CHAPTER 2

INTRODUCTION TO WAVES

2-1. The Wave Equation. A field that is a function of both time and
space coordinates can be called a wave. We shall, however, be a bit
more restrictive in our definition and use the term wave to denote a solu-
tion to a particular type of equation, called a wave equation. Electro-
magnetic fields obey wave equations, so the terms wave and field are
synonymous for time-varying electromagnetism. In this chapter we
shall consider a number of simple wave solutions to introduce and illus-
trate various a-c electromagnetic phenomena.
For the present, let us consider fields in regions which are source-free
(Ji = Mi = 0), linear (z and y independent of lEI and IHI), homogeneous
(z and y independent of position), and isotropic (z and yare scalar).
The complex field equations are then
V X E = -zH
(2-1)
V X H = yE
The curl of the first equation is
V X V X E = - sv X H
which, upon substitution for V X H from the second equation, becomes
V X V X E = -zyE
The frequently encountered parameter
k = V -zy (2-2)
is called the wave number of the medium. In terms of k, the preceding
equation becomes
v X V X E - k2E = 0 (2-3)
which we shall call the complex vector wave equation. If we return to
Eqs. (2-1), take the curl of the second equation, and substitute from the
first equation, we obtain
v X V X H - k 2H = 0 (2-4)
Thus, H is a solution to the same complex wave equation as is E.
37
38 TIME-HARMONIC ELECTROMAGNETIC FIELDS

The wave equation is often written in another form by defining an

operation
V2A = V(V • A) - V X V X A
In rectangular components, this reduces to
V2A = U z V 2A z + u V 2A + u V2A .
ll II z

where U z , UII' and u, are the rectangular-coordinate unit vectors and V2

is the Laplacian operator. It is implicit in the wave equations that
V·E = 0 V·H = 0 (2-5)
shown by taking the divergence of Eqs. (2-3) and (2-4). Using Eqs. (2-5)
and the operation defined above, we can write Eqs. (2-3) and (2-4) as
V2E + k 2E = 0
(2-6)
V2H + k 2H = 0
These we shall also call vector wave equations. They are not, however,
so general as the previous forms, for they do not imply Eqs. (2-5). In
other words, Eqs. (2-6) and Eqs. (2-5) are equivalent to Eqs. (2-3) and
(2-4). Thus, the rectangular components of E and H satisfy the complex
scalar wave equation or Helmholtz equation 1
V~ + k~ = 0 (2-7)
We can construct electromagnetic fields by choosing solutions to Eq. (2-7)
for E z, E y , and E; or Hz, H y , and Hz, such that Eqs. (2-5) are also satisfied.
To illustrate the wave behavior of electromagnetic fields, let us con-
struct a simple solution. Take the medium to be a perfect dielectric,
in which case fJ = jWE, Z = jwp., and
k = W yEP. (2-8)
Also, take E to have only an x component independent of x and y. The
first of Eqs. (2-6) then reduces to

d;~" + k2E" = 0

which is the one-dimensional Helmholtz equation. Solutions to this are

linear combinations of ei k z and e-i k z • In particular, let us consider a
solution
B, = Eoe- i ka (2-9)
This satisfies V • E = 0 and is therefore a possible electromagnetic field.
1 We shall use the symbol w/I to denote "wave functions," that is, solutions to
Eq. (2-7). Do not confuse these w/I's with magnetic flux.
 INTRODUCTION TO WAVES 39
The associated magnetic field is found according to
jwp,H = -v X E = uujkEz

which, using Eq. (2-8), can be written as

E,,=JH/I (2-10)

Ratios of components of E to components of H have the dimensions of

impedance and are called wave impedances. The wave impedance associ-
ated with our present solution,

11 E" = ~
= n, (2-11)
'1;
is called the intrinsic impedance of the medium. In vacuum,

110 = ~",.
'1;;; 12<hr ",. 377 ohms (2-12)

We shall see later that the intrinsic impedance of a medium enters into
wave transmission and reflection problems in the same manner as the
characteristic impedance of transmission lines.
To interpret this solution, let Eo be real and determine t and 3C accord-
ing to Eq. (1-41). The instantaneous fields are found as
Bz = V2 Eo cos (wt - kz)
= -V2 Eo cos (wt -
(2-13)
XU kz)
'1

This is called a plane wave because the phase (kz) of B and X is constant
over a set of planes (defined by z = constant) called equiphase surfaces.
It is called a uniform plane wave because the amplitudes (Eo and EO/fJ) of
Band 3C are constant over the equiphase planes. B and X are said to be
in phase because they have the same phase at any point. At some specific
time, B and X are sinusoidal functions of z, The vector picture of Fig. 2-1
illustrates t and 3C along the z axis at t = O. The direction of an arrow
represents the direction of a vector, and the length of an arrow represents
the magnitude of a vector. If we take a slightly later instant of time,
the picture of Fig. 2-1 will be shifted in the +z direction. We say
that the wave is traveling in the +z direction and call it a traveling wave.
The term polarization is used to specify the behavior of t lines. In this
wave, the t lines are always parallel to the x axis, and the wave is said to
be linearly polarized in the x direction.
The velocity at which an equiphase surface travels is called the phase
40 TIME-HARMONIC ELECTROMAGNETIC FIELDS

velocity of the wave. An equiphase plane z = Zp is defined by

wt - kz p = constant
that is, the argument of the cosine functions of Eq. (2-13) is constant.
As t increases, the value of Zp must also increase to maintain this con-
stancy, and the plane z = Zp will move in the +z direction. This is illus-
trated by Fig. 2-2, which is a plot of 8 for several instants of time. To
obtain the phase velocity dZp/dt, differentiate the above equation. This
gives

The phase velocity of this wave is called the intrinsic phase velocity Vp of
the dielectric and is, according to the above equation,
dz p ca 1
vp = - = - =-- (2-14)
dt k YEP
In vacuum, this is the velocity of light: 3 X 108 meters per second.
The wavelength of a wave is defined as the distance in which the phase
increases by 211" at any instant. This distance is shown on Fig. 2-2. The
wavelength of the particular wave of Eqs. (2-13) is called the intrinsic
wavelength X of the medium. It is given by kX = 211", or
X - 211" _ 211"v p
v _ p (2-15)
-T--;--7
where f is the frequency in cycles per second. The wavelength is often
used as a measure of whether a distance is long or short. The range of
wavelengths encountered in electromagnetic engineering is large. For
example, the free-space wavelength of a 60-cycle wave is 5000 kilometers,
whereas the free-space wavelength of a 1000-megacycle wave is only 30
centimeters. Thus, a distance of 1 kilometer is very short at 60 cycles,

Direction of travel ~

FIG. 2-1. A linearly polarized uniform plane traveling wave.

 INTRODUCTION TO WAVES 41

tAlt = 0
CJJt = -rr/4
CJJt = -rr/2

104------- ~ ------..1
FIG. 2-2. 8 at several instants of time in a linearly polarized uniform plane traveling
wave.

but very long at 1000 megacycles. The usual circuit theory is based on
the assumption that distances are much shorter than a wavelength.
2.2. Waves in Perfect Dielectrics. In this section we shall consider
the properties of uniform plane waves in perfect dielectrics, of which
free space is the most common example. We have already given a special
case of the uniform plane wave in the preceding section. To summarize,

E; = Eoe- i k • H'I/ = Eo e- i k z
11
_/- 2r W
where k=WVJl.E=-=-
X VI'
(2-16)
~=J
It is an x-polarized, +z traveling wave. Because of the symmetry of the
rectangular -coordin~te system, other uniform plane-wave solutions can
be obtained by rotations of the coordinate axes, corresponding to cyclic
interchanges of coordinate variables. We wish to restrict consideration
to +z and -z traveling waves; so we shall consider only the transforma-
tions (x,y,z) to (-y,x,z), to (x,-y,-z), and to (y,x,-z). This procedure,
together with our original solution, gives us the four waves

(2-17)
42 TIME-HARMONIC ELECTROMAGNETIC FIELDS

where the previously used Eo has been replaced by A, B, C, or D. The

superscript + denotes a +z traveling wave, and the superscript - denotes
a -z traveling wave. The most general uniform plane wave is a super-
position of Eqs. (2-17).
We have already interpreted the first wave of Eqs. (2-17) in Sec. 2-1.
This also constitutes an interpretation of the other three waves if the
appropriate interchanges of coordinates are made. We have not yet
mentioned power and energy considerations, so let us do so now. Given
the traveling wave

we evaluate the various energy and power quantities as

E
W. =2 82 = EEo2 cos" (wt - kz)
Wm = ~ JC2 = EEo2 cos" (wt - kz)
2
(2-18)
S = 8 X 3C = u, -2 E o2 cos" (wt - kz)
71
E o2
S = E X H* = Uz -
71

Thus, the electric and magnetic energy densities are equal, half of the
energy of the wave being electric and half magnetic. We can define a
velocity of propagation of energy Ve as
power flow density S
Ve = energy density
=- --
ui, + W m
(2-19)

For the uniform plane traveling wave, from Eqs. (2-18) and (2-19) we find
1
v, = VJJ.E
which is also the phase velocity [Eq. (2-14)]. These two velocities are
not necessarily equal for other types of electromagnetic waves. In gen-
eral, the phase velocity may be greater or less than the velocity of light,
but the velocity of propagation of energy is never greater than the velocity
of light.
Another property of waves can be illustrated by the standing wave

E% = Eo sin kz H II = J.s, k
- cos z (2-20)
71

obtained by combining the first and third waves of Eqs. (2-17) with
 INTRODUCTION TO WAVES 43
A -C = jE o/ 2. The corresponding instantaneous fields are

8z = y2 Eo sin kz cos wt Xu = - '\1'2-Eo

- cos kz sin wt
11
Note that the phase is now independent of z, there being no traveling
motion; hence the name standing wave. A picture of E and 3C at some
instant of time is shown in Fig. 2-3. The field oscillates in amplitude,
with 8 reaching its peak value when X is zero, and vice versa. In other
words, 8 and JC are 90° out of phase. The planes of zero 8 and X are
fixed in space, the zeros of 8 being displaced a quarter-wavelength from
the zeros of X. Successive zeros of 8 or of 3C are separated by a half-
wavelength, as shown on Fig. 2-3. The wave is still a plane wave, for
equiphase surfaces are planes. It is still a uniform wave, for its amplitude
is constant over equiphase surfaces. It is still linearly polarized, for E
always points in the same direction (or opposite direction when 8 is
negative).
The energy and power quantities associated with this wave are

W. = ~ 8 2 = eE o2 sin" kz cos" wt

Wm = ~ :JC2 = eE o2 cos" kz sin" wt

(2-21)
S = E X :re = - Uz
E 0 2 SIn
2;J • 2k z SIn
. 2wt

·E 2
S =E X H* = -n. J 2: sin 2kz
The time-average Poynting vector S = Re (8) is zero, showing no power
flow on the average. The electric energy density is a maximum when
the magnetic energy density is zero, and vice versa. A picture of energy

FIG. 2-3. A linearly polarized uniform plane standing wave.

