Question

[50 pts) Given: A 100-MHZ TM polarized wave with an amplitude of 1 V/m is obliquely incident from air (z < 0) onto a slap of lossless, nonmagnetic material with er = 25. The angle of incidence is 40°. Find: a) Determine the angle of transmission; b) determine the reflection and transmission coefficients and c) determine expressions for the incident, reflected, and transmitted fields. Solution: Answers: a) The angle of transmission is 7.37º; b) Itm = -0.588 and Ttm = 0.318. The fields are: Ē' = (0.7662 - 0.6422) e-;(1.34x+1.60z) V/m H = 2.65e-j(134x+1.602) = mA/m Ë' = (-0.45 – 0.382)e-j(134x–1.602) V/m A' = (-1.55)e=/(1.34x– 1.602); 2)ýmA/m Ēt = (0.315- 0.042) e-|(1.34x+10.362) V/m A' = 4.21e-1(1.34x+10.382) yn mA/m

Answer 1

sol a Er THE HY ot Oy 3 Bi (407 medium o medium Ein (M21E2) (Mti) cincident angle) Shall's law Seri Sinoi = Ser2 Sinot 01-400 Sin 40 = 125 Sinot Sinoto0.128557 tt = 7.370 Ota transmitted angle

reflection transmision officient Ercosoi Seri Cosot Ter, sot + [Era Cosoi Tim = o.gg - 5x0.766 o.gg +50.766 Tim - 2.84 4.82 im _0589 z transmlusion officert Tim Im = 2 Coso Terz I Cosott L cosoi Verz TTM = Cos 40 1 125 Cos 7.37 + bcos W540 Gm= 0.48 0.766 02*0.99 +0.766 Tim = 0.316 -JB,(x Sing;+z6500) electric field incident wave Eio (coso, n - Sindi 2) e Eid = 1 vim f = 100 MHz Cett da 3468 3m 27 321094 bis on -T%209+(x Singotz Ei= (los too sintof) e Curgo Ei= (0.766 % - 0.642€) e J (1-34% + 1.602) ñ

incident magnetic field Ti = Elo e nie hon 120 -J8 (x Sơn 2 + 250i) Ħ= 1 -J *2.094(x sinto+26840) e -I(1-35 x + 1 602) Hi= 2.65 e Î mot reflected electric field -jBifresinor-zorg È Erollosur +Singaz) e Dyzoia 40 Ero= Timxti. =-5881 Era -0.5BB -342-094(x singo Er = -0.588(Koss ton + Sindo 2 De - ZW340) -(1.344 -1.602) Er - (-0.45 sî -0.37792) é restected H = e magnetic fied -JBil & Sinar-z Gosor) Ero 1. - 3 * 2:094 (* Sin 4o - W5402), -1.55 e -(1:35 -1,62), (-1.55e ima cms m

transmitted electric field -J B2 (r. Sinat AZ COSE êt = Fto (cosotan - Sin 9402) e Ota 320 Etoa Ermitio 0:318x3 Etos 0.318 that 822 WfM2E2 ß2= 25 x 108 x 875 3x10 $2 = 5*2.094 B2 = 10.47 -J *1847(x S/n937 É = 0.318[6057:34ñ – Sin 7372] e +Z Z 605782) -5 (1.34% +10.38) : 315 € -0.0 -0.04²) ē transmitted magnetic field HE --JB2 ( x Sinot + Z C50t) - Eto e M2 3X10.47 ( * Sin 734° +2 Copyspy HK = 0.31845 12- Hea -J (1.34 n +10.382) 24.21 vim Et = e . e m