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Unformatted text preview: ECE329 Homework 8 Spring2008 Problem 1. The problem was a bit unclear. The solution assumes that the x < range is called material 1, has = 3 , and is on the left side. The boundary condition a n ( D 1 D 2 ) = s can be used to nd E 2 x at x = 0 . D 1 x = D 2 x 3 E = 5 E 2 x E 2 x E 1 x = 3 5 From the boundary condition a n ( E 1 E 2 ) = 0 and at x = 0 , E 2 z must be zero. Also from this boundary condition, E 1 y = E 2 y . E 2 y E 1 y = 1 At x = 0 , E 2 = E ( 3 5 a x + 3 a y ) and D 2 = E 0 0 (3 a x + 15 a y ) .  E 2   E 1  = q ( 3 5 ) 2 + 3 2 1 2 + 3 2 = 0 . 97  D 2   D 1  = 3 2 + 15 2 3 2 + 9 2 = 1 . 61 Problem 2. First nd H in both regions using Faraday's law with phasors. E = j H H 1 = a y E 1 1 e j 1 z H 2 = a y E 2 2 e j 2 z Use the boundary condition a n ( H 1 H 2 ) = J s at z = 0 . For this case with perfect dielectrics, H 1 y = H 2 y ....
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This note was uploaded on 04/14/2009 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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