Waveguides and resonant cavities

Transcription

1 Wavegudes and resonant cavtes February 26, 2016 Essentally, a wavegude s a conductng tube of unform cross-secton and a cavty s a wavegude wth end caps. The dmensons of the gude or cavty are chosen to transmt, hold or amplfy partcular forms of electromagnetc wave. We wll consder the case of a hollow tube a wavegude extended n the z drecton, wth arbtrary but constant cross-sectonal shape n the xy-plane. We consder possble wave solutons matchng these boundary condtons. 1 Wave equatons As we have shown, the homogeneous Maxwell equatons, gve rse to wave equatons, D = 0 B = 0 E + B t = 0 H D t = 0 E = 0 B = 0 We assume that the electrc and magnetc felds have spatal and tme dependence of the form E = E x, y e ±kz ωt B = B x, y e ±kz ωt for waves travelng n the ±z drecton. Note that E x, y and B x, y stll have components n all three spatal drectons. Unlke our plane wave solutons, the felds must satsfy boundary condtons n the x and y drectons, along the sdes of the wavegude. Separatng the del operator nto longtudnal z and transverse xy parts, the d Alembertan becomes = t + ˆk z, = µɛ 2 t = µɛ 2 t t + 2 z 2 1

2 Substtutng the assumed tme and z-dependence nto the wave equatons gves 0 = 2t µɛ 2 t z 2 E = 2 t µɛ ω 2 + k 2 E and, wth the correspondng result for B, 0 = 2 t + µɛω 2 k 2 E 0 = 2 t + µɛω 2 k 2 B 2 General soluton: Separatng transverse and longtudnal components We can smplfy the problem by separatng the transverse, E t, and longtudnal, ˆkE z, parts, then treatng E z and B z as sources for the transverse parts. We can use the z-component of wave equaton together wth the boundary condtons to solve for the sources, E z, B z. Along wth = t + ˆk z, we defne E = E t + ˆkE z ˆk E ˆk and smlarly for B, the source-free Maxwell equatons become t B t ± kb z = 0 t E t ± ke z = 0 t + ˆk E t + z ˆkE z ω B t + ˆkB z = 0 t + ˆk B t + ˆkB z + ɛµω E t + ˆkE z = 0 1 z The frst two equatons already have the form we are after, t B t = kb z t ke z and we turn our attenton to the curl equatons. The curl terms expand as t + ˆk E t + z ˆkE z = t E t + t ˆkE z + ˆk E t z + ˆk ˆk E z z and smlarly for the magnetc terms. Thus = t E t + î ˆk E z x + ĵ ˆk E z y + ˆk E t z = t E t ĵ E z x + î E z y + ˆk E t z t E t ĵ E z x + î E z y ± ˆk ke t ω t B t ĵ B z x + î B z y ± ˆk kb t + εµω 2 B t + ˆkB z E t + ˆkE z = 0 = 0

3 We separate the transverse and longtudnal parts of each equaton. Notng that t E t les n the z- drecton, the frst equaton separates nto whle second gves ĵ E z x + î E z y ± ˆk ke t ωb t = 0 transverse t E t ωb zˆk = 0 longtudnal ĵ B z x + î B z y ± ˆk kb t + ɛµω 0 transverse t B t + ɛµωe zˆk = 0 longtudnal We can smplfy the transverse equatons further. Snce they are transverse, we lose no nformaton by takng the cross product wth ˆk. Then the frst becomes 0 = ˆk ĵ E z x + î E z y ± ˆk ke t ωb t = î E z x + ĵ E z y ± ˆk ˆk ket ωˆk B t = t E z ke t ωˆk B t Smlarly, the transverse equaton for the magnetc feld crossed wth ˆk Now we may complete the separaton. 0 = t B z kb t + ɛµωˆk E t 3 Generc soluton for E t and B t 3.1 Solvng for the transverse components In the resultng par of equatons, ±ke t + ωˆk B t = t E z ±kb t ɛµωˆk t B z the transverse E t and B t are stll coupled. To separate them, solve the second for B t, and substtute nto the frst, B t = ± 1 t B z + ɛµωˆk E t k ±ke t + ωˆk B t = t E z ±ke t + ωˆk ± 1 t B z + ɛµωˆk t E z k ±ke t ± ω k ˆk t B z + ɛµωˆk t E z ±ke t ± ɛµ ω2 k ˆk ˆk t E z ω k ˆk t B z 3

