au10_312ps7_sol

au10_312ps7_sol - ECE312 Autumn 2010: Problem Set 7...

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Unformatted text preview: ECE312 Autumn 2010: Problem Set 7 Solutions Problem 1 : d = 8 mm, air-filled guide. (a) Cutoff of TM is 0 Hz. Cutoff of TM 1 /TE 1 is c 2 d = (3 10 8 ) 2(8 10 3 ) = 18 . 75 GHz. Cutoff of TM 2 /TE 2 is 2 c 2 d = 37 . 5 GHz. (b) For TE 1 at 20 GHz, cos (1) i = f 1 f = 18 . 75 20 = 0 . 9375 , sin (1) i = radicalBig 1 cos 2 (1) i = 0 . 3480 Note k 1 = /c = 418 . 9 rad/s. The electric field is E = yE e jk 1 x sin (1) i ( 2 j ) sin( k 1 z cos (1) i ) = ye j 145 . 8 x ( 2 j ) sin(392 . 7 z ) (c) The magnetic field is H = E e jk 1 x sin (1) i bracketleftBig 2 x cos (1) i cos( k 1 z cos (1) i ) 2 j z sin (1) i sin( k 1 z cos (1) i ) bracketrightBig = (0 . 00265) e j 145 . 8 x [ 1 . 875 x cos(392 . 7 z ) j . 696 z sin(392 . 7 z )] (d) The time-averaged Poynting vector is S av = 1 2 Re braceleftBig E H bracerightBig = 1 2 Re { j . 0053 sin(392 . 7 z ) [1 . 875 z cos(392...
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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_312ps7_sol - ECE312 Autumn 2010: Problem Set 7...

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