 TIME-HARMONIC ELECTROMAGNETIC FIELDS

Z
FIG. 2-4. Standing-wave pattern of two oppositely traveling waves of unequal ampli..·
tudes.

oscillating between the electric and magnetic forms can be used for this
wave. Note that we have planes of zero electric intensity at kz = na ,
n an integer. Thus, perfect electric conductors can be placed over one
or more of these planes. If an electric conductor covers the plane z = 0,
Eqs. (2-20) represent the solution to the problem of reflection of a uniform
plane wave normally incident on this conductor. If two electric con-
ductors cover the planes kz = nl1r and kz = n21r, Eqs. (2-20) represent
the solution of a one-dimensional" resonator."
A more general x-polarized field is one consisting of waves traveling
in opposite directions with unequal amplitudes. This is a superposition
of the first and third of Eqs. (2-17), or
E; = Ae-jk~ + Ceik~ (2-22)
H
11
=! (Ae-ik~ -
'1J
Ceik~)

If A = 0 or C = 0, we have a pure traveling wave, and if IAI = ICI, we

have a pure standing wave. For A ¢ C, let us take A and C real! and
express the field in terms of an amplitude and phase. This gives
_j tan- 1 (A - Ctan kZ)
E:x; = V A 2 + C2 + 2AC COS 2kz e A+C (2-23)
The rms amplitude of E is
VA2 + C2 + 2AC cos 2kz
which is called the standing-wave pattern of the field. This is illustrated
by Fig. 2-4. The voltage output of a small probe (receiving antenna)
connected to a detector would essentially follow this standing-wave pat-
1 This is actually no restriction on the generality of our interpretation, for it corre-
sponds to a judicious choice of z and t origins.
 INTRODUCTION TO WAVES 45
tern. For a pure traveling wave, the standing-wave pattern is a constant,
and for a pure standing wave, it is of the form [cos kzl, that is, a "recti-
fied" sine wave. The ratio of the maximum of the standing-wave pat-
tern to the minimum is called the standing-wave ratio (8WR). From
Fig. 2-4, it is evident that
SWR = A +C (2-24)
A-C
because the two traveling-wave components [Eqs. (2-22)] add in phase at
some points and add 180 0 out of phase at other points. The distance
between successive minima is 'A/2. The standing-wave ratio of a pure
traveling wave is unity, that of a pure standing wave is infinite. Plane
traveling waves reflected by dielectric or imperfectly conducting bounda-
ries will result in partial standing waves, with 8WR's between one and
infinity.
Let us now consider a traveling wave in which both E; and E y exist.
This is a superposition of the first and second of Eqs. (2-17), that is,
E = (uzA+ uuB)e-jkz (2-25)
H = (-uxB + u A ) ! e- jkz
7l
1J

If B = 0, the wave is linearly polarized in the x direction. If A = 0,

the wave is linearly polarized in the y direction. If A and B are both
real (or complex with equal phases), we again have a linearly polarized
wave, with the axis of polarization inclined at an angle tan"! (B/ A) with
respect to the x axis. This is illustrated by Fig. 2-5a. If A and Bare
complex with different phase angles, 8 will no longer point in a single
spatial direction. Letting A = IAlej a and B = IBlej b , we have the instan-

y y
e vibrates in e rotates in
this direction
-- this direction

~
-,
C1Jt = 0
/ X
/
t = ~/4
ClJt = 'IT/2

(a) (b)

FIG. 2-5. Polarization of a uniform plane traveling wave. (a) Linear polarization;
(b) elliptical polarization.
46 TIME-HARMONIC ELECTROMAGNETIC FIELDS

taneous electric intensity given by

8~ = V2 IA I cos (wt - kz + a)
8" = V2IBI cos (wt - kz + b)
A vector picture of 8 for various instants of time changes in both ampli-
tude and direction, going through this variation once each cycle. For
example, let IAI = 2 IBI, a = 0, and b = r/2. A plot of 8 for various
values of t in the plane z = 0 is shown in Fig. 2-5b. The tip of the arrow
in the vector picture traces out an ellipse, and the field is said to be
elliptically polarized. Depending upon A and B, this ellipse can be of
arbitrary orientation in the xy plane and of arbitrary axial ratio. Linear
polarization can be considered as the special case of elliptic polarization
for which the axial ratio is infinite.
If the axial ratio is unity, the tip of the arrow traces out a circle, and
the field is said to be circularly polarized. The polarization is said to be
right-handed if 8 rotates in the direction of the fingers of the right hand
when the thumb points in the direction of propagation. The polarization
is said to be left-handed if 8 rotates in the opposite direction. The special-
ization of Eq. (2-25) to right-handed circular polarization is obtained by
setting A = jB = Eo, giving

(2-26)

A vector picture of the type of Fig. 2-1 for this wave would show 8 and 3C
in the form of two corkscrews, with 8 perpendicular to 3C at each point.
As time increases, this picture would rotate giving a corkscrew type of
motion in the z direction. The various energy and power quantities
associated with this wave are

'Wm = ~ 3C2 = fE o2
(2-27)
2
S = 8 X :JC = Uz - E 02
1]
2
S = E X H* = u, - E 0 2
1]

Thus, there is no change in energy and power densities with time or

space. Circular polarization gives a steady power flow, analogous to
circuit-theory power transmission in a two-phase system.
 INTRODUCTION TO WAVES 47
As a final example, consider the circularly polarized standing-wave field
specified by
E = (u, +
julI)E o sin kz
(2-28)
+
H = (u.. jUI/) Eo cos kz
1]

This is the superposition of .Eqe. (2-17) for which A = -C = jE o/ 2,

D = - B = E o/2. The corresponding instantaneous fields are
8 = (u, cos wt - UII sin wt) V2 Eo sin kz
3C = (u, cos wt - UII
.- /- Eo cos ke
sin wt) v 2 -
'1J

Note that 8 and 3C are always parallel to each other. A vector picture
of 8 and 3C at t = 0 is shown in Fig. 2-6. As time progresses, this picture
rotates about the z axis, the amplitudes of 8 and :JC being independent of
time. It is only the direction of 8 and :JC which changes with time. The
amplitudes of 8 and 3C are, however, a function of z, giving a standing-
wave pattern in the z direction. The energy and power densities associ-
ated with this wave are

we = ~2 6 2 = EEo2 sin" kz

Wm = ~ 3C2 = EEo2 cos" ke

(2-29)
S=8X:JC=O
S = -Uz tE
1]
02 sin 2kz

It is interesting to note that the instantaneous energy and power densities

are independent of time. This field can represent resonance between two
perfectly conducting planes situated where E is zero. It thus seems that
the picture of energy oscillating between the electric and magnetic forms

z
y

FIG. 2-6. A circularly polarized uniform plane standing wave.

48 TIME-HARMONIC ELECTROMAGNETIC FIELDS

is not generally valid for resonance. However, the circularly polarized

standing wave is the sum of two linearly polarized waves which can exist
independently of each other. We actually have two coincident reso-
nances (called a degenerate case), and the picture of energy oscillating
between electric and magnetic forms applies to each linearly polarized
resonance.
2-3. Intrinsic Wave Constants. When the wave aspects of electro-
magnetism are emphasized, the wave number k and the intrinsic imped-
ance 11, given by
k = V-zy (2-30)

play an important role. The second equation is a generalization of Eq.

(2-11), obtained in the same manner as Eq. (2-11) when z and yare not
specialized to the case of a perfect dielectric. We can solve Eqs. (2-30)
for z and '0, obtaining
z = jk11 y =jk (2-31)
11
A knowledge of k and 11 is equivalent to a knowledge of z and '0, and
hence specifies the characteristics of the medium.
The wave number is, in general, complex, and may be written as
k = k' - jk" (2-32)
where k' is the intrinsic phase constant and k" is the intrinsic attenuation
constant. We have already seen that when k = k', it enters into the
phase function of the wave. We shall see in the next section that k"
causes an exponential attenuation of the wave amplitude. The behavior
of k can be illustrated by a complex diagram relating k to z and 'o.
This is shown in Fig. 2-7. In the
Jm expressions
'0 = U + WE" + jWE'
Z = wp," + jwp,'
U, E", and u" are always positive in
source-free media, for they account
for energy dissipation. The param-
Re eters E' and p' are usually positive but
may be negative for certain types of
atomic resonance. Thus, z and y
- zy usually lie in the first quadrant of the
complex plane, as shown in Fig. 2-7.
FIG. 2-7. Complex diagram relating k to The product - zy then usually lies
i and g. in the bottom half of the complex
 INTRODUCTION TO WAVES 49
1m

FIG. 2-8. Complex diagram relating ."

to zand g.

l/y

plane. The principal square root, k = V -zy, lies in the fourth quad-
rant, showing that k' and k" are usually positive. Even when E' or Il' is
negative, k" is positive; it is only k' that could conceivably be negative.
In lossless media, 0 = jWE, Z = jwp" and k is real.
The intrinsic wave impedance can be considered in an analogous
manner. Expressing 11 in rectangular components, we have
1] = <R + JOC (2-33)
where <R is the intrinsic wave resistance and oc is the intrinsic wave react-
ance. For a wave in a perfect dielectric, 11 is purely resistive and is there-
fore the ratio of the amplitude of 8 to :IC. We shall see in Sec. 2-4 that
oc introduces a phase difference between 8 and sc, The complex diagram
relating 1] to '0 and z in general is shown in Fig. 2-8. In source-free
regions, (J', E", and u" are always positive, and E' and p.' are usually posi-
z
tive. Thus usually lies in the first quadrant and 1/'0 in the fourth
quadrant. The ratio zlO therefore usually lies in the right half plane
and 1] in the sector ±45° with respect to the positive real axis. When
E' or p.' is negative, 11 may lie anywhere in the right half plane, but <R is
never negative. In lossless media, the wave impedance is real.
There are several special cases of particular interest to us. First, con-
sider the case of no magnetic losses. From the first of Eqs. (2-31), we
have
Z zk* jk*z
11 = jk = jkk* = - Izl 101

the last equality following from Eqs. (2-30). Now for z = jwp. = llzl,
we have
k*
'1 = 101 no magnetic losses (2-34)
50 TIME-HARMONIC ELECTROMAGNETIC FIELDS

TABLE 2-1. WAVE NUMBER (k = k' - jk") AND INTRINSIC

IMPEDANCE (71 = <R +jX = 171le i l")

k' k" <R OC

General Re v' -zy -1m V -zy Re~ Im~

k' k"
No magnetic losses Im yjwp,y Re Vjwp,y - -
Iyl 101

Perfect dielectric wy; 0

~ 0

Good dielectric wYP,E' ~"~!!:.E'

2 ~ l'~
2E' E'

Good conductor
Jf Jf ~ 20- ~ 20-

Separation into real and imaginary parts is shown explicitly in row 2 of

Table 2-1. A similar simplification can be made for the case of no elec-
tric losses. (See Prob. 2-13.) Three special cases of materials with no
magnetic losses are (1) perfect dielectrics, (2) good dielectrics, and (3)
good conductors. The perfect dielectric case is that for which

k = WV~ ~=J
This is summarized in row 3 of Table 2-1. A good dielectric is charac-
terized by = jWJl, Y = WE" + jWE /, with E'
z » E". In this case, we have

k = w ~/ll (1 - j ::) ~ w y /lE' (1

k* Ip. ( . E" )
'I = I11 I ~ '\j;; 1 + J 2E'
which is summarized in row 4 of Table 2-1. Finally, a good conductor is
characterized by z = jWJl, Y = a + jWE, with a » WE. In this case, we have
k = V -jWJl(U + jWE) ~ V -jWJlU
~ = ~~ ~ ~j:/l
The last row of Table 2-1 shows these parameters separated into real and
imaginary parts.
 INTRODUCTION TO WAVES 51
2-4. Waves in Lossy Matter. The only difference between the wave
equation, Eq. (2-7), for lossy media and loss-free media is that k is com-
plex in lossy media and real in loss-free media. Thus, Eq. (2-9) is still a
solution in lossy media. In terms of the real and imaginary parts of k,
it is
E z = Eoe-ikz = Eoe-k"Ze-ik'J (2-35)
Also, H is still given by Eq. (2-10), except that '1 is now complex. Thus,
the H associated with the E of Eq. (2-35) is

H = Eo e-ikz = Eo e-ite-k"'e-ik" (2-36)

11 '11 1111

where '1 = 1'11lei t . The instantaneous fields corresponding to Eqs. (2-35)

and (2-36) are
8z = 0 Eoe-k"z cos (wt - k'z)
(2-37)
X
y
= y2 Eo
1111
e-k"z cos (wt - k'z - r)

Thus, in lossy matter, a traveling wave is attenuated in the direction of

travel according to «<», and X is no longer in phase with 8. A sketch
of t and 3C versus z at some instant of time would be similar to Fig. 2-1
except that the amplitudes of t and 3C would decrease exponentially with
z, and 3C would not be in phase with 8 (3C usually lags 8). A sketch of
8 z versus z for several instants of time is shown in Fig. 2-9 for a case of
fairly large attenuation. A sketch of :ley versus z would be similar in form.

