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8. Wave Guides and Cavities 8A. Wave Guides Boundary Conditions at Perfect Conductors • • • • Suppose we have a region bounded by a conductor We want to consider oscillating fields in the non-conducting region For oscillating fields, changing B would imply non-zero E E B t But E must vanish in the conductor Ec Bc 0 Dc Hc • • • • • On the surface of the conductors, and J are present Which appear in two of Maxwell’s equations Therefore, D and H|| need not be continuous But E|| and B must be continuous So the correct boundary conditions must be E S 0, B S 0 D H J D t Cylindrical Wave Guides • Consider now a hollow infinite cylinder of arbitrary cross-section – We’ll make it along the z-direction • We want to find solutions moving along the cylinder D E , B H • Assume it is filled with a linear material: • We will use complex notation D 0 B – Time-dependence will look like e-it • Maxwell’s equations with no sources E B t • Use linearity plus time dependence H D E 0 B , E iB , B i E t • Take curl of either of the last two equations: 2 2 E E i B E 2 2 B B i E B 2 2 E E0 • Use the double cross-product rule V V V • And therefore 2B 2B 0 2 Transverse and Longitudinal Dependence • Since we have translation dependence along z, it makes sense to look for ikz it E E x , y e ikz solutions that go like e – Moving in the z direction B B x, y e ikz it 2E 2E 0 2B 2B 0 • Divide any derivative into t zˆ , t xˆ yˆ z x y transverse and longitudinal parts • Then we have, for example t2 E 2 k 2 E 0 t2 B 2 k 2 B 0 E Et E z , E z Ez zˆ , Et Ex xˆ E y yˆ . • Also break up fields into longitudinal and transverse parts: • We now want to write our Maxwell’s equations broken up this way Breaking Up Some Maxwell’s Equations t zˆ , t xˆ yˆ z x y E Et E z , E z Ez zˆ , Et Ex xˆ E y yˆ . • Let’s look at some Maxwell’s equations: E iB , B i E E t iBt zˆ Et t Ez zˆ iBt zˆ Et t Ez zˆ iBt z z E z i Bz zˆ E i B t t z z • But we assume fields look like E E x, y e ikz it B B x, y e ikz it • Therefore the last equation becomes • We similarly have zˆ ikEt t Ez iBt zˆ ikBt t Bz iEt TEM Modes • Can we find solutions with Ez = Bz = 0? zˆ ikEt t Ez iBt • Such modes are called TEM modes zˆ ikBt t Bz iEt – Because both E and B are transverse • Then we kzˆ Et Bt and kzˆ Bt Et would have 2 2 zˆ Bt ˆ ˆ ˆ k z z Et kz Bt • Multiplying the first by kzˆ k 2 Et 2 Et • But the left side is • We would therefore have k – Implies phase velocity equal to free waves • Recall also zˆ t Et z i Bz 0 t Et 0 • • • • E 0 t Et 0 And also This tells us finding Et is a 2D electrostatic problem Recall also that potential is constant on surfaces Only get non-trivial solutions if there are at least two conducting surfaces Sample Problem 8.1 A coaxial cable has inner radius a and outer radius b, and is filled with a material with electric permittivity and magnetic susceptibility . Find exact electric and magnetic field solutions for TEM modes. • We will start by finding the electric t Et 0 t Et field, which is transverse and satisfies • Since it has no curl, it is derivable from a potential 0 n 2 • Potential must be constant on the inner and outer surfaces Et E ρˆ • Symmetry implies that Et must be radial: 1 • No divergence C E C E 0 E E , ρˆ t tells us: C ikz it • Put back in the z- and t-dependence E ρˆ e • Where: k • The magnetic C ikz it Bt kzˆ Et Bt zˆ Et field is then B φˆ e • Take real