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Unformatted text preview: ECE 329 Fall 2009 Homework9 Solution Due: Oct. 29, 2009 1. We observe that σ ω = 4 2 π × 2 × 10 4 × 81 × 8 . 8510 12 1 , which meas water can be treated as a good conductor. Then α ≈ p πfμσ = p π × 2 × 10 4 × 4 π × 10 7 × 4 = 0 . 562 ( rad/m ) . Assume that the distance between the submarine and the ship is d . We have e αd > . 1% = 0 . 001 , ∴ αd > ln(0 . 001) , ∴ d 6 ln(0 . 001) α ≈ 12 . 3 ( m ) . 2. a) By comparing with the given expression and the general expression H = ˆ xH e αy cos( ω t β y φ ) , we nd that α = 1 ( rad/m ) , β = √ 3 ( rad/m ) , ω = 8 π × 10 6 ( rad/s ) , γ = α + jβ = 1 + j √ 3 ( rad/m ) . From γη = jωμ , we have η = jωμ γ = j × 8 π × 10 6 × 4 π × 10 7 1 + j √ 3 = 4 π 2 5 √ 3 + j (Ω) . b) ∵ γ η = 1 + j √ 3 4 π 2 5 ( √ 3 + j ) = 0 . 110 + j . 0633 ( S/m ) . ∴ = 1 ω Im γ η = 1 8 π × 10 6 × . 0633 = 2 . 52 × 10 9 ( F/m ) σ = Re γ η = 0 . 110 ( S/m ) . c) ˜ H = ˆ x 10 e ( 1+ j √ 3 ) y j π 3 ( A/m ) . d) ˜ E = η ˜ H × ˆ y = ˆ z · 8 π 2 5 · e j π 6 · 10 e ( 1+ j √ 3 ) y j π 3 = ˆ z 16 π 2 e ( 1+ j √ 3 ) y j π 6 ( V/m ) . e) < E × H > = 1 2 Re ˜ E × ˜ H * = 1 2 Re ˆ y 160 π 2 e 2 y + j π 6 = ˆ y 40 √ 3 π 2 e 2 y ≈ ˆ y 684 e 2 y ( W/m 2 ) . 1 ECE 329 Fall 2009 f) According to the law of energy conservation, the dissipated power P dis equals the net power owing into the cube: P dis = ˛ S < E × H > · d S ....
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This note was uploaded on 05/05/2010 for the course ECE 329 taught by Professor Franke during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 FRANKE

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