and phase - are found by applying the boundary condition (3-79). The details are more readily carried out in the complex time-harmonic form; hence, the sum of (6-1) and (6-2), in complex notation, takes the form of (2-115)

Ex(z)

+ E; (z) = E:'e- jpz + E~eiPz Vjm =

E; (z)

-

Sources

(z)

(or (:1z)

o

);

I I

I I

FIGURE 6-2. Reflection of normally incident planc wave from perfect conducting plane.

(6-3)

6-2 REFLECTION FROM A PLANE CONDUCTOR AT NORMAL INCIDENCE

345

The boundary condition (3-79), that the total tangential electric field must vanish at the surface of the perfect conductor, is written Ex(O) = 0; so (6-3) becomes 0 = +E;;', whence

E:.

E~ =

-E;;'

(6-4)

Thus total reflection occurs, with the reflected wave amplitude equaling the negative of the incident wave. Inserting (6-4) into (6-3), the total electric field at arry location to the left: of the conducting plane becomes (6-5) a result with a wave amplitude 2E~, just twice that of the incident wave. The dependence of (6-5) on z is unlike the traveling wave nature of either wave constituent in (6-3). 1t has instead a standing wave character, in view of the factor sin /lz. A graphical space-time sketch of this standing wave is facilitated on converting (6-5) to its real-time form by use of (2-74). Assuming the real amplitude E~, one obtains

[It(z)eiwtl = Re [ - j2E~ sin {3::: ejrot ] Re [e - j9002E~ sin {3::: ejwt 1 2E~ sin {h sin wt

EAz, t) = Re

(6-6)

A sketch depicting the dependence on Z at successive t is shown in Figure 6-3(a). The total magnetic field accompanying the electric field (6-5) is obtained directly by substituting (6-5) into Maxwell's curl relation (2-108). This was, in effect, already done in Section 3-6, however, in which it was shown in (3-98b) that magnetic field traveling waves are related to corresponding electric fields by the intrinsic wave impedanfi:e. Hence, to (6-3) correspond the two terms of the magnetic field

Hy(z) = II; (z) + H; (z) e- jf!z

Em

eifJ z A/m

(6-7)

1J in which 1J == (Il/E) 1/2 is, from (3-99a), the intrinsic wave impedance of the lossless region. If (6-4) is inserted into (6-7), the complex magnetic field reduces to

2£"+ m

1J

cos

IJz

(6-8)

The real-time form of (6-8) (with f;~ taken to be the pure real E~) becomes (6-9) another standing wave. It is plotted in Figure 6-3(b) for comparison with the electric field. A space phase shiji of 90° occurs hetween the peaks of the electric and magnetic field standing waves, with the maximum magnetic intensity appearing at the perfectly conducting surface z O. The magnetic field (6-9) cannot fall abruptly to zero on passing into the interior of the perfect conductor without inducing an electric swface current, predictable trom

346

WAVE REFLECTION AND TRANSMISSION AT PLANE BOUNDARIES (/J., t, "1

= 0)

(/J., t, "I = 0)

z

z

z

(a)

Region 1: (/J., t, "1

(b)

= 0)

( y)

(c)

FIGURE 6-3. Standing waves resulting from a plane wave normally incident on a pcrfect conductor. (a) Incident, reflected, and total electric fields. (b) Incident, reflected, and total magnetic fields. (e) Showing the vector electric and magnetic fields of (a) and (b).

the boundary condition (3-72). Observe that the induced surface current density J. is x directed and cophasal over the conducting plane as shown in Figure 6-3(c). One can see a close physical analogy between the electromagnetic standing waves of Figure 6-3 and the mechanical standing waves of displacements and tensions along a transversely oscillating string anchored at one end! as shown in Figure 6-4(a). In (b) is shown another example of standing waves resulting from the reflection of electro1 For example, sce D. Halliday, and R. Resnick. Physics for Studmls Wiley, 1962, p. 412.

~f

Science and Engineering. New York:

6-3 TWO-REGION REFLECTION AND TRANSMISSION

~ .. .'

Incident wave_ ~ Reflected wave

.....

Electromagnetic transmitting horn

Region far frorn horn: spherical waves nearly plane 1

1

1 1

I

------..

I

t=-~/ --~----. ~-.::-::~::;:-; l " Vibration source (wave generator)

347

):

,..+'"

I I ::;:::

I

J...--. .: _""'

.. , '\ ~ '. / Generator (a)

(b)

fIGURE 6-4. Experiments involving standing waves. (a) Standing waves on a string connected to a rigid body and a wave generator. Null locations are checked visually. (b) Electromagnetic standing waves ncar conducting plane. Waves may originate !i'om a distant source as shown. A neon bulb reveals maxima and nulls.

magnetic waves from a conducting plane. Although the waves emanating from the horn are essentially spherical in the vicinity of the horn, at suitable distances away and over a limited transverse region they are very nearly plane waves, so that the solutions (6-5) and (6-8) are applicable in the vicinity of the plane reflector. If sufficient power is available, a small neon bulb might be used for detecting the nulls in the electric-field standing waves, yielding a rongh measure of wavelength. 6-3 lWO-REGION REFLECTION AND TRANSMISSION

The wave problem of Figure 6-2 can be generalized by assnming region 1 conductive ((j, i= 0) instead oflossless, and region 2 with a finite conductivity instead of being a perfect reflector. The system is shown in Figure 6-5. An incident plane wave originating from the far left is given by the positive z traveling wave terms of (3-9Ib) and (3-98c)

j;+xl (7) = j;+ml e- Y'z ""

(6-10)

wherein fj 1 is specified by (3-99a) f()r conductive region I or equivalently by (3-111). The propagation constant of region 1 is 1'1> given by (3-89) (6-11 )

which ex and p are obtained from (3-90a,b), or equivalently from (3-109) and (3-110). The continuity of the tangential fields across the interface in Figure 6-5 (a) gives lise to another plane wave at the same frequency in region 2. This wave is not sufficient to satisfy the boundary conditions (3-71) and (3-79) at the interface, however. One more wave, reflected in region I, is required if the boundary conditions are to be met. The three waves are shown in Figure 6-5(a) in real-time, and as complex vectors ill Figure 6-5(b). Thus, in region 1, the reflected wave is required as follows

111

(6-12)

348


-

Motion

Transmitted: A" +

'~+

Ex2

=

71c2Hy2

(a)

E;'(Z)

l

z

0

Wave • -- - -- motion

l

l

• - - -..-. Motion

, Motion ....;---- •

E:Z(Z)

o

EX](Zi

- - - (x)

if;;. (z) = (b)

FIGURE 6-5. Plane wave normally incident on an interface separating two lossy regions. (a) Incident and reflected waves in region I, transmitted wave in region 2. (b) Vector representations denoting the fields of (0).

in which

iiI and Yl

are given by (3-99a) and (6-11). In region 2, the transmitted wave

IS

( ) - E"'+ -nz E"'+ ~'x2 Z m2 e

(6-13)

No reflected wave can exist in region 2, because that region is infinite in extent toward the right in Figure 6-5, whereas the only sources of the fields are to the far left in region 1. Satisfying the boundary conditions at the interface in Figure 6-5(b) requires setting the total tangential fields equal to each other at Z O. In region 1, the total electric and magnetic fields are given by the sums of (6-10) and (6-12) (6-14 )

349

0-3 TWO-REGION REFLECTION AND TRANSMISSlON

Tn region 2, they are simply (G-13) The boundary condition (3-79) requires the equality of the electric fields of (G-13) and (6-14) at z = 0; that is, + e- Y1Z [E~.. . . m1

+ E~-1111 e = £';+ e-Y,Z] z=o .1m2 i1Z

(6-15)

obtaining (6-16) The other boundary condition (3-71) requires the continuity of the magnetic fields there, obtaining (6-17) Thc linear results (6-16) and (G-17) involve the known impedances fil and fi2 of the regions, as well as the com plex amplitudes of the incident.:. the reflected, and the transmitted waves. Assuming the incident wave to be given (E;:;1 is known), the other amplitudes are pbtained from the simultaneous solution of (6-16) and (6-17). Rearranging them with £;:;! on the right yields ~-

~+

Em! - Em2 =

it;;'1

~+

+

Em2

~+

Eml

(6-18)

~+

Em! fi1

(6-19)

Their simultaneous solution obtains the complex amplitude of the reflected wave (6-20) Similarly, the transmitted wave has the amplitude (6-21) Additional confidence is gained in the results (6-20) and (6-21) on considering two special cases: (a) for which region 2 is a perfect conductor and (b) for which regions 1 and 2Jlave identical parameters (no interface exists). In case (a), with fi2 = 0, (6-21) yields E;:;2 = O,~a result ez:pected from the null fields within a perfect conductor; while (6-20) obtains E;;'1 = - E;:;I, agreeable with (6-4) as one should expect. In case ide~tjcal regions means fil = fi2' whence from (6-20) and (6-21), it;;'1 = 0 and = E;:;I, implying the reasonable conclusion that no reflection occurs if the region has no discontinuity.

IXAMPLE 6·1. A uniIorm plane wave with the amplitude £;1 = lOOeW Vim in air is normally incident on the plane surface of a losslcss dielectric with the parameters Ji2 = Jio, E2 = 4€o, and (J2 = O. Find the amplitudes of the reflected and transmitted fields.

350


The geometry is sho""n in Figure 6-5. Region 1 is air, so iii = 110 = = 120n n. For region 2, liz = Jlo/4Eo = 60n n. The complex amplitudes of the reHected and transmitted waves are given by and (6-21)

J

60n 120n k';l1 = \00 ~~--~~60n + 120n

P;~2

=

I 00

_~~~On) 60n

These amplitudes into (6-13) and (6-

£.<1

= 1OOe - Jilt Z

_

33.3 Vjm

.. 120n

+

66.7 Vim

provide the total fields in each region

E~2(Z)

33.3e ifltz

10O_jPt Z 120n

Hy\(Z) - - ' - e

(-33.3)

_ 66.7 p Hyz(z) = 60n e- 1 2Z

ejPtZ

120n

66.7 e - ill,z

in which Il =jf1t =j(J)J;;~o and IZ j{J2 =j(jJJll0(4E~), the values of which can be inserted into the wave expressions once (jJ is specified. Observe that setting .z = 0 produces continnous tangential electric and magnetic fields across the interface, as expected.

6·4 NORMAL INCIDENCE FOR MORE THAN TWO REGIONS An extension of the two-region problem or the last section to three or more regions leads to a multiplicity or rellected and transmitted wave terms that, in the sinusoidal steady state, yield single f()rward- and backward-traveling plane wa~es in eac~ region. Suppose the three-region system of Figure 6-6(a) has the wave E;;A(Z) = E;:;Ae- YlZ impinging normally on it as shown. A study orlhe suhsequent phenomena in the time domain, after the arrival of the incident wave labeled A in Figure 6-6(b), reveals the generation of an infinite sequence of forward and backward waves in the system. Thus, two time-harmollic waves designated Band C are established successively in regions 1

Region 2:

Region 1:

I Region 3:

(fJ.2, f2, (12) 1(fJ.3, E3, (i3)

(fJ.b fh (11)

Region 1: I Region 2: I Region 3: (fJ.l. fl, (il) I (fJ.2, f2, (i2) I (fJ.3, f3, (i3)

-

Incident field: A

Wave motion

_0~0- ' ____ _ -L-\7.=~_ 0

-----i>-(z)

z= d

_ ____ Q. ______ d ____ i>-(z) BCD

~

//'"

Incident field:

G

E:A = 111Hy~

-:.: I I I

I I

Interface 1

Interface 2

etc.

--E

"F'~H

:

....;--

etc:--

etc.

I

(a)

1

--

-J

etc.

(b)

FIGURE 6-6. Three-region system on which a uniform plane wave is normally incident. (a) Three-region system, showing the plane wave field incident on a thickness d of region 2. (b) Depicting the effects of the incident field on reflected and transmitted waves, with increasing time.

6-4 NORMAL INCIDENCE FOR MORE THAN TWO REGIONS

351

and 2, the ((lfWard wave C in region 2 striking the second interhce to produce a transmitted wave D, plus another reflected wave E returning to interl'ace 1. A continuation of this process, as time increases, produces an infinite sequence of reflected and transmitted waves, the linear sum of which obtains sinusoidal steady ,I-tate forward- and backward-traveling waves in the respective regions, Thus, in region I, the net positive z traveling electric field will consist only of the postulated x polarized incident wave A, denoted by

while the reflected wave in that region consists (Jfan infinite sequence ol' contributions of the waves B, G, ... ; that is,

Each wave term of the latter has a common factor eY1Z , so that the infinite sum, in the sinusoidal steady state, becomes

(6-22) reducing to a net reflected wave in region I designated by eY1Z E~"ml

(6-23)

in which £:;1 denotes its eomplex amplitude. Every term of (6-22) has an associated magnetic field related by the intrinsic wave impedance of region 1, yielding

E~l -- e YiZ

(6-24 )

it

The net, sinusoidal steady state f()rward and backward waves in region 1 are depicted in Figure 6-7. Similar arguments applied to the infinite sequences of waves in regions and 3 lead to the net field vectors shown. Region 1:

Region 2:

(ILl, fj, ITj) or ('n, 1/1)

(1L2, E2,
E\+ :t2 =

h3,

~:l)

;;+

m2~ -"I2 Z

. -.,...

'+

o

Region 3: (1L3, Ea,
HY2

Ext = -.".'1 2

nGURE 6-7. Simplification of the multiplicity of reflected and transmitted waves of Fignrc showing the net plane wave fields.

352


The sinusoidal steady state wave solution of the three-region problem of Figure 1i-7, ~th a kno:vn incident field amplitude [;;;;'1' involves finding the amplitudes [;;;;'1' E;;'2, E;;'2, and E;;'3, a total offour unknowns. The four boundary conditions, involving the continuity of the tangential E and 11 fields at the interfaces, are sufficient to generate four linear equations in terms of these amplitudes. To illustrate the procedure for Figure 6-7, the material parameters (/1, E, (J) of each region are given, permitting ')' and q of each to be calculated by use of (6-11) and (3-99a). The depth d of region 2 is also specified. The total fields in the three regions are ~ () E xl Z =

E~+ -Y1 Z "ml e

+ E~-m1 eYlZ

(6-25) Region 1

(6-26) (6-27) (6-28) (6-29)

E\

~e-Y3Z

Region 3

q3

(6-30)

The boundary conditions (3-71) and (3-79) are satisfied by eq uating (6-25) to (6-27) and (6-26) to (6-28) at Z = 0, and equating (6-27) to (6-29) and (6-28) to (6-30) at Z = d. Then~the rearrangement of the resulting four simultaneous equations, placing' the known E;;'l on the right, yields

E'-m1 -

Em2

[;;;;'1

Em2 q2

-A-+ 111

~+

E'-m2

~+

E'-m2

Eml

q2

q1

e-Y2d E~+ m2

+ [;;-m2 e Y2d -

[;;+ e- nd m3

E'+m1

(6-31 )

~+

=

0

(6-32) (6-33) (6-34)

This is suitable for solution by fourth-order determinants or Gaussian elimination, but it is a tedious process, to say nothing of the higher-order results obtained when three or more interfaces are present. An alternative procedure is described in the next section.

6-5 SOLUTION USING REFLECTION COEFFICIENT AND WAVE IMPEDANCE The system of Figure 6-7 is generalized illto n ,.regions in Figure 6-{,1yExcited by the normally incident, time-harmonic wave (E;b H;1) in region I, each region a~cquiles, in the sinusoidal steady state, the forward- and backward-traveling fields (E.:, 11;)

• 6-5 SOLUTION USING REFLECTION COEFFICIENT AND WAVE IMPEDANCE

EXl

n - 1 I

2

Region 1: (f..i h tl, <11 )

-+

+

L.

l-->-

~~on

'+

L~

L~

. _rt>tion

- +

Hyn

yJ =-;::~l

.....

i-~

EXl

Exl

Mot i~lE;i

E=--

Region n:

I ( J..1.ill fn, un

Mk, tk, Uk

-+

.

.l

\\

\\

"

\\

~l

..l

I I

353

= z (or (:Jz)

..:l

.. j

I

I

I

I

I I

xl

~l

--To

sources

I I

I I

I I

J:?IGURE 6-3. A multilayer syslem ofn layers, ou which a uniform plane wave is normally incident from the left.

and (E';, if;) except for the last (k = n) region, in which only the forward-traveling components f;;;n, H:~ appear. The total electric field for each region 2 becomes

in which fez) is caIled the reflection coefficient at any location z in the region, defined the complex ratio of the reflected wave to the incident wave as follows

(6-36)

The corresponding total magnetic field is

E,'-'"

~e

E;

2yz

J

E,'+

_ -'me -YZ[l

q

f(z) I

(6-37)

total-field impedance Z(z) is defined at any Z location by the ratio of the total electric (6-35) to the total magnetic field (6-37)

(6-38)

these results apply to any (kth) region, an additional k subscript should be applied to all quantities. simplicity, such subscripts have been dropped.

354


A converse expression for ['(z) in terms of Z(z) is obtained from (6-38) by solving for ['(z)

['(z) =

~(z) Z(z)

fj

+ fj

(6-39)

a form convenient for finding ['(z) whenever Z(z) is known.~ Another useful expression is one that enables finding r at any location z' in a region in terms of that at another position z. At z', the reflection coefficient is expressed by use of (6-36): [,(z') = (E;;,/E:')e 2YZ '. Dividing the latter by (6-36) eliminates the wave amplitudes, yielding the desired result

(6-40)

In the application of (6-35) through (6-40) to the wave system of Figure 6-8, one should note the following properties of ['(z) and Z(z) at any interface separating two regIOns. 1. The total field impedance Z(Z) is continuous across the interface; that is, at an interface defined by z = a

(6-41 )

evident from the continuity of the tangential electric and magnetic fields appearing in the definition (6-38). 2. The reflection coefficient ['Jz) is discontinuous across the interface. This fol~ows from (6-39), for, because Z(z) must be continuous across the interface, r(z) cannot be if the wave impedance fj is different in the adjacent regions. The procedure for finding the complex amplitudes ofthe forward- and backwardtraveling waves in a multilayer system like that of Figure 6-8 is illustrated in two examples.

EXAMPLE 6·2. A uniform plane wave is normally incident in air on a slab of plastic with the parameters shown, a quarter-wave )hick at the operating frequency f = 1 MHz. The x polarized wave has the amplitude E~l = 100eN ' Vim. Use the concepts of reflection coeHicient and total field impedance to find the remaining wave amplitudes. To obviate carrying cumbersome phase terms across the interlaces, ass LIme separate Z origins 0 1 ,02> and 0 3 shown in (b) of the figure. The wave amplitudes are referred to these origins. First, values ofq for each region arcJ~und by using (3-99a); thus, fil fi3 .Jfto/Eo = 120n Q; in the plastic slab, q2 = .JJ4J4Eo = 60n Q. The propagation constants Y = rx + jf3 are computed from (3-90a,b) or (3-109) and (3-110); thus, in lossless region 2,

6-5 SOLUTION USING REFLECTION COEFFICIENT AND WAVE IMPEDANCE 1: Air (110, EO) I

!

Eii = 100e- Y1 " iI+

~~::n -+

yl

--..

(z)

~motion

Plastic

= Exl -

A

13 Air (110, EO)

I I

j:;+ x3

Ex2

L--

-+

HY2

l

0 1 O2

E;l

_

j,E;2 A

(~,

A

Motion -

......

Hy:

~l

M!i~jE~l HYI

I

2: Plastic (110, 4Eo)

355

03

-(z)

-- fry2

=-fI;

(b)

(a)

!EXAMPLE 6·2, (al Uniform plane wave normally incident on a plastic slab. (bl Side view wave components in the regions.

Then finding the complex wave amplitudes proceeds as follows. (a) One begins in region 3, containing no reflected wave. [3(Z), [rom (6-36), is therein zero, yielding the total field impedance from (6-38) Z3(Z) = ry3(1 + 0)/(1 - 0) = ry3 = 120n Q. By (6-4)), the t<;!tal field impedance Zz(O) just inside region 2 has the same value, that is, Zz(d) = Z3(0) = 120n Q.

(b) By use of (6-39),

f2

at

Z

= d = Az/4 in Z2(d) Z2(d)'+

region 2 becomes

ry2 ry2

120n - 60n l20n + 60n

1

3

Equation (6-40) is employed 3 to translate f2(d) to the value plane of region 2. With z' = 0 and 1'z = jfJz = j2nlA 2

f 2(0)

=

f 2 (0)

at the input

r 2(d)e 2Y2 (O-d) = r 2(d)ei (4n/A2)t- 1 2/ 4 )

Z

(e) Steps (a) and (b) are repeated to find and f in the next region to the left. First, the use of (6-38) at Z = 0 in region 2 obtains

which from continuity relation (6-41) yields Z2(O) = 30n Q = tion coefficient at the output plane of region I, !i'om (6-39), is

(

ZI (0).

The reflec-

3 5

advantage of specifying the thickness of the lossless region in terms of wavelength (d 2 = A2 /4) is evident the determination of f 2 (0). Note, in view of y = j{J = /(211./ A) for a lossless region, that the product ~ z) appearing in the exponential factor does not require an explicit numerical value for {J.

356


The reflected wave amplitude il';;'r is now obtained, us.ing the definition (6-36) of reflection coefficient. Applied at z = 0 in region I, given £;:;1 = IOOe iO", it yields

£-ml =£+f (z)e- 2YlZ ] z=o =(IOOeiO")(-.,l)= rnl 1 5

60V/In

Then the total electric field in air region I, from (6-36), is

The total magnetic field is obtained by usc of (6-37) H~

(z) ,1

= 100

120n

'" e'- lpoz

-

(-60)" eJpOZ 120n

_._-

. - 1"",oz + 0.15geJ'p = 0,266e

0'

A/m

(d) The rest of the problem c<.:mcerns finding R;:;2, £;;'2' and [;;;:;3' For example, £;;'2 is obtained by specializing 1'-"'xl (z) to z = 0 at the interface, whence

in which the last equality is~evident f~om the eont~nui ty condition (3-79). The total electric field in region 2 is Exz(z) = E;:;2e-Y~p + 1 2(z)), from (6-35), but at Z = 0, all quantities in (6-35) are known except E;:;2; solving for it obtains

£;:;2 = Exz(O)

eO

1+12 (0)

I

= 40 - - - - = 60 V /m 1+(

1/3)

Then applying (6-36) at Z = 0 in region 2, £;;'2 = f 2(0)£;;'2 = (-})60 = -20 V/m, whence the total fields Ex2 (z) and [I,2(z) can be written, A similar procedure applied at the second interface then completes the problem,

EXAMPLE 6·3. An x polarized wave arrives from the left at f = I MHz with an amplitude il';:;l = 100e-W V /m, It is incident on a lossless slab an eighth of a wavelength thick, backed with a quarter wave lossy slab, with parameters as shown in the diagram. Find the remaining wave aniplitudes.

Region 1: (p.o, fO)

EXAMPLE 6-3

\.

6-5 SOLUTION USING REFLECTION COEFFICIENT AND WAVE IMPEDANCE

357

The origins assumed for the k)Ur regions arc noted in the diagram. A tabulation of Ct, {3, )., and 1/ obtained for the regions using (3-109), (3-110), and (3-111) is given here:

REGION

E"

/1,

E,

I 2 3 4

a(m-I)

p (m-I)

A(m)

~(Q)

0 0

0.0209 0.0296 0.0461 0.0209

300 212 136 300

377 266 159e122s 377

E'

0 0 I 0

1

2 4 I

0.019\ 0

(a) Beginning in region 4, which contains no reflection, Z3(d 3 ). yielding, 'from (6-38), Z4(O) 1/4 120n Q

t 4(0)

is by (6-36) zero,

(b) Inserting the latter into (6-39) obtains

r 3 (d)3 A

=

377 - 159e1 22s . '21 4" =0.45Ie-.1 . 377 +

to yield from (6-40) at the input plane,

z'

0, the result

t 3 (0) = = 0.45Ie- j2 1.4'e- Z(0.0191)34 e -j180·

= 0.\233e- jZ0 1.4" =

(e) Steps (a) and (b) are repeated, this time to find region 2. Tbus, (6-38) at.;: = 0 in region 3 yields ;;;,

A

.«3(0)

=

I

+ t3(0)

113 1'- t ~(O)

Z and t

at the output plane of

"'225,0.885 l::l9<" . -1.-11-5---'-)-'0.-04-5-0

=

= 126.2e127 . 9 ' = 111.5 + j59.1 and fi'om (6-41),

-0.1148 + jO.0450

Q

Zz(d z )' yielding !i'om (6-39) in region 2

r 2(dz ) = Z2(dz) 1/2 A

-A----

Z2(d z ) +

1/2

=

111.5+j59.1-266

-- - = 0.434<" 1503.

,

111.5 + j59.1 + 266

The latter lransfi)fl11S, by use of (6-40) at the input plane

z'

=

0, to

(d) The total field impedance there, from (6-38), is A

_

A

Z2 (0) - 11 (

I I

+ £\(0)

---A---

2

r 2 (0)

,

.I

= 266

60 +- 0.434e1 .3' ---.

which, by continuity across the interface, yields A

r 1 (0)

=

Zl (0) -

_

---0 -

I - 0.434<,,60.3

Zl (0) =

.142.9'

390e-

390el42 . 9 ' Q. From (6-39)

1/1 390el42 . 9 ' - 377 '87.1' = + 377 = 0.393<" ZdO) + 1/1

-;0:-'---

Q

358

WAVE REFLECTION AND TRANSMlSSION AT PLANE BOUNDARIES

The refiecl~d wave amplitude is obtained using (6-36); applying it at 0 yields E;;'1 = £;:;l r l (0) = (100) (O.393ei 87 . 1 ) = 39.3ei 87 . 1 ", whence the total fields in region I become, from (6-35) and (6-37)

39.3 e1"(",'1Z +87 . I") AIm 377 (e) The rcmaining task concerns finding i:;:;2' [;;;'2' i:;:;3, [;;;'3' and [;:'4' The procedure has aLready been outlined in part (d) of Example 6-2.

*6-6 GRAPHICAL SOLUTIONS USING THE SMITH CHART A convenient way to attack multiregion wave problems like those of Examples 6-1 through 6-3, or the generalized system of Figure 6-8, is by usc of the Smith chart, named for it§.. originator. 4 This chart enables finding, by graphical means, the total !.ie1d impedance Z(z) at any point in a region from the known reflection coeftlcient r (z) there, or vice versa, thereby providing graphical solutions to (6-39) or (6-10). Additionally, from a rotation about t}.le chart, (6-41) is also solved graphically, to permit finding~the reflection coefficient i(z'), at any desired location z', from the known value i(z) elsewhere in the region. The theoretical development of this graphical tool is given in Appendix D. If you are unfamiliar with the theoretical basis for the Smith chart, refer iirst to Appendix D, before proceeding with applications of the chart to wave-reflection and transmission problems involving multilayer regions. The latter is taken up in the remainder of this section, as follows. . To establish the desired normalized wave impedance 2C:::) required in applying the Smith chart to multiregion wave reflection problems, a divisionofexpn:ssion (6-38) by the intrinsic wave imp~dance fi of the region is needed. This bbtains "

Z~z)

==

17

2(z)

I

+ ['(z) ['(z)

(6-42)

an expression comparable to (D-I) in Appendix D. The normalized expression (6-4-2) (or its inverse) is solved i!,raphicaLIy by the Smith chart (see Appendix D); in addition, the translational expression (6-40) ['(z')

[6-40J

is also solved graphically by use of the chart, fi'om an appropriate rotation about the chart, as illustrated in the examples that follow.

EXAMPLE 6·4. Rework Example 6-2 by making use of the Smith chart. This problem concerns a plane wave of amplitudc 100 V 1m, normally incident in air on a quarter-wave Losslcss slab. 4See articles by P. H. Smith, "Transmission-line calculator," Electronics. January 1939; and "An improved transmission-line calculator," Elec/ronics. January 1944.

ti££

6-6 GRAPHIGAI" SOLUTIONS USING THE SMITH CHART Region 2: (/-iD. 4<0)

Region 1: Air (/-i(), EO)

359

Region 3: Air (/-io. fO)

:>-

(z)

~"')

I

I

i""---d2=i---~ I

1

~2 = 607rfl 'Y2

=

, }w-y/-iO 4fO

,27r

= J -\2

1

~3 = 1207f ,

n , 211'

1'3 = ;(3o =; ;A(I

(a)

___-_.....1' plane

~2(0)

= 0.5 + jO

(d)

IXAMPLE 6-4

tal In region 3 of (a), containing no reflection, the total ficld impedance from (6-38) i~ <:'3(Z) =)h t20n n. From (6-41), the impedance just across the interface is <:'z(dz) = <:'3(0) = 120n n. Normalizing the latter using ry2 = 60n n obtains

120n 60n

(

2 (=1 +Jx)

Thus i = 2 and a; = 0 at Z dz in region 2, entered onto the Smith char1;." as in part (b) of the accompanying figure. (Although the reflection coefficient rz(d z) can be found at the location of x2(d 2 ), it may be ignored if desired.) Th~ normalized impedance at z' = 0 in region 2 is obtained from a phase rotation ofr z (d z ) according to (6-43), in which z' = -},,2/4, z = 0, and A = },,2' The use of (6-43) is unnecessary because the rim smles are mlibrated in terms q[ the phase rotation given by (6-43), a negative rotation (toward the source) by the all}ount ( - z) = -0.25},,2 in this example. In (e), the rotation to the new value r(z') = r 2 (0) is depicted. At the s1ime point, X2(0) is found, becoming '~2(0) = 0.5 + JO. Denormalizing ryZX2(O) 60n(0.5) 30n n. obtains <:'2(0) (b) From (6-41), the impedance just imide region I has the same value: ZI (0) = 30n

n.

360


To obtain the reflection coefficient there, normalize ,2 (0) =

3' (0) _'\,1_

1

fit

=

30n 120n

=

ZI (0), obtaining

0.25

+

J

"0

The latter, cntered onto thc chart as in (d), yields

in agreemcnt with that obtaincd analytically in Examplc 6-2. Remaining details proceed cxactly as given in Example 6-2.

EXAMPLE 6-5. Rcwork Example 6-3, making use of the Smith chart. With region 3 in (a) a lossy material, an attenuative as well as a phase shift effect is associated with its waves.

Region 1: Air (/LO, EO) Region 2: (/LO, 2Eo)

Region 4: Air (/LD, EO)

I

ill { 1'1

= 1)0 = 120 7r !.l

=ji30 =) 211" 1-0

lih=607r1l : t1'2 =)w v1lo2to : I

I I

ib =

l

1'3

=

154e j22 .5"!.l 0!3

+ )(33

= 0.0191 +)

7]4

!

= 120 IT 11 .

.211"

1'4=)i30=;-"

211"

"0

"3

(a)

Negative rotation (towards sources)

"3

by amount "4 (b)

EXAMPLE 6-5

( c)

6-7 STANDING WAVES

Beginning in region 4, because of no reflected wave, from (6-41), Z3(d 3 ) = Z4(0) 120:rc il. is normalized using

q3

361

Pi" 377 il, and yielding

2.19 - jO.907

labeled A in figure (b). (The value of ['3(d 3 ) available at A is ignored if it is not desired.) To find the normalized impedance at the input plane of region 3 using the Smith chart, one must usc (6-40)

[6-40] noting that in moving to the left in a region, [' undergoes a decrease in magnitude due to exp [2a(z' - z)], besides changing its phase according to the complex The latter, in moving from Z = 23/4 to z' 0 entails a phase rotation exp [2 (j2n/2) (z' of 0.2523 clockwise around the chart, read off the outer rim scale as shown in figure (c). The effect of the doublc attenuation factor in (6-40) is determined using a3 = 0.0191 Npjm and Z z' = d3 = 23/4 = 34 m, obtaining

Thus ['3(23/4) in (b) is also diminished in magnitude by the factor 0.274, yielding ['3(0) at B on figure (c). The normalized impedance there is £3(0) 0.78 +jO.08 = 0.79ei 5 .40 • Denormalizing yields

The remainder of the problem involving the lossless regions 2 and 1 proceeds in the manner already detailed in the previous example .

..., STANDING WAVES standing wave produced by the total reflection of a plane wave normally incident perfect conductor was observed in Figure 6-3. The hasis for the term standing is seen from the composite diagram; the total field magnitudes have a stationary ranee in space, similar to standing waves on a vibrating string as in Figure The undulations, from maximum to null amplitudes every quarter wave, occur accordance with the sin pz or cos pz factors in the total field expressions (6-6) (6-9). , The example of Figure 6-3 represents a special case of standing waves produced plane waves of equal amplitudes move in opposite directions through a region. In general, an arbitrary percentage of the incident wave is reflected, dieu::rmined by the reflection coefficient amplitude at the interface. T!Ie region. may, be}ossy. An analysis of standing-wave behavior requires the total electric magnetic field expressions, given in time-harmonic form by (6-35) and (6-37)

(

(6-43)

(6-44)

362


in which y = IX + jp given by (3-90a,b), and fi is specified by (3-99a). In some standing wave discussions, only the wave magnitudes are of interest. The magnitudes of (6-43) and (6-44) are written

IEx(z)1 = IE.!le-azp + r(z)i

1£+1

_m_

e-azll

17

r(z)1

(6-45a) (6-45b)

noting that the magnitud~s of the phase quantity, Ie - j{Jzl, and of the angular factor of the wave impedance, 1e1°I, are both unity. QLpa£!i,£~I
E+

il

(z) = -'!:. y

17

ei- (Jz

(6-47)

Figure 6-9 displays a real-time plot of EAz, I) and Hy(Z, t), showing the incident and reflected wave terms of (6-46) and (6-47) at successive instants along a portion of the z-axis. An inspection of the total fields with varying t and Z in the lower diagrams shows how a standing wave is developed in the region;. there the total fields appear to be traveling waves with a changing amplitude as they move in the z direction. Thus IEx(z, 1)1 and IHy(z, t)1 change from a maximum to a minimum, and vice versa, every quarter wave (90°) along the z-axis, a consequence of the forward and backward wave terms becoming phase-aiding and then phase-opposing at that spacing as the waves move in their respective directions. The maximum of the total electric field envelope, l!,~ax = IEx(z, tll ma " is observed to coincide in space with the minimum H min IHy(z, tllmio, and vice versa, a result of the sign reversal in the reflected magnetic field term in (6-47). The so-called standing-wave ratio (SWR), associated with incident and reflected uniform waves in a lossless region as exemplified in Figure 6-9, is defined as the ratio of the maximum amplitude, E max , of tbe electric field envelope, to tbe minimum amplitude, 1<-'.nin, occurring a quarter wave away; tbat is, abbreviating SWR as S,

s IEx(z, t) Imax

IEx(z, tll m;:

Emax

(6-48)

Em;n

It is seen from Figure 6-9 that the envelope maximum Emax occurs wbere the amplitudes and E;;' are aiding, wbile Emin is produced a quarter-wave away wbere they are in opposition, such that

E.!

Emax =

E.! + E;;' = IE.! I + IE;;' I = IE,~I-IE,;;I

(6-49)

6-7 STANDING WAVES

Hy

i

-360' -270· -180· -90·

o·

90·

(z,

363

tJ = H/ + Hy

-360·-270·-180· -90·

o·

go'

Standing - wave envelope

Hmin~Hma, t_~~:~versusz

, ...-

o· (a)

90·

. ,....-..

--'

-360· -270· -180· -90·

o·

90'

(b)

Real-time diagrams of forward- and backward-traveling waves of Ex and Hy at .:1I:egi\re instants, in a region where reflection occurs. The composite standing-wave pattern is the result. (a) Electric field. (b) Magnetic field.

conclusions can be reached concerning Hmax and H min along the magnetic standing-wave envelope; thus SWR, from (6-48), becomes (6-50) exampfe, with the launching of a forward-traveling plane-wave field with 100 V 1m in some lossless region, and a reflection occurring such that 20 VIm, the standing-wave ratio, from (6-50), becomes SWR = W = 1.5. A region (with IE';; = 0) will have the minimum possible SWR of unity. There are advantages in analyzing standing-wave phenomena by use of the foqns of the fields. Since the standing-wave diagrams of Figure 6-9 are total magnitJdes plotted against z, it behooves one to reexamine the wave magnitude

I

364


expressions (6-45). For a lossless region (ex

0),

beeomc

jE.'+ j

lily(,::)j. = -~ 11 11

['(z)1

(6-51 )

results of special interest in that they involve the reflection coeflicient ['(,~), a quantity readily available {i'om the Smith chart. It is evident rrom (6-51) that the maximum wave magnitude, iE:x(z)imax, occurs in the lossless region where + r(z)j is maximal; that is, ~here it !!as the value 1 + 1['(z)l· Thus, Fmax = 1.£':'1(1 + jf'(z)ll. Similarly, Emin = IE:'I(I -Inz)j). Hence, the SWR defined by (6-50) becomes

II

(6-52a)

I

For example, the reflection-coefficient magnitud(~ 0.2 (20';\, reflection) yields fi'om (6-52a) the SWR = (I + 0.2)/(1 ~ 0,2) 1 as bd(}l'c, Since the reflection co) the SWR is limited to the eflicient magnitude has the range 0 S Irj s 1, from range 1 S SWR < 00. 'fhe Smith chart, from which the reflection coefficient is readily f(HInd, is also convenient for finding the SWR grapl1.ically. For a losslesR region containing the total fields (6-46) and (6-47), the locus of nz) versus is a circle as shown typically on the chart in Figure ftlO(a). This 12cUS is sometimes callt'd the SWR circit'. The complex quantities 1 + r(z) and I r(z) occur ()~ the SWR at the poiuts A and B, as in (b) of the figure. The quantities 1 + lr(z)1 and 1 I are evidently the distances o'e and 0']) in Figure 6-10(c), yielding fi:mn

S, -

0'(,' O'f)

(6-5~h)

The use of (6-52b) can be avoided, however, sillce the normalized im[wdance %, at the point in the figure, has a value equal to the SWR in question, a hle! proved

e

r(z) versus z (SWR circle)

Toward source (a)

(6)

FIGURE 6-10. Smith chart field interpretations versus z: the SWR circle. (oj The quantities I + locations where E(z) is maximum and minimum.

(e)

lossless and 1

6-8 REFLECTION AND TRANSMISSION AT OBLIQUE INCIDENCE

:

(·)1 ~versusz

365

IHv (t)1

: (versus z

----r--------

--;--"'--

;Hmax I

"

I

-\Ex-I..... "'-- ---- ~ '" '" '" "-

____

~~~

______

_L~~

__

~~

-_4~5_·_______H_m_m+_----~(~)

_________

-/.,

).

2

4

0:

(z)

I

FIGURE 6-11. Forward z and backward z traveling field magnitudes deduced from the Smith chart (above) and the corresponding standing-wave field magnitude graphs (below).

by applying (D-3) at that point. Since x

=

0 and I

ri = 0

+ 1~(z)1

there, (D-3) yields (6-53)

1- W(z)1 or just the SWR given by (6-52a). Thus the SWR circle can be drawn on the Smith chart by noting it must pass through the point -t = SWR on the positive real axis. The SWR circle on the Smith chart can be used to obtain the z variations of the eLectric and magnetic field magnitudes in a los'}less region. Fr~m (6-51), IEAz) and jHy(z)1 are proportional to the quantities 11 + f(z) and jl - f(z)l, respectively, but these are just O'A and O'B in Figure 6-IO(b), whence the relative field magnitudes versus yield the Smith diagrams of Figure 6-11. The lower graph shows the standing waves of the two field magnitudes obtained therefrom. The occurrence of H min at the position of Em.x> and vice versa, is noted as mentioned before.

I

6.8 REFLECTION AND TRANSMISSION AT OBLIQUE INCIDENCE I

This section concerns an extension of the normal-incidence wave reflection and transmission problem of previous sections, by considering the effects of the oblique incidence of the impinging wave on a plane interface separating two regions. At radio frequenoblique-incidence plane-wave solutions are applicable to the reflected and transmitted wave eHects at air-to-sea Of air-to-earth boundaries, for example, Of to the problem of wave incidenee fi'om below on the ionized atmospheric layer (ionosphere) located far above the earth's surface. The solutions also have extensive applications to optical d~vkes such as lenses, prisms, and fiber optic transmission lines, forming the \

366


basis for the laws of physical optics ray bebavior. Whereas field ;;olutions are obtained in the following for a plane wave incident on a planar interhtce separating the two regions, the solutions are nc:vl'rtheless nearly correct fc)r curved interlaces, as long as the radius of curvature of the interEH:c is large compared to the wavelength of the incident wave. This is often the case at optical frequencies. The treatment of the problem ofuni/imn plane-wave rdkction and transmission (refraction) at oblique incidence is both filcilirated and enhanced by attributing a vector character to the wave-phase factor (or "wave number"). This involves making use of the equation of a plane in vector notation.

A. Planes In Three-Dimensional Space The position vector (1-18) is useful t()r writing, in vector fixrn, tbe equation of a plane. In Figure 6-12(a), with Po(xo,)'o, given to be the point Oil the plane S nearest to the origin 0, the perpendicular distance from 0 to Po is the position vector ro nro axxo + ayyo + azzo. Let ,c) be any arbitrary point on S. Then r axx + ayy + azz is the position vectgr of P. The plane S is thus defined by t;,

\

(6-54 )

'\

evident from the definition of the dot prodnct and fl'Ol11 the fact that the projection of every position vector r (on S), on the fixed perpendicular line oro, is the dosest distance 1'0 fi'om () to the plane S. (f)-54) is thus called the /lector of the plane. Additional insight into is gailled by writing it in expanded scalar fi)rrr!. Thus, multiplying (f)-54) by TO to obtain ro . r = r~ and substituting the rectangular forms of ro and r yields XoX + ':0';: tlIe ('Cjn:llion of the plane. Dividing through by TO obtains (6-55a) equivalent to the direction-cosine equation of the plane X

cos 11

+ )! COg B +

cos C

ro

(6-55h)

(x)

a I I I

",;~)

\

~~~~"_Po(XO' Yo, r~=

zo)

nro

~~----------~-~ (z)

0')

c

b

0')

(a)

FIGURE 6-12. Geometry of a plane Sin rectangular coordinates. on S, and Po on S nearest the origin. (b) Direction angles A, B, C

(b)


367

n which A, B, C denote the direction angles between the normal and the coordinate [xes as depicted in Figure 6-12(b). The ir£tm:ejJl equation of the plaJ1(~ is obtained from lividing (G-55a) by TO once more, yielding the i(mn 7

(l

h

+"=-

(G-56)

c

in which a = and so on, denote the distances to the intercepts the coordinate axes, as shown in Figure 6-12(h).

(l,

b, and c with

3y + 12z = 10 and ax Gy + 24,:; = 15 arc parallel, find the distance between them and determine the direction angles A, 11, and C. The distances TO from the origin to each arc {(Jund by converting the ex pressions to the {(lrm of Dividing each expression rcspectivcly by 26 obtains

EXAMPLE 6·6_ Show that the planes 4x

+

+

~IQ ~

13

(I)

15

(2)

26

The plalll's arc thns parallel, in view of the identical direction cosines. The planes arc separated by the distance ro I '02 = From the codlicients of (1) and provide the direction cosines, yielding A arc cos ( = 72.03°, B 103.34°, C = 22.62'.

B. Plane Waves Traveling in Arbitrary Directions The j()regoing disclIssion of planes in thrt~e-dimcnsi(Jl1al space has an important application, ill the designation of cquiphasc planes, to plane waves traveling in arbitrary direclions. To this end, a restatement of the simple case of the z-traveling uniform plane wave of Figure 6-13 is in order. I ts electric field vector is given by (G-57)

(x)

z

ro

=

nor

plane

\

(z)

FIGURE 6-13. A uniform plane wave, z-traveling, showing an equiphase plane and vector notation.

368


Since from Figure 6-12 and by use of the perpendicular distance ro fI'om the origin to any plane is written ro = n • r, the perpendicular distance to the typical equiphase plane of the plane wave in Figure 6-13 is f()r this case simply TO == Z = n' r. (6-57) can thus be written (6-58) With fJ = nfJ seen to become a vector phase factor, its vector direction n = a z defines the wave direction of travel. Generalizing the wave expression (6-58) simply requires a rotation of the zpropagation axis of Figure 6-13 by the direction angles A, B, and C as shown in Figure 6-14(a), with the wave direction of travel labeled l. That positive-l traveling wave is now expressed as (6-59) with the unit vector a e employed to denote its vector direction, and the vector phase factor fJ = nfJ aligned with the ::' propagatioIl axis. To enable expressing (6-59) in terms of the coordinates /::) of allY point P(x,y, z) on the typical equiphase plane of Figure 6-14(a), it is seen b'om that nor 10=Z' can be replaced with (6-55b), yielding from (6-59) cos A ..t- Y cos B

+ z cos C)

(6-60a) (6-60h)

(xl

(x)

P(x. y, z) ! I n

~

rl~ I

(a)

(yl

'~

(b)

(y)

FIGURE 6-14. Uniform plane wave propagating in the general direction l. Showing the direction angles A, B, C that determine z' and the position vector r of any l'(x,y, on the typical constant-phase plane n' r = z' = constant. (b) Showing several equiphase planes used in defining wavelengths.

369


in which

is made of the abbreviations

{lx =

fJ cos A,

fJ y

The vectlJr phase constant in

=

fJ cos B,

thus consists

{lz

=

{l cos C

(6-61 )

or the components (6-62)

The companion magnetic field H, mutually perpendicular to both the of (6-60) and the unit vector n defilling the propagatioll direction, becomes

t

field

(6-63)

H(x,],z)

The physical implications of tht' vector phase constant (6-62) are depicted in Figure 6-14(b). Three equiphase planes, chosen to coincide with successive E-field maxima, are seen to product' spacings along defining the true wavekngth A given by the familiar

(6-64) The equiphase planes also intersect along the X,],.:; axes, yielding the "intersection wavelengths" related to the cornponents of the vector phase constant of (6-62) such that

2n

A

fJ cos A

cos A

2n

(6-65)

-

fix

These skewed wavelengths art' thus greater than the true wavelength A of (6-64), also evident from the geomet ry of Figure 6-14( b). Phase velocities are also associated with each wavelength. Along z', the true phase velocity lip = OJ/f3 is observed; but the apparent phase velocities sustained by the constant-phase-plane intersections along the y, Z axes are, by use of (6-61),

OJ Vx

= (J x

cos A

v Y

=

cos B

cos C

(6-66)

all seen to exceed the true phase velocity, in view o["the cosine divisors. This "stretching wavelength" along the coordinate axes in Figure 6-14(6), concurrent with the apspeeding up of the wave when observed along the axes, is much like the int":reased wave speed observable along a coastline or seawall on which ocean waves are obliquely incident. A simplified example f(lllows in which the wave direction of travel ) as depicted in Figure 6-14, is confined to the plane by making B = 90°.

IXAft1PLE 6·7. A uniform plane wave ill a )ossirss region travels in the x-z plane (B = 90°) as . shown in (a) of the figure, with its propagation direction l tilted by the angle C == 0 from the z-axis. The E field is polarized parallel to the plane, yielding the field components

II'

I

370

WAVE REFLECTION AND TRANSMISSION AT PLANE BOUNDARIES I (xl

"lE

/
"tLA

ae

/

n

C' O~

~- " " ~ -

-

- - -(z)

(b)

: (x)

I E I

: I I

'a

(z'l e

n

I

I

ax sin 0

+ az cos IJ

.

o~l(~ __ (Z)

(a)

(cl

EXAMPLE 6-7

(6-59) and (6-63) given here by

n

X

1]

E = a -R:,- e _'flz' J

1]

Y

(I)

with n denoting the normal to any equiphase plane. Express.lhe fields (I) completely in the rectangular coordinate system, Looking down onto the x-z plane as in (b), the reetangular components of the unit vectors in (I) become n

ax sin 0

+ a z cos 0

(2)

yielding the vector phase constant with two components (3)

The phase exponent (1-18), becomes

fh:

in the field expressions (I), with the position vector r given by

fh' = P . r

=

=

+ a z cos 0) fl(x sin 0 + z cos 0) fl(a x sin 0

• (axx

+ a y)' + azZ) (4)

yielding the desired expression from (I): E(x, z) = (ax cos 0

a z sin O)E:'e

i/l(xsinO+zcosO)

(5)

"'";+

H(x 2') = a , -

E-m e-'j/l(xsinQ+zcosO) Y

1]

(6)

The results (3), (5), and (6) reveal components of the phase factor to be flx = fl sin 0, fly 0, and flz = fl cos 0, in agreement with thc forms of (6-60) and (6-63), since the direction angles are expressed in terms of 0 in this example by A = 90° - 0 and C = O.


c.

371

Reflection and Transmission of Plane Waves at Oblique Incidence Assume a uniform plane wave obliquely incident on a plane interface separating two lossless regions with the parameters (Ill, El ) and (1l2, Ezl as shown in Figure 6-15. It is sufficient to consider the two cases of the parallel and perpendicular polarizations of the incident electric field (relative to the x-z plane of incidence), depicted respectively in Figure 6-15(a) and (b). The general case, for an arbitrary polarization of the incident wave, can be constructed from a superposition of these two cases. The paralleL-polari,,,ation case is considered ill de~ail. To satisfy the boundary~cOIlditions at the interface z = 0, a reflected wave E" Hr and a transmitted wave E t , HI will be required, depicted at the angles Or and Ot froIll the normal z-axis as shown in Figure 6-J5(c). The right-hand rotation from each E vector into the associatedydirected H vector yields the desired direction vector n, normal to the equiphase planes and related to the vecto~phase factor f3 = nf3 of each wave in regions 1 and 2. Assuming the incident wave Ei and its angle of illeidencll 0i to be known, it is desired to deduce the reflected and transmitted waves Er and E t as well as their angles of departure Or and Ot ii'om the interface. The thrce plane wave fields of Figure 6-15(c) arc now expressed in the notation of (6-59) and (6-63 l.

(z)

(a)

(e)

(d)

FIGURE 6-15. Geometries associated with a wave incident on a plane interface sqparating lossless regions, for (a) the parallel-polarization case; (b) pcrpendiL~I,r polarization. Showing also the reflected and transmitted field components with (c) parallel polarization; (d) perpendicular polarization.

372


E., = a.E.ei/1,Di' r "

(6-67a)

Hr = a y Er e-

j

/1,n r 'r

(6-67b)

/}1

(6-67c) if E;, i:" and itt are taken to mean the (complex) amplitudes of the three electric fields. The total electric and magnetic fields are needed to satisfy boundary conditions at the interface z = 0; these are, in region I (6-68) while in region 2 (6-69) The substitution of (6-67) into (6-68) and (6-69) yields the required total field expressions. To express them in terms of the rectangular coordinates z), the technIques of Example 6-7 are employed. Thus, the geometry of Figure 6-15(c) obtains for the incident wave (6-70) PIn;' r

+ a z cos Gi ) · = PI (x sin 0i + z cos Oil =

pda x sin 0;

(axx

+ ayY + azZ) (6-71 )

yielding from (6-67a) (6-72)

Similarly, the reflected

Er expression

of (6-671» is shown to become

Er(x, z) yielding, from the sum of (6-72) and (6-73), the total

(6-73 )

EI

of (6-68)

The geometry of Figure 6-15(c) is employed to convert the total (6-67c) to the result

E2

z) of (6-69) and

(6-7,,))


Similarly, the total magnetic fields to become

HI

and

H2

373

of (6-68) and (6-69) are shown

(6-76)

H2 (x, z) = a y itt e- jfh(x sin 0, + z cos 8,)

(6-77)

rt2

The boundary conditions (3-71) and (3-79) require that the total tangential electric and magnetic fields be continuous across the interface at z 0 in Figure 6-15(c). Equating x components of (6-71) and (6-75) at z = 0 yields (6-78) while (3-79) requires that (6-76) be equal to (6-77) at

l~i

e-jfJlxsinO,

+ it, e-JlhxsinOr rt1

rtl

z=

0, obtaining

e-jfJ2xsinO,

(6-79)

rt2

(6-78) and (6-79) are to hold as equalities for all values of x OIl the interface z = 0 regardless of the values of #1 and li2) bu t this can be so only if the phase arguments are all equal; that is,

#1 sin 0i = #1 sin Or = #2 sin at

(6-80)

This means physically that the x components of the vector phase factors PI and P2 must be the same, implying that the phases of the waves to either side of the interface z = 0 must keep in step. The first equality of (6-80) means (6-81 ) or simply that the angle of reflection equals the incidence angle in region I. The last

equality of (6-80) yields, with Or

= Oil

sin 0i sin Of

For nonmagnetic loss less regions, with III

(6-82a)

= 112

Ilo, (6-82a) can also be written

(6-82b) in which n 1 = ~ and Tl2 ; ; : ; are termed the "indices of refraction" of the two regions. The results (6-81) and (6-82) are known as Snell's laws of reflection and refraction for lossless regions. To provide additional physical insight, Snell's law (6-82) can also be derived from graphical considerations. In Figure 6-16(a) are shown the incident, reflected, and transmitted waves, each represented by their equiphase surfaces. It is seen that the lame interse~tion wavelength Ax applies to all three waves to either side of the interface

374

WAVE REFLECTION AND TRANSMISSION AT PlANE BOUNDARIES

(x)

P

(z)

(z)

P'

(a)

(b)

FIGURE 6-16, Incident, reflected, and transmitted waves represented by shown at wave crests, (a) Showing in-step condition of waves at the interface leading (0 Snell's law,

surfaces Geometry

0, a conditioll required so that the waves on both sides of the illterhlce may remain in Ilhase-step. With this constraint and /i'om the right triangles OPP I and OPP z that share the length Ax are obtained sin OJ )~l/Ax and sin Ot = A2/Ax, lrom which their ratio yields Snell's law (6-82). In view of (6-RO), the cancellation of the equal exponentiallilClorS in (G-78) and (6-79) yields tW<2 algebrllic boundary relations that, when solved sim!:lltaneollsly, yield expressions for Hr and f;, in terms of the incident-wave amplitude Hi; that is,

i;r

fl'j == r ll =

11 cos OJ - 112 fit COS OJ + 112

COS

COS

Or 0,

(6-83)

(6-R4) The symbols r ll and Til are respectively called the "relleclion codlicient" and "transmission coefficient," relating the reflecting-wave and transmitted-wave amplitudes P;r and 11't to the illcident-w~ve amplituck 'Ei' fi)[' this parallel-polarization case. Alternative expressions that make llSC the index of refraction n = of each region are also nseful, particnlarly fix Ilonmagnetic regions ({tl = fl2 = flo), Then (6-83) and (6-84) can he written

.JEr

or

n2 - cos 0i nl

(6-85)

+ cos 0, (6-86)


375

For the the of Figure 6-1 and (d) is used, with the known, incident electric field and resulting reflected and transmitted electric field vectors all assumed jJcrpendicular to the paper (y-directed) as shown. With the new directions of the magnetic field vectors properly accoulltcd for, a procedure closely rescm b\illg 11Ia t used li)[ the parallel-polarization case is employed. Applying the boundary cOI](litions of continuity to the total fields at the interface Z = 0, Snell's laws are once again obtained, with the simultaneous solution of the boundary results yielding reflection and transmission coellicicnts in this perpendicular polarization case as follows Y/2 cos 0i - Y/ t cos Ot

+ Y/t

cos 0t

(6-87)

(6-88) These should be compared with (6-20) and (6-21) for the normal-incidence case of Section 6-4. The alternative f(lfms, written {()r nonmagnetic regions in terms of their indices of refraction n = ~, become cos 0i

n2

cos 0t

Ttl

:2 cos 0;

T

(6-89)

(6-90)

An example of the reflection and transmission coefficients graphed as functions of the angle of incidence for both polarization ca~es is shown in Figure 6-17. Lossless, nonmagnetic dielectrics are assumed, with Ert" = 1 (air) and Er2 = 4, making (rt 1!n2) = 2. It is observed that total reflection (Irj= 1), and zero transmission are approached as the grazing condition (0; -> 90°) is achieved. A zero-reflection point on the r ll curve also observed at the incidence angle of 63.4 0 Ii:)r the given medium parameters. The giving 7:(':.I:0 reflection, seen to exist only i()r the parallel polarization case, is known as the Brs,:~.:!.~;r angle.

D. Brewster Angle To obtain an analytical expression f()r the Brewster angle, setting the reflection eoefficient r ll of ((i-83) to zero yieidsY/t cos 07 = Y/2 cos Ot or Y/i(I - sin 2 = 1 sin 2 0t), in which O~ denotes the desired zero-reflection angle. Using Snell's law to express Ot in terms of O~' leads to the result il)r the Brewster angle

on

sin

Of

(6-91 )

2i

376


i

0.8

C

.~ .~

-'" 0

u

c

0.6 --.-~

..--.

f-

0.4 0.2

0

'iii

'"E

0

'"c r:

-0.2

o

-0.4

'--- 10

0 _

20 O<~. 30

0

T" r:.::::: r--;;....

...........

..

r--..7

40 -

60~o

50°

""SlO'

~ <1l 'ji; 0::

80

0

-

"

90 0

0

'\

r----... r.L

.........

0

~.

"-

t-

c

_-

" "'" ~
"r--...r,,0

-

..

-.;;;;:::

~

-0.6 -0.8

"""

\ ~

<

\

K\-

~

:FIGURE 6-17. Reflection and transmission coefficient as a function of angle of incidence, where (nz/Tlll = 2.

For common nonmagnetic materials [Ez/(Ej +E2)]1 I Z, or

(111

112 = 110), (6-91)

becomes sin Of

(6-92)

This reveals the Brewster or zero-reflection angle, {()T' the remits graphed in Figure 6-17, to be O~ = are tan 2 = 63.4°. You may prove that no zero-reflection angle exists f()r the case of perpendicular polarization, if /11 = /12' In the event of an incident wave possessing both parallel and perpendicular electric-field polarization components, the wave reAected irom a dielectric surface will have no parallel-polarized electric field if the wave is incident at the Brewster angle. This zero-reflection phenomenon thus makes it possible, for the common case of randomly polarized light waves, to use polarizing eyeglasses to diminish the remaining perpendicularly polarized waves reflected from a roadway, for example, thereby reducing glare.

E. Total-Reflection or Critical Angle Under some conditions, an angle of incidence can be found such that

~alreflee

~i~n occurs, that is, the reflection coefficient magnitude is unity. Examining the mag-

nit~ (6-83) and (6-87) shows that unity reflection would be obtained if either cos 0i or cos Ot were zero. The former case is of no physical interest because 0i = 90° means that the oncoming wave is at grazing incidence. Putting cos Ot 0, though, means 8 t = 90°, implying that the transmitted wave is traveling parallel to the interface in region 2 as suggested by Figure 6-18(b), rather than providing any z-propagated wave as shown in (a) of that figure. The angle of incidence corresponding to Ot = 90° is called the critical angle, labeled 0i = Oc.


377

(z)

(a)

(b)

(c)

FIGURE 6-18. Equiphase surE,ec, [or the incident (solid), reflected (dashed), and transmitted waves with €l > €2' for the angle of incidence (a) less than, (b) equal (0, and (c) greater than the critical value eo. Wave details for region 2 in (c) arc to be discnssed.

The value of ec is obtained from Snell's law (6-82); obtains

et = 90° substituted into it (6-93)

e e

ifboth regions are nonmagnetic. This expression shows that if a real angle i = c is to exist, Ej > E2 is required. Thus, the primary wave must be incident on the interface from the region having the higher relative permittivity. Total reflection within a glass prism occurs in this way. Experiments show that total reflection occurs not only f()r i but also for ineidence angles exeeeding Oc as well. The nature of the wave transmitted into region 2 is predictable analytically on substituting the expression for the critical angle c into Snell's law, with resulting expressions for t inserted into the wave expressions obtained earlier f()f region 2. Thus, (6-93) into Snell's law (6-82) yields for sin t

e en

e

e

sin

e t

h 1

-

e

E2

This obviously yields real-angle values for

ei S; ee·

.

sm

ei = sin. e j

(6-94)

sm

e only if sin l1 t

t

s; l, which occurs only if

If the incidence angle exceeds the critical angle as suggested by Figure 6-18(c), or ei > then the ratio sin OJsin ec of (6-94) will exceed unity. Thus, et becomes a complex angle, and the implications of tbis on the behavior of the field transmitted into region 2 may be deduced as follows. With sin t (6-94) exceeding unity, cos Ot is written

en

e or

(6-95) The negative root of the imaginary result is chosen here to preserve the physical realizability of the wave in region 2 (to be clarified momentarily relative to its producing

378


a vanishing wave there as

z -+

co). Putting (6-94) into (6-95) yields

cos Ot = - j

(6-96)

and substituting the latter and (6-94) into the directed magnetic field in region 2

H2 (x, z) = =

=

H2 expression of (6-77)

provides they-

ay Et e- jP2(X sin 0, +. cos 0,) 112 a

Y

Et e!- i/J,[xJ. 1/'2 sin 6i -

Et

a y

jzJ(;, 1/'2) sin 2 iii - 1]

112 e -[1I2J(E-'/'2) sin 2 8i =1]ze - 11112./;'-0" sin O;]x

172

The form of (6-97a) shows that H2 is a wave attenuated in change in x in accordance with the form of

H~

-

2 -

Et

ay-e '12

z and

(6-97a)

exhibiting a phase

- az - jbx

e

(6-97b)

with attenuation and phase constants a and b defined by (6-98) T1ws, the wave function (6-97) represents the magnetic field transmitted in region 2 for angles of incidence 0i that exceed the critical value (6-93). It is a wave attenuated in the increasing z direction and shifted in phase in the increasing x sense in region 2, as shown in Figure 6-19. The reason for the choice of the negative root of the radical

I (x) I

Constant-phase plane

(a)

(b)

FIGURE 6-19. (a) A typical nonuniform plane wave produced in region 2 when the angle of incidence in region I \exceeds the critical angle. (b) Detail of Hy of (6-97) in region 2, at a fixed instant. ',--

PROBLEMS

379

in (6-96) is now evident; a positive root would make the z-dependent exponential factor in (6-97) grow indefinitely large as z -+ W, which is not sensible physically. The wave is thus "trapped" into traveling with pure phase change along x (parallel to the interlace), while being attenuated in amplitude as one moves away from the interface in region 2. This attenuation is clearly not associated with dissipation in region 2, which is a lossless region. In the foregoing discussions of parts C, D, and E of this section, only the obliqueincidence case involving two lossless regions was treated. If region 2 were made a conductive region, the penetration of the transmitted wave into region would be analyzed in much the same way as is done for the lossless case in part C, except for the replacement or E2 with the complex permittivity € defined in (3-103). This has the effect of injecting a "complex angle" interpretation into Snell's law (6-82a). Details of this case are found in Appendix A. Of special interest in Appendix A is the case for which region 2 is a good conductor. In that instance, it is shown that the transmitted wave enters region 2 with its direction of travel essentially normal to the interface, as illustrated by Figure A-2(b). This result has an important application, for example, to the penetration of electromagnetic fields into the conducting walls of rectangular hollow waveguides, a fact utilized later in Section 8-6.

REFERENCES LORRAIN, P., and D. R. CORSON. Electromagnetic Fields and Waves. San Francisco: Freeman, 1970. FANO, R. M., L. 1'. CHU, and R. P. ADLER. Electromagnetic Fields, Energy and Forces. New York: Wiley, 1960. RAMO, S., J. R. WHINNERY, and T. VAN DUZER. Fields and Waves in Commnnication Electronics, 2nd cd. New York: Wiley, 1984.

PROBLEMS

SECTION 6-2 6-1.

The total-reflection magnetic freld solution (6-8) was obtained by inserting the boundary result (6-4) into the general reflection result (6-7). Show that (6-8) can also bcfound by inserting the total-reflection electric field solution (6-5) into the appropriate time-harmonic Maxwell equation.

6-2.

Employ the boundary condition (3-72) to obtain the

expre&~ion

for the current density

Js induced by the magnetic field onto the perfectly reflecting plane of Figure 6-3(r). What depth of penetration of this curreut is expected into region 2? Explain.

"6-3.

Assume that the totally reflective syste!? of)<'igure 6-2 has a knowny-polarized incident uniform plane wave with the components (E;, H;) instead of the x-polarized wave shown. Thus, assume

(1) ir,,(z) =

(2)

which_E~ is assumed know}!. Apply the required boundary condition to determine the total and Hx fields in terms of A-:;. Compare your results with (6-5) and (6-8) of the x-polarized case.

380


6-4.

Concerning Problcm 6-3, sketch a figure resembling figure 6-3(c), showing the resulting Ey and Hx in real time at a few selected instants. Comment on the comparison with Figure 6-3(c).

SECTION 6-4 (a) Given, in air, a plane wave of complex amplitude E';;;!, impinging normally on a lossless, nonmagnetic dielectric material (region 2) of permittivity E ErEO' show from (6-20) apd (6-21) that the amplitude of the transmitted electric field into region 2 is 2/(E:/2 + 1) tiInes E;;;I, while that of the reflected electric field in the air region is (I - E: 12 )/(1 + E:12) times E;;;I' (b) If the incident wave in air has the amplitude 100e-i° V/m, find the complex amplitudes of the reflectcd and transmitted waves if region 2 is polyethylene (Er = 2.25); and then if region 2 is water (Er = 81).

6-5.

,,6-6.

The so-called "index of refraction", n, of a loss!css nonmagnetic dielectric is dcfined as

n = E:12. (a) Show that n also denotcs the ratio of the phase velocity of a uniform planc wave

in free space to that in the dielectric, or n = c/v p • What are the indices of refraction for the polyethylene and the water regions (assumed losslcss) of Problem 6-5'? (b) give12 are the conditions of the two-region loss less system of Problem 6-5.~ Dcnyting the ratio E;;,dE;;;1 by r (and givcn the name "reflection cocfficient") and the ratio E;;;2/E;;'1 by T (called the "transmission coefficient") write the expression for f and '1' in terms of the index of refraction of the dielectric region 2. Plot rand T versus n over the range I S II S 10, on the same graph. Is the difference of these curves, T - r, a function of n? Comment on how this latter result is influenced by the electric-field boundary condition at the interface separating this two-region system,

6-7.

Make use of the solutions obtained in Example 6-1 to show that these total tangential electric and magnetic fields satisfy the boundary conditions (3-71) and (3-79) at the interface .

.I{

6-8.

For the two-region system of Figure 6-5, a plane wave arrives in air at normal incidence, with amplitude 200 V /m at the frequency 50 MHz, Region 2 is water (Er ~ 64 at this fi'equency), assumed lossless. (a) Find the intrinsic wave impedance, propagation constant, and wavelength in each region at this frequency. (b) Make use of (6-20) and (6-21) to find the reflected and transmitted wave amplitudes. (c) Write the expressions f()r the total fields in the two regions, in the manner of (6-12), (6-13). Show that the tangential-field boundary conditions (3-71) and (3-79) arc satisfied by these fields at the interface.

SECTION 6-6 6-9.

A lossless three-region sy}tem I'('sembling that of Example 6-2 involves a uniform plane wave in air region I given by £;1(:;.) = 500e-'jfio z V/m at the frequency f = 300 MHz. Plastic slab region 2, of thickness 0.375).2, has parameters (I.to, 4Eo); those of region 3 are (flo, 16Eol. Assume z-origins as in Example 6-2. (al Sketch and label this system. Find the intrinsic wave impedances ~and wavelengths in each of the three regions. What magnetic field if;'r (z) is associated with £;1 in region I? (b) Determine the total field impedance and the reflection coefficient at the output plane z = d in slab region 2. (c) Find the values of the latter at the input plane z = O. (d) Deduee the refl...ection coefficient at the output plane 0) of region I. Find the reflected wave amplitude E;;'r in region I, cOITe3!ponding to the given incident-wave amplitude. j141 [Answer: (a) fi3 = 30n n (b) r2(d) = -t (d) E;;'1 = 234e' V/m]

ct-10.

Give details as needed to find the remaining electric-field complex amplitudes £;;'2 and E;;;3 for Example 6-2, obtaining the total fields in regions 2 and 3: 60e- jfJ2z - 20e-iP2Z V/m

£X2(Z)

=

ify2 (z)

= 0.318e- jP2z

+ O.l06e-iPzz A/m

6-11.'--eonvert the electric and magnetic field solutions obtained for Example 6-2 in Problem 6-10 to their real-time forms.

PROBLEMS

Region 1: Ai r (/lO, €o)

Region

2: {JIo,

4fO)

381

Region 3: Perfect conductor (0"3 co)

(z)

if

= 1 5 GHz)

PROBLEM 6-13

6-12. Complete Problem 6-9 by finding £;;'2' £;;'2' and total electric and magnetic fields in regions 2 and 3.

