HW3_Solution

HW3_Solution - ECE 201A Homework 3 solution 1. (a) E = E 0e...

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ECE 201A Homework 3 solution 1. (a) 0 jk r EE e = G G i G G Gauss’ law in a homogenous, linear, anisotropic and source free medium 0 E ∇= G i . Substituting () 00 0 0 0 jk r jk r jk r jk r E E eE j k e j k E e −− = ∇ = − = = GG G G G G ii i i G G G G G i i i . Hence 0 kE = G G i . (b) 22 0 ωμε ∇+ = G G and 0 jk r e = G G i G G . Substituting 2 2 2 0 0 0 0 2 0 0 j kr j j j j j jk r Ee E e E jk jk e k k = + = −+ =− + = G G i G G G G i G i Hence 2 kk = G G i . (c) In a lossy medium '" kkj k G . Assuming that k G is in the xz plane '' ˆˆ ' xx zz ak a =+ G and "" " a k a G . Then the general plane wave can be expressed as ' ' ˆˆˆˆ "' 0 0 0 x z x z xz jk jk r ka xa za jka jk r k r jk r kx kz jkx kz e E e E e e E e e e + − + + −+ − + == = = = G G G i i G G G G So the wave attenuates along x and z directions while propagating in x and z directions. Note that attenuation coefficients and phase constants in a different directions could be independent of each other hence directions of ' k G and " k G are not necessarily the same. (d) 2 = G G i when k G can be rewritten as 2 ' ' " " 2 ' " 2 ' " k jk k jk k k k k jk k k k jk k = −= G GG G G G G G G G G G G G i i i . hence 2 R e G G and 2 1 I m 2 G G i . (e) The k G vector of the incident wave is '"' " ' " i i i ix ix x iz iz z k k j k a k j k a =− = − + − G The k G vector of the reflected wave is " ' " rr r r x r x x r z r z z k k j k a k j k a G The k G vector of the transmitted wave is " ' " t t t tx tx x tz tz z k k j k a k j k a G
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(f) Due to phase matching the tangential components of the k G vector should be the same across the interface. So ''' ix rx tx kkk == and """ ix rx tx Furthermore using the result in part d () 22 2 '" R e kk ωμε −= G G or () () () () 2222 ''"" 2 11 Re rx rz rx rz kkkk +−−= 2 Re tx tz tx tz and 2 1 I m 2 =− G G i or 2 1 Im 2 rx rx rz rz kk kk += ii 2 1 Im 2 tx tx tz tz These eight equations can be solved for the eight unknowns ' rx k , ' rz k , " rx k , " rz k , ' tx k , ' tz k , " tx k and " tz k . (g) The Faraday’s law states Ej H ωμ ∇× G G or E H j ωμ = G G . Using 0 jk r EE e = G G i G G 0 00 0 jk r jk r jk r jk r k e Ee E e k Ee kE H jj j ωμ ωμ ωμ ωμ ωμ −− −×− ×∇ × × = = = G G GG G G i i G G G G G G G '"'" ˆˆ ˆ ˆ xx xzz z x x z z kk j k k j k a k j k a k a =− = − + − = + G . For TE polarization E G is y polarized 0 ˆ jk r jk r xy z zy x zz y y jk r jk r xz zy xy kE a kE a e ka aE e Ha E e a E e ωμ ωμ ωμ ωμ ωμ × = = G G G G i i G G G The wave impedance in z direction is 0 0 z jk r yy TE jk r z y e Z k Hk ωμ ωμ = G G i G G i
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Similarly if we start with the Ampere’s law Hj E ωε ∇× = G G or H E j ωε = G G . Using 0 jk r HH e = G G i G G we obtain Hk E ωε × = G G G For TM polarization H G is y polarized () 00 0 ˆˆ ˆ jk r jk r xyz zyx yy x x z z jk r jk r xz zy xy kH a kH a e aH e ka kk Ea H e a H
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This note was uploaded on 12/02/2009 for the course ECE 000 taught by Professor O during the Spring '09 term at UCSB.

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HW3_Solution - ECE 201A Homework 3 solution 1. (a) E = E 0e...

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