Waveguides and resonant cavities

Transcription

1 Wavegudes and resonant cavtes February 8, 014 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 extended n the z drecton, wth arbtrary but constant crosssectonal shape n the xy-plane. We consder possble wave solutons matchng these boundary condtons. 1 Wave equatons As we have shown, the Maxwell equatons, gve rse to wave equatons, B E + B t D H D t E B 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. Unlke our plane wave solutons, however, 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 parts, the d Alembertan becomes t + ˆk, µɛ t + µɛ t + t + Substtutng the assumed tme and z-dependence nto the wave equatons gves 0 t µɛ t + E t µɛ ω + k E 1

2 and, wth the correspondng result for B, 0 t + µɛω k E 0 t + µɛω k B 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. Defnng t + ˆk E E t + ˆkE z ˆk E ˆk and smlarly for B, the source-free Maxwell equatons become t B t ± kb z t E t ± ke z t + ˆk E t + ˆkE z ω B t + ˆkB z t + ˆk B t + ˆkB z + ɛµω E t + ˆkE z The frst two equatons have the form we are after, t B t kb z t ke z and we turn out attenton to the curl equatons. The curl terms expand as t + ˆk E t + ˆkE z t E t + t ˆkE z + ˆk E t + ˆk ˆk E z t E t + î ˆk E z x + ĵ ˆk E z y + ˆk E t t E t ĵ E z x + î E z y + ˆk E t and smlarly for the magnetc terms. Thus t E t ĵ E z x + î E z y ± ˆk ke t ω B t + ˆkB z t B t ĵ B z x + î B z y ± ˆk kb t + εµω E t + ˆkE z

3 Notng that t E t les n the z-drecton, we separate the transverse and longtudnal parts of each equaton, ĵ E z x + î E z y ± ˆk ke t ωb t t E t ωb zˆk ĵ B z x + î B z y ± ˆk kb t + ɛµω 0 t B t + ɛµωe zˆk We can stll smplfy the frst and thrd. Snce they are transverse, we lose no nformaton by takng the cross product wth ˆk. For the frst ths 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 ke t ωˆk B t and smlarly for the thrd. t E z ± ke t ωˆk B t 0 t E z ± ke t + ωˆk B t 3 General soluton: Separatng E t and B t 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 ω k ˆk ˆk Et t E z ω k ˆk t B z ± k ɛµ ω t E z ω k k ˆk t B z ± k ɛµω t E z ω k k ˆk t B z and therefore, ɛµω k ±k t E z ωˆk t B z 3

4 Substtutng 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 k ˆk ±k t E z ωˆk t B z ± 1 ɛµω t B z k ɛµω k ±kˆk t E z + ω t B z 1 ɛµω k ɛµω k kˆk ɛµω t E z ɛµω k ω tb z ± t B z ɛµω ɛµω k ɛµω k kˆk t E z ± ɛµω k ω 1 t B z ɛµω k 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 k ±k t E z ωˆk t B z ɛµω k ɛµωˆk t E z ± k t B z To have a complete soluton, we must check the remanng Maxwell equatons, The dvergence equatons become t Usng the reduced form of the wave equaton, t B t kb z t ke z t ωb zˆk t B t ɛµωe zˆk ɛµω k ±k t t E z ω t ˆk t B z 0 t + µɛω k E together wth t ˆk t B z t ε 3k t B z,k t 1 t B z t t 1B z the rght sde smplfes, t ±k k ɛµω k µɛω E z ke z as requred. 4

5 For the curl, t ɛµω k t ±k t E z ωˆk t B z ω ɛµω k t ˆk t B z Sortng out the double curl, [ ] t ˆk t B z 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 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 t B z The second term vanshes because t has no z component. Therefore, usng the wave equaton agan, t ω ɛµω k ˆk t B z ω ɛµω k ˆk k µɛω B z ωb zˆk 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 t + µɛω k E z 0 t + µɛω k B z wth the approprate boundary condtons, then use these solutons to solve B t ɛµω k ±k t E z ωˆk t B z ɛµω k 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. 5

6 . 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. Ths method sketched above works for TE and TM waves see below, but n the case where E z B z we have ɛµω k and ths soluton fals. We treat ths specal TEM case frst, then use the technque 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 B z In ths case, our orgnal equatons have zero source, t B T EM t E T EM t E T EM t B T EM so that the transverse electrc and magnetc felds satsfy the Laplace equaton of electrostatcs n - dmensons. Ths means that our general soluton above cannot be used because µɛω k. Instead, the relevant soluton to the -dmensonal Laplace equaton s determned purely by ts boundary condtons. Snce a closed conductor allows no feld nsde, TEM waves cannot exst nsde a completely enclosed, perfectly conductng cavty. We also have Combnng these last two equatons, ±ke T EM + ωˆk B T EM ±kb T EM ɛµωˆk E T EM k E T EM + ɛµω ˆk ˆk ET EM k ɛµω E T EM we see that we must have and the magnetc feld satsfes 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. 6