£'!3'

and obtain expressions

fiJI'

the

'( 6-13. The lossless nonmagnetic slab (region 2) with Er = 4 as shown has the thickness d = A2 /8 at the opcrating frequcncy 1.5 GI!z and is backed by the perfectly conducting region 3. The given incident wave in region I is Exl (z) = 200e- jfioz Vim. Assume z-origins as shown. (a) Find the thickness d of the slab region (in ern). What is the total field impedance and reflection coefficient at the output plane (z = d) of slab region 2? (b) Determine the total field impedance at the input plane (z = 0) of the slab. (c) Find the ref~ectioll coefficient at the output plane z = 0 of regioll I. Determine the complex amplitude E;;'1 of the reflected wave in region I. Write the~ expressiolls for tEe total electric ane!. magnetic fields in region 1. [Answer: (a) d = 1.25 em, r 2 (d) = I (b) <:2(0) =j60nQ (c) E;;'1 200ei 126 .8 'V/m! 6-14. (a) Find the total field impedance al the output plane of region I in Problem 6-13, if the slab thickness is increased to a quarter wave. (b) Repeat this time for a half-wave thick slab. II:

6-15. Repeat Problem 6-13, but now assume the slab region to be 19ssy, having also ~the loss tangent E"/E' = 0.5 at the given frequency. [Answer: (a) d 1.22 em, r 2(d) = I (b) <:2(0) = 178.3ej82 .5 " Q (e) it;'l 180.5ejl29.6" Vim 1 6-16. The three-region system shown is illuminated from the left in region 1 by the given plane wave. (Note that all three regions are, in general, lossy.) Make use of (6-38), (6-39), and (6-40) to derive the following expressions for the output plane impedance (at z = 0) in region

Region 2: (/l2, <2, 0"2)

or

(,2, ~2)

fi:+xl

_+

l

Motion

HYI '---,,",

PROBLEM 6-\6

Region 3: (M:l, 10:1, fJ:1)

or (,:1,;;:1)

,382


3 intrinsic impedance fi3 and parameters of region 2

I, expressed in terms of the

:7 (0)

-,--(fic::3_+,_',_,,'-""-----c--'-'-=-----,',="--~ '12 (''13 +

= '

""1

, fi3 cosh 'Yzd = '12 fi2 cosh 'Yzd

+ fi2 sinh 'Y2 d + 1]3 siuh "hd

(6-99a)

(6-99b)

Note that the latter makes use of the definitions of the hyperbolic cosine and sine functions: (eo + and sinh () = (eo cosh 0 Assume in Problem 6-16 that region 2 is lossless and a quarter-wave thick (d "'2/4). Sketch and label this system. Show that the total field impedance at the input interface is given by (0) = '1~ji]3'

&,17.

ZI

6-18. (a) Use the answer to Problem 6-17 to find the rdativc permittivity ErZ of a so-called "quarter-wave matching plate," employed, lor example, in the low-reflection coating of lenses and prisms. In particular, suppose that a loss less glass medium (region 3 in Problem 6-16) has the relative permittivity Er 3 = 2.56 (or index of refraction n3 1.(0). Determine the required I'elative permittivity of a quarter-wave (d "'z/4) lossless dielectrie coating (region 2), if the normally incident unifi)rm plane wave is not to be reflected. (b) Show (briefly) that this system obeys reciproci~y; that is, that this system is also nonreflective if waves go from region 3 to region I. [Answer: (a) Er2 = 1.60]

"T

Determine the required thickness of the quarter-wave matching plate called If)r in Problem 6-18 (with E r3 = 2.56 for region 3), if the operating frequency of the normally incident plane wave is (a) 50 MHz; (b) 5 GHz; (c) .') x lO14 Hz (ti'ce-space wavelength Ao = 0.6 11m in the visible light range).

&,19.

Assume lor the three-region system of Problem 6-16 that region 2 is losoless and a halfwave thick (d = A2/2). Sketeh and label this system. Show that the total field impedance at the input interface has the value fi3' (This means that the loss less slab region 2 appears "transparent" to the incident wave in the sinusoidal steady state, if the wave frequency is such that the slab is a half-wave thick.) What other slab thickness will yield exactly the same result"?

+- &,20.

SECTION 6-6 6-21. Prove that equating the real and imaginary parts of (D-3), in Appendix D, leads to the results (D::4) there. Show that appropriately manipulating (D-4a) leads to (D-5), eireles mapped onto the r-plane of Figure D-I (b) for constant-i values. Use a S.!?ith chart to find the values ofr (or x) corresponding to the following specified values of (or I). Check all answers, obtained graphically, by using (6-42) or (D-I).

&,22.

x

(a) r = 0.707e j45 " (b) r = -0.5

0.5e- j126"

(c) r (d) (e)

)

x = 0.6 - jO.S x = 2 + jl

(f) .% = 0 (g)

;

---+

co

x= [Answer: x = [Answer': x

l Answer:

I

+ j2]

0.333J 0.41 - j0.44]

[Answer: r = 0.5e- i90o ] [Answer: r = 0.447ei270 ] [Answer: r

-I]

[Answer: r= I]

Rework Problem 6-9, making full use of the Smith chart. (Sketch the system; then show labeled sketches of the Smith chart, roughly as done for Example 6-4, labeling entry and exit points as well as any rim-scale rotation within a region, as needed.) In particular: (a) Find the

&,23.

PROBLEMS

383

normalized total-field impedance at the output of as well as the reflection coefficient f 2(d) there. (b) From the required rim-scale rotation, find f 2(0) at the region 2 inpu~planc, and 22(0) there. (c) Deduce the normalized impedanc~ .£[ (0) in region 1, then find r [(0) there. Determine from this the reflected wave amplitude E;"I, and also expressions for the total fields E~x1 and HYI in region I.

6-24. Work Problem 6-13, making full use of the Smith chart. [See instructions at the heginning of Problem 6-23, adding the same parts (a), (b), and (e) to the present problem.J In a totally reflective, lossless layered system such as this, what is invariably the magnitude of the reflection coefficient throughout each of the loss\ess regions? 6-25. Make use ofthc Smith chart to answcr parts (a) and (b) ofProblcm 6-14. (Showappropriately labeled Smith chart sketches.) '\ 6-26. Work Problem 6-15, making full use of the Smith chart. (Apply the instructions given in Problem 6-23 to this prohlem.) Make not£ in your Smith chart analysis of the lossy region 2, of how the reflection coefficient magnitude Ir! is reduced from unity at the perf'xtly conducting output plane, spiraling toward the chart center as one moves toward the wave source.

:r

6-27. Refer to the three-region system of Problem 6-16. Use the Smith chart to demonstrate, on making the lossy region 2 sufficiently thick, how the total-field impedance at the inpnt plane Z = 0 approac;hes the value ry2, independently of the properties of region 3.

SECTION 6-7 6-28. Make usc of the solution details of Example 6-2 to obtain the standing-wave ratio in region I three ways: (a) by means of (6-50), using the electric-field forward and backward wave magnitudes; (b) from (6-52a), making use of the reflection coefficient magnitude; and (c) from the Smith chart results of Example 6-4, using the osculation point C denoted in Figure 6-10 as the basis. '" 6-29. (a) Employ the field solutions given in Problem 6-10 to find the SWR in regions 2 and 3 of Example 6-2. (b) What arc the values of Ema. and ~~lin in region 2? How far apart (in meters) are they located? Use the Smith chart result of Example 6-4 as the b;:sis for determining how far E min is from the output plane of region 2. (c) Sketch a graph of \E(z)\ versus z in region 2, labeling values of Emax and E min at their correct locations within the slab. 6-30. (a) Make use of the total electric-field solution found in Example 6-3(d) to determine the SWR in air region I. (b) Confirm the SWR value ohtaiued in part (a), this time using the magnitude of the reflection coefficient obtained.

SECTION 6-8 6-31. Ay-polarized uniform plane wave travels in air with its propagation vector fJ tilted 30° from the z-axis in the x-z plane. Show sketches depicting the E and H vectors, and so forth, in the manner of figures (b) and (e) of Example 6-7. Its frequency is 100 MHz. Express the electric and magnetic fields in terms of the x,y, z coordinates, with the appropriate numerical values inserted. Find the wavelength and the phase velocity of this wave, as well as the values of Ay , Az , and uP' Vz associated with the y and z directions. 6-32. Solve (6-78) and (6-79) simultaneously, making use of (6-80), to obtain expressions (6-83) and (6-84) for the complex reflection and transmission coefficients for the parallel-polarized case.

6-33. Derive the results (6-87) and (6-88) for the perpendicularly polarized case, corresponding to Figures 6-15(0) and (d). "" 6-34. A uniform plane wave is incident at the angle 0; = 30° on the large planar interface that separates regiou I (air) from region 2 (a lossless plastic with the parameters Po and 6Eo)· The frequency of the incident wave is 1000 MHz. (a) Show that the refraction angle is about 11.8°. Find the wavelengths and the phase constants in the two regions. (b) Find the reflection

384

--


~

-----'-,----

--

(Glass) (Air)

(Air)

PROBLEM 6-37

and transmission coefficients for both polarizations of the incident Jidd. 1.1' the incident dectric fidd is parallel to the plane of incidence and has the complex IOOe w Vim, lind the amplitudes of the refketed anel the transmitted electric fields. Repeat (c), except for perpendicular polarizatioll. l( 6-35.

(a) A two-region system separated by an infinite planc interface consists of air and a lossless dielectric having the parameters Po and 3€(J. Show that the Brewster
6-36.

6-37. Suppose a unilfmn plane wave is incident, in air, on a glass plate with parallel laces A and B as shown in the sketch. Show that if 0i is choscn to be the Brewster angle such that zero reRection preva.ils at inted~1Ce A, then no reflection will occur at the second interfilCc. (This effect is used in the gas laser, the glass tube of which is terminated in a window tilted such that zero reflection at the desired polarization is obtained. This is done to discriminate, by means of an externally located resonator, against a resonant buildup of oscillations at the unwanted polarization ill the randomly polarized optical waves.) U€2 = 2€0, compare the reflection coefficients r i and r" at the first interface in this system.

6-38.

Seawater, at some temperature and atf = 10 MHz, has the relative permittivity 78, a dissipation factor of 62, and nnity relative permeability. (Justify calling this water medium a "good conductor" at this frequency.) A uniform plane wave, at 10 MHz in air, is incident at 30° from the normal on a flat sea of this water. (a) Find the phase constant, attenuation constant, waVelength, phase velocity, and the intrinsic wave impedance associated with the waves in both regions. (b) Based on Appendix A, justify the assumption that the refractive angle -/1 into the water is essentially zero. Find a better approximation. If the incident electric field in the air f"egion h'1;5 the amplitude lOe jO" V1m, find the electric and magnetic field amplitndes E~, fin E" and H, of the reRected and transmitted waves. (d) Give the numerical expression !Clf the fields in the water region as functions of x and y.

I

II I

)( 6-39.

Calculate the depth of penetration of the wave into the Hat sea of Problem 6-38. Is this frequency suitable lor electromagnetic communication between submerged submarines?

I "

II

-.__----------------------------------------CHAPTER 7

The Poynting Theorem and Electromagnetic Power

Energy can be transported through empty space and within or along conductive or dielectric wave transmission devices by means of electromagnetic waves. The power flow through a dosed surface in the region occupied by such waves may be interpreted from the surface integration of a power-flux density vector r1J' == E x H, known as the Poynting vector. The validity of this procedure is justified from the point of view of a theorem developed by J. H. Poynting. Applications to the power flow associated with a wire carrying a direct current and with plane waves in lossless or conductive regions are considered. The related questions of time-instantaneous and time-average powerflux density and total power flux through surfaces are treated using the real-time form of the fields. Simpler expressions for time-average power-flux density are then shown to arise from the employment of complex, time-harmonic forms of the fields. 7·1 THE THEOREM OF POYNTING

It is shown tbat the flow of electromagnetic power tbrough a closed surface is obtained from a surface integral of the time-instantaneous quantity (7-1)

known as the Poynting vector. 1 The units of (7 -I) suggest a power-flux density interpretation of r1J'. Taking the divergence of r1J' obtains the two-term expansion

v . ,OJ) = V .

(E x H)

=H

.Vx E

E.Vx H

(7-2)

lI'irst defined inJ. H. Poynting, "On the transfer of energy in the electromagnetic field," Phil. Trans. Royal

Society, 175, 343, 1884.

385

386

THE POYNTING THEOREM AND ELECTROMAGNETIC POWER

in view of the identity (16) in Table 2-2. The appearance of V x E and V x H in (7-2) prompts the substitution of Maxwell's equations (3-59) and (3-77), V x E = oB/ot and V X H = J + aD/at, yielding V·{j1=

aB aD -H'--Eo at at

JoE

(7-3)

Using the rules of differentiation (2-6) and (2-7), and with B = a linear medium

a

at

(H B) ~ [H aBtJt + Bo~!!] = [H at aB = =

0

1 2

0

2

2

H·

0

JiH, one can write for

a(J~~X + JiH OH]

at

0

at

(7-4)

at

assuming Ji is not a function oftime. Similarly, with DEE, and E not a function of time,

~(EOD)=E.OD

at

(7-5)

at

2

Substituting (7-4) and (7-5) into (7-3) yields

[H' B Eo D] -JoE

a at

(7-6)

-2-+-2-

This result shows that the power-flux density vector ~ has a divergence in a region if at least one term on the right side of (7-6) is nonzero. Integrating (7-6) throughout an arbitrary volume region V obtains

1V • v

i'P dv

EOD] 1 JoE =--ata1[H'B -+2 2 dv -

v

v

dv

(7-7)

Assuming {j1 in (7-7) meets the conditions of the divergence theorem discussed in Section 2-4A, it can be reexpressed

~s •

~

0

a1[HoB + -EoD] + 1J .

ds = -

at

v

--

2

This is the integral form of the theorem to Figure 7-1 as follows

2

oJ Poynting,

dv

v

E dv W

(7-8)

interpreted physically in relation

1. The left side of (7-8) denotes the ingoing power flux over S, assuming ds outwarddirected. In subsequent discussions, the symbol Pet) is chosen to denote the timeinstantaneous, net, ingoing power flux as follows

P(t)

==

-#s

~.

ds W

Ingoing power flux

(7-9)

7-1 THE THEOREM OF POYNTING

387

Closed --"--S~-~

9=ExH - _

FIGURE 7-1. A typical volume in a region, depicting quantities associated with Poynting's theorem.

2. The first term of the right side of (7-8) denotes, at any instant, the time rate oJincrease ql total electromagnetic energy within the volume V enclosed by S, in view of (4-61 a) (an~ (5-77) for electric field and magnetic field energies defined under static c~ditions

U==r~ndv e

Jv

2

Um

==

--dv 2 i H·B

(7-10)

v

3. The last term of (7-8) represents the total dissipated or generated power within Vat any instant. If the projection of the current density vector J along E lies in the direction of E, the power is dissipated in the region. An example occurs in a conductive region to which (3-7) applies; the substitution of J = uE into (7-8) then identifies the last term as an ohmic power-loss term. In the event of a negative E directed projection of J along E in the region, the power obtained from the last term of (7-8) is interpreted as generated power, in view of the reven,al in the sign of the integrated result. To summarize the observations (1) through (3) just given, (7-8) states that the net inward power flux P(t) = [ff> • ds, supplied by the field over a closed surface S, must equal the sum of the time rate of increase of electromagnetic energy inside V, plus the total ohmic losses in V, assuming V contains no generators. If V contains power generators, the additional volume integral of Jg • E over the designated active current sources J 9 in the region permits writing (7-8)

-h

ar

[H'B E'D]

r

,f, f -ys[ff>'ds= otJv -2-+-2- dll+ JvJ'Edv+ JvJg'Edv If the latter is rearranged with the generated power term on the left to read

i

°i[H'B E'D] i

- v J' E dv = 9 at v -2- + -2- dll + v J . E dv +

~'S [ff> • ds

(7-11 )

result is interpreted physically as follows. The total instantaneous generated power in V, given by the left side of (7-11), equals the sum of the time rate of increase in

388


electromagnetic energy in V, the ohmic losses in V, and the outgoing power flux passing through the surface enclosing V. This form has some interpretive advantages when applied to an antenna, for example, in which case the last term, the integral of [jJ • ds over any surface enclosing the antenna, denotes the power flux radiated into remote regions of space. A. The Static Poynting Theorem In a static electromagnetic system carrying only direct currents, the operator is zero, reducing Poynting's theorem (7-8) or (7-11) to

-

Iv J . E dv W

~s [jJ • ds

a/at

(7-12)

Time static

assuming V contains no generators. Thus, in a dc system, the net power flux entering a closed surface S constructed about the current-carrying conductors is a measure of the ohmic losses in those conductors. The application of (7-l2) to a dc-carrying wire is considered in an example.

EXAMPLE 7·1. From (7-12), evaluate the total power flux entering the closed surfl.tee S embracing a length t of a long round wire carrying a d.irect current 1 as in (a) of the accompanying figure. Compare the result with the volume integral of (7-12). I I

E

t

H

Closed surface S

Detail at an endcap

fa)

(b)

End view (c)

"

EXAMPLE 7-1. (al Long round wire carrying a static current 1. (b) The E and H fields on the surface S. (e) Inward power flux associated with direct current flow in a wire.

389


dng hen °

ds

The closed surface Sis noted in (b). The Poynting vector iY' on the peripheral surface p = a is obtained from the known E and H fields, H being given by (5-11) of Example 5-1, whereas E is obtained from the currcnt dcnsity Jz ljA combined with (3-7)

lOte The Poynting vector at p = a on S is obtained from

a/at

·12)

iY'

=EX H=

(axl) erA

X

12

(a4>1)

-a

2na

p

2naAer

As seen in (b), iY' on the endcaps contributes nothing to thc inward powerflux, making the total inward power flux (7-9) over S p=

_J. iY" ds j"s

= -

ct Ch Jz=o J4>=o

(-a ~) p

2naAcr

°a p

adcpdz

rmg e of

NIre

em:om-

a result expressed in terms of the resistance (4-138) of the wire. From (7-12), the result j2 R is also obtainable from the volume integral ofJ ° E taken throughout the interior of S. Thus

C JoEdv= Jv C (erE)oEdv= Jv r erE;dv= Jz-o C_ C: r"_ er(~)2PdPdCPdZ Jv J4>-O Jp.-o erA 2

t

integrating to 12R as expected. The positive sign accounts for the actual inward sense of the power-flux P over S, as noted in (c).

B. Time-Instantaneous Poynting Theorem and Plane Waves Illustrations of the Poynting theorem in the time domain can be drawn from the of plane waves developed in Sections 2-10 and 3-7 .-Thus, the power-flux-density vector &' associated with a plane wave in a region is obtained by use of (7-1) applied the appropriate fields. In empty space, assume that a positive z traveling plane wave electric and magnetic fields inferred from (2-121 a) and (2-130a)

(7 -13) (7-14) Applying these to (7-1) obtains the time-instantaneous Poynting vector at any Z position

&'(z, t)

E XH =

[axE~ cos (wt -

Poz)]

X

[ay~: cos

(wt - Poz) ]

(E+)2

= a z ~- cos 2 '10

(wt - Poz)

(7-15a)

390


Inset: flux Plot} of &"

(z)

I I I I (Ito, fO)

__ ..---

--~ Wave

I __ J...._ 0 - - ___ _

(y)

motion

--------

----(z)

(a.)

\

\

-~

Motion

----(y)

(z) (1))

FIGURE 7-2. The Poynting vector associated with a plane .wave in empty space. (a) The vector.o/' = a/!i', versus <: at t = O. (b) The scalar .o/'z(':' t) at t = O.

The sketch of (7-l5a) in Figure 7-2(a) shows f!IJ everywhere positive z directed. Denoting f!IJ(z, t) by azfJi': (z, t), an alternative plot of the scalar is shown by the solid line in Figure 7-2(b). A double frequency variation off!IJ with t and z produced by the squared cosine function is evident from these diagrams. Using the identity cos 2 0 = ! + (!) cos 20 permits writing

fJi':

(7-15b)

a result useful when considering time-average power in the next section. The Poynting integral (7-8) applied to a region with no ohmic losses and no generators present red1.i!ces to

-

£ 7-1 THE THEOREM OF POYNTING

J.

1'(1)

:f~~

fY>. ds

391 (7-16)

signifying that the flux of:1' into a closed surface S in the lossless region is instantaneously a measure of the time rate of increase of the stored electromagnetic energy within S. In the example that follows, the validity of (7-16) is examined relative to a plane wave in free space. EXAMPLE 7-2. Given the plane wave defined by (7-13) and (7-14), determine the net power flux P( t) entering a closed box-shaped surface S having dimensions as in the accompanying figure. Show that the time rate of increase of the electromagnetic energy within the volume of the box provides the same answer. Because {YJ is everywhere z directed, the only contributions to power flux entering the box are on thc ends Sl and 5;2 shown, so (7-16) yields

1\ (I)

-

.- !is f .0/'. N,

= -

i

b

y~o

fa [a x=o

z

(E~)2 - cos 2 (wt '10 (7-17a)

'10 (7-17b) The net power flux entering S is therefore

'-Ps PI)· ds

1'(1) = 1\(1)

+

i-

ab[cos 2mt - cos 2(w/

=

flod)] W

2'10 the last being obtained by use of cos 2 0 =

EXAMPLE 7-2

! + (!) cos 20.

(7-18)

Ci

A

392


Equivalently, if the right side of (7-16) is integrated throughout the volume of the box, (7-18) should again be obtained. Substituting (7-13) and (7-14) yields

a '" vt

J. [1l0H2 EoE2] a {llo(E;:Y J.d J.b J.a [1 + cos 2(wl - Po,;:)] dxdyd,;: -2- + - dv = '" 2 J.d J.b J.. [1 + cos 2(wt - Po';:)].dxdyd,;: } + Eo(E;:Y 4 v

vi

--2-

41'/0

0

0

a {Eo(E;:Y = at 2 =

0

0

0

'

0

J.d J.b J.a [1 + cos 2(wt 0

-

0

'}

Po';:)] dx dyd,;:

0

wEo(E;:yab 2Po [-cos2(wt-Po,;:)1~ (7-19)

------ [cos 2wt - cos.2 (WI - Pod)] agreeing with (7-18) as expected. 2

For a plane wave traveling in a conductive (ohmic) region, the effects of the attenuation of E and H and the phase shift between them is expected to i~fluence the power-flux P(t) entering a closed surface S. For this case, the fields are'fven by real-time expressions inferred from (3-94) and (3-98c) \. (7-20) (7 -21 ) which (J is the angle of the wave impedance (3-99). The Poynting vector (7-1) thus becomes [#>(z, t)

=E

x H

=

(E+)2 a z _m_ e- 2az cos (Wi - (Jz) cos (wt

{Jz -

(J)

(7-22a)

1J

and the use of cos A cos B [#> = a z

2P;

=

(!)[eos (A

- B)] obtains

t)

e- 2az[cos

2P;

+ B) + cos (A

(J

+ cos (2wt -

2{Jz

(7-22b)

(z, t) versus Z at t = 0 is shown in Figure 7-3. Not only does the A graph of attenuation of and account for a doubly attenuated power-flux density but the effect of cos (J in (7 -22b), replacing the term unity in (7 -I5b) for the lossless case, to go negative over a portion of each cycle, ~an effect associated with the to cause

E:

H;

2P; ,

2P;

~In the course of obtaining (7-19), note that with the snbstitntion (Jio/'1l) = Eo, the two integrals in the first Jlep become identical, so they combine into one. The time differentiation is taken inside the integral to elimthe constant unity term, whereas in the last step, the identity (W€o/ Po) = 1/0 1 is used.

In


393

(x)

At t = 0

(y) '~(z)

FIGURE 7-3. The instantaneous Poynting vector traveling plane wave in a conductive region.

(z, t) associated with a positive ::.

phase shift 0 between the electric and magnetic fields and detracting from the average power transmitted in the z direction. EXAMPLE 7·3. If a plane wave exists in a conductive region, evaluate the net instantaneous power flux entering the box-shaped closed surface of dimensions as shown. Integrating (7-22b) over the ends S1 at Z = 0 and 8 2 at z = d yields the instantaneous power fluxes

= -

f

b

y~O

fa {a z (E~)2 - [cos e + cos (2m{ X~O

2'1

e)]}. (-azdxdy)

abl cos 0 + cos (2m/ - e)]

P2(1) = -~~- e-2~dab[cos 2'1

e + cos (2mt -

(7-23a)

2{3d

(/J.,t, o)

..

'

(y) (z)

EXAMPLE 7-3

e) ]

(7-23b)

394


so that the net power flux entering the box is their sum P(t)

=

(E+ m

211

ah[(l

e- 2ad ) cos fJ

+ cos

(2M - fJ) - e- 2ad cos (2wt - 2fJd

fJ)] (7-24)

From the Poynting theorem (7-8) it is evident that (7-24) is a measure of the time rate of increase of the stored electromagnetic energy within the volume plus the instantaneous ohmic loss occurring therein. One can expect that (7-24) will reduce to (7-18) if a lossless region (IT = 0) is assumed.

7·2 TIME-AVERAGE POYNTING VECTOR AND POWER

In a consideration of the electromagnetic power delivered by sinusoidally time-varying fields to a region or system, one's interest from the point of view of practical measurements leans toward the time average of the power flux rather than its instantaneous value considered in the previous section. Time-average power in electromagrl~tic fields is important for the same reasons as in circuit theory. The time-average po"ter)entering the terminals of a passive network, found by use of an electrodynamomete~pe wattmeter or from the knowledge of the amplitude and phase of the input voltage and current, is a measure of the average power dissipated as heat in all the resistive elements of the network. From the electromagnetic viewpoint, the time-average power flux entering a closed surface containing no generators is a criterion of the same thing: the heat-producing ohmic losses in the region. In laboratory measurements, the time average of a time-harmonic function is customarily taken over a time interval embracing many cycles '01' periods. Since for steady state sinusoidal functions all periods are alike, an average over one period will yield the same result as that taken over many such periods. The time average of the Poynting vector ~(Ul' Uz , U3' f), denoted by ~av' is defined as the area under the function ~ over a cycle, divided by the duration T (period) of the cycle, that is,

f!lJav(Ul' Uz, U3)

=

Area under f!IJ over a cycle 1 Base ( T sec) = T

iT 0

~(Ul' Uz, U3, t) dt

(7-25a)

if t is chosen as the variable of integration. One may alternatively choose wt as the angular integration variable; then (7-25a) is written with 2n as the base-divisor

(7-25b)

It is evident that the time-average Poynting vector is a function solely of position in space, the time variable having been integrated out over definite limits (in t or wt) in the averaging process.

7-2 TIME-AVERAGE POYNTING VECTOR AND POWER

395

A. Time-Average Poynting Vector and Plane Waves Illustrations of the time-average Poynting vector can be drawn from examples in the last section. Equation (7-ISa) denotes a time-instantaneous Poynting vector ~(z, t) = a~z (z, t) attributed to the wave of (7-13) and (7-14). Applying (7-2Sb) obtains its time average

PI' av (z)

az

(E~

2

2n 2110 (E~)2

=a z

2110

P" d(wt) + a z (E~)2 f Z1t cos 2(wt

Jo

2n 2110

Jo

fJoz)d(w/)

(7-26)

Thus the time-average result (7-26) is attributable wholly to the constant first term of the time-instantaneous expression (7-ISb). The double frequency term contributes nothing on the time average because it possesses canceling positive and negative areas over a cycle, evident from the f!Jz(z, t) diagram of Figure 7-4(a), which is just an extension of Figure 7-2(b) to successive instants in time t. The inset in Figure 7-4(a), showing the wave at the fixed Z = 0 location, yields an average Poynting vector (area divided by the base) that is one-half the peak power density (E~)z/110' or (7-26). If the region is lossy, ~ av becomes a function of Z due to the wave attenuation produced by the losses. The time-instantaneous Poynting vector, in this case expressed by (7 -22b), is depicted in Figure 7-4( b), an extension of Figure 7-3. In the insets are shown variations off!Jz (z, t) with t at two fixed z locations (z = 0 and A). Making use of (7-2Sb) leads to the time-average Poynting vector

(7-27) The result is doubly attenuated in z; it also retains the factor cos 0 produced by the electric and magnetic fields being out of phase by an angle 0, a factor analogous to the power factor of a two-terminal impedance of circuit theory.

B. Time-Average Form of the Poynting Theorem If the total time-average power flux through some surface S (not necessarily a closed surface) is desired, one must integrate Pl'av over S by use of

(7-28a)

in which S denotes an arbitrary surface, open or closed. With S closed, (7-28a) yields the net (or total) power flux leaving that surface. A negative sign must be included with the integral of (7-28a) if Pay is to signify the net time-average power flux entering the

:1

396


z t

="

z (or (:iz)

(or wt)

o

\

(a)

(2wt - 2{:i - 8)1

~

"

Wave motion

z= "

(b)

j<'IGURE 7-4. Poynting vector 9",(,,,, t) of forward traveling plane waves in lossless and lossy regions. (a) Plane wave in a lussIess region, Time variations at 0 afC noted in the lower inset. (b) Plane wave in a lossy region, Below are shown time variatiollS at 0 and A.

surface S. Another way to evaluate Pay is by averaging the total time-instantaneous power-flux P(t) through S. Thus, (7-25b) inserted in (7-28a) yields

f f!l'av' Js

ds

[_1

= f f27t f!l'd(wt)] Js 2n Jo

. ds

whence

P av

=

1

2n

12" P(t)d(wt) W 0

(7-28b)

7-2 TlME-AVERAGE POYNTlNG VECTOR AND POWER

397

The preference for (7-28a) or (7-28b) in evaluating Pay depends on the comparative convenience of the integration process.

EXAMPLE 7-4. Evaluate the net time-average power flux entering the closed surface of Example 7-2 iu a free-space region containing the given wave. The time-average power flux entering the box is found by use of (7-28a) or (7-28b). With ds denoting a positive outward surface-element, (7-28a) is written with a minus sign if the net inward flux is desired

-~ f1l>.y • ds

Pay =

(7-29)

With ,Cj>av given by (7-26), the average power flux entering Sl' Pay

,

1

= -

f

s,

f1l>av'

ds = -

~

b ~a a [(E~)2J (E~ z - - • (-azdxdy) =

0

0

21]0

21/0

ab

is positive beeause the true direction of the flux is into the box. A similar integration over S2 yields the negative of that result because the flux comes out of the box. The net timeaverage power flux entering the box is thus zero, that is, Pay = Pay,l + P av ,2 = 0, a result expected generally from closed surfaces embracing a lossless region and containing no sources.

For a sinusoidally time-varying electromagnetic field in a region possessing losses but no sources, the time-average power flux Pay entering a closed surface is a measure of the time-average ohmic power loss within the interior volume. This is demonstrated by beginning with the time-instantaneous integral form (7-8) of Poynting's theorem

P(t)

[7-8]

Assuming sinusoidal fields, the time-average of the left side of (7-8), given by (7-29), equals the time average of the right side, yielding

1 2n

12" -;,aU d(wt) + 1 12" aU 2n e

0

m

ut

0

d(wt)

(7-30) The stored-energy quantItIes Ue and Um are, from (7-10), obtained from volume integrals of E2 and H2 respectively, implying double frequency variations in time. Such time variations of Ue in a volume region are depicted in Figure along with its time derivative aUe/at. Its time average, given by the first integral of the right side of (7-30), is therefore zero. Similar arguments lead to a zero time-average of DUm/at, reducing (7-30) to the time-average form if Poynting's theorem:

f2"[rJv J'EdV]d(wt)W

'+'flJ'av. ds = I fs 2n Jo

(7-31 )

398


ue =j' j)·E dv =)' cE" dv v 2 V 2

A

U',"=f,·:,~"]\-A~Aj\ I

I

I

-----.-----'>-

t

I

I I

I I

I

I

I

I

I

I

I

(or wt)

I

I I

I I -"._--_.. _--

.-.-~---~.~

t (or wt)

FIGURE 7-5, Total electric fidd energy and energy lime rail' of time I'll" a volume region, assuming sinusoidal Ildds,

(:()n~tions

as a function

'. ",

) One concludes that the time-average power flux entering a closed surface S equals the average power dissipated as heat inside V bounded by S, provided there are no sources in V.

EXAMPLE 7·5. Compare the net time-average power flux entering the box-shaped surface of Example 7-3 with the time-average ohmic losses inside, assuming the same attenuated wave in the region, Anywhere in the region (YJav is given by (7-27) e- 20. cos

21]

e

(7-32)

Inserting (7-32) into (7-31) yields contributions to Fav over only the box ends at d as follows

z=

0 and

.~ =

Pay

== -rhs {YJav • ds ~ _

fa_ x-o

(E+)2 m

fb_

Jy-O

ab[l

= - fa

fb

x-oJy-O

(a z (E;;; )~- e- 2az cos

(az (E;;;)2 e-- 2az cos 21]

e- Zad ] cos 0

21]

e).

e) .(-a

z

dx d1!)'] Y

.-0

(azdx dy)] _

z-d

(7-33)

211 One can also obtain (7-33) by use of the right s.ide of (7-31) through the time-average ohmic losses in V. The integration simplifies if one puts

(7-34) stating that the time average of the volume integral ofJ . E equals the volume integral of


and making use of J = aE yields

the time average of J . E. Inserting E from

Iv Un I:" J' Ed(Wt)}V = Iv L~ Iozn a(E~)2e-2.z cos =

a(E+ )2 m

399

2

(wt - PZ)d(Wt)}V

ab[l _ e- 2 • d ]

(7-35)

4cr which equals (7-33) provided that cos ()

a

I]

2cr

(7-36)

It is left to you to prove the latter, usiug the appropriate definitions of cr, Section 3-6.

1],

and () from

C. Time-Average Poynting Vector and Complex Time-Harmonic Fields In the discussion of plane wave fields in Sections 2-10 and 3-6, it has been seen how the use of the complex f()rms eliminates t through the use of the factor dwt • Because in the course of problem-solving, field solutions are frequently obtained in complex form, it is useful to be able to find the time-average Poynting vector [lI'av directly from the complex solutions. Such results are obtained in this section, along with a version of the Poynting theorem (7-8) employing complex forms. Revising (7-25) in terms of the complex fields requires restating the Poynting vector in terms \?f the c!:?mp]ex fields. The real-time fields E and H are related to their complex forms E and H by (2-74); that is, (7 -37) (7-38)

E and H are expressed in complex polar form as follows (7-39) (7-40) designating the vector directions of the fields. The polar form in (7-39) substituted into (7-37) obtains the relationships between the complex and real-time forms 3 E(uu

UZ, U3,

t) = Re [E(uu uz, u3)d wt ] = a e Re [Ed wt ]

(7 -41)

= aeE cos (wt + (}e)

(7-42)

= ae

Re [Ed(wt+O

e )]

lThe physical meanings of the symbols a" E, and (J, arc clarified by comparing (7-41) with explicit solutions. For example, from (3-93), a positive traveling wave in a dissipative region is

Ca)mparison wi th

shows in this case that a e = ax> E =

and (Je =

-fh.

i2

400


Similarly, for the magnetic field ( 7-43) By use of the real-time forms (7-42) and (7-43), the time-instantaneous Poynting vector (7-1) can be written

(7-44 ) On integrating the latter by use of the definition (7-25), the time-average Poynting vector is seen to retain just the constant term cos (ee - eh ), yielding the general result (7 -45) It is ofJnterest how (7-45) :night alternatively~e obtained by .use of the complex tir~te harmonlj forms ofE and H gIven by (7-39) and (7~40). On definmg the complex Poyntzng vector f? by \ )

#

=

Ex H*

(7-46)

it is seen that the time-average Poynting vector expression (7-45) is obtainable from (7-46) by simply taking one-half its real part, or

f? av =

1. 2

Re

[E

x

H*]

(7-47a)

It is left to you to show, on substituting the complex expressions (7-39) and (7-40) into (7-47a), that precisely the result (7-45) is obtained. An alternative expression for f?av IS

f? av =

l2

Re

[E*

X

Hl

(7-47b)

These expressions, (7-47a) and (7-47b), for the time-average Poynting vector, make it possible to use the complex forms of the fields directly in the calculation of time-average power density or power, thereby obviating the need for converting the field solutions to real-time forms, as required for using the more cumbersome real-time integration (7-25). If the net time-average power flux entering a closed surface S is desired, inserting (7-47a) into (7-29) now obtains

p av

=

_r+, j:s f? av . ds

=

1 Re IE _r+, -2 ~s \

X

H*) . ds W

(7 -48)

EXAMPLE 7-6. Use the complex form of the attenuated plane wave fields (7-20) and (7-21) to obtain the time-average Poynting vector at any position in the region.

I

4J


The complex forms of

401

and (7-21) are (7-49)

H~ (Z) =

Ed; e-az e-j" e-jpz

ay -

(7-50)

IJ

and with these into (7-51b) f1'av =

1

2

~ Re (E

X

~ H*)

= 1Re ( ax x a y

'1

(7-51 ) which agrees with

The foregoing showed how the time-average electromagnetic power flux entering a closed surfilce is obtained using the complex E and H fields. This was seen, from the time-average Poynting theorem (7-31), to have the important interpretation of representing the time-average ohmic power loss in the volume enclosed, assuming no sources therein. An alternative versioll of (7-31) is obtained directly from t~e complex~ Maxwell 5:ctualiol1s. 'l)ms, beginning with (3-83) and (3-84), V X E -j(f)ftH alld V X H = J + j(f)EE, and forming the dot product of (3-83) with the conjugate of H, and the dot product of the conjugate of (3-84.) with E, obtains (7-52) (7-53) Subtracting (7-53) from (7-52) yields

the left side of which reduces, using (16) of Table 2-2, to yield (7-54 ) Integrating (7-54) throughout any volume V obtains, on applying the divergence theorem (2-34) to the left side, the following complex version of the Poynting theorem

If the current densities j in V consist partly of driven sources jg (generated currents), the additional volume integral ofj: . E over those sources converts (7-55a) to a result

402


that, when rearranged with the generated power term on the left, reads

- Jv f j*. Edv 9

=

jO) f [pH· H* - EE' E*] dv

Jv

+ Jv f j*- Edv + J. JS (E

X

H*) . ds (7-55b)

(, The choice of (7-55a) or (7-55b) thus depends on whether or not cl}rrent generators }9 are present in the volume under consideration. Their real-time counterparts are (7-8) and (7-11), developed in Section 7-1. The physical interpretations are rather diflerent, however, as seen from the following. Physical interpretations of the complex Poynting expressions become evident on equating the real and the imaginary parts of (7-55a) or (7-55b). Assume a source free, dissipative volume region with p, and E pure real and) = aE. Equating one-half the real parts of (7-59a) yields the following Poynting integral expression

Pay

== -

~ t Re (E

~

S

X

~ H*) • ds

=

i

v

aE2

2

dv W

(7-56)

l20ting that Re (j* . E) = Re (aE* . E) = aE2 , in which E denotes the magnitude of E according to (7-42), while from (7-48) the left integral of (7-56) is just Pay, the timeaverage power flux entering S. Therefore, (7-56) and (7-31) are entirely equivalent expressions.

The equality of one-half the imaginary parts of (7-55a) obtains

J. tim (E~ -Js

X

~

H*) . ds = 20)

r [PH'4 H* - EE 4. E*] dv Jv

(7-57)

The terms (!)pH' H* and (!)EE' E*, independent of time, denote the time averages of the stored energy densities of the magnetic and electric fields in V, a fact appreciated on reexamining Figure 7-5, showing the total instantaneous field energy of a sinusoidal electric field along with its time average in a typical volume region. Thus, in a volume region containing no sources, (7-57) states that the imaginary part of the col1iplex power flux entering the closed surface bounding V is a measure of 20) times the difference of the time-average energies stored in the magnetic and electric fields. 4 (This quantity is sometimes symbolized Q. when applied to Land C energy-storage elements of circuits, details of which are given further discussion in tcxts on circuit analysis.) The foregoing interpretations of the real and imaginary parts of the complex Poynting tfteorem (7-55a) can be extended to a region containing currcnt generators of density }g by a similar consideration of (7-55b). One-half the real parts then yields 4The counterparts of the volume integrals of (llJlH . H* and (l)EE . E* in a series or parallel RLC circuit are the quantities cllLll* and (llCVV*, which represent the time averages of stored magnetic and electric field energies of an inductor and a capacitor.

PROBLEMS

f Jv ! Re

(J:~ .E)~ dv

=

f (JE 2 ~ ~ ~ Jv 2 dv + fs! Re (E x H*) . ds W

403 (7-58)

The left ~ide denotes the time-average generated po~er in V, contributed by components ofE in phase with the current density sources Jg. The time-average generated power thus equals the sum of the time-average ohmic losses in V plus the time-average of the total power flux leaving the closed surface S that bounds V. This form of the Poynting theorem is useful when applied, for example, to generators of radiated power such as antennas. Thus, in free space (containing no losses), (7-58) states that the power flux emerging (radiated) from any surface S enclosing the antenna equals the power driving the antenna terminals, or simply a statement of the conservation of energy.

REFERENCES ELLIOTT, R. S. Electromagnetics. New York: McGraw-Hill, 1966. LORRAIN, P., and D. R. CORSON. Electromagnetic Fields and Waves. San Francisco: Freeman, 1970. PLONSEY, R., and R. E. COLLIN. Principles and Applications McGraw-Hill,1961.

ql Electromagnetic

Fields. New York:

PROBLEMS

SECTION 7-1 7-1. Carry out the details of the volume integration of (7-12) applied to the long, round conductor of Example 7-1 to verify the result 12 R.

7-2.

Given is the coaxial line shown in Example 5-13, carrying the currents I, -I in its two conductors. Sketch this system. Show details for finding the power loss of only the outer conductor, using an approach suggested by Example 7-1 and involving only the left side (surface integral) of the static Poynting theorem (7-12). (Show the closed surface used on your sketch.) Show that the surface integral equals 12 R, R being the resistance of the outer, hollow conductor.

7-3.

Repeat Problem 7-2, except in this case determine the power loss in the outer conductor by use of the right side of (7-12), the volume integration of (fE 2 dv, establishing that it equals

J

llR.

74.

The following specify, in real time, a negative wave in free spate

z traveling,

x-polarized uniform plane

Find the corresponding expression for the real-time Poynting vector, [11't), expressed terms of a double-frequency term plus a constant. Sketch a figure, as perhaps suggested by 7-2(b) shown at a fixed instant. (b) Given the hypothetical rectangular closed box of dirnellSi()ns (a, b, d) like that of Example 7-2, find the net time-instantaneous power entering its lurface. Show a labeled, relevant sketch. State the physical meaning of this result, relative to time-instantaneous form (7-16) of the Poynting theorem.

404


A negative z traveling, x-polarized uniform plane wave, moving through a lossy (attenuative) region, is given by the fields

7-5.

E

t) = axE;;' eaz cos (cot

= axE".;

+ pz) '\

E-

H =ayH;

t)

(I)

- a y ~ eaz cos (wt

Yf

+ pz

0)

in which the intrinsic wave impedance ~ is Yfei°. (a) :Find the corresponding time-instantaneous Poynting vector, showing in detail that at any Z location

f!I'(z, t)

t)

(3)

Sketch a figure, suggested by Figure 7-3, in relation to these fields and depicted at t = O. Discuss it briefly. (b) Given a box-shaped elosed surface like that shown in Example 7-3, find the net time-instantaneous power flux entering that surf~Ke. Given a physical interpretation to this answer, in light of the Poynting theorem (7-8).

SECTION 7-2 7-6.

Prove that the result (7-35) in Example 7-5 is identically

Employ (7-25) to find the time-average Poynting vector f!l'av of the negative z traveling plane wave in a lossless region as given in Problem 7-4. Show that the net time-average power flux, defined by (7-29) and entering the closed box-shaped surface of Problem 7-4·(b), is zero. Show a sketch that indicates the choices of the (outward) vector surface elements, on those sides of the box through which power flux passes. Does the result satisfy the -time-average form (7-31) of the Poynting theorem? Explain.

7-7.

Make use of (7 -25) to obtain the time-average Poynting vector of the negative z traveling wave in a lossy region as defined in Problem 7-5. Sketch the dosed box-shaped surface given, and find the net time-average power flux entering that surface, as defined by (7-29). Label properly the surface elements ds 011 the important sides of the box. (Note, in the averaging integration process that the sinusoidal time-funetion term averages to zero by inspection; no formal, detailed integration is Interpret the result physically by means of the time-average Poynting theorem.

7-8.

SECTION 7-3 7-9.

Use (7-47a) to find the time-average Poynting vector of the uniform plane wave specified (a) in Problem 7-4 and (b) in Problem 7-5. (Be sure to convert the given real-time fields to their equivalent eomplex forrns.)

H;-

Given in free space are the x-polarized plane wave fields I';: t) and t) of Example 2-11 (c). (a) Find the time-instantaneous Poynting vector, expressed as the sum of a
7-10.

E:

X 7-11.

,!I II

I

H: .

The attenuated plane-wave fields in the lossy region of Example 3-8 are

il: (z)

lOOOe--L9ze-j4,S8z Vim =

6.2ge -

1.9z e- j4.58z e - j,,/8

A/m

• PROBLEMS

405

(a) Express these fields in real-time form, and find the time-instantaneous Poynting vector, expressed as the sum of a double-frequency term plus a constant. (b) Find the time-average Poynting vector two ways: from the time-averaging integral (7-25a) and from (7-47) using the complex fields. (c) Obtain the net time-average power flux Pay entering a closed box-shaped surface like that of Example 7-3, with a b = d = 50 em. Show an appropriately labeled sketch. Interpret the answer on physical grounds, from the standpoint of the time-average Poynting theorem (7-31) or (7-56).

+ j,ji is 4:* = + B)* = A* + B*,

~ 7-12. ~ Jf the ~conjugate o~ ~he con:pl..cx quantity)' = _Ar

IAI2,

('}) AA* (b) (e)A -A* =j2A j •

(AB)*

= A*B*,

(c)

(A

Ar - j4;, PIove that (d)

A

+ A* = 2A"

7-13. (a) A lossless region possesses the complex incident and reflected plane waves given by (6-35), (6-37) as follows

(I) Using (I) in (7-47a), show in detail that the total time-average Poynting vector power density at any Z becomes (7-59) in which (7-60) or just the vector sum of the time-average power densities associated with the incident and reflected waves when considered individually. [Hint: l.n the expansion using (7-47a), some results in Probkm 7-12 may be usefuL] (b) Show that the expression for the net time-average power flux passing through some normal open suriace S (of area A) is simply (7-61 ) Then show that the positive ratio of the reflected to incident time-average powers through Sis just the sq uare of the reflection coefficient, or (7-62) The so-called "return loss" of this power reflection process, expressed in decibels, is defined

Return loss (dB) = 10 log

11'1I If::

= 10 log

r2

(7-63)

'.14:,..

If, in a lossles} air region, the incident and reflected amplitudes of (I) in Problem 7-13 are E;:; = 100 Vim, E;" = 40Ci60 " Vim, use results developed in Problem 7-13 to find (a) the reflection coefficient magnitude; (b) the net time average Poynting power density at any point, the net time average well as g>~~ and {YJ~~ associated with the ineident and reflected waves; and P a-v power flux passing through the normal surface S of area A = 4 m 2 , as well as passing through S. Find also the ratio of these positive powers and the return loss (in dB). What would the return loss be if 100%, reflection occurred? Zero reflection?

r::.,

406


1: air

r-81 1

--, I + 1;1'av<

~

I

I

o

~ .~ ..

, ,:52

---+

.'~a---:';ll

I ~-

I ~.

2: (ila, 4'0)

1

1:1'.+vI

~..---- (z)

)

1

,

---4

d-o>-j

PROBLEM 7-15

7-15.

Given is the two-region lossless system of Example 6-1, with incident, reflected, and transmitted plane-wave solutions as noted. Construct a hypothetical closed rectangular box with opposite sides S't and S2 parallel to the interhce and protruding into Ihe two regions as shown. (a) Using the complex field solutions obtained in Exa mple 6-1, evaluate the corresponding timcaverage Poynting vector power densities &>,,',.1' &>~v.l' &>:V,2 as labeled on the diagram here. Given that the x and)) dimensions or the rectangular box arc a = b 50 em, determine t.he net time-average power flux entering the closed surface of the box. [s the time-average Poynting theorem (7-31) or satisfied? Explain. (c) It is seen that the conclusion orpart (b) is independent of the length d this closed box, Why is this so? [Answer: (a) 13.26, 1.47, 11.79 W/m2]

~

In the three-region lossless syste!p or Examrje 6-2, the s(~utions in 'lJe three regions revealed the electric field amplitudes E:;'1 = 100, E;d = 60, £:;'2 60, E;"2 = 20, and = ~ j80 V 1m. Sketch this system. (a) Find the magnetic-field amplitlldes (iI:;, 1, etc.) a('colIIpanying each of the given traveling-wave eleclric held amplitudes. Duermine the incident and reflecting time-average Poynting vector power dcnsilies (&'a~,l' etc.) in each regiou. (b) Add a dosed rectangular box, as for Problem 7-15, to this system, snch that the surfaces SI and S2 extend into I ancl2, respectively. Sketch it on yoU!' diagram. With the x and_y dimensions as a 2 m, fiud the time-average power fluxes, Pt, p~, pi, and Pi. passing through and 8 2 in the two regions. Calling the power flowing into the dosed box positive, determine net power flow into the box, whence conclude whether the time-average Poynting theorem ) or is satisfied. (c) Add a second dosed rectangnlar box to the this time with it frontal surbec S I still in region I, and with the opposing snrf;lCc extended into region 3. Sketch it OIl the diagram, Determine the net time-average power flow into the box. Is the Poynting theorem (7-56) satisfied?

7-16.

lc' 7-17.

(a) Do Problem 7-13(a), except assume this time a lossy region [i.e., llse the general Ibrms (6-35) and (6-37) instead of their lossless region versions]. Employ these in (7-47a) to show that the total (or net) time-average vector power density can be written &>av

az

(£:+)2 2az 2~ e[(1

··

(7-64)

in which f = Ir + jIi and fi = 11, + jl1; = firJo have been assumed. To what result does (7-64) reduce in a purdy rct1cctionlcss region? (b) Show that the general result (7-64) reduces to its lossless region version of Problem 7-13, if appropriate assnmptions concerning y and fi made. (e) In Problem 7-13, the result (7-59) shows, in a lossless regioll, that the total power can be dissociated into the contributions &':V + &'a--' provided by the incident and re!lee ted waves when considered individually, From the form of fi)!' this lossy region, argne to why no such equivalent statement can be made here.

PROBLEMS

407

2: (1'0, 6'0,

'---

0.03)

1: air Perfect conductor

j,;+ xl

/Waveguide

Direct ray ___ {

S --?-

iI.ii· ray

-~ ------

(z)

01 02 .:--

..... -

'0 4

(a)

(b)

PROBLEM 7-19

7-18. In Example 6-3, the f(mr-region problem involves a showing a hypothetical closed SurLUT S of rectangular box reglOn and the opposite such that its frontal surhec .)1 projects just within losslcss region 2 surlZtcc 8 2 is located at = 0 just inside region 4. 8 1 and S2 have the area A 4 m each. (a) With the known electric-field magnitudes E;:;2 = 85.7, E;;'2 37.0, and 47AVjm, find the time-average vector power densities on S 1 aryl 8 z . Label these vectors on the sketch. Find also lhe time-average power fluxes into (or out: of) S. How mueh power loss occurs within the region 3 bounded by the closed surl~lec S? 7-19. A microwave oven consists of a metal oven enclosure !Cd li'om a magnetron source, usually operating in the S-band (about 2.45 GHz), a hollow metallic by figure Power coupling from to the oven occurs (see Chapter B) as at the coupling aperture A, with microwaves illuminating the by direct rays /i'om the aperture, or from indirect rays produced as oblique reflections from the oven walls as shown. The sarnple is spaced above the oven wall to enhance the heating cfl(:cts. A simplilied model of the microwave heat ing process roughly produced by the direct is shown in figure (b). Let the operating frequency he 2.5 GHz. (a) Find the wavelength in air regions I and 3, the gap width and iX, /l, q and 'Az regi~.m 2. (b) Find the reflection coemcient at z d3 and Z 0 in region :l, as well as and r 2 (ri 2 l. (c) AssUfItc three cases of lossy slab thickness: d and Calculate lin' each the impedance <:1 (0) by the arriving wave at the first iEterf;lCe. Which case is least reflective in region I? Choose dz = 4..)A z of . Letting (z) 3161e,find the incident time-avcr~gc I~)ynLing power density ·1j>~.I' [Answer: 1 a z 13.25 kW/m2] Find the wave amplitudes R~2' E,;.2 in region 2. Sketch a labeled graph depicting only the magnitude versus of the incident and reflected electric lidds in the lossy 0 and ·c = slab region, labeling the values at

E:

7-20.

For the same microwave heating model of Problem 7-19, let d2 4.5Az and consider only the average power injected into a sample cube region of the lossy slab, with cross-sectional (x andy) dimensions a b = Sketch a cubical closed surLlec with these dimensions, jnst embracing that amount oflossy sample and having input and outpnt SUr!;lCCS S1 and la) Find the net time-average power Ilowing through 8 1 into the volume Why does zero power flow occur through Sz? Use (7-31) or (7-56) to illf(~r the timc-averageJoulc heat generated within the cubc. Answer: Pay 644 W] How do you know that heat is being generated nonunilormly in this lossy sample? Where the maximum density of heat generation? (e) The heat

r

408


generated within the lossy sample in t sec is given by

( (1)

if Fav is the net time-average power flux injected into the sample. Assuming that the heat is generated nearly uniformly in the cube, the heat required to raise the temperature of the sample mass m by /',. TOC becomes Ch /',. T, if Ch is the sample "specitic heat capacity." Assuming this particular sample to have the constant specific heat capacity Ch = 0.50 calorie/gOe (= 2.09 J/gOC) and a specific gravity dm = 1.3 (giving this sample thc mass m = drn V = 13.94 kg), calculate the amount of heat U h and the time t needed to raise the temperature of this particular sample ii'om ambient (20°C) to 170°C, assuming the same net power input as in (a). [Answer: U h = 4.37 MJ t 113 min]

.Ir 7-21.

At the distance from the sun to the earth, the sun produces the timc-average electromagnetic power flux density of about 1340 W /m 2 . Its power is contributed by frequency components ranging ii'om radio frequencies through the ultraviolet region and beyond. (a) Supposing that this power density arrived at a single sinusoidal frequency, what electric and magnetic field amplitudes would be required to produce this power density? (b) Use a suit.able surEtee integration to calculate the total time-average power radiated {l'om the sun. The distance from the sun to the earth is about 148 Gm.

__---------------------------------------CHAPTER8

Mode Theory of Waveguides

(

In this chapter, the wave reflection problems of Chapter 6 arc extended to the theory of waveguides, regions of uniform cross section bounded by conducting walls parallel to the propagation direction. 1 Typical waveguide configurations are shown in Figure 8-1. To simplify the analysis, perfectly conducting walls are assumed, except in Section 8-6 in which the attenuative eHects of wall losses are analyzed. The boundary effects of the conducting walls, producing only normal electric and tangential magnetic fields there, favors a z direction of energy How, so the waves are said to be guided in the z direction. In this sense, the wave transmission systems are said to be waveguides, though this term is usually restricted to the hollow, rectangular and circular cylindlical systems of Figures 8-1 (c) and (d). Two-conductor wave-guiding systems exemplified by the parallel-wire and coaxial lines of Figures 8-1 (a) and (b) arc commonly called transmission lines; in the strict sense they are also waveguides. The mode theory of uniform waveguides is considered in this chapter, with particular emphasis on the rectangular hollow wavegu'irles shown in Figure 8-1 (c). A boundary-value-problem approach is used, that is, solutions of Maxwell's equations, subject to boundary conditions, are obtained. The complex, time-harmonic forms of Maxwell's equations are used, time dependence of the fields being assumed according to the usual factor !1m!, but because of the invariance of the guide cross section with respect to the propagation direction z, an additional exponential Z dependence factor e+ Yz is assumed, with y identified as a z-direction propagation constant. With t and thus absorbed in the factor YZ, the wave equation in terms ofE or H reduces to iOptionally, you may elect to ddtT the study of Chapter 8 and direclly to Chapters 9 and 10 on transmission lines. Possible advantages of taking up Chapter 8 are that the study of velocity 8-5) and conductor attenuatiun losses (Section B-7) arc simpler ji)t rectangular than transmission lint'S.

409

410

MODE THEORY OF WAVEGUIDES

(a)

(b)

(c)

(d)

FIGURE 8-1. Uniform transmission line or waveguide structures of common occurrence. (a) Parallel-wire transmission line. (b) Circular cylindrical coaxial pair transmission line. (c) Rectangular hollow waveguide. (d) Circular hollow waveguide.

a dependence on only the transverse variables x,y in the case of the rectangular waveguide (or in terms of p,

It has been seen that expressing Maxwell's equations in complex, time-harmonic form through a time dependence given by the factor eiwt eliminates t from the equations. Wave-guiding systems of uniform cross section, like those in Figure 8-1, permit an additional assumption of z dependence of the fields in accordance with the factor e+l'Z, inasmuch as any length t of the system will influence wave propagation in exactly the

8-1 MAXWELL'S RELATIONS WHEN FIELDS HAVE ei""+Yz DEPENDENCE

411

same manner as any other length t. The time and z dependence is therefore assumed to occur solely in accordance with the factor rl W1 + YZ , in which the - and + signs are identified with the positive z and negative z traveling wave solutions respectively.;. The E and H fields of Maxwell's equations are thus replaced with com/,lex functions {f and :it ofthe transverse coordinates Uj and U2, multiplied by the exponential factor as follows w1 YZ E(Ul' U2, Z, t) is replaced by,g± (UI' U2)rl + H(Ul' U2l

z, t)

wt

is replaced by :it±(Ul' uz)rl +

yZ

(8-1a)

assuming~genen:lized cylindrical coordinates (UlJ uz, z). The superscripts ± on the symbols {f and:Yt' denote the field solutions identified with the positive ~and negative z traveling waves in the waveguide. Once the complex solutions {f±(Ul' uz) and :itt (Ut' U2) are found, a restoration to their real-time form is obtained using

(8-1b) The dielectric region bounded by the waveguide conductors is assumed lossless, making 0 therein, so that Maxwell's equations (3-59) and (3-77) governing the fields in the dielectric are

J

aB

(8-2)

VxE=--

at

\

aD

(8-3)

VxH=-

at

With the replacement of the complex forms of (8-la) into the latter, assurmng m rectangular coordinates ~+

{f- (x,y) :itt (x,y)

v± = ax~x (x,y) + ay&'i (x,y) + az&'z = ax~; (x,y) + ay~: (x,y) + az~i (x,y) ~+

~+

(8-4)

one obtains from (8-2)

a az

a

ax

ay

=

a -± ~+ at La x Yf x + a yYf-y

-11-

r

~+. + a zYf-]e'W/TYZ T

Z

j±rlW1+ yz z

The exponential factors cancel, obtaining the simplified expansion of (8-2)

(8-5)

412


These results can be written in the compact form

V'

X

J±

(8-6)

provided one defines a modified-curl operator, V'

ax V'

X

J±

a ;j±x

X,

as follows

ay

az

a

oy +y ~+

!!;

(8-7)

~+

!!:;

You should regard (8-6) as the equivalent of the Maxwell equation (8-2), assuming the exponential t and z dependence of the fields noted in (8-1a). You may note that the operator V' X defined by (8-7) differs from the conventional curl operator V X of (2-52) to the extent of a replacement of a/az with +y, a consequence of the assumption of the z dependence of the fields according to the factor yz In a similar manner, with the substitution o[ (8-Ia) and (8-4), the Maxwell curl equation (8-3) yields the compact result

with the modified-curl operator V' X defined once more by (8-7). In the generalized coordinate system (u l , uz, u 3 ) it is seen that the Maxwell modified-curl relation (8-6), for example, becomes

V'

X

g± ==

al

az

az

hz

h}

h1h z

a

oU l ~±

h1!!l

a auz

+y

-jWp;ie±

(8-9)

h2;j i:

assuming v± tff (Ub uz)

.ie±(Ul' U2)

~±

=

~±

~±

+ az!!z (ul' uz) + az!!z (ul' uz) a 1il't(Ul, U2) + a2£'i:(11 1, uz) + az£';t(ub u2)

a l !!} (u 1 , uz)

(8-10)

Simplifications of the wave equations are also possible when field variations occur according to the [actor e iwt + yz. The simultaneous manipulation of the Maxwell relations (8-2) and (8-3), applicable to a current-free region, has been seen in Section 2-9

_,

'4$,

,~

_==_,_________________

..•

,~_

B-1 MAXWELL'S RELATIONS WHEN FIELDS HAVE el""+Yz DEPENDENCE

413

to lead to the homogeneous vector wave equations

(8-11 )

(8-12)

Using the definition (2-83) of V2 E applicable to the rectangular coordinate system, the vector wave equation (8-11), for example, expands into the three scalar wave equations (8-13a)

(8-13b)

(8-13c)

In the cartesian system, all three scalar wave equations are of identical forms, so their solutions are\)f the same type. From the definition (2-78) of the Laplacian of a scalar function, (8-13a) expands as follows

o

(8-14)

If the substitution of the complex exponential form of Ex, given by (8-1a), is made into (8-14), one obtains, after canceling the exponential factors,

Denoting

1'2

+

0)2 jJE

by the symbol (8-15 )

one may write the scalar wave equation

(8-16a)

414


Similar substitutions of

la) into (8-13b) and (8-13c) produce the simplifications (8-l6b)

(8-l6c) Beginning with the vector wave equation (8-12), a procedure identical with the foregoing evidently produces three similar wave equations in terms of the components of :ie±. Any of these six partial differential equations is useful in obtaining wave solutions f(:>r the rectangular hollow waveguide of Figure 8-1 (c), to be discussed in Section 8-3. Relationships pertaining to the mode character oftlle solutions are developed first.

8·2 TE, TM, AND TEM MODE RELATIONSHIPS A study of the expansions of the Maxwell modified-curl relationships (8-6) and (8-8) reveals that, for the TE and the TM modes, you can express the transverse components it-, Ji, ii't-, and ;.,~i explicitly in terms of the x and] derivatives of the longitudinal field components and £1'. These results form a basis for the mode description of the field solutions, relationships established in the following i.n rectangular coordinates. Beginning with the expansions of (8-6) and (8-8) in rectangular coordinates

$;

A,~+

0$;-

~+

--+y$-

oy -

~+

_

ffi+

+Y0 X- -

(8-17a)

y

0$;-

~ ± = -jWII£;'

(8-17b)

(8-17c)

(8-13a)

(3-18b) ~+~+

O£y 0£X. ~+ - - - - - = )WE$-

oX

oy.

(8-13c)

Z

one can see th~at the fiIst two of each of these groups of equations contain derivative terms in only and £;-; the other terms are algebraic. This makes it possible, for exto eliminate :it'i from (8-17b) and (3-18a) and solve for if;, yielding

$;-

(3-19a)

8-2 TE, TM, AND TEM MODE RELATIONSHIPS

in which yields

k; is defined by (8-15). Similarly eliminating 1

Successively eliminating .

.'

and

,.- +

415

and (8-18b)

[aJ± ait±] =+= y _z_ + j(;)/l--~ oy ax

(8-19b)

J;: from the same pairs of relations obtains the fol, . . . . ;+

lowmg ex presslOns for :If;; and :It i :

1 [

:It +- = -,,A

x

.

k;'}(;)E

aJ±z -+ Y ---ait±] z ax

(8-19c)

(8-19d) These results permit fmding the transverse field components ofa rectangular waveguide and are known. They also serve as whenever the longitudinal components a basis fi)r decomposing the fieldAso\utio!ls into classes known as modes, depending on which longitudinal component, or :!It';, is present. The modes of the uniform waveguides of Figure 8-1(b), (e), (d) are defined as

J;

it;

Iff;

it; J;

1. Transverse magnetic (TM) modes, for which = O. 2. Transverse electric (TE) modes, for which = O. 3. Transverse electromagnetic (TEM) modes, for which both

J;: = 0 and it;: = o.

Out of these definitions evolve properties of the modes as follows

it;: = 0,

1. TM Modes (Transverse-Magnetic Waves). With

the TMA mode in a = 0 into equations (8-19) produees the following expressions for the transverse field components in rectangular coordinates.

wav~l.{uide has five eomponents, as noted in Figure 8-2(a). Putting :It;

(8-20a)

(8-20b) TM (8-20c)

A

+

:If;

aJ z k; ax

j(;)E

= -

(8-20d)

in which k; denotes y2 + (;)2/lE . Sinee the factor and (8-20d), their ratio becomes B+

(f)x--

Y

it;: = +- jWE

which means

J; it:

aJi/ax is common to (8-20a)

y jWE

and

Y j(;)E

416


Rectangular

Circular

Rectangular

Circular

(b)

(a)

Parallel-wire line

Coaxial line (e)

FIGURE 8-2. Field components of TM. TE, and TEM mopcs in typical waveguides or transmission lines. (a) Field components of 'I'M mode, for which Jf'z O. (b) Field components of TE mode, for which i'z = O. (c) Field components of the dominant TEM mode of two-conductor systems.

Similar results, with changes in signs, are obtained li'om the ratios of (8-20b) to (8-20c); calling YUWE in each case the intrinsic wave impedance ofTM modes, denoted by the symbol1]TM' one produces the four ratios (8-21 )

The use of the latter makes it necessary to ohtain only two of the transverse field components from by means of (8-20); thc remaining two components are available in terms of the impedance ratios (8-21). In the detailed analysis ofTM modes carried out in Section 8-3, it is seen that the propagation constant Y appearing in (8-20a, b) and (8-21) is dependent OIl the waveguide dimcnsions and the wave frequency. Using the modified-curl relations (8-9) and (8-10) and following a procedure similar to the foregoing, modal expressions similar to (8-20) and (8-21), but applicable to waveguides in the circular cylindrical (p, cp, z) or the generalized cylindrical system (111' 112, z) can he found. This is left as an exercise for you. • 2. TE Modes (Transverse-Electric Waves). With = 0, the TE mode has the five components typified in Figure 8-2(b), so that equations (8-19) in rectangular

$:

$;

8-2 TE, TM, AND 'rEM MODE RELATIONSHIPS

417

coordinates (8-22a) . ]WM y

~

+

a,;Yf;

ax

(8-22b) TE

:if± - -l a:if; x - + k2 ax c

(8-22c)

(8-22d) An intrinsic wave impedance

ryTE

is evident from ratios of the latter as follows (8-23)

3. IFM ,\.fades (Transverse-Electromagnetic Waves). This mode, having neither nor ::If z field components, is the d<:!.minant mod~ of transmission lines having at least two conductors. Substituting tf'; = 0 and::lf; = 0 into the four relations (8-19) would appear to force all field components to vanish, thereby reducing the TEM mode to a trivial, nonexistent case. Inspection of the 1enominator k; in these relations reveals the flaw in this argument, for putting f/; = 0 simulta~+ ~ + 2 2 neously as one assumes tf'; = 0 and ::If; = 0 means y + W ME = 0, or

tfj'z

\

y = jW~ME = j{J fad/m

(8-24)

Comparison with (3-88) shows that the transverse field components of the TEM mode comprise a wave phenomenon possessing a phase constant (8-24) identical with that of a uniform plane wave propagating in an unbounded region of parameters tt and E. Substituting (8-24) into either wave impedance relation (8-21) or (8-23) further obtains the intrinsic wave impedance for the TEM mode ryTEM

== ~ = jwJJ;. ]WE

]WE

=

~

~'E

11(0)

Q

(8-25)

Comparing (8-25) with (3-99a) reveals an intrinsic wave impedance identical with that of uniform plane waves in an unbounded region. These similarities of TEMmode characteristics with those of uniform plane waves are appreciated when one realizes that the uniform plane wave is itself TEM. The TEM mode, the dominant mode of energy propagation on two-conductor lines, is of such importance in wave transmission along open-wire or coaxial lines that it is accorded a separate detailed treatment in Chapters 9 and 10. Generally speaking, hollow single-conductor waveguides are capable of propagating 'I'M and TE modes. In Section 8-4 it is shown that the so-called TE 10 mode of the rectangular waveguide is its dominant mode, that is, the mode propagating at the

418


lowest frequency in that waveguide. Two-conductor systems such as coaxial lines propagate all three mode types: TEM, TM, and TE, although only the dominant TEM mode is capable of wave propagation down to zero frequency. Signal transmission ir tt\e microwave region (at frequencies of about 1000 MHz and higher) by use of rect,angular waveguides is a practical reality. Because of their intrinsically high pass characteristics, hollow waveguides become physically too large and expensive at frequencies much below this range; at lower frequencies, coaxial lines or open-wire lines may be more practical. A rectangular waveguide designed to operate with its dominant mode at about 10,000 MHz will be shown to require an interior width of about 2.5 cm; at one-hundredth this frequency (100 MHz), the guide width is required to be about 2.5 m if waves are to be propagated and not cut off. Coaxial, two-conductor lines are the obvious choice at such lower frequencies. In microwave transmission, a rectangular waveguide is usually more desirable than one of circular cross section because the asymmetry of the rectangular cross section provides a deliberate control of the polarization of the transmitted mode, of importance when considering the excitation of the line termination (a crystal detector, an antenna, etc.). Circular waveguide is of more limited use, having applications to rotating joints that couple into spinning antenna dishes, to cylindrical resonant cavity frequency meters, and so forth.