Direction of travel ~

-.... -e: Envelope -

-- e - k"z

-- ----
z

-- --
wt = -rr/2
ClJt = -rr/4
ClJt = 0

FIG. 2-9. 8 at several instants of time in a linearly polarized uniform plane traveling
wave in dissipative matter.
52 TIME-HARMONIC ELECTROMAGNETIC FIELDS

The wave of Eq. (2-37) is still uniform, still plane, and still linearly
polarized. So that our definitions of phase velocity and wavelength will
be unchanged for lossy media, we should replace k and k' in the loss-free
formulas, or
~=21r=~ (2-38)
k' f
Then Vp is still the velocity of a plane of constant phase, and A is still
the distance in which the phase increases by 21r.
Two cases of particular interest are (1) good dielectrics (low-loss), and
(2) good conductors (high-loss). For the first case, we have (see Table
2-1)

in good dielectrics (E" « E') (2-39)

Thus, the attenuation is very small, and e and 3C are nearly in phase.
The wave is almost the same as in a loss-free dielectric. For example, in
polystyrene (see Fig. 1-10), a IO-megacycle wave is attenuated only 0.5
per cent per kilometer, and the phase difference between 8 and 3C is
only 0.003°. The intrinsic impedance of a dielectric is usually less than
that of free space, since usually E' > EO and J.I, = J.l,o. The intrinsic phase
velocity and wavelength in a dielectric are also less than those of free
space.
In the high-loss case (see Table 2-1), we have

kl=~
k" = ~w;cr
in good conductors (0" » WE) (2-40)
1771 = ~w:
r=~4

Thus, the attenuation is very large, and 3C lags e by 45°. The intrinsic
impedance of a good conductor is extremely small at radio frequencies,
having a magnitude of 1.16 X 10- 3 ohm for copper at 10 megacycles.
The wavelength is also very small compared to the free-space wavelength.
For example, at 10 megacycles the free-space wavelength is 30 meters,
while in copper the wavelength is only 0.131 millimeter. The attenuation
 INTRODUCTION TO WAVES 53
in a good conductor is very rapid. For the above-mentioned to-mega-
cycle wave in copper the attenuation is 99.81 per cent in 0.131 milli-
meter of travel. Thus, waves do not penetrate metals very deeply. A
metal acts as a shield against electromagnetic waves.
A wave starting at the surface of a good conductor and propagating
inward is very quickly damped to insignificant values. The field is
localized in a thin surface layer, this phenomenon being known as skin
effect. The distance in which a wave is attenuated to lie (36.8 per cent)
of its initial value is called the skin depth or depth of penetration B, This
is defined by k" ~ = 1, or

a= .J;;r; = k~' = ~; (2-41)

where ~m is the wavelength in the metal. The skin depth is very small for
good conductors at radio frequencies, for ~m is very small. For example,
the depth of penetration into copper at 10 megacycles is only 0.021
millimeter. The density of power flow into the conductor, which must
also be that dissipated within the conductor, is given by
S = E X H* = u z lHol 2f/m
where H 0 is the amplitude of H at the surface. The time-average power
dissipation per unit area of surface cross section is the real part of the
above power flow, or
watts per square meter (2-42)
where <R = Re (11m) is the intrinsic resistance of the metal. <R is also
called the surface resistance and n« the surface impedance of the metal.
Eq. (2-42) is strictly true only when the wave propagates normally into
the conductor. In the next section we shall see that this is usually so.
In most problems Eq. (2-42) can be used to calculate power losses in
conducting boundaries. (An important exception to this occurs at sharp
points and corners extending outward from conductors.)
More general waves can be constructed by superposition of waves of
the above type with various polarizations and directions of propagation.
For waves uniform in the xy plane, the four basic waves, corresponding
to Eqs. (2-17), are
H + = A e-k"ze-ik'z
11 1J

H~+ = -B e-lr,"ze-ik'z
'11
(2-43)
H II - = -C e!<".ei k '.
'I
H s - = D e""-ei"'.
'I
54 TIME-HARMONIC ELECTROMAGNETIC FIELDS

The preceding discussion of this section applies to each of these waves if

the appropriate interchange of coordinates is made.
A superposition of waves traveling in opposite directions, for example
Ez = Ae-k"ze-ik'z + Cek"zeik'z
H =!'1/ (Ae- k"Ze-ik'21 - Cek"21eik'21 ) (2-44)
II

gives us standing-wave phenomena. However, it is no longer possible

to have two "equal" waves traveling in opposite directions. One wave
is attenuated in the +z direction, the other in the -z direction; hence
they can be equal only at one plane. Suppose that the wave components
are equal at z = 0, that is, A = C .in Eq. (2-44). There will then be
standing waves in the vicinity of z = 0, which will die out in both the +z
and -z directions. This is illustrated by Fig. 2-10 for a material having
fairly large losses. Far in the +z direction the +z traveling wave has
died out, leaving only the -z traveling wave. Similarly, far in the -z
direction we have only the +z traveling wave. The standing-wave ratio
is now a function of z, being large in the vicinity of z = 0 and approaching
unity as Izi becomes large. For very small amounts of dissipation, say
in a good dielectric, the attenuation of the wave is small, and standing-
wave patterns are almost the same as for the dissipationless case.
Other superpositions of Eqs. (2-43) can be formed to give elliptically
and circularly polarized waves. In a picture of a circularly polarized
wave traveling in dissipative media, the "corkscrews" for 8 and :re would
be attenuated in the direction of propagation. Also, 8 would be some-
what out of phase with se, A circularly polarized standing wave would
be a localized phenomenon in dissipative media, just as a linearly polarized
standing wave is localized.
2-6. Reflection of Waves. We saw in Sec. 1-14 that the tangential
components of E and H must be continuous across a material boundary.

__0- ---- -- _ _ e- k" z

--- Z
FIG. 2-10. Standing-wave pattern of two oppositely traveling waves in dissipative
matter.
 INTRODUCTION TO WAVES 55
A ratio of a component of E to a com-
ponent of H is called the wave imped- Region (1) Region (2)
ance in the direction defined by the
cross-product rule applied to the two
components. Thus, continuity of tan- Incident ...
Transmitted
gential E and H requires that wave Reflected ~
impedances normal to a material bound- ....Ee-------
ary must be continuous.
The simplest reflection problem is
that of a uniform plane wave nor-
mally incident upon a plane boundary FIG. 2-11. Reflection at a plane di-
between two media. This is illustrated electric interface, normal incidence.
by Fig. 2-11. In region 1 the field will
be the sum of an incident wave plus a reflected wave. The ratio of the
reflected electric intensity to the incident electric intensity at the interface
is defined to be the reflection coefficient I', Hence, for region 1
Ez(l) = Eo(e-iklt1 + reik1t1)

H II (1 ) = Eo (e-ik,. _ reik,.)
'111
In region 2 there will be a transmitted wave. The ratio of the trans-
mitted electric intensity to the incident electric intensity at the interface
is defined to be the transmission coefficient T. Hence, for region 2
E z ( 2) = EoTe-i k l •
H'II(2) = Eo Te-ik1tl
112

For continuity of wave impedance at the interface, we have

E~(1) I 1 + rr
ZtI
I.=0 = H 'II (1)
z-O
= '111 1 _ = 112

where '111 and '112 are the intrinsic wave impedances of media 1 and 2.
Solving for the reflection coefficient, we have
r = '112 - 111 (2-45)
'112+ 111
From the continuity of E; at z = 0, we have the transmission coefficient
given by
T = 1 +r = 2'112 (2-46)
112 + 7]1
If region 1 is a perfect dielectric, the standing-wave ratio is
_ E~ _ 1 + Irl
SWR - E~.. - 1 - Irl (2-47)
56 TIME-HARMONIC ELECTROMAGNETIC FIELDS

FIG. 2-12. A plane wave propagating at an angle ~ with respect to the x-z plane.

because the incident and reflected waves add in phase at some points and
add 1800 out of phase at other points. The density of power transmitted
across the interface is

(2-48)

where £inc = E 0 2/ f/ l is the incident power density. The difference

between the incident and transmitted power must be that reflected, or
£refl = £inol rl 2 (2-49)

We have used an x-polarized wave for the analysis, but the results are
valid for arbitrary polarization, since the x axis may be in any direction
tangential to the boundary. Those of us familiar with transmission-line
theory should note the complete analogy between the above plane-wave
problem and the transmission-line problem.
Another reflection problem of considerable interest is that of a plane
wave incident at an angle upon a plane dielectric boundary. Before
considering this problem, let us express the uniform plane wave in coordi-
nates rotated with respect to the direction of propagation. Let Fig. 2-12
represent a plane wave propagating at an angle ~ with respect to the xz
plane. An equiphase plane z' in terms of the unprimed coordinates is
z' = z cos ~ + y sin ~
and the unit vector in the y' direction in terms of the unprimed coordinate
unit vectors is
Uu' = UII cos ~ - u, sin ~
 INTRODUCTION TO WAVES 57
The expression for a uniform plane wave with E parallel to the e = 0
plane is the first of Eqs. (2-17) with all coordinates primed. Substituting
from the above two equations, we have
E 2: = E oe-i k (u ain f+- cos f)
H = (u, cos ~ - u, sin~) Eo e-ik(ualnl+-ooef) (2-50)
11
The wave impedance in the z direction for this wave is
E% 17
Z.=-=-- (2-51)
H; cos ~

In a similar manner, from the second of Eqs. (2-17), the expression for a
uniform plane wave with H parallel to the z = 0 plane is found to be
E :::: (ull cos ~ - U~ sin ~)Eoe-ik(fl8in f+ZC08 f)

H2: = _ Eo e-ik(1I8inl+~co8f) (2-52)

'1

The wave impedance in the z direction for this wave is

Z ~ = - -E" = 11 cos ~ (2-53)

H2:
Thus, the z-directed wave impedance for E parallel to the z = 0 plane
is always greater than the intrinsic impedance, and for H parallel to the
z = 0 plane it is always less than the intrinsic impedance of the medium.
Now suppose that a uniform plane wave is incident at an angle ~ = 81
upon a dielectric interface at z = 0, as shown in Fig. 2-13. Part of the
wave will be reflected at an angle t = 1r - Or, and part transmitted at an
angle ~ = 8t • Each of these partial fields will be of the form of Eqs. (2-50)
if E is parallel to the interface or of the form of Eqs. (2-52) if H is parallel
to the interface. (Arbitrary polarization is a superposition of these two

Region (1) Region (2)

FIG. 2-13. Reflection at

a plane dielectric inter-
face, arbitrary angle of
incidence. z
58 TIME-HARMONIC ELECTROMAGNETIC FIELDS

cases.) For continuity of tangential E and H over the entire interface,

the y variation of all three partial fields must be the same. This is so if
k, sin Oi = k, sin 0,. = k 2 sin 0,
From the first equality, we have
Or = Oi (2-54)
that is, the angle of reflection is equal to the angle of incidence. From the
second equality, we have

where v is the phase velocity. Equation (2-55) is known as Snell's law

of refraction. The direction of propagation of the transmitted wave is
thus different from that of the incident wave unless E1J.L1 = E2J.L2. In
practically all low-loss dielectrics, P.l == P.2 = p.o. If medium 2 is free
space and medium 1 is a nonmagnetic dielectric, the right-hand side of
Eq. (2-55) becomes VEl/EO = V~, which is called the index of refraction
of the dielectric.
The magnitudes of the reflected and transmitted fields depend upon the
polarization. For E parallel to the interface, we have in region 1
Ez(l) = A (e- i k1Z C08 6.. + reik1Z C08 6r )

H 1/ 1) = :! cos Oi(e-i k1Z COB 6.. - reikl21 c08 6r )

711

where A includes the y dependence. Thus, tile z-directed wave impedance

in region 1 at the interface is
Z (1) = Ez<l) = -.!!!...- 1 + r
• Hy(l) cos 8i 1 - I'
This must be equal to the z-directed wave impedance in region 2 at the
interface, which is Eq. (2-51) with ~ = (Jt. Thus,
I' = 712 sec 8, - 711 sec 8i (2-56)
712 sec 8t + 711 sec 8i
Note that this is of the same form as the corresponding equation for
normal incidence, Eq. (2-45). The intrinsic impedances are merely
replaced by the z-directed wave impedances of single traveling waves.
It should be apparent from the form of the equations that, for H parallel
to the interface, the reflection coefficient is given by
I' = 712 cos 0, - 711 c_os (Ji (2-57)
'12 cos 8t + 711 cos Oi
 INTRODUCTION TO WAVES 59
In both cases we have standing waves in the z direction, the standing-wave
ratio being given by Eq. (2-47).
Two cases of special interest are (1) that of total transmission and (2)
that of total reflection. The first case occurs when I' = o. For E
parallel to the interface, we see from Eq: (2-56) that I' = 0 when
1]2 _ 1]1
cos 8t - cos Oi
Substituting for Ot from Eq. (2-55) and for the 1]'S from Eq. (2-11) we
obtain
. 8i
SIn =
~E2/EI - JJ.2/JJ.I
(2-58)
J.Ll/JJ.2 - JJ.2/ JJ.I

as the angle at which no reflection occurs. This does not always have a
real solution for Oi. In fact,
sin Oi - . 00
1'1--+1'2

For nonmagnetic dielectrics (ILl = IL2 = JJ.o) there is no angle of total

transmission when E is parallel to the boundary. For the case of H
parallel to the boundary, we find from Eq. (2-57) that I' = 0 when

(2-59)

Again this does not always have a real solution for arbitrary JJ. and E.

