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sp08hw7sol1 - ECE329 HW7 Spring 2008 Problem 1 α = 1/m β...

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Unformatted text preview: ECE329 HW7 Spring 2008 Problem 1. α = 1 /m, β = √ 3 /m ,ω = 8 π · 10 6 rad/s, v p = ω β = 8 π √ 3 · 10 6 m/s , δ = 1 α = 1 m γ = α + jβ = 1 + √ 3 j /m η = μjω γ = 15 . 85 e j. 524 Ω σ = Re γ η = 0 . 109 S/m = 1 ω Im γ η = 2 . 5 · 10- 9 F/m σ ω = 1 . 73 This material is somewhere in between a conductor and a dielectric because σ ω is close to 1. Problem 2. H = ˆ a x 25 e- z cos(8 π · 10 6 t- √ 3 z- π 3 ) E =- ˆ a y 25 e- z | η | cos(8 π · 10 6 t- √ 3 z- π 3 + τ ) E = ˆ a y 396 . 25 e- z | η | cos(8 π · 10 6 t- √ 3 z + 2 . 62) ˜ H = ˆ a x 25 e- z e- j ( √ 3 z + π 3 ) ˜ E = ˆ a y 396 . 25 e- z e j (- √ 3 z +2 . 62) Power ows in the ˆ a z direction. h P i = 1 2 Re ˜ E × ˜ H * = 4284 e- 2 z Thus, the time average power through a surface of area one meter squared in the z = z plane is 4284 e- 2 z . Poynting's theorem: I S P · dS =- Z V σE 2 dV- d dt Z V 1 2 E 2 dV- d dt Z V 1 2 μH 2 dV Either side of Poynting's theorem can be used to nd the average power dissipated in the cubic volume. ItEither side of Poynting's theorem can be used to nd the average power dissipated in the cubic volume....
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