8·3 TM MODE SOLUTIONS OF RECTANGULAR WAVEGUIDES An analysis of the TM mode solutions of rectangular hollow waveguides is described in the following. The cross-sectional geometry of Figure 8-3 is adopted, and the following assumptions are made: 1. The hollow rectangnlar conducting pipe is assumed very long (avoiding end effects) and of uniform transverse dimensions a, b as noted in Figure 8-3.

ICY) I

~~~~~~

(x)

"-,,-

(x) FIGURE 8-3. Geometry of a hollow, rectangular waveguide of uniform crosssectional dimensions a, b.

8-3 TM MODE SOLUTIONS OF RECTANGULAR WAVEGUIDES

419

2. The dielectric medium filling the pipe has the constant material parameters p., E and is assumed lossless, such that Pv = 0 and J = 0 therein. 3. The waveguide walls are assumed ideal perfect conductors, simplifying the application of the boundary conditions. 4. All field quantities are assumed to vary with z and t solely in accordance with the factor ejwt+YZ, in which the - and + signs are associated with positive z and negative z traveling wave solutions. The sinusoidal angular frequency of the fields is w, determined by the generator frequency. 5. For the TM modes under consideration, = 0, leaving at most five field components in the pipe as noted in Figure 8-3.

if;

Bearing in mind that the field relationships (8-20) and (8-21) are applicable to the TM case, it is convenient to ~egin with the wave equation (8-16c), in terms of the longitudinal field component

iff;

jj2j± z

+

[8-16c)

in which k; = y2 + w 2 P.€. This partial differential equation is to be solved by the standard method of the separation of variables. Thus, assume a solution of the product form

j-; (x,y)

X(x) Y(y)

(8-26)

in which ,¥(x) and Y(y) are, respectively, functions ofx andy only, and in general are complex. Substituting (8-26) back into (8-16c) yields

X"Y + XY" = -kc2 Xl'

,

in which obtains

~rimes denote partial differentiations with respect to x to y. Dividing by Xl' X" 9" , -+-=-p

X

9

<

(8-27)

If the two functions of x andy comprising the left side of (8-27) are to add up to the indicated constant for all values of x and y within the cross section of Figure 8-3, then both those functions must be equal to constants as welL That is, one must have and

k;

(8-28)

k;

and denoting those constants. With (8-28) inserted into (8-27), it is with seen that the interrelationship (8-29) must hold among the constants. The meanings of the so-called separation constants kx and ky are ascertained later from the application of boundary conditions at the walls. A

420


Since the two differential equations (8-28) are, respectively, functions of x and)! only, they can be written as the ordinary differential equations and

(8-30)

\

These have the solutions

)

(

X(x) Y(y)

=

+ (;2 sin kxx cos ky)! + (;4 sin kyY

(;1 COS tx

(8-3Ia)

(;3

(8-3lb)

in which (;1 through (;4 are constants of integration (complex, in general), to be evaluated from boundary conditions. The separated solutions (8-31), substituted back into the product expression (8-26), thus yield the desired particular solution of the wave equation (8-16c) as follows (8-32) The complex constants appearing in (8-32) are evaluated from boundary conditions as follows. The component If; of (8-32) is tangential at the {our walls x 0, x = a an~d y = 0, y b noted in Figure 8-3. For perfectly conducting walls the tangential If z just inside the dielectric waveguide region must vanish, so from the continuity relation (3-79) one obtains the boundary conditions

1.

(O,y) = 0

2.

(a,y)

0

3.

0)

0

4.

(x, b)

0

(8-33)

Boundary condition (1) applied to (8-32) yields

0= ((;1)((;3 cos ky)i whence

G\ = 0 if the latter is to

hold for all_y

+ C4 sin kyY) 011

the wall x = O. Then (8-32) becomes

j± z

(8-34)

Applying the boundary condition (2) to (8-34) obtains

which holds for ally on the wall x = a on setting sin kxa = O. The latter is valid only at the zeros of the sine function, so that kxa rnn, in which 1Jl = ± I, ± 2, ± 3, ... , which corresponds to an infinite set of discrete values for kx (hence to an infinite number of particular solutions, or modes) that satisfy the original wave equation. The value rn 0 is omitted because it produces the null, or trivial, solution. The negative A

421


values of m, moreover, add no new solutions to the set, so that one obtains A

kx

=

mn

m

a

i()[

= 1,2,3, ...

t (8-35)

making (3-34) become 'V+

'

Ji;; (x,y)

=

";.

C2

SIn

mn a

m = 1,2,3, . ..

(3-36)

The remaining boundary conditionii. (3) and~(4) of (8-33) are next applied to (8-36); from the similarity of the solutions X(x) and Y(y) appearing in (8-32), together with the resemblance of the boundary conditions (3) and (4) to (1) and (2), one may iufer by analogy with the preceding arguments that applying the boundary conditions and (4) to (8-36) must lead to the results A

and

nn

ky=t;

n = 1,2,3, ...

(8-37)

With these inserted into (8-36), the solution finally becomes ~

~

C2 C4 sin

mn

. nrc x sm - y a b

m, n = 1, 2, 3, ...

{)l:4 this solution.. Replacing (

s~t,

1)4

evidently denotes the complex amplitude of any member of which must include both positive z and negative z traveling waves. by the symbol mn yields

E:

--------------------------------~+

.

mn

. nrc

= Ez-.mn SIl1 a x Sln -b- y

m, n

1,2,3, ...

(8-38£1)

in which £:m1l or E~mn denotes the complex amplitude of any positive z or negative traveling ifz component associated with specific values of the mode numbers m and The solution set (8-38'1) describes the z-directed electric field component of the transverse-magnetic mode with mode integers m, n assigned, so the field component is said to belong to ~he TMmn mode, and assuming that the transverse dimena is the larger (a> b). Solutions (8-38a) satisfying the partial differential equation (8-16c) and the boundary conditions (8-33) are also called the eigerifunctions (proper !bnctiom, or characteristic functions) of 1hat boundary value problem. Examples of the field variations of Iffz predicted by (8-3~a) within the waveguide cross section are depicted in Figure 8-4, which shows how Iffz varies with x andy for two of the modes, TMll and TM z1 . These sketches show that the integers m and n denote the number of half-sinusoids of variations in $z occurring between the guide walls, with $z vanishing at the walls as required by the boundary conditions (8-33). The sketches of Figure 8-4 do not show the complete field configurations of those TMmn modes; the four remaining transverse field components denoted in Figure 8-3

422


TMll mode (m 1, n 1)

=

=

TM2l mode (m = 2, n = 1)

I (y) I I

I(y)

I I

FIGURE 8-4, Typic!:,l cross-sectional standing-wave variations ill the longitudinal electric tleld component E, of TMmn modes, for two cases.

are yet to be found. These are obtained by substituting mode relationships (8-20), whence

~+

$:;: (x,y) =

kc

J.

a'

~+

,

=

of (B-38a) into the TM

~+ mn . nn +"2 mn E:;mn cos - x sm - y

[ - ')Imn

= E;- mn cos

~+ (x,y) $;

i'1'

mn. nn x sin - y a h

[+')1"", nn ~+

E:;-mn

-A--'

fj;

"'+

b·'

.

b

a

(8-38b)

J

,mn x cos nn) ! sin a b

mn nn x cos - y a h

= Ey mn sm -

,

~!(x,y)

=

.iWE

[

nn ~+

(8-38c)

'-A--E:;mn k~ b '

J.

sm

mn nn xcos-y a b

~+ . mn rm H;-,mfl SIn a x cos - b y

~ +

£'y(X,y) =

~+ [. ±-A--= ~ $;-

- JWE

mn

I1™mn

kc

a

(B-38d) ~+

E:;'mn

mn. nn J cos-xsm-y b

a

~+ mn rm Hcos a x sin - b ~y V,mn

(8-38e) .

~±

~+

~+

11!e bracketed quantities denote the complex amplItudes Ex,mm Ei,mn, H;:'mn, ~a,!ld Hi,mn of the transverse field components, expressed in terms of the amplitude E:;'mn of the longitudinal component. Note further that the total electric and magnetic fields associated with any TMmn mode are given by (8-4), or the appropriate vector sums

w

fl-3 TM MODE SOLUTIONS OF RECTANGULAR WAVEGUIDES

423

of the components (8-38a) through (8-38e) (with the component
z direction

propagation constant y, becomes

2

(8-39b) in which the subscripts mn denote the dependence of Y on the choice of mode integers. Thus Ymn is a function of the wall dimensions a and h, the frequency W, the parameters Jl and E of the dielectric, and the specified mode numbers m, n. The bracketed quantities in the radicand of (3-39b) are both seen to be positive reaL Since the d~fference of these positive quantities is to determine Ymn, it is evident that Ymn becomes a pure 2 real quantity (an attenuation factor IX) if (rmr/a)2 + (nn/b)2 is larger than W JlE; with Ymn becoming pure imaginary (a phase factor (J) if the reverse is true. The transition between these two propagation conditions occurs at an angular frequency W w e.mn called the cutqff frcq uency, defined where the bracketed quantities of (8-39b) are equal; that is, )

=

W;.mnJlE

Solving the latter for j;.mn

= wc,mn/2n fc,mn

(many + (n;y

yields

= _ 1 [(~)2 C

(8-40a)

2,-/ JlE

a

+ (~.)2Jl!2 h

(8-40b)

With (8-40a) substituted into (3-39b), the propagation constant Ymn is expressible in terms offc,mn as follows

Ymn

(8-41 )

=

2With both 1" and E positive real, only the root of (8-39a) with positive real and positive imaginary parts need be chosen as the solution lor Y."" because the earlier assnmption of t and z dependence of the fields according to the factor exp (jwt 1'z) accounts properly for the presence of both positive z and negative traveling wave solutions.

+

424


From the latter one may infer, depending on whether a given TMmn mode in the rectangular guide is generated at a frequency f that is above or below the cutoff value !c.mn of (8-40b), that Ymn becomes either pure real or pure imaginary as follows

Ymn =

IXmn

== 0)\1cJ-lE

J(fc.mn)2 J

I Np/m

f

(8-42a)

f > j~.mn

(8-42b)

The factor (J)~ appearing in these expressions is a phase constant 13(0) identified with a uniform plane wave traveling at the frequency f in an unbounded region having the material parameters J-l and E, a value obtained from (3-90b) with (J = 0 or (3-110) with E"/E' = O. The quantity 13(0) thus serves as a convenient reference value with which the phase constant 13mn in the waveguide can be compared. From (8-42) it is evident that a rectangular waveguide carrying a TMmn mode acts as a highpass filter, allowing unattenuated wave motion characterized by the pure imaginary Ymn = j13mn if the generator frequency f responsible for the mode exceeds the cutoff frequency j~.mn' but attenuating the TMmn mode fields with Ymn = IXmn iff < j~.mn' An additional appreciation of the physical meanings of the real and imaginary results (8-42) for I'mn is gained if the wave expressions for the TMmn modes, including dependence on t and Z, are examined. For example, multiplying the component of (8-38a) by the exponential factor ejwt+Ymnz according to (8-la) produces field solutions that depend on whether the propagation constant I'mn of (8-42) is real or imaginary, as follows. Iff> !c.m,,, then Ymn = j13mn so that (8-38a), including exponential t and Z dependence, becomes

$;

~+ . If(x y)e1mt+Y~"z Z ,

~+ nn.( - fJ ) = Esin -mn z.rnn a x sin -by e1 wt+ ~nZ

m, n

=

I, 2, 3, ... and

f > fe.mn

(8-43a)

The traveling wave nature of this field component is clearly specified by the factor Z ei{
+

. ---

~

±

-

mn

nn.

tff;; (x, y) eJwt + Ymn Z = Emn ze + "-mn Z sin x sin _ y e1wt . a b

m, n = 1,2,3, ... and f < !c.mn

(8-43b)

The attenuation with z provided by the factors e-a",n Z or e"mnz is thus noted whenever the generator frequency I is too low. The mode will not then propagate as a wave motion; instead, the fields of the mode evanesce (vanish) with increasing distance from the generator or wave source. A mode at a frequency below its cutofrfrequency lc.mn is called ~n evanescent mode. The foregoing discussion was limited to the longitudinal component If;. The four remaining transvers~ components (8-38b) through (8-38e) are similarly propagated as waves along with If; iff> !c.mn or are evanescent iff < !c,mn-


425

The real-time forms of the field components can likewise be employed to illustrate the conclusions of the foregoing discussion. Using (8-1 b), the real-time form of tll!\ time-harrnonic field component (8-43a) is, j()r f > !c.mn E~± Z, t) = Re [tff;(x,)i)eJ(rot+PmnZ)], yielding

mn

+

nn

Z, t) = E;.mn sin a x sin b ) cos (OJt

+ PmnZ + ;;n)

f > J~.mn

(8-Ha)

the traveling wave nature of which is illustrated, in a constant) plane, in Figure 8-5(a). Similarly, for f below the cutoff fi'equem:y j;',mn Z,

t) +

= E~~mne

'+

mn

amn Z sin a x sin

nn + bY cos (OJt + ;;;,,)

f
(8-44b)

Note that the complex amplitudes in these expressions may include arbitrary phase angles ;;" according to E;'m" E;'mnei
E:

t= 0

Wave motion ,J,:-.--"----

(a) I(y)

I

I

I

11';

FIGURE 8-5. Field intensity variations of the lonfiitudinal component (x,J, t) of the TM 2b mode, shown over the planeJ = b/2. (a) The forward z traveling wave E;, iff> j;, (b) The evanescence of E; with increasing z, iff
426


Once the phase constant (8-42b) is obtained, other TMmn mode properties, such as wavelength in the guide, phase velocity, and intrinsic wave impedance, can be derived. Assuming the generator frequency f of a given TMmn mode to be above the cutoff value (8-40b), the wavelength .-1 of that mode, measured along the.;; axis as noted in Figure 8-5(a), for example, is found from the definition p.-1 = 2n. By use of (8-42b), this yields

f> J~,mn

(8-45)

in which .-1(0) denotes the comparison wavelength 2rrJP(0) of a uniform plane wave in an unbounded region with the same dielectric parameters J1 and E. The.;; direction phase velocity is obtained using vp wlP, yielding

f > J~,mn

(8-46)

wherein v~O) = wIP(O) = (J1E) 1/2. The intrinsic wave impedance for TMmn modes, specifying the ratios of transverse field components, is found from (8-21). Iff> !c,mn, one obtains the real result A

lJTM,rnn

jPmn = JWE .

n(O) 'f

f~ (J~j,m.n)2

f.>

!c,mn

(8-47)

in which lJ(O) = ~. For a TMmn mode generated at a frequency f below the cutoff value, the wavelength and phase velocities are not defined, in view of the purely evanescent character ofthc field distributions as exemplified in Figure 8-5(b). The intrinsic wave impedance for f < J;.,mn, however, from the substitution of (8-42a) into (8-21), becomes A

lJTM,mn

rx

= JWE .

. (0)

-JlJ

J(.rc,mn)2 f-

1

f < !c,rnn

(8-48)

This purely reactive result implies no time-average power transfer in the.;; direction for an evanescent mode because of the 90° phase between the transverse electric and magnetic field components. If the information contained in the five field expressions (8-38a) through (8-38e) is combined to construct the total fields E and H of the TMmn modes, complete flux sketches resembling those in Figure 8-6 can be obtained. Flux sketches of two modes, TMll and TM 21 , are illustrated. A knowledge of such flux configurations is useful, for example, if the electric or magnetic fields are to be probed or linked with a short wire antenna or loop, for purposes of extracting energy from the mode. I.n general, a large number of modes, propagating or evanescent, exist in the neighborhood of waveguide discontinuities such as bends and transitions. The analysis of such nonuniformities in a waveguide is beyond the scope of this treatment. The propagation of energy in a rectangular guide is usually accomplished, at a given frequency, by selecting the dimensions a, b so that only one mode (the dominant


427

c Section C-C

Section A

A

B (b)

Two low-order TM modes of a rectangular waveguide. (a) The TMIl mode. the TM21 mode.

propagates, to the exclusion of all higher-order modes thus forced to become evanescent. This procedure assures a well-defined single-mode field configuration in fI'om which energy can be readily extracted by use of suitable transition (for example, a waveguide-to-coaxial line transition). The discussion of the section, covering TE modes, reveals that of all the modes capable of propagating rectangular waveguide, TM and TE, the TE lo mode is the dominant one, a > b once again. IlAMPLE 8-1. A common air-filled rectangular waveguide has the interior dimensions a = 0.9 in. and b 0.4 in. (2.29 x 1.02 em), the so-called X-band guide. (a) Find the cutoff frequency of the lowest-order, nontrivial TM mode. (b) At a source frequency that is twice the cutoff value of (a), determine the propagation constant for this mode. Also obtain the wavelength in the guide, the phase velocity, and the intrinsic wave impedance. (c) Repeat (b), assuming J = .fc/2.

428


(a) From (8-40b) it is seen that the cutoff frequency has its lowest value for TM modes if m = I and n = I, the smallest integers producing nontrivial fields. Thus for the TM II mode, the given dimensions yield

The 'I'M 11 mode will thus propagate in this guide if its frequency exceeds 16,100 MHz. Below this frequency, the mode is evanescent. (b) ALI

32,200 MHz, (8-42b) yields =

In fi'ee space,

A(O)

=

eLl =

3 x L0 8 j32,2 x 10 9

585 rad/m

0,933 ern, so from (8-45) 0.933 0.866

= 1.076 em

while the phase velocity and intrinsic wave impedance, from (8-46) and (8-47), are

3 1'1'.11

(c) At I

=

8 X 10 0.866

~TM.ll =

37.7(0.8G6) = 32G Q

8.05 GHz, (8-42a) ohtains 2n(8.05 x 10 9 ) ~.~

291 Npjm

Below j~.ll' wavelength and phase velocity arc undefined, ill view of evanescent fields, but below cutoff, from (8-48)

~TM.1I =

_j1](OlJ(!jlY--]

-j377.j22-1 = -jG53Q

8·4 TE MODE SOLUTIONS OF RECTANGULAR WAVEGUIDES 'The analysis of the TE mode solutions of rectangular waveguides proceeds essentially along the lines employed for finding the TM mode solutions in Section 8-3, so only an outline of this boundary-value problem is given. The assumptions are as follows 1. The rectangular hollow pipe is very long and of interior dimensions a, b, as noted iIi Figure 8-7. 2. The lossless dielectric has the parameters fl, E, with Pv = 0 and J = O. 3. The waveguide walls are perfectly conducting.

8-4 TE MODE SOLUTIONS OF RECTANGULAR WAVEGUIDES

429

y=b

x=O x=a y=O ~

"

#--C;)-

(z)

FIGURE 8-7. Geometry of a hollow rectangular waveguide, showing field components corresponding to the TE modes.

y

4. All field quantities vary as eimt + ". 5. 0 for TE modes.

if;

Only the last assumption differs from those used in the derivation of TM modes in Section 8-3. The four TE field relations (8-22) suggest that a solution for might first be obtained, whereupon (8-22) can be employed to obtain the remaining tr<;l:nsverse components. Beginning with the scalar wave equation (8-16f) in terms of ft;

if;

[8-16f] by analogy with the separation-of-variables method applied to the wave eqnation (8-16c) in Section 8-3, a particular solution of (8-16f) is analogous with (8-32) such that (8-49) The boundary conditions at the perfectly condncting walls shown in FigurS-8-7 demand tpat the tangenlial components oft~e electric field vanish there, that is, $;-(O,y) 0, (a,y) = 0, 0) = 0, and 111'; (x, b) = 0, bnt the latter are converte1 into equivalent boundary conditions applicable to the longitudinal component of (8-49), on inserting them into the two TE modal relations (8-22a) and (8-22b), yielding

tt;-

$;

Yf;

I

a.~;]

. ax

2.

3.

°

x=o

aif;] = 0 ax x=a

ait±] --==0 ay y=o a.~±]

4. --"-

ay

y=b

=

°

(8-50)

430


g;

Applying these boundary conditions to the appropriate x ory derivative of the solution (8-49) can be shown to obtain the following proper solutions (eigenfunctions)

~

+

:R;; (x,y)

mn

~+

nn

0, 1,2, ...

m, n

JJ;;'mll cos a x cos-/; y

=

(8-51 a)

in which m, n are arbitrary integers designating an infinite set of TEmn modes. As in the TM mode case treated in Section 8-3, two separation constants, = mn/a and ky = nn/b, are related to == y2 + W2/lE by (8-29). The remaining field components of the TErnn modes are found using (8-22), yielding

t

k';

@±

(() x

=

JW/l nn ~+ JJ:;-mn [ k2 b " ' c --A-

J

mn x sm .. nn y

cos -

b-

(1

~+ mn. nn E;:'mn cos -;; x 8m b _.Y

-JW/l mn ~+ [ ~ a H;;'mll

==

"'+ .

Ey- mn SIn •

mn (1

J.

x cos

(8-5] Il)

mn

nn b

(8-51c)

y

Ymll mn ~ + rlrE,mll

==

~+

.

=[ mn

nn

sm~--xcosbY

--;; H~~m"

J.

mn

. nn

sm--;; x COSb~Y

nn

(8-51 d)

H;; mn SIn - x cos . a b

[

g± y ~+

Ymn nn .• + b

mn. nn

!/y- m" cos , a

X SHl -

b

!!z.... mn

J

mn

cos -

(1

Y

. nn

x SIn -

b

J'

(8-5Ie)

wherein (8-51 f)

implying a propagation constant Ymn given by an expression identical with (8-:39b) for TM modes (8-52)


431

The latter implies a cutolffrequency lor TErnn modes in a rectangular waveguide given by all expression like (8-40b) lor the comparable TMmn modes (8-53)

It lherei()!'e follows that the propagation constant Ymn of (8-52) is a pure real attenuation factor am» if I < J~,mn' or a pure imaginary phase factor jfimn if I > lc,mn; thus, I Np/m

Ymn

Ymn

jfimn

=jw~~

f-e 7Y c

rad/m

1< lc,mn

(8-54a)

I > J~,mn

(8-54b)

From (8-54b) one can inkr, lor a specified TErn» mode, a wavelength Amn and phase veloci ly vp,mn given by expressions identical with (8-45) and (8-4b) Ii)!' the comparable TMmn mode

I> lc,rnn

(8-55)

I > J~,m"

(8-56)

in which A(O) and 1J~0) an, the wavelength and phase velocity associated with plane waves propagating at the frequency I in an unbounded region filled with the same dielectric with the parameters J1 and E, A comparison of (8-21) with (8-23) shows that the intrinsic wave impedances of TE and TM modes are not the same; from (8-23) and (8-54b) one obtains for TErnn modes above cutoff

JWI'

1](0)

1]TE,mn = -:p-- = -;=======(;==;==)~2 n ) mn

1

_!c:

I > J~,mll

(8-57)

t

which deserves comparison with expression (8-47) for ~TM,m", If a TEmn mode is generated at a frequency below the cutoff value specified by (8-53), the propagation constant Ymn becomes the pure real amn of (8-54a), producing an evanescence of the field components (8-51) resembling that for TMmn modes below cutoff as shown in Figure 8-5(b), Although wavelength and phase velocity are undefined in the absence of wave motion for I < j~,m'" the intrinsic wave impedance lor a TErnn mode below cutotfis obtained from (8-54a) and (8-23), yielding

f < ,Ic,mn

(8-58)

432

MODE TREOR Y OF WAVEGUIDES

,50) 1 {:I«()

Allin

o Increasing

o

f

f Increasing

f

f= t~,mn FIGURE 8-8. Universal circle diagram (left) and qnantities plotted directly against freqnency (right), fix TM and TE modes.

From this result one may again sec, as from (8-48) for TMrnn modes, that whenever a mode evanesces (f < j~,mn) the wave impedance qTM or qTE becomes imaginary, showing that tor an evanescent mode, there is no time-average power How through a waveguide cross section. The common factor 1 - (!c,mnlf) 2 appearing in the various expressions (8-45) through (8-48) for TMmn modes, together with the comparable relations (8-54) through (8-57) for TErnn modes, permits graphing them as normalized quantities on the universal circle diagram shown in Figure 8-8. For example, the expressions (8-4·2b) and (8-54b) for the phase factor Pmn of TM or TE modes are normalized by dividing through by P(O) = wJjlE to obtain

J

Pmtl)2 + (!c.mtl)2 ( pt O) f the equation of a circle, considering PmnIP(O) and J~,mtllf as the variables. A discussion of the group velocity Vg noted in the diagram is reserved for Section 8-5. To the right in the figure is shown a graph of the same quantities plotted directly against frequency, which may have some interpretive advantages. Thus, the phase constant Pmn of a desired mode is seen to be zero at the cutoff frequency !c.mn while asymptotically approaching the unbounded space value P(O) = wJjlE represented by the diagonal straight line as f becomes sufficiently large. The expressions (8-51) feJr the five f-Ield components of the TE_ modes lead to flux plots of typical modes as seen in Figure 8-9. The electric field lines are entirely transverse in any cross section of the guide, as required for TE modes; they terminate normally at the perfectly conducting walls to satisfy the boundary conditions there. The magnetic lines, moreover, form closed loops and link electric flux (displacement currents) in the process, as required by Maxwell's equations. A comparison with Figure 8-6 points out the inherent diflerences between TM and TE mode field configurations in a rectangular guide. In Section B-3, the TM mode expressions (8-38) reveal that the lowest-order nontrivial mode of this group is the TM 11 mode. A similar inspection of the field expressions (8-51) shows that the lowest-order nontrivial TE modes are the TElO and TEO! modes, flux plots of which are depicted in Figure 8-9(a) and (b). Of these two,


2

433

2

Section 1-1

um \\I\\l\! !![ Section 2-2

(b)

(el

(d)

FIGURE 8-9. A few low-order TEmn modes of the rectangular waveguide. (a) TE!o mode. (b) TEo! mode. (e) TEll mode. (d) TE2! mode.

the mode I)aving the lowest cutofffi'equency is determined by whichever of the two trans~erse guide dimensions, a or b, is the larger. With m = I and n = 0 inserted into (B-5~»), the TE lo mode is seen to have a cutoff frequency

!c,10

=

a

2a

(8-59a)

a result independent of the b dimension because n = O. Thus (B-59a) states that the cutofHrequency of the TE 10 mode is the freq uency at which the width a is just one-half

434


a frcc-space wavelength. Similarly, the TEol mode has a cutoff frequency (8-59b) a value larger than!c.10 if a > b, the dimensional condition depicted in Figure 8-1O(al. From the identical cutoff frequency expressions (8-53) and (8-40b), all higher-order TE and TM modes exhibit cutoff frequencies higher than (8-59a), assuming the standard convention of a> b, to make the TElO mode the dominant mode of that rectangular waveguide. For example, the so-called X-band rectangular waveguide, assumed airfilled and of interior dimensions a = 0.9 in. and b = 0.4 in. (0.02286 x 0.01016 m) has a cutoff frequency obtained from (8-59a), yielding 3 X 10 8 !c.lO = 2(0.02286) = 6.557 GHz

(8-60)

X-band guide

while the cutoff frequency of the next higher-order mode, TE 2o , becomes J~.20 = 13.12 GHz, from (8-53). The TEol mode, from (8-59b), yields !c.01 = 14.77 GHz, while using (8-53) or (8-40b) obtains cutoff frequencies for the TEll and TM 11 modes that are even higher (!c,ll = 16.10 GHz). Their positions on a frcquency scale are portrayed in Figure 8-10(a), showing why the propagation of electromagnetic power via the single dominant TElO mode in a rectangular waveguide is possible by kecping the generated frcquency f above the cutoff frequency of the TElO mode, but below the cutoff frequencies of all other modes. This choice assures a traveling wavc TElO mode and the evanescence of all other modes, thereby justifying the designation dominant for the propagating TE 10 mode. For example, the band 8.2 to 12.4 GHz is chosen as the X band; frequencies that propagate only in the dominant TElO mode in a 0.4 in. x 0.9 in. rcctangular waveguide.

ie, 10 = 01

6.557 GHz

For a = 0.9 in" b

=0,4 in,:

(GHz)

~

~ For

TE10

1- = 2.25:

[:j

.1

°

Q

r :}

TE21 TM21 I

tl

:)

3

i(>,rnn

fe, 10

(a)

I

1:

'f 2

TEJO TEoI

For~=

TEll TEoII TM[1 TE 20 I I I

I

TEll TMJl I

t

TE 20 TE02 I I

TE21 TEI2 TM21 TMI2 I I

r

~I__________-L 'f ____~_____ L_ _ _ __ L_ _ _ _~~~

°

1

't'

2

3

"

JO

(b)

}'IGURE 8-10, Cutoff frequencies oflower-order modes in rectangular and square waveguides, (a) }'or alb = 2.25, Cutoff frequencies shown relative to !c,IO on lower graph" (b) For alb = L


435

It is evident that a square waveguide (a = b) will not possess only one dominant mode, for the TE 10 and TE01 modes then have identical cutofftrequencies from (8-59). Figure 8-1O(b) shows the positions of the cutoff frequencies oflower-order modes for a square waveguide on a relative frequency scale. A comparison ofF'igure 8-1O(a) with (b) reveals that the use of a rectangular waveguide, with a> b as in (a) of that figure, provides a desirable control over the E-field polarization of the propagated mode. Figure 8-11 illustrates the manner in which a microwave power source (a klystron, magnetron, etc.) is connected to a waveguide by using a small antenna wire protruding into the waveguide, such that the wire alignment agrees with the polarization of the dominant mode being launched. The power can similarly be extracted at the other end, if desired, by means of the center conductor of a coaxial line used as a receiving antenna (a waveguide-to-coax transition). The propagation of the energy down the waveguide via the dominant TElO mode thus assures the known field polarization necessary to the efficient launching and retrieval of the energy. Since signal power in a rectangular waveguide is commonly dispatched by use of the dominant TE 10 mode, its properties are for convenience collected separately· in the following. The expressions (8-51) for TEmn modes, with m = I, n = 0 inserted, reduce to three components ~±

;j{'z

(x)

.

~±

n

= Hz ,10 cos -a

~+

(8-6Ia)

X

[-JWlla ~+

]

.

n

"'+.

n

Si (x) = --n-- Hz-;l0 sm a x = Ei,10 sm -;; x

.m±( .:n x X ) --

-+ it -A-- -

'lTE.I0

±j2a ~ + 11.10

(8-61 b)

[±jf310a H~±z,10 ] Sln . ~x n a ]

.

n a

~+

.

n a

= [ - - 1 - Hz.10 sm - x = Hx.10 SlIl- X

(8-61c)

assuming J > !c.lO' The foregoing may for~some purposes be more conveniently expressed in terms of the complex amplitudes Ei,10 of the y-directed electric field (8-61 b), Microwave source (klystron, etc)

[W] TEiO mode

Coaxial-to -waveguide transition

FIGURE 8-11. Typical waveguide transmission system, showing launching of the dominant mode and a transition from waveguide to coaxial transmission line.

436


yielding ~+

~+

n

"

a

gy- (x) = £;-;10 sin - x

(8-62a)

o . n A

1'/TE,1O ~+

SIn -

~ ± _. Ei:1O/i. yt'z(X)-J (0) 21'/ a

1(0)

a

.

~+

x

=

Hx.l0 sm

n a

(8-62b)

x

,n _ ~ ± n cos-x-Hz10cos X a . a

(8-62c)

The remaining properties of the TE10 mode are related to its cutoff frequency specified by (8-59a). From the latter, the ratio !c. I o/f is

.!c,10

viOl

J~.lO

(8-63)

-p-

2af

f

to permit writing the propagation constant, wavelength in the guide, and phase velocity for the TE IO mode as follows

YlO = 1J(1O == /3(0)

C(O)Y 2a

j/3(O)

Y1O=j/310

-

I Np/m

I - C(OIY rad/m

2a-

f < !c,10

(8-64a)

f > !c,10

(8-64b)

f > j~.10

(8-65)

f >

(8-66)

A(O) rn

)'10

vIOl

-(~:)Y

Pm/sec

j~,10

in which /3(0) = (JJ~, A(O) = 2n//3(0) = v~ol/I, and v~OI = (/lE) 1./2 as before. The intrinsic wave impedance obtained from (8-57) or (8-58) becomes 11(0)

fiTE, 1 0 = -;:::1=_='==(=A=(==01==)==2 Q

I > j~,lO

(8-67)

I

(8-68)

2a

fiTE,lO

111

U

.ih


437

wherein 11(0) ~. Thus, the TE lO mode fields (8-61) propagating in a rectangular waveguidc at a frequency above cutoff involve inphase transverse field components and ,#'; related by the real impedance (8-67). Below cutoff, the imaginary result (8-68) assures no time-average power transfer by the accompanying evanescent mode, attenuated through (8-64a). The time and Z dependence of the TElO mode field expressions (8-62) are included by multiplying them by eiWI+YlOZ. Taking the real part of the resulting products obtains the real-time traveling wave expressions as follows, assuming f > !c.lO:

i;

1[

Z,

t)

E;lO . sin a x cos (wt

Z,

t) =

--(0-)-

+

H; (x,

Z,

t) =

- E+ + y,10

PI0Z

+
sin 1[ x cos ( .4(0»)2 2a a

11 -

=+=

Ei,lO.4(O) 1[. (0) cos - x sm (wt 2al1 a

(8-69a)

(wt =+= PlOZ

+
+

_

+ PlOZ +
(8-69b)

(8-69c)

in which
Ps

Js

C/m 2

[3-45]

n X HA/m

[3-72]

= n' D

in which n denotes a normal unit vector directed into the dielectric region. The electric field of the TE 10 mode develops a surface charge density Ps on only two walls of the rectangular guide, since the y directed E field produces a normal component n • D on only the lower (y = 0) and upper (y = b) walls. Thus, using the electric field (8-69a) in D = EE = EayEi, the boundary condition (3-45) yields the surface charge density as follows 11:

EEy\o . sin -a x cos (wt

=+=

PlOZ

+
(8-70)

in time-instantaneous form. One may similarly show that the surface charge density on the opposite wall (aty = b) is the negative of (8-70). Surface current densities given by (3-72) appear at all four walls of the guide, because tangential magnetic fields occur at every wall. For example, on the lower wall where the total magnetic field is the vector sum of (8-69b) and (8-69c), the

438


E;,

FIGURE 8-12. Sketches of the wave nature of the separate components H;, and comprising the TElO mode, plus a composite flux plot (below). All are shown at t = O.

H:

surface current becomes 3 cos

. n

sm

a

x cos (wt

=+

n a

x sm (wt

f3lOZ

=+

+ 4>1'0)

f310X

+ 4)[0) (8-71a)

'Based on the current-continuity relation (3-130), you might consider how the surface charge density p, on the waveguide walls can be found from the surface-current density results (8-71).

439

8-4 TE MODE SOLUTIONS OF RECTANGULAR WAVEGUIDES Detector probe

(0)

Interior surface current flux ( a)

(e)

FIGURE 8-\3. The surface currents induced by the tangential magnetic field of the TE 10 mode on perfectly conducting inncr walls of a waveguide and wall-slot configurations. (a) Flux plot of surface cunent, on waveguide inner walls. (b) Slots producing negligible wall-current perturbation. (el Slots producing significant wall-current perturbations.

On the side walls, the surface current density has but one component, being entirely

y directed. Thus, on the wall at x = 0 (B-7Ib) The densities at y = b and x = a are similarly obtained. A sketch of the wall currents (B-71) is shown in Figure B-13(a), useful if slots are to be cut in the walls. For example, a longitudinal slot centered on the broad wall of a rectangular waveguide carrying the dominant TE tO mode as shown in Figure 8-13(b) is useful in field-probing techniques for the detection of standing waves (slotted-line measurements). A slot that does not cut across wall current flux lines produces a minimal perturbation of the waveguide fields, permitting field detection schemes that yield measurements essentially the same as those expected without the slot. In Figure 8-13(c), however, are shown slots that interrupt wall currents significantly, producing substantial field fringing through the slot with power radiation into the space outside the waveguide. Such configurations form the basis for slot antennas or arrays using waveguide fields for excitation.

EXAMPLE 8·2. An air-filled, X-band rectangular waveguide carries a positive z traveling TElO mode ati = 9 GHz. (a) Find the phase constant, wavelength, phase velocity, ar:d intrinsic has the wave impedance associated with this mode at the given frcqueney. (b) If amplitude 104 Vim, determine the amplitudes of .i'; and .i':. What time-average power flux is transmitted through every cross-sectional surface of the waveguide by this mode? 8 (a) At 9 GHz, the wavelength in unbounded free space is 1(0) = 1?~O)/f = (3 x 10 )/ (9 x 10 9 ) 3.33 em. With a 0.9 in., the ratio J~,lo/f given by (8-63) is 1(0)/2a = 0.729, whereas {3(O) = WJttoEo = 2n/1(0) = 60n rad/TIl. By use of (8-Mb), the phase

$;

440


constant becomes

PlO

= P(O)

(~:)y = 60n-Jl

1-

- (0.729)2

60n(0.683)

=

= 128.8 m- 1

Thus, from (8-65), (8-66), and (8-67)

3.33 0.683

A10 = - - = 4.88 cm 3 x 108

4.39

0.683 120n

A

IJTE 10

,

(b) With

i:;'10 =

= 0.683

=

552

X

10 8 m/sec

n

104 Vim, the remaining amplitudes, from (8-62), are 104

18.1 Aim

552

jl04(0.033) ., 2(120n)-6:0229 =)19.3 The time-average power flux transmitted through any cross section is obtained using (7-48), in which the minus sign is omitted if it is agreed that power flux emerging from the positive .e: side of the cross section is desircd. Thus, with Pay = jlllOz S 1.) 2 Re [E X H*J· ds , in which E = a Y $+e-jPloz y' , H = [ax x + a z £)+]ez' , and $;, and .~: are supplied by (8-62), one obtains

J(

*+

*:,

Pav --

i 1" b

y=O

1. 2 Re

x=o

1i:\oI2 = _Y_, b

21JTE,10

f i:+ )(P+ )* a z \ Y,lO 'y,lO A

{

1J'~E,10

. 2 n rasm -

Jo

a

xdx

~+ Ey,lO

• 2 SlIl

11:

a

} x ( e - JfJiO Z ') ( e - JlilOZ)* • a z dx dry

12

1 --ab 4IJTE.IO

With i:;'10 10 Vim, IJTE,lO = 552 n, a = 0.0229 m, and b = 0.0102 m, the timeaverage transmitted power becomes Pay = 10.6 W. 4

*8-5 DISPERSION IN HOLLOW WAVEGUIDES: GROUP VELOCITY All previous discussions of wave phenomena in this text have been restricted to single-frequency sinusoidal waves. Whether with reference to plane waves propagating in lossless or lossy unbounded regions as described in Chapter 6, or in connection with waves traveling in hollow metal tubes as considered in the present chapter, z traveling single-frequency waves are characterized by functions of the f(Jrm (8-72)

in which A is any complex amplitude coefficient [possibly a function of (x,y)] and in which any equiphase surface is defined by mt - /3::: = '" = constant. This yields the

8-5 DISPERSlON IN HOLLOW WAVEGUlDES: GROUP VELOCI-[Y

phase velocity

vp

by setting dt/J/dt

441

= 0, whence (8-73)

a quantity that mayor may not be frequency dependent, depending on the phase factor 13. Thus, in the case of plane waves traveling in unbounded free space, f3 = 130 = W,jPoEo, to yield (2-125b) (8-74 )

a result independent of frequency. Free space is therefore termed dispersionless, in view of the constant vp regardless of the frequency. On the other hand, waves of a given TM or TE mode in a rectangular hollow waveguide have a phase velocity given by (8-46) or (8-56) (8-75)

a decidedly frequency-dependent result. Although the concept of phase velocity is applicable only to steady state sinusoidal fields (constant amplitude and frequency), the Fourier superposition of any number of sinusoidal steady state field solutions having diflerent frequencies can be used to construct modulated waves of variable amplitude or frequency. This important process leads to another concept known as the group velocity, or the velocity of the signal, or information, associated with the group of waves distributed over thc spectrum of frequencies comprising the modulated signal. This is considered in the following. No information or intelligence is transmitted by a steady state, single-frequency sinusoidal traveling wave as that illustrated in Figure 8-14(a)_ It can, however, become a carricr of information by inflicting on it the process known as modulation. The transmission of information via a carrier wave requires a modulation (or changing, in time), in proportion to the instantaneous value of a desired signal, of either the amplitude or thefrequen~y of the carrier, thereby yielding an amplitude-modulated (AM) or a frequency-modulated (FM) carrier. The present discussion' is limited to the AM carrier, examples of which are illustrated in Figure 8-14(b) and (c). As suggested by the name, in this type of modulation the carrier amplitude is forced to become proportional to the signal level at every instant t. The !rcquency spectra of signals used to modulate a carrier typically fall within the audio range (dc to about 15 kHz) for ordinary voice or music transmission, or in the video range (dc to several megahertz) I()!' television or coded-pulse transmission. The Fourier analysis of a high-frequency carrier, amplitude-modulated by a spectrum orIower signal frequencies, reveals what range offrequencies must be transmitted by the system containing perhaps waveguides, coaxial lines, filter circuits, antennas, and other elements. Such an analysis shows that the transmission system must he capable of passing the carrier frequency fo plus additional Jrequency components contributed by the signal spectrum of width 2 I'lf, components appearing in two adjacent freqnency bands termed sidebands of width I'lfjust above and below fo. For example, a 100 MHz carrier, amplitude-modulated by a video signal embracing frequency components from dc to 4 MHz, will require a transmission band fromfo - I'lf

442


:> t

t

Wo

>W

(a)

:>W

E(t)

(Em{1

+ mcoswst)lcoswot

(b)

Modulation signal

L------~t

~t

(c)

FIGURE 8-14. Amplitude modulation of a continuous wave (cw) carrier, showing time dependence (lift) and Fourier components (right). (a) The single-frequency ew carrier, shown at z O. (b) A single-frequency signal used to amplitude-modulate a carrier and frequeney spectra. (c) A pulse signal used to amplitude-modulate a carrier and frequency spectra.

to fo + I'l.f, namely 96 to 104 MHz, or an 8% bandwidth. On the other hand, if a 10,000 MHz carrier were modulated by the same video signal, only an 0.08% transmission band extending from 9996 MHz to 10,004 MHz would be required to handle the ± 4 MHz signal spectrum. Short-pulse-communication and other high information rate systems require a correspondingly wide frequency band, therefore pulse communication systems using many channels simultaneously must operate at carrier frequencies in the uhf or microwave regions, and more recently they have even gone into the optical range of frequencies. The generic example of amplitude modulation is illustrated in Figure 8-14( b), depicting the simplest case of a carrier at the sinusoidal frequency wo, amplitudemodulated by a time-harmonic signal at the single frequency wS. The carrier amplitude Em is modulated sinusoidally in time with a signal amplitude mEm, in which m is called the modulation factor, so that the real-time expression for an electric field carrier mod-

8-5 DISPERSION IN HOLLOW WAVEGUIDES: GROUP VELOCITY

443

ulated in this way becomes

E( t) = [Em (l

+ m cos wst)] cos wot

(8-76a)

The bracketed factor denotes the amplitude variations at the signal frequency Ws' Equation (8-76a) specifies field behavior in the reference plane z = 0, whereas the additional z dependence needed to provide its traveling wave behavior is included momentarily. The amplitude-modulated carrier (8-76a) possesses three terms in its Fourier series expansion, or spectrum. Thus, with the substitution cos A cos B = (t)[cos (A + B) + cos (A B)], (8-76a) yields E(l)

Em cos wot

+

mEm 2

cos (wo

mEm

+ ws)t + -2- cos (wo -

ws)t

(8-76b)

This is a three-term (finite) Fourier series, possessing a carrier frequency term of amplitude Em' plus just two sideband terms of amplitude mE"./2 at the sum and difference fi-equencies (wo + ws) and (wo - ws)' This spectrum of three frequency components is depicted in the diagram at the right in Figure 8-14( b). The expressions (8-76) can be taken as the amplitude-modulated electric field of a plane wave (at Z = 0) propagating ·in unbounded free-space, or denote a field component of a propagating mode inside a hollow waveguide or a coaxial line, or such Equation (8-76b) is readily rewritten to specify the spectrum of positive z traveling waves of an amplitudemodulated carrier moving through a lossless transmission region, simply by adding in the proper phase delay terms pz as follows.

mE+

Poz)

mE+

+ -{- cos [Cwo -

+ -{- cos [(wo + ws)t ws)t - P-z]

(8-77)

in which Po, P+, and P_ denote the z propagation phase constants at the respective frequencies wo, Wo + Ws> and Wo - ws' One is to examine (8-77) for its wave-envelope velocity, or so-called group velocity, for two classes of regions: a nondispersive region, in which all frequency components of a spectrum of waves move with the same phase velocity; and a dispersive region, in which the phase velocities of the spectral components arc frequency dependent.

A. Group Velocity in a Nondispersive Region Suppose the signal (8-77) denotes the amplitude-modulated field Ex of a plane wave propagating in free space. The phase velocity is then the constant Vp = (/loEo) 1 = c given by (2-125b), making free space a nondispersive region. Therefore (8-77) written with Po = wole, P+ = (wo + w.)/e, and p_ = (wo - ws)le yields

E: (z, t)

E:'

cos Wo

+ m~:'

(t - ~) + m~:' cos [(wo + w.) (t - ~)]

cos[(wo-Ws)(t-~)J

(8-78)

444


Since the three Fourier terms remain in the same phase relationship no matter how far the modulated wave travels, the wave envelope must move at a velocity identical with the phase velocity in a nondispersive region. The wave envelope velocity, also called the group velocity (from the spectral group), is thus

w

Vg

vp

= 7f

(8-79)

Nondispersive

for a nondispersive region. This result is correct no matter how complex the spectral structure of the wave. Hence, for the pulse-modulated signal of Figure 8-14(c), all terms of its Fourier series expansion will propagate through the medium at the same phase velocity vp' One thus concludes that a dispersionless region is also distortionless.

B. Group Velocity in a Dispersive Region A hollow waveguide is an example of a wave-transmission device exhibiting the phase-velocity dispersion characteristic (8-75), depicted as a function of frequency in the graphs of Figure 8-8. The different phase velocities of the Fourier terms that characterize a modulated traveling wave in a waveguide result in the wave envelope appearing to slip behind the carrier appearing under the envelope. This phenomenon arises from the group velocity being slower than the phase velocities of the Fourier components. Thus, while the phase velocities of the Fourier terms of a modulated wave in an air-filled hollow waveguide all exceed the speed of light, the speed of the transmission of the information (the wave envelope) at the group velocity is at a speed less than c. The foregoing remarks are proved using the example of an amplitude-modulated carrier signal operating in the dominant TElO mode in an air-filled rectangular waveguide. The applicable phase constant expression (8-64b) is rewritten as

/3 10 -- /3(0)

),(0))2

1 - ( 2a

=

W.Jl1oEo

J

1-

(w ~10 )2

[8-64b]

This is graphed showing /310 as a function of the wave frequencies w in Figure 8-15(a), redrawn, for convenience, as w versus /310, to yield velocities from slopes (rather than /

W

/

Wo+Ws

Wo

WO-W s

W,'.lO

o !.I.(3'~!.I.(3

(a)

(b)

FIGURE 8-15. The w-/3 diagram for the TE10 mode in a hollow waveguide and velocity interpretations. (a) The (J}-/3 diagram. (b) Constructions leading to the phase and group velocities for the TE10 mode.

445

8-5 DlSPERSION IN HOLLOW WAVEGUIDES: GROUP VELOCITY

inverse slopes). The departure of 1310 from the linear (dashed-line) asymptote 13(0)

W.JftoEo is noted.

Suppose, now, that the waveguide carries the simple amplitude-modulated signal depicted in Figure 8-14(b), operating in the TE lo mode and having the three-term Fourier spectrum characterized by (8-77). Its amplitude factor is given by (8-62a)

Then the three-term spectrum (8-77) is rewritten

+

--t m!f+

(8-80a)

cos [(wo - ws)t - f3-z]

in which 1370 denotes the value of the phase constant 1310 of (8-64b) at the carrier frequency w = wo' Figure 8-15(b) depicts, by the use of (8-64b), the phase-constant values 1310 that correspond to the carrier frequency Wo and the upper and lower sideband frequency Wo + Ws = Wo + L\.w and Wo - Ws = Wo - L\.w of this modulated wave. Calling these {31O values {3+ = {370 + L\.{3 and /L = {3~0 - L\.{3' ~ f3~o - L\.f3 respectively, as noted on the graph (and assuming small frequency deviation L\.w, to allow putting L\.{3 ~ L\.{3') , enables rewriting (8-80a) as the three Fourier terms

Mo,

o

(310Z)

+

--t m!f+

m!f; + -2cos [(wo + L\.w)t -

cos [(wo - L\.w)t - (f3~o - L\.f3)z]

0

(1310

+ L\.f3)z] (8-80b)

This can be shown to recombine into the following product form (8-80c) A comparison with (8-76a) shows that (8-80c) describes the amplitude-modulated wave delayed in phase from the z 0 reference plane by the amount {370z insofar as the carrier at the frequency Wo is concerned, whereas the bracketed factor specifies how the envelope progresses down the Z axis in time. Since any equiphase surface on the envelope is defined by L\.w • t - L\.f3 • z = constant, the envelope moves down the z axis with the group velocity '1.'g,10 = L\.w/L\.f3. With the signal frequency Ws L\.w small compared to the carrier frequency, L\..w/L\.f3 becomes the limit

dw (df3lO) - I '1.'g,10 = d{310 = dw

(8-81a)

The last form, written as an inverse, is the more useful since 1310 is given explicitly in terms of the frequency w by (8-64b). A comparison with '1.'p = W/f310, the defining relation for the phase velocity of any of the Fourier steady-state sinusoidal terms in (8-80b), shows that group and phase velocities are obtained from slope interpretations

446

MODE THEORY OF WAVE,GUIDES

of Figure 3-15(b). Thus, the group velocity v g ,10 is given by the tangent to the (O-P curve at point P; whereas the phase velocity 'up,lO is the slope of the line from the origin 0 to P. The constant slope of the dashed-line asymptote ((0 versus P(O» is the free-space plane-wave comparison value (J1.oEo) 1/2, falling between the Vg,lO and v p ,10 values for this TE 10 mode. By extending this analysis to the modulated wave of the form of (3-77) operating in any TErnn or TMmn mode, the group velocity becomes

Vg,mn

=(

dP )-1 -d;n

(8-8Ib)

It should furthermore be clear that the same analysis applies to any uniform wavetransmission configuration (whether a waveguide, a two-conductor transmission line as described in Chapter 9, or whatever), so that its group velocity relates in general to its phase constant P through (8-8Ic) Applying the result (8-81 b) to the expression (8-42b) or (8-54b) for the phase constant P obtains the group velocity (8-82) for any TM or TE mode in a hollow waveguide. A comparison with the phase velocity expression (3-46) and (8-56) shows that (8-83) revealing that the unbounded-space velocity v(O) is the geometric mean of the phase and group velocities for hollow waveguide modes. Figure 8-16 illustrates the phenomena of phase and group velocities relative to the upper and lower sideband frequency terms of an amplitude-modulated carrier propagating in a dispersive medium. (The carrier term is omitted to simplify the graphic addition of the waves.) Note the alternate constructive and destructive interference (i.e., amplitude modulation) produced by the sum of the waves. Hthe sideband components were propagating in a nondispersive medium, their identical phase velocities would produce the same envelope velocity (group velocity). In a dispersive region as shown, however, the upper sideband term has a phase velocity lower than that of the lower sideband term, as noted from the slope of OP in Figure 8-15 (b). This causes the point of constructive interference, or maximum amplitude on the diagram of Figure 8-16 (b), to slip behind both sideband terms with the passage of time, yielding an envelope velocity ('Vg) smaller than the phase velocity, that is, smaller than vIOl = (J1.E) tl2 by an amount such that (8-83) is satisfied. The example just given involves a simple Fourier spectrum ofjust three frequency terms, insufficient to exhibit the envelope-distortion effects that would occur if the carrier had been modulated with a short-duration pulse, such as that shown in Figure 8-14(c). In the latter event, the corresponding spectrum would possess many more frequency terms, as depicted. The effect of propagating this pulse-modulated carrier, in

8-6 WALL-LOSS ATTENUATION lN HOLLOW WAVEGUIDES

Envelope of El

447

+ E2, showing

_ / resulting amplitude modulation

f\ z

Motion

(a)

(b)

FIGURE 8-16. Group and phase velocities associated with an amplitude-modulated wave. (a) The sum of the two sideband frequencies of an amplitnde-modulated wave, showing beat effects. (b) Depicting phase and group velocities in the wave of (a), as time increases. The medium is assumed normally dispersive.

the TE mode, over a sufficient length of rectangular waveguide, is to distort the shape lo of the pulse envelope, the extent of the distortion depending on the length of the waveguide. The distortion is a consequence of the Fourier components having different phase velocities over the frequency band of the Fourier spectrum, such phase velocities being given by 'Up,IO = ill//310' This causes the sinusoidal components to arrive at their destination in a different phase condition than that occurring at the waveguide input, thereby producing the distortion. This phenomenon is therefore given the name

dispersion. *8-6 WALL-LOSS AnENUATION IN HOLLOW WAVEGUIDES In the previous discussions of wave propagation in rectangular hollow waveguides, it was assumed that the waveguide walls were perfectly conducting. Practical waveguides are necessarily made of finitely conducting metals (e.g., brass, aluminum, silver), and

448


waves moving down the interior will generate wall currents much like those depicted in Figure 3-12 for the dominant TE 10 mode. In the ideal, perfectly conducting case, the wall currents are restricted to surface currents characterized by a penetration depth of zero, the tangential magnetic field being a measure of the surface current density according to the boundary condition (3-72). The fields inside the perfect conductor are zero, to make the wall power losses zero for this idealized case. With finitely conductive walls, however, the continuity of the tangential magnetic field guarantees a time-varying magnetic field inside the conductor, producing therein an electromagnetic field rapidly diminishing with depth. The fields penetrate the conducting wall essentially at right angles to the surface. The ensuing ohmic power loss due to the transference of a small portion of the available transmitted mode energy into the walls results in a measurable attenuation of the propagated mode. For example, the wall-loss attenuation occurring in an X-band brass waveguide carrying the TE 10 mode at 10 GHz is of the order of 0.2 dB/m, a significant amount for long waveguide runs. It is the purpose of this section to outline a method for the approximate analysis of the wall-loss attenuation problem for hollow guides. In the propagation of a 'I'M or TE mode down an ideal (lossless) waveguide, the power flux travels unabated down the pipe, the same time-average power passing through every cross section of the guide. As shown in Figure 8-17(a), the positive z traveling, unattenuated fields arc designated in the usual complex notation

$+(u 1, u2 )e- jpz

(8-34)

--

(z)

(z)

(a)

I

(I»

[i'~'iii"drj ~ilrr mmj

Pav , [.

dz

Small Cy (exaggerated)

(c)

(d)

FIGURE 8-17. Relative to the wall-loss attennation in a waveguide of uniform cross section. (a) Unattenuated fields in a lossless, ideal waveguide. (b) The attcnuation of the fields due to power absorption by the walls. (e) Showing a small tangential Ey component at the walls, compared to the lossless mode configuration. (d) Volume regiou of length dz, lor comparing transmitted and wall-loss average powers.

8-6 WALL-LOSS ATTENUATION IN HOLLOW WAVEGUrDES

449

fields defining the unperturbed mode in a loss-free waveguide. In the event of a finitely conductive wall material, a portion of the transmitted power is diverted into the walls, leading to an exponential decay of the average power through successive cross sections of the waveguide, as suggested by Figure 8-17 (b). The wall-loss attenuation achieved ill this process is designated by a', and with the tleld distributions If(ull and :#'(Ul' uz) assumed unchanged from (8-84), the attenuated fields are written (8-85) The 1;+ and ;ie+ factors in (8-85) will differ by a small amount from lhose given in (8-84), a fact appreciated on inspecting Figure 8-17(c). Shown)s the tffy distribution for the TE lo mode of a rectangular guide, with a very small tffy component existing at the x = 0 and x = a walls due to the field penetration of the tangential magnetic field into the conductive walls, leading to a first-order analysis as follows. An expression is derived for the wall-loss attenuation factor a' in terms of the time-average transmitted power and the small fraction of this power that escapes into th~ walls in every length dz of the waveguide. It is shown, for a given mode, that

, a

=

1

dPav,L dz --Np/m

2 Pav,T

(8-86)

in which the meanings of the symbols are illustrated in Figure 8-17 (d). PaY, T denotes the average power flux transmitted by the mode through any cross section of the waveguide, whereas dPav,L is that lost into the walls through the peripheral strip of width dz. One can derive (8-86) by noting that if the volume slice of length tiz in Figure 8-17(d) contains no ohmic losses or sources, then by (7-31) or (7-56) the net timeaverage power entering (or leaving) the surface enclosing that volume is zero. Therefore, dPav,I, = -Pav,T + [Pav,T - (apav,T/aZ) dz], yielding dPav ,L

=

apav,T --az dz

(8-87)

The average power transmitted through the waveguide cross section is obtained from the cross-sectional surface integral of the time-average Poynting vector (7-47a), with the fields (8-85) inserted

f

1.

JS(c.s.) 2

e- 2a 'z

f

Re [(1;+ e -a'ze -

JS(c.s,)

!- Re [1;+

j[!Z) X

(;ie+ e -a'ze- j[!Z) *] . ds

x ;ie+*] . tis

Differentiating (8-88) with respect to Z obtains

(8-88)

450


and solving for

(x'

yields

(X' = 2 Pav,T but from (8-87), iJPav,T/iJZ can be replaced with -dPav,ddz, so

= ~ dPav,L

(X'dz

(8-89)

2 Pav,T

which is just (8-86), that which was to have been proved. To illustrate the use of (8-89) in finding (x' for a given waveguide mode, consider the dominant TElO mode. The average transmitted power Pav,T in (8-89) has already been found in Example 8-2 ~+

_ 1E y ,10

Pav

12

T -

,

411TE,10

b a

(8-90)

The power loss dPav,L in (8-89) arises from the electromagnetic wave induced inside the conductor. Just within the walls are tangential magnetic fields, identical, by continuity, with the magnetic fields of the known components (8-62) of the unperturbed TE 10 mode. Also appearing therein are electric fields, obtainable from the known magnetic fields by use of wave impedance expressions like (3-97), since the electromagnetic field propagates essentially at right angles into the conductors much like a localized plane wave. This fact is corroborated by the plane wave tilt incidence analysis in Appendix A, showing how field penetration is analyzed for obliquely incident plane-wave fields at sufficiently high frequencies. In Figure 8-18(a) is shown the contInuity of the knowp Ye; component (8-62c) t;?f the 1EIO mode. A small component Iffy is induced by Yez in the metal such that Iffy = f/Yez , and together they comprise essentially a plane wave traveling nearly perpendicularly into the conductor with a

(a)

(b)

FIGURE 8-18. ConcerniIlg the boundary condition on the tangential magnetic field, leading to wall-skin currents. (a) Continuity oftangentialYtz leads to induced if, inside conductor. (b) Showing cosine distribution of the tangential magnetic field on the top and bottom, uniform on side walls (TEIO mode).

8-6 WALL-LOSS ATTENUATION IN HOLLOW WAVEGUIDES

451

large attenuation. ifz is maximal at the x = 0 and the x = a walls, with a consinusoidal variation between these values existing along the y = 0 and the y = b walls as in Figure B-IB(b). The electric fields induced just inside the x = 0 and x = a walls thus become

jj]

Yx=O

= -

~if] zx=o

(B-91 )

in which ~ = (W/1/(J)1/2e i1t /4 from (3-112c), the negative sign properly accounting for the propagation of the wave into the metaL Similar expressions apply at they = 0 and y = b walls. The time-average power loss dPav,L in (8-89) is obtained by integrating over the four sides of the peripheral strip of length dz embracing a cross section; thus (B-92) the x

=0

strip, for example, making use of (8-91) obtains

(8-93) the wall-loss at y

= 0 - becomes (8-94 )

a result accounting for both tangential components if; and if: of (8-62b) and (8-62c), and making use of the identity (rc/a)2 + p2 = (fJ2/1E for the TE lo mode. Evaluating all four wall-loss contributions of (8-92) yields _ 1 \ E~+ y ,10 \2

dPav,L - 2

a

2

W 2 p,2

\£:'12 0\22 W

/1

fti[7 w/1

~

2rc 2b

+ aw

2

J

p,E dz

ro;;; W2p,E[2b (fc,10)2 .;. IJdZ

...J~

a

f

(8-95)

the latter making use of (j~.10/f)2 = rc2/w2/1Ea2 from (8-53). Inserting (8-95) and (8-90) into (8-89) yields the wall-loss attenuation for the TE mode

10

(8-96)

452

MODE THEORY OF WAVEGUIDES I

0.6 TE

0.5

I

[0.04 in.

:

r::

lO

0.6

(a = 0.9 in., b = 0.04in.) I I

I

0.4

E ......

co

:0 0.3 ~

0.2

01 -

o

0.1

5

10

15

20

25

30

o

5

10

Frequency (GHz)

Frequency (GHz)

(a)

(b)

15

FIGURE 8-19. Wall-loss attenuation versus frequency for copIX,r. (a) Attenuation versns frequency lor modes in rectangular waveguides. (b) Attenuation characteristic If)r a circular waveguide.

The b factor in the denominator of (8-96) shows that making the height too small results in a large wall-loss attenuation. This is a consequence, at a fixed field amplitude E;:10, and as seen from (8-90), of the smaller cross-sectional area through which the correspondingly smaller transmi.tted power Pav,T must flow, the wall-loss power remaining nearly the same as for a waveguide with a larger b height. It is also evident from (8-96) that as the cutoff frequency is approached, the wall-loss attenuation becomes indefinitely large. A graph of (8-96) versus frequency for two choices of b height is shown in Figure 8-19(a), along with the wall-loss attenuation characteris6c of the TM] 1 mode in a rectangular waveguide. 4 From Figure 8-19, it is evident that different modes undergo different amounts of attenuation in a given waveguide. It would appear that a way of reducing wall-loss attenuation is to minimize the exposure of the magnetic field component tangential to the wall. Nearly all modes in hollow waveguides have an increasing wall-loss attenuation with increasing frequency, with at exhibiting a minimum value at some optimum frequency, as already seen in Figure 8-19(a). It develops that a circular waveguide mode, the TE 01 , deserves special attention in that it exhibits an indefinitely decreasing at with increasing frequency, the result of a smaller and smaller component of the tangential H field at the metallic wall as the incidence of the wave hecomes more nearly grazing. 5 This mode, having the attenuation characteristic depicted in Figure 8-19(b), shows promise in long-range, low-loss transmission at superhigh frequencies in hollow metal cylindrical pipes, though problems are posed by the fact that the TEll mode, and not TE 01 , is the dominant mode in a circular wavegnide. 4A

further discussion of the wall-loss attennation factor associated with the remaining modes of rectangular waveguides can be found in S. Ramo, j. R. Whinnery, and T. van DuzeL Fields and Waves in Communication Electronics, 2nd ed. New York: Wiley, 1984, Chapter 8. 5Yor a discussion of the theory of the circular waveguide, see Ramo, S., j. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics, 2nd ed. New York: Wiley, 1984, Chapter 8, Circular waveguides have important applications to rotating joints used for feeding movable antennas and to tunable resonant cavities.

PROBLEMS

453

GINZTON, E. L. Microwave Measurements. New York: McGraw-Hill, 1957, JORDAN, E. C., and K. G. BALMAIN. Electromagnetic Waves and Radiating ~ystems, 2nd ed, Englewood CliRs, N.J.: Prentice-Hall, 1963. LANCE, A. L. Introduction to Microwave Theory and Measurements. New York: McGraw-Hill, 1964, RAMO, S., J. R. WIliNNERY, and T. VAN DUZER. Fields aud Waves in Communication Electronics, 2nd ed. New York: Wiley, 1934,

On inserting the replacements (3-1a) and (3-4) into the Maxwell relation (3-3) and expanding it in rectangular coordinates, show that the modified curl relation (3-3) is obtained, noting from (3-7) how the "modified curl operator" is defined. Repeat Problem 3-1, this time carrying out the details in generalized cylindrical coordinates. Use the rectangular coordinate expansion (2-33) of the Laplacian of a vector field to !!how that the vector wave equation (3-12) expands into three scalar wave equations analogous (3-13).

U

Expand the modified Maxwell curl relations in circular cylindrical coordinates, and from these obtain = -I

p

it+

;p

[±Yap' + j:;/l

=

k~ [~l

=

k~ [j~:E .vi

=

-1[.

)WE

o¢z]

vcp + j(WI'jJp 1= Y .. opt.

J

J

oiz± yoit;] op ± p-o¢

From the results, modal expressions analogous with (8-20) through (3-25), but applicable to tile TE, TM, and TEM modes of circular cylindrical waveguides and coaxial transmission lines, I:l:lay be found. Repeat Problem 3-4, exr:ept carry out the details in the generalized cylindrical coordinate ,ystem, obtaining

a check, show that these results reduce to (8-19) in the rectangular system.

454


SECTION 8-2 8-6. The results (8-19), relating the transverse field components in a rectangular waveguide to the longitudinal components, were found from the simultaneous manipulation of (8-17) and (8-18). Show the details of how (8-19b) is obtained.

8-7.

Repeat Problem 8-6, this time showing how (8-19c) is found.

8-8.

Use results given in Problem 8-4 to obtain expressions for the intrinsic wave impedances associated with the transverse field components of the TM and TE modes of uniform circular cylindrical transmission systcms, namely

j±

v

;yet

-.-'_.

+-;L =

_j:

+"';,?" oft;;

8-9.

Repeat Problem 8-8, but

j(lr

== ~TM

JW€

jW/1 _

A

= - _ . = IJTE }'

generalized cylindrical coordinates.

SECTION 8-3 8-10. Verify the expression (8-38b) for the x-component of the electric field of the TMmn modes by substituting the solution (8-38a) into the proper l~M modal relation (8-20). Leave the answer expressed in terms of the electric-field amplitude E;'mn'

8-11.

Repeat Problem 8-10, but this time verify (8-38e) for the y-component of the magnetic field. [Note in this instance that the modal relation (8-20d) need not be used.]

8-12. Sketch a diagram resembling Figure 8-4, showing tbe z-directed electric field component of the TM 12 mode. 8-13. Based on its cutoff frequency, determine the inside dimension of thc smallest air-filled square (a = b) metallic waveguide that will just propagate the lowest order TM mode (TM l1 ) at tbe operating frequency: (a) 5 GHz, (b) 5 MHz, (c) 5 kHz. lAnswer: (a) a = Ii 4.24 cm (b) 42.4 m (c) 42.4 km] 8-14. An air-filled rectangular waveguidc with interior dimensions a = 0.9 in. by b = 0.4 in. opcrates at tbe fi'equency f = 18 GHz in tbe TMlI mode. Find, for tbis mode in the given waveguide (a) tbe eutofffrequencyj~,ll' (b) the pbase constant /311, (cl tbe wavelength All in the guide, (d) tbc phase velocity v p .lI, (e) the intrinsic wave impedance l1TM.ll' (f) What does the propagation constant }'11 for this TMll mode become, on reducing the operating frequency f to 9 GHz? (g) Compare answers (b) through (e) with the values expected at this operating frequency for a uniform plane wave in free spaec. Comment on differences observed.

SECTION 8-4 8-15. Use the separation-of-variables method, applied to the wave equation (8-16c), to obtain tbe wave solution (8-49) for TE modes. 8-16. Cardully apply the four boundary conditions (8-50) to the z-directed magnetic field solution (8-49), showing that tbe proper solutions (8-5Ia) are obtained for the z-component of the magnetic field for the TEmn mode-set. 8-17. Verify the expression (8-51 c), for tbe.JI-component of the electricficld of the TEmn modes, by inserting the magnetic-field solution (8-51 a) into the proper T~ modal relation (8-22). (Lcave the answer expresscd in terms of the magnetic field amplitude Htmn") 8-18. Make use of the expression (8-52), the propagation constant ofTE mn modes, to show in detail how the cutoff frequency (8-53) is obtained. (In this regard, review tbe discussion following (1:)-39), the identical propagation constant expression obtained for the TMmn mode-seLl

PROBLEMS

455

1-19.

Given are six air-filled rectangular waveguides with the tollowing inside dimensions. Calculate their eutoff frequencies for the dominant TE IO mode: (a) L-band: 6.25 x 3.25 in. 5.875 x 8.255 ern), (b) S-band: 2.84 x 1.34 in. (7.214 x 3.404 ern), (e) C-band: 1.872 x in. (4.755 x 2.215 ern), (d) X-band: 0.9 x 0.4 in. (2.286 x 1.016 ern), (e) K-band: 0.420 x 0.210 in. (1.067 x 0.533 ern), (f) V-band: 0.143 x 0.074 in. (0.376 x 0.188 em). [Answer: 0.944, 2.073,3.152,6.557, 14.048,39.366 GHz.]

1-20.

Show in detail, for the so-called dominant TE IO mode, that the five field-component ex(8-51) reduce to just the three given in (3-61). (Leave the answers expressed in terms the amplitude fr;'IO of the longitudinal magnetic-field componenL)

1-21.

Given is the so-called X-band rectangular waveguide, designated to carry frequencies in the 8.2 to 12.4 GHz band in the dominant TE IO mode, with the inside dimensions a = 0.9 in. = 2.286 cm, b = 0.4 in. = 1.016 em. (a) Calculate its cutoff frequency for each of the following modes in this waveguide: TEIO' TE oI , TEl!' TE 20 , TE 21 , TM Il , TM I2 , TM 21 , TM 22 . Label these cutoff-frequency values and corresponding modes on a diagram as suggested by 8-IO(a), showing also the extent of the "X-band" on the frequency scale. (c) Which of modes will propagate as waves, and which will evanesce (decay), at the generator frequency (operating frequency) of 5 GHz? 10 GHz? 15 GHz?

1-22.

Show how the expressions for the dominant-mode (TEIO) fiele! components (8-61) can be rewritten in the form (3-62) (expressed in terms of the amplitude E~lo).

1-23.

Calculate the smallest a-width of an air-filled rectangular waveguide that will just propagate the electromagnetic TElO mode at the following frequeneies: 5 GHz, (b) 5 MHz, 5 kHz.

1-24.

An X-band rectangular air-filled waveguide with dimensions 0.9 x 0.4 in. carries the dominant TE IO mode at the source frequency f = 9.375 GHz. Determine, for this mode: (a) the cutofffl·equency,fc.lo, (b) the phase constant PIO, (c) wavelength }'10 in the waveguide, (d) phase velocity, vp.10, (e) intrinsic wave impedance 1JTE,IO' (f) What is the cutoff frequency for the TE 20 mode in this size waveguide? What do the propagation constant 1'10 and the intrinsic wave impedance I'fTE.IO become for the TE lo mode on reducing the operating frequency to 4.5 GHz? (g) Compare answers (b) through (e) with the values expected for a uniform plane wave in free space at the same operating frequency.

1-25.

E:

The amplitude of the field of the dominant TE lo mode in an S-band (2.34 x 1.34 in.) air-filled rectangular waveguide is 0.5 MV/m. (a) Determine the amplitudes of the J-C and field components, if the operating frequency is 3 GHz. (b) Based on the result derived in Example 3-2(b), what average power is being transmitted down the waveguide in this mode? (c) What maximum value of average power density exits within any cross section of this waveguide? Explain.

H:

8-26. Assume the same waveguide of Problem 8-24 to be connected to a generator operating at the frequency f = 4.5 GHz, the mode produced in the waveguide being the TE IO mode. Determine its (a) attenuation constant ()(IO, (b) intrinsic wave impedance IiTE,IO' (c) In view of the propagation constant 1'10 becoming the pure real ()(IO below cutoff, is the field produced in this waveguide a "wave", in the strict sense? See Figure 8-5(b) and comment. Calculate the <-distance required by this field to diminish to e-I 0.363 of any reference value. (d) Explain the meaning of the imaginary value of intrinsic wave impedance obtained in (b) for this field. Explain how it affects the propagation of average (real) power down the waveguide, beloweutoff.

8-27. An automotive tunnel is rectangular in eross section (6 m high by 15 m wide) and with metal walls. Determine the lowest radio frequency signal that will propagate as a nonevanescent wave through this tunnel. Which mode is this (TE or TM), and which electric-field polarization must it have (i.e., aligned with which dimension)? Show a sketch of the tunnel cross section, depicting the E field flux distribution for this mode. Will radio signals in the AM broadcast band (550 to 1600 kHz) travel in this tunnel as waves, or will they evanesce? In the FM band (88-103 MHz)?

f

4IiW"

I IIi

456


SECTION 8-5 8-28. Show that the expansion of the amplitude-modulated traveling-wave expression (8-80c) yields precisely the three terms of the preceding expression (8-80b). Discuss the meaning of this result in relation to the w-{J diagram of Figure 8-15(b) and the concepts of group velocity and dispersion. 8-29. Show the details of the differentiation of the {Jm. expression (8-42b) or (8-54b) for rectangular waveguide modes, obtaining (8-82) for the group velocity. 8-30. Find the phase and group velocities at the operating frequencies 8.2, 9, 10, II, and 12.4 GHz, for an air-filled X-band rectangular waveguide (0.9 x 0.4 in.) having the dominantmode cutoff frequency le,lO 6.557 GHz. Graph these results versus frequency.

SECTION 8-6 8-31. Making use of (8-89), carry out the remaining details in the power-loss expression (8-92) to verify the wall-loss power expression (8-95) for the TElO mode. From this, verify the wall-loss attenuation factor a'lO of (8-96). 8-32. For copper waveguide walls (0" = 58 MSjm), evaluate the wall-loss attenuation factor for tt1e TEw mode in an air-filled X-band rectangular waveguide at f = 8,2, 9, 10, II, and 12.4 GHz. What happens to a'lO as the operating frequency f approaches the cutoff value?

~~.------

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ CHAPTER 9

TEM Waves on Two-Conductor Transmission Lines

The previous chapter considered the TM and TE mode configurations of reetangular hollow (single-conductor) waveguides. Omitted from detailed discussion was the TEM (transverse-electromagnetic) mode, the dominant mode of transmission lines using two (or more) conductors. The parallel-conductor line, shown in Figure 4-14(b) and in Example 5-16, and the circular coaxial line, depicted in Examples 4-9 and 5-13, are commonly used in the transmission of this mode. It is seen that at least two conductors a range of frequencies are required to establish the TEM mode, transmittable extending all the way down to zero frequency (de). Although the TEM mode is by far the most important, TM and TE modes are also capable of propagating on two-conductor transmission lines. The latter modes, however, are evanescent below their cutofffrequencies, which occur for ordinary coaxial lincs in the upper microwave frequencies and beyond. The TM and TE modes on two-conductor lines thus have no useful applications to signal or power transmission, so they are omitted from detailed discussion. 1 Two-conductor uniform transmission lines of the coaxial or parallel-wire type, operating in the TEM mode and illustrated in Figure 9-1 (a) and ( b), are commonly used in power distribution and signal communication systems. Power transmission lines carry power in the megawatts up to hundreds of kilometers from generating stations to urban regions. Voice and pulse-data signals are carried over telephone lines, with signal amplification applied every few tens of miles if the information is to be carried over long distances. Power lines usually operate at 50 or 60 Hz, employing parallel-wire lines suspended on poles or towers, or using buried cables. Telephone lines

'. IHigher-ord"r mod~s on the coaxial line are discussed in S. Ramo, J. R. Whinnery, and T. Van Duzer. Fields and Waves in Communication Electronics, 2nd cd. New York: Wiley, 1984, p. 428.

457

458

TEM WAVES ON TWO-CONDUCTOR TRANSMISSION LINES

(b)

(a)

(e)

(d)

FIGURE 9-1. Two-conductor uniform transmission tines. (al The coaxial line. Goncentric conductors separated by air or a dielectric material. (b) Parallel-wire line. Usually separated hy air. (e) Generalized line. One conductor inside the other. (d) Generalized line. Conductors externally located.

are seen in pairs on poles, though many buried coaxial and multiconductor cables are in use. These may carry audio signals directly, or information transmitted as a modulation of the amplitude of a carrier frequency operating up to several megahertz, permitting the transmission of several modulated carriers simultaneously over the same transmission line, or the signals may be multiplexed using pulse-code modulation at high pulse rates to increase the information-handling capacity significantly. Coaxial lines are commonly used, for example, to interconnect a radio frequency transmitter to an antenna employed for launching electromagnetic waves into the atmosphere. At the high.er microwave frequencies, hollow waveguides can be employcd to connect a data transmitter or perhaps a radar to an electromagnetic horn or a dish-reflector antenna. Short sections of uniform transmission line, having low losses at the higher frequencics, can be used as the high Q.resonant (frcquency selective) elements of filters; they may serve as reflective elements in pulse-forming networks; they may be used to transport pulse data from one place to another with low distortion in high-speed computers. From this partial list of applications, it becomes apparent that a detailed study of transmission line behavior can be of substantial importance to the engineer and applied scientist. This chapter begins with a discussion of the properties of the electric and magnetic fields of the TEM mode on two-conductor lines. The related currents and voltages are developed next, to introduce the concept of characteristic impedance. The transmission line equations are deduced in terms of the distributed line parameters, first assuming ideally perfect conductors, and then for the physically realizable line employing finitely conductive elements. The chapter concludes with a real-time analysis

9-1 TEM MODE FIELDS BASED ON STATIC FIELDS

459

of voltage and current traveling waves of arbitrary waveshapes on ideal (lossless) twoconductor transmission lines. The time-harmonic (sinusoidal steady state) analysis of voltage and current on lines with arbitrary load impedances is covered in the next chapter.

'·1 TEM MODE FIELDS BASED ON STATIC FIELDS 2 A uniform two-conductor transmission line is represented in generalized cylindrical coordinates in Figure 9-1 (c) and (d). The pure TEM mode exists (ideally) on a line composed of perfect conductors. For conductors with finite conductivity, the z-directed currents in them account for a z component of the electric field at the conductor surfaces. The small z component of the E field required to sustain the electric field inside even good conductors, if longitudinal currents are to flow in them, gives rise to what might be called essentially TEM waves. Such waves produce internal resistive and inductive eHects in the conductors, considered later in Section 9-6. A pure TEM wave, associated with two perfect conductors comprising a uniform transmission line, has only transverse components of the fields. The TEM mode is defined by putting (9-1 ) In generalized coordinates, the TEM mode E and H fields are thus given by expressions with the z components absent; that is, (9-2) assuming all field components to be functions of (Ub (3-45) and (3-79) yield that E/=O

and

U2,

n -D

z, t). The boundary conditions

= Ps

(9-3)

at the perfect conductors, meaning that E is normal to the conductor walls, terminating there in the surface charge density Ps' The magnetic boundary conditions (3-50) and (3-72), moreover, imply that and

n x H =J8

(9-4)

at the perfectly conducting walls, so that H is entirely tangential at the walls, terminating there in the surface current Js. The Maxwell integral laws of Faraday and Ampere, (3-66) and (3-78), applicable to the TEM mode obeying the conditions just noted, can be written in the reduced forms

1,

~(c.s.)

E-dt=O

1,

H-dt= JsIJ-ds

~(c.s.)

(9-5)

i)

(9-6)

2Section 9-1 covers details of the electric and magnetic fields of the TEM mode. Field details are important, for example, in the design of lines for which considerations of maximum field strengths, concerned with corona and voltage breakdown, may be of interest. The reader interested in a more conventional approach starting with line voltages and currents may elect to go directly to Section 9-2.

460


if the line integrations are restricted to closed paths t confined to any cross section of the transmission line. The simplifications to (9-5) and (9-6) are evident from the definition (9-1), that no z component ofE or H can exist between the conductors of the uniform line. This means that no flux of either D or B can pass through the surface S restricted to the cross section by any closed path t, thereby reducing the surface integrals of D . ds and B . ds in (3-66) and (3-78) to zero. The specialized forms (9-5) and (9-6) of the Faraday and Ampere laws provide the following interpretations for the TEM mode. I. Faraday's integral law (9-5) is of the same form as (4-6) discussed in Chapter 4. It means that the E field of the TEM mode of a two-conductor transmission line is a conservative field, relative to any closed path t within a fixed cross section at a given instant. One can thus expect that the static E-field solution ofa uniform

two-conductor line can be used as the basis for the TEM-mode E-field solutions on that same line. It is, moreover, correct to assume that a potential relation of the form of (4-31) will serve as an adjunct to the finding of the E-field solutions. 2. Ampere's integral law (9-6) is observed to have the same form as the static form (5-5) considered in Chapter 5. This form will be useful in relating the transmission line TEM-mode current to the corresponding H field in any cross section. The special forms (9-5) and (9-6) of the Faraday and Ampere laws are put to use for the TEM mode as detailed in the following discussions.

A. Electric Field and Line VoHage of the TEM Mode By analogy with similar conclusions drawn for electrostatic fields in Section 4-5, the Faraday relation (9-5) for the E field of the TEM mode also guarantees zero curl of E in any transverse cross section of the transmission line, or (9-7) in whieh the subscript T denotes the curl taken with respect to the transverse variables (u j , U2) only. Thus, by analogy with (4-31), the electric field E of the TEM mode must be related, within any fixed cross section, to an auxiliary scalar potential function such that E

(9-8)

VT

wherein VT denotes the gradient operator with respect to (Ub U2) only. By analogy with (4-38a), there then exists for the TEM mode the potential <1>, at any point in any fixed cross section, given by the integral of E . dt from an arbitrary potential reference Po to the desired point P. By extension, the voltage V between the two conductors of the transmission line is analogous with (4-46)

V=

-

(PI E . dt]

Jp2

(c.s.)

(9-9)

in which (c.s.) denotes that the integration path is to be kept within the fixed cross section. The additional property of the potential of the TEM mode is that it satisfies, by analogy with (4-68), the two-dimensional Laplace's equation

Vi· = 0

(9-10)


461

in which the Laplacian V} is defined by (2-79) with respect to the transverse dimensions (Ub u2) only. All the preceding expressions apply to the fields B(u t , U2, Z, t) and (])(Uj, U2, Z, t) in the time domain. They can more usefully be converted to the time-harmonicphasor form by assuming that the dependence on Z and t is specified by the exponential factor exp (Jwt yz), as already discussed in Section 8-1 for any wave transmission system uniform in cross section. Thus, let

+

(9-11) (9-12) with + and -~ superss.ripts denoting the positive Z and negative Z traveling wave solutions, and If± and (])± signifying complex phasor functions of only the transverse coordinates (ub U2)' Then (9-7), (9-8), and (9-10) can be written, after cancelling the exponential factors

(9-13)

- V T q,±

(9-14)

(9-15)

The total electric field distribution between the conductors is expressed by use of the + Z and - z traveling-wave electric field solutions of (9-14). Adding these together after multiplying them respectively by e -yz and el'z yields

(9-16)

The further replacement of V in (9-9) with v,;;d wthz , in which V';; denotes the cornplcx amplitudes of the voltage waves that accompany the electric fields of (9-11), yields, after canceling the exponential factors,

f~l j± . dt

1..

(9-17)

The integration is taken from P2 on the reference conductor to 1\ on the more positive conductor within a fixed cross section, as denoted in :Figure 9-2 (a). The linear superposition of (9-17), on multiplying the voltage amplitudes respectively by Y", yields the total line voltage on the transmission line

(9-18)

462


(a)

/

/

/

(6)

FICCRE 9-2, Two-comiuctol" generalized in relation to eknric and magndic field" (a) Voltage 1~ defined on the positive conductor.

showing' voltage

V

t;

and current.

1";:;

between condu<"lors, (b) Current

EXAMPLE 9-1. A long, uni/()rm coaxial transmission line consisting of perfect conductors with the dimensions shown has a dielectric with the parameters /1, E, apd (L (al Usc the static potential field solution (j) to obtain the time-harmonic potelltial (j)± in any cross section. (Express (j)± as a functioll of the potelltial difference between tlw conducto}"s, taking the outer conductor as the zero reference.) Obtain the transverse electric field .g'±. (Ii) Verify that the total voltage relation (9-20b) correctly leads to the result (9-18). (a) From Example 4-12, the static potential field of the coaxial system with the potential difference V is (4-70) $(p)

iii,I

I

II :1\

I ,

V

Ii

Ii t~ta

P

--t~t

(9-19)


(/J.,

463

f, a)

EXAMPLE 9-l

a solution of Laplace's equation and the boundary conditions. The analogous solution applicable to TEM waves is

~

$±(p) =

if;;; t" ~ a

tn

b

(9-20)

P

in which denotes voltage amplitudes associated with positive z and nq!;ative z traveling wavc solutions. The corresponding electric field solutions are f(mnd from 14) to yield

J±

= -VT± = -apiJ±jiJp,

obtaining (9-21 )

a

(9-22a) (9-22b)

in which y is yet to be found, the amplitudes if~ and V;;' depending on the generator and possible reflections occurring down the line. (b) Inserting (9-21) into (9-17) yields

V(z)

= [-

S:d:]e-

V,\ t"

a

= V~e-YZ

+

or jllBt the expected result (9-18).

YZ

+ [- V;;'b t.!

a

Sba;]e

YZ

464

TEM WAVES ON TWO-CONDUCTOR TRANSMISS[ON LINES

B. Magnetic Field and Line Current of the TEM Mode Much in the way that Faraday's law (9-5) was used to develop the connection (9-17) of the TEM mode If± fields to their corresponding voltage-wave amplitudes V;;, the Ampere law (9-6) leads to the relationship between the TEM mode magnetic fields and corresponding current-wave amplitudes. Thus, on converting (9-6) to timeharmonic phasor farm; that is, letting B(u!, U2,

z,

t)

i(z, t)

be replaced by jl'±(Ul> uz)ejwHyz

(9-23)

be replaced by I;;e iwt + yz

(9-24)

obtains, after canceling the exponential factors, a measure of the forward- and backward-traveling-wave current amplitudes ~+ 1m

= ~(c.s.) f~f±·

dt

(9-25)

provided that the closed line t completely encloses either conductor of the twoconductor line, as d£picted by the typical closed lines t ~s chosen in Figure 9-2 (b). If the wave solutions ff± were known, their superposition in the same manner as (9-16) leads to the total magnetic field distribution between the conductors, expressed as the sum of the +z and -z traveling waves

(9-26)

The solut0ns for jl'±(Ub u2) in (9-25) and (9-26) will be seen to be expressible in terms of the If± fields, previously found from the potential relation (9-14), and from use of the Maxwell modified curl expressions (8-6) and (8-8). Before finding these, note that the linear superposition of the sinusoidal current-wave amplitudes of (9-25), on multiplying them respectively by yz, yields the total line current on the transmission line

I;;

II "7)' = j+ e - yz + \'v

In

j-m eYz

(9-27)

in which I:. and I;;, are related to the fields jl'± by (9-25). The magnetic field solutions jl'± needed in (9-25) and (9-26) are found from impedance results obtainable from the Maxwell modified curl relations (8-6) and (8-8) developed in Section 8-1. Thus, with no z components of the fields present, expanding (8-6) yields two algebraic relations

±yit =

-jWJ1£"f

(9-28) (9-29)

seen to provide the following intrinsic wave impedance relationship fill' the transverse field components of the TEM mode

it

+-~-

- .Yt' f

=

_ it

+-~-

.Yt' 'f

jWJ1 Y

=-

(9-30)

9-1 TEM MODE FlELDS BASED ON STATIC FlELDS

465

which qTEM = jW/1/Y denotes the intrinsic wave impedance ratio between the indicated transverse components of the electric and magnetic fields in the line cross section. The other modified curl relation (8-8) is here extended to the form that accounts Ibr a lossy dielectric in the tranymission line; that is, (8-8) is written in the form including the conduction term alf± as follows

a) ~±

-

If

W •

A

......

+

(9-31 )

=)WEIf-

in which the complex permittivity E defined in (3-103) appears. Expanding (9-31) the two algebraic equations (9-32) (9-33) which provide another intrinsic wave impedance expression for the TEM mode field components' (9-34) Equating (9-30) and (9-34) yields for the TEM mode

"I

==

a

"12

=

W

2

/lE, obtaining the propagation constant

+ jf3 = jw# = jm

J(E /l

j;)

(9-35)

This is identical with the result (3-88) applicable to unijlJrm plane walles in an unbounded region; it is also seen to be the lossy-region extension of (8-24), deduced in Section 8-2. It thus follows that the expressions (3-90a) and (3-109) for the attenuation constant a, as well as (3-90b) and (3-110) for the phase constant f3, are equally correct for the fields orthe TEM mode on a two-conductor transmission line with ideally perfect conductors. On a completely loss less line (dielectric also perfect), the special results follow:

a=O The phase

lip

f3

w~

(lossless line)

is found by use of the universal result

v

p

W

=--

f3

(9-36) 101) (9-37)

in which the phase constant f3 is once again given by the imaginary part of (9-35), yielding either (3-90b) or 110) of the analogous uniform plane wave problem. From the frequency dependence of f3, it is clear that the phase velocity (9-37) will, in general,

466


also have a frequency dependence, denoted by P(w). This gives rise to dispersion effects related to (but somewhat different from) those described for hollow waveguides in Section 8-6, and yielding the group velocity, analogous with (8-81) v =

(d P)-l dw

9

(9-38)

This, as suggested by Figure 8-16, is the speed of information transmission (an envelope velocity) associated with the group of Fourier frequency components that comprise a modulated carrier wave on the transmission line. On an ideal lossless line, with P of (9-36) inserted into (9-37) and (9-38), the phase velocity and also the group velocity of the TEM mode reduce to the frequency independent result (lossless line)

(9-39)

Next, substituting the y expression (9-35) into either of the ~TEM relations (9-34) or (9-30) yields the following expression for the intrinsic wave impedance associated with the TEM mode

r;;

A

1fTEM

(=ry)

= ~I

(9-40)

seen to be identical with (3-97), the intrinsic wave impedance ~ associated with uniform plane waves in a (lossy) unbounded region. It is therefore evident that the wave impedance expressions (3-99a) and (3-111) are also correct for the TEM mode fields of a two-conductor line having perfect conductors. On a lossless line, with E -+ E, the intrinsic wave impedance (9-40) simply becomes the pure real

l

1fTEM =

(lossless line)

(9-41 )

Finally, an alternative expressiop for the impedance relations (9-30) and (9-34), including vector information about B± and 3{'±, is obtained from the expansion of the modified curl relation (8-6), yielding here (9-42)

y

±-.-az JWf1,

~±

X

B

(9-43)


467

if (9-30) is used. The result (9-43) enables finding the magnetic fields ~± of the TEM mode in a two-conductor transmission line, once the electric fields If± are known. (9-43) shows, moreover, that those electric and magnetic field vectors are everywhere perpendicular to each other and to the longitudinal unit vector a z • An extension of Example 9-1 to the determination of the magnetic field in that coaxial line, as well as the accompanying line current, is exemplified in the following.

EXAMPLE 9·2. (a) Find the phasor magnetic fields :ii'± for the coaxial line o[ Example 9-1, and use their superposition to express the total phasor magnetic field H in the lines. (b) Obtain the real-time sinusoidal E(p, z, t) and H(p, z, t) for this line, assuming the dielectric to be lossl~ss. Show a flux sketch of only the positive z traveling wavs:' of these fields. (c) Use the Je± fields of (a) to deduce the phasor curr~nt amplitudes r!;, on the line. Use their superposition to obtain the total line current I(z) for this coaxial line. (d) Sketch the +z traveling-wave electric and magnetic fields in a line cross section, showing the related voltage and current senses.

(a) The solutions (9-21) inserted into (9-43) yield +a4> -

A

Yf tn

(9-44)

bp a

The total magnetic field is given by (9-26), a superposition of (9-44) after multiplication by e- yz and eYz (9-45a) =

atjl

V+I

V-I]

_m _ _ e-Yz _ _ _ m_

bp Yftn~

[ in which

~

A

A

eYz

(9-45b)

bp

YJtn a

and yare given by (9-40) and (9-35).

(b) For a losslessdielectric, (9-35) yields y = jw.JP~ = jp and (9-40) yields ~ The real-time forms of (9-22) and (9-45), by use of (2-74), become

E(p,z, t)

a Re P

ap

[V~ .!. tJ(w,-pz) + V';; tn

bp a

tn

I

bp

'I

~.

tJ(W'+Pzl]

a

V+ I I .] cos(wt-fh+4>+)+ cos (wt+Pz+4>-) bp P In [ In a a _m_

(9-46) V+ I H(p,z,t)=a4> __ m_-cos(Wf bp [ 11 tn a

Pz+4>+)

_V_ m_

I cos (Wf bP 'I tna

+ pz + 4>-)] (9-47)

V;;

assuming complex amplitudes of the form V;; = tJ4> ±. A sketch of the positive z traveling fields is shown.

468


--(z)

H; lines~~~~~~~=i~~~~~~:-::~~-~··~~:--:~~:-:-:·:-·;;~:~-:;; I

I

I

I

I

I

o

i3z =

I

I

Wave motion

I 7T

27T

EXAMPLE 9-2(b)

(c) Use is made of (9-25) to find the phasor current amplitudes given by (9-44)

~+

I;;; =

~

"(c.s)

~

+ • dt = ;H'-

I,!

from the

_ie±

fields

(a 4> - V;;;) ± i21t - - . a p d pfilna

+

v±

(9-48)

rn

fi b --In 2:n:

a

whence insertion into (9-27) yields the

+ z and

-

z traveling current waves of the

total line curTent

(9-49)

Normal E terminates on charges

Tangential H terminates on surface currents

~ ~~rface

~'-.(z)

Negative current: - I EXAMPLE 9-2(c)

(z)

9-2 CHARACTERISTIC IMPEDANCE

469

(d) In the accompanying sketch are shown the electric and magnetic fields (;?f only JlIe positive z traveling waves), along with their related voltage and current V+ and

r.

9·2 CHARACTERISTIC IMPEDANCE It is usually desirable to characterize TEM waves on a line in terms of their voltage and current waves rather than the electric and magnetic field quantities discussed in the foregoing sections. The advantages are evident from the fact that voltages and currents on a transmission line are readily measured scalar quantities at frequencies below I GHz or so, whereas the electric and magnetic fields must usually be inferred from such measurements. The comparison of the total line current (9-49) as evolved in Example 9-2, with the expression (9-27) for i(z), suggests writing the total line current i(z) in the equivalent form

(9-50)

In (9-50), the quantity be defined by

<:0 is called the characteristic impedance of the given line, seen to

7 = +

v±

(9-51 )

m

"-0 ----:;;;::;:-

1m

as is evident from the direct comparison of (9-50) with (9-27). Thus, the characteristic impedance <:0 denotes the ratio V~ Ii;;. of the voltage and current amplitudes associated with the +z traveling s~nm.?idal waves of voltage and current on the line, while further denoting the ratio - V;;' 11;;' relative to the - z traveling voltage and current waves considered separately. The direct comparison of (9-50) with the total line current ~expression (9-49) obtained in Example 9-2 shows that the characteristic impedance ..(0 of a coaxial line, for example, is given by ~

..(0 =

~

2n

b a

In-

(9-52)

assuming idealized perfect conductors. A summary of the TEM mode relationships developed in this section is given in Table 9-1.

470


TABLE 9-1 Summary Relations for TEM Waves

Magnetic fields arc obtained from j± solutions by usc of

Electric fields are found from quasi-static potentials [9-14] in which $± are solutions of Laplace's equation (9-15). The total electric field, including dence, is

E(u 1 ,

U2,

=

depen-

The total magnetic field, including dence, is

z)

H(u 1 ,

U2'

Z)

.it'+ (u l ,

j+

+ ;ir(ul>

[9-161

[9-26J Current wave complex amplitudes are

Voltage-wave complex amplitudes arc

7±m

[9-171

J. :ltd (u 1 , 'j"t'(c.s.)

proportional to

[9-25]

[9-27]

[9-181

i' and 7;'

. dt

making the total current

making the total voltage

With

depen-

V,=;,

(9-27) can be written

[9-50] in which the characteristic .impedance

Zo is

~

. dt

a

X Z

t(c.s)

i}

[9-51 ]

j± •

dt

EXAMPLE 9·3. A 10ss1ess coaxial line has the dimensions 2a = 0.1 ~n. and 2b 0.326 in., using a dielectric with Er = 2. Assume J = 20 MHz. (a) Find its Zo, fl, and 'rp' (b) If the disiectric had the small loss tangent (E"IE') (l.0002 at this fi'equency, determine how much Zo, fl, and 'Up change and how much attenuatioq is introduced. (a) The charactcristic impcpance is (flUnd using q = /lo/2Eo = 120nj:j2 = 266.5 n. Thus

requiring (9-'11) which yidds

.J

q 2n

In

b

266.5

a

2n

In

0.326

0.1

son

9-3 TRA:>ISMlSSION-LlNE PARAMETERS, PERFECT CONDUCTORS ASSUMED

From

471

and

f3

=

Ohj/lo(2Eol

=

211:(2 x W),fi 8 3 x 10 2.12

X

0.594 rad/m

10 8 m/sec

(b) The dielectrie with E"/E' = 0.0002 has from (9-40) or (3-111)

, 1] =

12011:/./2

[1

yielding, lImn

ei(1/2) arc tan

+

0.0002

~

266.5ei0.OOOl

~

266.5 Q

Zo with the same negligible angle

The constants ex and fJ are evaluated by use of the small loss approximations (3-11 and (3-113b); thus - [ I + 8:1 (E")2J fJ ~ W.//lE 7 = 0.594 [(0.0002)2J I + ---8-- ~ 0.594 rad/m

ex

wf (~)

0.~94 (0.0002) =

5.94 x 10

5

Np/m

The latter implies a wave decay to e 1 in a distance d ry;-l = (5.94 x 10- 5)- 1 16.8 km = 10.4 mL With fJ essentially unchanged, 1J p remains at 2.12 x 108 m/sec in the lossy dielectric.

9·3 TRANSMISSION·LlNE PARAMETERS, PERFECT CONDUCTORS ASSUMED Maxwell's equations can be used to derive a pair of coupled differential equations expressed in terms of the voltage and current on a transmission line carrying the dominant TEM mode. The development is carried out tirst in the real-time domain. The present discussion concerns a line composed of perfect conductors separated by a dielectric with parameters (E, fl, 0'). It is shown that at any Z cross section on the line, the voltage and current satisfy the differential equations

iW

(9-53)

oz of C

'e,

av at

gV

(9-54)

in which c, and g are distributed (per meter) inductance, capacitance, and conductance parameters to be defined. The equations are coupled in the sense that both dependent variables V(z, t) and I(z, t) appear in each.

472


A. Distributed Parameters and the Transmission Line Equations The transmission line equation (9-53) is derived using the Faraday law (3-78)

~ E . dt = - at-aLs B . ds =

-~m -V

t

(9-55)

at

applied to a thin, closed rectangle t of width i\z --+ dz in a typical doss section as shown in Figure 9-3(a), with a magnetic flux i\t/Jm passing through the surface bounded by t. The left side of (9-55), integrated about t in the 1-4-3-2-1 sense, yields voltages - V and V + (aV;oz) i\z over 2-1 and 4-3, with no contributions over 1-4 and 3-2 at the perfectly conducting walls. Thus the left side becomes

av ) = -av i\z

~ E . dt = - V + ( V + -az i\z t

(9-56)

oz

The right side of (9-55) involves the flux L1..t/Jm intercepted by t. It was shown from (9-6) that the time-varying magnetic field of the TEM mode, 'in a fixed cross section of the line, satisfies the same Maxwell equations and boundary conditions as does the static magnetic field produced by a direct current flowing in the line. Therefore the external inductance expression (5-88a) is applicable, becoming (9-57a) in which L1..Le denotes the static external inductance associated with any L1..z slice. By writing L1..Le = (leL1..z), implying

(9-57b)

I.

(z)

- (z)

Hl1z (b)

FIGURE 9-3. Geometric constructions relative to transmission-line equations. (a) Thin rectangle t of width L\z intercepting magnetic flux L\I/I",. (b) Closed surface 8 8 1 + 8 2 + 8 3 of width L\z intercepting currents i9/conductor and dielectric.

II I\!

'11

473

9-3 TRANSMISSION-LiNE PARAMETERS, PERFECT CONDUCTORS ASSUMED

(9-57a) becomes (9-57c) in which Ie denotes the static external inductance per unit length, or external distributed inductance parameter, of the line. From (9-5 7b) it is evident that the distributed external inductance parameter Ie = I1Le/I1Z is identical with that provided by the static methods described in Section 5-11; thus, write (9-57b) as I = I1Le

(9-57d)

e

For example, Ie of a coaxial line is given by (2) of Example 5-13. With (9-56) and (9-57c) substituted into the Faraday law (9-55), one obtains

av az

a at

--(11)

(9-58)

e

The parameter Ie is a constant in a rigid line having a dielectrie with a eonstant fl, so (9-58) becomes

av az

01

-I e

(9-59)

at

which is (9-53), that which was to have been proved. Similarly, (9-54) is derived from the current continuity relation (3-82a)

J, J' ds = _ oq

r.5

at

[3-82a]

applied to a dosed surface 8 of width I1z -> dz in the same cross section, as shown in Figure 9-3(b). The conductor at the assumed positive polarity is chosen for the constructioIl, where the positive 1 sense is taken to be z-directed. The right side of (3-82a) involves a surface charge increment I1q deposited at any t on the peripheral 8 3 shown, in view of the boundary condition (3-45). As seen from (9-5) and (9-3), the timevarying electric field of the TE M mode, in a fixed eross seetion of the line, satisfies the same Maxwell equations and boundary conditions as the static electric field between those conductors. The definition (4-47) of static capacitance can therefore be used to rdate I1q (0 the instantaneous voltage V between the conductors as follows I1q = (I1C) V

with I1C dcnoting the static capacitance ofthc I1z slice. Putting I1C

c

I1C I1z F/m

(9-60a) (cl1z), or

(9-60b)

474


(9-60a) becomes

llq

= (c

llz) V

(9-60c)

in which e denotes the static capacitance per unit length, or distributed capacitance parameter of the uniform line. From (9-60b) it is seen that the distributed capacitance parameter c is also the static capacitance per unit length Cil as discussed in Chapter 4; so write (9-60b) as (9-60d) For example, e for a parallel-wire line is given by CIt obtained from (4-107). The left side of (3-B2a) denotes the net current flux emergent from S at any t. The contributions through S1 and S2 in Figure 9-3(b) yield the net amount

I

+J+

01

01 oz llz

llz

(9-61 )

An additional current increment llJ leaves the peripheral surface S3 and enters the region between the conductors, assuming the dielectric has a conductivity (J. From (4-119), llJ is proportional to V, obeying

111 = (llG) V

(9-62a)

in which llG denotes the eonductance of the llz slice. By putting llG = (g llz), implying

llG g= llzU/m

(9-62b)

III = (g llz) V

(9-62c)

(9-62a) becomes

in which g defined by (9-62b) denotes the conductance per unit length, or distributed eonductance parameter of the line. It is evident from (4-121), from which llG = ((J/E) llC, that g is not an independent quantity; it is related to the distributed capacitance parameter 3 c on making use of (3-IOB)

g

(J(

Elf) U 1m

(J

= -E e = -WE we = -E' we

(9-63)

Inserting (9-60c), (9-61), and (9-62c) into (3-B2a) yields

01 0 ozllz+ (gllz) V = -ot (ellz)V 3The last forms for g in (9-63) involve the frequency

ill

and so apply to tho; time-harmonic case only.

9-3 TRANSMISSION-LINE PARAMETERS, PERFECT CONDUCTORS ASSUMED

475

reducing to the differential equation

oJ OZ

0

= --(cV) -gV

ot

If the parameter c is not a function of time, the latter becomes

oJ

c

oz

oV -gV ot

(9-64)

or just (9-54). B. Line Constants ,)" Zo in Terms of Distributed Parameters Many dielectric materials used in transmission lines have parameters 11, E, and (J that may be functions of the sinusoidal frequency ill of the fields, as seen from Table 3-3. From this point of view, the time-harmonic forms of the transmission-line equations may be of greater interest than the real-time forms (9-53) and (9-54). Thus, if into the latter .

V(z, t) is replaced with v(z)ei wt l(z, t) is replaced with l(z)ei wt

(9-65)

one obtains the time-harmonic transmission-line equations

~

dV

- = -Jill I J

dz

e

(9-66a)

(9-66b) These are also written

dl

-=

dz

-if

(9-67a)

-yV~

(9-67b)

on taking i and y to mean

y = g + jwc t5/m

Series-distributed impedance

(9-68)

Shunt-distributed admittance

(9-69)

476


With the substitution of (9-63), Y is written in terms of the dielectric loss tangent as follows

(9-70) For example, using a dielectric with a loss tangent of 0.00] yieldsy = (0.001 + j)wc, or very nearly jwc. The wave solutions for V(z) and i(z) of the transmission-line equat~ms (9-67) have been supplied by (9-20a) and (9-49). They yield expressions for y and <:'0 in terms of parameters and y as ftJllows. into (9-53) and (9-54)

z

If:

.

~+

-

V(z, t) is replaced with V;;;e1 w t+ yz ~+

.

-

I(z, t) is replaced with I;;;e1 wt + yZ

(9-71)

one obtains purely algebraic results

+y~ = -jwIJ;:; = -zl,,~

+yI;:; But from (9-50), the ratio

= - (g

± 17;:;/1;;

is

(9-72)

+ jwc) V;:; = -yV;:;

Zo, obtaining from

(9-73)

(9-72) and (9-73)

(9-74)

The last equality enables expressing y in terms of the distributed parameters (9-68) and (9-69)

y == a +j/3 =

vzy

= ~(j(t)le)(g

(9-75a)

+ jwc)

m-

1

(9-75b)

also written

y

(9-75c)

y can also be expressed in terms of the dielectric parameters as

(9-75d)

9-3 TRANSMISSION-LINE PARAMETERS, PERFECT CONDUCTORS ASSUMED

477

obtained by substituting E = E(l + (Jf.jWE) £i-om (3-103) into (9-35). Ifg = ((J/E)C of (9-63) is combined with (9-75c) and (9-75d), one obtains the special result

(9-76)

The use of (9-63) and (9-76) permits finding g and Ie of a line once c, for example, has been obtained, presuming the constants /l, E, and (J (or the loss tangent) of the dielectric are known. Inserting (9-75a) back into (9-74) obtains also in terms of the distributed parameters

Zo

(9-77a) (9-77b)

Putting (9-63) and (9-76) into (9-77b), one finds that ductors and dielectric losses can be expressed

-

A

<0 = '1

Ie /l

A

Zo of a line with perfect con-

E

= '1-

(9-77c)

C

If the line is completely idealized by assuming' no dielectric losses, (9-7 5c) reduces to y =

jwji;: = jW.ji;., implying f3

=

wji;: = w.jJlE

Lossless

(9-78)

Then (9-77 c) yields the pure real characteristic impedance <0

= 1}(O) ~ = 1}(O) ~ = Lossless /l c~-;;

&

(9-79)

in which '1{O) = ~ is the intrinsic wave impedance associated with the lossless dielectric. Equations (9-78) and (9-79) are useful for short transmission lines used at high frequencies, for which neglecting the small losses may not entail serious errors. Note that (9-77c), applied to the special cases of coaxial and parallel-wire lines with perfect conductors, produces the f()llowing results on making use of the static capacitances (4-51) and (4-107) A

1}

2n

P

b

lj{t'l-

a

Coaxial line, perfect conductors

(9-80a)

478


800

200

700 f-

~

t

--

600 f - f -

V

_._-

V

V

f-l--

300

V

200 100

1/

/

¥

L

-

f-

I

,--If.-

Y

vI/Coaxial

t----t

1--'

50 -1--

iIt-f--

!--I

t---

2 34 6

10

20 4060100 200 !!:. or l!. R a

FIGURE 9-4. Characteristic impedance

~

01

,/ ....-

v

Vc~__

_ .. f- _.---f- f -

r-- I- V kt'oaxial

fL+- 1--t--V I---- 1-- +- +-

l-"

l·- f - f··-

--I-- !-t--

1,.-/

V

f---

i/-

.."

I

---

II

--

f-t·-

wire

f--fI--

..

V

~

,--

-f-- +-

i--j--

1·_·- /-- VParaliel

f-

/

~400 ~

/

rl

-V'

--

1--

I--- ..-- f-i--

-7

I-- i - - - j - f-j-- ._- r-

150 f-

Parallel wire / 500

V

f--l--

1-

1.21.4

or lossless coaxial and

ff-l-- -

1.8 2 2.4 !!:. or l!. Ii

1·-

t--

i -t----+--

-

3

4

5

a

parallel-wire lines.

and

Parallel-wire line, perfect conductors

(9-80b)

If dielectric losses are neglected, the wave impedance becomes fj ....., 11(0) = 120n1JE~ with the reasonable assumption ofa nonmagnetic dielectric. Then (9-30a) and (9-30b) yield the real results

Z'o Z'o

=

60 b ~tn Er

120

a

[h J(h)2 -R J

tn. -+ R

1

Lossless coaxial line

(9-80c)

Lossless parallel-wire line

(9-80d)

and are graphed in Figure 9-4. They are usefLl1 approximations for transmission lines at high frequencies, lor which impedive efiects due to the field penetration into the conductors, described in Section 9-6 under the topic of skin effect, are neglected.

EXAMPLE 9·4. A coaxial line has perfect conductors but not necessarily a lossless dielectric. (a) Adapt the static capacitance of the Ene to find its parameter c. (b) Find g. (e) Use (9-57c) to derive Ie, the distributed (external) inductance parameter.

9-4 CIRCUIT MODEL OF A LINE WITH PERFECT CONDUCTORS

):

The static given by f:.Cjf:.z, whence from (9-60d) c=

This ratio

2nE

479 also

(9-81a)

b

Cn~

a

(Ii) The distributed conductance parameter is given by

g=

(J E

2n(J

(9-81b)

c=--

Cn

b

a

Thc ShUlit distributed admittance (9-70) is therefore

y

g

EO) We =7 (En) 2nE + j w --b

+ jwc = (-;; + j

(9-82)

Cna

(c) The defining relation for the distributed external inductance parameter Ie is

(9-83) in which f:.t/lm is the flux intercepted by a thin rectangle as in Figure 9-3(a). Thus f:.t/lm in.tercepted by C is B· ds in which B = ,uH, with (9-44) providing the solutions :Ye± Ic)r the coaxial line. Thus (9-83) in complex form becomes, as f:.z --> dz

Is

but

± V;'/I;!;

=

Zo from (9-51), so canceling dz obtains (9-84)

a result also directly obtainable by use of (9-76). Then from (9-68)

A · I ·=)W , u p[n~ l!

Z =)W

e

2n

a

(9-85)

*9-4 CIRCUIT MODEL OF A LINE WITH PERFECT CONDUCTORS

The transmission-line equations (9-53) and (9-54) can be used to establish a circuit model ofa two-conductor line. Such a model employing lumped circuit elements L, C, and R exhibits the same voltage and current characteristics as the line being modeled.

480


--7.J.-

~======================~l=,,=========/~)============~~~T=w=o=-c=o=n=du=c=to=r='~in~e' b)

L/2

L

L

L

" I

r'

' L L

CGCGCG:

:

_...l

L

c = cLlz

G

L/2

= g!:J.z

FIGURE 9-5. Circuit model of a Ene with perfect conductors and dielectric losses.

It assumes the configuration as in Figure 9-5, depicting the model of some length increment Llz of the line to consist ofa series inductive element L = 'eLlz and shunt capacitance and conductance elements C = c Llz and G = g Llz, in whieh c, and g are the line distributed parameters. As many of such sections as are needed to model a line length d are cascaded as shown. This procedure permits scaling a transmission line of any length into a circuit model suitable for tests within the confines of a laboratory. Transmission-line network analyzers, useful in the performance prediction of power line and telephone systems, can be built according to such a modeling technique. Another application is to pulse-forming circuits, using the wave delay and reflection properties of a transmission line. To show that the model of Figure 9-5 essentially obeys the transmission-line equations (9-53) and (9-54), note that 1 flowing through the series inductive element L = Ie Llz produces an incremental voltage drop given by Ll V = - a/at[le Llzl], in which the minus sign accounts for the polarity relative to the assumed positive sense of 1. This is written

'e,

(9-86) if Ie is not time-dependent. The finite difference form of (9-53) is (9-86), one of the desired results. Similarly, the current increment diverted through the shunt elements C = c Llz and G=gLlz in Figure 9-5 is Lll= -a/at[(cLlz)(V+LlV)] - (gLlz)(V+LlV). Assuming c a constant and neglecting the Ll V terms yields

c I';

I. ,

I!II ,'!

I

the finite-difference counterpart of (9-54).

av -gV at

(9-87)

9-5 WAVE EQUATIONS FOR A LINE WITH PERFECT CONDUCTORS

481

*9·5 WAVE EQUATIONS FOR A LINE WITH PERFECT CONDUCTORS The voltage and current on a transmission line obey wave equations analogous to those derived in Chapter 8 for the fields E and H ofTM and TE modes. In the present case of TEM modes, manipulating the transmission-line equations (9-53) and (9-54) leads to wave equations in terms of Vor 1. Thus, if (9-53) is differentiated with respect to z and (9-54·) substituted into the result, a wave equation in terms of V is obtained, whereas reversing the procedure yields a wave equation in terms of 1. These become

(9-88a)

(9-88b)

The third term of each equation is attributable to dielectric losses, for setting g reduces them to

0

(9-89a) Wave equations for lossless line

(9-89b)

The complex time-harmonic forms of the wave equations (9-88) are obtained by the usual substitution of the exponential functions (9-65), yielding

(9-90a)

o

(9-90b)

also written

d2 V dz 2 d2 f dz 1

zyV = 0 zyI = 0

(9-90c)

(9-90d)

482


z

if = j(J)le and y = g + j(J)c according to (9-68) and (9-69). Solving (9-90c) and (9-90d) produces the wave solutions (9-91 a) (9-91b) with the propagation constant y given by (9-92) These results are consistent, as expected, with the solutions (9-18) and (9-27) for V(z) and I(z), and with (9-75) for y, all obtained previously via a rather different route.

9·6 TRANSMISSION·LlNE PARAMETERS, CONDUCTOR·IMPEDANCE INCLUDED4\,,: In previous sections, the transmISSIon line carrying the TEM mode was discussed, assuming possible losses within the dielectric but with the idealization of perfect conductors (O'c ~ CD). In the real world, conductors with a high (though finite) conductivity O'c are used in the fabrication oflines, introducing two new problems into the study of this mode I. The distortion of the electric field in the dielectric from a true transverse (TEM) condition, brought about by the presence of an E z component at the conductor walls. . 2. The current penetration into the conductor interiors, given the name skin effect. The longitudinal current I(z, t) in the conductors gives rise to a z-directed electric fIeld component along the conductor wall whenever its conductivity O'c is finite, in view of the relation (3-7), J O'eE. The presence of such a longitudinal component at the dielectric conductor interface would appear to contradict the assumption (9-1) defining the TEM mode for the line. With a sufficiently high conductivity of the conductors, however, an almost negligible E z component is developed at the walls, compared with the transverse component present there. This condition is shown in Figure 9-6(a), using a coaxial line for illustrative purposes. A slight curvature of the electric flux between the conductors is produced by E z . The current penetration into the conductors of a line complicates the derivation of suitable transmission-line equations resembling (9-67a) and (9-67b). Infinitely conductive lines possess only surface currents on the conductor walls, but with a finite conductivity, current conditions like those of Figure 9-6(b) are obtained, with a relatively small penetration occurring at high frequencies, while a uniform current density prevails over the cross section under dc conditions. An exact solution entails satisfying the boundary conditions fbI' the fields, that is, matching the normal and tangential components of the fields at the interfaces separating the dielectric region from the two conductors. Some of these solutions, derived for particular line geometries, are discussed in the following. 41f you wish to omit the development of internal resistance and indu!'tance parameters in this section, to conserve time, just refer to the results (9-103) and (9~105) for y and <:0 and proceed to Section 9-7.

9-6 TRANSMISSION-LINE PARAMETERS, CONDUCTOR-IMPEDANCE INCLUDED

483

Almost-TEM electric field lines

(a)

,~~~~~~~

Region 2 Region 3 ., Region 1

Low frequency (w (0-=)

0)

Medium frequency (medium 0)

High frequency (small 0)

(b)

FIGURE 9-6. Eficcts of finite conductivity in a coaxial line. (a) Infiuencc of the E z field on the distortion of the TEM mode (shown exaggerated). (b) Illustrating current density variations at various frequencies.

A. General Line Equations and Distributed Parameters In d.eriving the transmission-line equation (9-59) or its time-harmonic form (9-67a), dV/dz = -zI, it was seen that the Faraday law (3-78) yielded only an external inductance contribution, = jwle . No resistive term was obtained because of the assumption of perfect conductors, allowing no tangential E component along the short sides of the rectangle, For physical conductors with a finite conductivity, the timevarying tangential H-field at the conduct2r wall, required by (3-71) to be conti quo us into the interior, generates an associated E component propagating along with H into the conductors as suggested by Figure 9-6(b), A current density field J accompanies this electromagnetic wave, in view of (3-7). An illustration of this process was found in the propagation of wall-loss currents into the metal boundaries of a hollow waveguide, as in Figure 8-18, In the following development it is shown that the form of transmission-line equations (9-67) are also applicable to a line whose conductors have finite conductivity, the series-distributed parameter Z acquiring two interr}at imfJeda!,:ce contributions Zil and Zi2, attributable to the tangential components Ezl and Ez2 developed along the conductor walls, To show this, Faraday's law (3-78), E· dt = -dt/lm/dt, is applied again to a thin rectangle t constructed as in Figure 9-3(a) in~a cross section of the line. The finite conductivity produces additional voltage terms L1V1

z

't

484


Conductor 2

Hflux

E flux Conductor

(z)

1

(a)

tY' : Poynti ng vector

(b)

(c)

FIGURE 9-7. Relative to internal conductor-impedance em~cts in Iim's. (a) The coaxial line, showing dosed line t (1234) and voltages along its edges. (b) Parallel-wire linc, with rectangle t. (e) An isolated conductor, with axial symmetry.

and dV2 at the;: segments as depicted in Figures 9-7(a) and (b). Along the edge 4-1, using time-harmonic quantities, (9-93) is obtailled fl'om the integral of -E . dt along d;:. Continu~us into the conductor surface is E z, where it is proportional to the current density ]z through (3-7), while Jz is in turn proportional to the current I carried in the conductor, related to the H field

by (9-94)

I',J

One can therefore write (9-93) as

I I

(9-95)


485

in which the proportionality constant Zil is called the internal distributed (a surface impedance) parameter, in view of its units (ohm per meter). A similar argument concerns the voltage along the edge 2-3, so define internal impedance parameters for each conductor by (9-96)

The integral (3-78) taken about the four sides of t in Figure 9-7(a) or (b) thus obtains

~ + (~V +

V

af!) ~ ~ llz + (Zi1 llz)! + (zizllz)!

The right side of (3-78), moreover, in time-har~onic form, becomes -jw(lellz)J, based on the external magnetic flux lll/tm (lellz)! linking t as already considered for Figure 9-3(a). Thus (3-78) becomes

which, after canceling llz, yields the differential equation

(9-97)

The total series distributed impedance Z is therefore (9-98a) the sum of the internal parameters (9-96) plus the reactance of the external inductance parameter Ie defined by (9-57b). In the discussion to follow, it develops that Zii and Zi2 consist of resistive and inductive parts, to permit writing (9-98a)

= (ril =

r

+ jwlil ) + (ri2 + j w1i2) + jw1e

+ jwl !lIm

(9-98b)

in which r and I denote the total series distributed resistance and inductance parameters of the line.

486


The othcr transmission-linc equation, obtainable from the continuity relation (3-82a), maintains the samc forms (9-64) or (9-67b) as for the line with perlect conductors

(9-99)

in which thc shunt-distributed parameter y dcnotes

Y=

g

+ Jwe

= (

EN 7 +.i)

(9-l00)

we U/m

with g and e dcfined as usual by (9-60b) and (9-63), or by (9-70).

B. General Line Constants ,)" Distributed Parameters

Zo and

The simultaneous manipulation of (9-97) and (9-99) further lcads to wavc equations cornparable to (9-90) obtained for thc perfect conductor case, becoming

(9-101a)

zyl

0

(9-10lb)

Their solutions are evidcntly (9-102a) (9-102b) if y denotes the complex propagation constant

y

== Ct + J/3 =

JIJ = .j(r + Jwl)(g + Jwe) m

1

(9-103a)

~+ ± . and V;;; and 1m are the usual complex amphtudes of the forward and backward waves. If the line parameters r are small compared with Jwl and Jwe rcspectively, + gJlj-;;) + a useful simplification of (9-103a) can be shown to yield y= (i)

(rFc/i


487

Jw-Ji~, obtaining the attenuation and phase constants

rJ.

rA gA

'C::: -

- 2

-

I

+-

- Np/m

2

e

p ~ wJlc rad/m

(9-I03b)

Low-loss line

r« wi

(9-103c)

g« we

The f(xm of (9-103b) is seen to separate the attenuative effects into contributions due to the series and shunt loss-parameters rand g. ~ ~ That (9-1 02b) can be written in terms of V~ and V,~ according to (9-104)

is demonstrated by inserting the solution (9-102a) back into (9-97), yielding

Tbis becomes (9-104) on inserting (9-1 03a), provided the characteristic impedance is written

Zo

V,;; ~+

~

=

J' z

(9-105a)

r;:+Jwl Q

(9-105b)

ZO=±J~=

j

~g+Jwc

The circuit model illustrated in Figure 9-8 can be shown to he valid for the transmission line having both dielectric and conductor losses. Applying the method of Section 9-5, finite difference versions of the transmission-line equations (9-97) and (9-99)

(z)

Qr:=U}HV -

R- ,,6.z

L= 16.z

'

\C=g6,z

FIGURE 9-8. Circuit model of a line with dielectric (shnnt) and condnctor (series) losses.

488


hold for the model. A comparison with Figure 9-4 reveals that the effect of conductor losses is to insert the series resistance element r ~z into each section of the model, while the series inductance element I ~z must include internal inductance contributions. The details leading to analytical expressions for the distributed parameters and y of particular uniform transmission lines are considered in Appendix B. In part A, the so-called "skin effect" in a round wire due to its finite conductivity, leading to a surface impedance interpretation of the internal distributed impedance parameter r + jwl, is taken up in detail. This leads, in part B, to the determination of the distributed parameters of the parallel-wire line as functions of frequency. An extension to the coaxial line is taken up in part C. Examples are included.

z

z=

9· 7 WAVES OF ARBITRARY SHAPE ON LOSSLESS LINES Although a good share of this chapter has involved the complex time-harmonic relationships between sinusoidal line voltage and current waves and their TEM fields, the detailed application of these phasor results to two conductive transmissions lines fed by generators operating in the sinusoidal stead)! state will be delayed until Chapter 10. In this section, attention is focused primarily on the voltage and current-wave solutions of the lossless-line wave equations (9-89) in the real-time domain. A generator of an arbitrary voltage waveshape Vg(t) is assumed applied to the input of a lossless line, and the voltage and current responses, V(z, t) and I(z, t), will be examined at any location z on the line. It is to be shown that whatever voltage disturbance V(O, t) is applied by a generator source to the lossless-line input (at Z = 0), will subsequently appear as a voltage V(z, t) of identical waveshape at the location z further down the line, except it will be delayed in time by the amount z/v seconds, a time delay determined by the intervening line distance and the wave velocity v. Two methods for relating nonsinusoidal voltage and current waves on a lossless line are discussed.

t. Direct solution 'If the wave equation. A method for determining the general form of the voltage and current waves launched on a lossless line by the generator of an arbitrary voltage waveshape uses a direct attack on the time-domain wave equation (9-89a) or (9-89b). Consider (9-89a):

82 V

I 82 V - -2 - = 0 v 8P

[9-89a]

in which lee in view of (9-39) and (9-76), is replaced with I wherein the phase velocity vp of (9-39) is here replaced with a wave speed, v, shown to be identical with Vp on a lossless line. I t is to be demonstrated that the general form of the time-domain voltage solution V(z, t) of (9-89a) consists of the linear superposition of arbitrary incident and reflected voltage-wave functions of (t ± z/v) as follows (9-106) In (9-106), V+(t z/v), the forward z traveling voltage wave, denotes anyJunction whatsoever of the variables t, Z in the combination (t - z/v), the waveshape of which is determined by the generated voltage source attached to the line

9-7 WAVES OF ARBITRARY SHAPE ON LOSSLESS LINES

489

input. 5 V- (I + is the reflected voltage wave function of (t + determined by the nature of the load termination, to be discussed later. To show that the voltage function V+ (t - z/v) is a solution of the wave equation (9-89a), let the variable t - z/v be denoted by u. Since au/at = 1 and au/az = - l/v, observe that

av+

av+ au ---

az

au az

{l2V+

a2 v+

{lz2

TuZ

1 av+ ----v au (9-107a)

Similarly, it is shown that (9-I07b)

On substituting (9-107) back into the wave equation (9-89a), an identity is obtained, thus verifying that the arbitrary function V+ (t - z/1)) is a solution. A similar proof verifies that the arbitrary reflected-wave function V- (t + z/t}) is also a solu tion of (9-89a), thereby proving the correctness of the general voltagewave solution (9-106). The real-time voltage-wave function V+ (t - z/1)) is easy to interpret physically. At Z = 0 (the line input), the wave has the function form V+(t). Further down the line, at any arbitrary Z location, the wave becomes functionally V+ (t - z/t}), of precisely the same waveshape as V+ (t) at the line input, except for being delayed (or shifted) in time by z/1) seconds. This confirms the remark made at the start of this section, that "whatever voltage disturbance is applied by the arbitrary voltage source to the loss less line input will subsequently appear at the location Z further down the line, except it will be delayed in time by z/1) seconds." This conclusion is not strictly correct, however, when the line has losses, for then the wave equatio~gffi,~ingthls~voltage wave behavior typically acquires a loss term like that seen in (9-88a), for example, to alter the solution. The general form of the time-domain current wave solution I(z, t) in a lossless line is of the same form as the voltage solution (9-106), or

I(z, t)

(9-108)

in view of the wave equations (9-89a) and (9-89b) being identical in form. The simple relationship between the incident and reflected current wave functions in (9-108) and their voltage wave counterparts in (9-106) is discussed next. 5 An important function of the space-time variable (t z/v) encountered earlier is the sinusoidal travelingwave functiou, A cos (WI - (h), also written A cos w[t - (/3/w)zl = A cos w(t z/u) (see Figure 2-11). Depending on one's observation point z along the z-axis, this function will replicate the wave A cos wt occurring at z 0, except it will be delayed by z/v seconds at the new z. Thus, z/v denotes a time-shift, or delay, a characteristic of this class of wave functions of t - z/v.

'i 490


2. Fourier method. Suppose that a voltage generator of voltage Vg(t) of arbitrary waveshape and with some series internal resistance R o' is connected to the input terminals z 0 of a lossless two-conductor line, developing the input voltage V+(O, t) there. An example of the consequence of this is shown in Figure 9-9(a), showing a generated triangular wave applied to the line input, launching onto the line the positive z traveling triangular wave shown. From the wave solution (9-106) obtained in part (I) by the direct method, this positive z traveling wave is precisely V+ (t - z/v). At the line input (z = 0), if the triangular voltage V+ (t) there were represented by its Fourier series of sinusoidal harmonics, these would proceed down the line as individual odd-harmonic sinusoidal traveling waves given by A1 cos w(t - z/v), A3 cos 3w(t - z/v), As cos 5w(t - z/v) , and so OIl. Each sinusoid travels with exactly the same phase velocity VI = V3 = Vs = v in the wave equation (9-89a) and given by (9-39) and (9-76)

v=v

=--=-p

JJ;.

.ji;

[9-39]

confirming the conclusion of part I, that no waveform distortion of V+ (t can occur on this lossless line, in view of this precisely maintained phase relation-

(1I)

(b)

(e)

FIGURE 9.9. Features of nonsinusoidal wave propagation on loss and lossy lines. (a) Typical recurrent nonsinusoidal wave, synthesized using Fourier series of sinnsoids. (b) Singly occurring wave; may be synthesized using Fourier integral. (e) Singly occurring wave on a lossy line, dispersion of Fourier components produces waveform distortion.

491


ship among all the Fourier sinusoids,making up this voltage wave as it travels down the line, The positive z traveling current wave 1+ (t - z/v) in (9-108) that accompanies any arbitrary voltage traveling-wave function V+ (t - z/v) on the lossless line, is also of interest. This positive z traveling-current wave can be shown to be of precisely the same waveshape as the voltage wave, being given by

(9-109a) in which Ro Jij;: denotes the pure real characteristic resistance of this lossless line, obtained from (9-79) (or from (9-105b), if r = 0 and g = 0). The simple proportionality (9-1 09a)Js easily proved by inspection, on noting that an identical relationship, I~ = V~/Ro, obtained from (9-51), connects the amplitudes of all corresponding sinusoidal harmonic terms in the Fourier expansion of V+ (t - z/v) and the related expansion obtained for the current wave 1+ (t - z/v). Extending this argument to negative z traveling waves of voltage and current of arbitrary waveshape on the line, in the event of wave reflections from the load, a ratio similar to (9-109a) applies

(9-109b)

Having accounted for both positive z (incident) and negative z traveling (reflected) waves of voltage and current on this lossless line, one can now write (9-109) for the general voltage-wave solution

(9-11Oa)

in addition to the general current-wave solution (9-108), which becomes

(9-110b)

On a line with losses, the phase velocities of the harmonic Fourier terms vary with fl-equency, in view of the phase constant f3 given by the imaginary part of (9-103a). Waveform distortion is expected on a TEM line with losses, becoming more severe with longer line lengths because of the increase in the phase disparity among the Fourier terms. This phenomenon is depicted in Figure 9-9(c).

492


Rg

Vg(tJ

'---V-(-\l,,{D

o

z={ (a)

(b)

FIGURE 9-10. A lossless line fed by a nonsinusoidal generator. (a) Lossless line terminated arbitrarily. (b) Typical load terminations.

Consider a class of problems typified in Figure 9-1O(a). The source Vg(t) of arbitrary waveshape (i.e., a pulse, a ramp function, etc.) has an internal resistance R g . The details are given in the following in terms of (A) the input conditions resulting in positive z traveling waves of voltage and current and (B) the reflected waves obtained from the load conditions.

A. The Line Input Conditions and the Forward-Propagated Waves On applying Vg(t) to the system of Figure 9-1O(a), the. voltage and current waves

consist initially ofonly the forward-traveling terms V+ and r of (9-110a) and (9-1 lOb ), from the physical fact that, with active sources only at the generator end, backward waves V- and 1- cannot appear until the incident waves V+ and 1+ reach the load (and then only if a load mismatch exists there). So before reflectigIls appear, (9-1 lOa) and (9- J lOb) are written . .-~

(9-11la)

t< I(z, t) = 1+

(t

t (9-111 b)

The analytical form of V+ (t z/v) depends on the input voltage V(O, t) developed at A-A, found by writing (9-111) in a form valid at the input z

°

V(O, t)

(9-112a)

z=

°

and 0< t <

t v

(9-112b)

493


but Kirchhoff's voltage law around the generator circuit and across A-A yields

Vg(t)

= Ri(O, t) + V(O, t)

(9-113)

so combining (9-113) with (9-112) yields the input voltage and current at Z V(O, t)

= v+ (t

-~) = Vg(t)---'--v Rg + Ro

=

°

(9-1l4a)

(9-114b) (9-114) suggests the equivalent input circuit in Figure 9-11(a). Thus V+(t-O/v) developed at A-A sees the characteristic resistance Ro of the line, to permit finding O/v) by use of (9-114b). the input current 1(0, t) == The forward waves V+ (t z/v) and (t - z/v) launched on the line are a direct consequence of (9-114) appearing at A-A; that is, they are simply (9-114) delayed by the retardation-time z/v. For example, if the applied voltage V(O, t) = V+ (t - O/v) obtained from (9-114a) were a steady voltage obtained by switching a battery onto the line at t = 0, the traveling wave V+ (t - z/v) would be a wavefront moving down the line with the velocity v = (JlE) 1/2, spreading the constant voltage distribution on the entire line after a time lapse of t/v seconds. For an arbitrary applied voltage V(O, t) like that of Figure 9-11 (b), the resulting wave V+ (t - z/v) is a consequence of the voltage V(O, t) appearing at A-A and moving down the line with the velocity 'I), illustrated at successive instants 0, t 1, t2 , t3 in the figure.

ru -

r

Rg

Vg(t) ~(z)

Rg

n(O,t) v,

"'O-)V(o."

~ rv=~+(t-ZIV)

t3:---~lL-Lr----'----z= (a)

°

~ (z)

(b)

FIGURE 9-11. Nonsinusoidal input phenomena for a lossless line. (a) Equivalent input circuit for determining forward waves. (b) Launching of forward voltage wave produced by V(O, t).

494

TEM WAVES ON TWO-CONDUCTOR TRANSMISSION LfNES

EXAMPLE 9·5 The generated voltage Vg(t) of I flsec duration and having the trapezoidal shape shown is applied at t = 0 to the lossless 50 Q air line. The generator has 20 Q resistance. Find the voltage and current developed at A-A, and the forward waves established after t = 0, up to the moment that they reach the load. Sketch the results. At A-A, by use of (9-114) and the input circuit of (b) 50

V It\

;>7-· V9(t)

+ 50

20

g\ }

(1 )

J(O, t)

(2)

The resul ting waves are (1) and (2) delayed by V(z, t) = V+

C

:) c =

,

~V (t 9

=

-If

z C

t:

in the air line), yielding

)

(3)

~)

V+ (

I(z

sec (with

t) = _-"-t_ ' -

(4)

50

The latter are sketched at typical instants in (e).

h

20

10

°

:-s

(Ro=50Q,v=c) (Air)

Vg (t)

I

Q---'--'-----,

(t)

V; (t) g

(a)

'I'

I

20 _ -1>10 ""

V(O, t) ! 7 5 V; g

=}

""j)'-

° 0.5

R

_

To load

1 fJ, sec

Vg (t)

OA] 201L I(O,t)

R g = 20Q

g(t)

01(0,

. V A _.

0

(b)

20 Q

V; (t) g

t)

~

(Ro, v)

A t \ _ _ _ _ _ _ _ _ _ _ _ _ _. A

_

,

(t) 1 fJ,sec

yeo, t) == (t-o/v)

Load

V(O, t)

,

z

::t

.+-----;---:----------,--- __ (z) /If·z t

r"

= 0.5 fJ,sec __ ___ l_ t = 1 jlsec --------

t=!-------c

(z)

z =t (c)

EXAMPLE 9-5


495

B. Reflected Waves from Load Boundary Conditions Next are found the reflections produced when the incident waves arrive at the load. The general solutions (9-110a) and (9-llOb) incorporating both incident and reflected waves are required. At Z = t these are

~) +V- (t +~)

V(t, t)

t)

+(

I(t, t)

=V

(9-115a)

1/

IV

t - -;;

(9-l15b)

Ro The reflected waves in (9-1] 5) depend on the load, examples of which are suggested in Figure 9-10(b). The two relations (9-115) contain three unknowns, V(t, tl, I(t, tl, and V- (t + tlv) , since the forward wave V+ (t tlv) is presumed known from part l. A third equation is thus required at z = t. It can be established once the load configuration is assigned. For the representative loads shown in Figure 9-10(b), their Kirchhoff voltage relationships are

l(t, t)

j ~1I

VIt."

vet, t)

(9-1l6a)

RI(t, t)

I( t, t) )

v(t, t);~

T

Ie T

I(t, t)

cdV(t, t) dt

(9-116b)

R

vet, t) = RI(t, t)

+ L dI~; t)

(9-1l6c)

L

e I(t, t)

I( t. t)

vet, t)

~

+ C dV(t, t) dt

(9 -116d)

(R02. v) RL

V(t, t) I

I

z=t

=

R02

V(f, t)

RozI(r, t) (9 -116e)

496


Considered first is the simple though important case of a resistive load, depicted in Figure 9-12. At the load position z t, (9-115a), (9-115b), and (9-116a) must be satisfied, so with the incident waves V+ (t zlv,) and r (t arriving when t tlv, one may write for any t after that instant Vet, t) =

I(t, t) =

v+

(t - f) + v- (t + f)

v+

(t - ~)

v-

(9-117a)

(t + f) z

1/

Ro

Ro

t,

Vet, t) = RI(t, t)

t;;>:

t v

(9-117b) (9-117c)

Defining a real time domain rtiflection coefficient reflected to incident voltages

r

Vr(t, t) =

V+

at the load as the ratio of instantaneous

( +-;;t) t

(

t

(9-118)

t) v

permits writing (9-117a) and (9-117b) Vet, t)

= v+

(t - f) [I + r(t,

t)J

(9-119a)

[1 - r(t, t)]

(9-1 19b)

t)

V+ ( t-I(t, t) =

1/

Ro

From (9-117c), the ratio of the load voltage Vet, t) to the current I(t, t) is R, yielding from the ratio of (9-1 19a) to (9-119b) R-R 1 + ret) ' - () 1 - r(t)

V+(t -~)-'>oRg

Yg (t)

~

L

~V-(t+~)

(9-120)

I' 1

[(t, t)

r_(_R_O'_V_) _ _ _ _ _ _ _ _ _ V..:....(t..:....'....... t) c

~--- --~ z

z=O FIGURE 9-12, Resistive terminated line.

! z=t

~

R


497

Solving for 1(t) obtains

(9-121)

1(t)

in which the notation 1(t, I) is altered to read just 1(t). Equation (9-121) is useful for finding V-(t + tlv) at the resistive load whenever the incident wave V+(t tlv) is known. It is emphasized that the pure real1(t) in (9-121) is a consequence ofthc output exprf:ssion (9-116a) being purely algebraic. 1\ time domain reflection cpefIicientjs undefined'll)r reactive loads corresponding to (9~116b, c,.aI~d d), except in the asymptotic limits for which the derivative terms in the load differential equations become negligible. EXAMPLE 9.6. If the line of Example 9-5 is terminated in a short circuit (R = 0), what reflected wave is produced by the incident trapezoidal voltage wave? Sketch the results. From (9-121), qt) = -1. 'fhen from (9-118)

(I) holding for all t> tic after V+ (I zlc) first appears at the load. To obtain the desired V" (t + ,:/c) reflected to theleft, the reflected wave (I) must be delayed in time by (t - z)le (a time delay incurred by wave motion from the load back to any Z location toward the generator), yielding

t

Rg ~(O,t) V(t,t) = 0 A:>--'-,.L£~-----"'''"'''''~--r . _/ --A ~------------------~ I

U

v. + g-

I Z

;"t

1=0

o t= 1. (

L - -_ _ _ _ _

o

L -_ _ _ _ _ _ _

--

-."j..J~ . .•.•i-Ir-:~~T~ __ I

o I=l! 2 (

...

~~= ~----~ --

•

I,.....f:.:'-_ ..

:~t:E;:J-:"'-" "..-----~---+--V---(-I....J:~:~ _v~,,-~ l--'f~m - .. -.

EXAMPLE 9-6

(z)

..,.... __

,b Mfl 498


As a check, observe that (2) becomes V+ (t 2t/c) at the input z = 0, the reflection arriving there after a delay of 2t/c. These results, shown in the sketch, reveal the echo V- (t + z/c) as a mirrored replica of the incident V+, inverted by the effect of the minus sign in (2) (maintaining zero volts across the shorted load). The echo apparently originates from an image location z = 2t, due to the time-delay 2t/c needed for the reflection to reach to the input. The corresponding reflected current wave 1- (t + z/t) is obtained by substituting (2) into the second term of (9-1 lOb), yielding

V-

(t + ~)

=

0.02V+

Ro

(t + ~ _ c

2t) A c

(3)

In the foregoing example, a second echo (re-reflection) occurs when V- (t + z/v) reaches the input at z = O. It can be found by use of the arguments of that example applied to the generator circuit. With a generator internal resistance Rg R no Ol re-reflection occurs. In general, an internal resistance Rg provides the time-domain reflection coefficient at the line input (z = 0)

vt (t ~v) _ Rg -

Ro

0) - Rg + Ro

--~-.....;:-

V1- ( t

vi

(9-122)

+v

with denoting the forward wave re-reflected from the generator to the load. These processes repeat when the wave in its turn arrives at the load, causing a third echo V;:(t + t/1-», and so on. The total wave solution is the superposition of all waves obtained in this way.

vt

EXAMPLE 9-7. A lossless 50-Q eoaxial line 200 m long using dielectric with E, = 2.25 is terminated in 100 Q and fed from a l50-V dc source having Rg = 25 Q as shown. With the source switched on at t 0, Vg(t) 150u(t) V, a step function. Find the voltage and current waves on the line after t = O. The equivalent input circuit of (b) yields at A-A V(O, t)

J (0,

'I

V+

~

(t -~)

l50u(t) _5_:-- = 100u(t) V

50

r(, -~) ~ v.~: ~) ~

The incident waves are (1) and (2) delayed

''itl

A

(I)

(2)

z/v sec

~)V 1-'

t

(3)

t
(4) wherein

'U

=

(J.loEoE,) -li2

= c/-li: =

2 x 10 8 m/sec.


Rg = 25 n

~~~: \

o

Switch

<~"

Switch

______ (R_O_'V_)______

--Z-~

=n

B~Bj ~~"t J:

-=-150V

g

50n

:

I

z={

I I

z=O

(a)

(b)

25fl

R Vg = 150 V -=-

Ro=

(Ro = 50fl.)

