HW4_Solution

HW4_Solution - ECE 201A Homework 4 solution 1. kr Er Ei ki...

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ECE 201A Homework 4 solution 1. i E G Beam splitter z x y i k G r E G r k G i H G r H G t E G t k G t H G (a) 0 ˆ jkz ix i E aE e = G , 0 ˆ jkz i iy E Ha e η = G . 0 ˆ jkx rz i E aj rEe =− G , 0 ˆ jkx i jrE H ae = G . 0 ˆ jkz tx i E atE e = G , 0 ˆ jkz i ty tE Ha e = G . k ωεμ = and μ ε = . (b) () 0 2 * 0 0 11 ˆˆ ˆ Re Re 22 2 i i jkz jkz i ii x y i z E E PE H a a E e e a −+ ⎡⎤ ⎛⎞ = × = ⎢⎥ ⎜⎟ ⎝⎠ ⎣⎦ GG G ( ) 0 0 * 0 ˆ Re Re 2 i i jkx jky rr r z y i x rE jr E H a a j r E e e a ⎛−⎞ = × = G 0 * 0 0 ˆ Re Re 2 i t jkz jkz i tt x y i z tE tE H a a t E e e a = × = G Due to power conservation irt PPP =+ GGG . Hence 00 0 2 2 2 2 i E ηη or 1 rt += as expected
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(c) in E G Beam splitter z x y k G in H G out E G out H G Mirrors Mirrors k G '" kkj k =− G d d The output wave consists of two parts. The first part is due to direct transmission through the beam splitter. So choosing the origin at the middle of the beam splitter 10 ˆ jkz out x in E atE e = G . Note that the magnetic field always remains y directed. Therefore the direction of the electric field at any point can be found using the Poynting's vector arguments. From this description it is clear that the electric field at the output will be polarized as shown in the figure. The second part is formed when the reflected beam from the beam splitter reflects from the four mirrors and reflects out from the beam splitter after a round trip. Since the mirrors are perfectly reflecting round trip propagation only introduces a phase delay and attenuation. Therefore, after one round trip () ( ) 4 12 4 " 4 ' 20 0 ˆˆ jk j k d jkz j k d j k d jkz out x in x in E aE jr e arE e e e −− == G . The part of this wave transmitted through the beam splitter reflects out from the beam splitter after one round trip again. Hence ( ) ( ) ( ) 4 4 2 28 " 8 ' 0 k d k d jkz j k d j k d jkz out x in x in E t e artE e e e G One can easily see that the infinitely many output waves will be created this way. Their sum will be an infinite summation as ( ) 24 " 4 ' 4 ' " 8 ' ˆ 1 .... jkd j k z out x in E aE re e te e e e + + + G Defining 4" 4' xt e e = , the infinite summation can be expressed as ' 2 ' 0 0 1. . . . j k z i j k z out x in x in i E e x x e e x e = ⎛⎞ + + + ⎜⎟ ⎝⎠ G Since 1 e =< the infinite summation converges. This summation can be found as
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2 1 .... Sx x =+ + + , 2 1. . . . x −= + + , ( ) 2 1 1 ... x x x S −= + + + = . Hence 1 1 S x = . Therefore 24 " 4 ' 20 4" 4' 1 ˆ 1 jkd j k z out x in E aE re e e te e −− =− G . The total output field is found as ' 12 0 ˆ 1 jkz out out out x in re e EE E a E et te e ⎛⎞ =+= ⎜⎟ ⎝⎠ GG G . This result can be rearranged as ( ) 22 4 ' ' ' 0 0 ˆˆ 11 jkz jkz out x in x in ttre e tt e e r e e Ea E e a E e te e te e −+ == G . But since tr + =1 0 ˆ 1 jkz out x in te e E e te e = G . (d) out E G can be zero if 0 e . Since t is real this condition requires = and 1 e = or 4' 2 kd m π = or ' 2 = . This condition is known as a resonance at which all the incoming power is absorbed by the resonator. Hence the arrangement works as a band stop filter.
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2.. (a) We can solve the problem using the steady state analysis we developed.
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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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HW4_Solution - ECE 201A Homework 4 solution 1. kr Er Ei ki...

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