450fall09hw7_Sol

# 450fall09hw7_Sol - ECE 450 Fall 2009 Homework 7 Solutions...

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ECE 450 Fall 2009 Homework 7 Solutions Due: Tuesday, Oct 13 th , 2009 1. We are given that this is a TE mode ( polarization) problem and thus the electric field is polarized. Referring to the figure, medium 1 Æ ( ˆ y ) 0 , o μ ε , medium 2 Æ ( ) 02 , ro εε a. To find ε we start with the following: * Snell’s law: 2 r 11 2 2 sin sin kk θ = 12 21 1 2 sin sin sin 1c o s cos 1 Equation (1) r r k k == = −= ε ε ε ε 2 1 22 1 2 2 1 2 sin s i n sin r r r r r r θθ ε ε ε ε * The TE Fresnel reflection coefficient: 1 1 cos cos cos cos yr yi E 2 2 E η θη θ Γ= = + () 1221 2 1 cos cos cos cos cos 1 Equation(2) cos 1 r r ηθ ⎛⎞ ⎜⎟ −Γ ⎝⎠ ε ε Γ+ = Substitute Eq (1) into Eq (2) 2 1 1 2 1 1s i n 1 cos 1 r r r r ±− = ε ε ε ε 1

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2 1 2 1 12 2 21 2 2 11 2 2 2 1 2 1 1s i n 1 cos 1 1 1 cos sin 1 1 cos 1 r rr r r r θ ⎛⎞ = ⎜⎟ −Γ ⎝⎠ = + εε ε ε ε ε Using , 1, and 45 3 r Γ= = = ° ε , 2 1.372 r ⇒= ε b. Go back to Snell’s Law: 1 2 1 2 sin sin 1 sin sin sin sin 45 37.1 ° = 1.372 r r r r θθ −− = == ° ε ε ε ε c. Find τ 42 33 ⊥⊥ =+Γ=+ = d. Find 2 S Assume that the incident field magnitude is yi E , then we have: 2 22 2 ˆ 2 yt k η = S E ± With 0 222 2 ˆsin and z 2 ˆ ˆ and cos r yt yi kx τθ = EE ±± + ε , 2
() 2 4 0 m Æ ( 2 2 2 2 22 2 42 3 1.171 W ˆ ˆ ˆ 0.7976 0.6032 120 2 1.372 yi yt yi E EE kx z π ηπ ⎛⎞ ⎜⎟ ⎝⎠ == + = S 2. We are given that the reflection coefficient is zero and in reference to the figure, medium 1 ) o 0 , μ ε , medium 2 Æ ( ) 0 4, o ε . and 0 a. Since 12 1 2 η ≠⇒≠ Γ = p and εε , then the incident wave is a TM wave or & olarization. With Γ the incident angle is at the Brewster’s angle. 0

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## This note was uploaded on 02/21/2010 for the course ECE 450 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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450fall09hw7_Sol - ECE 450 Fall 2009 Homework 7 Solutions...

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