G

>---------U R = 100 fl z=f (z) _ _ _ _ ..J _ _ _ _ _ _ _ _ .J __ +_

-

2f

~1). 33.3

3t

Image source _____r:.qC:,,":;~.?!."=_'=_~~_

r(o)

t

-21

-I

=_1.. 3

z=o

--..L..,,-- _____ L ____ - - - -

---.---.,

--

--

------------

~-~

Frornz - - - -

""-4( ---- ___ _

EXAMPLE 9-7. (a) Transmission-line system. (b) Equivalent input circuit. (c) Echo diagram.

499

500


After a delay of x 10- 8 = 1 psec, (3) and (9-121), 1(t) = t yielding the reflected voltage at z = t

arrive at the load. From

(5)

Equation (5) is delayed by (t - z)/v to produce the first reflection toward the generator

V-.1 . ( t+;z)

=

333 .u ( t- .t -t-u

v

z)

"33 ( t+; z

=J.U

~~)

(6)

incorporating the delay 2t/v = 2 psec. The accompanying wave diagram shows that (6) appears to originate from the image position Z 2t, arriving at the load after 1 psec and reaching the generator after 2 psec. When (6) reaches the generator, (9-122) yields 1(0) = (25 - 50)/(25 + 50) = -!, yielding from (6) at Z = 0 the second reflection

(t--;,-2t)

(7)

0, must be delayed

seconds as it moves toward the load

ILlu The latter, specified at

Z =

2t) v

(3)

The accompanying diagram shows how echo (8) originates from the image position Z =

-2t.

Extending the preceding details, (8) produces at the load a third echo, appearing to originate from Z = 4t. This process continues indefinitely, approaching the steady state 120 V (obtainable from the generator-load circuit in the absence of the line). The edw diagram shown catalogs these results. The total voltage (9-11Oa), given by the sum of all the positive z and negative Z traveling waves, thus becomes

V(z, t)

(I -;) +V- (t +;) [Vi C ;) + vi C-;) + ... J+ [Vi- C+ ~) + V2 (t + ;) + ... J = [IOOU(t -::...) - ll.lu(t _ z _ 2t) + 1.2u(t _::... ~)- ... J =

V+

=

v

/[)

v

V

In the preceding examples, resistive loads were assumed, permitting the use of the time-domain reflection coefficient (9-121). More generally, loads with capacitive or inductive elements as illustrated in Figure 9-10 (b) may prevail. Analyzing their effects requires satisfying (9-l1Sa) and (9-11Sb) in addition to the appropriate Kirchhoff

9-7 WAVES OF ARBITRARY SHAPE ON LOSSLESS LINES V+(t-~)·--'>--

l(t,t)

--E-V-(t+~)

RO

B 2V+ (/ -

Load V(t,t)

B

BI({,t)

U]

f)

I

-

Load

'B

V(t,l)

I

501

I I

I I

I

z =1

z =t

FIGURE 9-13. Equivalent load circuit correspondiug to an arriving voltage wave v+ (t - t/v).

relationship (9-116) (in general a differential or integro-differential equation). Eliminating the reflected voltage term V- (t + t /v) from the load relations (9-115) obtains

V(t, t)

= 2V+

(t -~) -

(9-123)

RoI(t, t)

seen to correspond to the load terminal Thevenin equivalent circuit, illustrated in Figure 9-13. On combining (9-123) with a load relation selected from (9-116), the rcsulting differential or integro-differential equation obtaincd is solved for the unknown load voltage or current, yielding the reflected wave. An example illustrates this procedure.

EXAMPLE 9·8. The 50-Q lossless line in (a) is terminated in a capacitor and fed from a 150-V de source, assumed switched on at t O. To eliminate reflections at the generator, it is in series with a resistance such that Rg = Ro = 50 Q. Find the waves on the system after t O. The input equivalent circuit of Figure 9-11 (a) obtains, at A-A, V(O, t) = - O/v.). Delayed it yields the forward wave

0< t<

f v

(Il

At the load position z = f, V(f, t) of (9-115a) consists of (I) plus an unknown reflected voltage V- (t + fjv). To find the latter, combine (9-123),

(2) applicable to the equivalent load cireuit of (b), with the current voltage relationship (9-1I6b) for the capacitive load

,dV(f, I) C - - - - = I(f t) dt ' (I) and (3) into

(3)

obtains the differential equation

f) dV(f, t) 150u ( t - - = RoC-q~ dt

+ V(C', t)

(4)

502


c

o (a)

(b)

(Ro.»)

C

L..-_.....Jy-------------9L..JT o

=t

z

-;0..

t= 0

------,-----,-'---1_ _ _ _ _JI V+(t z) t

LI

o t =}

-f I

(

t

__

_n _____ J --52

L

z

+;;)~

2t 75

n'_/_-___-___~-~~-_-J

I

-75):'-

75_

."

L _ _ _ _ _ _ _ _ _ _ _ _~L~ ,

'-

L---________________

0 2 v

h

V-It

0

t =~ t

_

75 -~

o t=1;

