au10_312ps6_sol

au10_312ps6_sol - ECE312 Autumn 2010: Problem Set 6...

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Unformatted text preview: ECE312 Autumn 2010: Problem Set 6 Solutions Problem 1 : 3 GHz plane wave E i = (3 x- 2 y- 4 z ) e jA (4 x +3 z ) (a) By examining e jA (4 x +3 z ) = e j 5 A ( 4 x 5 + 3 z 5 ) = e jk 1 ( x sin i + z cos i ) We find sin i = 4 / 5, cos i = 3 / 5, so that i = 53 . 13 (b) From part (a), 5 A = k 1 = , therefore A = 2 (3 10 9 ) / (5(3 10 8 )) = 4 rad/s . (c) In our coordinate system, y components of the electric field are perpen- dicular polarized because they are perpendicular to the plane formed by the incident vector k = x sin i + z cos i and the surface normal z . Thus E i =- 2 ye j 4 (4 x +3 z ) (d) Parallel polarized part of E i lies in the xz-plane, E i bardbl = (3 x- 4 z ) e j 4 (4 x +3 z ) (e) The magnetic field can be found by ( 377) H i = 1 k E i = 1 parenleftbigg 4 x 5 + 3 z 5 parenrightbigg (3 x- 2 y- 4 z ) e j 4 (4 x +3 z ) = 6 x + 25 y- 8 z 5 e j 4 (4 x +3 z ) Problem 2 : Fiber optic = 2 . 2 , f = 420 THz....
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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_312ps6_sol - ECE312 Autumn 2010: Problem Set 6...

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