But in the nonmagnetic case

Oi
• _
= SIn 1
~
-
E2
---
EI
= tan-1
+ E2
J;E2
-
EI
(2-60)

There is usually an angle of total transmission when H is parallel to the

boundary. The angle specified by Eq. (2-60) is called the polarizing angle
or Brewster angle. If an arbitrarily polarized wave is incident upon a
nonmagnetic boundary at this angle, the reflected wave will be polarized
with E parallel to the boundary.
The case of total reflection occurs when \r\ = 1. We are considering
lossless media; so the 'I1'S are real. It is apparent from Eqs. (2-56) and
(2-57) that [I'] ¢ 1 for real values of 8i and 8t • However, when EIILI >
E2JJ.2, Eq. (2-55) says that sin 8, can be greater than unity. What does
this mean? Our initial assumption was that the transmitted wave was
a uniform plane wave. But Eqs. (2-50) specify a solution to Maxwell's
equations regardless of the value of sin~. It can be real or complex.
All that is changed is our interpretation of the field. To illustrate, sup-
60 TIME-HARMONIC ELECTROMAGNETIC FIELDS

pose sin ~ > 1 in Eqs. (2-50) and let

k sin ~ = {3
(2-61)
k cos ~ = k VI - sin" ~ = ±ja
If we choose the minus sign for a, Eqs. (2-50) become
E~ = Eoe-ifJJle-cu

H = - (U i; + U. ~) ~o
II rifJlle-a.
(2-62)

which is a field exponentially attenuated in the z direction. Note the

90° phase difference between E~ and H 1I ; so the wave impedance in the z
direction is imaginary, and there is no power flow in the z direction. A
similar interpretation applies to Eqs. (2-52) when sin ~ > 1. Returning
now to our reflection problem, from Eq. (2-55) it is evident that sin 0,
is greater than unity when sin Oi > V E2JJ.2/CIJJ.l. Thus, the point of tran-
sition from real values of Of, (wave impedance real in region 2) to imaginary
values of Of, (wave impedance imaginary in region 2) is

(2-63)

The angle specified by Eq. (2-63) is called the critical angle. A wave
incident upon the boundary at an angle equal to or greater than the
critical angle will be totally reflected. Note that there is a real critical
angle only if EIJJ.l > E2J.l.2 or, in the nonmagnetic case, if El > E2. Thus,
total reflection occurs only if the wave passes from a "dense" material
into a "less dense" material. The reflection coefficient, Eq. (2-56) or
Eq. (2-57), becomes of the form
R -jX
r = R +jX
when total reflection occurs. It is evident in this case that Irl is unity.
Remember that the field in region 2 is not zero when total reflection
occurs. It is an exponentially decaying field, called a reactive field or an
evanescent field. Optical prisms make use of the phenomenon of total
reflection.
All the theory of this section can be applied to dissipative media if the
TJ'S and O's are allowed to be complex. Of particular interest is the case
of a plane wave incident upon a good conductor at an angle Oi. When
region 1 is a nonmagnetic dielectric and region 2 is a nonmagnetic con-
ductor, Eq. (2-55) becomes

sin B, = k 1 ~ /jWE
sin e, », "J U
 INTRODUCTION TO WAVES 61

L R

t--- dz ----1·...1
I
t---dz--....-.t
(a) (b)

Flo. 2-14. A transmission line according to circuit concepts. (a) Physical line; (b)
equivalent circuit.

This is an extremely small quantity for good conductors. For most prac-
tical purposes, the wave can be considered to propagate normally into the
conductor regardless of the angle of incidence.
2-6. Transmission-line Concepts. Let us review the circuit concept
of a transmission line and then show its relationship to the field concept.
Let Fig. 2-14a represent a two-conductor transmission line. For each
incremental length of line dz there is a series voltage drop dV and a shunt
current dI, The circuit theory postulate is that the voltage drop is
proportional to the line current I. Thus,
dV = -IZdz
where Z is a series impedance per unit length. It is also postulated that
the shunt current is proportional to the line voltage V. Thus,
dI = -VYdz
where Y is a shunt admittance per unit length. Dividing by de, we have
the a-e transmission-line equations
dV dI -VY
-dz = -IZ dz = (2-64)

Implicit in this development are the assumptions that (1) no mutual

impedance exists between incremental sections of line and (2) the shunt
current dl flows in planes transverse to z, The transmission line is said
to be uniform if Z and Yare independent of z,
Taking the derivative of the first of Eqs. (2-64) and substituting from
the second, we obtain
d 2V d 2[
- 2 - ZYV
dz
= 0 - 2 - ZYI
dz
=0 (2-65)

which are one-dimensional Helmholtz equations. The general solution

62 TIME-HARMONIC ELECTROMAGNETIC FIELDS

TABLE 2-2. COMPARISON OF TRANSMISSION-LINE WAVES

TO UNIFORM PLANE WAVES
.. ,
Transmission line Uniform plane wave

d 2V d2E%
- 2 - 1'2V == 0
dz
--
dz 2 + k E % == 0
2

d 2[ d 2H1/
- 2 - 1'2[ == 0
dz
--
dz + k
2
2H
1/
== 0

l' == YZY jk == yiO

V == Vo+e--Y· + Vo-e'Y· Ex == Eo+e- i k• + Eo-e i b

[ = [o+e--Yal + [o-e-Y· H y = Ho+e- i k• + Ho-e i k•

z, = V o+ =_ V o- = ~ ~ = ~:: = - ~:~ = ~
1 0+ [0- Y

P = VI* S. = ExH:

is a sum of a +z traveling wave and a -z traveling wave, with propaga-

tion constant
'Y = YZY (2-66)
Choosing the +z traveling wave
V+ = Voe-'Y~

we have from Eqs. (2-64) that

V+ Z 'Y
/+ =:y = Y

Substituting for 'Y from Eq. (2-66), we have

Zo = V+
/+
= fZ
\jV (2-67)

which is called the characteristic impedance of the transmission line. The

imaginary parts of Z and Yare usually positive, and it is common practice
to write
Z = R +jwL Y = G +jwC (2-68)
The equivalent circuit of the transmission line is then as shown in Fig.
2-14b. The reader has probably already noted the complete analogy
between the linearly polarized plane wave and the transmission line.
This analogy is summarized by Table 2-2.
In the circuit theory development, we assumed no mutual coupling
 INTRODUCTION TO WAVES 63
between adjacent elements of the transmission line. From the field
theory point of view, this is equivalent to assuming that rio E~ or H.
exists. Such a wave is called transverse electromagnetic, abbreviated
TEM. This is not the only wave possible on a transmission line, for
Maxwell's equations show that infinitely many wave types can exist.
Each possible wave is called a mode, and a TEM wave is called a trans-
mission-line mode. All other waves, which must have an E. or an H.
or both, are called higher-order modes. The higher-order modes are
usually important only in the vicinity of the feed point, or in the vicinity
of a discontinuity on the line. In this section we shall restrict considera-
tion to transmission-line, or TEM, modes.
For the TEM mode to exist exactly, the conductors must be perfect,
or else an E. is required to support the z-directed current. Let us there-
fore specialize the problem to that of perfect conductors immersed in a
homogeneous medium. We assume E~ = Hz = 0 and z dependence of
the form er». Expansion of the field equations, Eqs. (2-1), then gives
'YE" = -~Hz 'YH" = yEz:
'YEz: = ~H" 'YH:t: = -'DE"
aE u _ aE z =0 aH u _ aH:t: = 0
ax ay ax ay
It follows from these equations that
l' = jk (2-69)
The propagation constant of any TEM wave is the intrinsic propagation
constant of the medium. The proportionality of components of E to
those of II expressed by the above equations can be written concisely as
1
E = flH X u. H = - u, X E (2-70)
fI

Thus, the z-directed wave impedance of any TEM wave is the intrinsic
wave impedance of the medium. Finally, manipulation of the original six
equations shows that each component of E and H satisfies the two-
dimensional Laplace equation. We can summarize this by defining a
transverse Laplacian operator
a2
+ aay
2
V," = ax 2 2 (2-71)

and writing V,2E = 0 V t 2H = 0

The boundary conditions for the problem are

E, = O} at the conductors (2-72)

H" = 0
Thus, the boundary-value problem for E is the same as the electrostatic
64 TIME-HARMONIC ELECTROMAGNETIC FIELDS

problem having the same conducting boundaries. The

boundary-value problem for H is the same as the mag-
netostatic problem having "anticonducting" (no H.)
boundaries. It is for thiS reason that U static" capaci-
tances and inductances can be "used for transmission
lines even though the field is time-harmonic.
To show the relationship of the static L's and C's to
the Zo of the transmission line, consider a cross section
of the line as represented by Fig. 2-15. In the trans-
mission-line problem, the line voltage and current are
FIG. 2-15. Cross
section of a trans- related to the fields by
mission line.
V={E.dl (2-73)
101
where C 1 and C 2 are as shown on Fig. 2-15. From the second of these and
the second of Eqs. (2-70) we have

I r u, X
=!11 JOt E· d1 =!11 jrc, En dl
But in the corresponding electrostatic problem the capacitance is

C = !l. =.!.
V V
r
JOt En dl
Thus, the characteristic impedance of the transmission line is related to
the electrostatic capacitance per unit length by
V E
(2-74)
Zo = T = 11 C
Similarly, from the first of Eqs. (2-73) and (2-70) we have

V = 11 (
lOI H X u, · dl = 11 (
lcl H n dl
In the corresponding magnetostatic problem we have

L = tI =!!:. { Hndl
I JOI
Therefore, the characteristic impedance of the line is related to the
magnetostatic inductance per unit length by
V L
Zo = -I = 11-
p,
(2-75)

Note also that Land C are related to each other through Eqs. (2-74) and
(2-75). The electrostatic and magnetostatic problems hrve E and H
everywhere orthogonal to each other and are called conjugate problems.
 INTRODUCTION TO WAVES 65
TABLE 2-3. CHARACTERISTIC IMPEDANCES OF SOME COMMON TRANSMISSION LINES

Line Geometry Characteristic impedance

Two wire 0 ~ Zo ~
." 2D
-log- D »d
~D~ r d

Coaxial @b, Zo =
."
-log-
2...
b
a

Confocal elliptic (iii)

~2b~
~2c ~ Zo = -log
2...
." b + Vb 2 - at
a+~

~w--+t:l b
Parallel plate b z, ~.,,- w»b
w
T

Collinear plate
-
-+iw~
J+-D--+f
- Zo ~
." 4D
-log-
r w
D»w

..t.