n

~~~, -:~~==~~==~

___ _

~~------------------------

0 t=

2!...v

v<

0

150

t_co 0 (c)

EXAMPLE 9-8. Figure 9-13. (c)

Transmission line {""ding a capacitor. (b) Equivalent load circuit f1'om and reflected voltage waves.

the solution of whieh is

Vet, t) = 150U(

(5)

the voltage aeross C. Equation (5) into (9-l15a) then yields the reflected voltage at <:

Vet,

t) - v+ (t

t

!..) ~v

t)

--. 11 ~I'

(6)

The reflection at any V-

( + z) t

~ 11

IS

= 75u

PROBLEMS

503

~) e -- [/ + (zlv) --(U!V)JIRoC

(7)

delayed by (I

(Z t+

~ V

2/)

- --V

150u t +

(

z 1}

The total voltage on the line is thus the superposition of (I) and

shown in (c).

REFERENCES JORDAN, E. C., and K. G. BALMAIN. E'lectromagnetic Waves and Radiating ,Systems, 2nd ed. Englewood Cliff" N.J.: Prentice-Hall, 1968. MAGID, L. M. Fields, },,'rtergy and Waves. New York: Wiley, 1972. RAMO, S., J. R. WHINNERY, and T. VAN DUZER. Fields and Waves in Communication Electronics, 2nd cd. New York: Wiley, 1984.

PROBLEMS

SECTION 9-1A 9-1. Using the replacements (9-11) and (9-12), show how the TEM mode expressions (9-7) through 10) are converted to their sinusoidal steady state limns the factors eiro' 'f yz canceled out).

through

(with

G\ 9-2.

A long parallel-wire linc in air carries the TEM mode and consists of ideally perfect conductors of radius R separated 2h as shown. Make usc of the static potential field solution given by and (4-97) in Chapter 1 to show that the time-harmonic scalar potential solutions arc given by

121)

in which Ii = from (4-103). [Hint: Express the static potential expression (4-96) in terms of the voltage between the two conductors by usif!:g q = CV and the capa~itance expression (4-107). Then employ the usual replacements of V~ illr the static V and ± ll1r the static
(x)

PROBLEM 9-2

504

o


9-3. Make use of (9-124) in Problem 9-2 to obtain the electric field in the region exterior to the parallel-wire line, showing that

~±

tff' (x,y)

V';;

= 2t

h

R

n

+ in which d= Jhz

_R2 from

+d

{[ x - d ~_ d)z + y2 -

ax

a{(X _{~2 +

y2 - (x

+

(x

x +d + d)z + yZ

J

~z + yzJ}

(9-125)

(4-103).

SECTION 9-1B 9-4. Expand the Maxwell modified curl relations (3-6) and (8-3) of Section 3-1 to obtain, assuming no z-field components present, the pair of results (9-23), (9-29) and (9-32), (9-33) relating the transverse electric and magnetic field components of the TEM mode. From these, deduce the intrinsic wave impedancc ratios (9-30) and (9-34) connecting the field pairs (if, .i?f) and (if, .il'f). 9-5. (a) From the equality of the results (9-30) and (9-34), deduce the propagation constaut y for the TEM mode, obtaining (9-35). Comment on its comparison to the propagation COllstant associated with uniform plane waves in an unbounded region. (b) Use the propagation constant y of part (a) in either (9-30) or (9-34) to obtain the expression for the intrinsic wave impedance 'hEM associated with the TEM mode of the uniform two-conductor transmission line, showing that (9-40) is the result. Comment on its comparison with the intrinsic wave impedance of uniform plane waves in an unbounded region. G 9-6. Employ (9-43) to show that the magnetic fields accompanying the elcctric field solutions (9-125) obtained in Problem 9-3 for the parallel-wire line become

~/±(x,y)

±;

, +d 21Jotn.~

{ax[-(~ + ~2+7 - (;~-d~T+y2J (9-126) x +d ~-+ d)2+ y2

with d =

oJ}

JIz2 - RZ.

9-7. Show that the line integral (9-17), with the electric field~solution (9-125) for the parallelwire line of Problem 9-3 inserted, yields the expected result (the sinusoidal amplitudes of

V;;

(x)

(a)

PROBLEM 9-7

\ \

PROBLEMS

505

(b)

PROBLEM 9-8

the voltage differences bctween the conductors, associatcd with thc forward and backward traveling voltage waves). To case the integration, the path t on the x-axis, between P 2 and P l as shown in the figure, is suggested.

9-8. Substitute the magnetic field (9-126) of the parallel-wire line into the cl.?sed-linc integral (9-25) to deduce the expression f()!' the traveling-wave current amplitudes F;' on this line in terms of V;'. Inserting the results into (9-27), show that the total phasor line current, in terms of the forward and backward traveling current waves, is expressed

v~

[v-) = 110

n

tn

h + d e-

(9-127)

Yz

R

wherein Ii .jT R2 from (4-103). [Note: The integration path should encompass one conductor, as noted in Figure 9-2(b). In this parallel-wire line problem, two suitable paths tl and t z arc shown in the figure given here. t z is suggested for case of integration; it is valid to assume no contribution to the integral about the semicircle at infinity.]

SECTION 9-2 9-9. From (9-1 in Problem 9-8, show that the characteristic impedance of an ideal parallel-wire line in air is given by

<:'0

110 1 h h + Ii - cosh - . - '" J20 t n - - -

n

R

R

(9-128)

with Ii = .ji1'. R2. Show that if three different air lines have the hlR ratios 2, 6, and 32, their characteristic impedances become about 160, 300, and 600 ohms, respectively. Sketch these lines in cross-sectional view, dimensioned appropriately. <=)

9-10. 1\ particular idealized (Iossless) coaxial line, as shown in Example 9-1, has an air dielectric and the dimensions a = I em, b = 3 em. It contains only positive z traveling waves, the electric field being given by

(I)

(the prop':).gation constant y of (9-35) reducing toj/3o). (a) Comparing (I) with (9-22a), identify in (I). Make use of (9-43) to show that the corresponding magnetic field can be the field

If:

; §.iii

506


expressed

(2) (b) Find lhe sinusoidal voltage amplitude V;:; associated with the electric field (1), making use of - S~~ 4+ . dt of (9-17) and denoting a suitable integration path t between P z and PI' sketched on a sectional view of the coaxia~line. (c) Determine the sinusoidal current amplitude I;:; related to the magnetic (2), using ft;H'+ • dt of (9-25). Show your closed integration path t on the same seetional view. Based on this result and that of part (b), what is the value of the characteristic impedance of this coaxial line? [Ails: (b) 1099 V (c) 16.65 A, 660]

9-11. Ths. coa~ialline of Problem 9-10 has its fields :E and H specified by (I) and (2). (a) Employ Re [E X H*] of (9-47a) to evaluate the time-average Poynting-vector power density as a function of p within the coaxial line. Sketch the vector fY'av, showing its vector sense at the typical point P(p) between the conductors. (b) Make use of the integral SS(e.s.) fY' av • ds over the line cross section to find the time-average power Pay flowing through any cross section of this lossless line. [Ails: (a) a z l325/pz W/m 2 (b) 9.15 kW]

t

SECTION 9-3 9-12. (a) Employ the defining relation (9-57c) to derive the distributed external inductance parameter Ie of a parallel-wire line with an air dielectric as diagrammed for Problem 9-2, showing that I

/10

e

It

+d

=-tn-1L R

(9-129)

Jil"-

if d = Ri. [Note: The integration for the magnetic flux increment /1t/J .. is simplified by using a rectangle bounded by the closed rectangular contour t as suggested in the figure.] (b) Compare the distributed external inductance parameter (9-129) with the static external inductance per length t deduced from (5-90) in Chapter 5, commenting on the approximation used in the latter. (c) Determine the distributed inductance parameters of three air-dielectric, parallel-wire lines with the It/R ratios of2, 6, and 32. Assuming a fixed-wire separation, comment on the effect on Ie of making the wires thinner.

G'l 9-13. (a) Use (9-60d) to adapt the static capacitance per length t of a parallel-wire air line, as found in Section 4-1 I of Chapter 4, to obtaining the distributed capacitance parameter of that line earrying TEM waves, showing that . 1LEo

C

=---'--It+d tJt--

R

PROBLEM 9-12

(9-130)

PROBLEMS

if d = .ji1 2~R2. (b) If this line were immersed in a region of conductivity tributed conductance parameter becomes

(j,

507

show that its dis-

n(j

(9-131)

g

(c) Find the c parameters of three air lines with the hiR ratios of 2, 6, and 32. Comment on the effect on c of making the wires thinner, assuming a fixed separation.

SECTION 9-3B 9-14. Prove the equality (9-76), based on an appropriate comparison of the expressions (9-75e) and (9-75d) for the propagation constant l' on an idealized TEM line with a lossy dielectric. 9-15. (a) Verify that the equality (9-76) is satisfied by the distributed constants (9-129) and (9-130) developed for the parallel-wire line in Problems 9-12 and 9-13. (b) Verify, on using the parallel-wire line distributed constants (9-129) and (9-130) in each of the three expressions for ,(0 in (9-79), that the result (9-128) is obtained in each instance. 9-16. A particular parallel-wirc air line has round wires 1 mm in diameter, separated 1.2 em center-to-center. Sketch a labeled cross-sectional view. Assuming ideal perfect conductors, use (9-129) and (9-130) to determine the distributed inductance and capacitance parameters. Make use of these in (9-79) to determine the characteristic impedance of this line. Cheek the lattcr by use of (9-80d). 9-17. (a) Employ (9-52) and (9-79) to determine the characteristic impedance and the distributed inductance and capacitance parameters of three ideal (los81es8) air-dielectric coaxial lines for which the b/a ratios are 3.5, 6, and 12. (b) Repeat assuming this time a lossless polyethylene dielectric. Compare the phase constant {J, the wavelength A, and the phase velocity 'up in an ideally lossless air-dielectric coaxial cable, carrying the TEM mode at the frequency f = 100 MHz, with the values obtained if the dielectric were polyethylene (Er = 2.26).

Q\ 9-18.

(,) 9-19. Three different, ideal coaxial lines are to have the characteristic impedances 50, 70, and 100 ohms. (a) Determine their bla ratios if the dielectric is air. With b = 1 cm, find the radius of the inner conuuctor (in mm) for each case. (b) Repeat (a), assuming the dielectric to be polyethylene (Er = 2.26). 9-20. Assuming only that an ideal (lossless) coaxial line has a polyethylene dielectric and a 50-ohm characteristic impedance, find its distributed parameters c, and g.

'e,

e" 9-21.

Shown in (a) of the figure is a printed-circuit configuration known as a microstrip trans.. mission line, a two-conductor line consisting of a thin conducting strip of width w separated from a conducting ground plane by a dielectric slab of thickness h. In figure (d) is depicted the low-frequency (quasi-static, pure TEM) view of the transverse electric field flux produced by an applied sinusoidal voltage-difference. This shows the relractive effect, at the air-dielectric interface, of electric flux emanating from the top side of the strip. This may be comparcd with the field distribution of (c) in the absence of the dielectric slab, and with (b), which ignores (incorrectly) the presence offield fringing ncar the strip edges. (a) Based on the parallel-plate result (4-52), neglecting field-fringing as in (b), show that the stripline distributed capacitance is roughly approximated by c ~ Ew/h. Employ (9-76) to show that the low-lrequency distributed external inductance parameter is then roughly approximated by Ie ~ Jiohjw, assuming a nonmagnetic dielectric. Use these results to deduce an approximate characteristic impedance expression for this microstrip configuration. (b) Compare the field maps of figures (b) and (d) (which assume equal voltage diflerenees between the conductors), arguing whether thc approximate static c deduced in part (a) is large or smaller than the "correce' value deducible from

508


(a)

(z)

(b)

PROBLEM 21

(e)

(el)

~~o

~

Zn

'''0 ~ p"am"~n d'~'''''i,ti, im. ~

, fidd pi", '"en'''ling hgun (d). I, 'he "ppm,im,,, dcd",.", in "'" H (,," h'g' '" 4 for wi,hthis b ll1icrostrip O.I25;n. linc. 3.13 mm and" 4, d",nninnh"'"ugh

m'aU? Why? I.e"ing approxinlations to c, Ie> "'Ih and

o

'-22. 10 Pcoblem 9-21 a 'n"gh 'naIY'i, of (h, dimibu"'d 'nd p,d"", of, mi,eo'('ip lin, (, giv'n. d,i, line h" in fidd. ""n(ained ''''"Y in 'b, did'uri, >O.,"oo, '" hcre: ""d 6'he bne ebac"'''i''i,' im",danee, yidd ;"g an be((" !h'n 2

'h,

B",,,,,

'I',

'ex"

Ere

= z( 1 Er +

I ).J +:2 (Er

-

(10 h)

I) I + __ _

"""'''y

1/2

lei

t;!

h ( 13 --

w

170

=

~

w)

+ 0.25--

lei

h

W

~. [ h +

~/h ~

(9-132)

for-- ~ J h

10

1.393

+

O.GG7

t ";; + (

1.444 )]

(9-133a)

-1

w /or-3 It

(9-133b)

in w bid, " i, 'h, ,d"i" p. nni((iviry "«h, di,"'tit, 'lob. (a) A mi«o",ip lin, "''' n did,,,", "ah wi'h " 4 =d 4. Sk''',h a «"W-."",in"", vi,w. Find ,b, dftt"" eel"iv, p"mit!ivi'y and d" lin, eham'''"''ie imp,d,,,ce. (b) A, 'he vinn,"id,1 "peeaUng heq,,"",,, .f I GH" whO( i, 'h, WaV'-ph ..,. lin 'h, lio, of part ('II Wha( iy 'h, ph,,, co"'!a,,!?

<,

p'"'

ve/~i'y

~

'd,~",",

'>'0. n dig", '''nlnMass: Artech '''''''''on, .ee 60.-6.5. K. C. G"P!a, " ,I. ("_w.Aickd Microwave Circuits.vnd, Dedham, House, 1981, pp.

D,n" ,I

509

PROBLEMS

The wavelength? Compare the wavelength obtained on this line with the value obtained if the dielectric were all air (no slab); if it were all dielectric material of Ey = 4.

9-23.

Repeat Problem 9-22, this time assuming wjlt = 0.5.

SECTION 9-4 9-24. Manipulate the transmission-line differential equations (9-53) and (9-54) (the-so-called "tclegr
Repeat Problem 9-24, tbis time obtaining (9-88b).

SECTION 9-5 9-26. If r « wi and g « wc, show that the general propagation constant expression (9-103a) reduces to the two results (9-103b) and (9-103c), employing the binomial approximation (1 + a) 1/2 ~ I + for a « 1. Show also for this case that (9- J05b) reduces to ~ (llc) 1/2.

Zo

APPENDIX B 9-27. Starting with (B-2.\) in Appendix B, verify the dc limiting results (B-22) (as w the internal distributed resistance and inductance parameters of a long round wire.

-->

0) for

9-28. Beginning with (B-24) in Appendix B, verify the high-frequency expressions (B-27) for the internal distributed resistance and inductance parameters of a long round wire.

o

9-29. For the parallel-wire telephone line of Example B-3 in Appendix B, show that its distributed parameters (assuming g 0) becom(' the following (a) Ati = 10 kHz, c = 5.10 nF/km, r = 7.46 lljkm, I = ~.~6 mH/km. (b) At 1= 100 kHz, c = 5.10 nF/km, r = 21.8 Q(km, 1= 2.21 mH/km. 9-30. Make use of (9-103b) to determine, lor the coaxial line described in Example B-4 of Appendix B, what percentages of the attenuation factor IX are attributed to conduetor wall losses and to dielectric losses. Comment.

e

9-31. (a) Use the giV(,Il answ('rs to Problem 9-29(a) to determine the values of the propagation constant, attenuation constant, phase constant, and characteristic impedance of that airdielectric parallel-wire line at 10 kHz. Are (9-103b) and (9-103c) useful for this case? (b) Find the values of the phase velocity and wavelength on this line at lO kHz. Comment on the comparison of this vp with the free-space phase velocity for uniform plane waves. 9-32.

Repeat l'rohlcm 9-31 this time for the sinusoidal frequency

I

= 100 kHz.

9-33. Evaluate the distributed constants r, I, c, and g of the coaxial cable of Example 8A described in Appendix B, but for the frequency lOO times as large (f 2 GHz), noting how r and the external inductance contribution increase with frequency. (b) Employ the results of to evaluate the attenuation and phase factors and the characteristic impedance of this line, as well as the phase vdoci ty and wavelength 011 this line. 9-34. Use (9-J03b) and results developed in Appendix B, Section B-3, to express the attenuation constant of a coaxi
SECTION 9-7 9-35. Denoting the variable (t + tive <; traveling-wave function V- (t + function whatsoever of the variable (t

+

_",JUSk

s. SLii £££

510 @

o


A 70-ohm coaxial cable, with losses and a polyethylene dielectric (Er 2.26), is of length t = 50 m. A single square-topped pulse, of 40 V amplitude and O.l flsec duration, is switched onto this line at t = 0 from a pulse source of JOO n internal resistance. (a) Wbat is the velocity of propagation of this pulse on the given line? Find the amplitudes of the forward traveling voltage and current waves initiated onto this line at = 0, sketching an appropriate equivalent input circuit for it as by Figure 9-11 (b) Express the positive traveling voltage and current waves V+ (t and f+ (t terms o/" the applied pulse V.(t), wilh appropriate delays properly indicated relative to Vg(t). Sketch the voltage waveforms devdop(;d on this line at successive instants, just as the wave leading edge is (I) at the line input, (2) halfway to the load, (3) at the load. Denote what those time instants arc. 9-36.

9-37. With the line of Problem 9-36 terminated in a short circuit, what is the time-domain reflection coefficient at the load? (a) Sketch and label the forward and backward voltage waves developed in this line (including any superpositions required), shown at typical instants (such as t = 0, tlv, I and 2t/u, for example.) (b) Repeat this time relative to the fc)rwalTi and backward curren! waves on this line.

9-38. A so-called time-domain rejlectometer (TDR) uses a step-voltage generator developing the voltage V(t) 2u(l) volt, in which u(t) denotes the Hcavisidf~ step function {defined by u(l) 0 lor 1 < 0, u(l) = I It)r t > 01. This generator has the internal resistance Rg 50 n, with its terminal voltage monitored by a last-rise-time calibrated oscilloscope (CRO). Sketch a block diagram, showing . (a) Let a section of 50-ohm air-dielectric coaxial cable, oCiength t = 2 m, be connected to the TDR terminals. With the cable output terminals shortcd, sketch a laheled waveform of the display observed on the CRO screen [showing Vet) versus tl, assuming the step voltage to be switched on at t = O. (b) Repeat (a), this lime with the cable output terminated in R = 100 n. @ 9-39. The voltage Vel) 10011(1) volts is applied, in series with a 500-ohm resistor, to a 200 m having the characteristic resistance Ro = 250 nand openlow-loss air line oflength t circuited at its load end. What are the time-domain reflection coefIicjents the load) and (at the line input)? (a) With Vet) switched on al I 0, find the line voltage V(z, t) as a function of time, during thc interval needed to produce three successive reflections from the load. (Give details, using an echo diagram to depict the remIts). (b) vVhat asymptotic value is approached by I) on this line, as t -> oo? (e) Repeat (a), this lime relative to thf~ line current, fez, t).

reO)

4

9-40. The transmission-line system of Example 9-B is now terminated in the pure inductance L. Carellllly determine the voltage waves V+ (t and V- (I + developed on the line after switching OIl the applied voltage at t = O. (Show the applicable "equivalent load circuit" required to determine V(C', t) at the load, and by usc of (9-11 thc reflection V-- (t + tlv) produced at thc load.) Sketch waveforms on the line at typical illStan ts (for example, as suggested by the wavct(mn diagrams of Example 9-B).

,

1

;~:

.~

Phasor Analysis of Reflective Transmission Lines

The introductory paragraphs of Chapter 9 cited applications of two-conductor transmission lines, emhracing the transmission or at the lower end of the frequency spectrum to transmission at frequencies of many megahertz. That chapter coven~d the determination of line parameters and propagation characteristics [rom the line geometry and materials, in addition to relating the electric and magnetic fields of the line to its voltage and current waves. This chapter continues with the analysis of such transmission lines when tenninated ill arbitrary load impedances. In engineering practice, a communication line used jt)r signal transmission is usually terminated in its characteristic impedance, unless the load value is fixed by the physical nature ortbe load (e.g., an antenna). Then it may be necessary to employ a load-matching scheme to adjust the input impedance of the combination of the value of the line-characteristic impedance. Power transmission lines, OIl the other hand, invariably operate under load-mismatch conditions, in view of the variable loading depending on power demand. At their low operating frequency (usually between 50 and 400 Hz), however, power lines are usually electrically short (l« A), so the analysis can often be simplified through lumped-element, equivalent circuit methods. These techniques are omitted from discussion here. This chapter begins with analytical methods f()r determining voltage, current, and line impedance conditions on a two-conductor transmission line l(~d from a sinusoidal source and terminated in an arbitrary load impedance. Use is made of the reflection coefllcient and line impedance technique, developed in Chapter 6 felf uni{()rm plane waves at normal incidence to plane inlerfaces. The logical application of the Smith chart f(lllows, wi th emphasis on both the impedance and admittance versions or lhe chart. There /clllows an analysis of standing waves of current and voltage on mismatched lines, making further use of the Smith chart. Analytical expressions for line

511

; ;

ii! !! ! f !!

512

PHASOR ANALYSIS OF REFLECTIVE TRANSM1SSION LINES

input impedance under arbitrary termination conditions are developed next. Impedance matching of a mismatched line by use of reactive stubs is considered.

10-1 VOLTAGE AND CURRENT CALCULATION ON LINES WITH REFLECTION In this section, a connection is established bctween the forward and backward wave amplitudes V;:; and V;;: .Jt is found that the reflected voltage amplitudc V;;: relalive to an incident amplitude depends on the disparity ~between the load value Zl. terminating the line and its chara£teristic impedance Zo, no reflection occurring on a line properly terminated with Zl. = Zo- In a manner analogous to the methods of Chapter 6 concerned with plane wave reflections in multilayer systems, the cascaded line system of Figure 10-1 (a) is analyzed by using reflection coefficien t and impedance concepts. A simple extension of these ideas permits analyzing transmission-line systems such as the branched arrangement of Figure 10-1 (b). From the developments of Sections 9-3 and 9-6, the total voltage and current on a line in phasor time-harmonic form are (9-102a) and (9-104)

V;:;

V;:;e- YZ + 17-me

V(~t:) i(::.)

=

V,~

17-m

e~Yz

yz

[9-102a]

eYz

[9-104]

The propagation constant and characteristic impedance are related to the line parameters by (9-103a) and (9-105) y(

=0:

+j{J) =

"!fP

=

Zo( =±if)= jj=

J(r +jml)(g r

g

+ jwl + jmc

+ jmc)

[9-103a] [9-105 J

A comparison of (9-102a) and (9-101) with the electric and magnetic fields (6-29) and (6-31) reveals the analogy of the waves of voltage and current on a transmission line with plane waves normally incident on multilayer systems as described in Section 6-6. Thus the cascaded line systeIlls shown in Figure 10-1 (a) can he analyzed by techniques already described in Chapter 6. Equations (9-102a) and (9-104), applicabJ.c to any line section of Figure 10-1, can be written in terms of a reflection coeliicient r(,z) as fc)llows (10-1 )

( 10-2) with r(z) defined in a manner analogous with (6-36)

(10-3)

10-1 VOLTAGE AND CURRENT CIRCULATION ON LINES WITH REFLECTION

513

Junction or interface

A

(b)

FIGURE to-I. Generators connected to loads by use oflincs with different and y values. (a) Cascad!~d system of transmission lines ofdijrer~nt y and Zo, connected between a generator and load ZL' (b) Generator feeding two loads through a branched system.

A transmission-line impedance, defined as the ratio of the line voltage (10-1) to the line current (10-2), is analo?;ous with (6-33)

(10-4)

Solving for ['(.~), one has conversely

( 10-5)

the analog of (6-39). The reflection coefficient at any other location z', in terms of [' (z), is obtained from (10-3) in the way used to obtain (6-40), yielding

['(z)e 2Y(Z'-Z)

(10-6)

514

PHASOR ANALYSIS OF REFLECTIVE TRANSMISSION LINES

Coaxial lines

Parallel-wire lines

Line 2

Line 1 Line 1 6

yi(z+)

,,

z FIGURE 10-2. Continuity of characteristic impedances.

z

V and f

at junctions s('parating lines with diHcrcnt

The item completing the analogy concerns the continuity of the impedance Z(,c) of (10-4) at the junction of different lines (e.g., in Figure 10-1 (a), the junctions A, B, and C). From the continuity of the voltage and the current to either side of the common plane between the lines, as in Figure 10-2, it is required that

(10-7)

the analog of (6-41). While frs>m (10-7) Z(,c) is...continuous at a junction of two lines, it is evident from (10-5) that 1(<:) is not, since <:'0 is different on the two sides of the in terface. Table 10-1 gives a summary of the foregoing relations, along with analogous relations of Chapter 6 for plane waves. The application of (10-1) through (10-7) to line problems is illustrated in examples to follow. In transmission-line systems it is noteworthy that, besides the dominant TEM mode, TM and TE modes can also exist on two-conductor lines. The latter modes, however, as for hollow waveguides treated in Chapter 8, are ordinarily highly attenuated below their cutoff frequencies, occurring for typical coaxial lines at the upper microwave frequencies and beyond. When a desired signal is dispatched in the dominant mode down a two-conductor line or a waveguide, a partial conversion of the signal power into higher-order modes will occur at discontinuities (sudden dimensional changes, sharp bends, etc.). Becanse of their high attenuation (evanescence), the higher modes vanish to negligible levels a short distance away. Accompanying the presence of the higher modes, however, is the development of an unwanted reflection in the dominant mode. For example, in joining two coaxial lines of dinerent radial dimensiems, the discontinuity at the junction may be shown to generate a reflection equivalent to that produced by a small capacitance C shunted across the lines at their junction, even if their characteristic impedances are the same. I At all frequencies except those approaching the microwave region, however, this dkct is ordinarily very small (C is of the order of a few picofarads). It is ignored in the present treatment.

1 For the application of higher modes to coaxial-line discontinuities, seeJ. R. Whinnery, and H. W.Jamieson. "Equivalent circuits for discontinuities in transmission lincs," Prot. I.R.E., 32, February 1944, p. 98; or R. N. Ghose, Microwave Cirruit Theory and Analysis. New York: McGraw·-Hill, 1963, Chapter I L


515

EXAMPLE 10-1. A transmitter, operated at 20 MHz and developing Vg = 100ei°' V with 50 n internal impedance, is connected to an antenna load through 6.33 m of the line described in Example B-4 of Appendix B. The antenna impedance at 20 MHz measures ZL = 36 + j20 n. (a) What are Zo, rJ., and Pof this line, and how long is}t in wavelengths? (b) Determine the input impedance of the line when terminated with ZL' (e) How mueh power is deli>:.ered to t~e line? (d) Compute the load current and time-average power absorbed by ZL' (e) I1~ZI, 50 n, what is the input impedailce and how much average power is delivered to ZL?

ZO

3 (a) From Example B-4, = 50 n, rJ. = 1.97 X 10- Np/m, p = 0.595 rad/m. With A = 10.55 m, { in wavelengths becomes {IA = 6.33/10.55 = 0.6.

(b)

Zin is obtained by first finding f

Thcn

f

at the load using (10-5)

at A-A, by usc of (10-6), becomes 2(0.00197)6.33 e - j4n(O.6)

=O.2765ejI1l9'O.975e-J432' ~ 0.212+jO.177

n il?

II'

Vg = 100e V

50 Q

j (0)

cio."'()

A::= ~

i~(t)

-

I'-z-=-O----------z-=-(

~- . _ - - t = 6.33 m

---I

(a)

=

.lin i

(0)

=

70.5 + j28fl

(b)

'"

Vg =100e

10" .

V

(c)

EXAMPLE 10-1

(2)

516


TABLE 10-1. Transmission-line Analog of Plane Wave Propagation in

Multilayered Regions

Multilayer regions with plane waves

A:

Region 1

Region 2

:J - - - - --- --

+

A

Hy

• -.....

--..

---

---

-

Region 3

Region 4 (terminal region)

--;.-.

--;0..

------- ----- ..... ---

----

(ii 4 , 1'4) To source

I

I I

~--

Interface A

I

I I

B

C

(z)

Total fields:

+

Ex(z) = E,;ie- YZ [l Hy(z)

1"(z)]

E,;i e-yz[l -

I\z)]

[6-29] [6-31]

with [6-30] Total transverse field impedance: Z(z)

== ~x(z)

=

Hy(z)

.ry 1 + 1"(z) 1 - 1"(z)

[6-32]

making 1"(z)

= ~(z)

- ~ + 1]

Z(z)

[6-33]

At another location z': 1"(z') = 1"(i)e 2Y (z, -

z)

[6-34]

Continuity of Z(z): Z(z-)

Z(z+)

[6-35]

Smith-chart use, normalizing (6-32): t(z)

== Z~z) TJ

=

+ i'(z) 1 - 1"(z)

1

[6-361


TABLE 10-1. continued

Cascaded transmission lines

Line 1

"--v-l~ Source or: generator

(201 ,1'1)

i I I

Line 2

Line 3

Line 4 or , a lumped : load

--;0..

~-~

I

(.2 02 ,1'2): (.2 03 ,1'3)

:

I I

' I

I

I

I

Junction A B C

Total voltage and current:

+

V';e- yz [1

V(z)

V';

~

I(z) = - , Zo

e-YZ[l

fez)]

,

r(z)]

-

[10-1] [10-2]

with

-

V,;

[10-3]

I'(z) == -~- e 2YZ

v.;i

Line impedance: Z(z) ==

~(z)

=

Z 1

l(z)

0

+

fez)

1 - f'(z)

[10-4]

making fez) =

~(z)

+

Z(z)

~o

[10-5]

Zo

At another location z':

fez')

=

f(z)e 2Y (Z'-Z)

[10-6]

Continuity of Z(z): Z(z-) = Z(z+)

[10-7]

Smith-chart use, normalizing (10-4): £(z) ==

Z~z) Zo

=

1 + fez) 1 - fez)

[10-8]

517

518


in which the I~lctor e = 0.975 is approximated as unity in what follows (line losses are ignored). into (10-4) then yields

75.8ei21.0' = 70.8

+ )27.1 n

~n is obtained from tbe equivalent input circuit of ------- =

(3)

(b)

O.813e- j12 . 6 ' A

(4)

Thus, the average input power becomes 1~*)-.1R,(7' 1~1A*)_1[> P P avjn --LR'(V~ 2 ( in in - 2 (. ". . . . in in in - 2 '"in in

=

1(70.H) (O.B!

=

23.4 W

(5 )

(d) I}y (I 7in 7(0) I,;;f°[I - f(O)L in which all quantities a~c known except 1:'. Solving f(H' it yields the l(lIward-travding currcnt-wave amplitude: O.813e - jI2,6'

7+ In

Then

7J"

1-0.212

(6)

)n.l77

written in terms ofJ,~ using (10-2), becomes

7L = itt) =

7:'e- jPt LI

- f(t))

=

I.Ok j216 l1

= 1.14e-j229.1" A

+ 0.103 -JO,257] (7)

The load average power is thell (8) agreeing with (e) With

cuit,

because of negligible line losses.

Z:L = ]:0 ~ 50!}, <'in = Zo = 50 n also. Then Ii'om the equivalent input cirlin =

VY/(,"(:g

+ (:o)

= I A,

yielding from figurc (e) (9)

From a well-known theorem of circuit theory, (9) represents the maximum power available from Vq •

In Example 10-1, the load current iL itt) was found irom t:.,he line inpu!. current by first solving, in (d), for the forwar2 currerlt wave amplitude 1:', whence I(l) "Yas obtained. A convenient way to find IE from lin, eliminating the need for finding 1:', is simply to fOrIn the ratiu of the load current to the input current 011 the line, by making use of (10-2) as follows

l:'e-Yt[l - T'(tl] i:';°[1 - T'(O)]--

(10-9a)

in which, from (10-6), T'(O) ~ t(l . (10-9a) enables finding the load current IL whenever the input current lin is known, or vice versa. Thus, in a comparison of the

10-]

519

VOLTAGE Al'\D CURRENT CIRCULATION ON UNES WITH REFLECTION

phasor input and output currents, the knowledge of the line output and input reflection coefficients, related by (10-6), plus the line input-output wave amplitude and phase changes associated with the e- yt factor, are all that arc needed. The current ratio (10-9a) can be generalized, if desired, by fcwming the ratio of the line current I(z), at any position z on the line, to the input (or output) current lin' The latter, by usc of (10-2) and (10-6) once more, obtains the result _~_

=

lin

l(z)

== e

~"",-

_

z

I - [(0)e 2Yz

'Y

1(0)

(10-9b)

""

I - flO)

This fc)rm is 1:seful, for example, in the graphing of both amplitude and phase of the line current l(z) as a function of z on the line. Additional results resembling (10-9) can be fC)fIned of various ratios of the desired line current or line voltage, as given by (10-1) and (10-2), to some known voltage or current on the line. An application of (10-9a) is found in the next example.

EXAMPlE 10-2. miles of the line of Example B-3 are connected between a generator (developing, at I kHz, Vg 20eiO" V with Z"g = 7000) and the 1000-0 load shown. (a) What are ,,~(» ct., and fJ and what is C in wavelengths? (Ii) Determine Z"in at A-A. (e) What is Pay into A-Ae (d) Determine iL and P av •L ' 'With the line terminated in Z"O and the ,gcm:rator adjusted It)r a conjugate match how much power is delivered to A-A and the load? 703e-)13.2' 0, ct. 0.0083 Npjmi, and fJ 0.035 radjmi. (a) From Example B-3, 179 mi, obtaining CjA = = 0.335. The latter yields ;, (b)

Zin

is {(JUnd by first obtaining I'" at the load from (10-5), yielding f(C) (ZL + Z"o) = 0.20ge j32 .4 ". With (10-6), f at A-A hemmes

flO)

['(t')e- Zyt = [,(/)e- 2at e -)2W (O.20ge)32.4")e'< 2(0.0083)60 e- j4n(0.335)

no

Z/Z = 700 n r~---

v~ ~ ;~,v

-

f= 60 ml

= 966 km--

flo =701,-'" ,_ Q

1000(1

2(0)

(b)

EXAMPLE 10-2

+ jO.0372

d~--~9Z' =

z=l

(a)

v~ = 20V

0'()678

----1

,.

00083+} 0035 ml- 1)

-

(1)

520


yielding the line input impedance, from (10-4)

~

«0) =

1 + f(O) frO)

<0 1 _

= 6l3.8e- j8 . 9 " =

606.4 - j95.2

n

(2)

(e) The line input power at A-A is found from the equivalent input circuit of figure (b): ~n = 1(0) = 20/(700 + 606.4 - j95.2) 0.01527ei4 . r A, yielding (3)

(d) To find ~_ = i(t) from the line input current, (1O-9a) yields, with itO) from part (c), and e- Yc e"xCe-jPt 0.608e"j120.6"

I = 1(t) =

~~L-yt

1(0) 1

L

1 - rtO)

= 0.OI527ei 4 . 2 " 0.823

122 0.608e - j120.6' l.068 - jO.0372

=

7 .18e - j106.6' mA

(4)

This yields the average load power (5)

(e) With a matched termination ZL

Zg

Z~

684 +j160.5

n,

Zo, the input impedance becomes Zo; also with 20/2(684) 0.01461 A. Then

=

~n

(6)

To find the load current on this reHectionless line evaluated in part (el), 7(t)

=

I(O)e- Yt

(f =

0), (l0-9a) yields, with

e- yt

= 0.01461 (0.608e - j120.6') = 8.SHe - j120.6· rnA

to obtain (7)

10-2 GRAPHICAL SOLUTIONS USING THE SMITH CHART A convenient way to solve transmission line problems Ii ke those of Examples 10-1 and 10-2 or those suggested by Figure 10-1 is through the use of th~ Smith chart. This convenient chart enables finding graphically the line impedance Z(z) at any location z on a transmission line from the known reflection coefficient there, or vice versa, providing graphical solutions to expressions (10-4) or (10-5). In addition, from a rotation about the <;hart, (10-6) is also solved graphically, to permit finding the reflec!ion coefficient r(z'), at any desired location z', from a known reflection coefficient r(z) elsewhere on the transmission line. The theoretical basis for the Smith chart is given in Appendix D. The reader unfamiliar with its theoretical development is advised at this point to stUll); Appendix D first, before proceeding with its applications to wave reflection and transmission

10-2 GRAPHICAL SOLUTIONS USING TIlE SMITH CHART

521

problems on transmission lines. Such applications are considered in the remainder of this section. To establish the desired normalized line impedance x(z) needed [or Smith chaIt applications, a division of expression (10-4) by the line characteristic impedance is required,

<0

1

+ r(z)

(10-10)

1 - r(z) The normalized expression (10-10) (or its inverse) is solved graphically by the Smith chart (see Appendix D); additionally, the translational expression (10-6), r(z')

= r(z)e 2Y(z'-Z)

[10-6]

is also solved graphical01, from an appropriate rotation about the chart as illustrated in the examples that follow. Although some accuracy is admittedly lost in such graphical solutions, the Smith chart is both a time saver and, with practice, a valuable tool capable of displaying many transmission-line solutions at a glance.

EXAMPLE 10-3. Rework (b) of Example 10-1, using the Smith chart to obtain the input impedance. ~ ~ The load impedance ZL = Z(t) is normalized using (10-10)

Z(t)

== -~~ =

Zo

36

+ j20

- - - - = 0.72

50

. + )0.40

(1)

a result entered into the chart at P in the figure. The rotation by 0.6A., the line It:ngth, toward the generator obtains the normalized input impedance £(0) = 1.41 + jO.56, shown at Q, Denormalizing obtains

+ jO.56)50 =

70.5

+ j23 = 75.3 e.i217 Q

the same as (3) in Exampl,:: 10-1. If desired, values of r at the output and input of the line can also be read at P and Q, obtaining

['(t) = O.28ei 112' agreeing with (1) and (2) obtained analytically in Example 10-1.

~(O)

1.41

.!.}, = 0.6

EXAMPLE 10-3

+j

= 0.56

522


~= 0.335

EXAM PLE 10-4

EXAMPLE 10-4. Rework (b) of Example 10-2, using the Smith chart to obtain line.

Zin of the lossy

Normalizing ZL == Z(t) by usc of (10-10) yields x(t) Z(t)/Zo 1000/703e- jl3.2- = 1.384 + jO.324, entered onto the chart at P in the figure. One obtains .£(0) from a phase rotation from P to Qplus a decrease in reflection coefficient amplitnde in accordance wi th (10-6) t(t)e 2(<>: + }[J)(O

t(O) =

t(t)e-2<>:t'e-J2[Jt' =

£)

t(t)e-2(O.0083 x 60)e-j2fJt'

0.369tCt)ej2pr

j2pt The phase rotation oft(t) by eneed not be evaluated, being obtainedtrom the rim scale on the chart, an amount 01'0.335 wavelengths clockwise (toward the generator) required by the line length. The real factor e' 2aC accounts for a decrease in the amplitude of r by e - 2aC 0.369 in rotating from P to Q, Then x(O) is read ofr the chart at Q, it is x(O) = 0.87 + jO.07. Denormalizing yields

in agreement with (2) of Example 1O-2(b).

EXAMPLE 10-5. An antenna with a measured impedance 72 + j40 nat 20 MHz is to be driven by a transmitter 27 fl: away. All that is available li)r this purpose are two coaxial lines with the characteristics Line I: Z01

=

Line 2: Z02

= 50 n, 12 =

70 n, t1

= 15 ft = 12 ft

4.57 m, air dielectric

= 3.66

m, dielectric

Er

=

2

Ignore losses for these relatively short lines, connected as dsricted in(:z). (al Ifxpress line lengths in terms of wavelength on each line. (b) With Vg = 100el° and Zg = 100 n, Ilse the Smith chart to find the impedance at A-A, and t.he average power delivered at

ZL'

(a) Line I is lossless, so by usc 01'(9-36) and (3-100), ).1 = 27[//30 = II = 4.57 m = (4.57/15),1.1 = 0.3052 1 , Similarly, for line 2, ,1.2 12 = 0.346,1.2'

=

15 m, whence 10.6 m, yielding

(b) The origins 0 1 and O2 arc located as. shown in (a). The normalized load impedance is x2(12 ) Z2(t2 )/Z02 (72 + j40)/50 = 1.44 + jO.B, which entered at A in figure (b) and rotated 0.346,1.2 toward the generator yidds X2(0) = 0.5 + jO.IB,

10-2 GRAPHICAL SOLUTIONS USING THE SMITH CHART Line 2

Line 1

(201

523

= 70~)

(202

= 50~) = 2)

(t r

(Air)

(a)

(cl

(b)

EXAMPLE 10-5

shown at B. From (10-7) it is the aetualline impedanee (not normalized) that is continuous at the junction, so .0enormalizing %2(0) ~obtains Zz(O) %2(OlZ02 = (0.5 + jO.13)50 = 25 +j9 = ,(1 (ttl. Normalizing ,(1 (tl ) yields

n

Zl (ttl 25 + j9 "'l(td =-~- = - - = 0.353 A

,(01

.

+ )0.123

(I)

70

entered at point C in figure (c). Rotating 0.305 Jo j yields the normalized input impedance 21 (0) = l.l - j1.07, whence at A-A

Zin = Zl(O) =i?l(O)ZOJ

= 73 -

)73.5

(1.1 - j1.07)70

= 107 .2e ~ j43.rn

From the equivalent input circuit one obtains O.513ei 224" A, yielding from figure (d)

(2)

l;n =

100/(173 - j73.5) =

(3) With both lines lossless, this is also P av •L delivered to the antenna.

524


i

£ chart

chart

(b)

(a)

nGURE 10-3. Reciprocal aspects of the Smith chart. (a) Reciprocal of rotation. (b) Normalized impedance and admittance charts.

x through

180"

In the cascaded system of Figure 10-1 (a), the line impedance appearing on the load side of any opened-up junction is the impedance seen by tbe line at the other side of the junction, evident from the continuity relation (10-]). If the junction at O2 in Example 10-5 were opened, for example, the impedance Z2(0) = 25 + j9 Q optained looking into line 2 is that seen by line 1 on closing the junction. Similarly, Z1 (0) = 78 - j73.5 Q into line I is seen by the generator when connected to those terminals. In a system with branched lines as in Figure 1O-l(b), the impedance seen by the line 1 at B-B, where line 2 and lipe 3 are pa~allel-connected, is just the parallel com bination of their input impedances Z2 (0) and ,(3 (0). To find the impedance of two elements and connected in pa'rallel, the expression

Z'l

applies. By use of their admittances

Z'2

1\

can be employed. If~an a,!tainable 1 graphic accuracy is adequate, the Smith chart is useful for finding Y == Z-I. This is possible through a property of the chart yielding the reciprocal of any complex number from its 180° rotation about the chart. 2 Thus, entering k = 1 + jl onto the chart and rotating it 180 as in Figure 1O-3(a) yields 11 = 0.5 - jO.5, the reciprocal of k. In the case of a complex number with a large or small magnitude compared with unity, however, an arbitrary normalization forcing its magnitude near unity improves the accuracy of this process. For example, Z = 150 + jlOO is very close to the point 00 on the Smith c!lart; thus its reciprocal falls near the diametrically opposite zero point. Normalizing Z through a division by 100 to obtain k = 1.5 + jl, however, yields from the chart its reciprocal 11 = x-I = 0.4§ - jO.3l, and denormalizing yields Y 0.0046 jO.Oml, the desired reciprocal of Z. The foregoing reciprocal property leads to another version of the Smith chart, the normalized admittance chart, shown in Figure lO-3(b) alongside the more usual 0

2This property can be proved by use of (10-31) in Section 10-4.

10-2 GRAPHICAL SOLUTIONS USING THE SMITH CHART

525

normalized impedance form. By simple relabeling the i and circles of the chart with f1 and 6, respectively, and rotating the chart 180°, one obtains the normalized ~ad mittance chart on which = f1 + j6 = £-1. Thns, a given reflection coeffIcient r(z) depicted at P yields £ = i + jx there on the £ chart, while simultaneously displaying the corresponding = f1 + j6 at the same point Oil the admittance chart. chart is especially useful in the analysis of systems involving parallel-connected components as depicted in Figure 10-1 (b); fe)r example in the application of stub sections of line to impedance matching taken up in Section 10-5.

y

y

They

EXAMPLE 10-6. Suppose the line and load described in Example 10-1 arc connected to the lines of Example 10-5, forming the branched system in (a). With the aid of a Smith admittance chart, determine the following. (a) Find the input admittance into line 2. (b) Repeat, but lor line 3. (c) Determine the admittance produced at B-B the parallelconnected lines. Find the line 1 input admittance. (d) What power is delivered into A-A?

Zg

100 n

ZL 0,305 (Z01

~1

= 70 ill

=

72 + )40 n

B

(a)

Zg

vg~~0-1----~I~I--------:~

ZL

= 36 + j20 ()

(b)

~ 1'1(0)

Vg~

or ZI(O)

(c)

(d)

(e)

EXAMPLE 10-6. (a) Branched line system. (b) Eqnivalence at B-B. (c) Equivalence at A-A. (d) i-chart on line 2, (e) On line 1.

526

PHASOR ANAIXSIS OF REFLECTIVE TRANSMISSION LINES

With the normalized on line 2 by = 1.44 + jO.30 = j29 1.647 e . I', its reciprocal yields ,o/-L XL I = 0.607e = 0.531 -jO.295 = ? The latter entered at PIon the admittance chart (d) and rotated 0.3461 2 (from 0.444 to 0.444 + 0.34·6 = 0.500 + 0.290, or to 0.290 on the rim yields at P z , 1.33 - jO.90, making

(b) Similarly, on line 3, £~ 0.72 + j0.40 = 0.824ei29 . 1', so if~ = (~) -I = 1.061 jO.590 = ? Entering this on the admittance chart and rotating 0.600 on the rim ,('ale yields (0) = 0.60 - jO.21, so the inpnt admittance into line 3 becomes

Y3 (0) = Y03 Y.1(O)

0.02(0.60 -JO.21) = 0.012 - jO.0042S

(2)

(e) As in figure (b), at B-B is seen the parallel combination oflinc 2 and 3 inputs, yielding th~re YBB Y2 (0) + Y3 (0) = 0.0486 - jO.OZ22.s:: Normalizing the latter using YO! = 70- 1 of line I yields ;;;I(t Il = YBB!}';)! = 3.40 - j1.55. Entered at Pion the if-chart of figure (e) and rotated by t l 0.3051 1 obtains ifl (0) = 0.30 +j0.47 = {I whence the admittance at A-A YI(O)

=

Yolifl(O)

=

70 1(0.30 +jO.47)

= 0.0043 + jO.0067 = O.0080ei 573

S

in which Gin = 0.0043 S and Bin = 0.0067 S denote the parallel-connected elements of the line admittance seen at A-A. The corresponding line input impedance seen by the generator at A-A is the reciprocal of (3), or '~I (0) = "'~AA = 125e- j57.3" 68 - j107 Q = Rin + jXin , the latter denoting the series-cqnnccted elements of the line seen at A-A. (d) Irom !hc iI1put~circuit of figure (e), the generator delivers into line 1 the current lin Vg/(Rq + ZAA) = 100/(100 + 68 - j1(7) 0.50ej3ZO A, to yield the average power input at A-A

10-3 STANDING WAVES ON TRANSMISSION LINES TllC ref!ected waves of voltage and current occurring at the mismatched termination «L ¥= of a line produce standing waves in a manner analogous to that process described fi:)r plane waves in Section 6-7. Voltage and current waves on a line with reflections are given by (10-1) and (10-2)

<0)

17+m e- YZ + 17-m eYZ = 17+m e - 1+ ['(z)J

V(z)

I(z)

=

j?+

___m.._

<0

e- yz

17-

~"'..e)'Z '7

""0

e

['(z)]

(10-11 )

( 10-12)

The real-time origins of standing waves established by f()rward- and backwardtraveling waves in a region is perhaps best visualized from the careful study ora diagram iike that of Figure 6-9. It is immaterial whether the waves depict electric and magnetic

10-3 STANDING WAVES ON TRANSMISSION LINES

~

M

527

~

r-:J, M r-:J, ~~b~~~ (Zo,jtJ)

(Zo,"()

o

z

=

-I

l

=

0

(al

FIGURE 10-4. Depicting {()rward and backward waves ofvoltag,· or current in real time (above) on a line of length t. Resulting sianding waves of voltage or current magnitudes (middle diagrams).

II fl

Smith-chart interpretations of the related quantities + (voltage magnitude) and (current magnitude) (below). (a) Line with losses. (b) A lossless line.

II ~ fl

flclds as in that figure, or voltage and current waves as apply here; the phenomenon of standing waves is the same, I t should be emphasized that the usage standing wave reiers to the changes in only the magnitude of the composite-wave oscillations with a change in z, Thus, as the incident and reflected waves move by in time, the observer sees the magnitude distribution "standing" in space with its characteristic undulations along z, as shown by the dashed-line standing-wave diagram at the bottom 0(' Figure 6-9, or as in the middle diagrams of Figure 10-4 (shown in the latter for both the lossy and lossless Ii ne The presence of i()rward and backward waves gives rise to standing waves of voltage and currcnt magnitudes as depicted in Figure lOA. In (a) is seen the effect of the factors e- az and eaz on the forward and backward vo!tilge or current waves with losses present..:, This pn~duces the standing-wave behavior shown as curves of the magnitudes \ V(.c) \ andlf(z)i plotted against distance, with the undulations becoming

528


smaller as the reflected wave diminishes with increasing distance from the load. The current standing wave dips where the voltage standing wave is maximal, a direct result of the minus sign in the reflected term of (10-12). On examining the magnitudes of (10-11) and (10-12)

(10-13)

(10-14 )

it is seen that their graphs are readily obtained with the aid of a Smith chart. Thus, with the r;.yaluation off at the load position by use of (10-5), the quantities 11 + fl and 11 appearing in (10-13) and (10-14) are obtained directly from the Smith chart as shown in Figure 10-4(a). Thus f is retarded in phase by ei 2fJt and diminished in length by according to (10-6) in going to ,::; t toward the generator. Then with 11 + fl and 11 fl multiplied l;y e -(LZ anel scaled by the factor V:; and Vr~ /,(0 according to (10-13) and (10-14), \V(z) and 11(z)1 f()llow cm:ves typified in Figure 1O-4(a). For a lossy line of sufficient length, the spiraling of r(z) toward the chart center reduces the undulations nearest the generator to practically zero, in view of the reduction of the reflected wave
q

I

(10-15) li(;:) 1

( 10-16)

The latter are analogous with (6-51) relative t~ the reflection of plane waves discussed in Chapter 6. In the absence of attenuation, r(zL varies ~nly in phase along the line in accordance with the 1(:!ssIess version of (10-6): r(z) = r(0)o:J 2 fJ z . This provides the familiar circular locns of r(z) on the Smith chart shown typically in Figure lO-4(b) and termed the SWR circle. From it one can find the volt!ige and currenl along the line, obtainable by use or (10-15) anc! (10-16), with + r(;:)1 and r(z)1 found graphically from the charts of Figure 1O-4(b). The analogy of this process with that of Figure 6-11 fc)r a region with plane waves is evident. In such lossles;; systems, the standing-wave maxima and minima occur 90° (or ,1./4) apart, with Vmax (or equivalently Emax) at the location of I min (or H min ), and vice versa. The standing-wave ratio (SWR) associated with the voltage and current magnitude diagrams of Figure 1O-4( b) is defined for a lossless line by

II

SWR:=S

IV(z)lmax Vmax IV(z)lmin := Vmin

II -

Imax

I min

(10-17a)

10-3 STANDING WAVES ON TRANSMISSION LINES

529

analogous with (6-50). From the Smith chart representations in Figure 10-4(b), Vmax and fmin on the line are seen to occur respectively at the locations of I + and I -Iq on the SWR circle. Thus (1O-17a) can be written in terms of the reflection coefficient magnitude as follows

ItI

S=

1+

jf('<;)1

I

Ir(z) I

(1O-17b)

~

having the inverse ~

lr(z)1 =

S- I

( 1O-17c)

S+ I

By arguments analogous with those used for plane waves, the SWR circle of a lossless line with reflections is centered on the Smith chart such that it passes through the SWR ,< point on the real axis of the chart, as depicted in Figure 6-10(c) for plane waves. Additional details concerning graphic interpretations of the forward and backward voltage and current waves can be developed from figures analogous with the electric and magnetic field diagrams shown in Figure 6-1 L

EXAMPLE 10.7. Find the SWR on the loss less line 2 in Example 10-5. Where afC Vmax and VOlin located) The magnilude of the reflection codficient obtained from thc Smith chart in Example 10-5 is Irj = 0.3b. The SWR using (IO-l7b) is therefore S

1+

ifl

1 + 0.3b

-If! = -j-O.-36

1.36 0.64

2.12

an answer also obtained horn the 1 value intercepted by the SWR circle along the positive real axis of the chart as shown in (a). Vmax occurs at N on the SWR circle, located d = 0.06}'2 toward the generator, or d = 0.06(10.b) = 0.63b m from the load as shown in (b). Vmin is at M, an additional quarter wave towards the generator as shown.

EXAMPLE 10-7

530


r

Slotted line se. ction

Movable voltage probe

c£J IV

.

Source:

i i ··

·l

Movable=ik...-'" probe! ....

..... --- •--- -.~

~ :

-..

-

I

,

k",

-,,~

L __________

~

_ _ _ _, __ _

,

~)

----~~~

°

:t: ~. . Y. fJ

~_· ::.t'L

___-",,-__ P (Proxy

0:

. load position) _ _ _ _ _ _ _ _ _~Short

Sectional view (a)

(b)

FIGURE 10-5. Slotted line and impedance measurements. A coaxial slotted-lines section and voltage (electric field) probe. (b) Determination an unknown impedance from standing-wave measurements.

At the higher frequencies, above 100 MHz or so, impedances can be inferred from standing-wave data obtained iI-om an instrument known as the slotted line, illustrated in Figure 10-5 (a). A slotted line may be a rigid section of air dielectric line having a precision slot milled lengthwise through the outer conductor to accept a movable voltage-sensing probe. The latter travels along an externally mounted carriage to permit measuring, usually by usc of a detector and amplifier system, the relative voltage anywhere along the slot. Position measurements are facilitated by use of an attached scale. The probe is permitted to penetrate only a short distance into the slot to minimize the distortion of the electric field being measured. When the slotted line is connected between a generator and a load as in Figure 10-5(h), the voltage standing wave developed within the slotted section is measured by the detector output. The impedance of the load can be inferre~ from measurements on the standing wave as follows. With an unknown load .(L attached to the slotted line, the SWR and the V min position are recorded_ The corresponding SWR circle is drawn on a Smith chart, with Vrnin (denoteg M) occurring at the intersection of the SWR circle and the negative real axis. If -(L is replaced with a short circuit in the load plane, the standing wave produced has nulls spaced by half wavelengths as shown in Figure 1O-5(hl- Each null location can be regarded as a proxy load position, a position where the load impedance is replicated when the line is once again terminated in (This property of a lossless line reproducing an impedance every half wavelength is evident from the Smith chart, since moving a half wavelength corresponds to a full rotation about the SWR circle.) Thus, if the proxy load position P were located a distance d II-om the Vmin location M as in Figure lO-5(h), the impedance at P would be obtained from the Smith chart by a rotation dl A on the SWR circle fi'om M to P. Denormalizing % there by use of the line .(0 obtains the unknown impedance at P, and hence .(L' .

ZL-

10-4 ANALYTICAL EXPRESSIONS FOR LINE IMPEDANCE

531

d A

(b)

(a)

EXAMPLE 10-8

EXAMPLE 'IO-S. An unknown impedance ZL is

to be measured at 500 MHz by use of a 50-Q slotted air line. Because of the location of ZL, it is cOIllC!ccted to the slotted line using an additionallcngth oflossless 50 Q cable as in (a). With'(L in place, the measured ~WR is 3.2, Vmin occurring at the scale position 19.4 cm along the slotte~ line. Replacing ,(1. with a short, a null is observed at the position 11.2 cm. Determine ,(L' Drawing the SWR = 3.2 circle on the Smith chart as in (b), Vmin is at M. The shift from Vmin to the proxy load position Pis d = 19.4 - 11.2 8.2 em toward the load, making d/2 = 8.2/60 = 0.137. Rotating by this amount to P yields XL = 0.65 - jO.93; denormalizing obtains

ZL = XLZO = (0.65 -)0.93)50 = 32 - j46 Q

1 0-4 ANALYTICAL EXPRESSIONS FOR LINE IMPEDANCE From previous sections, it was seen t~at the input impedance of a section of line is a function of the co~mplex termination ZL, the line length, and the line parameters that determine)' and Zoo One can consolidate these effects into a single expression for input impedance, if desired, noting first that the input impedance of the terminated line illustrated in Figure 10-4 is expressed by (10-4)

;; 1 + f(O)

(l0-18)

-<"0 -1-f(O)

The input reflection coefficient f(O) is transformed to its load value f(t) by usc of (10-6)

(10-19) in which f(t) is written in terms of the load value Z(t)

=

/ZL by

(10-5) (10-20)

'l~

532

[

The substitution of the latter and (10-19) into (10-18) obtains

1'1 PHASOR ANALYSIS OF REFLECTIVE TRANSMISSION LINES

(10-21a)

yt

Zin Zo ~~--~~~---'--;.:::----;-~-, n yt

(I 0-21 b)

On collecting like terms, (10-21 b) can also be written in terms of the hyperbolic cJ>!'ine and sine functions, obtaining sinh yt sinh yt

n

( 1O-2Ic)

if the definitions cosh yt

+ e- yt

== - - - -

(10-22)

2

are employed. Note that if each of the expressions (! 0-21) is examined fl1r tl.!e inPllt impedance obtained if the load impedance equals ,.(0, the expected result ,.(ill = ,.(0 obtains. EXAMPLE 10-9. Usc one of the expressions (10-21) to find the input impedance of the 60 mi ofline terminated in 1000 Q described in Example 10-2. Substituting into (1O-2Ia) the values of ct., (3, and t obtained from Example 10-2 yields

Zo, ZL'

which agrees with (2) of Example 10-2.

One can simplify (10-21) for the special case of a lossless line. With y = the pnre real characteristic impedance Zo, (10-21 c) reduces to

-

,.«(0)

_ ZL cos pt + j,.(o sin f3t

= Zo

,.(0 cos

."'. n pt + J,.(L sm pt

Lossless line

jp

and

(10-23)

10-4 ANALYTICAL EXPRESSIONS FOR LINE IMPEDANCE

533

noting from (10-22) that cosh (jpt)

sinh (jpt)

cos pt

=j

sin

1M

(10-24)

In impedance calculations for lossless coaxial or parallel-wire lines, particularly at high frequencies, one is reminded that the Zo expressions (9-80c) and (9-80d), graphed in Figure 9-4, are useful in lossless line expressions such as (10-23). Additional special cases of (10-21) can be generated lor particular line ~engths and loads. Of interest are the short-circuit and open-circuit load cases. If ZD = 0, the input impedance (10-21c) reduces to Zin,sc

=

Zo tanh yt Q

Short-circuit load

(10-25)

with the hyperbolic tangent function defined by tanh yt With the line lossless (y =j{J, Zin,sc

sinh yt

e ~ 2yt

= --- = cosh yt

~--~

(+e~2Yt

(10-26)

Zo pure real), (10-25) becomes

jZo tan Ilt Lossless line, short-circuit load

( 10-27)

since from (10-24) and (10-26), tanh (jpt) = j tan pt. Equation (10-27) sbows that tpe input impedance of a shorted loss less line is a pure induetive or capacitive reactance Zin,sc = jX L or - jXn taking on all the positive and negative values of the tangent function with varying line length. In Figure 1O-6(a) is shown a graph of (10-27), together with its Smith chart interpretation. Entering the S!:,llith chart at the shorted load value 0 (eorresponding to the reflection coeflicient [' 1 there), a rotation on the rim by tjJc provides the desired input reactance predicted by (10-27). Thus, a quarter wave lossless shorted line has an infinite input impedance. A shorted length of low-loss line is called a stub; if its length is variable through the use of telescoping conductors, it is an adjustable stub. Stubs are often used at high frequencies as the reactive elements in narrow band impedance-matching schemes, as described in the next section. 1\ section of transmission line with open load terminals (ZL -+ 00) has an input impedance ohtained from (1O-21c) Zill,OC

=

Zo cotb yt Q

Open-circuit load

(10-28)

wherein the hyperbolic cotangent function, coth yl, means (tanh yt) ~ 1. Equation (10-28) reduces, for a lossless line, to Zill,QC

= - jZo cot Pl

Lossless line, open-circuit load

(10-29)

The unlimited range of capacitive and inductive reactance values provided by an open-circuited stub is depicted similarly in Figure IO-6(b). Of additional interest is the ~nput impedance of a lossless, one-half or one-quarter wave line with an arbitrary load ZL' For the line one-half wave long (pt = J[ = 180°), (10-23) reduces to Zin

=

ZI_

Lossless line,

~ long

(10-30)

+111/ <

"1'[ 'Ii

'I

534


j4

j4

j3

j3

j2

j

jl

o -jl -j2

J

L7T I

I I I I

-j3 -j4

I

/

1~? /~

90· X 4

I

I

J

j2

/

jl

270· 3X

"4

fit

t

9~

0

/i-

-jl

I

X

2

fit

7f /

t

I

I I

~<-f-----

1I

-j4

I I I--t-I

270·

1~0·

If

-j2 -j3

I

J

"----

I

/

i I

~""*-O

O---~A ~-jZo 0 0--

___

~

SL= 0 (Short)

8

~

~ ====~~

00

Xin, oc

(a)

(b)

FIGURE 10-6< Graphs of the input impedance of short- and Below arc Smith chart interpretations. (a) Shoft-circuited stub.

This result is not unexpected, since from the Smith chart, impedances on the SWR circle of a lossless line repeal themselves every half wavelength along the line, corresponding to a fiill rotation around the chart. For a line one-quarter wavelength long (/3t n/2 = gOO), (10-23) becomes ,{

Lossless line,

~

4

long

(10-31 )

lOA ANALYTICAL EXPRESSIONS FOR LINE IMPEDANCE

535

Feed line

EXAMPLE 1O-1O

Tn view of (10-30), adding any integral number of half-wave sections of Insoless line to the input of a quarter wave line still yields (10-31), making the latter valid fix any line length totaling an odd number of quarter wavelengths. A lossless line obeying (10-31) is called a quarter wave transformer, a name arising from its use in matching a high or a low impedance load to a transmission line, from the insertion of a quarter wave section ofloss\ess line having a properly chqsen characteristicjmpedance. Thus, if a given load '(,L is to be li~::1 from a line with a Zo different {roll' ZL, a quarter wave transf()rmer connected to /(,L will have an input impedance Zin that matches the feed-line charactcristic impedance if the transformer section has the characteristic impedance, fri)ITl (10-31), given by (10-32)

In practice, a quarter wave transformer is us~d at high frequencies to connect a resis-

tive load to a lossless line (with a pure real Zo). Because the method depends on the transj()rmer scction being a quarter wave long, the degree of impedance match is necessarily frequency-dependent. The fi'equency bandwidth of a matching scheme is COIlveniently specified in terms of the frequency deviation, to either side of the design frequency, over which the SWR on the feed line departs from unity by not more than some specified amount; a limit of 1.5 or so is often an acceptable criterion. It can be shown that an increased bandwidth of the quarter wave matching scheme is realizable if the impedance transformation is made in two or more stages, with lransf(Jrmations made to intermediate resistive values. The limit of stepped systems such as this is the transmission line, made several wavelengths long, providing a slowly varying characteristic impedance starting at the Zo of the input line and tapering to the load resistance value. The result is an extremely broadband matching device. Details of the bandwidth analysis of relatively narrow band matching devices wch the quarter wave transformer and of stub-matching systems described in the next section are {(lUnd in a number of sources. 3

EXAMPLE 10-10. A dipole antenna having; a measured terminal impedance of 72 n at 150 MHz is driven from a parallel-wire lirH~ baving a 300 characteristic imped8l1cc. The ICed-line conductors arc spaced 2h = 0.75 in. Design a quarter wave section of parallelwire air line that will match the 72 load to the 300 line at this frequency.

n

n n

3For example, see H. J. Reich, P. F. Ordung;, H. L. Krallss, and J. G. Skalnik. A1icrowave Theory and Techniques. New York: Van Nostrand, 1953, Chapter 1.

536


The characteristic impedance of the quarter wave transformer is obtained {i'oIn its load impedance and required input impedance, <:'in = 300 Q, by use of (10-32)

(i) Using the graph of Figure 9-4, (h/R) = 1.85 yields <:'0 = 147 Q for a loss less parallel-wire line in air. Choosing the spacing 2h = 0.75 in. f()f the quarter wave transformer, the conductor diameter becomes 2R = 0.4·05 in. At 150 MHz, A on the air dielectric transformer section, assumed loss1ess, is obtained using (9-34b) and (9-38); that is, ). = cU' = 2 m, yielding the required length ,1,/4 = 0.5 m lor the quarter wave section.

10-5 IMPEDANCE-MATCHING: STUB-MATCHING OF LOSSLESS LINES In communication systems, the matching of line terminations to line characteristic impedances results in no power reflections, important in maximizing the power transfer to the load. Just as vital to system performance is the considera lion that, if a load impedance were not rnatched to a line, the generator at the feed end would se diflerent impedances at the various frequencies within the information-carrying bana, a result of the frequency sensitivity exhibited generally by the input impedance (10-21) of an improperly terminated line. Thus, if pulse data were being transmitted, an improper termination would yield different reflections at the various frequencies within the Fourier spectrum of the pulse. The result is pulse-shape distortion, correctible by properly matching the load over the desired frequency band. Practical impedance-matching arrangements are shown in Figure 10-7. At lower frequencies, transforrners as shown in (a) of that figure can be employed for impedancematching, with untuned iron core transformers useful at power or audio frequencies. At radio frequencies not too high to eliminate the use of lumped or printed circuit elements, the L, T, and n configurations of Figure 10-7(b) are useable. Thus, antenna impedances can be matched with such schemes well into the VHF band (up to frequencies of the order of 100 MHz or so). At still higher frequencies wavelengths up

l

JIIO

RL

~o--rooor

oT

00

(Step up)

To

. (Step down)

L sections

Iron core

Single stub matcher

Air core (secondary tuned)

~

Tsection

~

Double stub matcher

M

FIGURE 10-7. Impedance-matching schemes. (a) Transformers as impedance matchers. (b) Narrow band matching sections using lumped reactors. (e) High-frequency matchers using translcmncrs or stubs.

10-5 IMPEDANCE-MATCIlING: STUB-MATCHING OF I.OSSLESS LINES /J

537

=0

Coaxial line

system

Parallel wire

system

(b)

(a)

(d)

(e)

FIGURE 10-8. Details of single-stub-matchiug on a lossles> line. (a) Use of a shorted stub in impedance-matching. (b) Usc of the} chart in impedance-matching. (c) Adjustment of d and t for it match. (d) Determining thl' stub length t.

to a few meters), lumped clements are physically too small or inefficient (low Q.), so replacing them with low-loss transformer or stub sections as depicted in Figure 1O-7(c) might be desirable. The object of the present section is to examine an impedancematching technique making use of reactive stn b elements introduced along the transmission line. The single-stub-matching arrangement of Figure 10-8(a) is analyzed. Similar schemes employ double- or triple-stub combinations. With the prl?per adjustment of the length t of the stub and its position d from the arbitrary load ZL, it is shown how a low-loss line can be matched to the impedance produced by the parallel cornbinatipn of the stub and the remaining length d of line terminated in the mismatched load ZL' The Smith chart is an important time-saver in the analysis. Because of the parallel connection of the stub and the transmission line, it'is advantageous to employ the admittance form of the Smith chart described relative to Figure 10-3(b). The line and stub are both considered lossless, each having the same pure real characteristic~admittance Yo = 0 1. With the known load Yr. = L I, normalized it becomes ih YdYo, yielding an SWR circle passing through itL at some point such as P in Figure 10-8(b). Moving toward the generator a distance d such that the intersection with the fI = 1 circle at Q.is obtained, the input admittance into that d length becomes it = 1 as depicted in Figure 10-8(c). Another intersection with fI = 1 is obtained farther toward the generator at R on the SWR circle; there the line admittance is 1 [fat either Q.or R the line is shunted with the susceptance =+= (provided by 6 an adjustable shorted stub), there results a cancellation of the susceptive part of the line admittance, yielding the parallel admittance it 1 at Q. or R. A matched impedance is thus obtained at Q. o[ R on reattaching the line to the left.

Z

Z

+J161

J161.

JI61

±ji i

538

PHASOR ANAI"YSIS OF REFLECTIVE TRANSMISSION LINES (

6'

;1'>:;; = 0.411

= -1.6

I

/---- ...........

/

/

"-

\

.L

! (a)

(b)

(c)

EXAMPLE 1O-1l

-j161

The remaining task is to find the stub length t needed to provide at Q." or at R. Hthe stub is attached to Q.,as in Figure lO-8(c), the positive (capacitive) susceptance of the input admittance if = 1 + must be canceled by the negative (inductive) susceptance of the shorted stub oflength t. Its length is obtained as the distance tlA shown in Figure lO-8(d), measured as a rotation toward the generator from the susceptance 6 -> OCJ at the short to the required susceptance 161 at S on the chart rim.

+J!6!

JI61

JI61

(

EXAMPlE 10-11. A transmitter operated at 150 MHz (,10 = 2 m) feeds a 72-0 antenna load through 12 m of a lossless, 300-0 parallel-wire air dielectric line. Determine the position d from the load at which a shorted stub should be connected and the required stub length, to match the load to the line as shown in (a). Assume the stub to be made of the same 300-0 line. The load admittance being (72) 1 0.014 and with the line r'o (300)-1 OJ)0333, the normalized admittance becomes 1!L (0.014)/(0.00333) = 4.17, shown at P on the Smith chart in (b). Rotating the lattcr by 0.07lA on the SWR circle provides an intersection with fI = 1 at if = 1 - j1.6, so the stub must be located d = 0.07 lAo = 14.2 cm from the load. To cancel the inductive susceptance = -j1.6 there, the required stub length t is given by the clockwise rotation {jAo from its short to the normalized susceptance (f = 1.6 as shown in (e), yielding {jAo = 0.411, so { = (0.411)2 = 0.822 HI. If an open-circuited stub had been used, its length producing the same input susceptance is ,1/4 shorter than that of the shorted stub, corresponding to the distance o to S on the rim.

-jldl

REFERENCES JOHNSON, VV. C. Transmission Lines and Networks. New York: McGraw-Hill, 1950. JORDAN, E. C., and K. G. BALMAIN. Eleetroma,l!,netic Waves and Radiating Systems, 2nd ed. Englewood Cliffs, N.J.: Prentice-Hall, 1968. REICH, H . .J., P. F. ORDUNG, H. L. KRAUSS, and.J. G. SKALNIK. Microwave Theory and Techniques. Princeton, N.J.: Van Nostrand, 1953. STEVENSON, W. D., .JR. Elements

of Power Systems

Analysis. New York: McGraw-Hill, 1962.

PROBLEMS

SECTION 10-1 10-1. A sinus'2idal generator, operating atI = 50 MHz, has the internal resistance Rg = 500 l!.nd generates Vg 200 V (sinusoidal peak). It is connected to an antenna load impedance Z1- = 100 + j50 0 through 3 m of coaxial line. The line, assumed lossiess, has a polyethylene

PROBLEMS

539

dielectric (Ey = 2.25) and the characteristic impedance Zo = 50 Q. Assume the z-origin at the input. Sketch and label this system. Show by use of (9-30) that the propagation constant l' on this line is jp = jn/2 rad/m, and the phase velocity is two-thirds the speed of light. Find the value of ..:l on this line. Show that the line length is t = 0.75..:l at this frequency. (b) Find both the load and line-input reflection coefficients. What percentage voltage (or current) reflection is occurring at this mismatched load? (c) Find the line input impedance, and determine the current = 1(0) and the average power delivcred to the line input. (d) Deduce the load current IL = I(l) and average power kd to the load. In view of the Poynting theorem, explain why this power and that found in (c) should be the same. [Answer: (b) frO) = 0.447e-j153.4" (c) 1(0) = 2.8281', 81 A (d) I(t) = 1.265ei7 1.5T A]

I.n

10-2. In Problem,to-I, the load average power~was found to be 80 W, obtained from the line current magnitude 1(t) flowing into the load Zr. = 100 + j50 Q. Show that the same load powcr can be found from the algebraic sum, Pa~ + P a-;', of the incident and reflected average power How through any cross section on thjs 1o.'11ess coaxi31llil1e and obtained h'om the incident and reflected voltage and current waves (V~, J~) and (V';;, 1';;), respectively. lHinl: Show, for example, that the average power carried by the + Z tr31ve~ng waves at any line cn~~s section = } Re [V~ (!~) *J W, where V~ = Zo1~.] (ineluding the load terminals) is given by

F:v

10-3. Suppose a matched load (ZI" = 50!1) is now used to terminate the lossless 50-Q line described in Problem 10-1. Sketch this system. (a) What r~Hection coefficient is expected at this load? Elsewhere on the line? What is the line imped:mce Z(z) anywhere of} this line? (b) Write only the symbolic expressions for the line voltage V(z) and line current J(z) on this matched line. From an inspection of these expressions, at what location yn the line do the line voltage and line current become just the complex amplitudes V~ and J~? (e) Calculate the line input current lin 1(0), whenc~ deduce I~ and V~. Find the line voltage and current as functions of ;c. (d) Calculate the average power input to the line; the power to the load. 10-4. Recalculate the load current IL of Example 10-1, making usc of ( 10-9a) and the value of the line input current obtained in part (c). Compare your result with (7) of part (d). 10-5. (a) In a manner similar to that used to obtain (10-9a), show that the ratio of the load voltage to the input voltage of a section of line of length t can be expressed

(10-33) (b) Rcpeat (a), except find the expression 'tor the ratio of the load voltage to the input currmt, showing that (10-34)

10-6. In Example 10-1, the solution given ignores the ctfect of the line losses associated with its attenuation factor rx. Take now the effects of rx into account. (a) Recalculate the input reflection coefficient, showing that f(O) is about 2.5% lower when losscs are accounted for. (b) Find the line impedance and input average power, comparing the results with the lossless case. Deduce the load current and load average power. [Answer: (b) 23.2 W] 10-7. The same source and load as described in Example 10-1 are connected to an identical line except for its length, which is now 11.5 m, making t/..:l = 1.09 at the operating frequency. Sketch this system. Neglect the effects of the small line attenuation in the following (assume rx = 0): (a) Find the load reflection coefficient and the line input impedance. (b) Determine the line input current and the average power delivered to the line at A-A. (c) Find the line current at the load and the average power fed to the load. 10-8. Repeat Problem 10-7, except include the effects of the line attenuation factor rx = 0.00197 Np/m in this case.

540


10-9. In the lossless line problem of Exal.!:lple 10-1, the load average powe! was found to be 23.4 W by making use of the line current I(t) driving the load impedance ZD' Show that the same load power can be deduced from the algebraic sum of the forward and backward power flow, P:v and P;:v> associate~ wilh the for~ar~ and backward traveling voltage and current wave complex amplitudes (V;:;, I;:;) and (V;;;, I;;;) respectively. (See hint in Problem 10-2.) 10-10. The same source and load are connected to the lossy line described in Example 10-2, but the line length is reduced to 30 mi, making tlA = 0.167. Assuming the origin at the input, sketch and label this system. (a) Find bOlh the load and input reflection coefficients and the input impedance. (b) Find the current 1(0) delivered to the line by the generator and the average input power to the line. (e) Determine the load current let) and the average power fed to the load. (d) If}he lipe had been impedance-matched = and the generator were conjugate-matched (Zg = Z(l') for maximum power transfer, determine both the average power delivered to the line and to the load.

(ZL Zo)

APPENDIX D 10-11.

Work Problem 6-21 in Chapter 6.

10-12.

Work Problem 6-22 in Chapter 6.

10-13. Use the Smith chart to find the complex reflection coefficient f (in polar {()rm) corresponding to the following normalized impedance: (a) 21 = 0.4 + jO, (b) 22 = 3 + jO, (c) ;3 = 0.8 + jO.6, (d) ;4 = 0.25 - jO.55. Show a Smith chart sketch, labeling these x locations thereon, along with appropriate phasors (arrows) depicting the complex reflection coefficients with labeled magnitudes and angles. Check the answer to part (c) analytically, using the normalized version of (10-5). (Also label the proper rr and r, axes on your charL) [Answer: (a) f 1 = 0.43ei 180 (d) 0.68e- j120 o

]

SECTION 10-2

Zo

10-14. Use the Smith chart to determine the input impedancc of a lossless line with = 500 and length t = 0.25A, assuming the following load impedances to tcrminatc this line: (a) 100 n, (b) j100 0, (c) -j100 0, (d) 100 j50 O. Show a simplified Smith chart sketch depicting the solution of (d) only, labeling thc entry (XL) and exit (Xin) values, plus the rim-scale rotation employed. [Answer: (a) 250 (b) -j25 n (d) 20 + jlO 0] 10-15. (a) Make use of the Smith chart to obtain graphical solutions for Problem lO-l. In particular, use the chart to determine thc load reflection coefficient. Show details of this on a labeled chart, or sketch of the chart. (b) Graphically determine the line-input reflection coefficient and impedance. Add relevant details of this to the chart sketch, taking care to denote the entry and exit points of the rim-scale rotation used. The reflection coefficients should be displayed as labeled phasors (arrows) on the chart (including their angular arguments). 10-16. (a) Employ the Smith chart to obtain graphically the input impedance of the 50-ohm lossless line of 1.09A length described in Problem 10-7, Provide details relative to a labeled chart or sketch, showing also details of the rim-scale rotation required. (b) Find graphically the reflection coefficients at the load and the line input, labelingthe magnitudes and angles of these phasors on the chart. 10-17. Repeat Problem 10-16, except apply the Smith chart to that line with the small attenuation factor rx = 0.00197 Np/m taken into account. By what factor is the reflection coefficient phasor diminished in length as it is rotated from the load at z = t to the input at Z = 0 on the chart? Is this rotation (i.e., the rimoscale entry and exit points) affected by this attenuation? 10-18. Apply the Smith chart to the lossy lil1c of lcngth 0.1.67). described in Problem 10-10. (a) Find graphically the reflection coeffieient ret) at the load, entering labeled results on the chart or chart sketch. (b) Employ thc required rotation via the rim scale to obtain the reflection coefficient and line impedance at the input terminals, giving details and verifying that thc

PROBLEMS

reflection coeftlcicnt magnitude in this rotation must be diminished value. Label these results on the chart.

541

about 6] (ir, [rom its load

10-19. Alter the cascaded lossless line lengths of Example 10-5 such that now t 1 = 3 m = 0.2Al and t 25m = 0.4 72A 2. Make usc of the Smith chart to find the lille impedance seen by the generator, supplying details and labeled charts or sketches that indicate appropriate entry and exit points and rim-scale rotations. (b) Find the average power delivered by the generator to the line input at A-A. Why is this also the average power delivered to the load? Use this observation to deduce quickly the load current magnitude. 10-20. Use the Smith chart to obtain graphically the normalized admittances corresponding to the following normalized impedances: (a) I + j2, (b) 4 (c) 3, (d) j4, (e) O. lAnswer: (b) 0.16 +jO.121 10-21.

The load impedance terminating the transmission line in Example 10-1 is

ZI. =

36

+

j20 a. Using an appropriate nonpalization, employ the Smith chart to transfi)rm this impedance graphically into its reeiprocal, YL . Check your result analytically. 10-22. Make usc of the Smith chart to find graphically the complex reflection codlicicnt f (in polar corresponding to the following normalized admittance value. (a) Yl = 0.4, (b) Y2 = 3, (C)Y3 = 0.8 + jO.6, (d)Y4 = 0.25 - jO.55, (e)ys = 1.4 + jO.8, (f)Y6 = O. Show a Smith chart sketch, on which label these,y points and their corresponding values (the latter as complex phasor arrows). Label the rr and r i axes 011 yonry-chart sketch, as depicted in Figure 10-3(b). Check the aI1sweI to part (e) analytically, using f = (I - y)/(I [the normalized form of (10-5) with Z = Y- 1 inserted]. 10-23. In the branched linc system of Example 10-6, prove that the ratio of the awrage powers injected into lines 2 and 3 at their common input at B-B is (l'av,2/ P ((;2/(;3)' where (;2 and (;3 denote the conductive (real) parts of their line input admittances. Then, fi'om tJ:c know!l average power inpllt into line I, find the average power reaching each of the loads ZI. and Z~, on lines 2 and 3. [Answer: Pav,2 = 6.40 WJ 10-24. The branched lossless line system of Exam pie 10-6 is rearranged by a simple interchange of the lines I and 3, the generator anclloads being left as shown. (a) Usc the Smith chart as an admittance chart to find the input admittances of lines 1 and 2. Then find the admittance looking into line 3 at A-A. Determine the average power delivered by the s"ouree to the input at A-A. Usc the of froblem 10-23 to determine the average power reaching each of the branched loads and Z~·

SECTION 10-3 10-25. Make use of the Smith chart results obtained for the lossless line of Example 10-3 to determine the following. Find the SWR on this line, obtained in two ways: (I) hom the from tbe osculation point of the SWR circle reflection coefficient magnitude on the line; with; (= SWR) circle lchcck Figure 6-10(c)]. (b) Usc the Smith chart in the manner of Figure 10-4 to locale the first Vmax and Vmin to the left of the load plane (express the distances in deeimal wavelengths and in centimeters). Locate 1m.x and {min as well. Depict all these on a labeled sketch of the graph of line voltage and current magnitudes versus z·

ZL

10-26. A Zo = 50 a lossless line is terminated in the load impedance (a) 5011, (b) 25 a, (c) 100 (cl)J25 (e) -jIOO (f) 20 + jl0 (g) () (short), (h) 00 (open). Normalize each load value and usc a Smith chart to determine graphically the reflection coefficient magnitude and the SWR on the line produced by each termination. What is the SWR on any lossless line terminated in a pure reactance? Explain.

a,

a,

a,

a,

10-27. Standing-wave measurements on an essentially lossless, 50-a slolled air line reveal all SWR value of 4.00 when it is terminated in an unknown load. The voltage minimum with the load in place is seell to shift 0.150A toward the generator when a short cifeuit replaces the Ipad at its load plane. Show a sketch depicting these details, and find the value of the unknown ZL with the aid of a Smith chart.

542


10-28. An unknown load impedance is connected to a 50-Q slotted air line (considered lossless), operated at 600 MHz. The load produces a measured SWR = 3.5 with the standing-wave Vmin occurring at the scale position 15.20 ern, the scale having its zero rd(~rence nearest the generator. On replacing the load by a short circuit, a null occurs at the seale location 9.60 cm. Ske~ch a line diagram depicting these details, and employ the Smith chart to find the unknown ZL' 10-29.

On connecting a mismatched load to a losslcss line, it will yield a standing wave on that line with its Vmin located according to one of four cases. Vmin will be located (a) at the load plane; (b) a quarter wave from the load; (c) between the load plane and a quarter wave from it; or (d) between a quarter wave and a half wave away from the load plane. For eaeh case, use a Smith chart to deduce the type of load: that is, whether it is an inductive or a capacitive impedance or a pure resistance (in which case argue as to whether it is larger or smaller than Zo). Be dear.

10-30. The SWR reading S = 2.55 is provided by a slotted line known to be terminated in a pure resistance. Use a Smith chart to deduce what two resistance values the load might have. Find in each ease the distanee [rom the load plane to the first Vmin on the standing wave.

SECTION 10·4 10·31.

Make use of the hyperbolic [unction definitions (10-22) to obtain (10-21 c) from (1O-21b). Show [rom the latter how (10·23) follows for the losslcss case and that it ean also be written ( 10-35) or, in normalized form, as ( 10-36)

10-32. Use the input impedance expression (10-21 b) to derive the shorted load version (10-25), from which deduce its lossless version (10-27). 10-33.

A section of coaxial line with negligible losses and 50-Q characteristic impedance is terminated in a short circuit (a "shorted stub"). (a) Find its input impedance by use of (10-27) if it has the length (I) t = 0.05;[; (2) t = 0.15;[; (3) t = 0.35;[. (b) Verify the results of part (a) by use of the Smith chart. (Show a sketch.) (c) Neglecting losses, what is its input impedance if the stub length is a quarter wave? A half wave? 10-34.

Make use of(10-23) to prove the input impedance expressions (10-30) and (10-31) for the special cases of haU:'wave and quarter-wave lossless lines that are al-bitrarily loaded.

At the operating frequency J = 400 MHz, a quarter-wave-Iong section of line with negligible losses and polyethylene dielectric (E, = 2.26) is terminated in the load impedance (a) 25 Q; (b) 25 + j25 Q; (c) 100 Q. Find the input impedance in each rase, as well as the length of this line. Let Zo = 50 Q.

10-35.

10-36. A particular antenna load operated at 500 MHz has the measured input impedance of 104 Q. It is to be fed li'om a 50-Q coaxial line through a lossless quarter-wave section of line (a "quarter-wave transformer"), with its Zo chosen such that the input impedance to the quarter-wave section matches the load to the 50-Q feed line. Sketch this system, and find the required Zo of the quarter-wave line section and its physical length, if the line dielectric is polyethylene (E, = 2.26). 10-37. A lossless, 50-Q slotted line is attached to an unknown load <:1. = <:(0), located at the load plane z = O. The mismatch there produces a measured standing-wave ratio S on the slotted

PROBLEMS

2:

543

totr) (z)

f----I z = d

~Y\i

/~

(z)

PROBLEM 10-38

line, with the voltage minimum located to the left at z = - d. Sketch this system. (a) Prove that the unknown load impedance at = 0 can be exprcssed in terms of the measured SWR and the distance d as l()lJows

~

~

=

jS tan (2n~)A <0 (d) S - j tan

(10-37)

2n~

[Hint: Using the Smith chart, s~ow that the reflection codhcicnt at the Vmin location = has the negative-rcal valn<.: with its magnitude obtained trom (10-17).] (b) the load produced total reflection (\q = I), what would the SWR then becomc'? Show for this case that (10-37) red uces to

IrI,

d)

-j
=

What kinds of load impedances produce total reflection?

1()"38. The unknown Er of a low-loss dielectric sample can be measured as f()llows. A closcfitting cylindrical plug of dielectric material of measured length d and unknown Er is slid into the end of a coaxial slotted air line terminated in a short circuit as showu. (The dielectric is assumed losslcss and nonmagnetic.) A<;sume the z-origin ilt the air-dielectric interface. The short at z = d causes total reflection into both regions I and 2, with a null measured hy the slotted do to the left of the interface as noted. The generator, to the left, operates at the line at measured frequency j; and is not shown. (a) Use (9-80a) to show that the ratio
(10-39)

544


(Hint: Use (10-38) of Problem 10-37 for Zl(O).] (b) With the generator operating at the I:equency J '" 3 GHz, and a lossless dielectric plug of depth d = 2.35 em inserted, the slotted lme shows a null at z - do -0.89 em. Use numerical methods on (10-39) tD t1.nd EOr of tillS sample.

SECTION 10-5 • 10-39.

An unspecified impedance is connected to a section oflossless coaxial 50-Q air line. A 50-Q slotted air line is connected between a 500-MHz source and the load. Probed voltage measurements along the slot reveal an SWR of 3.50, with a voltage minimum loea t('d at the 50-em mark on the slotted line. Sketch this system. (a) Find, by use of an admittance Smith chart, where an air-dielectrie adjustable-length shorted stub should be located relative to the Vmin posltlOn. Find also the required length of the stub, choosing the position toward the load from Vmin for the point of attachment. Assume the stub to have a 50-Q characteristic impedance also. (b) What is now the SWR value on the lint' between the generator and the stub? Between the stub and the lOad? (e) Repeat (a), but this time choose the stub locati011 to be toward the generator frOm Vruin.

10-40. In single-stub matching, prove that locating a shorted stub where theAille normalized admi.ttan.ce has the value 1 + will yield a shorter stub length than lhat obtained when !ocatmg it where it has the value 1 What are your conclusions if an open-circuited stub IS used?

jlbl

jlbl.

~

__________________________________________ CHAPTER 11

Radiation from Antennas in Free Space

The problem of the radiation of electromagnetic energy from a transmitting antenna to a receiving system is of considerable interest to the communications engineer. Transmitting antennas are devices used in terminating a transmission line or waveguide with the intent of efficiently launching electromagnetic waves into space, and they may be regarded as sources of such waves in space. This chapter is concerned with the analysis of the radiation fields obtained Ii-mn typical antenna sources, important examples of which are the linear wire antenna and the electromagnetic horn illustrated in Figure II-I. First the physical E and B fields arc described in terms of the scalar and vector auxiliary potentials
t

~

-;,..

Generator ~\=::;:===~

Horn antenna

Linear

~antenna Aperture

Transmission line)

(a)

..0<;-

t (b)

FIGURE II-\. Examples of ant~nnas. (a) Linear wire antenna, center-fed by a transmission line. (b) Electromagnetic horn, excited by nse oj' a waveguide.

545

546

RADIATION FROM ANTENNAS IN FREE SPACE

special case, with an integration leading to the radiated fields of a thin center-ted antenna of arbitrary length. The extension of Maxwell's equations to a symmetrical set using postulated magnetic charges and currents, together with their boundary conditions, forms the basis for predicting the radiation fields of electromagnetic horns and related aperture-type antennas.

11-1 WAVE EQUATIONS IN TERMS OF ELECTROMAGNETIC POTENTIALS As an aid in computing the radiation fields of antennas, frequently auxiliary functions (potential fields) are helpful in systematizing the mathematics. In particular, one may recall that the electric and magnetic fields of charge and current sources have already been related to potential functions by (5-22) and (5-48)

B=VxA

( 11-1 )

oA ot

E = -V--

(11-2)

in which is the scalar electric potential and A the vector magnetic potential field at any point in a region. It is recalled from Chapter 5 that the potential fields at any field point in free space may be expressed by integrations over the charge and current sources as given by (4-35) and (5-28) as follows

(x,y, z) A(x

,y,

)=

z

r

dv'

Jv~~-

r

Jv

P,oJ(x',y', z') dl/ 4nR

[4-35 J

[5-28aJ

assuming quasi-static conditions, that is, that the source densities Pv and J vary sufficiently slowly in time so that the finite velocity of propagation of the field effects (time retardation) can be neglected. For radiation problems in which the fields many wavelengths from the sources are usually desired, the time retardation eHects are of such significance that the potential integrals (4-35) and (5-28) become useless. Revised forms can be obtained by showing that and A in general satisfy inhomogeneous wave equations, and from these are found integral solutions for free space. One must first find wave equations satisfied by the potentials and A, by usc of (11-1) and (11-2) substituted into the Maxwell relations (3-24), (3-48), (3-59), and (3-77)

v·n =

Pv

( 11-3)

V'B=O VxE=

( 11-4)

oB at

VXH=J+

DD

ot

(1 \-5)

( 11-6)

11-1 WAVE EQUATIONS IN TERMS OF ELECTROMAGNETIC POTENTIALS

547

Note that there is no further need to use (11-4) and (11-5) here, for the potential relationships (II-I) and (11-2) were obtained from these two Maxwell equations originally. [Putting (II-I) and (11-2) into (11-4) and (11-5) merely leads to identities.] With E assumed a constant in what follows, write (11-3) as V . E = pjE, and substituting (11-2) into the latter yields V . (- V
V 2 <1>

+

a(v· A) at

P

= ---"-

(11-7)

E

if V . (V<1» = V2 <1> from (13) in Table 2-2 is used. From Section 2-3 it is recalled that the specification of both the divergence and the curl of a vector function assures its uniqueness (within an arbitrary constant), but only the curl of A, by (11-1), has thus far been established. The divergence of A is therefore still arbitrary, so put V·A=

a
( 11-3)

p,E-

at

whereupon substituting the latter into (11-7) obtains

pv

(11-9)

E

Comparing this result with the form of (2-100) reveals that (11-9) is a scalar wave equation in terms of the potential
The identity (21) of Table 2-2 permits writing V X (V X A) = V(V' A) - V A, and further substituting the Lorentz condition (11-3) produces a cancellation of terms containing <1> in (11-10) to yield (11-11)

the desired vector wave equation expressed in terms of A. A comparison of the wave equations (11-9) and (Il-II) with the Poisson-type differential equations (4-67) and (5-26)

pv

[4-67]

E

[5-26]

548

RADIATION FROM ANTENNAS IN FREE SPACI':

shows that the latter are just special cases of the wave equations, subject to the timestatic assumption a/at = O. The integrals (4-35) and (5-28) given earlier are the timestatic solutions of (4-67) and (5-26) in free space. In the next section, comparable integral solutions of the wave equations (II-g) and (11-11) arc derived. Complex, time-harmonic fields are used to accomplish this. Thus, with A(u 1 , U2, u3 , t) in the time domain replaced with A(u 1 , U2, U3)ei wt in the manner of (2-~7), and simjlarly for the remaining fields, the Lorentz condition (11-8) becomes V - A = - jWJlE(j) , yielding

~

V-A

(j) = - - jWJlE

Lorentz condition

(11-12)

The collected results (/1-1), (11-2), (II -g), and (11-11) in time-harmonicform hecome

(11-13) ~ ~ ~ V(V E = - V(j) -jwA = .

_

-

}WJlE

in which and

A are

jwA

(11-14)

solutions of the wave equations

E

(11-15)

(II-Hi)

11·2 INTEGRATION OF THE INHOMOGENEOUS WAVE EQUATION IN FREE SPACE It is shown that a solution of the vector wave equation (11-16) in free space can, 1:>e represented as the following integral over the time-harmonic current sources J(U'b u~, U3)

(11-17)

a result closely resembling the integral (5-28a) fe)r direct current sources in free space, except for the additional time retardation factor exp ( - jPoR). The geometry of a generalized system of current densities in free space is shown in Figure 11-2.,The integral (11-17) over a system of such current sources leads to the vector magiletic

11-2 lNTEGRATlON OF TilE lNllOMO(;ENEOUS WAVE EQUAT10N IN FREE SPACE

549

System of,../f currents

FIGURE 11-2. Generalized field point P at which the

source distribution in free space, and {(lUnd using (11-17).

potential A at the field point P, wbence :B and E fields are then obtained using ( 11- 13) and (11-14). A formal proof of (11-17), as the particular integral solution of the vector wave equation (11-16), is given in Appendix C. The integral (11-17) has extensive applications to the determination of the radiation fields of current distributions on conductors in free space. Examples are shown in Figure 11-3(a). Insight into the radiation fields of such devices can be acquired initially From a study of the infinitesimal dipole element illustrated in Figure li-3(b), sillce an end-to-end superposition of such elements can be used as a basis lor constructing any of the antennas in Figure 11-3(a), and hence their fields as well. The field integral ( 11-1 7) is simply an expression of such a superposition.

Linear antenna

Loop antenna

Rhombic antenna

Linear array

(a)

p p

R

(b)

(c)

(d)

FIGURE 11-3. Antenna configurations amenable to field analysis by use of (11-17). (a) A few antenna configurations of physical interest. (b) An infinitesimal oscillating current-element. (c) Linear antenna as a superposition of current-elements. (d) Array of linear wire antennas.

550


dt' Volume-element

du'=ds'dt'

~-.l,

,',

•

++ ~ j +A+ q = lW

,, ,, ,, '' ,, :'

A

,'r-dl=Jds'

!.l.±t! ~t

-.9_- -Ij

ds' (a)

p

,,

(z)

, A

(f)

,,

()

,:~

'

H'

it '

---- 0

----

0

(b)

) (c)

