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ELEG413lec3 - ELEG 413 Lecture#3 Mark Mirotznik Ph.D...

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Unformatted text preview: ELEG 413 Lecture #3 Mark Mirotznik, Ph.D. Associate Professor The University of Delaware Email: [email protected] SUMMARY m B D J t D H M t B E ρ ρ = ⋅ ∇ = ⋅ ∇ + ∂ ∂ = × ∇- ∂ ∂- = × ∇ m B D J D j H M B j E ρ ρ ϖ ϖ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ = ⋅ ∇ = ⋅ ∇ + = × ∇-- = × ∇ ) ~ ~ ( ˆ ~ ) ~ ~ ( ˆ ~ ) ~ ~ ( ˆ ) ~ ~ ( ˆ 1 2 1 2 1 2 1 2 =- ⋅ =- ⋅ =- × =- × B B n D D n J H H n E E n s s ρ ) ( ˆ ) ( ˆ ) ( ˆ ) ( ˆ 1 2 1 2 1 2 1 2 =- ⋅ =- ⋅ =- × =- × B B n D D n J H H n E E n s s ρ E J H B E D c ~ ~ ~ ~ ~ ~ σ μ ε = = = E J H B E D c σ μ ε = = = σ ϖμ 2 ) 1 ( j jX R Z s s s + = + = , ] [ ] 2 1 [ , ] 2 1 [ ) ( 2 2 2 = = = ⋅ = ⋅ × = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ = = v d v i i v e v m s s dv E P dv J E P dv E W dv H W ds H E P σ ε μ ~ 2 1 , ] ~ ~ 2 1 [ ] ~ 4 1 [ , ] ~ 4 1 [ ) ~ ~ ( 2 * 2 2 * = = = ⋅ = ⋅ × = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ = = v d v i i v e v m s s dv E P dv J E P dv E W dv H W ds H E P σ ε μ Frequency Domain Time Domain Example Problem 2 2 , μ ε 1 1 , μ ε Two lossless half spaces z j o x inc e E a E 1 ˆ ~ β- = z j o x r e E a E 1 ˆ ~ β Γ = z j o x t e TE a E 2 ˆ ~ β- = Problem: Using Maxwell’s equations + constituent equations + boundary condition find expressions for β 1 , β 2 , Γ and T z x y E j H H j E B j E ~ 1 ~ ~ ~ ~ ~ × ∇- = ⇒- = × ∇- = × ∇ ϖ μ ϖ μ ϖ (1) Start by finding H using z j o y z j o z y x t z j o y z j o z y x r z j o y z j o y z j o z y x inc z y x z y x Te E a Te E z y x a a a j H e E a e E z y x a a a j H e E a e E z j a e E z y x a a a j H E E E z y x a a a j H 2 2 1 1 1 1 1 2 2 2 1 1 1 1 1 1 1 ˆ ˆ ˆ ˆ 1 ~ ˆ ˆ ˆ ˆ 1 ~ ˆ ) ( 1 ˆ ˆ ˆ ˆ 1 ~ ˆ ˆ ˆ 1 ~ β β β β β β β ϖ μ β ϖ μ ϖ μ β ϖ μ ϖ μ β ϖ μ ϖ μ ϖ μ = ∂ ∂ ∂ ∂ ∂ ∂- = Γ- = Γ ∂ ∂ ∂ ∂ ∂ ∂- = = ∂ ∂- = ∂ ∂ ∂ ∂ ∂ ∂- = ∂ ∂ ∂ ∂ ∂ ∂- =---- H j E E j H D j H ~ 1 ~ ~ ~ ~ ~ × ∇ = ⇒ = × ∇ = × ∇ ϖ ε ϖ ε ϖ (2) See if we can find β uisng 1 1 1 1 1 2 2 1 1 1 2 2 1 1 1 2 2 1 1 1 1 1 1 1 1 ˆ ˆ ˆ ) ( 1 ˆ ˆ ˆ ˆ 1 ˆ ~ ˆ ˆ ˆ 1 ~ 1 1 1 1 1 1 ε μ ϖ β ε μ ϖ β ε μ ϖ β ε μ ϖ β ϖ μ β ϖ ε ϖ μ β ϖ ε ϖ ε β β β β β β = ⇒ = ⇒ = ⇒ = ∂ ∂- = ∂ ∂ ∂ ∂ ∂ ∂ = = ∂ ∂ ∂ ∂ ∂ ∂ =------ z j o x z j o x z j o x z j o x z j o z y x z j o x inc z y x z y x e E a e E a e E a e E z j a e E z y x a a a j e E a E H H H z y x a a a j E 2 2 2 ε μ ϖ β = Similarly: Combine the two last results: z j o y t z j o y r z j o y inc Te E a H e E a H e E a H 2 1 1 2 2 1 1 1 1 ˆ ~ ˆ ~ ˆ ~ β β β ϖ μ β ϖ μ β ϖ μ β = Γ- = =- 2 2 2 1 1 1 ε μ ϖ β ε μ ϖ β = = z j o y z j o y t z j o y z j o y r z j o y z j o y z j o y...
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ELEG413lec3 - ELEG 413 Lecture#3 Mark Mirotznik Ph.D...

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