20102ee161_1_Homework06_solution

20102ee161_1_Homework06_solution - E16’\ 5 W6 Reagan if...

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Unformatted text preview: E16’\ 5 W6 Reagan if 1: a»? _ rinigaixii $2..an RM Em l f» , “MW 3wa Problem 8.43 A hollow cavity made of aluminum has dimensions a : 4 cm and d : 3 cm. Calculate Q of the TE-ml mode for (a) b : 2 cm, and (b) b = 3 cm. Solution: For the TE101 mode, fun is independent of b, c 1 ‘ 1 ‘ f10'1 — E + i 3 X '108 1 2 i '1 2 _ 2 4 ><'10’2 3 >~<110’2 = 6.25 GHZ. For aluminum with (EC : 3.5 X 107 S/m (Appendix B), "l 65 : — : 1.08 ><10’6 m. VJTmeOGc (a)Fora:4cm, b:2cmandd:3cm, )7 l abd(a2+d2) 9 — 55 [a3(d+2b) +d3(a+2b)1 : 8367. (b)Fora:4cm_, b:3cm_,andd:3cm_, Q : 9850. Problem 9.1 A center-fed Hertzian dipole is excited by a current I0 = 20 A. If the dipole is K/ 50 in length, determine the maximum radiated power density at a distance of '1 km. Solution: From Eq. (9.14), the maximum power density radiated by a Hertzian dipole is given by S _ 110181522 _ 377 X (Mn/AV x 202 x (1/50)2 0 _ 321W? _ 327:2(103)2 = 7.6 x 10*6 W/m2 2 7.6 (yW/mz). Problem 9.4 Repeat Problem 9.3 for an antenna with FCB _ Sinzecoszq)’ forOflBSTEand in/zgogn/zj ’ — 0 elsewhere. 1 Solution: The direction of maximum radiation is the +it-axis (where 9 : 11/2 and q) : 0). From Eq. (9.23), 4TE D : — _ 4TE for: sin2 8cos2 q) sin 8 d8 dq) i 4TE C032 for: Sin3 9 7 4n 2 - I fit/2%“ +cosZ¢jd¢f_11(1_x2)dx 47: 4n :.1 1- TE/2 A 1:1 43 26:7'8dBj j(q)+Egmmnlair/2(xix3/3H—1 EEC L) 4111 sr 4TE sr 2 91) : D : 6 : En (Sf). In the x-Z plane, q) : 0 and the half power bearnwidth is 90°, since sin2 (450) : 541120350) 2 g. ...
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20102ee161_1_Homework06_solution - E16’\ 5 W6 Reagan if...

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