FIGURE 11-4. The oscillating current-clcmtnt. (a) Limit of the volume current-element Jdv', becoming the linear current-element IdC' as ds' -> O. (b) Geometry relative to a current-element at the origin. (c) Components of electric and magnetic fields at P.

11-3 RADIATION FROM THE INFINITESIMAL CURRENT-ELEMENT The usefulness of (1) 17) lies in the fact that it yields, at any field point in free space, !:..he total potential A due to a system of time-harmonic current sourc~s. The physical E and H fields of such sources can then be derived fi'om the known A field by use of (1 1-13) and (I 1-14). It is instructive to find the fields of the most basic current source: lhe irifinitesimal current-element (or elementary ~dipole) illustratcd in Figures 11-3(b) and 11-4. Th!:: current-element is defined by J dv' appearing in the potential integral (11-17), with J denoting the complex lime-harmonic vector current density at some volume-elcment dv'. For present purposes, the transverse area ds' of the volume-element is assumed to vanish as suggested in Figu~e 11-4(a), with the current source carrying a finite (rather tha.!l infinitesimal) current I. This permits expressing the volume current-element Jjv' as J dt', a linear current-element. It is seen horn Figure 11-4(a) that the current I is accompanied by charge accumulations ± fj at the ends. The relation connecting a timeinstantaneous current flow i with real-time charge accumulations ± q is, by (3-82a), i = dq/dt. The corresponding time-harmonic form is I =jwfj

( 11-18)

Because the elementary current source involves charge displacements ± q to the ends of the element, it is often called an oscillating electric dipole. The vector magnetic potential of an oscillating current-element located at the origin of a spherical coordinate system is obtained with reference to Figure 11-4( b). Since only an infinitesimal current-element is assumed present, no integration of (11 17) is required, to yield the diflerential potential at P

(11-19)

11-3 RADIATION FROM THE INFINITESIMAL CURRENT-ELEMENT

551

The electric and fields corresponding to (11-19) are found by usc of (II-I) and (11-2). Equation (II is given in mixed coordinate systems, so it becomes desirable to express the z-directed potential in spherieal coordinates. From the geometry in Figure 11-4 (b) (1l-20) Then the

Ii field

of the elementary dipole becomes, from (11-1) ar r2

diI=

dB

Vx

lio

110

{LO

= a.p j dz'

e - j 'p or

471:

1 2"

r

r

r

r SlIl

a ar

a ao

0

dAr

rdA o

0

fPo + J. e _.

a.p

ae

sin 0

sm

(11-21 )

A/ m

The electric field E is obtainable two ways. One is by use of (11-4) ~ V(V . dA) ~ dE = . - jUJdA )UJlioEo

( 11-22)

into which the substitution of (11-19) and (11-20) yields the desired dE at P. An alternate method involves the use of the Maxwell equation (3-85), which in the freespace region becomes

dE

,

)UJEo

V

X

(dill

(11-23)

and into which the substitution of (11-21) obtains the desired electric field, Either method yields (11-24) in which

I dz'- e- J'P or [2Y10 dE~ = r 471: r2

j UJE or3

I dz'- e- jfior [jUJlio - - + Ylo dE"O;e -_ 4n

r

J

2 + ---

cos

e

J'

+ -,-1 ] UJEor3

(11-25)

SIll

e

(11-26)

and Ylo denotes the intrinsic impedance -./lio/Eo for free space encountered in Chapter 2 in connection with uniform plane waves. In Figure 1l-4(c) are shown the vector field components of the oscillating current element at the typical field point P(r, 0, 4») and given by (11-21) and (11-24).

552


The real-time forms of the tlelds of an oscillating current-element are found by use of (2-74); thus, from (11-21) and (1l-24)

. . [1

~ Re [dEre-'wtJ = Re -dz'. - e-'(rot - por) (2110 -4n r2 =

21 dz' [110 2 cos (wt - flor) 4n r

+ - -13 sin wEor

1 dz' [ wf..lo . dEe = - - - - - sm (wt - /301) 4n r

+ _1_3 sin wEor

(wt - /3or)] sin

+ 110 2 r

C

2e --j90 +-

)

wEor3

cos ()

(wt - /3or) ] cos

]

e

(] 1-27)

cos (wt - /3or)

e

)

1 dz' [ I /30 . dHq, = - - 2: cos (WI - /3or) - - 8m (WI 4n r r

(11-28)

/3or)] sin

e

(11-29)

assuming the current amplitude 1 to bc the pure real r. These real-time results are useful in sketching the flux fields of the oscillating dipole, depicted in Figure 11-5. Only the electric field lines are shown, since their components dEl) and dEr lie in the plane of the paper; the flux of dHq" from (11-29), consists of an azimuthally oriented system of circles about the z axis of the figure. The ficlds close to the dipole, termed the nearzone fields, resemble the electric flux of a static charge dipole discussed in Example 4-8, in contrast with the farzone, or radiation, fields that become impqrtant at distances of a few wavelengths or more from the source.

(z)

Wave motion ----..;,.. Nearzone region

~

Farzone region

FIGURE 11-5. Electric field flux of an oscillating current-element at a fixed instant.

INFINITESIMAL CURRENT-ELEMENT

II

553

fields warrants a look at the simregions, distinctions made possible 3 1, r- 2 , and r- in the field expressions For example, from the magnetic field inverse r to inverse r2 terms is obtained

The plifications in by a comparison and facilitated

( 11-30) From this it is term predominates" source (r« and (11-26) l/r 2 to the nearzone region (I' only the

from the dipole source (r» Ao) the l/r

of (11-21) is far more important near the to the electric field expressions (11-25) the l/r to the l/r 2 terms as well as the (11-30). One concludes that in the 1~I,ementary dipole are well approximated by of dEr and dEe as follows (11-31) Nearzone: r« Ao (11-32) a static charge dipole shows that (11-31) limit (w ---* 0), on substituting fj for J/jw magnetic field (11-32) reduces, as w ---* 0, to Savart law (5-35b) applied to a differen tial

curren t-elemen t. With the are on I y those

theJarzone region, the important field terms

1-21), (11-25), and (1/-26) to sin

e

(11-33) Farzone: r» Ao

itlor

These fields are in-phase, remote regions. Their ratiu

sin

e

(11-34)

become important in the radiation of energy into real intrinsic wave impedance

~ == '10 ~ 377 n V-;;;

( 11-35)

identical with (2-130), the wave impedance associated with uniform plane waves in free space treated in Section 2-10. This is not an unexpected result if one realizes that the spherical waves (11-33) and (11-34) are TEM waves. They are essentially uniform plane waves over a small portion of the surface of a large sphere of radius r centered at the radiating elementary dipole. The factor sin appearing in those field expressions

e

554


I

I I

I I

I I

()

/

I

,

~/

/r

(~~1C~~) I

IdEe

(a)

(b)

FIGURE 11-6. Relative to the elementary dipole. (a) Field pattern of the elementary dipole. Plot of sin versus Note the axial symmetry. (b) Large spherical surface S(r» ,to) enclosing dipole [or finding the radiated power by use of farzone fields.

e

e.

is called the field pattern factor of the elementary dipole. I t accounts for maximum field intensities in a direction broadside to the elementary dipole as shown in Figure 11-6(a), tapering to zero along the dipole axis. The time-average power radiated from any surface S enclosing the elementary dipole, depicted in Figure II-6(b), is obtained by use of the Poynting theorem (7-58). No ohmic dissipation occurs in the free-space region enclosed by S, making (7-58) ( 11-36)

The volume integral denotes the time-average power generated by active sources driving the elementary dipole, seen to equal the time-average power flux leaving (radiated from) the enclosing surface S. It should be realized that arry surfaee S whatsoever may be used to enclose the dipole source, but by use of a sphere of large radius r, requiring only the farzone fields (1l-33) and (11-34), one eliminates the need for incorporating all the terms of (11-21) and (11-24·). The additional contributions to the time-average power are found to be zero anyway because of the phase condition of the nearzone terms. Inserting (11-33) and (11-34) into (I 1-36), one may show that the time-average power radiated from the elementary dipole becomes P av

I

II

I

,r, 'jS 1- Re

~

~

[dE x dH*] • ds

'101[[2 =3

(rlz')2 ..1,0 W

(11-37)

On increasing the differential length dz' of the current element to a value It (though yet small compared to the wavelength ..1,0) it is seen from (11-37) that the radiated power is proportional to the square of the length. Even so, an electrically short currentelement is incapable of radiating much power. For example, a wire antenna 3 cm long operated at 100 MHz has a length It = O.OU o , making (h/Ao) = 0.01. If one could excite the wire with I A of current, its radiated power, from (11-37), would be only 50 mW. Aside from the difficulty of exciting an electrically short antenna with very much current, this is still substantially less than the power obtainable ii'om a half wave

CENTER-FED THIN-WIRE ANTENNA

11-4

555

iI'om the next scction, A detailed conimpedance, a subject not considered in impedance to current flow is offered by radiation,

linear antenna sideration of the this text, reveals that electrically short antelmas,

11.4 RADIATION FlaDI Of A UNIAR CENTER·FED THIN·WIRE ANTENNA The thin-wire antenna, point along the wire, radiated at remote an oscilla ti ng It involves the (11-17) or the electric the antenna, accoul1tillg fixed field point P, 011 is suggested by the

source applied at a gap located at some for years, The prediction of the fields antenna, making use of the known fields of section, is the subject of this discussion, (~ither the differential radiation potential 1 or (11-34) over the finite length of "'''''''",',,"., of the differential field contributions at a R to source points on the antenna. This . The field integrals arc seen to contain

(z)

(a)

(b)

(z)

(c)

(d)

FIGURE 11-7. Relative to antennas and their currcnt distributions. (a) The summing of field contributions infinitesimal current-elements along an antenna. (b) Linear antenna current obtained from a deformation of an open-circuited transmission line. (c) Loop antenna current standing wave, obtained from a deformation of a shorted transmission line. Pertaining to the distribution of a current standing wave along a thin wire, as a tilncti"!l

556


\ ~

/

(- the current /, a quantity that must be known or specified at each source position l along the ~antenna if the integration is to be carried out. The antenna current distribution /(z') may be found experimentally or by analytical means. Experiments in which antenna currents are probed show that their amplitudes along the antenna wire are very nearly sinusoidal standing waves. Qualitatively, at least, an open-ended linear antenna may be regarded as an opened-out section of an open-circuited parallel-wire transmission line, possessing sinusoidal standing waves of current tapering to zero at the wire ends as depicted in Figure 11-7 (b). Similarly, the circular loop antenna shown in Figure 11-7(c) has a current distribution on the wire that deviates only slightly from the current standing wave on the conductors of a shorted parallel-wire transmission line, as noted in the same figure. Although the effects of power radiation from an antenna tend to produce deviations from these simplified standing-wave pictures, a comparison of measured antenna field patterns with calculated results based 011 assumed sinusoidal current standing waves reveals that the assumption is quite suitable for tarzonc field calculations. A much better current distribution approximation is required, on the other hand, for predicting the terminal impedance of a linear wire antenna; this latter task is omitted from the present discussion. An analytical proof of the fact that a sinusoidal standing wave of current is a reasonable assumption for linear wire an tennas was original! y provided by Pocklington. 1 His conclusion is demonstrated from an expression for the vector magnetic potential A developed at t~e surface of a thin wire, using the integral (II 17). Assuming the antenna current 1(:::.') to be concentrated wholly on the wire axis as in Figure 11-7(d) does not significantly altel· the potential A(a, z) at the typical, fixed field point P the wire surface. A(a, z) is obtained from (11-17) with ide' = aJ(z') dz'

on

( 11-38)

in which the distance R from a source point P'(O, point P(a, z) on the wire surface is

on the wire axis to the fixed field

R=

( 11-39)

It is sufficient to assign the integration limits in (J 1-38) over only a rather small neigh borhood (~: - t, /: : + t) of the field point P, in view of the close proximity of the field point to the axial current sources. In otber words, current sources located at more remote points (;;;' - z) » a are too far away to alter the integral appreciably. Then the phase factor e-j/JoR in (11-38) takes on values near unity, since PoR will not depart greatly fimn zero. Also, the current 7(l) being investigated acquires an average value 7(Zl over the neighborhood (z - t, z + t) ~b()ut the Jixed field point P, or essentiaIly a unifixm value over that z' range. Then f(z) call be removed fforn the integral of ( II to obtain fz+1'

Jz~=z-I'

Thus that is,

at any field point dose to the wire axis is proportional to the local current 7(z); ( 11-40)

lB. E. Pocklington, "Electrical oscillations in wins," Cambro Phil. Soc. Prot., 9, On. 25,1897, pp.

324~-332.

557

11-4 RADIATION FIELDS OF A LrNEAR CENTER-FED THIN-WIRE

In an essentially axially symmetric system such as this, the field al position p a on the ~ire is dependent on z only. The wave equation \ 11-1 only in terms of the A z component, therefore reduces to

radial

for anyficld point Pta, z) on the wire surface,~but (11-40) substituted into yields a wave equation in terms of the current I(z) on the wire axis

11-41)

2~

() I

2

~

+Wf.1oEoI=O

(II

This homogeneous second-order differential equation evidently has the solution (11-43) a sum of iorward- and backward-traveling waves of current on the wire. With the current at the open ends of the linear antenna of Figure 11-7 (b) required to vanish, (11-43) agrees with the experimental evidence that the time-harmonic current distribution on a thin-wire antenna is essentially a sinusoidal standing wave. 'The phase factor 130 applicable to the eurrent standing wave on the wire implies that the wavelength associated with the current distribution is the same as that fi)f the field in the free space surrounding the antenna. In Figure II-8(a) are shown examples oflinear antennas driven by a sinusoidal source amI developing sinusoidal standing waves of current in accordance with the foregoing remarks. The following rules are applicable 1. The current through the driving source must be continuous. This follows from the requirement that just as much current must leave one generator terminal as enters the other terminal. 2. If the antenna ends are open (the wire does not form a continuous loop), the current at the ends must vanish. This follows from the conservation of charge.

: (z)

: (z)

t' ,

t11,

t\ t '-

,,

t "

t/ -t /

Center-fed

'\ \

\

\

I

I

---I I

I

o "'-:.:f.. :) 0 /

\

--~I

I I

I

\

\

---\1m1

t

\

\

\

I

t---~1 I m

\

\

L-~1 Tim

o '" --~\

,

,

\

I

I

I

I I I

'" ,/

1m2

I

:

/

I

'", 0 \

-t2

I"

II

\

Off-center fed (a)

(b)

FIGURE 11-8. Standing-wave current distributions found on thin linear antennas driven by a single sinusoidal generator. (a) Straight linear antennas: centerfed (lejt) and off-center fed. (h) Examples of curved linear antennas: open-wire (lejt) and loop antenna.

558


3. The current distributions to either side of the generator Illust be sinusoidal standing waves with a free-space phase constant /10 = wSoEo. The distribution is specified by (11-43), satisfying the boundary conditions of rules I and 2. The foregoing rules, ;(;~iied to the center:fed straight-wire antenna of Figure 11-8(a) lead to the following current distributions on the upper and lower halves of the antenna

1«) = I«)

=

7", sin /30 (t - <) Im sin /1o(t + <)

o«
(11-44)

!m

in which denotes the ~urrent amp~itud,e occurring at tl~y maximum along the standiQgwave. In general 1m may be assumed eornplex; that is, 1m = Imel'" iran arbitrary phase angle ¢ is desired to be included. The standing waves (J I -44) are seen to be continuous at the generator location < = 0 in accordance with rule I; whereas they further vanish at the antenna ends < = as required by rule 2. If the generator is placed off center as in Figure 11-8(a), nonsymmetrical current standing waves are obtained as shown. The rules applied to this case yield current standing waves

±t

/«) = 7",1

sin

/30(tl -
0< < < tj

I«) = Im2

sin

/30(t2 + z)

-t2 < ':: < 0

(11-45)

you may~verify what additional relationship between the standing-wave amplitudes 1m2 needs to be satisfied if current continuity through the generator is to prevail. T'he current standing waves are illustrated ft)r center-f(:d antennas of typical lengths in Figure 11-9(a). Shown in (b) is the variation of the antenna currents with time; the standing character of the waves is evident from the real-time bebavior.

1m! and

\

\

\

\

Im

~

,

I

I I I I

,V

I

\

--1

\

Im \

I I

Im\

--1

I

/

/

I

\

I

/

(Short dipole)

I

I I

it

I

--<

\

t

It

/

/

/

\

t

--<\

U --""21::.t

\

/

!

/

t= 0

T

8

T

"4

u= 1.3 AO (a)

(b)

FIGURE 11-9. Currents on center-fed linear antennas. (a) Current standing waves on antennas ofditferent lengths, operated at the same frequency. (b) Variations or the current standing wave with time t, for a three-quarters wavelength antenna.

11-4 RADJATJON fIELDS OF A UNEAR CENTER-rED 'I'IIlN-WlRE ANTENNA

559

Having seen ii-om the preceding developmen! that only local or neighborhood current-elements appreciably aHect the potential A near the suriace of a linear antenna, Olle should also expect sinusoidal current standing waves to exist on a linear antenna even when it is curved as depicted in Figure 11-8(b). For the open-wire antenna oflength 2l in that figure, the distribution (11-44) may still be taken as a approximation to the current standing wave, if,::: is replaced by a variable' delloting distance measured along the curved wire axis. A departure from the sinusoidal wave would be expected at sharp corners or regions of wire curvature small with a free-space wavelength. Finally, olle may note the efl(~ct of bringing the ends of the antenna together, f()lTning the "loop antenna" of Figure 11-8(b). The current standing wave in this case is predictable frorn a change in rule 2: instead orthe current vanishing at the previously open ends, it must be continuous and well-behaved at the midpoint M, and symmetric about A1. Armed with the knowledge of a reasonable approximation (11-44) of the current distribution on the center-ted linear antenna of Figure 11-9(a), one can with the evaluation of its farzone electromagnetic field. Two methods arc at this potential A at stage of the development: (a) the integration for~the vf~·t()r any f~m:one field point by use of (11-19), whence E and H may be l()llud (11-31) and (II 14); and (b) the direct integration for the faxzone electric field means of (11-33). Approach (h) is employed, appealing to the geometry of Figure J 1-10. The COlltrilmtion to the total electric field at the field point P, offered eurrentelement Idz' located at P' on the antenna wire, given by (11

. {3 0 JdZ' - jfJo}{.· iJ : _)1]0 dE"0 e sIn v 4nR

( 11-46)

With P(r, 0, ¢) in the filfzone and the antenna centered at the origin, the distance R from the source pc)int P' to P is essentially parallel to r, so that R:;:;r

( 11-47)

z'eose

dA z ,t A

I

dHq,_ x ...- P(r, (J,

4:»: field

point

y

-/ z'cos6

:FIGURE 11-10.

of center-fed linear antenna in relation to the determination of

560


assuming R very large compared with the antenna length 2t. Using (11-47), electric field at P is found by integrating (11-46)

·13'"

sin 0 4·n

=) 0'(0

.

e-Jfior

it

cos

z' =

-/ r -

z'

(1

(11-48)

cos 0

The ;:;' dependence of the current is expressed by (11-44)

i(z') = im sin Poll =

z')

0<;:;' < t

im sin Poll +

t
(11-49)

The phase factor exp (jPoz' cos 0) appearing in (11-48) accounts for the out-or-phase condition of the contributions dEe a~rtving at P, making the integral quite sensitive to the variable phase delay due to the variable distance R. In contrast, theJactor r - ;:;' cos 0 in the denominator in (11-48) affects only the amplitudes of the dEo contribution at P, permitting there the substitution r - z' cos ~ r. Therefore (11-48) sim plifies to

e

which the integration is by parts or by substituting the exponential expression 1 [eja e-jaJ for the sine functions. Thus sin ex

111

(11-50)

From the latter are deduced the following properties of the radiation fields ora centerfed linear dipole. 1. The fields E~ and ifq, in the [;irzone region are outgoing TEM spherical waves, related by the real intrinsic wave impedance 1]0 like the fields of uniform plane waves in free space. The fields have spherical equiphase surfaces (observed b·om putting r = constant in the spherical wave phase factor , whereas the amplitudes vary as r - i, 2. The fields are directly proportional to the excitatioll current amplitude 1m. 3. Eo and ifq, are independent of the azirnuthal angle

(/lot cos 0) sin

e

- cos

Pot

(II-51 )

11-4 RADIATION FIELDS OF A LINEAR CENTER-FED THlN-WIRE ANTENNA

561

(b)

(a)

(d)

(e)

(e)

(f)

FIGURE 11-11. Field patterns F(O) for several center-fed linear antennas. Cnrrent standing wave distribntions arc realized. (a) Field pattern of half wave dipole = Ao/2), for any q,. (b) F(O) as a solid pattern of revolution about the antenna (e) F(O) represented by relative intensity contonrs on an r constant sphere. (d) Field pattern of a fnll wave dipole Ao). (e) Three-halves wavelength dipole pattern (2t = 3Ao/2). (1) Field pattern very short dipole (2t -+ d;;'J.

F(O) is called the field pattern of the center-fed linear antenna. Some features of this factor are discussed in the following.

e

The pattern-factor F( 0) how the farzone fields vary with over a sphere. In Figure 11-11 ((I), (b), and are shown three ways to depict the field pattern of a center-led antenna a half wavelength long so-called half wave dipole). The pattern is axially symmetric, so the sectional-cut pattern of Figure II 11 ((I) adequately represents F(O). From (d) and (e), as the antenna length 2t is increased, the pattern acquires additional lobes, between which nulls or dead spots in the transmitted fields

562


occur. This is a consequence of phase interference effects, more pronounced {or longer antennas. Thus the pattern of the three-halves wavelength antenna in Figure 11-11 (e) shows how the phase effects of the field contributions in the broadside (0 = 90°) direction yields a relatively small lobe maximum there, a consequence of partial phase cancellation. For the limiting case ofa very short dipole having essentially a triangular standing-wave distribution (the tips of sine waves), F(O) reduces to the simple sin 0 function of Figure 11-11 (f). In all cases, a null occurs in the pattern along the antenna aXIS.

The total time-average power radiated from a linear antenna may be found in the same way as for the infinitesimal dipole, by making use of (11-36). The farzone fields (11-50) thus yield the radially outward-directed, time-average Poynting vector I'fh

_

iT av -

1. 2

Re

X

{jl1o I m 2n:r

a [ -j7;, 4>

2n:r

e

- j/3or

(cos

((Jot

cos 0) . ()

cos

8m

~flor (cos ((Jot co~ 0)

(Jot) ae

- cos (Jot)]}

sm 0

(II-52) The latter substituted into (11-36) and integrated over a sphere of radius r, on which 2 ds ards a r r2 sin 0 d(} d, obtains the time-average radiated power

Pay =

#8 &'av • ds = £"'=0 S::o &'av • a

= 110 1;;' fn: 4n:

Jo

[cos

((Jot

cos sin 0

r r2

sin 0 dO d

dOW

(11-53)

This expressiOIl is not integrable in closed form, though it can be evaluated by power series substitutions or by use of the tabulated sine and cosine integral functions Si(x) and Ci(x);2 also it is readily computerized or solved by graphical methods. Closely related to the time-average radiated power (11-53) of a linear centerfed antenna is its so-called radiation resistance, R rad . It is defined such that, on multiplying it by the square of the rms value of the current amplitude 1m associated with the antenna current distribution (11-44), the radiated power (II-53) is obtained; that is, R rad1?;j2 Pay, making

(II-54) Thus R rad denotes a fictitious resistance that, on carrying the rms current 1m1j2, dissipates the same amount of power as that radiated by a center-fed linear antenna possessing a standing wave of current with the amplitude 1m. It is emphasized that the radiation resistance of a lineal antenna is in general nol the same as the resistance part of the antenna impedance Za seen by a generator attached to the antenna terminals. This fact is related in part to the current amplitude 1m appearing elsewhere 2For example, see S. Ramo,J. R, Whinnery, and T. van Duzer. Fields and Waves in Communicationl,'iectronics, 2nd ed. New York: Wiley, 1984, p. 599.

I'~QtlATlONS AND THEIR VECTOR POTENTIALS

563

antenna terminals, as seen in Figure 11-8(a)~ the time-average power radiated from the of which may be recalled from Figures favorable two-lobed pattern without an exhalf wave dipole also has a desirable terminal is essentially a pure resi~tance, virtually view of the maximum value 1m of its standing terminals as noted in Figure II-II (a). The half wave dipole can he calculated by use of

1[/2)

,.~,.~~---~,~~

The radiation

dO

~

36.5/; W

Half wave dipole

(11-55)

therefore 73 Q

Half wave dipole

(11-56)

*11·5 SYMMETRIC MAXWal!S EQUATIONS AND

THEIR VECTOR POTENTIALS: THE FIELD EQUIVALENCE THEOREM Section 11-4 demollslrawd how the current distribution I(z') over a linear antenna leads to the fit'lds in the surrounding region of free space, regarding electric current deml'nt~ tilt, field sources. The external fields of aperture antennas such as the electromagnetic horn or 11-1 (b), can similarly be found from the integration over the electric currents, these occurring mainly on the inner conductive surfaces of the horn~ The radiation fields of parabolic reflectors might be obtained in the same way. However, bt:cause the irregularly shaped surfaces occupied by the currents are generally not in one of the common coordinate systems, the integration could be a tedious For aperture antennas such as horns and reflecting paraboloids, it is simpler more natural to regard the electromagnetic fields in the aperture plane as the sources of the exterior fields. This section describes a field equivalence theorem developed by Schelkunoff-3 that puts this idea to use. A brief discussion of the history of this subject, which had its beginnings in the theory of optical diffraction, is related first to help provide some additional insight into this problem. The radiation fields of aperture antennas may be explained by adopting a point of view not unlike that proposed by the Dutch physicist Christian Huygens in 1678. He regarded each differential surface-element of a wavefront in an electromagnetic (optical) aperture as the source of a spherical wavelet disturbance, the total effect of which, at any field point further along, was simply regarded as the phased superposition, or sum, of all the spherical wave contributions there. 4 This approach was 3S. k Schelkunoff, "Some equivalence theorems of electromagnetics and their application to radiation problems," Bell Syst. Tech. Jour., 15, January 1936, pp. 92~ 112. 4In this regard, one may sec that the integral (11-17) for the vector potential field ora system of current sources represents nothing more than a superposition of elementary spherical wave functions, e ~ jPoR / R, proportional in strength to the current sources J dv', and summed up (integrated) in proper phase at the field point P of Figure 11-2. The Huygens picture differs, however, in that it deals only with scalar wave phenomena.

564


refined later by Augustin Fresnel (1788-1827). The Huygens-·Fresnel principle was placed on a firm mathematical basis for acoustical wave phenomena by H. von Helmholtz in 1859 and for optical waves by G. Kirchhoff in 1882. Kirchhoff used essentially the Green's identity (2-92) to derive a field integral solution of the appropriate scalar wave equation. 5 A generalization of the Kirchhoff method, extended to the vector electromagnetic wave equation, was provided by Love 6 and subsequently extended by Stratton and Cbu. 7 In the latter method, the usc of potential fields is avoided by employing a vector version of Green's identity to yield direct integrals for the E and H fields in terms of electric and magnetic current and charge densities in an arbitrary volume region in free space. A vector field integral approach deduced earlier by Schelkunofr8 is entirely equivalcnt to the Stratton-Chu method, but it possesses the possible advantage of additional physical insight. SchelkunotT's method is to be described here and applied to problems involving aperture-type sources of electromagnetic radiation fields. It makes use offield potentials (both magnetic and electric vector potentials), and it incorporates a concept termedjield equivalence, involving the replacement of electric and magnetic fields on an arbitrary closed surface (principally on the aperture) with equivalent electric and magnetic currents and charges. The concept of field equivalence is seen to make it desirable to postulate fictitious magnetic charges and currents in the region in question.

A. Symmetric Maxwell's Equations clhd Their Vector Potentials Calling the magnetic charge density Pm and the magnetic current density 1m provides the following sJ1mmetricai f(mn of Maxwell's diflerential equations for free space, written here in complex time-harmonic form (11

(11-57 b)

VxE

(11-57 c) (11-57d)

Suppose diat the electric and magnetic fields fi = EoE and B ,uoH of the latter are resolved into the contributions fie = EoEe and Be = ,uoHe attributahle solely to the eJectric curzents and charges p" and j, plus the additional contributions = EoEm and Bm = ,uOHm related to only theJictitious magnetic currents and charges Pm and 1m in the region. The total fields in (11-57) can then be written as superpositions of the two sets of fields

Pm

(11-58) (II-59) SAn account of Kirchhoff's method is given in M. Born, and E. VI/olf. Principles of Optics. Elmsi(ml, N.Y.: Pergamon, 1964, pp. 375-382. 6A. E. H. Love, "The integration of the equations of propagation ofdcctric waves," Phil. Trans., (A), 197, 1901, p. 45.

7J. A. Stratton, and 8SeheikunoJf, op. cit.

L.J.

Chu. "Diffraction theory of electromagnetic waves," P~ys. Rev., 56,1939, p.99.

EQUATlONS AND THEIR VECTOR POTENTIALS

Maxwell equations (I as follows

v-n

V~

(e)

V

V-B m =pA m v x Em = - jm -

V

V x Hm

V

into two groups,

=0

m

565

jWEoEm

(f) ( 11-60)

jWfloHm

(g)

(h)

yield simply (11-57) again. Equations equations, identical with (3-24, 3-48, and magnetic fields associated with distri$.hjU,w'~Ml)rdensitiesj and PV' Equations (e) through (h) fields de;::cloped in a region presuming densities 1m and Pm are also present. In electric and magnetic, the superposed satisfy the Maxwell relations (11-57). (11-60) (a) through (d) has already vector magnetic potential A. Recall the (11-14), and (11-12) in this regard

VxA

(11-61 ) (11-62)

V
V-A

in which
(11-63)

A (11-64 )

(11-65 ) It was shown

is an integral of the wave equation (11-65)

r floj(r')eJv

j JoR /

dzi

(11-66)

4nR

The additional fictitious magnetic current and charge distributions in the frete-space Ill' deducted by analogy for the systtem of equations (11-60), (e) through (h), on the dualities that texist between pairs of quantities in the first and second columns of Maxwell equations (11-60). Analogous pairs of tequations are

566

RADIATION FROM ANTENNAS IN FREe SPACE

(a) and (f), (b) and (e), (c) and (h), and

and

, so there exist the j()Jlowing dualities

Ee is the dual ofH m

He is the dual of

Em

Eo is the dual of flo flo

is the dual of Eo

p" is the dual of Pm j is the dual orjm

,I

(11-67)

Since the fields Eeynd He are related to the vector rnagnetic potential A anci the sca~u' electric potential through (11-61) and (11-62), the c()rreSpond~lg fields Hm and Em are analogously connected to the so-called vector electric potential, F, and scalar 0/, such that

A is the dual ofF
of\f1

A substitution of the quantities in the second columns of (11-67) and (II duals in (11-61) through (1/-65) leads to the potential relations

:Om =

V X

I'

( 11-(8)

{c)r

their

(11-69) ( 11-70)

Vol' (/1-71 )

in which

0/

and

I' satisfy

wave equations analogous with (1 I-(4) and (II

(II (II 'I:hen the ve£tor electric potential dlstriQution lm(r') at P(r) is

I'

produced by the electric current density source

( 11-74) analogous with (11-G6), " Jl rom }he f()!'egoing it is seen that if both electric and magnetic cu!rents of denSHIes 1 and 1m exist simultaneously in a free-space region, the total field E produced at any field point P(r) hecomes the sum (11-58) of Ee and Em;from (II-62) added to

EQtJATIONS AND THEIR VECTOR POTENTIALS

567

(1

_ v<1> - jwA -

1 V X Eo

F

1 ~ ---VxF

(11-75)

Eo

from (11-61) and (11-70) substituted into

1

~

- jwF + - V

~

(11-76)

X A

flo

flJr once :E (r) has been found using ( 11-75) version of Maxwell curl relation (11-57 c) ~ources to which the foregoing potential exmoreover, the electric and magnetic curlimn ofsUlJace densities and jsm) in which j

1.

_:~::.C-_-'-_ _

ds'

(11-77)

( 11-78)

theorem described in the following.

The latter

theorem to be described requires the associated with the symmetric Maxwell of the latter, in real-time form, are

use of the equations (

c

(11-79a)

(11-79b)

11", Wb

_ i Jm'

Js

H'

i

Js

ds -

J' ds +

~ i B • ds V dt Js

diD' ds A dt Js

(I1-79c)

(11-79d)

568


The corresponding boundary conditions are derived by the methods of Chapter 3, using the devices of the Gaussian pillbox or tbe thin, dosed rectangle constructions typified in Figures 3-4 and 3-10. The results are ( 11-80a) n' n X J,

=

Psm Wbjm 2

(II-80b)

(El - E2l =

-Ism Vjm

( 11-80c)

(B 1

-

Bll

'

H

I'

f

2)

=l,Ajm

( 1I-80d)

in which n denotes as usual the normal unit vector directed frorn region 2 into region 1. A comparison oC (II-80b) and (II-80c) with (3-50) and (3-79) reveals the additional effects of Iictitious magnetic sur/iice charge aod surfilce current densities Psm and Jsm at an interface. The fieJd equivalence theorem is concerned with a boundary interface, to one side of whicb the fields are assumed nullilied. Witb the assurnption of nut! fields in region 2, the boundary conditions (I 1-80) specialize to n'

D[ = Ps C/m2

n . Bl

-n X E J

= Psm Wbjm 2

Ism

Vjm

n X Hl = Js Aim

(11-81 a) ( 11-81 b)

( 11-81c) (II-8l d)

(11-81 a) and (11-81 b) slate that the normal components of electric and magn(~tic fields may undergo an abrupt jump to zero from region I to 2 only if a surt~lCe electric charge Ps and a surfilCC. magnetic charge (Jsm are present iri strength equal to the normal ])"1 and Enl componeptihrespeclivdy. According to (Il-Slc) and (II-BId), moreover, abrupt transitions limn finite values to zero, of the tangential components E~I and Iftl' are allowable only ifthc respective magnetic current and electric current surfa.ce densities Jsm and Is prevail at the inlerbce. Although the free magnetic charge and magnetic current densities of (I I-8Ic) and (ll-8ld) have not been proved to exist physically, they are an important mathematical concept in the field equivalence theorem. Suppose lirst that known distributions of electric current and charge densities exist in some portion of a £I'ee-space region, as in figure 11-12(a). '['he fields at P(r) could, as usual, be /(mnd by use of the potential integral (11-17), whence E and 8 follow from (11-l3) and (11-14). The fields at P(r) can also he obtained, however, Irom equivalent currents established over an arbitrary surface 8 1 enclosing all !.he sou r<.:es , as depicted in Figure 11-12 ,the enclosed sources producing the fields El and HI on SI as shown. Suppose the sources inside 8 1 are consider-cd nullified, but the fields E1 and 8 1 just outside 8[ are maintained at their previous values. This condition is mathematically allowable only if the four boundary conditions (II-Bl) are upheld, implying the sinwltaneous existence of electric and magnetic charge and current densities p" Psm' Jsm' and Js on the surface :2'1' as sugges~ed by f]gure 1 1-12(e); with the establishment of the equivalent sources Ism = - n X El al~.d Js =~n X HI on 8 1 , the integrals (11and (11-78) can be employed to find A and F at any field point P(r) in the source free-volume region V. )fhese potential solutions, inserted into (11-75) and (11-76), then yield E and 8 at

f(r).

Il-Cl SYMMETRIC MAXWELL'S EQUATIONS AJ';D THEIR VECTOR POTENTIALS

(a)

(b)

F(r) P(r)

R

(e)

A(r)

(d)

FIGURE 11-12, Development of the field equivalence principle, Arbitrary doss,d sl!..rface S 5\ + S2 excl':!,dcs sourceS from ~cgio!, V. The sources establish Ej> Hi> on SI' (c) Sources J assumed nullified in 8 1 with E I , H at P(r), (d) Application offield equivalence}o ~lectromagnetic hur!', !
1: 1 , HI, on SI' (b)

An illustration of this technique relates to Figure II-12(d), Shown is a rectangular horn f(~d Ii'om a rectangular waveguide carrying the dominant TE 10 mode, Assuming a reasonably small horn taper, the field distribution over the horn aperture differs negligibly {i'om that over the waveguide cross section, Then from assumed null fields imide the surJ;tce S\ just embracing the horn and feed system as shown, the aperture fields are replaced with equivalent current and charge surface densities over the aperture as given by (11-81). This field eq uivalence process is examined in the following example.

EXAMPLE 11·1. A pyramidal horn of aperture area ali is fed from a rectangular waveguide carrying the TE lo mode as in diagram (a). Find the equivalent electric and magnetic surface curren! distributions over the sUlface SI enclosing the horn and its f(~ed system. (Neglect fields OIl the exterior conducting surfaces of the horn, assuming the field distribution over the horn aperture to be essentially that in the waveguide cross section,) The of coordinates is assumed at the horn aperture center, as in The tangential in tlie horn aperture (at Z = 0) arc required; from the TEIO mode

570

RADIATION FROM ANTENNAS IN FREE SPACE I

£r__-

J. m flux

Surface 81

I(Y)

E I Yf1-'" ---_--...,..:.:..---,----r---n-..JJ++.J \ -_">--_L---I

E

r::::~~::::-::;;:;:r;o;:::-----::::~rtt

<..

_

enclosmg sources

:::=1"1

r<___

_/

1

'-- __

n-/

==,.; --(x)

-:;

-"" '\

1 I'I~~!

'\

1_ ",

r

I '"

== -n X E Vim

-r~~1 I '-t1J--L- I ,

-4-t"'-H,

n -tI ~I--'

I

't-..J-

J. ==

(a)

J.m

I In

J1tti< Lfj::r!-'-'- "'-J.

IH

,

(z)

r:~1

n)(

flux

H Aim

(b)

EXAMPLE 11-1. (a) Pyramidal horn, showing aperture fields and magnetic surface currents on St: J8m, J,<

expressions (8-62), the positive D+

L'y

z

fix (b) Equivalent electric <

traveling wave fields become

"+

0) = E y ,10 cos n a

0)=

Ey,

x

(I)

_ 1f;(x,O)

~ 'ITE,10

(2)

in which a denotes the horn width in the H plane as shown. (The cosine distributions are the result of placing 0 at the aperture center.) Thus, the equivalent magnetic and electric surface current densities over the aperture become, by use of (II-8Ic) and (1l-8Id)

(3)

1s = n

x

H1 = a z

X

a}i;(x, 0)

(4)

flux sketches of which are depicted in (b). You may verify what equivalent magnetic charge density fJsm exists over the aperture, using (II-8Ih).

The electromagnetic fields exterior to the boundary surface 5\ enclosing the electromagnetic horn in the previous example can now be obtained by use of the potential integrals (11-77) and (11-78) taJs..en oVt;[ the aperture equivalent source curren ts (3) and (4), from which the fields E and Hare fi)Und by usc of ( 11-75) and (11-76). This process is illustrated in the somewhat different example f()llowing, which considers the radiation (difIraction) fields of a rectangular aperture in an absorbing screen illuminated with a uniform plane wave. EXAMPLE 11·2. Suppose a rectangular aperture of dimensions a and b is cut into a thin, flat, perfectly absorbing (black) screen excited with a uniform, plane wave as in the accompanying figure. Assume that the fields are perfectly absorbed (without reflection) cvery-

11-5 SYMMETRIC MAXWELL'S EQUATIONS AND THEIR VECTOR POTENTIALS

571

where on the screen except in the aperture, and that the tangential tields are zero on the screen just to the right of it (at Z = 0 + ). (a) Find the equivalent surface current distributions over the elosed surface SI (consisting of the plane ,t 0+ located just to the right of the screen and a hemisphere of indefinitely large radius encompassing the entire righthalf space). (b) Evaluate Ar) and F(rL at anyJarzone field point l'(r) in the region z > O. Use spherical coordinates. (c) Derive E and H Irom the potentials at P(r) in the farzone. (d) Sketch the farzone diffraction field patterns in the principal planes 1> 0 and 1> n/2, if a = 5Ao and b = lOAD· A+ Ex

I (x)

I I

r--Thin. black screen

1""",'"',

.

.+

A+ Hy

aperture -----_._.-

(y)-

(z)

a

-----

~

Black screen

(/J.o, fO)

(b)

(a)

($ = 0)

/ P(r) (field poi

----- --------.

'J -----...

= ,'.m =0 on black screen A

\\\, - ------J \'

____...

\

($ =.-)

(c)

-~

r'~ I

----1 EXAM.PLE 11-2. (a) Edge view of aperture in black screen. (b) Showing uniform plane wave Equivalent source currents in the aperture and field point field in the rectangular ~perture. geometry in spherical coordinates. (d) Graphic constructions leading to the normalized field pattern of a rectangular aperture, in the I/! = 0° principal plane. (e) The field pattern of a Side lobes are shown only in the principal planes. rectangular, 5Ao by lOA"

572


--,r-_ _ _ _ _ _+_~---

. . \:' I

\ \ \ \ \

\\ . \

\

\

\

,

I I

""

\:

6 ',,'----_

(

~

,..)

" / ",

\

.~

( = 0)

1r)

U versus II (U= 57r sin 9)

I

)/904

80'2

001 = 11'

8--

0

(1)

/

\1I \1-/ )/ 803

__1--

--

I

\, I1

,

___o_-u--;..

Rectangular plot of farzone pattern (solid curve)

I

0

( = 0)

~002 ~

.' / / /

Incident ------>wave

a =5

0)

(

804 003

" 1 1 0 \ "" "01 =

h{~~=----7-n---n

~ot

Black screen1

Polar of farzone pattern: (

1

+ cos 2

0) I

Sin (5,.. Sin

5rrsinO

(1) 0)

~

0.".)

I

Cd)

= O· prinyiPat-plane pattern

I

--:--:.-__._/,,:://-01 -- (z) --~ --~ r = constant .

(e)

573

EQUATIONS AND THEIR VECTOR

m

. _ . " (~ol1ditions (ll-Sle) and (1 0) become

(1)

DxH 1 =a z

g+

xa---'"~ y 110

The geometry in figure (e) shows P(r) in a with the origin at the center of the aperture. The the source poiut P' (r') is approximated (3)

siu 4»

r - sin O(x' cos 4>

and r in (3) is significant in the phase exponent of the bnt negligible in the denominator provided P(r) is a becomes

• •

";+

-a/2

- lloa x E m e~.ipo[r- sin e(x' cos q,+y' sin qI)] dx' dy' 4n11of

e- jpor sin U ~Ln V

,

U

(4)

V

(S)

flob. . -_ .. sm () sm 4>

(6)

2

foE;;; ab 4nr

(II

e

_ 'PoP sin U sin V J ----~-[f V

(7)

usc of (11-7S) into which (4) and (7) arc subwith f no faster than lir are retained, the result sin U sin V

U

V

lae cos 4>

a,p sin

4>]

(11-32)

components i£e and R,p in the farzone. the free-space version of (ll-S 7c). If only the u·',lThmeu. the results an~ irn.n.Wo'"''

(II thus related by the intrinsic wave impedance 110

574


(d) Graphs of the farzone field pattern of the rectangular aperture of figurc (bJ arc usually desired in its "principal planes," the two symmetry planes that include the z axis and that slice normally through the aperture. Thus, the complete verticalfJrincipal filane of the aperture, as shown in figure (c), is defined by the = 0 and the 1) n semi-infinite planes, with the polar angle 0 having the nonnegative range 0 :s:; 0 ~ 0 90 in each. The taxzone E-fielel in that vertical plane is given by thc single expression

E(r, 0,
jfJoR~.. abe- J'p or

~ ao~~-

[(I + 0) cos

2m

sin 0)

sin

.~ ..-~-

----~~~

2

5n sin 0

J

( 11-(4)

sincc from (5) and (6),

U(O,O)

= 5n sin 0,

v=O

5n sin 0,

U(O, n)

(8)

in that complete vertical principal plane, to yield (11-84). jHlttem in the 'I 'he bracketed (actor of ( 11-84) is termed the complete vertical principal plane. In Jigure (d) is shown a graph of just the filctor lJ)/U plotted versus U. From (8) it is evidcnt, since 0 has the range of (0,90 0 ) within the semi-infinite planes offigmc (e), that the visible range of U in this graph extends over ( 5n, 5n). In the diagram to the right in (d) is shown the graph of the normalized field pattern plotted versus 0, with the curves to the right and to the kft of the origin o corresponding to the upper and lower semi-infinite planes 0 and n, respectively, in figure (c). There is a slight tapering effect of the factor (I + cos 0)/2, called the Huygensfactor, which has the di('ct of reducing the side-lobe amplitudes to values somewhat below those values predicted from the basic (sin lJ)/U function. The fidd pattern in the other complete horizontal principal plane, defined by = n/2 and = 3n/2, is analyzed in a similar manner. The aperture width b = lOA o applies in this case, yielding 18 side lobes (instead of the eight obtained in the (p 0 principal plane). The beamwidth of the principal beam of such an aperture is usually defined as the angular width meawred between the amplitude points. Since (sin U)/U has the value 0,707 at U o 1.39, one can write an expression lilr beam width (20 0 ) in a prim'ipal plane of the copbasal, unifbrmly illuminated rectangular aperture ill the f(lrln 2 sin -1 0.443

(~; )

rad

(11-85)

If the aperture width a is sufficiently large compared to Ao , (11-85) is approximated by

a

rad

~

Ao

50 -- deg

a

(11-86)

from which it is seen that an aperture width a = IOAo produces a beamwidth 20 0 = 5°, whereas a = 100A o pro~des an associated beamwidth 20 0 = 0.5°, and so forth. This result shows that the beaJ11width in the principal plane of a unifbrmly cophasally excited aperture is inversely proportional to the aperture width measured in that principal plane.

Another important characteristic of the diffraction patteI'll of an aperture source is the relative strength of its side lobes in relation to the level of the principal beam. The (sin U)/U diffraction patteI'll of the uniformly illuminated case treated in the preceding example and shown in figure (d) l1as a first side-lobe level that is 2 L 7% of the mainbeam maximum, or about 13 dB doWfi. The side-lobe level achieved in a

575

11-6 AN'I'ENNA DiRECTIVE GAIN

TABLE 11-1. Field-Pattern Characteristics of Large AperturesQ

Type of aperture illumination

Pattern beamwidth (deg)

Sketch of amplitude Rectangular aperture (a

=

width) 50° a/.A o

(1) Uniform

~

(2) Cosinusoidal

~

68° a/.A o

(3) Cosine squared

~

82° a/.A o diameter) 58° D/.Ao

~a----l

Circular aperture (D

(4) Uniform

~

(5) Parabolic

~

(6) Parabolic squared

~

a

L---D--1

=

First sidelobe level (dB)

13.2 23 32

17.6

72° D/.Ao

24.6

84° Dj.A o

30.6

By large aperture is meant one whose principal dimensions are large compared to

"0.

given aperture antenna design is dependent on the functional nature of the aperture field distribution. In particular, if the aperture field excitation remains cophasal but has a cosinusoidally tapered amplitude over the aperture, the first side lobe will be 23 dB below the mainbeam maximum, or 10 dB lower than that obtained with uniform aperture excitation. Side lobe suppression by means of aperture illumination tapering is usually obtained, however, only at the expense of an increase in the beamwidth. For example, a cophasally and uniformly excited aperture of lOA width has a beamwidth of about 5" in the principal plane that includes that width. With cosinusoidal tapering, the beamwidth increases to about 6.8°. A sumrnary of the effects that aperture amplitude tapering has on beamwidth and side-lobe level is given in Table 11-1 for the cases of rectangular apertures and axially symmetr'ically excited circular apertures. The proof of these results can be established in the same manner as described in Example 11-2 for the case of a cophasally and uniformly illuminated rectangular aperture. The equivalent sources over a horn aperture with a cosinusoidal tangential field distribution in one dimension has been consider'ed in Example 11-1. The char'acteristics of its farzone field pattern in the wineipal x-z plane are found in (2) of Table II-I. Additional details of the circular-aperture diffraction problem are found on p. 192 of the book by Silver listed in the refi'Tences.

11-6 ANTENNA DIRECTIVE GAIN To reduce transmitter power requirements in point-to-point transmit-receive communication systems, it is advantageous to use an antenna that will direct most of its radiated power within a relatively small solid angle. Figure 11-11 shows how the

576

RADIATION I'ROM ANTENNAS IN FREE SPACE

angular field patterns of simple axially symmetric linear antennas of diflerent lengths rndnage to enhance their radiated power densities in certain O-directions, while reducing them in other directions. This property of power-density enhancement, through the proper design of the antenna or antenna system, is even more pronounced in systems of antennas called arTl~Ys, as well as in aperture antennas sllch as horns or reflectors. Such antennas are generally capable of enhancing the power densities relative to both spherical coordinate angles 0 and 4>, thereby concentrating the radiated power density in a desired direction even further in some point-to-point communication link. A property of an antenna indicating how effectively its radiation pattern concentrates its power density in a given angular direction (0',4>') is known as its directive gain, denoted by D(O', 4>'). The directive gain of a given antenna is defined as the ratio of the power density fYJaAr, 0' , ') at some distance r, to the total power radiated by the antenna averaged over the surrounding sphere S of area 4nr2; or

D(O', 4>')

fYuv(r, 0', 4>') 1

(11-87 a)

J, £!Pav • ds ~

This expression can also be written

D(O', 4>') Power radiated

antenna

Power radiated by given antenna

(11-87b)

suggesting an alternative interpretation of directive gain. The numerator of (11-87b) is seen to denote the power radiated from a fictitious isotropic antenna comparison antenna), defined to radiate the same power density 'O»av(r, (J', (/>,) in all diredions (its power-density pattern is a perfect sphere), with Of, 4>') denoting the power density of the given antenna in the particular direction (0',4>') for which its directive gain is being defined. To illustrate this concept, Figure 11-13(a) shows how the directive gain of the half:wave linear dipole is obtained by use or (II-87b). Its axially symmetric power density pattern is independent of 4>, shown as the pattern H (in sectional view) in the figure, so suppose that one desires the directive gain D(O') in the particular direction 0= ()' shown. The spherical power density pattern of the comparison isotropic antenna, labeled f, is drawn so that its power density at 0., with the power density ;J}>av(r,O') of the given dipole. Then, as given by (II-87b), the ratio of the radiated power 4nr2;J}>av(r, 0') of the isotropic source to that radiated by the given dipole, Pay = ~s 'o/>av • ds, yields the desired directive gain D(()') in the direction 0'. The radius r denotes the radius of the integration sphere S', with r arbitrarily located in the f;irzone region of the given antenna. ]\Tore commonly, the directive gain of the given antenna is desired in the direction of its maximum power density, since this is the direction in which the maximum possible power can be captured by a distant receiving antenna. I n this case, the coincidence point of the power density patterns of the given antenna and the comparison

11-6 ANTENNA DIRECTIVE GAIN

577

Sphere S /'

I (z)

/'

/'

/

/

"\ \

I

/

/

/

\

/

!

I

I

I

\

\

\

\

\ \

(b)

(a)

antenna, (a) the direction 110 of its dipole; I is the power

It)r the example of a half-wave linear direction II and (b) the Ii is the power density field isotropic antenna, both at the distance r.

isotropic antenna yielding for lht'

0) of that given antenna maximum, the given antenna

FIGURE 11-13.

(II-87c)

in which

maximum of its power density pattern at a fixed Figure II-B(h) /()r the special case of the in the following example,

gain of the half,wave dipole antenna. maximum of the POW!'f density pattern of the half: souree is shown at (Lo, where 00 = 90°, as in density rJ'a,(r, ()o) in (11-El7c), at the pattern (II-52), with flol = 90° (rjJ being absent becanse fixed range r, (II-52) yields the maximum power

[COS (90" 90u~J2= 151~

(I)

Sill

use of (II-55) m the denominator of (11-87c) 4nl2 (151~/nr2) 36.51~ 9 Maximum

1.64

(2)

578


the desired answer, showing that the comparison isotropic source of:Figure 11-13(b) would need to be driven with 1.64 times as much power as the given half-wave dipole to produce the same power density &'av at 90 0 • [Note that (2) can also be evaluated usiug the radiation resistance Read defined in (II if Pay !RradI; = t(73)1; = 36.51; is employed in (11-87e).] Expressed in decibels, (2) can be written

eo

D(900)dB

= =

10 log [D(900)]

lO log 1.64

= 2.15 dB

(3)

A second way of expressing the directive gain D(O',
dPav = W'av· ds = PJ>av(r, 0,

sin OdOd
With this, the directive gain (II-87b) becomes, in terms of the time-average radiation in tensi ty (1l-88) I , 4n$(O',
(11-89)

This is also wri tten

4n$(O' A.,) D( 0',
( 11-90a)

sinee sin 0 dO d
(11-90b)

in terms of the maximum radiation intensity $(0 0 , 4)0)' The directive gain results (11-87) and (11-90) can be written in a third form, on noting from (11-52), the time-average power dellsity expression for the example of the linear center-f{~d dipole of arbitrary length 21, that the power density PJ>av, and therefore also the radiation intensity $ defined by (11-88), are quantities that are proportional to the square of the electric and magnetic field-pattern factor F(O) defined by (11-51). For antennas in general, the field pattern factor F will be a function of both . I

ll-i TRANSMIT-RECEIVE SYSTEMS, RECEIVING ANTENNA

579

:(z)

}<"'2(0')

(b)

(a)

FIGURE 11-14. Graphical basis for finding, for the example of the half-wave linear dipole, (a) the directive gain D(O') in the arbitrary direction ()' and (b) the directivity D(Ool. Only the squared held pattern 1"2(0) is used. H denotes the halfwave dipolc pattern; J, the comparison isotropic source pattern.

eand ¢, denoted henceforth by F(O, ¢). Since both the numerator and denominator of (11-87) and of (11-90) are proportional to F2(O, ¢) through common proportionality constants that cancel out, one can write the directive gain (l1-87b) or (lI-90a) in the general form

D(O' A,') , 0/

4nF2 (e' A.') = ___ ,_0/_

fs F2 dQ

(11-9Ia)

if dO. denotes the usual differential solid angle dO. = sin 0 dO d¢ on the sphere S or integration. The directivity, or maximum directive gain, of an antenna having the field pattern factor F(fJ, (1)), therefore becomes

(Il-91b)

with F(Oo, ¢o) denoting the maximum of the radiated electric or magnetic field pattern factor F(O,
11·7 TRANSMIT-RECEIVE SYSTEMS: RECEMNG ANTENNA The electric or magnetic field pattern, power density pattern, directive gain, and input impedance are important characteristics of an antenna when connected to a signal generator and used as a transmitter of radiated power; but they are of equal importance when the antenna is used as a receiving antenna, that is, the receptor of a very small {i'action of the radiated power of a remote transmitting antenna. In the faI'Zone region of

580


a transmitting antenna, the radiation field arriving at the location of a receiving antenna is an essentially uniform plane wave. To maximize the power accepted fi'om this wave by the receiving antenlla and its attached circuit, that antenna must be oriented (or polarized) in relation to the polarization of the arriving wave (see Section 2-11), whereas the attached circuit must provide a matched load to the receiving antenna if a maximum power transfer to it is to occur. Using the principle of reciprocity in connection with the equivalent circuit used to model the behavior orthe transmitting antenna, considered to be extremely weakly coupled to the remote receiving antenna circlli!, the l()lIowing comparative properties of transmitting and receiving antennas can be proved.

1. The effective receiving area (or "power-acceptance area") of the receiving antenna is shown in the following to be proportional, thr!mgh the universal constant A~/4n, to the directive defined fell' that antenna when transmitting. 2. The equivalent internal (Thcvenin) impenance of an antenna, when used as a reeeiving antenna, is the same as its input, or terminal, impedance when used as a transmitting antenna. 3. The measured field pattern of' an antenna is the samc, whether it is being used as a transmitting antenna or as a receiving antenna. These comparable £('aturcs of any antenna, the direct consequence of the principle o/'reciprocity, mean simply that no essential distinctioIls need to be made between its important functions when acting as a transmitting antenna or as a receiving antenna. Affecting the relative signal picked up by a receiving antenna is its polarization relative to the wave arriving II·om the transmitting source. A discussion of this aspect of a transmit-receive link is considered first.

A. Antenna Poiarization 10 In general, the of an antenna, whet her transmitting or ITcelvmg, is taken to b(~ specified by the polarization of its radiation field when it is transmitting. In what /C)lIows, it is assumed that the transmitting and receiving antennas of any communications link are polarization-matched, and that the radiation field of the transmitting antenna is linearly polarized. An illustration of polarization matching is shown in Figure II 15(a) and ) depicting fell' simplicity a transmitting dipole I located a large distance T from receiving dipole 2. These dipoles arc r~olarizat~m-matched becaust~, if both were transmitting, their radiation electrie fields ~~Il and E 02 , as given by (11-50)) would then lie in a plane common to both dipoles (the plane of this paper). With the dipoles parallel to each other, oriented such thaI 0 01 = 90° and 0 02 = 90", and only dipole 1 transmitting as in Figure 11-15(a), tlle essentially uniform plane wave arriving at receiving dipole 2 has its electric fidd EOI aligned with dipole 2, thereby inducing a maximum voltage aeross its output load. Even if the dipoles were tilted in their common plane by the arbitrary amounts 8'1 and O~ relative to their separation distanee r as sh(~wn in Figure II-IS (a). they remain polarization-matched, al though the electric field Eo 1 arriving at the site of dipole 2 now induces less voltage across its load, due in part to the reduced value of the dipole I pattern factor FdO) given hy (11-51). A polarization mismatch would occur if the receiving dipole 2 of Figure II-IS (a), say, were tilted as shown in (c) of that figure by some angle r away from the plane of the paper (the polarization-mateh plane). Then the received signal would decrease lOScc Section 2-11 in Chapter 2 fbI' a discussion of wave polarization. A discussion or elliptical polarization is also given in R. E. Collin, and F. J. Zucker. Antenna Theo~y. Part I. New York: McCraw-Hill, 1969, p. 103.

11-7 TRANSMlT RECEIVE SYSTEMS: RECEIVING ANTENNA

581

FIGURE 11-15. A ft'e~-space transmittinf!; and receivinf!; dipole link. Two polarization matched systems: dipoles parallel-polarized wit h 0 90" for maximum signal; (b) with z-axes tilted by 02 0', and if1 relative to distance T. In (e), showing the depolarizing dfect of the tilt angle c.

from its previous polarization-matched value by the factor cos T, evident from only the component EOI cos T of the arriving plane wave now aligning itself with the receiving dipole. It is thus evident that a T = 90° polarization tilt or one of the antennas relative to the other would result in zero power captured by the receiving antenna, assuming a linearly polarized transmitted wave. If the transmitted wave were circularly or elliptically polarized, the receiving dipole would capture power for all angles of depolarization tilt,

T.

B. Effective Receiving Area To enable a quantitative assessment of the average power captured by a receiving antenna, an effective area is defined for it, denoted by Ae( 0, cp). The effective receiving area is dehned such that the power P avr removed by the receiving antenna from the incident plane wave and delivered to its attached (matched) load, is simply the product or Ae(O, cp) times the average power density £Ylavt of the wave arriving from a transmitting antenna, or (11-92)

582


Transmitting antenna

ReceIvIng antenna

FIGURE 11-16. A transmit-receive link, showing received power Pm interupted by a conceptual "effective receiving area,'~ Aero

This situation is illustrated in Figure II-Hi. It is later shown ill Part C that the effective area Ae(O, 4» of a receiving antenna is directly proportional to its directive gain D( 0, 4» when used as a transmitting antenna; hence, a dependence on the angles (0, 4» is noted in (11-92). It is important to realize that Ae is an area in concept only; it is in general not related to the antenna dimensions. A connection between the received power and the fi'ee-space distance r to the transmitting antenna is readily established. If the transmitting antenna, radiating P avt W, were for the moment postulated to be a fictitious alltellna, then its power density :c1Pavt at the range r in Figurc 11-16 would become Pavt/4rcr2 at the receiving antenna location. However, the actual transmiu\ng antenna, having the directive gain f),(O, 4» in the direction (0,4» of the receiving antenna, would increase its power density at the receiving antenna location by the amount of that directive gain, becoming (I 1-93)

Thus, the power P avr deliver'ed to the receiving antenna matched load becomes, from (11-92), (11-94) in which the positioning of the two antennas, relative to their common spherical radial distance r, is denoted by the subseripted positions (° 1,4>1) and (° 2,4>2)' A refinement of the result (10-94) is developed in the next part, in which an equivalent circuit of the transmit-receive link leads to some useful conclusions.

c.

The Transmit·Receive Antenna Link

A typical transmit -receive communication link in free space, generally employing different kinds of transmitting and receiving antennas, is depicted in Figure II-17(a). With the sinusoidal voltage Vi (at input port I) driving the input current 11 into the transmitting antenna I as sh2wn, the objective is to determine the average power P av2 accepted by the impedance ZL2 loading the receiving antenna 2. The latter is located

11-7 TRANSMIT RECEIVE SYSTEMS, RECF.lVING ANTENNA

583

Antenna 1

_ I

+

II

I

q;J~_1~~_~_1 rl I ;' " I I i / I

I I

I

I

I

iPort 1

(a)

Port 1: (1))

t2J

\ \ rc,

:

-

I

Port 11I

Vee2

i-+

I Port 11

,-' I

(e)

rv+

I

rc,

- -

= II Z3-

~

z:[:

+0= 0+ Z2

I

I

ZL2

Vz

iPart 2

I

I"

I

I I

I

-

4-

I

rc,

V''

-'

Port 2:

(if)

FIGURE 11-17, A transmit receive link, (a) Transmitting, receiving network representation, (c) Simplification for large antennas, (b) Equivalent separatioll, (d) Reciprocal version of (c),

in the firzone region of antenna I, r meters distant, such that the arriving wave is l~cally an essentially uniform p~ne wave there, producing the current 12 in the load ZL2, across which the voltage V2 (at output port 2) is thereby developed as shown. It is sufficient to represent the system of Figure 11-17 (a) by means of a two-port (four-terminal) network; for example, by use of an equivalent 'It or T network. The latter equivalence is chosen as shown in Figure 11-17 (b), with corresponding voltages and currents at the input port 1 and output port 2 as noted. The mesh-voltage equations of Figure 11-17 (b) are (11-95a) (1l-95b) with

Z3 being the common (mutual) coupling between the two meshes.

If the separation r between the antennas I and'J. of Figure 11-17 (al is presumed sufficiently large, then the mutual coupling element becomes very small compared

<3

584


ZI

Z2'

to and thereby enabling the partial d~sociation of the transmit and receive meshes as suggested by Figure 11-1 7 (c). Wi th <:"3 very small, (11-95) can be rewritten (11-96a)

:7)72 + '\.,L

(11-96b)

in which the Z3!2 term of (11-95b) is discarded because 72 «71 ; while Z)1 is retained in (11-96b) since this term amounts to an equivalent, open-circuit (Tht~vellin) driving voltage, <:"311 =~Voc2' required to drive the l'Cceiving antenna current 12 into the attached load <:"L2' Knowing the _value of ,(3 would evidently enable finding the received power reaching th(~ load ,(1.2' This is accomplished as follows. From the simple, series-circuit (Thevenin) equivalences of the separated transmitting antenna and recei'::.ing antenna circuits of Figure II 17(£'), it is evident~that (a) The impedance ,(1 is the transmitting-antenna terminal impedance <:"1 Rl + jX 1 seen by the driving terminal voltage Vl and responsible for the power radiated by that antenna, given by

1>avl (b) The irnpedance

-

I

-"2

in the

Re [I> 1-*] - 112R "' 1 1 - 2" 1 1

( 11-(7)

~quivalent

series circuit of the receiving antenna of that~antenna. On adjusting the load impedance so~tbat i~absorbs maximum power from Vue making it the complex cgnjugate Of,(2' or <:"1, R2 jX2 ), then the receiving antenna currcnt becomes 1L Voc l/2R2' yielding the power absorber! by the receiving antenna load

2 is simply the terminal impedance <:"2

R2

+ jX2

I

( 11-98)

The ratio of the re('(~iving-antenna load power to the transmitted power transmit-receive link thus bccomes, from (11 and (11-98)

III

this

(11-99)

This ratio should be compared with that given by (11-94) antenna effective area, expressed here as

Hl

terms of the receiving

[11-94]

in which the dependence of ])1 and Ae2 on the angular orientations (0,
585

11-7 TRAC:SMIT RECE1VE SYSTEMS, RECEIVI1'
the roles of the transmitting and receiving antennas of Figure 11-17 that is, replacing with the load impedance and with the d~ivillg ~oltage and assuming that receiving antenna 1 is now conjugate-matched (ZLl Zf), then Figure 11-17 (d) becomes applicable, and a result analogous with (11-IOOa) is obtained.

Zu,

t:\

Zu

(11-100b)

From their equality is obtained the simple ratio ])\ =

~~

1)2

Ae2

(11-101)

showing tbat, klr any antenna whatsoever, the directive gain D(e,4» is proportional to its effective receiving area Ae(O, 4»)· One can therefore write

(11-102)

A,(O, q» =KJ)(O, cp)

To evaluate the universal constant K in (11-102), the simplest example of mentary dipole receiving antenna is considered.

th(~

ele-

EXAMPlE 11-4. Find the maximum crfective area of the elementary dipole of Figure 11-6, when acting as a receiving antenna. Use this to deduce the universal constant K in (ll-l02). Assume, at the origin as in figure (a), a dipole oflcngth I{Z with a small gap, the terminals of which are connectn\ to it conjugate-matched load Zl., = R rad jX, as noted ill the circuit of (b), the equivalent series receiving-antenna circuit inferred il'om To obtain a maximum effective receiving area, the E field lllust be aligned Figure I with the conductor, 1e,! the arriving unitCmll plane wave be as Hl (Il), with the components , U;) given the complex = E~e'

~+ H

jPoY

x

=

Em"

jPoY

(1)

110

coming from a remote transmitter along the ,y-axis, From

, this plane wave has the

time-average y-directed power density 1

f} av

~

= - Re [E 2.

X

~ (1':,;, H*J = a ---v 2110

(b)

EXAMPLE 11-4

(2)

586


The open-;:ircuit voltage Voe developed at the dipole gap by the incident field at z = 0 is simply Voc E;' dz. With a matched load, the load current magnitude becomes

E;' dz 2Rrad

Voc 2Rrad

(3)

The radiation resistance Rrad of the dementary dipole is obtained by equating its definition (11-.54), Pay =!zRradl~, to the di.pole radiated power (11-37), obtaining (Elementary dipole) Then the power absorbed by the attached load valent circuit (b), becomes by use of (11-103)

(11-103)

ZL2 in Figurc (a), deduced from its equi16n170

By use of (11-92) and (2), the received power can also be expressed in terms of the (maximum) effective receiving area Ae as (.5) whence equating (4) and (.5) obtains the maximum effective area (Elementary dipole)

(11-104)

It is remarkable that, despite the infinitesimal size of the dipole, its eRective receiving area is finite. The maximum directive gain of the elementary dipole can be shown, from ( 11-91 b), to be D(Oo)

)

D(900)

1..5

(Elementary dipole)

( 11-10.5)

whence, from (t 1-102), the universal constant K, correct for any antenna, is

3Ail/8n

K

1..5

4n

(6)

Thus the relationship (11-102), enabling finding the eRective area of any antenna from its directive gain, is seen to become

Ae(O,
=

A2 4~ D(O,
(For any antenna)

(11-106)

Relative to the transmit-receive system of Figure 11-17, the insertion of (11-106) into (11-94) now produces the ratio of the received to the transmitted power expressed in terms of the antenna directive gains

(1l-107)

11-7 TRANSMIT-RECEIVE SYSTEMS: RECEIVING ANTENNA

587

a result called the Friis transmission formula ll . It assumes that both antennas are polarization-matched, that the receiving antenna load is matched, and it ignores the internal antenna losses (which have the effect of reducing the values of]) 1 and ])2)' The inverse-r 2 -dependence in (11-107) shows that, with the transmitted power Pay 1 fixed, the received power Pay2 is reduced by one-quarter each time the range r between the antennas is doubled (a 6-dB loss). It is also evident that the effect of increased range r on the loss of received power can be overcome by the use of antennas with higher gain, or by increasing the transmitter power P ay1 . EXAMPLE 11·5. In a

transmit~reeeive link, two identical rectangular horn antennas, operated at! = 3 GHz, have maximum directive gains of26 dB each and are located 5-km apart in free space, each directed toward the other and polarization-matched. The transmitting horn (1) is fed with 0.5-W average power at 3 GHz, and the receiving horn (2), using a waveguide-to-coaxial-line transition, is terminated in a matched 50-!! load. Find the received power that reaches the load and the load voltage. With power gain D expressed in decibels by D [dB1 \0 log D, the horn maximum directive gains become DI = /)2 = 1026110 = 400. The wavelength is Ao = 10 cm at 3 GHz, yielding Ii-om (I] - \07) the received power

0.5/400) • ;

2(-411:(5)0.1-\03-)2

0.20 {IW

(1)

Sinee antenna losses are ignored, the power P'V2 is absorbed by the 50-!! matched load of the receiver, with p. v2 Vi/2R L2 . Thus, the voltage at 3 GHz developed across the matched load becomes

V2

= .j2R LP' V2 = J2(50)O.20(10

6)

= 4.5 mY

(2)

EXAMPLE 11·6. Find the maximum directive gain (directivity) of the uniformly illuminated rectangular aperture antenna analyzed in Example 11-2. Make use of (11-87c) [1I-87c) From figure (e) in Example 11-12, the peak power density 9. v (r, 00 , 4>0), at a fixed farzone distance r, occurs along the z-axis wJlere (10 .... 0 and with 4>0 arbitrary (choose 4>0 = 0). By use of (11-82), the electric field E(r, 0, 0) at any farzone distance r along the z-axIs IS

(1) in which the values of sinc U and sine V in (11-82) become unity for 0 = 0, 4> = O. The corresponding magnetic field, from (11-83), is if", Eo/110' obtaining the maximum radiated power density needed in the numerator of (1I-87c)

rY-'.v(r, 0, 0)

I

= "2

-

Re [E

X

-

H*)

= a,

Ei 2110

=

(E~aW

a, - - ,-

211oAijr2

llH. T. Friis, "A, Note on a Simple Transmission .Formula," Proc. I.R.E., 34 (1946),

254~256.

(2)

588

RADlATlON FROM ANTENNAS IN FREE SPACE

"Rectangu lar aperture

EXAMPLE 11-6, of a unif,mnly illuminated aperture, '#'"v(r, 0,
with flo = In tbe denominator of (11 the total radiated power can be [(lUnd from the integration of the f~!rzone power density I/f'avlr, 0, cfJ) • ds over the cnclosing sph!,:f(' S of radius r as depicted th!': flgHfe,

~~,

Re [E(r, 0, cfJ)

X

H*(r, 0, cfJl • ds

.it and H of (I and (I yet to be inserted, The resulting integral is not simple; its numerical integration would be prohibitive without the aid ora computer. However, the radiated power Pay is much more {,)Und li'OIll the integration oU#'"v . ds taken over the a/Jerture, since from the Poynting theorem the total power Pay keding the aperture must he that radiated through the enclosing sphere S. Thus, with flP.v = az(F:;' on the aperture, the time-average power fi:Tding it becomes

with thehelds

)

r

JS(apcr)

,i.fPw ' ds •

f~(apef) [

Substitming (2) and (4) into (11

obtains the desired directivity

D(O, 0)

For example, a unif()rmiy illuminated

b = 5,1.0 has the directivity Dm = 4n(aj,1.0) is Dm [dB] 10 log Dm 28.0 dB.

(11-108)

aperture with dimensions a lOA o , 200n = 628, Expressed in decibels, this

PROBLEMS

589

the method of Example 11-4, one can similarly show, for the TEJO-modefied rectangular horn antenna illustrated in Example 11-1, that its directivity becomes D(O,O)

Dm

32ab _ 12 nAo

(~) 4n 2 ab n

( 11-109)

A comparison of the latter with the directivity of the uniformly illuminated rectangular aperture shows that the effect of the sinusoidally tapered (TEIO mode) illumination ill the aperture is such as to reduce the directivity by the factor 8/n 2 , or to about 81 %) of the directivity obtained fi'om a rectangular aperture of the same dimensions if it were uniformly illuminated. The directivity of the rectangular pyramidal horn is in practice somewhat less than that specified by (11-109), which assumes (ideally) that the field in the horn apertun:' is cophasal over the flat aperture sur!itce S. A bowed-out curvature of the cophasal surface in the aperture is usually inevitable in physical horn designs, resulting in a broadening of the main beam and a reduction in the directivity. Considerations of the effects of such phase deviations, li'om the idealized in-phase condition of the fields over the flat aperture suriiu:e, yields modified expressions for the directivity of the pyramidal horn considered ill more detail elsewhere. 12

REFERENCES JORDAN,

E. C., and K. G.

BALMAlN.

tflfllIl.."fleI.U.

~Vaves

and

2nd ed. Englewood

Cliffs, NJ.: Prentice Hall,IlJ6!l. RAMO,

S.,

J.

R. WHINNERY,

and T. V AN

DUZLR.

Fields and '"Valles in Communication Electronics,

2nd ed. New York: Wiley, 1984.

Microwave Antenna McUraw Hill, 1949.

SILVER. S.

and

Radiatinn

.rlf)l)1f1l0rv

SerifS, vol. 12, New York:

PROBLEMS

SECTION 11-1 11-1. Beginning with Ampere's law (11 show how (11-10) is obtained, whence supply details leading to (II-I I), the wave eqnation in terms of A. Make snbstitu tiolls appropriate to converting (11-11) to its tinw-harmonic lrJl'ln (11

APPENDIX C 11-2. Prove that the free-space Green's functioll, scalar wave equation , noting that R is specified

satisfies the homogeneous rectangular coordinates by (4-12).

SECTION 11-3 11-3. (a) Show, from geometrical detail added to Pin Fignre 11-4(b), that the expressions (1l-20) are correct. (b) Make use of the potential components (11-20) in the curl expressioll (J I-I) to show in detail that (11-21) is the magnetic field of an elementary dipole in free space. 12For example, see S. Silver, Microwave Antenna Theory McGraw-Hill, 1949, p. 587; or C. A. Balanis, Antenna Theory, 1982, p. 570.

Radiation Laboratocy Series, vol. 12, New York: and Design. New York: Harper & Row,

590


11-4. Shuw tbat the wave equation (11-16). 11-5.

side of (11

that of (1

by

and (11

Usc magnetic field result (11-21) in Maxwell's relation (11 the electric field of an elementary dipole in free space.

11-6.

Repeat Problem 11-5. bn! this time make use of (1

use

or (I 1-1)

and

todcrive in detail (11-25)

11-7.

From its complex time-harmonic form (11-21), derive in detail tile field expression (11-29) of the elementary electric-current dipole.

real-tim(~

magnetic

11-8. (a) Usc (11-30) to determine the distancc 1 fi'om an elementary dipole at which the inversc-r tcrm in (11-21) or (11 is 20 times as large as the invcrse-r 2 term, at the three frequencies: J MHz, 100 MHz, 10 GHz. (Assume here that (Jor 20 defines the "f~Hzone" region of the elementary dipole.) (b) Repeat (a), this time t{)r as that {Jor 2\) defines its "ncarzonc.") (c) Comment. on the dl<:ct of Jll the fil.rzoflc and fl(,arz(Jne of an dipole. Taking ratios of the magnitudes (a) or the inverse-I to the invcrsc-r 2 terms of ( I 1-26) and (b) of the inverse-r 2 to the inverse-r 3 terms of dEe in (] I and of dl~~ in (I show that results comparable to (11-30) arc obtained. Comment on terms are important in the nearwnc (r Ao), and in the farzone (I' Ao ).

11-9.

11-10.

Determine the rcal-time forms of the Elrzonc phasor (II dipule. your results with the f~lrZOf1(' terms in (I I

of the

and (II and (I

11-11.

cmTcnt-carrying dipole, with its length relaxed to the incremental carries the current j 5 A at the fi-cqucncy j 3()O M Hz. How At the phase distance :20 rad II'om this dipole, vt'rify (II in the " ~What distallce r (in m) is this? (Il) Find the Luzone electric and magnetic fields and dH,p at this same range in tl1<' broadside (0 90") direction. Determine the power al 1he loca t ion in .(b).

11-12. Prove ill<: time-average radiated power rcslli t (I I j()l" the elementary dipole, assuming a sufliciclltly sphere of integration that only the Euzonc ficlds (11 and (I I arc required. 11-13.

Prove (I terms in (11-:21) and (11

all

of

size, such that aft field

aT(' required.

11-14.

Assuming the same incremental dipole current excitation, and frequency as Problem II-II, determine its tolal radiated powcr. What valw, would the radiated power have if the dipole length were doubled? Halved? 1Il

SECTION 11-4 11-15.

Sketch the responding analytical (a) 2t 101B, (b) '2t =

and usc (11-4/1) to supply ('01'I()r thin-wire antennas of the t()llowing lengths: (d) '2t = Ao, (e) 2t

11-16. For the following thin-wire distrihutions II (b) t1 = (e)

()l!~("('ntcr-fC.d

antennas, sketch their the distributions /2 (d) (1

curren!

11-17. A circular thin-wire loop antenna, as by Figures 11 and 1I-8(b), has the cireumkrential length of 0.751 0 between the terminals of the applied generator. Sketch the standing-wave current distribution on the wire loop, as approximated the Cllrrent standing wave on the shorted transmission line of Figure 11-7(b). Express the current standing wave as a function of the angle 1> about the loop ccnter, assuming 4) = 0 at the current maximum M in Figure II What is the em~ct on the current distribution of making the loop length small (com pa red to

PROBLEMS

11-18. Beginning with (11 the center-led linear antenna

in detail the Figure 11-10.

f~uz()ne

electric Held

(\

591 for

11-19. Make use of (11-:) I) to calculate and plot t he polar Held pattern F( 0) of each of the /i)Uowing center-ted linear antennas: (a) hall~wave dipole, (0) five-eighths-wave dipole, (c) fullwave dipole. 11-20. Repeat Problem 11-19 Illr the lilllowillg antennas: (a) three-halves-wave dipole, (b) two-wavelength dipole. 11-21. For a half~wavc dipole in free space, usc (11 10 km. if the sinusoidal driving current amplitude is 5 there.

to find Filld

at the broadside range of the time-average power

11-22. For the hall~wavc dipole in free space, integrate (l 0-55) either graphically or by usc of a computer, to lind its radiation resistance. 11-23.

Repeat Problem 11-22 t(lr a three-halves-wave dipole in free space.

SECTION 11-5 11-24. For the plane-wavc-cxcited unii()J'mly illuminated rectangular aperture problem of Example 11-2, prove the lilfzonc vector magnrtic potential result (4) by carrying out a detailed integration. 11-25. Verify the filrzonc electric field result (11-82) for the uni/clfInly illuminated rectangular aperture, ii'om the substitution of the potential functions (4) and (7) into (11-75). tHirlt: use the coordinate lranshmnations (1-79) to express the poteutials A and F in the desired spherical coordinate system; If)r example, .4', = sin 0 cos rP, F~ = F~ sin 0 sin rP. On expanding V . A and V x F, discard all terms that decrease Elster than I/r ill the t;trzonc region.] 11-26. A unifrmnly illuminated rectangular aperture, as described in Example 11-2, has the dirn(']Jsiotls a and h Plot the normalized field pattern versus 0 in the rP () plane. Show (he plot in rectangular as well as polar limn, as suggested by . Find the locatiolls of the nulls 0 01 , . . . of the pattern, determine its \xamwidth plane. and its lirst side-lobe level in decibels beluw tile principal beam As figure (d) of Example 11-2, the dIt:ct of the slowly tapered can be ignored, except as () exceeds :ZO° or so.] 11-27.

Repeat Problem 11-26, this time for the

rP

90° principal plane.

11-28. Repeat Problem 11-26, in this case f(lr an aperture with the dimensions a h (a "slit aperture," producing a Jan-beam pattern). [Note: With the much-diminished dimension Il, in determining the normalized field pattcrn in the rP = 0 principal plane, the Huygcns factor must be cardlllly acounted {or. With this narrow aperture width, (11-8G) is no applicable. Beamwidth may be determined directly ii'om the pattern plot. I 11-29. Make usc or the equivalellt aperture surface currents , determined in Example 11-1 (assuming only the dominant 'I'E1O forward-traveling mode in the horn aperture shown) to derive the fiJliowing Euzone field results lilr the rectangular horn. Show that the vector electric and potentials at any farzone point become

592


AY ~

Fx

Jl/:

rI

P(r, 0, 1»

'! I

o

/

/

I I

/

/

/

(xl (y)

PROBLEM 11-29

in which A

/Job.

/3

2

3m

0

., Sill

t:p

(b) l;se thc latter potential solutions to derive the electric and rnagnetic fields in the jiuzone, ohtaining

~ 0, 1») = E(r,

[(ao.sm 1».('IJIO cos 0 + I ) + a,p cos 1> (#10 -- + cos 0)] 13 0

x

jfJOlC~,l 0

8

,

lio

.Ipor sin 15 cns A - - - - -,,-~:----

r

B

(n)2 - /1 2 2

(c) Compare the f~uzone fidd results !(n' the rectangular horn with those /i)Und in Example 11-2 for the unifi:mTl pJane~wave-cxcited rectangular aperture having the same dimensions a, b. Comment especially on the similarities or differences between the field patterns in the principal planes.

11-30. From (11-85) was dCTived in eHen a "5 and 10 rule" fiJI' a rectangular aperture illuminated unifi:lrlTliy in amplitude and phase. This rule implies that a 5)'0-wide aperture has a 10" heamwidth in that principal plane whereas making it lOAo reduces the bcamwidth to 5°, and so

PROBLEMS

593

on. Using the results compiled in Table 11-1 for other types of illumination but with the amplitude tapered in the manner shown) over rectangular and circular apertures, determine what comparable rule applies to the remaining cases in the table.

I.) se Table 11-1 to compare the beamwidths of the farzone field patterns of each of the 11-31. following circular apertures. Assume uniform aperture illuminations. (a) A parabolic dish having an aperture diameter of 10 wavelengths. (b) The same as (a), except 100 wavelengths diameter. (c) An optical laser having an aperture or5 mm diameter, transmitting red light at a wavelength of about 7000 A = 7 x 10 -7 m = 0.7 11m. (What is the aperture diameter in wavelengths?) (d) By what factor will each beamwidth in the foregoing increase, if the aperture illumination is assumed to taper parabolically in each case?

SECTION 11-6 11-32. dipole in

Employ (11-33,34) to show that the power density available from an infinitesimal broadside direction is

(1 dZ)2

[5I>,v(r,90 o ) = a, 110 - - , 2 2AoT

(1 )

Make usc of (11-37c) and tbc result (11-37) to prove that the maximum directive gain of an infinitesimal dipole is Dm D(900) = 1.5. 11-33. Use (11-33) and the result of Problem 11-32 (a) to find the expression for maximum radiation
SECTION 11-7 11-34. A particular hall~wave dipole, at 500 MHz, has the measured terminal impedance = 73 + 40 Q and is terminated in ZL = 73 - j40 Q. A distant source produces, at the site of dipole, a polarization-matched uniform plane wave of the knowll power density, 10 n\'Vjm 2 • Sketch this system, depicting the arriving wave and the terminated receiving antenna. \'Vhat amplitude E:' is associated with the arriving uniform plane wave? (b) Use the value of the directivity of this half~wavc dipole to determine its maximum efTeetive receiving area Aem at this frequency. Find the total time-average power accepted by this receiving dipole, based on (11-92). (e) Making usc of the equivalent (Th(~venin) receiving antenna circuit of Figure 1I-17(c) (sketch it), and with the received power of part (b) observed to be that power accepted by just the receiving-antenna load impedance, fmd the voltage developed across the load at this !J'equency. 11-35. Three difIcrent haH:wavc dipoles arc operated at the li'cqllencies (a) 100 MHz, (b) I GHz, (c) 10 (a-h. Find the physicallcngth (2l) of each antenna, and determine the maximum effective area of each. Provide a labeled sketch of each antenna, showing a sq uare surface superposed on each, the latter depicting the rdative size of the maximum cftcctive receiving area of each antenna in relation to the antenna length. 11-36. Use (11-103) and (11-\09) to obtain for the maximum eflective receiving areas oflhe aperture antenna of Examplc 11-:2, and o[the rectangular horn antenna sbown in Example II-I. Compare each result with the area of the appropriate aperture, and comment. 11-37. Two high-directivity aperture antennas used in a microwave communication link are separated 50 km, arc aligned tllr maximum directive gains, and polarization-matched. Each antenna has 30-dB at the operating liTquency f= 7 GHz. antenna losses. Use the Friis transmission tllflnuia (II to determine how much time-average transmitter 10- 8 W = 10 nW over power is required at this f1'cqucllcy to obtain the received power P'V2 this link. (Express P avl in rnW; and in dBm.) (b) Employ the equivalent receiving-antenna circuit of Figure 11-17 (c), assuming a 50-Q load connected to a matched voltage source (a series 50-Q equivaknt source resistance). Sketch this equivalent receiving circuit. With the received

594


It'zl

power of ]0 nW from part (a), find the voltage developed aeross the receiving antenna load at this frequency. (c) [f the range of this link were increased to 500 km (as in an earthto-satellite system, for example), what new value of transmitter power would be required to provide the same received power as in (a)? (Express P av1 in W; in dEm.)

11-38. The farzone electric and magnetic fields of a transmitting antenna obcy the impedance relationship E/ll = l1tb whence the transmitted time-average power density is expressed

4>W

IE(r, e, 2110

(ll-llO)

Equating this to (11-93), show that the t~lrzone electric field magnitude, in any direction (B, 4», can be expressed in terms of the total radiated power P avt and its directive gain D,(e, 4» as

(11-111)

11-39. A transmitting half-wave dipole radiates the to}al time-average power P avt 10 W. (a) Employ (11-111) to find the electric field magnitude at a 10-km distance from this dipole in its broadside plane (e = 90°). [Answer: 3.24 mV /ml (b) Compare the numerical answer to (a) with that obtained using the dipole electric-field expression (11-50), showing first, in symbolic terms, that

lEI

60 [2Pavt r R.

J1 12

(1l-112)

[Hint: Express the dipole current 1m in (II-50) in terms of the real power delivered to the antenna terminal impedance, Ra + jXa.l

_ - - - - - - - - - - - - - - - - - - - - - APPENDIX A

Oblique Incidence: Region 2 Conductive

A·i. REGION 2 SOLUTION BY ANALOGY Even if region 2 in Figure 6-15 were made a eonductive (lossy) instead of lossless dielectric, a wave analysis closely akin to that of Section 6-8e, is seen to apply. The difl'ercl1ces that arise are from the effeets of the added conductivity parameter (J 2· Thus, a complcx propagation constant ')'2 replacesjp2 in region 2, as defined by (3-88); while the complex intrinsic wav(~ impedance q2l as given by (3-99h), takes the place (~f the pure rca I 112. With these simple changes, the forms (6-67(") of the fields Et and H, transmitted into region 2 still apply, the ineident and reHected waves (6-67a) and (6-67b) in region I remaining unaltered in {()I'm. This modification or the earlier probl(~m applies to either polarization of the incident uniform plane wave. The analogy extends equally well to the boundary conditions (3-71) and (3-79) concerning the continuity of the total tangential fields at the interface between the regions. Then Snell's law equivalents of (6-81) and (6-82) are obtained ie)r this case with region 2 lossy, yielding (A-I) sin 0i sin Ot

with Ez seen to be the permittivity of 2 as defined by (3-103). The angle or incidence 0i in region 1 of Figure 6-15 is by definition a real angle, fe)!· which 0 ::::: 0i ::::: goo. The effect of the complex Ez in Snell's law of refraction (A-2a)

595

596

OBLIQUE INCIDENCE: REGION 2 CONDUCTIVE

is thus to make the quantity sin 0, c()mj!iex. Writing (A-2a) in the form

JW.J;;f; JWJfllEl

Y2

(1\-2b)

J/31

with the use of (3-88) and (3-89), allows expressing the complex sin Ot as _iLsinOi

sin Ot

(X2

(1\-3)

+ J/3z

in which the real attenuation and phase constants (X2 and found from (3-90a,b). The corresponding complex cos Ot =

"f

/32

of region 2 arc as usual sin 2 Ot is thus written

..J!"-

°i

/3i 5in 2 1+----2 ((X2

+ J/32)

(A-4a)

For convenience, write (1\-4a) in its complex rectangular form, denoted by COS

Ot = A

+ JB

(1\-4b)

These modifications, which account for the lossy condition of region 2, can now be applied to either field expression (6-75) or (6-77) in region 2 for the lossless parallelpolarization case. The simpler expression (6-77) is chosen, in view of its having only one (fly) component. Thus, with the phase factor J/32 in (6-77) replaced with Yz = (X2 + J/32' with sin 0, and cos Ot given by (A-3) and (A-4), and the complex ~2 inserted for 112, (6- 77) is seen to become ay

Ht

e -- (<2 + j{h)(x sin U, + z cos 0,)

A

112

'

Multiplying out the factors in the exponent yields (1\-5a)

(1\-5b) wherein the quantities

(xz,

fix,

and

/32 B ,

/3z,

defined by (1\-6)

signify, through (xz and /3z, both attenuation and phase variations of this wave normally away h·orn the interface. Furthermore, the wave-phase changes along the x-axis at the interface at .<; = 0 are seen to be in phase step with those occurring ill region I, ip view of the presence of precisely the same x-directed phase factor /3t sin 0; in the Hl expression (6-76).

A-\, REGION 2 SOLUTlON BY ANALOGY

597

I(~)

I Region 1

\ Region 2

I I

Equal-arnpli tude plane (b)

(<1)

FIGURE A-l. Showing the equiphase and equal-amplitude plam" of the nonuniform wave produced in regioIl 2: (a) as viewed along thc.y·axis; (b) seen in perspective.

Equiphase and equal-amplitude planes, discernible from a graphing of the wave (A-5), are of intcrest, leading to an example of nonunil()rrn plane waves. The equalamplitude planes are characterized by setting the real-exponential factor (a2A - f52 B )Z in (A-5a) equal to a constant, yielding simply ,z = constant surfaces that are parallel to the in ter[;tce, as shown in Figure A-I (a) _ The attenuation with Z in region 2 occurs at the rate nfthe ElCtor a2 A f52B neper/m, showing a depth of penetration 62 therein given by its reciprocal. Note, in general, that 62 diners from for a uniform plane wave. The equiphase planes in region 2 are defined hy setting the imaginary phaseexponent of (A-5) equal to a constant, or fixx + fizZ = constant, in which f5x and [J z , the x and Z components of the applicable phase constant of region 2, are given by (A-6). The normal to the equiphase planes is inclined at the angle!p denoted in Figure A-I (a). This angle is not the Ot of (lossless) region 2 depicted geometrically in Figure 6-16, since in the present example involving the lossy region 2, the angle Ot was absorbed into the cornplex angle manipulations involving Snell's law (A-3)_ The tilt angle !/I is evaluated from the equation of the equiphase planes: f5xx + {)zz = constant. Dividing it through by the phase-constant magnitude () = ((); + /3;) 1/2 obtains the standard li.mn of the equiphase plane

(A-h)

in which the coefficients ofx and.( are the direetion cosines cos A and cos C of (6-55). From the geometry of Figure A-I (a) it is evident that C = !/I, A 90° - "', and B = 90°, making (A-7a) or the f()rm

x sin ~I

+ z cos ~f

= ro

(A-7b)

598


A comparison with (A-7a) reveals that the tilt angle Ij; is given by

cos Ij;

(A-B)

*A·2. REGION 2 A GOOD CONDUCTOR A special case of the Part A-I is generated on making region 2 a "good conductor," implying that (J2/WE2 1. Then tile ,implificd expressions 112) j()r a l , fil' and 112 are applicable. The complex transmission angle 00 moreover, tends toward zero, since applying (A-3) to a good conductor obtains

(A-9)

becomes sutficicntly large. This means also that cos fJ t -+ 1; whcrc;,ts from (A-4), I and B -+ O. Then the expression 1(lt' the magnetic field Hz in region 2 reduces to the good approximation, fi)r region 2 a good conductor.

as

(J2

A

-+

(A-IO) It is usdi.ll to note li'OIl! (A-1O) that the {-directed phase constant fJ2 is much larger than the phase constant fJl sin ()i applicable along x, parallel to the iuterEice. This is evident from the ratio

very large)

III

At an air-conductor interfiH'e, with regioJ] 2 having a conductivity (J 2 of the order of 10 7 , even f(lr frequencies into the microwave range (fof the order of 10 10 Hz), the ratio (A-II) becomes no less than 3000 or so, becoming eVClllarger at [ower frequencies. From it is thus evident, with A -+ I, B -+ 0, and (fll/fJ2) --+ 0, that the tilt angle Ij; of the wave transmitted into region 2, as depicted ill Figure A-I, approaches essentially zero for any of incidence (Ji' This is also clear geometrically from Figure vVith the phase constant #2 ill conductive region 2 so much larger than 13 1 , the wavelength A2 is necessarily much smaller thall )'1 in region J. In satisfying the boundary conditions at the interfiln:, the waves in the two regions mallage to keep in phase step at the interfilce only if' there is severe refraction of the incident Ot toward a very small exit angle t/J. For region 2 a good eonductor, the reflected and tl1UlSmitled complex wave amplitudes, relative to a known incident wave amplitude E i , are of interest. To this

A-2. REGION 2 A GOOD CONDUCTOR

599

: (x) I

"

>~~~0\\\

,,-

"',, '0,

"

/~

\\

\

\ \\

U

~

I~·. o,:~:; -" ~

\. \\

