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Unformatted text preview: 24 Waveguides 24—1 The transmission line In the last chapter we studied what happened to the lumped elements of circuits
When they were operated at very high frequencies, and we were led to see that a
resonant Circuit could be replaced by a cavity with the ﬁelds resonating inside.
Another interesting technical problem is the connection of one object to another,
so that electromagnetic energy can be transmitted between them. In lowfrequency
c1rcu1ts the connection is made With Wires. but this method doesn‘t work very well
at high frequenCies because the c1rcu1ts would radiate energy into all the space
around them, and it is hard to control where the energy Will go. The ﬁelds spread
out around the Wires; the currents and voltages are not “guided” very well by
the Wires. In this chapter we want to look into the ways that objects can be
interconnected at high frequencies At least, that’s one way of presenting our
SUbJCCt. Another way is to say that we have been discussmg the behav10r of waves in
free space. Now it is time to see What happens when oscillating ﬁelds are conﬁned
in one or more dimensions. We Will discover the interesting new phenomenon
when the ﬁelds are conﬁned in only two dimensions and allowed to go free in the
third dimensmn, they propagate in waves. These are “guided waves”—the subject
of this chapter. We begin by working out the general theory of the transmission line. The
ordinary power transmiSSion line that runs from tower to tower over the country
side radiates away some of its power. but the power frequenCies (50—60 cycles/sec)
are so low that this loss is not serious. The radiation could be stopped by surround
ing the line With a metal pipe, but this method would not be practical for power
lines because the voltages and currents used would requ1re a very large, expenSive,
and heavy pipe. So Simple “open lines" are used. For somewhat higher frequenCies—say a few kilocycles~radiation can al
ready be serious However. it can be reduced by using “tWistedpair” transmtssron
lines. as is done for shortrun telephone connections. At higher frequencies, how
ever, the radiation soon becomes intolerable, either because of power losses or
because the energy appears in other Circu1ts where it isn‘t wanted For frequenCIes
from a few kilocycles to some hundreds of megacycles, electromagnetic Signals
and power are usually transmitted Via coaxial lines consisting of a Wire inside a
cylindrical “outer conductor” or “shield " Although the following treatment will
apply to a transmiSSion line of two parallel conductors of any shape, we Will carry
it out referring to a coaxial line. We take the Simplest coaxial line that has a central conductor, which we sup
pose is a thin hollow cylinder, and an outer conductor which is another thin
cylinder on the same ﬁlms as the inner conductor, as in Fig. 24—1 We begin by
ﬁguring out approximately how the line behaves at relatively low frequenCies
We have already described some of the lowfrequency behaVior when we said
earlier that two such conductors had a certain amount of inductance per unit
length or a certain capaCity per unit length. We can, in fact, describe the low
frequency behaVior of any transmission line by givmg its inductance per unit
length, L0 and its capacity per unit length, C0. Then we can analyze the line as
the limiting case of the LC ﬁlter as discussed in Section 22—6. We can make a
ﬁlter which imitates the line by taking small series elements LOAx and small
shunt capaCities C0 Ax, where Ax is an element of length of the line. Usmg our
results for the inﬁnite ﬁlter. we see that there would be a propagation of electric 24—1 24~l The transmission line 24—2 The rectangular waveguide
24—3 The cutoff frequency 24—4 The speed of the guided waves
24—5 Observing guided waves 246 Waveguide plumbing 24—7 Waveguide modes 24—8 Another way of looking at the
guided waves \"h: _ _ Fig. 24—l. Acooxioltronsmissuonline. was I 1‘11 IE Ax)
2'23:
/' 1\
V(x)‘l l V(x+Ax)
I
I
WIRE 2 \ _ ,/
x x + Ax
Fig. 24—2. The currents and voltages of a transmission line. signals along the line. Rather than following that approach, however, we would
now rather look at the line from the point of View of a differential equation. Suppose that we see what happens at two neighboring points along the
transmission line, say at the distances x and x + Ax from the beginning of the
line. Let’s call the voltage difference between the two conductors V(x). and the
current along the “hot” conductor I(x) (see Fig. 24—2). If the current in the line
is varying, the inductance will give us a voltage drop across the small section of
line from x to x + Ax in the amount AV = V(x + Ax) — V(x) = —L0Axg Or, taking the limit as Ax ——> 0, we get 6V 61. ~ = —L0 at (24.1) The changing current gives a gradient of the voltage. Referring again to the ﬁgure, 1f the voltage at x is changing. there must be
some charge supplied to the capacity in that region. If we take the small piece of
line between x and x + Ax, the charge on it is q = C0 Ax V. The time rateof
change of this charge is C0 Ax dV/dt. but the charge changes only if the current
1(x) into the element is different from the current I(x + Ax) out. Calling the diner
ence A], we have dV
AI —— —C0 Ax —$ '
Taking the limit as Ax —> 0, we get
61 6V
5; _ —Co .67. (24.2) So the conservation of charge implies that the gradient of the current is propor
tional to the time rateofchange of the voltage. Equations (24.1) and (24.2) are then the basic equations of a transmission
line. If we wish, we could modify them to include the effects of resistance in the
conductors or of leakage of charge through the insulation between the conductors,
but for our present discussion we will just stay with the simple example. The two transmission line equations can be combined by differentiating one
with respect to t and the other with respect to x and eliminating either V or I.
