au10_312ps5_sol

au10_312ps5_sol - ECE312 Autumn 2010: Problem Set 5...

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Unformatted text preview: ECE312 Autumn 2010: Problem Set 5 Solutions Problem 1 : E 1 = ( x j 2 y ) e jkz , and E 2 = (2 + j ) ye jkz . (a) The time average Poynting vector S av = 1 2 Re { E H } . First find the magnetic field H = 1 k E = 1 ( z ) ( x j 2 y ) e jkz = ( j 2 x y ) e jkz S av = 1 2 Re { E H } = 1 2 Re braceleftbigg ( x j 2 y ) ( j 2 x y ) e jkz e jkz bracerightbigg = 5 z 2 = z 6 . 63 10 3 W/m 2 Note this is k | E 1 | 2 / (2 ) which applies for all single plane waves in lossless media. (b) For E 2 = (2 + j ) ye jkz , S av = k | E 2 | 2 2 = 5 z 2 = z 6 . 63 10 3 W/m 2 (c) E 1 + E 2 = ( x + (2 j ) y ) e jkz , so S av = k | E tot | 2 2 = z (1 + 5) 2 = z 7 . 96 10 3 W/m 2 Note S tot ,av negationslash = S 1 ,av + S 2 ,av because Maxwells equations are linear in terms of fields not power. Fields add but powers do not....
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This note was uploaded on 01/11/2012 for the course ECE 312 taught by Professor Johnson,j during the Fall '08 term at Ohio State.

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au10_312ps5_sol - ECE312 Autumn 2010: Problem Set 5...

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