20102ee161_1_Homework05_solution

20102ee161_1_Homework05_solution - 53:: Problem 8.36 A TE...

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Unformatted text preview: 53:: Problem 8.36 A TE wave propagating in a dielectric-filled waveguide of unknown permittivity has dimensions a : 5 cm and b : 3 cm. If the x-component of its electric field is given by Ex : —36cos(407tx)sin(100rty) -sin(2.4rt X 101% — 52.9w). (V/m) determine: (a) the mode number, (b) 81. of the material in the guide, (c) the cutoff frequency, and (d) the expression for Solution: (a) Comparison of the given expression with Eq. (8.1'10a) reveals that H? TE _ : 401T, hence m : 2 a ? : 1007:, hence I? : 3. Mode is TE23. (1)) From sin(oor * [32), we deduce that o) = 2.4:: >< 1010 rad/s, B = 52.97: rad/m. Using Eq. (8.105) to solve for 81-, we have 5; [Ba (TV + (W = 2.25. (C) _ c _3><108 "p" 7 V2.25 2 2 2 2 (a) db) : 10.77 GHZ. : 2 x 108 m/s. (d) ZTE = Z'U/V '1 — (53/02 _ 7 i 81. 3 . 2 7 1 — fl : 569.9 9. ..F 12 Hence, H _ Ex- .1" _ ZT = —0.063 c05(407tx) sin('1001't_v) sin(2.4rt >< '1010r— 52.9112) (A/m). Problem 8.40 A narrow rectangular pulse superimposed on a carrier with a frequency of 9.5 GHZ was used to excite all possible modes in a hollow guide with a : 3 cm and Z7 : 2.0 cm. If the guide is 100 m in length, how long will it take each of the excited modes to arrive at the receiving end? Solution: With a : 3 cm. I) : 2 cm. and Hpo : c : 3 X 108 m/s, application of Eq. (8.106) leads to: f10 : 5 GHz f01 = 7.5 GHZ fll : 9.01 GHZ f20 = 10 GHZ Hence, the pulse with a 9.5-GHZ carrier can excite the top three modes. Their group velocities can be calculated with the help of Eq. (8.1 14), “g : CV 1* (flint/.flzs which gives: 0.856 = 2.55 X '108 m/s, for T1310 Hg : 0.616 : 1.84 X 108 Ms. for T1301 0.320 : 0.95 X 108 rn/s. for T1311 and TMH Travel time associated with these modes is: 0.39 ,us. for TEN : 0.54 ,us. for T1301 1.05 ys. for TEN and TM-l 1. d 100 T : — : Hg Hg ...
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20102ee161_1_Homework05_solution - 53:: Problem 8.36 A TE...

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