write and run a MATHLAB code for a sinsoudial wave as shown above where Zn is on the x axis and lEl max and E is on the y axis as shown above, additional information can be found above

Medion Media Media Medium 1 3 . We have 4 medium our Mo = 1-3 2 s = 22 23 Given f. 11GHz Pau- 30dB We use 1% of the frequencys value. Medium 1 Medium 2 aur Porypropylene Medium 3 Medium 4 E fiberglass 2.1% HE E. Ho Ho We need to find the final E4 and H4 F 11 x 109 xl Il x 107 Hz ; K=w WIME 100 K2 = 2xx 10 x 107 142x107x 14 x 109 x 2-1 0.9667 367 K3 = 2x x10 x 107 142x107x 1/362x107x4 -207m; ke 22 = 1.538 x 01662 Kz 1-32 Air: ki=k4 = 0-672 x = X4 21 - 2-987 0-677 between 1 and 2 medium 7. 7 = 174 = Molso = 7 = 74 1207 2 = 16 Mo 2160 82-87 12-1/362x10 13 - 12-16 = 82.87-1202 = -0-1834 82.82 +1202 = 20 27 x 10 x107 = 2xf 3 x 10 Y 42x107 -4 142x107 = 602 14/362x10-9 -0-159 2 2+1 - |- 0:1834 = 08 65 2 = 13-12 2 602 82-82 13+1+ 602 +82-85 3= 14-15 z 1207-607 te t13 120 + 60 7 0.333 Pau= 2 ET 30 1202 30 0.7072 14 za 6 (0.840 1207 30 = 0.70726, 1207 30 = @z 1877 x 1036 G 36 X103 = 126.42% 1-877 G r 126-42 cos Cost - K42) ; wa 2af k4= 0.672 6 = 6, 126-2 cao C222 x bk- 6-622) Hi I an xi 14 n = 2z 1207 - Zn = ; n = 1, 2, 3, 4 2n to @z xx 126-42 cos (227 x 1076 -0-6722) H4 = 0-3353 Qy cos C222 *174-0.6722) 4 , C, o3353 cos C224 X 107 0-674) lo+ - n da i 2 14 13 the other mazing of IEI located at 2k4 2 = -207 if na!, Zm-1.49 =1 1 Eyl126-42 ns2, 212--2-98 ns31213-4.47 $4=2.98m n=4 12n4 = -5.96 are Zn = = 1 Elmer M 2.0 -5.96 -441 -2.97 -1:49