part to get actual fields All We Need is the Z-Direction of the Fields zˆ ikEt t Ez iBt • Now consider non-TEM modes zˆ ikBt t Bz iEt – Ez or Bz are non-zero • Multiply second equation by i and substitute the first one zˆ ikzˆ ikEt t Ez it Bz 2Et • For a transverse vector, • We therefore have • Solve for Et: • Also recall zˆ zˆ At At k 2Et ik t Ez i zˆ t Bz 2Et Et Bt i k 2 2 kt Ez zˆ t Bz 2 kt Bz zˆ t Ez i k 2 • Normally you have just Bz or Ez – Modes with Ez= 0 are called TE modes (transverse electric) – Modes with Bz = 0 are called TM modes (transverse magnetic) Finding Non-TEM modes Et i k 2 k t Ez zˆ t Bz 2 Bt i k • All modes of E and B satisfy t2 2 k 2 0 • Let us define 2 2 k 2 t2 2 0 • Then our equations become Et i 2 k t Ez zˆ t Bz Bt i 2 k t Bz zˆ t Ez • We must also satisfy our boundary conditions E S 0, B S 0 2 2 kt Bz zˆ t Ez TE Modes 2 k 2 2 Et i 2 k t Ez zˆ t Bz 0 2 t 2 Bt i 2 k t Bz zˆ t Ez Case 1: TE modes (transverse electric) • Search for solutions with Ez = 0, so everything comes from Bz t2 Bz 2 Bz 0 • The transverse fields are then Bt ik 2t Bz , Et i 2 zˆ t Bz • We must also satisfy boundary conditions E • These imply tBz must be parallel to the walls of the cylinder Bz n S S 0 B S 0 • Solve the eigenvalue equation subject to the boundary conditions TM modes 2 k 2 2 0 2 t 2 Et i 2 k t Ez zˆ t Bz Bt i 2 k t Bz zˆ t Ez Case 2: TM modes (transverse magnetic) • Search for solutions with Bz = 0, so everything comes from Ez t2 Ez 2 Ez 0 • The transverse fields are then Et ik 2t Ez , Bt i 2 zˆ t Ez • We must also satisfy the boundary conditions E S 0 B S • These imply that Ez must vanish on the walls Ez S 0 • Solve the eigenvalue equation subject to the boundary conditions Sample Problem 8.2 (1) A hollow cylindrical waveguide has circular cross-section of radius a. Find the relationship between the frequency and wave number for the lowest frequency modes for the TE and TM modes. • The frequencies are given by 2 k 2 2 • The 2 values are eigenvalues of the equation t2 2 0 0 n2 – Where is Bz (TE) or Ez (TM) modes • Makes sense to work in 2 1 1 2 2 0 cylindrical coordinates 2 2 2 • Rotational symmetry im , R e implies solutions of the form • Substitute in: d 2 R 1 dR m2 2 R R0 • Let x 2 2 d d • Then we have d 2 R 1 dR m2 1 2 R 0 2 dx x dx x Sample Problem 8.2 (2) • • • • • A hollow cylindrical waveguide has circular cross-section of radius a. Find the relationship between the frequency and wave number for the lowest frequency modes for the TE and TM modes. 2 2 d R 1 dR m 2 2 2 x k 1 2 R 0 2 dx x dx x This is Bessel’s Equation Solutions are Bessel functions R x J m x J m 0 n 2 Recall that represents Bz or Ez Bz S 0 or Ez S 0 And we have boundary conditions n We therefore must have TE modes: J m a 0, TM modes: J m a 0 • Let xmn be the roots of Jm and let ymn the roots of its derivative • Then the formula for 2 2 y x 2 2 mn the frequencies will be: TE: 2 k 2 mn , TM: k a2 a2 Sample Problem 8.2 (3) A hollow cylindrical waveguide has circular cross-section of radius a. Find the relationship between the frequency and wave number for the lowest frequency modes for the TE and TM modes. 