7 4. Transverse electrc, TE, modes and transverse magnetc, TM, modes 4..1 Boundary condtons Modes drven by nonzero E z and/or B z fall nto two categores. 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. The full boundary condtons are therefore n D Σ n H K n E n B 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 so that E z 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 Et n t B z k n B t ɛµωˆk E t n B z n The left sde of ths equaton vanshes by the boundary condtons, so we must have B z n at the surface as well. Therefore, we seek solutons to the -dmensonal wave equatons wth boundary condtons 0 t + µɛω k E z 0 t + µɛω k B z E z S B z n S Snce each of these components also satsfes the orgnal wave equaton, 0 t + µɛω k E z 0 t + µɛω k B z 7

8 we have 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 everywhere whle B z satsfes the boundary condton Bz n S.. TM waves: The magnetc feld s transverse so that B z everywhere whle E z satsfes the boundary condton E z S. A general soluton for the feld n a wavegude or cavty s a superposton of TE, TM and TEM waves. 4.. The transverse felds Frst, we smplfy our solutons for transverse felds n the TE and TM cases. For TM waves, we set B z. Then B t Takng the curl of the transverse electrc feld, so that ˆk For TE waves, E z we get a smlar result, B t so substtutng the second equaton nto the frst, so takng the cross product wth ˆk, ±k ɛµω k te z ɛµω ˆk t ɛµω k E z ±k ɛµω k ˆk t E z ± k ɛµω B t ± k ɛω H t H t ± ɛω k ˆk E t ω ˆk ɛµω k t B z ±k ɛµω k tb z ω ˆk ɛµω k t B z ω ˆk ɛµω k ɛµω k B t ±k ω ±k ˆk B t ˆk ω k ˆk ˆk B t ± ω k B t 8

9 so that H t ± k µω ˆk E t Both of these relatonshps, for TM and TE waves, have the form where, usng k 0 εµω, Z { k The quantty Z s called the wave mpedence. Ths gves solutons of the form: H t ± 1 Z ˆk E t ɛω k µ k 0 ɛ T M modes µω k k0 µ k ɛ T E modes k ± ɛµω k te z H t ± 1 Z ˆk E t wth Z k k 0 µ ɛ for TM and ω ˆk ɛµω k t B z H t ± 1 Z ˆk E t wth Z k0 k µ ɛ for TE 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 µɛω k Then the reduced wave equaton for ψ becomes an egenvalue problem, wth boundary condtons t ψ γ ψ ψ S T M ψ n T E S Snce these boundary condtons are perodc, the wave equaton wll have a dscrete spectrum of allowed values for the constant, γ γ λ. These correspond to dscrete values of the wavelength, or equvalently, wave number k, gven by k λ µεω γ λ 9

10 The Laplacan wll have wave modes rather than exponental modes only for γ > 0. Ths leads to a cutoff frequency, ω λ, Expressng k λ n terms of frequency, we have µɛω > γλ ω ω λ γ λ µɛ k λ µɛω γλ µɛ ω γ λ µε µɛ ω ωλ so the wave vector becomes magnary, leadng to attenuaton, for frequences lower than ω λ. Now consder a fxed frequency, as we allow ω λ to range over the possble egenvalues. There s some maxmum allowed cutoff frequency, correspondng to some mnmum value of k λ and therefore some maxmum wavelength that can propagate at that frequency. As a result, only a fnte number of possble wavelengths can propagate. It s possble to choose the dmensons of the wavegude so that at the desred frequency or frequency range, only a sngle wavelength can propagate. Once we have solved for ψ, we can fnd the transverse felds from the expressons above. 4.3 Example: TE modes n a rectangular wavegude Suppose we have TE modes n a rectangular wavegude. Let the cross-secton of the gude run from x to x a, and from y to y b. Then, wth γ H z,we solve wth boundary condton 0 t + γ ψ x + y + γ ψ ψ n T E S The boundary condton n each drecton may be satsfed at the orgn by a cosne, so we have ψ 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 follows from γ mn π m a + n b and β nπ b. Therefore, the ω mn γ mn µε π m 1/ µε a + n b 10

11 for the varous modes. If we want to desgn a wavegude wth only one allowed mode, we want the lowest values for m, n. If a > b, the smallest value of ω mn occurs for n and m 1, For ths mode, we have the felds, ω 10 and therefore, for waves movng n the +z drecton, π a µɛ H z ψ 10 H 0 cos πx a B t + k γ tb z ka µ π t H 0 cos πx a kaµ π H 0 sn πx a e kz ωt From the mpedence equaton, so crossng wth ˆk H t + k µω ˆk E t µω k ˆk H t µω ka k π H 0 sn πx e kz ωt ĵ a ωaµ π H 0 sn πx a ekz ωt ĵ 11