~
TJ 4h
Wire above ground plane h d Zo ~ -log- h »d
2... d
~

Shielded pair
D
+Id~
17G-!f
~ 1+
8
., ., e ZO ~ ;; log d
8 D 1
D2 +
- 8
1)
8'
D »d
8 »d

Wire in trough
I~i:~: Zo ~ ., ew
- log
2r
-
rd
tanh -Th)
w
h »d
w» d

Once the electrostatic C or the magnetostatic L is known, the Zo of the

corresponding transmission line is given by Eq. (2-74) or Eq. (2-75). Table
2-3 lists the characteristic impedances of some common transmission lines.
When the dielectric is lossy but the conductors still assumed perfect,
all of our equations still apply. Zo (proportional to 7]) and 'Y (= jk)
66 TIME-HARMONIC ELECTROMAGNETIC FIELDS

become complex. The most important effect of this is that the wave is
attenuated in the direction of travel. The attenuation constant in this
case is the intrinsic attenuation constant of the dielectric (Table 2-1,
column 2, row 4). When the conductors are imperfect, the field is no
longer exactly TEM, and exact solutions are usually impractical. How-
ever, the waves will still be characterized by a propagation constant
'Y = a + il3. Hence a -l-z-traveling wave will be of the form
V = V oe-(a+itJ)z V
1=-
Zo
and the power flow is given by

PI = VI* = li~2 e- 2... = P oe- 2a•

or, in terms of time-average powers,
{pJ = Re (PI) = Re (P o)e- 2a z
The rate of decrease in {PI versus e equals the time-average power dissi-
pated per unit length {Pd, or
-
CPd = -
d{PJ
- =
2-
aCPI
dz
Thus, the attenuation constant is given by
{Pd
a =- (2-76)
2 <PI
While this equation is exact if {Pd and {PI are determined exactly, its
greatest use lies in approximating a by approximating (Pd. For example,
attenuation due to losses in imperfect conductors can be approximated by
assuming that Eq. (2-42) holds at their surface. We shall carry out
such a calculation for the rectangular waveguide in the next section.
2-7. Waveguide Concepts. The
waves on a transmission line can be
x viewed as being guided by the con-
ductors. This concept of wave guid-
ance is quite general and applies to
many configurations of matter. In
general, systems which guide waves
are called waveguides. Apart from
transmission lines, the most com-
monly used waveguide is the rectan-
b y gular waveguide, illustrated by Fig.
FIG. 2-16. The rectangular waveguide. 2-16. It is a hollow conducting tube
 INTRODUCTION TO WAVES 67
of rectangular cross section. Fields existing within this tube must be char-
acterized by zero tangential components of E at the conducting walls.
Consider two uniform plane waves traveling at the angles ~ and - t
with respect to the xz plane (see Fig. 2-12). If the waves are x-polarized,
we use Eq. (2-50) and write
E:I: = A (e-ik'll ain E - e1k'll sin E) e-ik~ eoa f
= - 2jA sin (ky sin ~) e-ikzcoa f
Let Eo denote (-2jA) and define
k; = k sin ~ 'Y = jk cos ~

In view of the trigonometric identity sin" ~ + cos" ~ = 1, the parameters

'Y and k; are related by
'Y 2 = ke2 - k2 (2-77)
The above field can now be written as
E:I: = Eo sin (keY) e-'Y~ (2-78)
Let us see if this field can exist within the rectangular waveguide. There
is only an E:I:; so no component of E is tangential to the conductors x = 0
and x = a. Also, E:I: = 0 at y = 0; so there is no tangential component
of E at the wall y = o. There remains the condition that E:I: = 0 at
y = b, which is satisfied if
k = n1r n = 1,2,3, . . . (2-79)
e b
These permissible values of k; are called eigenvalues, or characteristic
values of the problem.
Each choice of n in Eq. (2-79) determines a possible field, or mode.
The modes in a waveguide are usually classified according to the existence
of z components of the field. A mode having no E~ is said to be a trans-
verse electric (TE) mode. One having no H~ is said to be a transverse
magnetic (TM) mode. All the modes in the rectangular waveguide fall
into one of these two classes. The modes represented by Eqs. (2-78)
and (2-79) have no E~ and are therefore TE modes. The particular modes
that we are considering are TE on modes, the subscript 0 denoting no
variation with x, and the subscript n denoting the choice by Eq. (2-79).
The complete system of modes will be considered in Sec. 4-3.
For k real (loss-free dielectric), the propagation constant 'Y can be
expressed as
k > n1r
b
'Y= (2-80)
k < n1r
b
68 TIME-HARMONIC ELECTROMAGNETIC FIELDS

where a and {j are real. This follows from Eqs. (2-77) and (2-79). When
'Y = j~, we have wave propagation in the z direction, and the mode is
called a propagating mode. When 'Y = a, the field decays exponentially
with z, and there is no wave propagation. In this case, the mode is
called a nonpropagating mode, or an evanescent mode. The transition
from one type of behavior to the other occurs at a = 0 or k = nr/b.
Letting k = 21rf VEp" we can solve for the transition frequency, obtaining
n
Ie = 2b VEP (2-81)

This is called the cutoff frequency of the TE on mode. The corresponding

intrinsic wavelength
xe = 2b
n
(2-82)

is called the cutoff wavelength of the TE on mode. At frequencies greater

than I, (wavelengths less than ~e), the mode propagates. At frequencies
less than i, (wavelengths greater than ~e), the mode is nonpropagating.
A knowledge of I, or ~e is equivalent to a knowledge of k e ; so they also
are eigenvalues. In particular, from Eqs. (2-79), (2-81), and (2-82), it is
evident that
2r _ I-
ke = ~c = 2rfe V EP. (2-83)

Using the last equality and k = 2rl yEP. in Eq. (2-80), we can express 'Y
as

I> Ie
(2-84)
I < I,
Thus, the phase constant {j of a propagating mode is always less than the
intrinsic phase constant k of the dielectric, approaching k as f ---+ 00.
The attenuation constant of a non propagating mode is always less than
k e , approaching k; as f ---+ o. When a mode propagates, the concepts of
wavelength and phase velocity can be applied to the mode field as a
whole. Thus, the guide wavelength Xg is defined as the distance in which
the phase of E increases by 2r, that is, {3X g = 21r. Using {3 from Eq.
(2-84), we have
(2-85)
"A g = VI - (/e/f)2
showing that the guide wavelength is always greater than the intrinsic
wavelength of the dielectric. The guide phase velocity Vg is defined as the
 INTRODUCTION TO WAVES 69
velocity at which a point of constant phase of 8 travels. Thus, in a
manner analogous to that used to derive Eq. (2-14), we find
w Vp
(2-86)
Va = ~ = V1 - (fe/f)2
where Vp is the intrinsic phase velocity of the dielectric. The guide phase
velocity is therefore greater than the intrinsic phase velocity.
Another important property of waveguide modes is the existence of a
characteristic wave impedance. To show this, let us find H from the E of
Eq. (2-78) according to V X E = -jwp,H. The result is
E z = Eo sin (keY) e--Y·
H; = ;L Eo sin (keY) e-r·
JWP, (2-87)
H. = ~ Eo cos (k.y) e-'"
JWP,

where E; has been repeated for convenience. The wave impedance in

the z direction is
z = E., = jwp, (2-88)
Z H" 'Y

This is called the characteristic impedance of the mode and plays the
same role in reflection problems as does the Zo of transmission lines. If
we substitute into the above equation for 'Y from Eq. (2-84), we find

z; = Z. = Iv! -~ JTJ
(1.//)2

v(fe/f)2 - 1
f > I,

I «t.
(2-89)

Thus, the characteristic impedance of a TE on propagating mode is always

greater than the intrinsic impedance of the dielectric, approaching 71 as
f ~ 00. The characteristic impedance of a nonpropagating mode is
reactive and approaches zero as f ~ o.
All our discussion so far has dealt with waves traveling in the +z
direction. For each +z traveling wave, a -z traveling wave is possible,
obtained by replacing 'Y by -'Y in Eqs. (2-87). The simultaneous
existence of +z and -z traveling waves in the same mode gives rise to
standing waves. The concepts of reflection coefficients, standing-wave
ratios, etc., used in the case of uniform plane-wave reflection, also apply
to waveguide problems.
The mode with the lowest cutoff frequency in a particular guide is
called the dominant mode. The dominant mode in a rectangular wave-
guide, assuming b > a, is the TE oI mode. (This we have not shown, for
70 TIME-HARMONIC ELECTROMAGNETIC FIELDS

x z
e----~~ Lines into paper x x x

c!J{-----~ Lines out of paper • • •

FIG. 2-17. Mode pattern for the TE ol waveguide mode.

we have not considered all modes.) From Eq. (2-82) with n = 1, we

see that the cutoff wavelength of the TE ol mode is he = 2b. Thus, wave
propagation can take place in a rectangular waveguide only when its
widest side is greater than a half-wavelength. I A sketch of the instan-
taneous field lines at some instant is called a mode pattern. The mode
pattern of the TE ol mode in the propagating state is shown in Fig. 2-17.
This figure is obtained by determining E and 3C from the E and H of
Eqs. (2-87) and specializing the result to some instant of time. As time
progresses, the mode pattern moves in the z direction.
It is admittedly confusing to learn that many modes exist on a given
guiding system. It is not, however, so bad as it seems at first. If only
one mode propagates in a waveguide, this will be the only mode of
appreciable magnitude except near sources or discontinuities. The
rectangular waveguide is usually operated so that only the TE ol mode
propagates. This is therefore the only wave of significant amplitude
along the guide except near sources and discontinuities.
Because of the importance of the TE ol mode, let us consider it in a
little more detail. Table 2-4 specializes our preceding equations to this
mode and includes some additional parameters which we shall now
consider.
The power transmitted along the waveguide can be found by integrat-
ing the axial component of the Poynting vector over a guide cross section.
This gives
PI = hh
G b
E:&H: dx dy = IEol220::
which, above cutoff, is real and is therefore the time-average power trans-
mitted. Below cutoff, the power is imaginary, indicating no time-average
1 We are referring to the intrinsic wavelength of the dielectric filling the waveguide,
which is usually free space.
 INTRODUCTION TO WAVES 71
TABi..E 2-4. SUMMARY OF WAVEGUIDE PARAMETERS FOR THE DOMINANT MODE
(TE oI ) IN A RECTANGULAR WAVEGUIDE

Ez Eo sin 1ry e-~~

==
b
Eo . Try
Complex field H = - sin - e-"·
JI Zo b
Eo It: Try
H = - - cos - e-r·
• il1 I b

1
Cutoff frequency t :=--
e 2bV;

Cutoff wavelength x, = 2b

Propagation constant
{jfJ = jk VI - (j./f)1 f »s.
'Y = 2Tr
a = -VI
x, - (file)! f «s.

Characteristic impedance iWIJ

Zo=- = { 111''1'1 - u.o» I »s.
'Y iT/IV (fe/f) 2 - 1) I <t.

Guide wavelength
x
~g =
VI - (fell)!

VI'
Guide phase velocity Vg =
VI - (felf)!

Power transmitted p = \E ol2ab

2Z o

Attenuation due to lossy dielectric ad - "

= WE 11
VI - (felf)2
2

Attenuation due to imperfect conductor Ci e =

a71 VI m- (felf)2 [2a
1 +b
~.y]
f

power transmitted. (The preceding equation applies only at z = 0 below

cutoff unless the factor e- 2a z is added.) It is also interesting to note that
the time-average electric and magnetic energies per unit length of guide
are equal above cutoff (see Probe 2-32).
In contrast to the transmission-line mode, there is no unique voltage
and current associated with a waveguide mode. However, the amplitude
of a modal traveling wave (Eo in Table 2-4) enters into waveguide reflec-
tion problems in the same manner as V in transmission-line problems.
72 TIME-HARMONIC ELECTROMAGNETIC FIELDS

To emphasize this correspondence, it is common to define a mode voltage V

and a mode current I such that
V
z, = T P = VI* (2-90)

From Table 2-4, it is evident that

V = Eo~e-'Y' V
1=-
Zo
(2-91)

satisfy this definition. Remember that we are dealing with only a +z

traveling wave. In the -z traveling wave, I = - V /Zo. When waves
in both directions are present, the ratio V / I is a function of z, Other
definitions of mode voltage, mode current, and characteristic impedance
can be found in the literature. These alternative definitions will always
be proportional to our definitions (see Probe 2-34).
Our treatment has so far been confined to the ideal loss-free guide.
When losses are present in the dielectric but not in the conductor, all
our equations still apply, except that most parameters become complex.
There is no longer a real cutoff frequency, for 'Y never goes to zero. Also,
the characteristic impedance is complex at all frequencies. The behavior
of 'Y = a + j~ in the low-loss case is sketched in Fig. 2-18. The behavior
of 'Y for the loss-free case is shown dashed. The most important effect
of dissipation is the existence of an attenuation constant at all frequencies.
In the low-loss case, we can continue to use the relationship

+ j(3 ~ jk ~1 - (J) 2

"y = a

provided f is not too close to i: Letting k = k' - jk" and referring to

FIG. 2-18. Propagation

constant for a lossy wave-
guide (loss-free case
kc
shown dashed).

o Ie f
 INTRODUCTION TO WAVES 73
Table 2-1, we find

(2-92)

This is the attenuation constant due to a lossy dielectric in the guide.