4 and multplyng by k, ± k 2 ɛµω 2 k t E z ωˆk t B z 2 Therefore, as long as k 2 ɛµω 2 0, we may solve for the transverse electrc feld n terms of the longtudnal felds, ɛµω 2 k 2 ±k t E z ωˆk t B z Now substtute ths back nto the expresson for B t, B t = ± 1 t B z + ɛµωˆk E t k = ± 1 ɛµω t B z k ɛµω 2 k ˆk 2 ±k t E z ωˆk t B z = ± 1 ɛµω t B z k ɛµω 2 k 2 ±kˆk t E z + ω t B z = 1 ɛµω k ɛµω 2 k 2 kˆk ɛµω t E z ɛµω 2 k 2 ω tb z ± t B z = ɛµω ɛµω 2 k ɛµω 2 k 2 kˆk t E z ± ɛµω 2 k 2 1 t B z = ɛµω 2 k 2 ɛµωˆk t E z ± k t B z and we have solved for the transverse felds n terms of the longtudnal ones: B t = ɛµω 2 k 2 ±k t E z ωˆk t B z ɛµω 2 k 2 ɛµωˆk t E z ± k t B z 3 Ths soluton for the transverse felds holds n all cases except when k 2 ɛµω 2 = Checkng the remanng Maxwell equatons To have a complete soluton, we must check the remanng Maxwell equatons, and the longtudnal parts of the curl equatons, The dvergence the transverse electrc feld s t Usng the reduced form of the wave equaton, t ke z t B t = kb z t ωb zˆk t B t = ɛµωe zˆk ɛµω 2 k 2 ±k t t E z ω t ˆk t B z 0 = 2 t + µɛω 2 k 2 E 4

5 on the frst term on the rght, together wth t ˆk t B z = ] t ε jk [ˆk t kb z j jk = jk t ε jk δ j3 t kb z = k t ε 3k t kb z = k ε 3k t t kb z t 1 t 2 t 2 t 1 Bz to see that as requred. For the curl of the electrc feld, t t = = k = 0 ±k k 2 ɛµω 2 k 2 µɛω 2 E z = ke z Sortng out the double curl, [ ] t ˆk t B z ɛµω 2 k 2 t ±k t E z ωˆk t B z ω ɛµω 2 k 2 t ˆk t B z = ] ε jk t j [ˆk t B z k j,k = ε jk t j εk3m t mb z j,k,m = ε jk ε 3mk t j t mb z j,k,m t ˆk t B z = δ 3 δ jm δ m δ j3 t j t mb z j,m = δ 3 t j t jb z t 3 t B z j = ˆk 2 t B z The second term vanshes because t has no z component, t 3 = 0. Therefore, usng the wave equaton, 0 = 2 t + µɛω 2 k 2 B, agan, the curl of E t becomes t = ω ɛµω 2 k ˆk 2 2 t B z ω ɛµω 2 k ˆk 2 k 2 µɛω 2 B z = ωb zˆk 5