2 2 y x TE: 2 k 2 mn2 , TM: 2 k 2 mn2 a a • Maple is happy to find roots of Bessel’s equation > for m from 0 to 3 do evalf(BesselJZeros(m,1..3)) end do; x01 2.4048, x11 3.8317, 0 n 2 x21 5.1356, x02 5.5201, x31 6.3802, • With a little coaxing we can also get it to find the y’s > for m from 0 to 3 do seq(fsolve(diff(BesselJ(m,x),x), x=(n+m/2-3/4)*Pi),n=1..3) end do; y11 1.8412, y21 3.0542, y01 3.8317, y31 4.2012, y41 5.3175, Comments on Modes 2 2 y x TE: 2 k 2 mn2 , TM: 2 k 2 mn2 a a y11 1.8412, y21 3.0542, y01 3.8317, y31 4.2012, y41 5.3175, x01 2.4048, x11 3.8317, x21 5.1356, x02 5.5201, x31 6.3802, ynm a TE modes • Note that for each mode, there min is a minimum frequency xnm a TM modes • If you are at a limited , only a finite number of modes are possible – Usually a good thing – Ideally, want only one mode • The lowest mode is usually a TE mode • Note that the lowest modes often have m > 1 – Actually represent two modes because modes can be eim 8B. Rectangular Wave Guides Working out the Modes • Let’s now consider a rectangular wave guide – Dimension a b with a b 2 2 2 0 • We will work in Cartesian coordinates 2 2 x y • Boundary conditions: Bz S 0 or Ez S 0 n mx ny • For TE modes, we Bz x, y B0 cos cos want waves of the form a b • For TM modes, we mx ny want waves of the form Ez x, y E0 sin sin a b • In each case, we have 2 2 m n 2 2 2 2 b a b a Restrictions on Modes mx ny mx ny Ez x, y E0 sin Bz x, y B0 cos sin cos a b a b 2 2 m n 2 2 2 2 2 2 2 k b a • For the TM modes (Ez), we must have m > 0 and n > 0 • For the TE modes (Bz), we can’t have m = 0 and n = 0 – Since then Bz has no transverse derivative • Hence TE modes have m > 0 OR n > 0 • Degeneracy between TE mode and TM mode if both positive • Lowest frequency mode: TE mode 10 a with (m,n) = (1,0) • Second lowest modes: modes with 2 , b 20 01 a b a (m,n) = (2,0) or (0,1) • If a 2b, then factor of two difference between lowest and next lowest modes • Normally make a 2b The TE10 mode • We can work out all the fields for the TE10 mode – Put back in the z and t dependence • We then work out all the other pieces using Bt ik 2t Bz , Et i 2 zˆ t Bz x ikz it Bz B0 cos e a • We therefore have Bt x ikz it B0 xˆ sin e a iak x ikz it B0 yˆ sin e a • The factors of i represent a phase shift between the various modes Et ia b a 8C. Cavities Cylindrical Cavities • Let’s now cap off the cylindrical cavity, make the ends conducting • Let it be along z with z from 0 to d • On the end, we must have E 0 B S S • This means at z = 0 and z = d, we have Et • We have up until now used modes with Bz 0,c 0 Bz 0,c or Ez x, y eikz it • We no longer want eikz, instead we want • We will have to take linear combinations of our previous solutions sin kz 1 2i cos kz 1 2 d ikz ikz e e e ikz e ikz TE Modes in Cavities • For TE modes, we had t2 Bz 2 Bz 0 2 k 2 2 Bz Bz x, y eikz it Bt ik 2t Bz , Et i 2 zˆ t Bz • But we also need Bz 0 Bz d 0 ikz ikz • We therefore must have Bz to be like sin kz 21i e e • To get it to vanish at z = d, must have p k , p 1, 2,3, • So Bz is given by d Bz Bz x, y sin kz eit d • To get the transverse components, take same linear combinations: Bt ik t Bz x, y 2 1 2i e Et i zˆ t Bz x, y 2 ikz 1 2i e e ikz ikz e e Bt 2kt Bz x, y cos kz eit it ikz e it Et i 2zˆ t Bz x, y sin kz eit TM Modes in Cavities • For TM modes, we had t2 Ez 2 Ez 0 Ez Ez x, y eikz it • • • • • 2 k 2 2 Et ik 2t Ez , Bt i 2 zˆ t Bz But we also need Et 0 Et d 0 We have to make the two contributions to Et cancel at z = 0, d This can be done if Ez is like cos kz 12 eikz e ikz To get it to work at z = d, must have p So Ez is given by k , p 0,1, 2, d it Ez Ez x, y cos kz e d • To get the transverse components, take same linear combinations: Et ik t