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Unformatted text preview: ECE 329 Fall 2010 Homework 8  Solution Due: Nov. 2, 2010 1. a) We check the value of σ ω . For 10 7 6 f 6 5 × 10 9 , σ ω ranges from 1 . 44 × 10 8 to 7 . 19 × 10 10 . So the lossy medium can be regarded as good conductor. b) We assume the wave propagates in the lossy medium from point B to A , and the magnitude of the electric eld at these two observation points are E B and E A . Consider the attenuation constant α , we know E A = e αd E B , where d denotes the distance between B and A . As for the timeaverage power, we have P A = e 2 αd P B . If we choose d as unit length, then the attenuation of the timeaverage power (in dB) will be 10log 10 P A P B = 10log 10 ( e 2 α ) = 2 α × 10log 10 ( e ) = 8 . 686 α. c) Based on b), we know the attenuation in dB per unit length is 8 . 686 α . Also, the de nition of the skin depth is δ = 1 α . Therefore, the attenuation in dB per skin depth is 8 . 686 αδ = 8 . 686 , a constant irrelevant to frequency. d) We know the penetration depth of lowfrequency component is larger than that of highfrequency component. As a result, we only need to make the lossy material layer thick enough to screen all the lowfrequency components. The attenuation required by the problem is 20log 10 10 6 . 1 = 100 ( dB ) . At the lowend frequency, α ≈ p πfμσ = p π × 10 7 × 1 . 25663706 × 10 6 × 4 × 10...
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 Spring '08
 Kim
 Polarization, Gate, Wave propagation, arbitrary integer, lossy material, electric 1Celd

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