6 so the equatons for the electrc feld are dentcally satsfed. The correspondng dvergence and curl of B t are left as exercses for the reader. 4 Characterstcs of solutons Our general method of soluton s now to solve 0 = 2 t + µɛω 2 k 2 E z 0 = 2 t + µɛω 2 k 2 B z wth the approprate boundary condtons, then use these solutons to solve B t = ɛµω 2 k 2 ±k t E z ωˆk t B z ɛµω 2 k 2 ɛµωˆk t E z ± k t B z for the transverse parts. We consder three specal cases, dependng on one or both of E z and B z vanshng: 1. If both E z and B z vansh, then both the electrc and magnetc felds are purely transverse. These solutons are called TEM waves Transverse Electrc and Magnetc. From eq.2 and the correspondng equaton for B t, we see that TEM waves have ether ɛµω 2 k 2, or the felds must vansh entrely, B t = If E z vanshes, then the electrc feld s purely transverse. These solutons are called TE waves Transverse Electrc. 3. If B z vanshes, then the magnetc feld s purely transverse. These solutons are called TM waves Transverse Magnetc. Generc waves are a combnaton of all three. The method sketched above works for TE and TM waves see below, but n the case where E z = B z = 0 we have ɛµω 2 k 2 = 0 and ths soluton fals. We treat ths specal TEM case frst, then use the generc soluton to dscusss TE and TM waves. 4.1 Transverse electromagnetc waves: TEM Transverse electromagnetc waves n a wavegude are those wth no z-component to the felds, E z = 0 B z = 0 and accordng to eq.2 ths requres ɛµω 2 k 2. The generc soluton, eqs.3, does not hold and we return to the Maxwell equatons, eqs.1 wth E z = B z = 0. Settng E T EM and B t = B T EM and separatng the longtudnal and transverse parts of the curl equatons, t E T EM = 0 t B T EM = 0 t B T EM = 0 t E T EM = 0 ±kˆk E T EM ωb T EM = 0 ±kˆk B T EM + ɛµωe T EM = 0 6

7 The frst four of these mmedately mply the 2-dmensonal Laplace equaton for both the transverse electrc and magnetc felds. The relevant solutons to the 2-dmensonal Laplace equatons s then determned purely by ther boundary condtons. Snce a closed conductor allows no feld nsde, TEM waves cannot exst nsde a completely enclosed, perfectly conductng cavty, but TEM waves may exst n wavegudes. Combnng the fnal two equatons by substtutng B T EM = ± k ˆk ω E T EM nto the fnal one, ±kˆk ± kω ˆk E T EM + ɛµωe T EM = 0 k 2 + ɛµω 2 E T EM = 0 so these are satsfed by the specal case condton, together wth k = k 0 = ɛµω B T EM = ± ɛµˆk E T EM TEM waves are the domnant mode n a coaxal cable: nner and outer cylndrcal conductors held at opposte potental lead to a radal electrc feld, whle opposte currents on the conductors lead to an azmuthal magnetc feld. These are transverse to the drecton along the cable, so waves propagate along the cable between the conductors. 4.2 Boundary condtons for longtudnal modes: transverse electrc and transverse magnetc modes Modes drven by nonzero E z and/or B z fall nto two categores: transverse electrc TE and transverse magnetc TM. To see why, consder a perfectly conductng boundary. For a perfectly conductng wavegude, we fnd the boundary condtons usng the assumpton that free charges move nstantly to produce whatever surface charge densty, Σ, and surface current densty, K, are requred to make the electrc and magnetc felds vansh nsde the conductor. Wth n normal to the conductor hence perpendcular to k as well the full boundary condtons are then n D = Σ n H = K n E = 0 n B = 0 There can therefore be no tangental component of the electrc feld at the surface, and no normal component of the magnetc feld. For the longtudnal electrc feld, the boundary condton s n ˆk E z = 0 so that E z = 0 at the boundary. For the boundary condton on B z we start wth our separaton of the Maxwell equatons nto longtudnal and transverse components, where we found kb t ɛµωˆk t B z Consder the normal component of ths equaton, kn B t ɛµωn ˆk E t = n t B z k n B t + ɛµωˆk n E t = B z n 7

