Power Loss. dp loss = 1 = 1. Method 2, Ohmic heating, power lost per unit volume. Agrees with method 1. c = 2 loss per unit area is dp loss da

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1 How much power is dissipated (per unit area?). 2 ways: 1) Flow of energy into conductor: Energy flow given by S = E H, for real fields E H. so 1 S ( ) = 1 2 Re E H, dp loss /da = ˆn S, so dp loss = 1 µc ω [(1 da 2 2σ ˆn Re i)(ˆn H ) H ] = µ cωδ 4 H 2 = 1 2σδ H 2 Method 2, Ohmic heating, power lost per unit volume 1 2J E = J 2 /2σ, J = σe c = 2 δ H e ξ/δ, the power loss per unit area is dp loss da = 1 δ 2 σ H 2 Agrees with method 1. 0 dξ e 2ξ/δ = 1 2δσ H 2. TEM, TE, 1 The 1, Re, * discussed in lectures B H. 2

2 In terms of surface current K eff = Thus 0 = ˆn H. dξ J(ξ) = 1 δ ˆn H dξ (1 i)e ξ(1 i)/δ dp loss da = 1 2σδ K eff 2. 1 is surface resistance (per unit area) σδ is the surface impediance Z. 0 E K eff = 1 i σδ TEM, TE,

3 For electromagnetic fields with a fixed geometry of linear materials, fourier transform decouples, we can work with frequency modes, E( x, t) = E(x, y, z) e iωt B( x, t) = B(x, y, z) e iωt Actually the fields are the real parts of these complex expressions. If ρ = 0, J = 0, Maxwell gives TEM, TE, Then E = B t = iω B, E = 0, B = 0, B = µ H = µ D t = µɛ E t = iωµɛ E. 2 E = ( E)+ ( E ) = (iω B) = ω 2 µɛ E.

4 similarly for B, so we get Helmholtz equations ( 2 + ω 2 µɛ ) E = 0, ( 2 + ω 2 µɛ ) B = 0. Consider a waveguide, a cylinder of arbitrary cross section but uniform in z. Fourier transform in z E(x, y, z, t) = E(x, y)e ikz iωt B(x, y, z, t) = B(x, y)e ikz iωt TEM, TE, k can take either sign ( a sting wave is a superposition of k = ± k ). The Helmholtz equations give [ 2 t + (µɛω 2 k 2 ) ] ( E(x, ) y) = 0, B(x, 2t := 2 y) x y 2.

5 Decompose longitudinal transverse Let E = E z ẑ + E t B = B z ẑ + B t with E t ẑ B t ẑ ( E) z = ( t E t ) z = iωb z, ( E) = ẑ E t z ẑ te z = iω B t. For any vector V, ẑ (ẑ V ) = V + ẑ(ẑ V ), so for a transverse vector ẑ (ẑ V t ) = V t. Taking ẑ last equation, E t z t E z = iωẑ B t. (1) Similarly decomposition of B = iωµɛe gives ( t B ) t = iωµɛe z z B t z t B z = iωµɛẑ E t. (2) TEM, TE,

6 Divergencelessness: t E t + E z z = 0, t B t + B z z = 0. Equations (1) (2), with the fourier transform in z, give ik E t + iωẑ B t = t E z (3) ik B t iωµɛẑ E t = t B z (4) TEM, TE, Solving 4 for B t plugging into 3, then the reverse for E t, give E t = i k t E z ωẑ t B z ω 2 µɛ k 2 (5) B t = i k t B z + ωµɛẑ t E z ω 2 µɛ k 2 (6) Unless k 2 = k 2 0 := µɛω2, E z B z determine the rest.

7 We have seen that E z B z largely determine the fields, these satisfy the two-dimensional Helmholtz equation ( 2 t + γ 2) ψ = 0 with γ 2 = µɛω 2 k 2 (7) If the walls of the waveguide are very good conductors, we may impose the perfect conductor conditions E 0 B 0 on the boundary S of the two-dimensional cross section. E z is parallel to the boundary so E z S = 0. Also the component of E t parallel to the boundary vanishes at the wall, so E t is in the ±ˆn direction. Then from the ˆn component of (2) (normal to the boundary) TEM, TE, ˆn B t z ˆn t B z = iωµɛˆn (ẑ E ) t = 0 B z n = 0, where / n is the derivative normal to the surface. So we have Dirichlet conditions on E z Neumann conditions for B z.

8 In general, nonzero solutions exist only for discrete values of γ, those values are generally different for Dirichlet for Neumann. So we need to consider, with E z (x, y) = B z (x, y) 0. That is, there are no longitudinal fields, both electric (E) magnetic (M) fields are purely transverse to the direction z of propagation. TE modes, E z (x, y) 0, the transverse fields are determined by the gradiant of B z = ψ, a solution of (7) with Neumann conditions. TM modes, B z (x, y) 0, the transverse fields are determined by E z = ψ, a solution of (7) with zero boundary conditions. TEM, TE,

9 With E z (x, y) = B z (x, y) 0, (5) (6) = everything vanishes or the denominator vanishes, k = ±k 0 with k 0 = µɛ ω Wave travels z with speed 1/ µɛ, same as for infinite medium. No dispersion. t E t = 0 t E t = iωb z = 0, so Φ E t = t Φ (though Φ might not be single valued) 2 Φ = 0. As E S = 0, Φ = constant on each boundary. If cross TEM, TE, section simply connected, Φ = constant, E = 0 No on simply connected cylinder Yes on coaxial cable, or two parallel wires.