~~~::'{\\\\ \\\\\\\\ \

\

\

/'.' J\\\ \\\ I

.~;\ ~2

\

(a)

(b)

FIGURE A-2. Tilt angle '" of the normal to equiphase planes in n'gion 2 lor (a) region 2 moderately conductive; (b) region 2 a good conductor.

end, the reHe(l(ioll and transmission coefI1cient (6-83, 84) and (6-87, 88) are applicable, 011 making use of the simplification «111 for region '2 a good conductor. With this, 84) iC)f the parallel-polarization case are seen to be well approximated by

\qz\

(A-12a)

((J 2 very large) (A-12b)

The result (A-l'2a) means that nearly total reflection oceurs from a good Rr ~ R;, one has also that Hr ~ Hi in region 1, since the ratios RrlHr and fo'JH; are both lJi. From (A-l'2b) it is seen that the eleetric field transmitted into region 2 has (he amplitude (at the inteda.ce) given by

cond~ct2r. With

(A-13a) a~ very s!.Tlall amplilude, i~l view Of\q2\ « '11 fi)!' region '2 a good conductor. On inserting IIi for EilY/l and lIt for k'tlY/2 into (A-13a), one obtains

(A-I3b) stating that the magnetic field amplitude H, just inside the good conductor has twice the amplitude of the incident magnetic field, in this parallel-polarization case. These

600


(x)

(a)

(b)

FIGURE A-3. The parallel-polarization case, region 2 a good conductor. (a) Depicting Geld vectors in both regions. (b) Time-instantaneous, ]-directed magnetic Geld in region 1, showing both standing-wave behavior with z, and traveling waves along x, at a fixed instant.

conditions are depicted in the vector diagram of the field components as shown in Figure A-3(a). The foregoing simplifications lead directly to the field eKpressions in both regions. The full expressions for the total electric and magnetic fields in region I are alrea<;ly given by (6-74) and (6-76)lix this case, so with the substitution of (A-12a) Ie)]' E;" one can reduce (6-74) to

Z)

a x (j2E'; cos

Oil sin

(fJtZ cos 0i)e-ifJlxsinOi

-a z (2E; sin 0i) cos (fJl'~ cos

Oi)e-jfJlxsinOi

(A-14)

while (6-76) becomes (A-15) A graph of the time-instantaneous form of this ]-directed magnetic field is shown in Figure A-3(b). The exponential function in (A-15) clearly exhibits the traveling-wave nature of this field along the x direction; while the cosine function shows its standingwave behavior with z, the result or essentially equal amounts of reflected and incident wave amplitudes in region 1. Observe that assuming the special case of normal incidence (0; = 0) reduces the expressions (A-14) and (A-15) to (6-5) and (6-8), the standing-wave fields obtained in the normal-incidcllce case discussed in Section 6-2. Wi~hin the good conductor (region 2), the electric field:E z can be found by subsituting Hz of (A-lO) into Maxwell's equation (3-85); or it can be inferred fi'om (6-75). Choosing the latter method, on repiacingjpz in (6-75) by the complex Yz = ()(z + jpz of this good conductor region and making use of the cornp!!~x Snell's law result (A-3) combined with the reduction cos 0/ ~ 1 and (A-13a) for E t , one obtains

A-2. REGION 2 A GOOD CONDUCTOR

601

the good approximation ~

EZ(x, z)

~ ax

2ryz ryt

A

Ei

sin(h

(A-lEl)

The accompanying Hz field has already been found as (A-IO). It might alternatively have been 1()Und by use of (6-77), or, if desired, from (A-16) and Maxwell's curl relation (3-84c). The result is repeated here. (A-17) It is evident that the ratio Ex/fly given by the foregoing expressions isjust the intrinsic wave impedance q2 of the good conductor. The details of plane-wave reflection and transmission, for the incident wave perpendicularly polarized, proceed along lines that closely resemble the parallel-polarized case just treated. They are left as an exercise lor you. The analysis given has revealed that the fields in the lossless region I remain essentially uncbanged, on comparing the reflection from a good conductor with tbat obtained from one assumed ideally perfect. This suggests that, in solving a boundaryvalue problem in which a lossless dielectric is separated from a good conductor by a suitable interface, the problem may be simplified by first finding solutions in the dielectric region b'L~ed on the assumption of a perft~ctly conducting boundary. Then, on relaxing the wall conditions by assuming a good conductor instead of a perfect one, tbe fields within the conductor can be inferred from the continuity of the tangential magnetic field component across the interface. The electromagnetic fields penetrating the metal walls of transmission devices such as hollow rectangular waveguides and cavities, j()r example, can be obtained in this manner. An instance of this method is detailed in Section 8-6.

__________________________________________ APPENDIXB

Transmission line Parameters

In the following, Iwo examples of transmission lines arc analyzed f()r their line pal'amdel's: (a) the parallel-wire line, assuming a separation sufTicienl to neglect proximity effects; and (b) the coaxial line, examined only fe)r its high-frequency and de behavior. The analysis of the internal distri bUled impedance of an isolated wire is usehd f()r simplifying both problems, so this is taken up first.

B·1. CURRENT PENETRATION IN ROUND WIRE (SKIN EFFECT): INTERNAL DISTRIBUTED PARAMETERS The isolated round conductor shown in Figure is to be regarded as an element of a two-conductor transmission line, so to the associated time-harmonic electric and magnetic fields are attributed the usual filctors r;iWl+ . Only positive z traveling waves need he considered, thus requiring only ei wt - yz • The internal impedance contribution Zi defined by (9-96) is desired for this conductor. Considered as one of a pair of conductors canying the qua~i-T£M mode, its magnetic fi~d is assumed to be the axially symmetric component Yt~e-'wt-yz. 'The continuity of £4) into the conductor interi~r generates an electric field component therein, related to its axial current density Jz by (3-7)

(B-1 ) The internal impedance Zh defIned by (9-96), denotes the voltage drop per unit length A VI A/~ =
602

B-t. CURRENT PENETRATION IN ROUND WIRE (SKIN EFFECT)

603

that is,

(B-2)

d
The evaluation oU i is facilitated iL~ is expressed in terms of Sz. The desired relationship is obtained from the m?dified-<:yrl expression (8-6), reducing to the following from axial symmetry and only aJ'z and ::IC'
p

,it = ai'~ =

.lwPc

U

up 0

a

az

p

0

-'l

0

Sz

=a

.lwPc

oSz op

(B-3)

In the latter, satisfies a wave equation ofthe felffn of (2-96) which, fo~ the condu~tor with COllstants (p", Eo and in time-harmonic form, becomes V 2 E + w2PcEcE = JWPc(J Combining terms in E yields

J;.

A

V2E

+

2 (j)

A

11cEc( 1 - j(JjwEc)E

=0

but in a good conductor, (JjWE c » 1, reducing it to y2E - jWJIc!{cE = O. With only Ez prc:;;cllt, one obtains the scalar wave equation V 2 Ez -.fW/lc(JcEz = O. One expresses V 2 E z in circular cylindrical coordinates by (2-80), symmetry requiring Djo
(B-5) multiplying by p2 obtains (B-6)

a result known as tbe Bessel differential equation of zero order. Its solution is obtainable by assuming a power series solution,l leading to the current density

1 For details of the power series method and Bessel functions, see C. R. Wylie, 2nd ed. New York: McGraw-Ifill, 1960, Chapter 10; Of S., Ramo, j. R. Whinnery, and Fields and Waves in Communication Electronics, 2nd cd. New York: Wiley, 1984, pp. 360~-371.