8 The left sde of ths equaton vanshes by the boundary condtons, so we must have B z n = 0 at the surface as well. Therefore, we seek solutons to the 2-dmensonal wave equatons wth boundary condtons 0 = 2 t + µɛω 2 k 2 E z 0 = 2 t + µɛω 2 k 2 B z 4 E z S = 0 B z n = 0 5 S These two boundary condtons are suffcent for a unque soluton, snce the current and charge densty modfy freely to satsfy the remanng condtons. The system, eqs.4 and 5, consttute a well-defned egenvalue problems for E z and B z. Snce the transverse drecton s a bounded regon, we expect a dscrete set of egenvalues. Because the boundary condtons are dfferent but the equatons the same, unqueness guarantees that the spectrum of allowed values wll be dfferent for E z and B z. Ths means that at a gven resonant frequency, n general only one or the other source feld wll be excted. Ths dvdes the solutons nto two types, 1. TE waves: The electrc feld s transverse,.e., E z = 0 everywhere whle B z satsfes the boundary condton Bz n S = TM waves: The magnetc feld s transverse so that B z = 0 everywhere whle E z satsfes the boundary condton E z S = 0. A general soluton ncludng a superposton of frequences for the feld n a wavegude or cavty s a superposton of TE, TM and, for wavegudes, TEM waves. 4.3 The transverse felds Now we may study the transverse felds n the TE and TM cases. transverse felds n each of the two cases. Frst, we smplfy our solutons for TM waves For TM waves, we set B z = 0. Then the soluton eqs.3 satsfes B t = ±k ɛµω 2 k 2 te z ɛµω ˆk ɛµω 2 k 2 t E z Takng the cross product of ˆk wth the transverse electrc feld, ˆk ±k ɛµω 2 k 2 ˆk t E z = ± k ɛµω B t = ± k ɛω H t 8

9 so that H t = ± ɛω k ˆk E t TE waves For TE waves, E z = 0 we get a smlar result, B t = so substtutng the second equaton nto the frst, = ω ˆk t ɛµω 2 k 2 B z ±k ɛµω 2 k 2 tb z ω ˆk t ɛµω 2 k 2 B z ω ˆk ɛµω 2 k 2 ɛµω2 k 2 B t ±k = ω k ˆk B t so takng the cross product wth ˆk, ˆk ω k ˆk ˆk Bt = ± ω k B t so that H t = ± k µω ˆk E t Wave mpedence Both of these relatonshps, for TM and TE waves, have the form H t = ± 1 Z ˆk E t where, usng k 0 = εµω, { k Z = ɛω = k µ k 0 ɛ T M modes µω k = k0 µ k ɛ T E modes The quantty Z s called the wave mpedence. Ths gves solutons of the form: k ± ɛµω 2 k 2 te z H t = ± 1 Z ˆk E t 6 wth Z = k k 0 µ ɛ for TM and ω ˆk t ɛµω 2 k 2 B z H t = ± 1 Z ˆk E t 7 wth Z = k0 k µ ɛ for TE. 9

10 4.3.4 The egenvalue problem We want to fnd solutons for TE and TM modes. These wll dffer from the TEM mode due to the presence of ether nonzero E z or nonzero B z, whch provde the source for the transverse felds. Denote ether of these source felds by ψ, { Ez T M modes ψ = T E modes and defne B z γ 2 µɛω 2 k 2 Then the reduced wave equaton for ψ becomes an egenvalue problem, wth boundary condtons 2 t ψ = γ 2 ψ ψ S = 0 T M ψ n = 0 T E S Because these boundary condtons are perodc, there wll be a dscrete spectrum of allowed values for γ, γ 1 < γ 2 < γ 3 <.... For each value of n, we have k 2 = µε ω 2 1 µɛ γ2 n Snce we have assumed waves of the form e ±kz ωt, waves wll propagate only f k 2 > 0, and therefore f ω ω n γ n µɛ Now fx a frequency, ω 0. As n ncreases, γ n and ω n ncrease, so for some n 0 we have ω 0 < ω n0 wll no longer propagate, nstead decayng as exp µɛ ω0 2 ω2 n 0 z = exp µɛ ωn 2 0 ω0 2 z and waves Ths means that, for a gven frequency ω 0, only a fnte number of wavelengths are possble, lyng n the range µɛ ω0 2 ω1 2 k 2 µɛ ω0 2 ωn 2 0 and equal to k 2 = µε ω0 2 1 µɛ γ2 n for any n n the range 1 n n 0. If we choose ω 1 < ω < ω 2 then only the wavelength wth wave number k 2 = µε ω0 2 ω1 2 wll propagate n the wavegude. More commonly we desre a partcular frequency or frequency range, and we choose the dmensons of the wavegude so that at the desred frequency, only the sngle value of n = 1 s allowed and only a sngle wavelength propagates. Once we have solved for ψ, we can fnd the transverse felds from the expressons above. 10