Then we have either 62V 62V 6x2 _ COL" at? (243)
OF 621 621 $5 _ COLO W (24_4) Once more we recognize the wave equation in x. For a uniform transmissron
line, the voltage (and current) propagates along the line as a wave. The voltage
along the line must be of the form V(x, t) = f(x — wt) or V(x, r) = g(x + (it),
or a sum of both. Now what is the velocity 1v? We know that the coefﬁcient of
the 62/6t2 term is just 1/02, so I = . 24.5)
x/LOCO ( U We will leave it for you to show that the voltage for each wave in a line is
proportional to the current of that wave and that the constant of proportionality
isjust the characteristic impedance 20. Calling V+ and 1+ the voltage and current
for a wave going in the plus xdirection, you should get V+ Z 201+. (24'6) 24—2 Similary, for the wave going toward minus x the relation is
V_. = —'Zol_. The characteristic impedance——as we found out from our ﬁlter equations——is given by
iL
Z0 = F: a (247) and is, therefore, a pure resistance. To ﬁnd the propagation speed v and the characteristic impedance 20 of a
transmission line, we have to know the inductance and capaCity per unit length.
We can calculate them easily for a coaxial cable, so we will see how that goes. For
the inductance we follow the ideas of Section 17—8, and set éLI2 equal to the mag
netic energy which we get by integrating eucsz/Z over the volume. Suppose
that the central conductor carries the current I; then we know that B = I/27r60c2r,
where r is the distance from the EMS. Taking as a volume element a cylindrical
shell of thickness dr and of length l, we have for the magnetic energy eoc2 b I 2
U = T a (276063,) lZ7rrdr, where a and b are the radii of the inner and outer conductors, respectively. Carry
ing out the integral, we get 121 b
U — ZW€0C2 In a ‘ (24 8) Setting the energy equal to %Ll2, we ﬁnd I lnl—J L = o
27T€062 a (24.9) It is. as it should be, proportional to the length [of the line, so the inductance per
unit length L0 is _ 131(b/a)
L0 — 2m“,  (24.10) We have worked out the charge on a cylindrical condenser (see Section 12—2)
Now, diViding the charge by the potential diﬁerence, we get 27reol C = 1n (b/a)‘ The capacity per unit length C0 is C/l. Combining this result With Eq. (24.10),
we see that the product LOCO is Just equal to l/cz. so u = l/x/LOCO is equal
to c. The wave travels down the line with the speed oflight. We point out that this
result depends on our assumptions: (a) that there are no dielectrics or magnetic
materials in the space between the conductors, and (b) that the currents are all on
the surfaces of the conductors (as they would be for perfect conductors). We will
see later that for good conductors at high frequencies, all currents distribute
themselves on the surfaces as they would for a perfect conductor, so this assump
tion is then valid. Now it is interesting that so long as assumptions (a) and (b) are correct, the
product LHCO is equal to l/c2 for any parallel pair of conductors—even, say, for a
hexagonal inner conductor anywhere inSide an elliptical outer conductor. So long
as the cross section is constant and the space between has no material, waves are
propagated at the veIOCity of light. No such general statement can be made about the characteristic impedance.
For the coaxial line, it is
_ 1n (b/a) . Z _
0 2760c (24.11) 24—3 Fig. 24—3. Coordinates chosen for
the rectangular waveguide. (b) o x Fig. 24—4. The electric field in the
waveguide at some value of z. —> Vph (D) Fig. 24—5. The zdependence of the
ﬁeld in the waveguide. The factor l/eoc has the dimensions of a resistance and is equal to 1207r ohms.
The geometric factor In (b/a) depends only logarithmically on the dimenSions, so
for the coaxial line—and most lines—the characteristic impedance has typical
values of from 50 ohms or so to a few hundred ohms. 24—2 The rectangular waveguide The next thing we want to talk about seems, at ﬁrst sight, to be a striking
phenomenon: if the central conductor is removed from the coaxial line, it can still
carry electromagnetic power. In other words, at high enough frequencies a hollow
tube will work just as well as one with wires. It is related to the mysterious way in
which a resonant circuit of a condenser and inductance gets replaced by nothing but a can at high frequencies. Although it may seem to be a remarkable thing when one has been thinking
in terms of a transmission line as a distributed inductance and capacity, we all
know that electromagnetic waves can travel along inside a hollow metal pipe.
