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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 θ (1) i = f 1 f = 18 . 75 20 = 0 . 9375 , sin θ (1) i = radicalBig 1 − cos 2 θ (1) i = 0 . 3480 Note k 1 = ω/c = 418 . 9 rad/s. The electric field is ˜ E = ˆ yE e − jk 1 x sin θ (1) i ( − 2 j ) sin( k 1 z cos θ (1) i ) = ˆ ye − j 145 . 8 x ( − 2 j ) sin(392 . 7 z ) (c) The magnetic field is ˜ H = E η e − jk 1 x sin θ (1) i bracketleftBig − 2ˆ x cos θ (1) i cos( k 1 z cos θ (1) i ) − 2 j ˆ z sin θ (1) i sin( k 1 z cos θ (1) i ) bracketrightBig = (0 . 00265) e − j 145 . 8 x [ − 1 . 875ˆ x cos(392 . 7 z ) − j . 696ˆ 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 . 875ˆ z cos(392...
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