11 4.4 Example: TE modes n a rectangular wavegude Egenfunctons and egenvalues Suppose we have TE modes n a rectangular wavegude, so that E z = 0 and we solve the wave equaton for H z. Let the cross-secton of the gude run from x = 0 to x = a, and from y = 0 to y = b. Then, wth ψ = H z wth boundary condton 0 = 2 t + γ 2 H z 2 = x y 2 + γ2 H z n = 0 T E S H z The boundary condton n each drecton may be satsfed at the orgn by a cosne, so we have H z = H 0 cos αx cos βy and fttng the boundary condtons at x = a and at y = b requres α = mπ a egenfunctons are ψ mn = H 0 cos mπx nπy cos a b wth egenvalues The cutoff frequency s γ 2 mn = π 2 m 2 a 2 + n2 b 2 and β = nπ b. Therefore, the 8 ω mn = γ mn µε = π m 2 1/2 µε a 2 + n2 b 2 for each choce of m, n Selectng a wavelength Suppose for concreteness that b = a. Then the lowest frequency mode s, wth subsequent modes, 2 3 ω 10 = π a µɛ 3 ω 01 = 2 ω 10 ω 20 = 2ω 10 5 ω 11 = 2 ω ω 21 = 2 ω 10 ω 02 = 6ω 10 11

12 and so on. If we want to desgn a wavegude wth only one allowed mode, we may choose a frequency n the range Ths restrcts the wavelength to le n the range µɛ ω0 2 ω10 2 ω 10 ω 0 < ω 01 > k 2 > µɛ ω0 2 ω01 2 µɛω0 2 π2 a 2 > k 2 > µɛω0 2 3π2 2a 2 Furthermore, the wavelength must be such that k 2 = µε ω0 2 1 µɛ γ2 mn = µε ω0 2 ωmn 2, and ths gves the possble values k1 2 = µε ω0 2 ω10 2 k2 2 = µε ω0 2 ω01 2 and only the frst of these les n the propagatng range Soluton for the felds For ths mode, we have the felds, and therefore, from we fnd for the electrc feld and. H z = ψ 10 = H 0 cos πx a ekz ωt B t = H t = ω ˆk t ɛµω 2 k 2 B z ±k ɛµω 2 k 2 tb z ω ˆk t ɛµω 2 k 2 B z = ω 0 ˆk γ10 2 î H 0π a = ω 0aH 0 π k µω 0 ˆk E t B t = k ω 0 ω 0 ah 0 π = kah 0 π 4.5 Example: Half-crcular tube πx sn a ekz ωt sn πx a ekz ωt ĵ sn πx a ekz ωtˆk ĵ sn πx a ekz ωt î Consder TM modes propagatng n a perfectly conductng wavegude wth semcrcular cross secton of radus R, and the remanng sde flat. Pck a frequency such that only a sngle wavelength wll propagate, and fnd that wavelength. Fnd the resultng electrc and magnetc felds. 12