Even more important is the attenuation due to imperfectly conducting
guide walls. Our solution is n<? longer exact in this case, because the
boundary conditions are changed. The tangential component of E is
now not quite zero at the conductor. However, for good conductors, the
tangential component of E is very small, and the field is only slightly
changed, or "perturbed," from the loss-free solution. The loss-free
solution is used to approximate H at the conductor, and Eq. (2-42) is
used to approximate the power dissipated in the conductor. Such a
procedure is called a perturbational method (see Chap. 7). The power per
unit length dissipated in the wall y = 0 is

iJ>d 1_0 = CR foa IH.12 dx = CRIEol2 (;;Y ta dx

= CRlEol2a (~;Y
and an equal amount is dissipated in the wall y = b. The power per unit
length dissipated in the wall x = 0 is

«'d I~O = CR [b (IH1I12

jo
+ IH.12) dy
= CRIEol2fob[8in2i1Ib) + (;;)2 C0827] dy

= CRIE l2[2;02+ (~iY ~]

o

and an equal amount is dissipated in the wall x = a. The total power

dissipated per unit length is the sum of that for the four walls, or

iJ>d = CRIE ol2 [;02 + (~iY (2a + b) ]

Equation (2-76) is valid for any traveling wave; so using the above «'dJ
arid «', = P of Table 2-4, we have

a. = ~~o [;02+ (t;Y (2a + b)]

= al1 VI ~ (f.I!P [I + 2: GYJ (2-93)

This is the attenuation constant due to conductor losses. When both

74 TIME-HARMONIC ELECTROMAGNETIC FIELDS

dielectric losses and conductor losses

x
need to be considered, the total
attenuation constant is
(2-94)
for by Eq. (2-76) we merely add the
two losses.
2-8. Resonator Concepts. InSec.
a 2-2 we noted a similarity between
standing waves and circuit theory
resonance. In the loss-free case, elec-
tromagnetic fields can exist within
b Y a source-free region enclosed by a
FIG. 2-19. The rectangular cavity. perfect conductor. These fields can
exist only at specific frequencies,
called resonant frequencies. When losses are present, a source must
exist to sustain oscillations. The input impedance seen by the source
behaves, in the vicinity of a resonant frequency, like the impedance of an
LC circuit. Resonators can therefore be used for the same purposes at
high frequencies as LC resonators are used at lower frequencies.
To illustrate resonator concepts, consider the "rectangular cavity"
of Fig. 2-19. This consists of a conductor enclosing a dielectric, both of
which we will assume to be perfect at present. We desire to find solu-
tions to the field equations having zero tangential components of E over
the entire boundary. The TE ol waveguide mode already satisfies this
condition over four of the walls. We recall that standing waves have
planes of zero field, which suggests trying the standing-wave TE ol field.
For E; to be zero at z = 0, we choose
E., = E.,+ + E.,- = A sin 1r: (e-i/h - eiflz )

= Eo sin (1r:) sin I3z

For E z to be zero at z = c, we choose pc = 1r, which, according to Table
2-4, is

1r = ck ~1 - GY = c21rfVEP ~1 - (2b JEpf)2

Solving for the resonant frequency f = i-, we have

_ 1 ~b2 +
c2 (2-95)
fr - 2bc EJI.

When a is the smallest cavity dimension, this is the resonant frequency of

 INTRODUCTION TO WAVES 75
the dominant mode, called the TEo!! mode. The additional subscript 1
indicates that we have chosen the first zero of sin {3z. The higher zeros
give higher-order modes, that is, modes with higher resonant frequencies.
Setting {3 = 1r/ e in the above expression for Ex and determining H from the
Maxwell equations, we have for the TEo!! mode
· 1ry • 1rZ
E x = E 0 SIll b SIll e-
fl.y = jbEo . «u 1rZ
SIll - cos - (2-96)
11 yb + e2
2
b c
jeEo 1ry • 1rZ
Hz = - cos - SIn -
11 yb + 2
e b2 c
Note that E and Hare 90° out of phase; so S is maximum when 3C is
minimum and vice versa. A sketch of the instantaneous field lines at some
time when both 8 and 3C exist is given in Fig. 2-20. Also of interest is
the energy stored within the cavity. From the conservation of complex
power, Eq. (1-68), we know that Wm = We. Thus, the time-average
electric and magnetic energies are

W m = W. = ~ Ilf IEI
cavity
2
dT = i IE l2abc o (2-97)

We also know from conservation of energy, Eq. (1-39), that the total
energy within the resonator is independent of time. If we choose a time
for which 3C is zero, W m will be zero, and Wl, will be maximum and twice
its average value. Therefore,

w= 2W. = :i IE l2 bc o a (2-98)

is the total energy stored within the cavity.

1
x x x x /,,----~---,
x x x x
x x x ~ - / / ... ---~--""
.,,---~ - - , . ~ I
'\
-
--
r I
I ,. \ I
-::
--
b I,· I.
I I \
•••
•• •
.:.1
I I
I

J
I
• · ·• ·•--- \ \ '---:.:--...,.1
\ "
, ,---~----'/• • • /
J I
J

·· ·· · ·
--_/ ----~ /

La-J
e---~~ .!J[--~-

FIG. 2-20. Mode pattern for the TE oll cavity mode.

76 TIME-HARMONIC ELECTROMAGNETIC FIELDS

When the resonator has losses, we define its quality factor as

w X energy stored wW
Q = average power dissipated = <Pd (2-99)
by analogy to the Q of an LC circuit. If the losses are dielectric losses,
we have

(2-100)

so the Q of the resonator is that of the dielectric, Eq. (1-79). This is

valid for any mode in a cavity of arbitrary shape. Usually more impor-
tant in determining the Q is the loss due to imperfect conductors. This
is determined to the same approximation as we used for waveguide
attenuation. We assume H at walls to be that of the loss-free mode
and calculate <Pd by Eq. (2-42). To summarize,

s, = <R 1[> IHI2 de = 27]2~!~2 c2) [bc(b2 + c2) + 2a(bS + CS) ]

cavity
walls

Substituting this, Eq. (2-98), and Eq. (2-95) into Eq. (2-99), we have
rl1 a(b 2 +c 2)%

Qc = 2<R bc(b 2 +c 2) + 2a(bS + c 3) (2-101)

From the symmetry of Qc in band c, it is evident that b = c for maxi-

mum Q. For a "square-base" cavity (b = c), we have
1 1.1111
fr = b y2EIl Qc = <R(l +
b/2a)
(2-102)

The Q also increases as a increases, but if a > b we no longer have the

dominant mode. As an example of the Q's obtainable, consider a cubic
cavity constructed of copper. In this case we have
Qc = 1.07 X 109 /V! (2-103)
which, at microwave frequencies, gives Q's of several thousand. This
idealized Q will, however, be lowered in practice by the introduction of a
feed system, by imperfections in the construction, and by corrosion of
the metal. When both conductor losses and dielectric losses are con-
sidered, the Q of the cavity becomes
1 1 1
Q = Qd + Qc (2-104)

which is evident from Eq. (2-99).

 INTRODUCTION TO WAVES 77
2-9. Radiation. We shall now show that a source in unbounded space
is characterized by a radiation of energy. Consider the field equations
v X E = -jwp,H v X H = jWEE +J (2-105)
where J is the source, or impressed, current. These equations apply
explicitly to a perfect dielectric, but the extension to lossy media is
effected by replacing jwp, by z and jWE by '0. In homogeneous media,
the divergence of the :first equation is
v·H = 0
Any divergenceless vector is the curl of some other vector; so
H = V X A (2-106)
where A is called a magnetic vector potentiai.' Substituting Eq. (2-106)
into the first of Eqs. (2-105), we have
V X (E + jwp,A) = 0
Any curl-free vector is the gradient of some scalar. Hence,
E + jwp.A = -V4> (2-107)
where 4> is an electric scalar potential. To obtain the equation for A,
substitute Eqs. (2-106) and (2-107) into the second of Eqs. (2-105).
This gives
v X V X A - k 2A = J - jWEV~ (2-108)
which, by a vector identity, becomes
V(V· A) - V 2A - k 2A =J- jWEV4>
Only V X A was specified by Eq. (2-106). We are still free to choose
V • A. If we let
V·A = -jWE~ (2-109)
the equation for A simplifies to
V2A +k 2A
= -J (2-110)
This is the Helmholtz equation, or complex wave equation. Solutions
to Eq. (2-110) are called wave potentials. In terms of the magnetic wave
potential, we have
= -jwp.A
E J- V(V· A)
+ JWE (2-111)
H=VXA
1 In general electromagnetic theory it is more common to let A be the vector poten-

tial of B. In homogeneous media the two potentials are in the ratio p., a constant.
78 TIME-HARMONIC ELECTROMAGNETIC FIELDS

z obtained from Eqs. (2-106), (2-107),

and (2-109). The principal advan-
tages of using A instead of E or H
are (1) rectangular components of A
r have corresponding rectangular com-
ponents of J as their sources and (2)
A need not be divergenceless.
Let us first determine A for a cur-
rent I extending over an incremental
y
length l, forming a current element or
x electric dipole of moment Il. Take
FIG. 2-21. A z-directed current element this current element to be z-directed
at the coordinate origin. and situated at the coordinate origin,
as shown in Fig. 2-21. The current
is z-directed; so we take A to have only a z component, satisfying

everywhere except at the origin. The scalar quantity A z has a point

source Il and should therefore be spherically symmetric. Thus, let
A 21 = Az(r), and the above equation reduces to

This has the two independent solutions

_1 e-JOk r _1 e'Ok r
r r

the first of which represents an outward-traveling wave, and the second

an inward-traveling wave. (In dissipative media, k = k' - jk", and the
first solution vanishes as r ~ 00, and the second solution becomes
infinite.) We therefore choose the first solution, and take

AZ = Q
r e-
ikr

where C is a constant.! As k ~ 0, Eq. (2-110) reduces to Poisson's

equation, for which the solution is

A IS =-!i
41rr
1 To be precise, C might be a function of k, but the solution must also reduce to
the static field as r -7 O. Hence, C is not a function of k,
 INTRODUCTION TO WAVES 79
Our constant C must therefore be

c= Il
41r
and hence A =!i e-i kr (2-112)
• 47rr
is the desired solution for the current element of Fig. 2-21. The out-
ward-traveling wave represented by Eq. (2-112) is called a spherical wave,
since surfaces of constant phase are spheres.
The electromagnetic field of the current element is obtained by substi-
tuting Eq. (2-112) into Eqs. (2-111). The result is

E; = Il e-i kr
21r r2
(.!. + JWEr
_._1_)3
cos 8

Eg = Il e-i k r
41r
(iWp.r + .!.r + JWEr
_._1_)3 sin 6
2
(2-113)

Il
H ~ =-e- Ok ,r(i-k + -1) SIn· 6
41r r r2
Very close to the current element, the E reduces to that of a static charge
dipole, the H reduces to that of a constant current element, and the field

I
is said to be quasi-static. Far from the current element, Eqs. (2-113)
reduce to
E 9 = 11 2Xr
jIl e-1orkSIn
· (J
r»X (2-114)
H~ jIl Ok •
= 2Xr e-1 r sin 8
which is called the radiation field. At intermediate values of r the field is
called the induction field. The outward-directed complex power over a
sphere of radius r is

PI = 1ft E X H* · ds = f02" dq, fo" ae r 2

sin 0 EBB:

= '1 ~ Ii
l
r[
1 - (~)3] (2-115)

The time-average power radiated is the real part of PI, or

_
CP/=113" ~
21r I Ill2 (2-116)

This is independent of r and can be most simply obtained from the radi-
ation field, Eq. (2-114). The reactive power, which is negative, indicates
that there is an excess of electric energy over magnetic energy in the
near field.
80 TIME-HARMONIC ELECTROMAGNETIC FIELDS

r-r'
FIG. 2-22. Radius vec-
:-=-.:..-----:JJIII""" (X1YI Z )
tor notation.

y
x

To obtain the field of an arbitrary distribution of electric currents,

we need only superimpose the solutions for each element, for the equa-
tions are linear. A superposition of vector potentials is usually the most
convenient one. For this purpose, we shall use the radius vector notation
illustrated by Fig. 2-22. The "field coordinates" are specified by
r = UzX + UuY + u.z
and the "source coordinates" by
r' = uzX' + uuY' + u.z'
In Eq. (2-112), r is the distance from the source to the field point. For
Il not at the coordinate origin, r should be replaced by
[r - r'l = V(x - X')2 + (y - y')2 + (e - Z')2
Note the direction of the vector potential is that of the current; so Eq.
(2-112) can be generalized to a current element of arbitrary orientation
by replacing Il by II and A. by A. Thus, the vector potential from
current element of arbitrary location and orientation is
II e-jklr-r'l
A = 41rlr - r'l
To emphasize that A is evaluated at the field point (x,y,z) and n is situ-
ated at the source point (x',y',Z') , we shall use the notation A(r) and
II(r'). The above equation then becomes
II(r')e-jklr-r'l
= 41I"r
A(r) I - r 'I (2-117)