Mathematics, . van Duzer.

604

TRANSMISSION LINE PARAMETERS

in which the factor

-jwflc
in (B-6), denoted by

k?,

implies

1!2JW fl c
V;

(B-8)

It is seen, on takingAthe principal square root ofj-l/Z as (1 - j)/)2, that if: of (B-8) can also be written k = (I - j)/(j. The symbol () used in (B-8) evidently means (B-9)

which from (3-95) and (3-112a) specifies the skin depth of penetration of a plane walle into a conductor with the same parameters Pc,
Jo(ul =

I

(B-1O)

m=O

If a = kp = (I - j)p/(j, and with the latter into (B-IO) and grouping the real and imaginary terms, one obtains

(B-IIa)

(B-llb)

The symbols bel' and bei (Bessel-real and Bessel-imaginary) thus denote the real and imaginary parts of Jo, and are tabulated in numerous rderences. 2 The Bessel Iimetion No(u) in (B-7), the second solution of (B-6), is discarded here because ofa singulari ty 3 at p 0 (No(O) -+ -00); ~hat is, the wire center p = 0 is within the region of discussion of the CUrr(,llt density ]z, so No(u) is of no physical use. Therei()re (B-7) can be written with (;z = 0 and (B-Ilb) inserted, obtaining

(H-12) 2For example, see H. B. Dwight, 'Tables 'if Integrals and Other Mathematical Data, revised cd. New York: Macmillan, 1961. 'Sec Ramo, S., et a!., op. cit., p. 366, for a graph.

B-1. CURRENT PENETRATION IN ROUND WIRE (SKIN EFFECT) a

°

i= I-

I- a T

I

For small

V~

1-

8=1

0.8

I- 1----

I/'

1--- 1--

l-

I

I--- I--

I---

It

r-- - 1-%=2

V!

I

0.4 --

j

/

1"- i--

0.2 ~t:-r _1:

o! I

~~;=

II /-:1

~

5

j,

1H/J.1__ wV0

1

I.

0.2

0.4

0.6

t=~

0.2 -

I---

0.1

IJz(a) Jz(O) I

-

f\

0.8 a

0.8

~

""-- Wire axis

0.5

\

--- --- i - I---

i\ --1--

0.006 -- -I---0.004 -- f--- f--

i-

-

L\

i-- i--1-- --

I-

i\

i- e_

1\

0.002 0.001

1.5

1---

0.06 1-- re- t-0.04 I- I---t--0.01 -- i-I----

-~ = 10

i

0.9

0.02

/

~

0.6 I--0.4 1-

605

\

°

2

4

6

8

10

--From exact (B-13) - - - From plane wave formula (B - 17) (a)

(b)

FIGURE B-1. Current density in round wires (a) Current density magnitude versus normalized radial distance pia. Q'Lshcd lines show plane wave approximations. (b) Ratio of current density magnitude at the center to that at the snrface, as a function of wire radius in plane wave skin depths.

Putting p = a into (B-12) expresses C1 in terms of the density lz(a) at the wire surface, obtaining C1 = lz(a)/[her (J2a/b) + j bei (J2a/b)J. The latter into (B-12) yields the current density distribution in the wire

(B-13)

Using tabulations of ber and bei, or resorting to the series definitions (B-1 I ), one can plot (B-13) as the skin elfect curves shown as solid lines in Figure B-I(a). The curves are universalized by expressing the argument J2plb in (B-13)

(B-14)

This permits using the normalized radius pia with a range (0, I) froIll the wire center to the surface, while all) expresses the wire radius a in terms of the plane wave comparison skin depth b given by (B-9). Both graphs show that the axial current density

606


is depressed noticeably below the surface value if the wire radius exceeds a plane-wave skin depth or so (ali> > I). For example, if ali> 2, iJz(O)i is seen to be about 62'1,) of the value at the surface. Figure B-1 (a) shows (dashed) the amplitude attenuation of plane waves in a conductor, for comparison with the round-wire current density (solid curves) given by (B-13). Plane wave attenuation was described in Chapter 3 with ref<:;rence to Figure 3-17, implying that a z-directed current density plane wave, traveling in the x direction in a good conductor, is given by (B-15)

J m denotes a reference amplitude at x =

0, and~attenuation and phase factors 112a) and (:-1-11 . That.7Ap) of(B-13) in a round wire reduces essentially to (B-15) is shown using the asymptotic limits of the Bessel flmctions. Assuming the argument sufficient large (u > 10 or so), the following asymptotic form holds: 4

in which

rx

=

f3 = 15- 1 are given by

(B-16a) beeoming exact as u -> co. With the complex argument u =j-l/2/2pli5, (B-16a) becomes n;8)

(B-16b)

for large pli>. Using (B-16b) in the numerator of (B-13) and the same with P = a in the denominator yields the t()llowing approximate current density in a round conductor.

-p)lb

(B-17)

essentially correct if the conductor radius is several plane-wave skin depths (ali5 > 10 or so). Comparing (B-17) with (B-15) thus verifies that ifali5 > 10 or so, i> can be used to denote the depth ifpenetration in a round conductor. The solid and dashed curve" in Figure B-1 (a) indicate how well the plane wave expression (B-15) approximates the exact round wire relation (B-17).

EXAMPLE 8·1. (a) At 60 Hz, what maximum radius should a copper wire have if}. on the wire axis is not to be less than 90% of the surface value? (b) If J is raised to 6 GHz (in the microwave range), what maximum radius meets the same criterion?

4For a tabulation of asymptotic forms, sec for example S., Ramo, J. Whinnery and T. van Duzer. Fields and Waves in Communication Electronics, 2nd ed. New York: Wiley, 1934, p. 212.

.

607

B-1 CURRENT PENETRATION IN ROUND WIRE (SKIN EFFECT)

(a) From Figure B-1

,if \]z(O)/]z(a)\ = 0.9, then ali5 = 1.16, implying a plane-wave skin depths, but at 60 Hz, 15 from (B-9) heeomes

1.16

= 0.00853 m = 8.53 mm

so the wire radius must not exceed (b)

II

= 1.1615 = 9.89 mm

0.389 in.

If f were increased by 10 to 6 X 10 Hz, 15 of (B-9) becomes 0.00853 x 10-4, yielding II = 1.1615 = 0.989 x 10- 3 mm = 0.989 Jim 0.039 mil, a very small wire. Barrctter wires, used as resistance clements in microwave power bridges, are made this thin to yield de and rf resistances that arc the same, permitting substitution methods (rf lor de power) to be used. 8

9

EXAMPLE B·2. A copper conductor is ofa ImD! radius. At what frequency is the radius 10 skin depths? (b) What is 15 iff is increased by 104 ? o

(11) Pu uing (j forfyiclds

]

015 (or

a/l!

= 10) means From (15-9) that

f

so solving

(j

200

2n(]O 6)4n x 100.432

X

10 6 = 432 kHz

At or above this frequency, the skin depth ill the wire is essentially that for a wave in this conducting material.

(b) Increasing f to 4.32 GHz decreases 15, froIll (B-9), by 10-oJ mm = 1 pill.

4 = I ()

, yidding 15

The impedance parameter Zi or an isolated round wire is now found by use of (B<2), Substituting

(l3-1~?)

into (B-1) yields

ber ( ber (

J2 ~) + j

bei (

J2 ~)

J2 i) + j bei ( J2~)

(B-18)

Also needed is it~ at the wire surface, obtained Irom (B-18) by use of (B-3); hence

=jo~~c~?a~P) l=a = jW~/JC ~i~ll=a =jW~c;~ (-;) ~J~~P) 1="

blT(J2 i)+jbei(J2 g) /Jw~c~c J2~) + J2 i) ]z(a)

ber (

In (B-19) a change to the variable u

(B-19)

j bei (

J2p/b

has been made, the primes signifying

608

TRANSM1SSION LINE PARAMETERS

differentiations with respect to u. With (B-18) and (B-19) i~to (B-2), the symmetry of :Ye", permits a simple integration around the wire to yield I, obtaining

(£-20)

z;

One can decompose into r; + jOJI; by rationalizing the denominator, obtaining internal resistance and inductive contributions as follows

(B-2Ia)

_

1

wi; - 2na

/mJZ ber ( vf2 i) ber ( J2 i) + bei (fi i) bei ( 12 i) '1/ -;;: (Jc

[ber(J2

i)J2 + [ ()J2 J2 i bei

(B-21b) O/m

That the latter are correct is appreciated from the zero frequency limits. Noting that J2a/b w approaches zero as OJ -'> 0, all but two terms, one real and one imaginary, of (B-ll) can be discarded, yielding ber w -'> 1 and bei w -'> w 2 /4 as w -'> O. Similarly, derivatives of the power series obtain ber w -'> w 3 /16 and bei w -'> w/2 as OJ -'> O. These into (B-21a) yield the dc resistance (B-22a) a result seen to agree with the static result (4-138)

The zero frequency inductance obtaincd from (B-2 I b) is

(B-22b) 4 agreeing with the static result Lit of (5-82).

q 609

B-1. CURRENT PENETRATION IN ROUND WIRE (SKIN EFFECT)

10 8 6

---

,-

---- c--

,/

,--- - -

--

--

4

ri

r:--;;-" ~r' t. /'

- --

2

I

V

0.8 - - - -0.6

V/ /

I

dc,----_

"

1-

1--- ..._- r---

0, 2

I I! I

--

--

JL~

Ii,

I

From (B 27b)

t--~~

~

From (B- 27a) (high-frequency approximation)

/~

i

1

1I

C

'"

1';

--".-

"

r----- 1--O. 1 0.2

2

0.4 0,6

=

4

6 810

~ 20

FIGURE B-2, Internal resistance and inductance parameters for an isolated ronnd wire.

A graph of ri and Ii for the isolated conductor, expressed as ratios to the de values (B-22), is shown as solid curves in Figure B-2. Also shown dashed are highfrequency approximations to the internal parameters, approaching the exact curves for alb sufficiently large as discussed in the following. Asympt9,tic approximatiolls can be found for Zj from (B-2), but it is convenient to reexpress :if", in terms of Jo(kp), instead of using the ber and bei functions. From (B-3) and (B-12) (B-23a) wherein

k -J -jWJ.l(J

112.j2IO and 0 is given by (B-9). Also

OA -;--- Jolkp) = up

0

A

o(kp)

~A-Jo(kp) -~:1 =

o(kp)

up

A koJ (kp) A,

in which the prime denotes the derivative with respect to the argument 1/2 .j2pll5. Then (B-23a) yields, at p = a

r

kp =

(B-23b) Using (B-23b) in (B-2), the internal distributed impedance becomes (B-24)

610


in which the Bessel function of order unity Jl(V).5 The asymptotic (orm ten' It(ka) IS

obtained from the identity

IS

J~(u)

=

(B-25)

and the latter with (B-16b) into (B-24) yields

1/2 jill;,: -

~

~(allJ-1C18)

Zi --> ----.--

J2na

20"c

=-

1

2na

(1

.

+ J)

JWfJc - - !l/m 20"c

(B-26)

valid for sufficiently large a/b. The real and imaginary parts of (B-26) yield

ri

-->

1 2na

JWfJc

1 20", =2~aO"tb

a

a =

2b

ri(dc)

!lIm

J large

(B-27a)

a -large

(B-27b)

b

C

For instance, if alb = 5, the asymptotic expression (B-27a) can be used in lieu of (B-21a) with an error orabout 10'1,,, decreasing to zero error {()I' alb sufficiently larger. The asymptotic result (B-26) is seen to contain the quantity ~ of (3-112c), that is, the intrinsic wave impedance for a plane wave in a conductive region. Thus (B-26) yields tbe ratio lor a round wire a (j

large

(B-28)

One concludes tbat for current penetration small, the impedance ratio izj:it~ at the surface of a round wire becomesq, the same as the ratio of electric to magnetic plane wave fields in a conductor. B-2. DISTRIBUTED PARAMETERS OF A PARALLEL-WIRE LINE, CONDUCTOR IMPEDANCES INCLUDED The isolated wire internal impedance results of the previous discussion can be applied directly to the parallel-wire line of Figure 9-7(b), with the object of tin ding (9-98b), the distributed pararneter z. Proximity em~cts6 arc neglected, which assumes fields in each wire undisturbed from the axially symmetric configuration attained when isolated, a 5See S. Ramo, J. Whinnery, and T. van Duzer. Field, and 111alles in Communicatiolls Electronics, 2nd eeL New York: Wiley, 1934, p. 370. 6An analysis of the cf1(,cts of the proximity of the wires on the increase in internal rcsistdllcC is found in A. H. M. Arnold, "The alternating-current resistance of paralid conductors of circular cross-section," Jour. lEE., 77, 1935, p. 49.

611

B-2. DISTRIBUTED PARAMETERS OF A PARALLEL-WIRE LINE

reasonable assumption if the axial separation is greater than about 10 conductor diameters. With effects of proximity neglected, the internal parameters of the parallel-wire line of Figure 9-7(b) are double the results (B-21) obtained for the isolated conductor, in view of the impedance encountered twice along the edges Az of the rectangle t. The series parameter (9-98b) therefore becomes

z = 2ri + jw(2li + Ie)

(B-29)

= r + jwlOjm

'i

in which ri and are given by (B-2l) or (B-27). In (B-29) Ie is related to the magnetic field exterior to the wires, permitting the use of that obtained in Problem 9-12 (a) for the perfect conductor case

Ie = J.l t n h + d Hjm

(B-30)

a

1t

wherein a is the wire radius, 2h the separation, and d = --.lh 2 - a2 . The expression (9-100) for the shunt parameterj = g + jwe is the same whether or not the conductors are perfectly conducting; thus, from Problem 9-13

y

g+jwe

- + (Ell)

j we =

(Ell) + j

-

E'

E'

W1rE Ulm tnh+d' a

(B-3l)

For an air dielectric, the assumption g = 0 is appropriate. In telephone lines using poles for support, the insulator leakage is often reduced to an equivalent distributed loss effect along the line, yielding a parameter g determined by the number of poles used per mile or kilometer.

EXAMPLE B-3. A telephone line consists of 0.104 in. (0.264 cm) diameter hard-drawn copper wires (o-c 5.63 x 10 7 Vim) separated 12 in. (305 em) in air. Neglect leakage due to the supporting insulators. (a) Compute the distributed constants r, I, g, and c at 1 kHz. (b) Find Zo, iX, (3, .Ie, and 1/p at 1 kHz. (a) With 2a

= 0.104 in. and 2h 1CE

12 in., It/a

= 115.5, so (B-31) yields

10- 9 /36

c=-tn-[~-l+-J-::-(=~=y=···=I-] ~ t. [2311 ~ 5.10 pF(m ~ 0.00822 ,F(m; in whieh the conversion 1609 mimi is used. If leakage is neglected g = O.

612


With z given by (B-29), evaluating r, and Ii requires first expressing the wire radius as a funetion of 0 given by (B-9) 0.00213 m 0.999. Thus, with

making alb 0.622. From Figure B-2, rjr i •de 1.003, Irl1i,de r j = L003r j • de , (B-22a) in (B-29) yields

2r j

r

or 11.47 Q/mi. Similarly, using (.de of (B-22b) in (B-29) obtains 2/i 8n = 0.999 ,uH/m = 0,161 mH/mi., and tl'om (B-30)

Ie

It + d 4n x 10 - 7 t" (231) = 2.18 ,uH/m nan

,uo

= ~tn~- =

=

2(0.999),u0l

3.51 mH/mi

so the total distributcd ind uetanee in (B-29) becomes

1= 21i

+ Ie =

2,28 flH/m

= 3.67 mH/mi

(b) Onc obtains /~o using (9-105) '7 _

,,"-0-

By usc of

~r + jwl _ "--.-g +Jwc

03a) y=

yielding

(X

+ jwl)(g + jwc) = 0.0083 + jO.035 rni I

= 0.0083 Np/mi, f3 = 0.035 A = 2n

f3 up =

w

2n 0.035

7i = fA =

=

rad/mi. From (9-37)

179 mi = 288 krn

179,000 rni/scc = 2.88 x 10 8 m/scc

from thc valuc of (x, a wavc 011 thc line attenuates to e- I = 36.8'/,) ol'its input valuc in d (X I = (0.0083) - I 120 mi at f 1 kHz.

B-3. DISTRIBUTED PARAMETERS OF A COAXIAL LINE, CONDUCTOR IMPEDANCES INCLUDED Figure 9-6 shows the current distributions obtained in the outer conductor ofa coaxial line at low, medium, and high fr·equencies. At the higher frequencies the currents concentrate toward the TEM fields responsible for those currents, namely, toward the outer wall of the inner conductor and toward the inner wall of the outer conductor. This latter property of the coaxial line, like hollow waveguides, makes it an excellent shielding device at high frequencies, the essentially zero currents on the outer wall eliminating the possibility of a small tangential electric field being coupled outside the outer conductor.

B-3. DISTRIBUTED PARAMETERS OF A COAXIAL LINE

613

Poynting vector

I,

;, (a-) Conductor 1 ;; flux

(a)

(b)

FIGURE B-3, Relative to the distributed internal impedance or a coaxial line. (a) Contin':!.ity of tangential magnetic fields into conductors. (b) Longitudinal electric field induced by .1f
The isolated wire, internal impedance results (B-2 I ) are, in view of the axial symmctry, directly applicable with no approximation to the center conductor of t~e coaxial line. An additional internal impcdance is associated with the continuity of :ffi/J on the inner wall of the outer conductor (at p b), as suggested by Figure B-3(a), Defmed by (B-2), it is written (B-32) which requircs expressing if<{J in terms of the induced Sz just)nside the outer conductor, already accomplished through (B-3). The solution for €!z(p) in the outer conductor, moreover, must satisfy the differ~ntial equatipn (B-6), yielding once again solutions (B-7). Both Bessel functions Jo(kp) and No(kp) must be rctained to satisfy boundary conditions on the outer conductor. (Thus, from the range b < p < c within the outer conductor, the singularity at p 0 of No does not become troublesome,) The cxact field solution in the outer conductor is, however, not pursued further here; only a high-Irequency approximation to the internal impedance for the outer conductor is employed in the following. In determining the distributed impedance parameters of the coaxial line, £I-om axial symmetry the high-frequency approximation (B-26) for an isolated conductor is the same for the inner conductor of the coaxial line; hence A

Zil

I

.

-l> -2 ( 1 + J)

1W

A Wile

-2-

(B-33)

(J e

Thus, the internal resistance and inductance contributions are (B-34a) High-frequency approximations (B-S4b)

614


The outer conductor internal impedance defined by (B-32) can be heuristically expressed in the high-frequency limit by use of the wave impedance 3-112c

iz(b)

-~--~ ->

.Yt'
A

11

= (1 +.1) J"wJ1.c '-,•

(B-35 )

2ac

a ratio approaching that for plane wave fields in a good conductor, at frequencies sufficiently high to make (j small compared to the radius b. Thus (B-35) applied to (B-32) yields (B-36)

-

with real and imaginary parts analogous wi th (B-34)

Ti2

->

1 2nb

fJ,'2a

OJJ1.c

(B-37a)

c High-hequencyapproximations

(B-37b) Adding (B-34), (B-37), and (9-84) thus obtains, from (9-98)

Z

=

ZII

+ Zi2 + jOJlc =

Til

+ Ti2 + jOJ(lil + li2 + Ie)

->~ (~ +!) j~~J1.c + JOJ 2n a b 2a c

•

[_1,(~ +~) w2n

a

f0i. +~ tn ~J 2n a

b >J2~

= T + jwl n/m I ligh-freq uency approximations

(B-38)

The series distributed parameters of a coaxial line, from (B-38), have the following properties in the high-frequency approximation L The resistive part!" increases as the square root of the frequency, and it decreases inversely with c• 2. The inductive part has two contributions. The first is the inductive part of the internal impedance, behaving like T. The second is the external inductance (9-34), providing the major contribution to I in practical coaxial lines.

J'a

Finally, the shunt parameter (9-100), Y = g + jOJe, remains unaltered Irom (9-82), applicable to the line with perfect conductors. This conclusion follows as usual from the dependence of g and e on only the electric field in the dielectric region. Thus, A

y =g

(Elf.) We (Elf.) E' +.1 W2nE - - (JIm b

+ .1we = -? +.7 •

=

In

a

valid at all frequencies, and not just a high-frequency approximation.

(B-39)

B-3. DISTRIllUTED PARAMETERS (W A COAXIAL LINE

615

The general derivation of the series parameters of the coaxial line at an arbitrary frequency ill is omitted here, but at low frequencies they are readily obtained in the absence of the skin effect. Then the dc inductance results obtained in Example 5-13 are applicable, while the resistance parameters are obtained from an adaptation of 4-133), assuming uniform current densities over the conductor cross sections. It is left for you to carry out the details. EXAMPLE B·4. Assume the cable in Example 9-3 has the 1ielectric loss tangent of 0.0002, hut this time copper conductors are used. Recaleulate ZO, IX, and fj from the distributed

pal'ameters y and z, atf = 20 MHz. _._ Note first the plane wave b obtained from (3-114) is b -J2/CO~IPc = 1.48 x 10 - 5 m, = 0.0148 mm, sufficiently smaller than a = 0.05 in. 1.27 mm such that the high-frequency approximatioll assumed !()r (B-38) applies. The series r parameter then becomes r

I

(~+

2n

a

1,))(;12 b 2u c

=~, (}03. + 211

1.27

3

10 _) 4.14

r.;;(2~}~)411 ~li:-..~ 7

-V'"

2 x 5.8

X

10

(I)

= 0.191 Q/m

The inductance parameter has internal and external contributions

bJ.[

. .[r+w--tn" /Lo )col=) 2n a

7

l

,7 x 100.191+2n(2x 10 ) ..4n ·---tn3.26 2n

=)

= j29.8 Q/m

z

in which Ie contributes almost wholly to jwl. Thus, == r + jcoi 0.191 + j29.8 Q/m. The shunt parameter y is found from (B-39), whence c = 2nE/tn (bfa) =2n(2 x 10 9/36n)/1.18=94.2 pF/m, yielding y=g+jwc +j)wc (0.0002 + j)())c= ,i1.l84 x 10" 2 Vim. Dielectric losses are negligible in this example. The characteristic impedance is found by lise of (9-105) (3)

or essentially that of Example 9-3

f()f

the lossless case, expected since r« col

g « wc. The 191 + j29.8)j1.184 x 10-

obtained = 0.595ei 89 . 85 " m

(J = 0.595 sin 89.8,'j"

~

~wle

and

Jfi

I,

Ji'om (9-103a) IS ')' yielding from the imaginary part

0.595 rad/m

(4)

The latter is also obtainable using (9-103c), since r« col and g« wc are satisfied. To find IX, (9-103b) provides good accuracy

r

gZo

""- + -- = 2Zo 2 ~

1.97

X

10"3 Np/m

0.191

" -..~

2 x 50

2.37 x 10- 6

+ ~---,- .10 2

1.71 x 10- 2 dB/m

and while still a low value, it represents an increase of ovcr 30 times that obtained in Example 9-3 with conductors assumed. Waves thus attenuate to 35.8'1,) of the input value ofalcngth d=IX- 1 = (1.97 x 10- 3 ) I =508mofthislineatf=20MHz. At 20 MHz, (9-37) yields A 2n/f5 10.55 m and vp w/f5 2.12 x 10 8 results essentially those of Example 9-3, in view of the small losses.

_____________________________________________ APPENDIX C

Integration of the Inhomogeneous Wave Equation

The formal proof that (11-17) is a solution of (1l-16) is given here. It is convenient to expand (II 16) in rectangular coordinates into the scalar wave equations

(C-l) making use of (2-83). Denote any of (C-l) in the unsubscripted form (C-2) Recall the Green's symmetrical theorem (2-92) g

~~)dS

an

(C-3)

correct for any pair of well-behaved functions f and g in and on a volume V bounded by the closed surface S. It is shown that (C-3) leads directly to the solution (11-17) if suitable choices are made for the functions f and g. Choosef as the field A in (C-2), assuming A means any component of A located at a source point pt. Thus, A A(u'u u~, u~) = A (r'), in whicl.!c r' denotes the position vector ofP'o The wave equation (C-2) is certainly satisfied by A (r') at all such points in space, and the Green's theorem

616

INTEGRATION OF THE INHOMOGENEOUS WAVE EQUATION

617

P'(r')

on 81 Enlarged

view of 81

(b)

(a)

FIGURE C-1. Geometries relative to the derivation of (11-17). (a) Current distribution at source points P' and a fixed field point P enclosed by surface S. (b) Showing small sphere 8 1 excluding P( r) from the region of integration.

(C-3) is written (C-4)

in which dv and ds are primed, since they are identified with locations P' in V and on S. For reasons about to be clarified, the function g in (C-4) is chosen as

e- i/loR

(C-5)

g=--

R

In (C-5), flJr the free-space region to which the constants /10' Eo apply (C-6)

Po = w)/1oEo

and R = Ir - r ' \ denotes the d~tance between the source point P' (r') and any fixed field point P(r), at which A is desired to be expressed in terms of the sources. The geometry is shown in Figure C-l(a). The integration (C-4) in V and on S being taken over all points P' and with P an arbitrary fixed point in the region, it is seen that R = () at P (i.e., P is the origin of R). Moreover, a property of the function g is that it satisfies the scalar wave equation,

(C-7) With (C-5) inserted into (C-4), the latter becomes

rh [~ 0 = Ys A(r') on

(e-.iPOR) e- jPoR OA(r')] --R- - R --an- ds'

(C-3J

618


From (C-7), - p~ can replace V 2g in the first term of (C-8), obtaining for the volume integral

r[

Jv

A(r')p~

J

e - j(JoR R

=

e - j(JoR 2 R V A(r') dv'

r _ e-j(JoR [V A(r') + pf;A(r')]dv' = r J1.o](r')e- i (JOR dv' 2

Jv

Jv

R

(C-9)

R

the latter following from the use of (C-2). Next, an inspection of the right side of (C-8) reveals that if the field point per) is to be inside the volume region V of integration as in Figure Col (a), then the requirement of Green's theorem that the function g = e-j(JoR/R be well-behaved in V is violated at P (where R = 0), in view of the singularity in g there. The point P is excluded from the volume region of integration by constructing a small sphere S1 of radius R = R1 about P as shown in Figure Col (b). Then V is bounded by the closed surface S = S1 + S2 as noted in the figure, whence (C-8) is written symbolically

r [. ] dv' Jv

=

r JSl

[] ds' + JS2 r [] ds'

(C-lO)

in which the brackets denote the volume and surface integrands of (C-8). The contribution of the integral on S1 in (C-lO) is now shown to approach the value 4nA(r), or just 4n times the potential A at the field point P, as the sphere S1 vanishes. From Figure Col (b) it is evident that %n = -%R on S1, since the normal is directed outward with respect to the volume region V (meaning in the negative R sense). Th us ~ , 0 (e- j(JOR) A(r) -oR R

r [] ds' = JSl r [ JSl =

l { ~ (r), (I-Rl + . A

]Po

S,

e - j(JoR oA (r') ]

+-----

)e-i(JORl Rl

R

oR

R~Rl

+ e-j(JORlOA(r')J -R1

oR

ds

t

R=Rl

}' ds (C-II)

By definition A(r') is well-behaved in the vicinity of the fixed point P(rl;. Allowing the radius Rl of the sphere S1 to become arbitrarily small, the value of A(r') on S1 can be replaced by its mean value
The integral ISl ds' is just 4nRi, so as R1 ~ 0, (C-II) becomes lim

r [] ds' ~ 4nA(r)

(C-12)

R1-.O JSl

or just 4n times the potential A at the field point per), that which was to be proved. On putting (C-9) and (C-12) back into (C-lO) and solving for A.(r), the desired integral


619

(External current sources) Fixed P(r) I,'

1.

~

r,

I , ......

I

II /I

R

H ,I

(V)

f I

IJ I.

(a)

P'(r')

(c)

(b)

FIGURE C-2. Three cases applicable to the Helmholtz integral (C-14). (a) Sonrces outside S. A is fcmnd by usc of (C-l5a). (b) Sources inside S only. A(T) is obtained using (C-15b). (c) General case. Sources inside and outside S; (C-14) applies.

solution for the scalar wave equation (C-2) is obtained

~ A(r) =

! J.lo.7(r')ev

jfJo

4nR

R

dv

,

1 ~ [oA(r') e- jfioR

+4n

S

-an-

~

A(r')

R

-i(iOR)] ds' ana (e~

(C-13)

in which the subscript on S2 is discarded since Sl is no longer present. Since (9-13) is a solution of each scalar wave equation in (C-l), three such solutions, for AAr), Ay(r), and Az(r), can be added vcctorially to provide the desired solution for the vector wave equation (11-16)

~

A(r)

! J.loJ(r')e-i(JoR dv, +-1 ~ [oA(r') - - e- {3oR j

=

v

4nR

4n

S

an

R

~

a (e-i(JOR)] ds.'

-A- - R

an

(C-14)

a result due originally to Helmholtz. The meaning of the integral solution (C-14) is clarified on referring to Figure C-2, and three cases of physical interest can be identified. CASE A. Suppose no current densities exist inside the region V, bounded by a finite closed surface S as in Figure C-2(a). With J = 0, (C-14) yields only the surface integral

~

iiloR

I ~ [oA(r') e---S an R

A(r)=4Jr

~ A(r')

-iilOR)] ds' ana (e-R-

(C-15a)

620


VVith no sources inside V, the potential A(r') on the closed surface S must be attributed to current sources lying entirely outside S. This type of integral was extensively investigated by Kirchhoff in optical diffraction problems.

B. Suppose current densities exist within a finite distance from the origin. Then the closed surface S in (C-13) can be expanded indefinitely toward infinity, to reduce the contributions to the potential at the field point P to the volume integral (11-17)

CASE

(C-ISb)

as depicted by tbe geometry in Figure C-2(b). This is the reduced form of (C-H) commonly encountered in free-space antenna radiation problems.

If current density sources exist both inside the closed surface S and outside it, the general form (C-14) is applicable, amounting to a superposition of(C-lSa) and (C-1Sb). This case .is largely academic, for the better expedient is usually to expand S until all current sources are enclosed, in which event Case B applies.

CASE C.

In a manner analogou~to the arguments leading to (C-15b), one call show that the scalar electric potential
(r)

f ----..__... dv' Jv

(C-lti)

assuming the closed surface S to have been expanded inddinitely toward infinity. In antenna radiation problems to follo~, no llse is made of (C-lti), since the Lorentz condition (11-8) yermits expressing E and B, via (11-13) and (11-14), solely in terms of the potential A.

__- - - - - - - - - - - - - - - - - - - - A P P E N D I X D

Development of the Smith Chart

In the following are given the details of the theoretical basis for the Smith chart, a necessary background for its intelligent application to solving reflected wave problems. The chart provides graphical solutions of wave or transmission-line relationships involving impedance and reflection coefficient; for example, (6-38) in Chapter 6, concerned with uniform plane-wave fields in layered regions treated in Sections 6-5 and 6-6; or such as (10-4) in Chapter 10, concerned with voltage and current reflection on two-conductor transmission lines taken up in Sections 10-1 and 10-2. Additionally, by means of simple angular rotations about the Smith chart, it yields graphical solutions to the related translational relationships (6-40) and (10-6), concerning the effect on the reflection coefficient of a z-position change in the region or transmission line in question. The foregoing properties of the Smith chart become possible because the chart represents an overlay, or mapping, of constant-resistance and constant-reactance components of the impedance of (6-38) or of (10-4), onto the complex-reflect ion-coefficient plane, as developed in the following. ~ Suppose one begins with the total-field impedance Z(z) defined in Chapter 6 by (6-38) (applicable to a multiregion wave system), by normalizing it through a division by the intrinsic wave impedance of the region in question.

(6-42]

in which the dimensionless ratio Z(z)/ff, symbolized £(z), is called the normalized totalfield impedance of the region. Alternatively, beginning with the line impedance Z(z)

621

622

DEVELOPMENT OF THE SMITH CHART

defined in Chapter 10 by (10-4) (applicable to a uniform tran5rr~ission line), normalizing it through a division by the line characteristic impedance <:0 obtains a result of a form identical with (6-42)1

Zi z ) == x(z)

<:0

=

1 + ['(z) I

[lO-IO]

in which the ratio Z(z)IZo, also denoted by x(z), is called the normalized line impedance of the region. On solving (6-42) or (10-10) for ['(z), their inverse is obtained: ['(z) = - , - - - -

(D-I)

+1

x

Now writing and [' in the rectangular complex forms

x = -z + jx

(D-2)

yields, on substituting these into either (6-42) or (10-10), the following

1+ rr + jri

-z+jx

~-~--.---

I

rr - jr i

(D-3)

Rationalizing the denominator and eq uating the real and imaginary parts of each side yields -z

1

r; - rf

=-------~--c~

(I

(D-4a)

(D-4b)

x

Setting"t and x equal to any real constants in (D-4) yields two mappings (transformations) from coordinate straight lines in the complex ~lane into circles in the complex r plane, as depicted in (a), (b), and (el of Figure D-l. This is proved as fi)llows. With -z = constant in (D-4a) and manipulating it into the form

x

(D-5) one obtains in the [' plane a family of circles wi th radii 11 (,z. + I) and centers displaced horizontally to the positions Po(-z/(It + 1),0). For example, the It = 2 line in Figure D-l (a) transforms, by use of (D-5), into the circle la beled ·i = 2 in Figu re D-l (b), with a radius of and its center at 0).

t

pori,

the theory of complex variables, (6-42) or (10-10) art' bilinear Iransflnmalions, having the property of translarming circles (or straight lines) in the plane of one of its complex variables into circles in the plane of the other complex variable. See also R. Y. Churchill, Complex Variables and Applications. New York: McGraw-Hili, 1948.

1 From

2Since th~ 1 = constant and x constant lines of Figure D-I (a) intersect orthogonally, their maps as circles onto the r plane of Figure D-l (d) also intersect at righ t angles. This is the conformal property of a bilinear transi()rmation.


623

«-plane)

o -~-~~-

21

31 41

51

61

71

-1

~t---r-i-t-t-1~

-

-+-+-+~+~~~+I I I I I I

-2

-3

~+~+~+-+~+-+

I

I

I

I

I

I

(a)

,= 0

r plane 4

r plane

= 05

_-+__-Ic---+-+--c"=----k-->-l'r

(c)

(b) ,,-=2

r

r = 0.5 + jO.5 = O.707e

j45 '

(Corresponding to" 1 + j2)

plane

.r=-2

Cd)

FIGURE D-1. Development of the Smith Chart. (a) Typical i-constant and x-constant coordinate lines in the plane. (b) Circles of cons~ant i mapped by (D-5) onto tbe plane. (c) Circles of constant ,,; mapped by (D-6) onto the r plane. (d) Complete mapping of -i-constant and x-

r

constant lines onto the

r plane: the Smith chart.

Similarly, any::]'; constant line in Figure D-I(a) maps into the circles in (e) of that figure, evident from manipulating (6-40b) into (D-6)

This is a family of circles with radii I/o: and centers at thelocations Po(l, 1/0:). Typical circles corresponding to a; = 0, ±O.5, ± 1, ±2, and ±4 are shown in FigUl:e D-l(c).

624


f

plane

~

"";!-

~

~-r "'tv

~

1:

0."

sOIJ

rCe

FIGURE D-2. Smith impedance ch~rt, with a dashed line overlay shown, to denote magnitude and angle of the reflection coeflicient r. (Note that lhis overlay is missing on commercial charts, to make the -; and circles more readable.)

The superposi tion of the circles of Figures D-l (b) and (e) yields (d), the so-called Smith chart. Any point (r" r i ) denotes solutions to (10-10) or (6-42) or, conversely, to (D-I . Thus, if some reflection cocflicicnt fl.::) rr +.JTi If(.::)lejiJ is located at p(r" ;) on tlu~ chart, that point also provides the coordinates read off the .z = constant and = constant circles,~yielding the solution .£ = i +J:'l' of that was sought. For example, given that f(.:::) 0.5 +jO.5 = 0.707ei 45 ', (r" = (0.5,0.5) on the Smith chart produces (t, ,r) (I al that same implying the solution .; 1 = I + j2 2.23ei63 .4". Thus = O. and = 2.23ei62 .4' are a solution-pair seen to satisfy the normalized impedance relations and (10-10), or their converse, (D-l). More accurate solution-pairs satisfying and (10-10) are available from the enlarged Smith chart of l:igure D-2. Besides the usual oveday of circles of constant .z and constant x onto the r plane, Oil the chart rim are three scales, the inner of which denotes the angle e of the reflection coefficient, a q uilntity usually expressed in j the polar form, f = ifle (/. The radial dashed line overlay on th~ chart gives the angle O. The concentric dashed .line circles denote the magnitude with a range from zero to one.

el, ,

Ill,

DEVELOPMENT OF TIlE SMITH CHART

625

The two outer rim seales Oll the chart pertain to the angular rotation associated with the exponential filctor of (~ither (6-40) (in the multi region wave problem) or of (10-6) (in the transmission-line prohlem):

[6-40 j, [l0-6\ going from a location to a !lew position l in a region. For the case of a lossless region, wherein the propagation constant y is the pure imaginary 'Y = jp j2njJe, (6-34) simplifies to

111

-z)

(D-7)

stating tha.t the reflection coellicicnt at the location .:' is found multiplying at some other positioll times the pure phase farlor exp [2(j2njA) z) j. The factor 2 denotes that the phase shift that undergot's is twice that with the wave motion in going from to . Both outer rim scales are calibrated to include this factor of two, so the user needs only to read the displ
r

0)

(D-8)

In this case, besides the phase Elctor of (1)-7) appearing in (1)-8), there is also the double-altenuation factor exp !2ex(l - z)], implying a decrease in the magnitude of the reflectioll coeflicicnt as ont moves toward the source.

30 ne may note that if the direction of motion in going ironl ,:: to is toward the SiJUtFf,\ of the wave') in a region (to the lefi in Figure 6-8), the quantity is then more negative than z, making the corresponding

phase shift produced by the exponential betor of (D-7) it negative, or dockwi,lc angular shill around the f-planc of the Smith chart. The (Juter rim scak of Figure D-:! i, calibrated to dellote lhat phase shin (in decimal in moving toward the source (generator). The middle scale is used when moving away waves,

_____________________________________________ INDEX

waveguide, below cutoff, 424, 431, 436 Absolute potential, 191 waveguide wall-loss, 451 Acceleration, 13 Average power: Addition, vector, 3 definition, 394, 400 Adjustable stub, 537 Poynting'S integral for, 402 Admittance, 524 radiated by antenna, see Radiated power A field, 269. See also Vector magnetic potential in terms of complex Ilelds, 400 Air gap, magnetic circuit, 265, 266, 293, 328 transmitted in waveguide, 440 Ampere's circuital law: in free space, 35 Band theory, 115 in material region, 134, 259 Bandwidth, 442, 535 for static Ilelds, 35, 259, 263 Barrier potentials, 290 Ampere's differential law, 85, 132 Beamwidth, antenna, 574,575 Analogy: Bei, ber functions, 604, 607 capacitance and conductance, 232, 236 Bessel functions: magnetic and electric circuit, 264 asymptotic forms, 606, 610 scalar and vector Poisson equation, 270 complex argument, 604 Anisotropic, 165 differentiation, 607 Antenna: B field, 28, 30, 35, 40, 75, 126. See also Magnetic aperture, 567-575 field, from A center- fed, 555, 558 B-H curve, 142 curved linear, 557 Bilinear transformation, 622 directive gain, 575-579 Biot-Savart, 275 directivity, 577 Black screen, 571 effective receiving area, 586 Boundary conditions: elementary dipole, 550-555 of electrostatics, lSI half-wave dipole, 558, 561, 563, 577 of magnetostatics, 259 horn, 545,569, 589 for normal Bn> 127, 139 impedance, 563 for normal Dn, 123, 148 isotropic, 576, 579 for normal}n, 172 linear, 545, 549, 555-563 for normal Pn' 124 loop, 555, 557 at perfect conductor, 124, 136, 344, 459 polarization, 580-581 for rectangular waveguide, 420, 429, 450 receiving, 579-582 table of, 147 transmit-receive link, 582-589 for tangential E t, 14S Antenna pattern: for tangential H t , 135 aperture, 572 for tangential M t, 136 elementary, 554 in terms of complex permittivity, 17 I horn, 592 at transmission line conductors, 459, 602 linear, 561 Boundary-value problems, 204, 209-215, 342, Antiferromagnetic, 145 347,418,428,512 Aperture antenna, 567-575 Bound charge, II 7-119 Argand diagram, 99 Bound current, 127-132 Armature, 294 Brewster angle, 375 Associative law, 3 Assumed solution, 209, 419 Cable, coaxial, see Coaxial line Atomic currents, 127 Capacitance: Attenuated waves, 154-160 approximation, 225 Attenuation constant: of coaxial line, 198, 203, 231 from distributed parameters, 476, 482, 486-487 of concentric spheres, 199, 235 for low-loss line, 487 of current analogy, 238 plane waves, 154, 159, 162, 163 dellnition, 197 transmission line, 465, 476, 486, 612, 615

627

628

INDEX

Capacitance (Continued) energy definition, 203 of field- cell, 230 by field mapping, 228-232 of parallel plates, 199 of parallel wires, 225-228 Q.of, 235 of transmission line, see Transmission line parameters of two-dimensional systems, 228 Cartesian (rectangular) coordinates, 4 Characteristic impedance: coaxial line, impedive conductors, 615 perfect conductors, 478 definition, 469, 470 line: impedive conductors, 487 perfect conductors, 477 parallel-wire line: impedive conductors, 612 perfect conductors, 478 Charge: bound, 117·-119 cloud, 33 conservation of; 150 differential, 23 electric, 23, 49 {()fce on, 28, 189 free, 112, 119 line, 24, 33, 184, 193 magnetic, 45, 564 planar, 34 point, 31, 182,220 spherical, 33 surface, 23 volume, 23, 33 Charge density: on coaxial line, 188 definition, 23 of image system, 221 on parallel plates, 149 on perfect conductor, 124 Circle diagram, waveguide, 432 Circuit model, 291, 294, 295, 317,327,480,487 Circular cylindrical coordinates, 4,5,8,9 Circulation, 76 Closed line integral, 35, 41, 81,190,459 Closed surface integral, 44, 72 Coaxial capacitor, 197, 203, 231 Coaxial line, capacitance of, 197 characteristic impedance of; 469, 478 with different dielectrics, 188, 250-251 distrihuted parameters of, 478 electric field of, 187

magnetic field of; 302-303 propagation constant, 471 self inductance of, 303 TEM mode, 458-47 I TE and TM modes, 5 I 4 as transmission line, see Transmission line waves in, 461-469 Coercive force, 143 Commutative law: of addition, 3 of multiplication, 15 Complex amplitude, 98, 154,345,421 Complex angle, 377,596 Complex conduCtivity, I 15 Complex dielectric constant, see Complex permittivity Complex permittivity, 160 - 161, 171 Complex phasor notation, see Time harrnonic Complex Poynting theorem, 401, 402 Complex time harmonic, see Time harmonic Components, vector, 5, 7, 8 Conductance, 241 Conductance analogy of capacitance, 232, 234, 236 Conductance by field mapping, 239 Conduction current, 36 Conduction model of capacitance, 232-239 Conductive region: classification of; 160 current density in, 113, 156, 605 parameters, table, I 68 skin depth in, 156, 163,604 Conductivity: defInition, 113 Drude model, I 15 table, 168 Conservalion of charge, 150 Conservative field, 23, 66, 188 Continuity, see Boundary conditions Convection current, 36 Coordinate lines, 6 Coordinate points, 4, 6 Coordinate surfaces, 6 Coordinate system(s), 4, 8 circular cylindrical, 4, 10 generalized orthogonal, 7 rectangular (Cartesian), 4, 10 spherical, 4, 10 Coulomb's force law, 33, 181 Coupled circuits: energy oC 319-321 mutual inductance of; 320-325 Crank method, 158 Critical angle, 376

INDEX

Cross (veaor) product, 17 Curie temperature, 145 Curl: definition, 77 of electric field, 84, 146 of magnetic fleld, 85, 132 of velocity fleld, 77 Curl operator, 79, 90 modified, 412, 465, 466 Current, flux oC 27 linear, 271 magnetic, 564 Current density: definition, 26 dielectric polarization, 121 discontinuity, at interface, 170 displacement, 36 magnetization, set Magnetization current density on parallel plates, 149 on perfect conductor, 136 in plane conductor, 156 in ronnd conductor, 605 sudace, 136, 149, 271 in waveguide, 439 Current loop, inflnitesimal, 127 torque on, 128 Current sheet, 38, 40 Curvilinear coordinates, 7 Curvilinear squares method, 230 Cutoff freqnency, waveguide, 423, 431, 433 Cylindrical coordinates: circular, 4, 10 generalized, 412 Del operator, 66, 70, 80 Depth of penetration: plane waves, 156, 163 in round wire, 605 D field, 119 Diamagnetic, 140 Dielectric, anisotropic, 165 Dielectric boundaries, 122 breakdown, 120 complex permittivity, 153, 160, 161, 171 constant, see Relative pennittivity losses, 160-163 permittivity, see Perminivity polarization, 116 polarization current density, 121 properties, table, 168 susceptibility, 119 Differential, total, 64 Dipoie:

629

static electric, 195 static magnetic, 274 Dipole antenna: center- fed, 555 infinitesimal, 550 Dipole moment, 116, 127 Dipole moment per unit volume, 116 Directivity: aperture, 588 rectangular horn, 589 Dirichlet boundary condition, 207 Discontinuity: at charged surface, 124, 170 in D n fleld, 123 in H t field, 135 transmission line, 514 waveguide, 426 Dispersive,' 444 Displacement current, 36 Dissipation !actor, 160 Distributed parameters, set Transmission line parameters Distributive law, 3, 15 Divergence: deflnition, 68 of electric field, 75, 119 of magnetic field, 76, 126 Divergence theorem, 72 Domain wall, 141 Dominant mode, 410, 417, 427, 434 Dot (scalar) product, 14 Drift velocity, 112 Drude model, 115 Dry ·cell, 290 Duality: between electric and magnetic dipoles, 274 electromagnetic, 564-567 Echo diagram, 499 Eddy currents, 143 Effective area, 581-582, 585 E field, 28. See Electric field Eigenfunction, 421 Electric charge, 23. See also Charge Electric dipole: field oC 195 moment, 116 moment per unit volume, 117 receiving area, 586 Electric field: of charged aggregate, 182 of charged cloud, 33 of coaxial line, 188, 463 conservative, 41

630

INDEX

Electric field (Continued) definition, Lorentz force, 28 energy, 199--204 flux, 31 induced by magnetic field, 43, 278 of line charge, 33, 183 of parallel plates, 125 of parallel-wire line, 504 at perfect conductor, 148 of planar charge, 34 of point charge, 32 from potential fields, 189, 288, 546 Electric permittivity, see Permittivity Electric polarization, 116. See also Volume polarization density Electric susceptibility, 119 Electrolytic tank, 238 Electromechanical generator, 293 Electromotive force (emf), see Induced voltage Electronic polarization, 116 Electrostatic energy, 20 I Electrostatic forces and torques, 241 Electrostatics, equations of: 181 Elliptic integrals, 3 I 4 Emf, see Induced voltage Energy: of capacitor, 203 of coupled circuits, 319-321 electric field, 20 I of inductive circuits, 296--308 magnetic field, 298-301, 306, 308 summary, 316 in terms of inductance, 299, 320 Energy density: electric, 202, 387 heat loss, 297, 387 magnetic, 301, 387 Equipotential surface, 193, 220, 223 Equivalence theorem, 567 Equivalent conduction loss mechanism, 161 Evanescent mode, 424 E waves, see TM mode External inductances, 30 I, 308 Faraday's law: applied to moving circuit, 280 applied to transformer, 278 differential form, 85, 146 integral f()rm, 29, 40, 146 static form, 41 Farzone field, 553, 560 Ferrimagnetism, and ferromagnetism, 145 Ferrite, I4 6 Ferromagnetic alloys, 144 F field, 566. See also VeCtor electric potential

Field: conservative, 23, 66, 188 defined, I mapping of, 228, 236 non conservative, 23 scalar, I solenoidal, 76 temperature, I, 66 vector, I Field cell, 229 Field mapping, 228, 236 Field pattern, see Antenna pattern, aperture Field- point source- point concepts, 181-184 Filter, waveguide as, 410, 424 Finite- difference method, 215 Fluid-velocity field, I Flux: current, 27 definition, 25 electric, 31 magnetic, 127 partitioning, 306 plots, 71, 229 power, see Power flow Flux field plot: of antenna, 552 of coaxial line, 468 of waveguide modes, 427, 433 Flux linkage, 306, 322; 323 Flux plotting, 228-232, 236 Flux tube, 228 Force: between charges, 181, 244 between parallel plates, 244 on current-carrying wire, 284 electric Held, 28, 242 Lorentz, 28,116,127 magnetic field, 28, 329 on moving charge, 28 from virtual work, 242, 329 Fourier series, 213 Gap voltage, see Induced voltage Gauss' law: for electric field, 29, 121 for magnetic fields, 44, 126 Generator: electrochemical, 290 electromechanical, 293 Gradient: detlnition, 64 of potential field, 189 of temperature field, 66 Grad operator, 64 Graphical flux plotting, 228-232

INDEX Green's theorems, 93,616 Group velocity, 444, 445 Guided waves, see Transmission line; Waveguide Half-wave dipole, 561, 563, 577 Half-wave line, 533 Heat power, see Ohmic (heat) loss Helmholtz equation, 619 H field, see Magnetic intensity Highpass filter, 410, 424 Homogeneity, 167 Horn antenna, 545, 570, 589 H uygens- Fresnel principle, 563 H waves, see TE mode Hyperbolic functions, 532 Hysteresis, 142 Identities, vector, 92 Image method, 219-225 Impedance: of antenna, 563 charaneristic, see Characteristic impedance internal, of isolated wire, 608 internal distributed, 485, 610, 611, 614 imrillSic, see Intrinsic wave impedance; Characteristic impedance measurement of, 530 normalized, 521, 621 total field, 353 total series distributed, 485 transmission line, 513 Impedance matching: of coated lens, 382 methods, 536 quaner wave line, 535 stub, 537 tapered section, 535 Imperfect (lossy) dielectric, 160 Incremental permeability, 143 Index of refraction, 374 Induced voltage (emf): of circuit in motion, 280-286 of coupled circuits, 326 of electromechanical generator, 294 from energy, 317 from time-varying A, 286 from time-varying B, 279 of transformer, 278 Inductance, see Mutual inductance; Self inductance; Transmission line parameters Infinitesimal current element, radiation from, 550 Inhomogeneity, 167 Inhomogeneous wave equation, 94 Input impedance, transmission line, 515-526, 531-536

631

Insulator, 116, 162. See also Dielectric Integral form, Maxwell equations, 29, 86, 147 Integral solution: for static A, 270 for static B, 275 for static IP, 189, 190 for time-retarded A, 548, 619, 620 for time-retarded IP, 620 Integration: line, 21,36,41 surface, 27, 33, 43, 74 volume, 25, 33, 74 Internal (surface) impedance, 485, 610 Internal inductance, 30 I, 309 Intrinsic wave impedance: conductive region, 157, 163 free space, 102, 553 lossy dielectric, 162, 163 plane wave, 102, 157 TEM mode, 417, 466 TE mode, 417, 431 TM mode, 416,426 transmission line, 464, 465 Iron core, inductor, 137, 261, 308 transformer, 324, 325 Irrotational field, 76 Isolated wire, cunent penetration (skin effect), 605 internal (surface) impedance, 608, 610 static magnetic field of; 37, 272, 275 Isotropic, 165 Iterative process, 218, 268

J

Held, see Current density Joule heating, see Ohmic (heat) loss J unction, between transmission lines, 514 Kirchhoff voltage law: for coupled circuits, 327 for de circuit, 240 from energy considerations, 317 for inductive circuit, 292, 295 Klystron, 435 Laminar core, 145 Laplace's equation, 205, 209, 217, 460 Laplacian: of scalar Held, 89, 205 of vector Held, 90, 92, 94 Leakage flux, 263 Ledanche cell, 290 Length element, venor fmill, 11 Lcnz's law, 44 Linear: antenna, 545, 549, 555-563 current, 272, 276

632

INDEX

Magnetic potential, see Vector magnetic potential Magnetic properties, table, 144 Magnetic susceptibility, 133 Magnetic torque, 128, 329 Magnetization current density: surface, 130-132, 136-137 volume, 129 Magnetomotive force (mmf), 264 Mapping of fields, 228 Matching, see Impedance matching Material parameters, table, 168 Matrix, 165,218 Maxwell equations: differential limn, free space, 75, 86 material region, 119,126,132,146,147 with e1 w1 'Fyz dependence, 411-412 integral form, free space, 29, 86 material region, 121, 126, 146, 147 static form, 36, 41, 181, 258--259 summal), tables, 86, 147 Magnetically coupled circuits, sec Coupled circuits time-harmonic form, 86, 88, 153 Magnetic charge, 45, 564, 568 Mean free time, 113 Magnetic circuit, 262-269, 330 Median path, 239, 263, 264 Magnetic core, 137,262,263, 265, 278, 308, 311, Metric coeflicients, 10 M field, see Volume polarization density, magnetic 324 Magnetic dipole: Microstrip line, 507, 508 field of, 274 Mixed boundary value problem, 207 moment, 128 Mks system of unils, 49 moment per unit volume, 129 Mm!,264 torque on, 128 Mobility, 113 Magnetic energy, see Energ), Mode: Magnetic field: definition, 415 from A, 272 relationships, 414--417 of circular loop, 274 TE, 416, 428-440 of coaxial line, 261, 302 TEM, 417, 459 of current distribution, 270-271, 275 TM, 415, 418-428 definition, Lorentz force, 28 Model: of flat current sheet, 39 electrolytic tank, 238 resistive paper, 238 of long isolated wire, 38, 272, 276 of long solenoid, 40, 43, 137 Modified curl operator, 412,465,466 of magnetic circuits, 262-269 Modulation, of carrier, 442 of parallel-wire line, 59, 504. See also Parallel wire Moment: line, capacitance dipole, electric, 116 of toroid, 39, 262 magnetic, 127 with gap, 267 Moment of force, 19. See also Torque Magnetic flux: Motional emf, 282. See also Induced voltage from A, 286 Multilayer system in plane wave propagation, from B, 263, 286 353 Mutual inductance: Magnetic force, 28, 127,329 from flux linkages, 323 Magnetic intensity, 102, 132 Magnetic materials, 140 from magnetic energy, 321 Neumann integral for, 321 Magnetic moment, 138 per unit volume, 129 reciprocity fill', 320 in terms of coupling coefficient, 325 Magnetic permeability, 133 of toroidal transformer, 324 Magnetic polarization, 127

Linear (Continued) region, 164 Linearity, 164 Line current, 464, 469, 482, 486, 491 Line impedance, 513, 517, 521 Line integral, 21-23, 35, 41, 81. See also Ampere's circuital law; Ampere's differential law; Faraday's law Line voltage, 461, 482, 486, 491 Linkage, flux, 306, 323 Loop antenna, 555, 557 Lorentz condition, 547, 548 Lorentz force, 28,116,127,140 Lossless dielectric, 123 Lossless transmission line, 466, 477, 481 Loss tangent, 160, 162 table of; 168 Lossy dielectric, 162-163. See also Conductive region

INDEX Nearzone field, 553 Networks, matching, 536 Neumann boundary condition, 207 Neumann formula, for mutual inductance, 321 NOllconservative field, 23 Nondispersive, 441, 443 Nonlinearity, 120, 164 Nonsinusoidal waves: echo diagram, 499 forward propagated, 492 011 lossless lines, 488-503 with reactive load, 501 reflected, 495 Normal incidence, see Plane waves Normalized admittance, 524 Normalized field impedance, 358, 621 Normalized transmission line impedance, 521 Normal to surface, 10 Ohmic: (heat) loss: in circuits, 297, 315 from Poynting theorem, 387 time-average, 397,402 Ohmic region, see Conductive regioll Ohm's law, point form, 113 w-fJ diagram, 444 Open-circuit voltage, sa Induced voltage (emf) Orientational polarization, 116 Orthogonal curvilinear (generalized coordinates), 7 Orthogonality, of trigonometric functions, 214 Paddle wheel, 77 Parallel line charges, 222 Parallel plates: capacitance, 199 force, 244 Parallel wire line: capacitance, 225 characteristic impedance, 478, 612 distributed parameters, 611 electric field, 504 electrostatic potential, 222 external inductance, 506 internal inductance, 611 magnetic field, 504 time harmonic potential, 503 as transmission line, 459, 478, 484, 503,

GiO Paramagnetic, 140, 145 Parameters, see Transmission line parameters Partial derivative, 62 Pattern, see Antenna pattern Penetration, depth 01: 156, 163

633

Perfect conductor, 124, 148, 156 surface charge on, 124 surface current on, 136 Perfect dielectric, 123 Period, 100, 159 Penneability: of free space, 30, 36, 133 incremental, 143 of materials, table, 168 relative, 133 Permeance, 265 Permittivity: complex, 153,160,161,171 of free space, 30, 50, 120 of materials, table, 168 relative, 120 P field, 116. See also Volume polarization density, electric Phase constant, from distributed parameters, 476, 477,482,486,487 for low-loss line, 477,487 for plane wave, 99, 154, 162, 163 forTEM mode, 417, 465, 476, 477,482,486, 487 for transmission line, lossless, 465, 477,487 lossy, 465, 476, 482, 486, 487 vector, 368 lor waveguide, 424, 431, 436 Phase velocity: apparent, 369 plane wave, 101, 159 transmission line, 465 waveguide, 426, 431, 436 phasor notation, see Time harmonic Plane, equation of; 366 plane waves: in conductive region, 152-163 in empty space, 96-103 normally incident, multiple region, 352-358 perfect conductor, 344-347 two regions, 347-350 oblique incidence, 365-366 power relations, 389-394, 395, 397-399, 401 Pocklington's theorem, 556 Point charge, 32, 42, 182, 199 Point in space, see Coordinate points Poisson equation: scalar, 204, 215 vector, 269 Polarization: circular, 105 electric, 116 elliptical, 104, 105 linear, 103, 104 magnetic, 127

634

INDEX

Polarization (Continued) parallel, 371 perpendicular, 375 Polar molecule, 116, 117 Position vector, 11 Potential: reference, 190, 197 retarded, 548 scalar electric, 189, 288 for time varying fields, 288, 548 vector magnetic, 269, 288, 546 Potential difference, 197 Potential field: in coaxial line, 462-463 complex form, 461 of dipole charge, 195 of elementary antenna, 550-551 ofline charge, 193 of parallel-wire line, 222, 503 of point charge, 194 Power density, see Poynting vector Power flow: in plane wave, 389, 392 in transmission line, 515, 518 in waveguide, 440 Power loss, see also Ohmic (heat) loss plane wave in conductive region, 393, 394, 399, 401 time-average, 397, 398 in waveguide walls, 447-452 ill wire, 388 Poynting theorem: complex form, 401, 402 real-time form, 386, 387 time-average form, 397, 402 Poynting vector: complex representation, 400 definition, 385 for plane waves, 390, 392 time-average, 394, 400 time-instantaneous, 385 Primary cell, 291 Probe, slotted line, 439, 530 Product: scalar (dot), 14 vector (cross), 17 vector with scalar, 4 Product solution, 209, 419 Projection of vector, 7, 14 Propagating mode, 424, 437 Propagation constant: for plane wave, 154 for transmission (TEM) line, 417, 465,476,477, 482, 486 for waveguide modes, 423, 424, 430, 431, 436

Proper (eigen) function, 421, 451 Q(pertaining to complex Poynting theorem), 402 Q(quality factor), 235 Quarter-wave line, 534 Quarter-wave transformer, 535 Quasi-static fields, 43, 276-277 Radiated power: from elementary dipole, 554 from half-wave dipole, 563 Radiation: from aperture antenna, 567-575 from elementary dipole, 550-554 from linear antenna, 555-563 Radiation resistance, 562 Radiation zone, see Farzone field Rationalized mks units, table, 50 Real amplitude, 99 Real part of complex field, 88 Real-time field, 87, 99, 155,344,425,437,467 Rectangular (Cartesian) coordinates, 4, 5, 8, 9 Rectangular waveguide modes, see Mode Reflection: at dielectric interface, 347 of nonsinusoidal waves, 495 from plane conductor, 344 at transmission-line junction, 512 Reflection coefficient: lor oblique incidence, 374, 375 for plane waves, 353 for transmission lines, 512 Refraction: of current flux, 172 of electric flux, 148 of magnetic flux, 139 of wave at oblique incidence, 373 Relative permeability: definition, 133 table, 144, 168 Relative permittivity: definition, 120 table, 168 Relaxation time: charge, 152, 234 drift velocity, 114 Reluctance, 264 Remanent (residual) magnetic field, 142 Resistance: analogy of magnetic circuit, 264 radiation, 562 ofthin circuit, 240-241, 264 transmission line, see Transmission line parameters

INDEX Resistive paper model, 238 Retardation effects, 546, 548 Retarded potential, 548, 619, 620 Rhombus, 15 Right-hand rule, 18,35,41,77,82 Round wire, 37, 81,272, 302, 309, 388, 484, 602-610 Saturation magnetization, 142 Scalar, 2 Scalar electric potential, see Potential Scalar field, 1 Scalar (dot) product, 14 Secondary cell, 291 Self inductance: from A and], 299 of circuit in free space, 300 of circular loop, 315 of coaxial line, 303 from energy, 299 external and internal, 301, 308-309, 313 from flex linkages, 306 from integration throughout space, 301 internal, of round wire, 302, 309 from Neumann integral, 312 of parallel-wire line, 305, 310 summary table, 316 of toroid, 311 Separation constants, 209-210,419 Separation of variables, 209, 419, 429 Series solution, 213 Sidelobes, 572, 574 Single-stub matching, 537-538 Sinusoidal steady state, see Time harmonic Skin depth, 156, 163 Skin effect: plane wave in conductive region, 156, 163 in round wire, 605 Slotted line, 530 Slot in waveguide, 439 Smith chart: derivation, 621 phase rotation on, 358,359,521,625 plane wave problems, 358-361 transmission line problems, 520-526 Snell's law, 377, 595 Solenoid, 40, 44, 137,278,280 Solenoidal field, 76 Source-point field-point concepts, 182-184, 189 Speed oflight, 49 Sp herical capacitor, 198, 235 Spherical charge, 32-33 Spherical coordinates, 4, 5, 8, 9 Spin, 127

635

Standing wave ratio: for plane waves, 362-364 from Smith chart, 364 on transmission line, 527-529 Standing waves: plane wave, 345-347,361-365 on transmission line, 526-529 Static fields: electric, properties of, 181 magnetic, properties of, 258-259 Step function, 498 Stokes' theorem, 81 Stored energy, see Energy Stub matching, 537 Surface, outward normal on, 10, 11, 73 Surface current density, 136:149, 151, 271 Surface divergence, 151,170 Surface free charge density, 24, 32, 34, 122-124, 125,148,149,151,169,171 Surface impedance, see Impedance Surface integral, 26, 27, 29, 33, 74, 75, 83,84 Surface polarization charge density, electric, 124-126 Surface polarization current density, magnetic, 130-132, 136, 137 Surface vector element, 10, 11 Susceptibility: electric, 11 9 magnetic, 133 SWR, see Standing wave ratio Symmetry: in field mapping, 232 about a line (axial), 33, 37 about a plane, 34, 38 about a point (spherical), 31, 33 Tangential component, see Boundary conditions Taylor's expansion, 69, 78, 216 TEM mode, 417, 459-469 TE mode, 416-417, 428-440 TE 10 mode, rectangular waveguide, 433-440 Temperature field, 2, 66 Time constant, see Relaxation time Time-domain fields, set Real-time field Time harmonic: Maxwell equations, 86, 88, 153 phasor notation, 87-88 Poynting theorem, 397-402 wave equation, 95 Time instantaneous, see Real-time field TM mode, 415, 418-428 Toroid, with gap. 267 magnetic field of, 39-40, 261-262 self inductance o( 311

636

INDEX

Torque: electrostatic, 242 on infinitesimal current loop, 128 magneto static, 330 Total differential, 64, 190 Total field impedance, 353, 516 Total reflection, M4, 376 Transmission band, 441, 535 Transmission coefficient: parallel polarization, 374 perpendicular polarization, 375 Transmission line, attenuation constant, see Attenuation constant characteristic impedance, see Characteristic impedance circuit analog (model), 264, 480, 487 coaxial, see Coaxial line current waves, 464,469,482, 486, 491 diHerentiai equations of (transmission line equations), 471, 475, 4805-486 electric field of; 460-461, 463 fields of, 460-468 hall~wave, 0533 input impedance, 515, 532 intrinsic wave impedance, 466 lossless, 466 magnetic field of, 464, 465, 466, 467 microslrip, 507-508 mode (TEM), 417, 459 non sinusoidal waves on, 488-503 normalized im pedance, 521 parallel wire, see Parallel wire line parameters, see Transmission line parameters perfect conductor, 459-482 phase constant, see Phase constant phase velocity, 465 propagation constant, 417, 465, 476, 477, 482, 486 quarter-wave, 534 quasi- static potential field of, 460, 462 reflection coefllcient, 496, 05 12-513 Smith chart, applied to, 520-526 standing waves, 526-529 table, summary relations, 517 tapered, 535 voltage waves, 461, 463, 470,482,486,488, 491-503, 512, 526-530 wave equations, 481, 486, 488 Transmission line equations (telegraphist' 5), perfect conductor line, 471 conductor impedance included, 485-486 Transmission line parameters: coaxial line, perfect conductors, 478-479 conductor impedence included, 612-6 I 5

parallel wire line, perfect conductors, 506-507 conductor impedance included, 610-612 perfect conductor line, 471-479 Transmit-receive link, 582-589 Transverse electric, see TE !node Transverse electromagnetic, see TEM mode Transverse magnetic, see TM mode Traveling wave, 99-100 Trigonometric series, 213 Two-conductor line, see Transmission line Uniform plane wave, see Plane waves Uniqueness, 93, 206, 269 Units, mks, table, 50 Unit vector, 3, 5, 6, 8 Universal circle diagram, 432 Vector: acceleration, 13 complex, 87-88 displacement, 3, 11-12 negative, 3 position, II Poynting, see Poynting vector unit, 3, 5, 6; 8 velocity, 13 Vector algebra, 3, 4, 14-20 Vector calculus, differentiatioll, 61-63 integration, 20-23 Vector component, 5, 7, 8 Vector diHerential operator, see Curl; Gradient Vector electric potential, 566 Vector field, I conservative (irrotational), 41, 42 nonconservative, 76 solenoidal, 76 Vector identities, table, 92 Vector length element, 11 Vector magnetic potential: static, 269 time-varying, 288, 0546 of wire loop, 274 Vector product, 17 Vector sum, 3 Vector surface clement, 9, 10 Velocity: drift, 112 group, 444-446 phase, see Phase velocity Virtual work: electrostatic, 242 magneto static, 328 Voltage, induced, see Induced voltage (emf) Voltage generators, 290-296

INDEX

Voltage standing wave ratio, see Standing wave ratio Volume charge density, 23-25, 27, 29-30 Volume current density, 26, 27, 36-38,113, 156 Volume element, 9, 10 Volume integral, 20,25,33,74, 75 Volume polarization density: electric, 116-122, 124-126 magnetic, 129-134, 136-138 VSWR, see Standing wave ratio Wall loss: attenuation, 451-452 waveguide, 447-452 Wave equation: with e1wt'fyz dependence, 413-444, 419, 429 scalar, 95, 98 time harmonic, 95, 98 transmission line, 481, 486 vector, 94, 95, 96 Waveguide: attenuation below cutoff, 424, 431, 436 boundary conditions, 420, 429 cutofffrequency, 423, 428, 431, 433-436 group velocity, 444-445 as highpass filter, 424 mode relations, 415-417 modulated signal in, 445 phase constant, 424, 431, 436

637

phase velocity, 426, 431, 436 propagation constant, 409, 423,424, 431, 436 rectangular, 418-452 TE mode solutions, 428 TM mode solutions, 418 wall losses, 447 wave equation for, 419, 429 wave impedance, intrinsic, 416, 417, 426, 431, 436 Wave impedance, see Intrinsic wave impedance Wavelength: free space, 101 in lossy region, 157, 160 plane wave, 101, 157 waveguide, 426, 431, 436 Waves: incident and reflected, 344, 347, 351, 353, 361, 411 Ilonsinusoidal, 488r489, 492, 495 plane, see plane waves spherical, 552, 553, 560 standing, see Standing waves transmission line, 461-468, 488-491 traveling, 100 X band waveguide, 427,434,439 Zero potential reference, 190 Zero reflection angle, 376

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