MIT OpenCourseWare http://ocw.mit.edu 6.013/ESD.013J Electromagnetics and Applications, Fall 2005 Please use the following citation format: Markus Zahn, 6.013/ESD.013J Electromagnetics and Applications, Fall 2005. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms� � � � � � 6.013 - Electromagnetics and Applications Fall 2005 Lecture 15 - Dielectric Waveguides Prof. Markus Zahn November 3 , 2005 I. TM Solutions ∂ = 0 ∂y We are considering solutions with ⎧ ⎪⎨ Re A2e−α(x−d)ej(ωt−kz z) x ≥ d Ez(x, z, t) = Re (A1 sin(kxx) + B1 cos(kxx)) eRe A3e+α(x+d)ej(ωt−kz z) j(ωt−kz z) x| | ≤ d x ≤ −d ⎪⎩ k2 + k2 = ω2�µ (in dielectric) x z −α2 + kz 2 = ω2�0µ0 (in free space) For propagation in the dielectric and evanescence in free space k2 = ω2�µ − k2 = ω2�0µ0 + α2 z x k2 < ω2�µ, k2 > ω2�0µ0 ⇒ ω2�0µ0 < k2 < ω2�µz z z 1 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.� � ����A. Odd Solutions [Ez(x, z, t) = −Ez(−x, z, t)] ⎧⎪⎨ ⎪⎩ ⎧⎪⎨ ⎪⎩ A2e−α(x−d) x ≥ d A1 sin(kxx) −A2eα(x+d) |x| ≤ d x ≤ −d Eˆz(x) = ∂Eˆx � · E ¯ ⇒ ∂x − jkzEˆz = 0 −jkz A2e−α(x−d) x > d α Eˆx jkz A1 cos(kxx)kx = < d − |x| x < −d ⇒ jkz A2eα(x+d) α− = −µ� × E ¯ ∂∂t H ¯ ⇒ Hˆx = 0, Hˆz = 0 1 ∂EˆzHˆy = −jωµ −jkzEˆx − ∂x = ⎧ ⎪⎨ ⎪⎩ α−jω�0 A2e−α(x−d) x ≥ d jω� −kx A1 cos(kxx) |x| ≤ d x ≤ −djω�0 A2eα(x+d) α− Boundary Conditions Ez(x = d+) = Ez(x = d ) A1 sin(kxd) = A2−⇒ Hy(x = d+) = Hy(x = d )−⇒ + jω�0 α A2 = + jω� kx A1 cos(kxd) A1 1 kx�0 �0kx = = α = tan(kxd)A2 sin(kxd) α� cos(kxd) ⇒ � Critical condition for a guided wave occurs when α = 0. At this point kxd = nπ, kz 2 = ω2�0µ0 �nπ �2 + ω2�0µ0 = ω2�µ ω2 =(nπ/d)2 , n = 1, 2, 3, . . . d ⇒ �µ − �0µ0 For real frequencies (ω2 > 0), �µ > �0µ0 B. Even Solutions [Ez(x, z, t) = +Ez(−x, z, t)] Eˆz(x) = ⎧ ⎪⎨ ⎪⎩ B2e−α(x−d) x ≥ d B1 cos(kxx) B2eα(x+d) x| | ≤ d x ≤ −d ∂Eˆx Eˆz = 0 − jkz∂x ⎧ ⎪⎨ −jkz B2e−α(x−d) x > d α ˆjkzEx = B1 sin(kxx) < d |x| x < −d ⇒ kx jkz ⎪⎩ B2eα(x+d) α 2� ����jω�0 B2e−α(x−d)−α x ≥ d jω� Hˆy = ⎧ ⎪⎨ ⎪⎩ kx B1 sin(kxx) jω�0 B2eα(x+d) α x| ≤ d x ≤ −d | Boundary Conditions Ez(x = d+) = Ez(x = d ) B2 = B1 cos(kxd)−⇒ Hy(x = d+) = Hy(x = djω�0 B2 = jω� B1 sin(kxd)−) ⇒ −α kx B2 �α �0kx B1 = cos(kxd) = −�0kx sin(kxd) ⇒ α = − � cot(kxd) Critical Condition: α = 0 ⇒ kxd = (2n + 1)π 2 , kz 2 = ω2�0µ0 �2(2n+1)π 2d ω2 n = 0, 1, 2, . . . = �µ − �0µ0 II. TE Solutions A. Odd Solutions ⎧ ⎪⎨ ⎪⎩ A2e−α(x−d) x ≥ d Hˆz = A1 sin(kxx) −A2eα(x−d) ≤ d |x| x ≤ −d ⎧ ⎪⎨ H ¯= ∂Hx + ∂Hz ∂Hˆx Hˆz = 0 ∂x − jkz� · ⇒ ∂x ∂z −jkz A2e−α(x−d) x > d α Hˆx jkz A1 cos(kxx)kx < d = − |x| x < −d ⎪⎩ jkz A2eα(x+d) α− ¯ ∂E ¯ Eˆx = 0, Eˆz = 0 H = � � × ∂t ⇒ ⎧ ⎪⎨ jω� Eˆy = −jkzHˆx − ∂Hˆz ∂x jωµ0 A2e−α(x−d) α x ≥ d jωµ Eˆy = A1 cos(kxx) |x| ≤ d x ≤ −d kx⎪⎩ jωµ0 A2eα(x+d) α Boundary Conditions jωµ0 ⇒ α A2 = jωµ kx A1 cos(kxd)Eˆy(x = d+) = Eˆy(x = d−) Hˆz(x = d+) = Hˆz(x = d−) ⇒ A2 = A1 sin(kxd) A2 µα µ0kx A1 = sin(kxd) = µ0kx cos(kxd) ⇒ α = µ tan(kxd) 3����B. Even Solutions ⎧⎪⎨ ⎪⎩ ⎧⎪⎨ ⎪⎩ ⎧⎪⎨ ⎪⎩ B2e−α(x−d) x ≥ d Hˆz B1 cos(kxx) B2eα(x+d) = ≤ d |x| x ≤ −d −jkz B2e−α(x−d) x > d α jkzHˆx B1 sin(kxx)= < d |x| x < −d kx jkz α B2eα(x+d) jωµ0 B2e−α(x−d) α x ≥ d jωµ Eˆy = B1 sin(kxx) ≤ d − |x| x ≤ −d kx jωµ0 B2eα(x+d) α− jωµ0 ⇒ α jωµ kx B1 sin(kxd)Eˆy(x = d+) = Eˆy(x = d−) B2 = − Hˆz(x = d+) = Hˆz(x = d−) ⇒ B2 = B1 cos(kxd) B2 αµ µ0kx B1 = cos(kxd) = −µ0kx sin(kxd) ⇒ α = −µ cot(kxd)
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