Finally, for a current distribution J, the current element contained in a

volume element dr is J dr, and a superposition over all such elements is
-
A(r) - 41r J~
1 1~ ry J(r')e-jklr-r'l
Ir - r'l dr
,
(2-118)
 INTRODUCTION TO WAVES 81
The prime on dT' emphasizes that the integration is over the source
coordinates. Equation (2-118) is called the magnetic vector potential inte-
gral. It is intended to include the cases of surface currents and fila-
mentary currents by implication. We therefore have a formal solution
for any problem characterized by electric currents in an unbounded homo-
geneous medium. The medium may be dissipative if k is considered to
be complex.
2-10. Antenna Concepts. A device whose primary purpose is to
radiate or receive electromagnetic energy is called an antenna. To illus-
trate antenna concepts, we shall consider the linear antenna of Fig. 2-23.
It consists of a straight wire carrying a current I(z). When it is ener-
gized at the center, it is called a dipole antenna. The magnetic vector
potential, Eq. (2-118), for this particular problem is
A =...!... jL/2 I(z')e-ilclr-r/l dz'
(2-119)
z 41r -L/2 Ie - e'l
where [r - r'l = vr + 2
Z'2 - 2rz' cos () (2-120)
The radiation field (r large) is of primary interest, in which case
[r - r'] ~ r - z' cos 0 r» z' (2-121)
e-jlcr jL/2
and Az~-- I(z')eikz'cos6dz' r»L (2-122)
47rr -L/2
Note that the second term of Eq. (2-121) must be retained in the "phase
term" e-iklr-r/l, but not in the "amplitude term" [r - r'I-I. To obtain
the field components, substitute Eq. (2-122) into Eqs. (2-111) and retain
only the 11r terms. This gives
E fJ = J·WJl.
1
sin () Az} r large
z
H4J =-EfJ r-r'
11
(2-123)
L/2
This result is equivalent to super-
imposing Eqs. (2-114) for all ele- dz'
ments of current.
To evaluate the radiation field,
we must know the current on the
y
antenna. An exact determination
of the current requires the solution X
to a boundary-value problem. For-
tunately, the radiation field is rela-
tively insensitive to minor changes in -L/2
current distribution, and much use- FIG. 2-23. The linear antenna.
82 TIME-HARMONIC ELECTROMAGNETIC FIELDS

ful information can be obtained from an approximate current distribution.

We have already seen that on transmission lines the current is a harmonic
function of kz. This is also true for the principal mode on a single thin
wire. The current on the dipole antenna must be zero at the ends of the
wire, symmetrical in z, and continuous at the source (z = 0). Thus, we
choose
(2-124)

The vector potential in the radiation zone can now be evaluated as

Az = 1
m
41fT
e-
iTer
jL/2
-L/2
sin [k (~ -
2
IZ'I)J e ik. ' co8 e dz'

Ime-jkr2[COS(k~COSO) - COS(k~)]
=~ ksin 2 8
From Eq. (2-123), the radiation field is

E 8
= i"11me-ikr [cos (k ~ cos ~) - cos (k ~)] (2-125)
21fT SIn 8
with H", = E 8/ "1. Note that the radiation field is linearly polarized, for
there is only an E e• The density of power radiated is the T component
of the Poynting vector

_ *_ "1IImI2[COS(k~COSO) - COS(k~)]2
S,. - EeH", - - ( 2
1rT
)2
SIn
. 8 (2-126)

The total power radiated is obtained by integrating Sf over a large sphere,

or
(PI = fo2" fo" s. r 2 sin 0 s» dq,

= 7111mI21" [cos (k ~ cos~)

2r 0 SIn
- cos
8
(k~) dO r (2-127)
The radiation resistance R, of an antenna is defined as
{J>,
Rr = fIT2 (2-128)

where I is some arbitrary reference current. For the dipole antenna,

the reference current is usually picked as L«. Hence,

(2-129)
 INTRODUCTION TO WAVES 83
280
r
240
I,.......",. /
FIG. 2-24. Radiation re- r
200
R 160
120
V
7 ,
~
)
1/
sistance of the dipole
80 J ~V
antenna. [7
40
l/ I I

o >"/2 3>"/2 2>..

L

This integral can be evaluated in terms of tabulated functions (see Probe

2-44). A graph of R; versus L is given in Fig. 2-24.
The radiation field pattern of an antenna is a plot of lEI at constant r
in the radiation zone. For a dipole antenna, the radiation field pattern
is essentially the bracketed term of Eq. (2-125). This is shown in Fig.
2-25 for kL small (short dipole), kL = 1r (half-wavelength dipole), and
kL = 21r (full-wavelength dipole). The radiation power pattern, defined
as a plot of ISrl at constant r, is an alternative method of showing radi-
ation characteristics. When the radiation field is linearly polarized, as
it is for the dipole antenna, the power pattern is the square of the field
pattern. The gain g of an antenna in a given direction is defined as the
ratio of the power required from an omnidirectional antenna to the power

FIG. 2-25. Radiation field patterns for the dipole antenna.

84 TIME-HARMONIC ELECTROMAGNETIC FIELDS

required from the actual antenna, assuming equal power densities in the
given direction. Thus,
(2-130)

For L ::; A, the maximum gain of a dipole antenna occurs at (J = 1r/2.

From Eqs. (2-126) and (2-128), we have

kL)2 ( kL)2
( ~) = 1Il I ml ( 1 ~
2
cos 2 = 11 1 - cos 2
g 2 1r(J>j 7r R (2-131)
r

In the limit kL --:; 0, we have g(7r/2) = 1.5; so the maximum gain of a

short dipole is 1.5. For a half-wave dipole, we can use Fig. 2.24 and
calculate a maximum gain of 1.64. Similarly, for a full-wave dipole,
the maximum gain is 2.41.
The input impedance of an antenna is the impedance seen by the source,
that is, the ratio of the complex terminal voltage to the complex terminal
current. A knowledge of the reactive power, which cannot be obtained
from radiation zone fields, is needed to evaluate the input reactance.
The input resistance accounts for the radiated power (and dissipated
power if losses are present). We define the input resistance of a loss-
free antenna as
(2-132)

where (P, is the power radiated and I« is the input current. If losses are
present, a "loss resistance" must be added to Eq. (2-132) to obtain the
input resistance. For the dipole antenna,
· kL
I i = I mS1n2
and the input resistance is
Rr (2-133)
R i = sin? [k(L/2)]

In the limit as kL is made small, we find

R. = 11(kL)2 L «A (2-134)
, 241r
The short dipole therefore has a very small input resistance. For exam-
ple, if L = A/10, the input resistance is about 2 ohms. For the half-
wavelength dipole, we use Fig. 2-24 and Eq. (2-133) and find
A
R; = R,. = 73.1 ohms L=-2 (2-135)
 INTRODUCTION TO WAVES 85
For the full-wavelength dipole, Eq. (2-133) shows R, = 00. This incor-
rect result is due to our initial choice of current, which has a null at the
source. The input resistance of the full-wavelength dipole is actually
large, but not infinite, and depends markedly on the wire diameter (see
Fig. 7-13).
2-11. On Waves in General. A complex function of coordinates
representing an instantaneous function according to Eq. (1-40) is called a
wave function. A wave function y;, which may be either a scalar field or
the component of a vector field, may be expressed as
1/1 = A (x,y,z)e;~(x,tI,z) (2-136)
where A and tI> are real. The corresponding instantaneous function is
V2 A (x,y,z) cos [wt + tI>(x,y,z)] (2-137)
The magnitude A of the complex function is the rms amplitude of the
instantaneous function. The phase tI> of the complex function is the
initial phase of the instantaneous function. Surfaces over which the
phase is constant (instantaneous function vibrates in phase) are called
equiphase surfaces. These are defined by
4>(x,y,z) = constant (2-138)
Waves are called plane, cylindrical, or spherical according as their equi-
phase surfaces are planes, cylinders, or spheres. Waves are called uni-
form when the amplitude A is constant over the equiphase surfaces.
Perpendiculars to the equiphase surfaces are called wave normals. These
are, of course, in the direction of Vtl> and are the curves along which the
phase changes most rapidly.
The rate at which the phase decreases in some direction is called the
phase constant in that direction. (The term phase constant is used even
though it is not, in general, a constant.) For example, the phase con-
stants in the cartesian coordinate directions are
cJtI> dtl>
{j=--
tI ay (2-139)
dZ
These may be considered as components of a vector phase constant defined
by
(2-140)
The maximum phase constant is therefore along the wave normal and is
of magnitude /vcPl.
The instantaneous phase of a wave is the argument of the cosine func-
tion of Eq. (2-137). A surface of constant phase is defined as
(JJt + cf>(x,y,z) = constant (2-141)
86 TIME-HARMONIC ELECTROMAGNETIC FIELDS

that is, the instantaneous phase is constant. At any instant, the sur-
faces of constant phase coincide with the equiphase surfaces. As time
increases, ~ must decrease to maintain the constancy of Eq. (2-141), and
the surfaces of constant phase move in space. For any increment ds the
change in ~ is
a~ a~ a~
V<I> • ds = -
ax dx + -ay dy + -az dz
To keep the instantaneous phase constant for an incremental increase in
time, we must have
ca dt + V<I> • ds = 0
That is, the total differential of Eq. (2-141) must vanish. The phase
velocity of a wave in a given direction is defined as the velocity of surfaces
of constant phase in that direction. For example, the phase velocities
along cartesian coordinates are

(2-142)

The phase velocity along a wave normal (ds in the direction of - V<I» is
w w
vp = - IV4>1 = ~
(2-143)

which is the smallest phase velocity for the wave. Phase velocity is not a
vector quantity.
We can also express the wave function, Eq. (2-136), as
(2-144)
where e is a complex function whose imaginary part is the phase ~.
A vector propagation constant can be defined in terms of the rate of change
of e as
y = -va = a +j~ (2-145)

where ~ is the phase constant of Eq. (2-140) and a is the vector attenu-
ation constant. The components of a are the logarithmic rates of change
of the magnitude of 1/1 in the various directions.
In the electromagnetic field, ratios of components of E to components
of H are called wave impedances. The direction of a wave impedance is
defined according to the right-hand" cross-product" rule of component E
 INTRODUCTION TO WAVES 87
rotated into component H. For example,

Ex - Z + - Z (2-146)
H 7I - X71 - •

is a wave impedance in the +z direction, while

- : : = Z:&/I- = Z_. (2-147)

is a wave impedance in the -z direction. The wave impedance in the

+z direction involving E u and H:z; is
(2-148)

The Poynting vector can be expressed in terms of wave impedances.

For example, the z component is
S. = (E X H*). = ExH: - E 7IH:
= ZX7l+IH7I12 + Zuz +IH z12 (2-149)
The concept of wave impedance is most useful when the wave imped-
ances are constant over equiphase surfaces.
Let us illustrate the various concepts by specializing them to the uni-
form plane wave. Consider the x-polarized z-traveling wave in lossy
matter,
Ex = Eoe-k"·e-ik'~
H = Eo e-k".e-ik'.
71 1/

The amplitude of Ex is Eoe-k"lS and its phase is -k'z. Equiphase sur-

faces are defined by -k'z = constant, or, since k' is constant, by z = con-
stant. These are planes; so the wave is a plane wave. The amplitude
of E:z; is constant over each equiphase surface; so the wave is uniform.
The wave normals all point in the z direction. The cartesian compo-
nents of the phase constant are {3z = {311 = 0, {3. = k'; so the vector phase
constant is (J = u.k'. The phase velocity in the direction of the wave
normals is V p = wlk'. The cartesian components of the attenuation con-
stant are a:z; = au = 0, a. = k"; so the vector attenuation constant is
a = u.k". The vector propagation constant is
y = a + J{J = u.(k" + jk') = uJk
The wave impedance in the z direction is Z. = Z Z1J+ = E xlH f/ = 'TJ. Note
that the various parameters specialized to the uniform plane traveling
wave are all intrinsic parameters. This is, by definition, the meaning of
the word "intrinsic."
88 TIME-HARMONIC ELECTROMAGNETIC FIELDS

PROBLEMS

2-1. Show that E~ == Eoe-;1c~ satisfies Eq. (2-6) but not Eq. (2-5). Show that it
does not satisfy Eq. (2-3). This is not a possible electromagnetic field.
2-2. Derive the "wave equations" for inhomogeneous media

v X (i-IV X E) + ~E =0
V X (~-lV X H) + iH = 0
Are these valid for nonisotropic media? Do Eqs. (2-5) hold for inhomogeneous
media?
2-3. Show that for any lossless nonmagnetic dielectric

k == ko v;:.

where Er is the dielectric constant and k o, T]o, Ao, and. c are the intrinsic parameters of
vacuum.
2-4. Show that the quantities of Eqs. (2-18) satisfy Eq. (1-35). Repeat for Eqs.
(2-21), (2-27), and (2-29).
2-6. For the field of Eqs. (2-20), show that the velocity of propagation of energy
as defined by Eq. (2-19) is

v, == _1_ sin 2kz sin 2wt < _1_

W 1 - cos 2kz cos 2wt - W

2-6. For the field of Eqs. (2-22), show that the phase velocity is

_ (A + C 2 k + AA +
- C . k )
Vp - W1 A _ C cos z C SIn
2
z

2-'1. For the field of Eqs. (2-28), show that the z-directed wave impedances are

Z:eJ/+ == E:e = -iT]tankz

H'II
-E
ZJ/:e+ = _ _J/ = -iT] tan ke
H:e

Would you expect Z:e'll + = Z 'II:e + to be true for all a-c fields?
2-8. Given a uniform plane wave traveling in the +z direction, show that the wave
is circularly polarized if
s, ±.
E'II == J

being right-handed if the ratio is +i and left-handed if the ratio is -j.

