sp08hw7sol1 - ECE329 HW7 Spring 2008 Problem 1. = 1 /m, = 3...

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Unformatted text preview: ECE329 HW7 Spring 2008 Problem 1. = 1 /m, = 3 /m , = 8 10 6 rad/s, v p = = 8 3 10 6 m/s , = 1 = 1 m = + j = 1 + 3 j /m = j = 15 . 85 e j. 524 = Re = 0 . 109 S/m = 1 Im = 2 . 5 10- 9 F/m = 1 . 73 This material is somewhere in between a conductor and a dielectric because is close to 1. Problem 2. H = a x 25 e- z cos(8 10 6 t- 3 z- 3 ) E =- a y 25 e- z | | cos(8 10 6 t- 3 z- 3 + ) E = a y 396 . 25 e- z | | cos(8 10 6 t- 3 z + 2 . 62) H = a x 25 e- z e- j ( 3 z + 3 ) E = a y 396 . 25 e- z e j (- 3 z +2 . 62) Power ows in the a z direction. h P i = 1 2 Re E H * = 4284 e- 2 z Thus, the time average power through a surface of area one meter squared in the z = z plane is 4284 e- 2 z . Poynting's theorem: I S P dS =- Z V E 2 dV- d dt Z V 1 2 E 2 dV- d dt Z V 1 2 H 2 dV Either side of Poynting's theorem can be used to nd the average power dissipated in the cubic volume. ItEither side of Poynting's theorem can be used to nd the average power dissipated in the cubic volume....
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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.

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sp08hw7sol1 - ECE329 HW7 Spring 2008 Problem 1. = 1 /m, = 3...

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