13 4.5.1 Solvng the wave equaton Choose the zaxs along the axs of the gude. Then wth B z = 0, we must frst solve the egenvalue problem for the electrc feld, 2 t + γ 2 E z = 0 wth boundary condtons E z r, ϕ S = 0: E z r, 0 = 0 0 r R E z r, π = 0 0 r R E z R, ϕ = 0 0 ϕ π Wrtng the wave equaton n polar coordnates we have 1 r E z E z r r r r 2 ϕ 2 + γ2 E z = 0 Settng E z = E r e nϕ, ths becomes Now dvde by γ 2 and defne ρ = γr, to get the Bessel equaton, wth solutons 1 r 1 1 d γ 2 r dr E z = r r E z + r γ 2 n2 r 2 E z = 0 r de 1 + dr γ 2 γ2 1 n 2 γ 2 r 2 E = 0 d 2 E dρ 2 n= Imposng the boundary condtons + 1 de ρ dρ + 1 n2 ρ 2 E = 0 a n J n γr + b n K n γr e nϕ The boundary condton at the orgn mmedately shows that b n = 0. Also, at r = 0, ϕ = 0, we have 0 = a n J n 0 n= = a 0 J 0 0 so we must set a 0 = 0. Now consder the full set of boundary condtons, E z r, 0 = 0 0 r R E z r, π = 0 0 r R E z R, ϕ = 0 0 ϕ π Notce that the frst two condtons wll be satsfed f our seres depends only on sn nϕ. To acheve ths, we note that J n ρ = 1 n J n ρ, so that we may wrte E z = an J n γr e nϕ + a n J n γr e nϕ = n=1 J n γr a n e nϕ + 1 n a n e nϕ n=1 13

14 and the boundary condton shows that we need an e nϕ + 1 n a n e nϕ = 2a n sn nϕ and therefore Absorbng the constant 2 nto a new constant c n, a n = 1 n+1 a n E z = c n J n γr sn nϕ n=1 The remanng boundary condton, E z = c n J n γr sn nϕ n=1 must hold for all ϕ n the range, so γr must be a root of the Bessel functon.let J n x nk = 0 gve the roots of the Bessel functons, where k = 1, 2,.... Then at r = R, we must have Restrctng the wavelength γ nk R = x nk γ nk = x nk R The frst roots of the varous Bessel functons are ncreasng, that s, x n+1,1 > x n,1. Therefore, the lowest root of all the x nk s x 11, and we want to select for the correspondng γ 11. Let the cutoff frequences be defned by ω nk = 1 ɛµ γ nk and choose ω = ω 0 n the nterval ω 11 < ω 0 < ω 12, ω 21, etc. for all remanng n, k. Then the wave number, k nk = ɛµ ω0 2 ω2 nk s magnary except for the lowest cutoff frequency, k 11 = ɛµ ω0 2 ω2 11 and the feld n the wavegude s gven by only J 1, where the wavelength s E z r, ϕ = E 0 J 1 x01 r R 2π λ = ɛµω0 2 x11 R 2 sn nϕe ±kz ωt 14

15 4.5.4 The electrc and magnetc felds For TM waves the felds are gven by k ± ɛµω 2 k 2 te z H t = ± 1 Z ˆk E t where γ 2 = ɛµω 2 k 2 and Z = k µ k 0 ɛ. Wth E z = E 0 J x01r 1 R sn nϕe kz ωt, the transverse dervatve gves ± k γ 2 x01 r t E 0 J 1 sn nϕe ±kz ωt R = ± ke 0 ˆr γ 2 r + ˆϕ1 x01 r J 1 sn nϕ r ϕ R = ± ke 0 γ 2 and the magnetc feld follows as H t = ke 0 Zγ 2 x01 R ˆr J 1 x sn nϕ + ˆϕ n x r J 1 ˆϕ x 01 J 1 x sn nϕ ˆr n R x r J 1 e ±kz ωt x01 r cos nϕ R e ±kz ωt x01 r cos nϕ e ±kz ωt R 15