2-9. Show that the uniform plane traveling wave of Eq. (2-25) can be expressed as
the sum of a right-hand circularly polarized wave and a left-hand circularly polarized
wave.
2-10. Show that the uniform plane traveling wave of Eq. (2-25) can be expressed as

E == (El + jE 2 )e- ;I:·

 INTRODUCTION TO WAVES 89
where E 1 and E 2 are real vectors lying in the xy plane. Relate E 1 and E 2 to A and B.
2-11. Show that the tip of the arrow representing 8 for an arbitrary complex E
traces out an ellipse in space. [Hint: let E == Re (E) +
j Im (E) and use the results
of Probe 2-10.]
2-12. For the frequencies 10, 100, and 1000 megacycles, determine k == k' - jk"
and" == CR + j~ for (a) polystyrene, Fig. 1-10, (b) Plexiglas, Fig. I-II, (c) Ferramic A,
Fig. 1-12, Er == 10, and (d) copper, (T == 5.8 X 10 7 •
2-13. Show that when all losses are of the magnetic type (eT == E'l == 0),

2-14. Show that for nonmagnetic dielectrics

Q» 1

where Q is defined by Eq. (1-79).

2-16. Show that for nonmagnetic conductors

k' ~ ~(1 +~)

k" ~ ~6J~ (1 -~)
Q« 1
at ~ ~';'; (1 +~)
X ~ ~';'; (1 -~)
where Q is defined by Eq. (1-79).
2-16. Show that for metals

7J == <R(1 + j) k == !a (1 - j) <R==.!.
eTa

where at is the surface resistance, a is the skin depth, and iT is the conductivity.
2-17. Derive the following formulas

at (silver) == 2.52 X 10- 7 vi

at (copper) == 2.61 X 10- 7 Vl
at (gold) == 3.12 X 10- 7 Vl
at (aluminum) == 3.26 X 10-7 Vl
at (brass) == 5.01 X 10-7 Vl

where f is the frequency in eycles per second.

90 TIME-HARMONIC ELECTROMAGNETIC FIELDS

2-18. Find the power per square meter dissipated in a copper sheet if the rms mag-
netic intensity at its surface is 1 ampere per meter at (a) 60 cycles, (b) 1 megacycle,
(c) 1000 megacycles.
2-19. Make a sketch similar to Fig. 2-6 for a circularly polarized standing wave in
dissipative media. Give a verbal description of £ and 3C.
2-20. Given a uniform plane wave normally incident upon a plane air-to-dielectric
interface, show that the standing-wave ratio is
SWR == V; == index of refraction
where Er is the dielectric constant of the dielectric (assumed nonmagnetic and loss-free).
2-21. Take the index of refraction of water to be 9, and calculate the percentage of
power reflected and transmitted when a plane wave is normally incident on a calm lake.
2-22. Calculate the two polarizing angles (internal and external) and the critical
angle for a plane interface between air and (a) water, Er == 81, (b) high-density glass,
Er == 9, and (c) polystyrene, Er == 2.56.
2-23. Suppose a uniform plane wave in a dielectric just grazes a plane dielectric-
to-air interface. Calculate the attenuation constant in the air [a as defined by Eq.
(2-61)] for the three cases of Probe 2-22. Calculate the distance from the boundary
in which the field is attenuated to lie (36.8 per cent) of its value at the boundary.
What is the value of a at the critical angle?
2-24. From Eqs. (2-66) and (2-68), show that when R «wL and G «wC

a ~ R + G v'LlC
2 yLIC 2
fJ~wVLC
where "y = a + j{j.
2-26. Show that It and C of a transmission line are related by

G = ~"C == WE " 71
E' ZO
when the dielectrIc Is homogeneous. Show that R of a transmission line is approxi-
mately equal to the d-e resistance per unit length of hollow conductors having thick-
ness a (skin depth) provided H is approximately constant over each conductor and the
radius of curvature of the conductors is large compared to a.
2-26. Using results of Probe 2-25, show that for the two-wire line of Table 2-3
a» a
D »d
and that for the coaxial line
R ~ <R a +b a » a
211" ab

and that for the parallel-plate line

R ~ 2<R w» b
w
2-27. Verify Eqs. (2-70).
2-28. Consider a parallel-plate waveguide formed by conductors covering the planes
y == 0 and y == b. Show that the field

n == 1, 2, 3, . . •
 INTRODUCTION TO WAVES 91
defines a. set of TEn modes and the field

n1rY
Hz z::: H o cos b e-'Y· n == -0, 1, 2, . . .

defines a set of TM" modes, where

in both cases. Show that the cutoff frequencies of the TEn and TMn modes are

Show that Eqs. (2-83) to (2-86) apply to the parallel-plate waveguide modes.
2-29. Show that the power transmitted per unit width (x direction) of the parallel-
plate waveguide of Probe 2-28 is

p = l:;12 ~1
b - (yy
for the TEn modes, and

for the TM n modes (n ¢ 0).

2-30. For the parallel-plate waveguide of Probe 2-28, show that the attenuation
due to conductor losses is

for the TEn modes, and

2<R
ae = b71 VI - (felf)?
for the TM n modes (n ¢ 0).
2-31. Show that the TM o mode of the parallel-plate waveguide as defined in Probe
2-28 is actually a TEM mode. Show that for this mode the attenuation due to con-
ductor losses is
<R
ae = ~

Compare this with a obtained by using the results of Probs. 2-26 and 2-24.
2-32. For the TE ol rectangular waveguide mode, show that the time-average elec-
tric and magnetic energies per unit length are

Can this equality of W. and W m be predicted from Eq. (1-62)1

2-33. Show that the time-average velocity of propagation of energy down a rec-
tangular waveguide is
0. == ~ == _1 ~1 _ ff!)2
W V; \J
for the TE ol mode.
92 TIME-HARMONIC ELECTROMAGNETIC FIELDS

2-34. For the TE ol rectangular waveguide mode, define a voltage V as IE · dI across

the center of the guide and a current I as the total z-directed current in the guide wall
x = O. Show that these are

Show that P ¢ V I·. Why? Define a characteristic impedance ZVI = V / I and

show that it is proportional to Z 0 of Table 2-4.
2-36. Let a rectangular waveguide have a discontinuity in dielectric at z = 0,
that is, El, PI for z < 0 and E2, JJ.2 for z > O. Show that the reflection and trans-
mission coefficients for a TE oI wave incident from z < 0 are

r = Z02 - ZOI T = 2Z 02
Z02 + ZOI Z02 + ZOI

where ZOI and Z02 are the characteristic impedances e < 0 and z > 0, respectively.
These results are valid for any waveguide mode.
2-36. Show that there is no reflected wave for the TE ol mode in Probe 2-35 when

where leI is the cutoff frequency z < O. Note that we cannot have a reflectionless
interface when both dielectrics are nonmagnetic. This result is valid for any TE
mode.
2-37. Take a parallel-plate waveguide with EI, JJ.l for z < 0 and E2, JJ.2 for z > o.
Show that there is no reflected wave for a TM mode incident from z < 0 when

For nonmagnetic dielectrics, this reduces to

L = .. lEI + E2
Icl 'J E2

Compare this to Eq. (2-60). These results are valid for any TM mode.
2-38. Design a square-base cavity with height one-half the width of the base to
resonate at 1000 megacycles (a) when it is air-filled and (b) when it is polystyrene-
filled. Calculate the Q in each case.
2-39. For the rectangular cavity of Fig. 2-19, define a voltage V as that between
mid-points of the top and bottom walls and a current I as the total x-directed cur-
rent in the side walls. Show that
V = Eoa

Define a mode conductance G as G = <Pd/IVlt and show that

G == CR[bc(b 2 + c + 2a(b + c
2
)
3 3
) ]

a 2(b t + c 2)
21]2

Define a mode resistance R as R == <Pd/III2 and show that

r 2CR[bc(b2 + c2 ) + 2a(b + c 3 3) ]

R == 32(b t + C2) 2
 INTRODUCTION TO WAVES 93
1-4:0. Derive Eqs. (2-123).
2-4:1. Consider the small loop of constant current I as shown in Fig. 2-26. Show
that the magnetic vector potential is

Act» = All I .-0

= la
4:Jr J0
(2... f cos q,' dq,'

where
f = exp (-jk vT2+ a 2ra sin 8 cos q,') 2
-

vr 2 + a 2ra sin 8 cos q,'

2
-

Expand f in a Maclaurin series about a = 0 and show that

A 4>
a-+O
Ira"
--+ - -
4r
e-'Okr (irk + r1) .
- -
2
SIn 8

The quantity I rat = 18 is called the magnetic moment of the loop.

z
,.

FIG. 2-26. A circular loop

of current.

x.

2-42. Show that the field of the small current loop of Probe 2-41 is

H r = 18
-2r e- , k r ° (ik + -1 ) cos 8
- 2
r r3
k +i k + 1).
2
H 8 = -I S e-'Ok r ( - - - -
3
SIn 8
4r r r r 2

E 4> = -T/IS e- Ok r (k-

2
-
i- k ) .
SIn 8
411'" ' r r2

Show that the radiation resistance of the small loop referred to I is

2-43. Consider the current element of Fig. 2-21 and the current loop of Fig. 2-26
to exist simultaneously. Show that the radiation field is everywhere circularly
polarized if
It = kIS
2-44:. In terms of the tabulated functions

SI·()
X=
.
f 0
% sin-z z
-
z
d Ci (e) == _
J%
t: cos
Z
Z dz
94 TIME-HARMONIC ELECtrkOMAGNETIC FIELDS

show that Eq. (2-129) can be expressed as

n, = ir [ C + log kL - CikL + sin kL (~2Si2kL - SikL)

+ ~ cos kL ( C + log k ~ + Ci 2kL - 2Ci kL ) ]

where C = 0.5772 . . . is Euler's constant.

2-46. If the linear antenna of Fig. 2-23 is an integral number of half-wavelengths
long, the current will assume the form

l(z) = i ; sin k (Z + ~)
regardless of the position of the feed as long as it is not near a current null. Such an
antenna is said to be of resonant length. Show that the radiation field of the antenna is

E8 = J",
.I
m
cos (~ cos
e- iTer _ _~. _
0) n odd
211"r SIn 0

",I _" SIn

• (n1l"
2 cos 0 )
E8 = -211"rm e ITer - - - : - . - - -
SIn 0
n even

where n = 2L/>.. is an integer.

2-46. For an antenna of resonant length (Prob. 2-45), show that the radiation
resistance referred to L; is

Rr = 4: [C + log 2n1l" - Ci(2n1l")]

where n = 2L/X, C = 0.5772, and Ci is as defined in Probe 2-44. Show that the input
resistance for a loss-free antenna with feed point at z = a>.. is

R. _ Rr
, - sin 211"(a + n/4)
Specialize this result to L = X/2, a = 0 (the half-wave dipole) and show that
R« = 73 ohms.