If the pipe is straight, we can see through it! So certainly electromagnetic waves
go through a pipe. But we also know that it is not pOSSible to transmit lowfre
quency waves (power or telephone) through the inside of a single metal pipe. So
it must be that electromagnetic waves will go through if their wavelength is short
enough. Therefore we want to discuss the limiting case of the longest wavelength
(or the lowest frequency) that can get through a pipe of a given size. Since the
pipe is then being used to carry waves, it is called a waveguide. We Will begin with a rectangular pipe, because it is the Simplest case to
analyze. We will ﬁrst give a mathematical treatment and come back later to look
at the problem in a much more elementary way. The more elementary approach.
however, can be applied easily only to a rectangular gu1de. The basic phenomena
are the same for a general guide of arbitrary shape, so the mathematical argument
is fundamentally more sound. Our problem, then, is to ﬁnd what kind of waves can eXist inside a rectangular
pipe. Let’s ﬁrst choose some convenient coordinates: we take the z—axis along the
length of the pipe, and the x and yaxes parallel to the two sides, as shown in
Fig. 24—3. We know that when light waves go down the pipe, they have a transverse
electric ﬁeld; so suppose we look ﬁrst for solutions in Which E is perpendicular to
2, say with only a ycomponent, E,,. This electric ﬁeld will have some variation
across the guide; in fact, it must go to zero at the sides parallel to the yaxis, because
the currents and charges in a conductor always adjust themselves so that there is
no tangential component of the electric ﬁeld at the surface of a conductor. So
E,’ Will vary with x in some arch, as shown in Fig. 24—4. Perhaps it is the Bessel
function we found for a cavity? No, because the Bessel function has to do with
cylindrical geometries. For a rectangular geometry, waves are usually simple
harmonic functions, so we should try something like sm kxx. Since we want waves that propagate down the guide, we expect the ﬁeld to
alternate between positive and negative values as we go along in 2, as in Fig. 24—5.
and these oscillations will travel along the guide with some velocny I’. [f we have
oscillations at some deﬁnite frequency co, we would guess that the wave might vary
with 2 like cos(wt — kzz), or to use the more convenient mathematical form.
like ellwhl‘iz”. This z—dependence represents a wave travelling with the speed
1) = w/kz (see Chapter 29, Vol. I). So we might guess that the wave in the guide would have the following
mathematical form: E, = E0 sin march”. (24.12) Let’s see whether this guess satisﬁes the correct ﬁeld equations. First. the
electric ﬁeld should have no tangential components at the conductors. Our ﬁeld
satisfies this requirement; it is perpendicular to the top and bottom faces and is
zero at the two side faces. Well, it is if we choose k, so that onehalf a cycle of 24—4 sin erjust ﬁts in the width of the guide—that is. if kxa = 1r. (24.13)
There are other poss1bilities. like kxa = 27r, 371', . . . , or, in general.
kg = 1171', (2414) where n is any integer. These represent various complicated arrangements of the
ﬁeld, but for now let’s take only the Simplest one, where k1 = 7r/a. where a is
the width of the inSide of the guide. Next. the divergence of E must be zero in the free space inside the guide,
since there are no charges there. Our E has only a ycomponent, and it doesn’t
change With y, so we do have that V  E = 0. Finally. our electric ﬁeld must agree With the rest of Maxwell’s equations in
the free space inside the guide. That is the same thing as saying that it must
satisfy the wave equation 3215., + 62Ey + 62E” _ 1 62Ey
6262 ay2 622 c2 012 ~ 0. (24.15) We have to see whether our guess, Eq. (24.12), will work. The second derivative of
E,, With respect to x is just —kEEy The second derivative with respect to y is
zero. Since nothing depends on v. The second derivative With respect to z is —ka,,,
and the second derivative with respect to t is ~w2Ey. Equation (24.15) then says that 0.12 1:35., + kal, — 72 Ey = 0. Unless E" is zero everywhere (which is not very interesting), this equation is correct
if (.02
E2— kf + k3 — = 0. (24.16)
We have already ﬁxed k1, so this equation tells us that there can be waves of the type we have assumed if k2 is related to the frequency co so that Eq. (24 I6) is
satisﬁed—in other words, if kz = «m. (24.17) The waves we have described are propagated in the zdirection with this value osz
The wave number k2 we get from Eq. (24.17) tells us. for a given frequency w.
the speed with which the nodes of the wave propagate down the gurde. The phase veloaty is
w v — k—Z (24.18) You will remember that the wavelength A of a travelling wave is given by A = 27rv/w, so k2 is also equal to 2717A”, where M is the wavelength of the oscrlla
tions along the zdirection—the “gu1de wavelength." The wavelength in the gu1de
is different ,of course, from the freespace wavelength of electromagnetic waves
of the same frequency. If we call the freespace wavelength M, which 18 equal to 27rc/w. we can write Eq. (24.17) as = __3:0__~.
