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329fall10hw8sol

# 329fall10hw8sol - ECE 329 Fall 2010 Homework 8 Solution Due...

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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 time-average power, we have P A = e- 2 αd P B . If we choose d as unit length, then the attenuation of the time-average 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 low-frequency component is larger than that of high-frequency component. As a result, we only need to make the lossy material layer thick enough to screen all the low-frequency components. The attenuation required by the problem is 20log 10 10- 6 . 1 =- 100 ( dB ) . At the low-end frequency, α ≈ p πfμσ = p π × 10 7 × 1 . 25663706 × 10- 6 × 4 × 10...
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329fall10hw8sol - ECE 329 Fall 2010 Homework 8 Solution Due...

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