329fall08hw9sol

329fall08hw9sol - t = 10 2 T e-j 2 z y = 1 9 e-j 2 z y A m...

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ECE-329 Fall 2008 Homework 9 — Solution October 27, 2008 1. Region 1 ( z < 0 ) is free-space while region 2 ( z > 0 ) is occupied by a dielectric with permittivity 4 ± o . A plane wave in region 1 is propagating in + z direction and is incident on z = 0 plane. The phasor of the incident field is given by ˜ E i = 10 e - 1 z ˆ x V m . a) The reflection and transmission coefficients at the given interface are R = η 2 - η 1 η 2 + η 1 and T = 1 + R . Since the intrinsic impedances of each region are η 1 = q μ o ± o = 120 π and η 2 = q μ o 4 ± o = 1 2 η 1 = 60 π , we can find that R = - 1 3 and T = 2 3 . In consequence, the reflected and transmitted electric field phasors are ˜ E r = 10 R e 1 z ˆ x = - 10 3 e 1 z ˆ x V m , ˜ E t = 10 T e - 2 z ˆ x = 20 3 e - 2 z ˆ x V m where β 1 = ω ± o μ o and β 2 = ω 4 ± o μ o = 2 β 1 . b) The incident, reflected, and transmitted magnetic field phasors are ˜ H i = 10 η 1 e - 1 z ˆ y = 1 12 π e - 1 z ˆ y A m , ˜ H r = - 10 η 1 R e 1 z ˆ y = 1 36 π e 1 z ˆ y A m , ˜ H
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Unformatted text preview: t = 10 2 T e-j 2 z y = 1 9 e-j 2 z y A m . c) The time-average Poynting vectors of the incident, reected, and transmitted waves are P i av = 1 2 Re n E i H i * o = 50 1 z = 5 12 z W m 2 , P r av = 1 2 Re n E r H r * o =-50 1 R 2 z =-5 108 z W m 2 , P t av = 1 2 Re n E t H t * o = 50 2 T 2 z = 10 27 z W m 2 . Notice that the power density transported by the incident wave is distributed between the re-ected and transmitted waves, such that P i av = | P r av | + P t av and therefore, it can be shown that R 2 + 1 2 T 2 = 1 . 1...
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