\/1 — (40/20)2 BeSides the electric ﬁelds there are magnetic ﬁelds that will travel with the
wave. but we Will not bother to work out an expressmn for them right now. Since
62V X B = 6E/61, the lines of B will circulate around the regions in which
aE/al is largest, that is, halfway between the max1mum and minimum of E. The loops of B Will lie parallel to the xzplane and between the crests and troughs of
E, as shown in Fig. 24—6. M (24.19) 24~5 Fig. 24—6.
waveguide. The magnetic ﬁeld in the Fig. 24—7. The variation of Ey with
z for u: << we. 24—3 The cutoff frequency In solving Eq. (24.16) for k, there should really be two roots—one plus and
one minus. We should write k2 = i V (wZ/CZ) —— (72/a2). (24.20) The two Signs simply mean that there can be waves which propagate With a nega
tive phase velocity (toward —z), as well as waves which propagate in the positive
direction in the guide. Naturally, it should be pos51ble for waves to go in either
direction. Since both types of waves can be present at the same time, there will be
the possibility of standingwave solutions. Our equation for k2 also tells us that higher frequencies give larger values of
k, and therefore smaller wavelengths, until in the limit of large to, k becomes
equal to w/c, which is the value we would expect for waves in free space. The
light we “see” through a pipe still travels at the speed c. But now notice that if we
go toward low frequencres, something strange happens. At ﬁrst the wavelength
gets longer and longer, but if 0.: gets too small the quantity inside the square root
of Eq. (24.20) suddenly becomes negative. This Will happen as soon as w gets to
be less than 7rc/a—or when A0 becomes greater than 2a. In other words, when
the frequency gets smaller than a certain critical frequency wr = 7rc/a, the wave
number k, (and also )ig) becomes imaginary and we haven’t got a solution any
more. Or do we? Who said that k: has to be real? What if it does come out
imaginary? Our ﬁeld equations are still satisﬁed. Perhaps an imaginary kc also
represents a wave. Suppose to Is less than we; then we can write k~ = =tz'k’, (24.21) where k’ is a positive real number: k’ = x/(ir2/a2) — (w2/62). (24.22) If we now go back to our expression, Eq. (24.12), for E,,, we have E, = E0 sin kzxetwlk’z), (24.23)
which we can write as
151] = E0 sin k,xeik’zew‘. (24.24) This expression gives an E—ﬁeld that oscillates With time as em but which
varies with 2 as ei’”. It decreases or increases with z smoothly as a real exponent
ial In our derivation we didn’t worry about the sources that started the waves.
but there must, of course, be a source someplace in the guide. The sign that goes
with k’ must be the one that makes the ﬁeld decrease With increasmg distance
from the source of the waves. So for frequencies below w, = 7rc/a, waves do not propagate down the guide;
the oscillating fields penetrate into the guide only a distance of the order of l/k’.
For this reason, the frequency w, is called the “cutoff frequency” of the guide.
Looking at Eq. (24 22), we see that for frequencies just a little below cop. the num
ber k’ is small and the ﬁelds can penetrate a long distance into the guide. But if
m is much less than we, the exponential coefﬁcient k’ is equal to 7r/a and the ﬁeld
dies off extremely rapidly, as shown in Fig. 24—7. The ﬁeld decreases by l/e in the
distance a/7r, or in only about onethird of the guide width. The ﬁelds penetrate
very little distance from the source. We want to emphasize an interesting feature of our analysis of the guided
waves—the appearance of the imaginary wave number k2. Normally, if we solve
an equation in physics and get an imaginary number, it doesn’t mean anything
physical. For waves, however, an imaginary wave number does mean something.
The wave equation is still satisﬁed; it only means that the solution gives expo
nentially decreasmg ﬁelds instead of propagating waves. So in any wave problem
where k becomes imaginary for some frequency, it means that the form of the wave
changes—the sine wave changes into an exponential. 24—6 24—4 The speed of the guided waves The wave velocity we have used above is the phase velocity, which is the speed
ofa node of the wave; it is a function of frequency. If we combine Eqs. (24.17) and (24.18), we can wri...
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 Spring '09
 LeeKinohara
 Physics

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