10 TE TM modes Equations (5) (6) simplify for TM modes, B z = 0, γ 2 Et = ik t E z, γ 2 Bt = iµɛωẑ t E z, so H t = ɛωk 1 ẑ E t. TE modes, E z = 0, γ 2 Bt = ik t B z, γ 2 Et = iωẑ t B z, so E t = ωẑ B t /k = ẑ H t = kẑ E t /µω. TEM, TE, In either case, H t = 1 Z ẑ E t, with { k/ɛω = (k/k0 ) µ/ɛ TM Z = µω/k = (k 0 /k) µ/ɛ TE

11 To Summarize Solutions given by ψ(x, y), with ( 2 t + γ 2) ψ = 0, γ 2 = µɛω 2 k 2, by TM: E z = ψe ikz iωt, Et = ikγ 2 t ψe ikz iωt with ψ Γ = 0 TE: H z = ψe ikz iωt, Ht = ikγ 2 t ψe ikz iωt with ˆn t ψ Γ = 0 TEM, TE, By looking at 0 = A ψ ( 2 t + γ 2) ψ we can show γ 2 0. There are solutions for discrete values γ λ, so only certain wave numbers k λ for a given frequency can propagate: k 2 λ = µɛω2 γ 2 λ, only frequencies ω > ω λ := γ λ / µɛ can propagate, k λ < µɛ ω, the infinite medium wavenumber. Phase velocity v p = ω/k λ is greater than in the infinite medium.

12 : Circular Wave Guide Jackson does rectangle. You should too. Needed to do homework. We will consider a circular pipe of (inner) radius r. Of course we should use polar coordinates ρ, φ, with 2 t = 1 ρ ρ ρ ρ ρ 2, try ψ(ρ, φ) = R(ρ)Φ(φ), φ2 ( 2 t + γ 2) ψ = ( ) 1 ρ ρ ρ R ρ + γ2 R(ρ) Φ(φ) + 1 Φ(φ) ρ 2 R(ρ) 2 φ 2 = 0. Divide by R(ρ)Φ(φ) multiply by ρ 2 : ( 1 ρ ) R(ρ) ρ ρ R ρ + γ2 ρ 2 R(ρ) Φ(φ) Φ(φ) φ 2 = 0. TEM, TE,

13 : Circular Wave Guide Jackson does rectangle. You should too. Needed to do homework. We will consider a circular pipe of (inner) radius r. Of course we should use polar coordinates ρ, φ, with 2 t = 1 ρ ρ ρ ρ ρ 2, try ψ(ρ, φ) = R(ρ)Φ(φ), φ2 ( 2 t + γ 2) ψ = ( 1 ρ ρ ρ R ) ρ + γ2 R(ρ) Φ(φ) + 1 ρ 2 R(ρ) 2 Φ(φ) φ 2 = 0. Divide by R(ρ)Φ(φ) multiply by ρ 2 : ( 1 ρ ) R(ρ) ρ ρ R ρ + γ2 ρ 2 R(ρ) = C 1 2 Φ(φ) Φ(φ) φ 2 = C. TEM, TE,

14 Solving it Φ first: 2 Φ(φ) φ 2 + CΦ(φ) = 0 Φ(φ) = e ±i Cφ. Periodicity = C = m Z. Now R(ρ): (ρ ρ ρ ρ + γ2 ρ 2 m 2 ) R(ρ) = 0 TEM, TE, Bessel equation, solutions regular at origin are R(ρ) J m (γρ), so ψ(ρ, φ) = m,n A m,n J m (γ mn ρ)e imφ. γ mn is determined by boundary conditions...

15 Boundary conditions: For TM, ψ(r, φ) = 0 = J m (γr) = 0, so γmn TM = x mn /r where x mn is the n th value of x > 0 for which J m (x) = 0, given on page 114. For TE, ˆn t ψ(r, φ) = 0 = djm dr (γr) = 0, so γmn TM = x mn/r where x mn is the n th value of x > 0 for which dj m (x)/dx = 0, given on page 370. J 0 J 1 J2 J 3 TEM, TE, Thus the lowest cutoff frequency is the m = 1 TE mode, with x 11 = while the lowest TM mode or circularly symmetric mode has x 01 =

16 For a waveguide 5 cm in diameter, with air or vacuum inside, the cutoff frequencies are f = ω 2π = 3.5 GHz for the lowest TE 4.6 GHz for the lowest TM modes. TEM, TE,