Ez x, y 2 1 2 e ikz Bt i zˆ t Ez x, y 2 1 2 e e ikz ikz e e Et 2kt Ez x, y sin kz eit it ikz e it Bt i 2 zˆ t Ez x, y cos kz eit Rules for Cavities Summarized 2 k 2 2 k p • For both TE and TM modes, we have d – k > 0 for TE, k 0 for TM 2 2 t 0 • We have to solve the equation – represents Bz for TA and Ez for TM S 0 S 0 • Boundary conditions for TE: n • Boundary conditions for TM: d Bz x, y sin kz eit • For TM modes, we then have Bt 2kt x, y cos kz eit Et i 2zˆ t x, y sin kz eit • For TM modes, we then have Ez x, y cos kz eit Et 2 kt x, y sin kz eit Bt i 2 zˆ t Ez x, y cos kz eit Box-Shaped Cavities • Consider a cavity of dimensions a b d • Frequency is given be 2 k 2 2 • Where k is given by p – p = 1, 2, 3, … for TE k d – p = 0, 1, 2, … for TM d 2 2 2 2 • In both cases, we had m n 2 mn 2 2 – m > 0 OR n > 0 for TE a b – m > 0 AND n > 0 for TM 2 2 2 m n p • Therefore, our frequencies are: 2 mnp 2 2 2 a b c • At least two of m, n, p must be non-zero – If m = 0 or n = 0, it is a TE mode – If p = 0, it is a TM modes – If all three are non-zero, it can be a TE or TM mode b a Sample Problem 8.2 (1) A conducting box-shaped cavity has dimensions a b d. Write explicitly the form of all fields for the TE1,0,1 mode, and find the energy in electric and magnetic fields as a function of time. • For TE modes, we first solve 2 2 0 t – Where is Bz n x • The 1,0 solution is B0 cos a – I arbitrarily 2 included 2/2 2 2 • is given by a • Next find Bz: Bz x, y sin kz e • Where k is given by • So we have x z it Bz B0 cos sin e a d it k S 0 d 0 n 2 d b a Sample Problem 8.2 (2) . . . Write explicitly the form of all fields for the TE modes, and find the energy in electric and magnetic fields as a function of time. x B0 cos a • Now let’s get Bt: 2 2 a2 k d Bt k 2t x, y cos kz eit a2 x z it a x z it Bt B0 2 t cos cos e ˆ B0 x sin cos e d a d d a d • And finally, let’s go for Et: Et i 2zˆ t x, y sin kz eit ia 2 i a x z it x z it ˆ Et 2 B0 z t cos B0 yˆ sin sin e sin e a d a d • We will also need the 2 2 2 2 2 frequencies, which are given by k 2 2 2 a d Sample Problem 8.2 (3) . . . find the energy in electric and magnetic fields as a function of time. i a x z it E B0 yˆ sin sin e a d 2 2 a 2 2 d2 a x z • Keep in mind that we need to take ˆ E B0 y sin sin sin t a d the real part of these expressions 2 2 • The electric energy a 2 2 x 2 z 2 2 1 uE 2 E B0 sin density is: sin sin t 2 2 a d • Total electric energy is a2 d 2 2 2 x 2 z 2 U E u E dx dy dz B sin sin sin t dx dy dz 0 2 2 d a c a2 d 2 2 a d 2 UE B b sin t 0 2 2 d 2 2 UE a a2 d 2 b 8 d B02 sin 2 t Sample Problem 8.2 (4) . . . find the energy in electric and magnetic fields as a function of time. a x z it x z it Bz B0 cos Bt B0 xˆ sin sin e cos e d a d a d • Again, take the real part • Now let’s go for the magnetic energy density: 2 1 2 1 x z a 2 2 2 2 x 2 z 2 uB B B cos sin sin cos cos t 0 2 2 2 a d d a d U B uB dx dy dz 1 2 2 x 2 z a2 2 x 2 z 2 B0 cos sin sin cos cos t dx dy dz 2 2 a c d a c 1 2 a d a2 a d 2 B0 b 2 b cos t 2 2 2 d 2 2 UB a a2 d 2 b 8 d B02 cos2 t Sample Problem 8.2 (5) A conducting box-shaped cavity has dimensions a b d. Write explicitly the form of all fields for the TE1,0,1 mode, and find the energy in electric and magnetic fields as a function of time. UE a a2 d 2 b 8 d B02 sin 2 t • It is pretty easy to see that UB a a2 d 2 b 2 d UE UB B02 cos2 t d a a2 d 2 b 2 d 0 n 2 B02 b • Of course, this is just conservation of energy a