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Feynman Physics Lectures V2 Ch24 1962-12-20 Waveguides

Feynman Physics Lectures V2 Ch24 1962-12-20 Waveguides - 24...

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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 low-frequency 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 “tWisted-pair” transmtssron lines. as is done for short-run 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 low-frequency 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 L-C 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 24-6 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 rate-of- 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 rate-of-change 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 x-direction, 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 z-dependence 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 low-fre- 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 y-axes 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 y-component, E,,. This electric field will have some variation across the guide; in fact, it must go to zero at the sides parallel to the y-axis, 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 one-half 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 y-component, 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 z-direction 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 z-direction—the “gu1de wavelength." The wavelength in the gu1de is different ,of course, from the free-space wavelength of electromagnetic waves of the same frequency. If we call the free-space 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 xz-plane 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 standing-wave 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 one-third 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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