309-2008-Solutions10

309-2008-Solutions10 - ECE 309 — Electromagnetic Fields...

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Unformatted text preview: ECE 309 — Electromagnetic Fields University of Virginia Fall 2008 Homework # 10 Solutions 1. Propagation in Moist Earth With the given constitutive parameters, σ = 0 . 01 S/m, r = 30 , and μ r = 1 , we first check to see if moist earth can be considered a “good” conductor at 100kHz, σ ω = . 01 2 π · 10 5 · 30 · 8 . 854 × 10- 12 = 60 which is significantly larger than unity. Thus, we can assume moist earth is a good conductor and use the appropriate approximate formulas. k = ω √ μ 1- j σ ω . 5 ’ r ωμσ 2 (1- j ) = β- jα α = β = p ωμ σ/ 2 = 0 . 063m- 1 (a) 10log( e- 2 α‘ ) =- 3dB ⇒ ‘ = 5 . 5meters (b) λ = 2 π β = 99 . 7meters (c) v p = ω β = 9 . 97 × 10 6 me / s (d) Z TEM = ωμ k = ωμ r 2 ωμ σ 1 1- j = r ωμ 2 σ (1 + j ) = 6 . 23(1 + j )Ω 2. Complex Wavenumber For a conducting medium, the complex wavenumber k is given by, k = β- jα = ω √ μ 1- j σ ω 1 2 The last factor can be expressed, 1- j σ ω 1 2 = r 1 + σ ω 2 exp(- jφ ) 1 2 where φ = tan- 1 σ ω Thus, 1- j σ ω 1 2 = 4 r 1 + σ ω 2 exp(- j φ 2 ) = 4 r 1 + σ ω 2 n cos φ 2- j sin φ 2 o noting that cos 2 φ 2 = 1 2 (1 + cos φ ) , and sin 2 φ 2 = 1 2 (1- cos φ ) where, cos φ = 1 p 1 + ( σ/ω ) 2 we have, k = ω r μ 2 4 r 1 + σ ω 2 h 1 + 1 p 1 + ( σ/ω ) 2 1 2- j 1- 1 p 1 + ( σ/ω ) 2 1 2 i = ω r μ 2 h r 1 + σ ω 2 + 1 i 1 2- jω r μ 2 h r 1 + σ ω 2- 1 i 1 2 Thus, β = ω r μ 2 h r 1 + σ ω 2 + 1 i 1 2 α = ω r μ 2 h r 1 + σ ω 2- 1 i 1 2 3. Plane Wave Incident on Dielectric Slab Let us first apply the boundary conditions at z = 0 (the rightmost interface); this leads to the same ratio for...
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This note was uploaded on 10/16/2010 for the course ECE 309 taught by Professor Weikle during the Spring '08 term at UVA.

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309-2008-Solutions10 - ECE 309 — Electromagnetic Fields...

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