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Waveguide Theory Maxwell Equations: ∂ ∂ ∂ ∂ H B E D Μ Ε

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Waveguide Theory Maxwell Equations: ∂ ∂ ∂ ∂ H B E D Μ Ε Waveguide Theory Maxwell Equations: ∂ B ∇ × E = − ∂ t ∂ D ∇ × H = ∂ t ∇ ⋅ D = 0 ∇ ⋅ B = 0 D = εE B = µH Boundary Conditions: 1 tangential E continuous - Et1 = Et 2 or k × H continuous ε normal D continuous - ε11EEnn= ε 22 As a review, consider the parallel-plate capacitor with V and d. If the left side is air and the right side is dielectric, what is the E-field? If the capacitor is filled with air (top) and dielectric (bottom), how is the voltage divided? tangential H continuous - ∇ × E continuous or k × E continuous normal B continuous - To simplify vectors, consider TE and TM separately. Wave Equation: 1 ∂ 2 E ∇−2 E =0 c2 ∂ t 2 = ikx()−ω t Wave along +x axis - EEeo = −+ikx()ω t Wave along -x axis - EEeo 2π k = , ω = 2πf λ sinusoidal in time - ∝ eitω Helmholtz Equation: ∇+222 = EknEo 0 The refractive index determines properties of the EM wave. Waves in metallic structures Standing wave between two parallel plates - 1D problem Traveling wave between two parallel plates - 2D problem Traveling wave in a hollow tube - 3D problem Metallic 1D problem - two plates separated by 2a ∂ 2 E + k 2n2 E = 0 ∂ x 2 o E ∝ e±ikx = ⋅ k ko n Standing wave π 3π E = E coskx ka = , ,etc o 2 2 = π π Eo sin kx ka ,2 ,etc 2D problem - Reflection ∂ 2 E ∂ 2 E + + k 2n2 E = 0 ∂ x 2 ∂ y 2 o Incidence angle equals reflected angle. Incidence wave: + Aei(kx x k y y) Reflected wave: − Bei(kx x k y y) Boundary condition - zero tangential E-field For TE wave, A = - B Question: How about TM? 2D problem - Planar metallic guide with a separation of 2a ∂ 2 E ∂ 2 E + + k 2n2 E = 0 ∂ x 2 ∂ z 2 o Use separation of variables and assume a periodic wave along z. = iβ z β = E E(x)e , neff ko ∂ 2 E + k 2 (n2 − n2 )E = 0 ∂ x 2 o eff sin or cos solutions with 2 − 2 1/ 2 ⋅ = π = ko (n neff ) 2a m , m 1,2,3,etc. π 2 = 2 − m 2 neff n ( ) 2ako The geometrical interpretation Magnitude of the wave vector: 2π k = = k n λ o z-component θ = β = ko nsin neff ko where the angle is between the beam and the surface normal. x-component θ ko ncos Periodic condition for a round-trip along x ⋅ θ = π 4a ko ncos 2m From the above two equations, you will find π 2 = 2 − m 2 neff n ( ) 2ako Rectangular metallic guide - 3D problem ∂ 2 E ∂ 2 E ∂ 2 E + + + k 2n2 E = 0 ∂ x 2 ∂ y 2 ∂ z 2 o Apply separation of variables = iβ z β = E E(x)E(y)e , neff ko Therefore, ∂ 2 E + k 2 E = 0 ∂ x 2 x ∂ 2 E + k 2 E = 0 ∂ y 2 y 2 + 2 = 2 2 − 2 k x k y ko (n neff ) Dielectric boundary - normal incidence (1D) ∂ 2 E + k 2 n2 E = 0 ∂ x 2 o E = Aeikx + Be−ikx , x < 0 E = Ceikx + De−ikx , x > 0 = ⋅ k ko n The wave impinges from left, D=0. Et continuous, A+B=C Ht continuous, − = ko n1 (A B) ko n2C Therefore, 2n C = 1 A + n1 n2 n − n B = 1 2 A + n1 n2 Power flux equals energy density (Poynting vector) times the speed of light: ε 2 c E ⋅ o 2 n ε = n2 Question: Can you prove that the power is conserved? Dielectric interface with incidence beam at an angle - 2D problem plane of incidence, incidence angle, reflect angle, refract angle Incident beam: ⋅ ⋅ θ − ⋅ θ Aeikon1 r = Aeikon1 ( x sin in y cos in ) Reflected beam: ⋅ θ + ⋅ θ Beikon1 (x sin in y cos in ) Refracted beam: ⋅ θ − ⋅ θ Ceikon2 ( x sin r y cos r ) TE case - E || z Continuity of tangential component Ez at y=0: ⋅ θ ⋅ θ (A + B)eikon1 ( x sin in ) = Ceikon2 ( x sin r ) Hence, we get the Snell's law: sinθ n r = 1 θ sin in n2 Continuity of tangential component of Hx at y=0: k × E i j k θ − θ kon1 sin kon1 cos 0 0 0 E − θ + θ = − θ kon1Acos in kon1B cos in kon2C cos r Bn continuous. Try to find the y component of k × E After applying the Snell’s law, you will find: A + B = C From two equations, we can solve B and C in terms of A. n cosθ − n cosθ B = 2 r 1 in A θ + θ n2 cos r n1 cos in Question: Can you find C as a function of A? After using the relation between cosθ and sinθ and applying the Snell’s law, you will find: n2 n 1− 1 sin2 θ − n cosθ 2 n2 in 1 in B = 2 A 2 − n1 2 θ + θ n2 1 2 sin in n1 cos in n2 Question: Can you confirm the result for normal incidence and grazing incidence? θ = o in 0, 90 When n1>n2, the following situation may occur: 2 − n1 2 θ ≤ 1 2 sin in 0 n2 or θ ≥θ = −1 n2 in cr sin n1 If we define: = θ k1 kon1 cos in γ = θ i 2 kon2 cos r You will find: iγ − k B = 2 1 A γ + i 2 k1 The numerator and the denominator are complex conjugate, therefore, the power reflection coefficient is: | B |2 R = = 1 | A |2 φ B = Aei2 TE γ φ = − −1 2 TE tan k1 This is the Goos-Hanchen phase shift associated with total internal reflection. The phase shift is a function of the incidence angle or mode and is different from the metallic guide. Question: Apply the boundary condition and prove that for TM polarization: γ 2 φ = − −1 2n1 TM tan 2 k1n2 A planar dielectric waveguide can be formed by a three-layer structure, substrate, guiding layer, and superstrate. If the index of the substrate is different from that of the superstrate, we have an asymmetric guide. Otherwise, it is a symmetric guide. If the thickness of the guiding layer is d, the periodic condition becomes the mode or characteristic equation: ⋅ θ + φ + φ = π 2d kon1 cos 2 12 2 10 2m Comparison between TE and TM: Snell’s law – same (independent of the polarization) Critical angle – same Reflection below critical angle – different For the TM-polarization, there is the Brewster angle at which R=0. Reflection above the critical angle – phase shift is different Polarization states Linearly polarized Circularly polarized Elliptically polarized The polarization state is crucial for beam interference. A transmission medium must have different indices for two orthogonal polarizations, i.e., breaking of symmetry, in order to preserve the polarization state. Otherwise, a polarization restoration device is needed to control the polarization state. Summary of equations for dielectric interface Power reflection coefficient: TE: n2 n 1− 1 sin 2 θ − n cosθ 2 n2 in 1 in B = 2 A 2 − n1 2 θ + θ n2 1 2 sin in n1 cos in n2 TM: n2 n 1− 1 sin 2 θ − n cosθ 1 n2 in 2 in B = 2 A 2 − n1 2 θ + θ n1 1 2 sin in n2 cos in n2 Critical angle – both TE and TM: θ = −1 n2 cr sin n1 Brewster angle – applicable to TM only: θ = −1 n2 B tan n1 Goos-Hanchen phase shift: TE n2 n 1 sin 2 θ −1 2 n2 in φ = − tan −1 2 TE θ n1 cos in TM n2 n 1 sin 2 θ −1 1 n2 in φ = − tan −1 2 TM θ n2 cos in Characteristic or mode equation: θ +φ + φ = π kon1d cos 12 10 m Numerical example: Interface between glass with an index of 1.5 and air with an index of 1.0: Reflection coefficient at normal incidence: 4% Critical angle: 41.8o Brewster angle: glass-air 33.7 o ; air-glass 56.3 o G-H phase shift angle at 80o-incidence: TE –76.5o and TM –83.9o Concept of effective mode index Projection of the magnitude of wave vector along the direction of propagation: β = θ = ko nsin neff ko For total internal reflection of occur: θ < θ < o cr 90 Therefore, < < n2 neff n1 The effective index represents an average between indices of the guide and the cladding layer. In general it is a function of ko d or d λ o The dependence of effective index becomes the dispersion curve. Each mode has its own dispersion curve. For a symmetric guide, the fundamental mode doesn't reach cut off but all higher order modes can. For an asymmetric guide, even the fundamental mode may reach cut off. In the region where only one mode is supported and all other modes are cut off, we have a single-mode waveguide. The geometrical considerations may be extended to multi-layer waveguides. You can establish characteristic mode equations by making the sum of propagational delay and Goos-Hanchen phase shift to be integer multiple of 2π. The double core structure is commonly used by fiber amplifiers. There is one thing that the geometrical consideration can't provide, namely, the transverse field pattern. Only the solution to the Helmholtz equation can lead to mode patterns. Consider a symmetric waveguide formed by a guiding layer of thickness d and index n1 surrounded by a medium with index no. ∂ 2 E ∂ 2 E + + k 2n2 E = 0 ∂ x 2 ∂ z 2 o Use separation of variables and assume a periodic wave along z.
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