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# Class Note for ECE 6341 with Professor Jackson at UH 4

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Date Created: 02/06/15

ECE 6341 Spring 2009 Prof David R Jackson ECE Dept Notes 31 Example Far field from an aperture 2 Given Ex y0 Exyz 1 TOTEQXky0e f39kzze 139kxxkyydkxdky 272 7w w Example cont Approximate this result for Use spherical coordinates x r 5m 6 cos yrsin6lsin z r0056 and define kxx kx kor sm lcos 1 kx I ky I k2 xk O yk O zk O kyykyk0rsm6lsm kzz 12 kor cos 6 Example cont We then have Exyz My 02010ng Egg 0 where Example cont 00 00 Exyz 22 kg 1 JENzazao e J39korl2 cos 9l c sin6cos le sin sin dkx dky Identify the following terms and functions Example cont 5 20 SPP 0 akx 6 E 0 aky Hence kx0 1 cos 6 sin 6 cos 1c056 sin sin Example cont Squaring both sides we have 30 cos2 6 sin2 6 cos2 1 130 20 cos2 6 sin2 6 sin2 1 130 Iy20 Add these two equations Iago lgy20cos2 6sin2 61 130 2 yo 2 2 2 or kx0kyO sm 6 Example cont Hence from the first of the two equations 30 cos2 6 sin2 6 cos2 1 sin2 6 sin2 6 cos2 cos2 6 Q kxo sin6cos Similarly kyo Sin6sin 1 E 0056 yO Also E0 1 130 2 Example cont Z a 139415 Physical interpretation PW that radiates at angles 6 x x09ky0 Example cont 1 2 2 2 2 a sec 49 cos 651n 49005 2 1 2 2 o 2 2 sec 9cos 651n 95m 7 sec2 9 sin2 49 sin cos details omitted Example cont Hence Also 2 7 1 4 2 2 2 2 2 2 a T Zsec 6cos 631n 600 cos 631n 95111 N isec4 6 sin4 6 sin2 cos2 1 4 cos4 6 cos2 6 sin2 631112 cos2 2 360 6 1114 6 cos2 sin2 sin4 6 sin2 cos2 11 12 Example cont 2 1sec 6 0054 6 cos2 6 sin2 6 4 2 1sec2 6 cos2 6 sin2 6 4 2 1sec2 6 4 Hence This problem thus fits into the special case Example cont Recall that for this special case Choose sign since 05gt 0 and 6gt 0 2 Also recall that foamg E EEO Example cont Therefore we have 2 awr j Example cont 1 gl xoly0 1 lfo 430 cosQl x0 sichos le0 sianingb 1 1 sin2 65 cos 6 sin 6 cos 2 sin 6 sin 2 cos2 6 sin2 6 1 Hence engkxOky0 e jkor Example cont Also Example cont Hence returning to unnormalized variable notation Example cont Assume that only the tangential aperture field is known Etz ExZEy Z0 The uniqueness principle indicates that this should be sufficient to determine the far field It should be possible to find the farfield components E6 and E in spherical coordinates from these 18 19 Example cont To find EZ use VEO 8E 8EX yaEZ 0 6x 6y 62 Assuming a planewave field in the far field jkx0EfF jky0EfF ijOEZFF 0 Hence or Example cont x0 x yO y EFF kik EFFk EFF Example cont In the farfield EfF EfF sin EfF cosg E51 EfF cos6cos EfF cos6sin EfF sin6 To eliminate E2 in the second equation above we use sin 6 EfF sin6 sin6cos EfF sin6sin cos 6 Hence sin2 6 cos 6 E EfF cos6cos cos6 39 2 FF sm 6 cos Ey COS6SIn mnq EfF cos EfF singi 5606 21 Example cont The farfield spherical components are thus expressed in terms of the farfield rectangular components as We now use the result from the 2D stationaryphase method which is jk0r FF 1 6 Ex2yj1 cos r jEx ykx0kVOO 0 22 Example cont Final result y Introduce new unit vectors icos izsin I 2 2 sin 3gtCOS 23 Example cont The final simplified form is

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