450fall09hw8_Sol

# 450fall09hw8_Sol - ECE 450 Fall 2009 Homework 8 Solutions...

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ECE 450 Fall 2009 Homework 8 Solutions Due: Tuesday, Oct20 th , 2009 1. Observe the fine details in the problem statement; such details outline the necessary steps needed to solve the questions. (1) It is observed that the electric field phasor at the origin, i.e., 3 (0,0,0) 2 1V ˆˆ ˆ 2m i xj y z ⎛⎞ ++ ⎝⎠ =− E ± . Given the standard form of a plane wave, · 0 ˆ j eE e = kr E ± , the evaluation of the field at the origin gives us polarization and amplitude information of the field. Rewriting the above yields, 1 ˆ 3 2 ˆ 2 i ˆ x zj y + + = E ± . This clearly shows that the electric field is a linear combination of a TE and TM wave, thus all the governing relationships you’ve learned for each mode apply. (2) It is stated that the reflected field is linearly polarized, despite the fact that the incident wave is circular polarized (CP). Thus instead of the reflected wave being CP as well, something special happens—that being the TM component of the incident wave is fully transmitted ( ) and thus is incident at the Brewster’s angle . We know it the TM component because the reflection coefficient cannot be zero in the TE case when 0 Γ= & 1 2 12 and μ =≠ εε . a. To find the angle of incidence w.r.t. the normal in the x-z plane, use geometry and what the plane wave forms of Maxwell’s Equations dictate: form a triad and are perpendicular to each other. See the figure below. The incident electric field direction in terms of its polarization components are drawn at the origin and a projection of the field in space, incident of on the interface. Note the polarization of the incident field does not ,, a d ˆ n k EH 1

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change. (Why can’t we choose in the y direction?) Since the is a unit vector, from the geometry we have, ˆ k ˆ e 31 and si co n 30 22 s θθ θ == = ° 90 180 ii 60 ++= ° = 00 , μ ε , r εε ˆ e b. As discussed previously, we know the incident field is impinging at the Brewster’s angle and thus we can solve directly in terms of the incident angle. Given that pi = , then 2 2 1 21 tan tan pr p 2 2 1tan 60 3 r r =⇒ = = ε ε ε c. Use the special relationship between the Brewster’s angle and the transmission angle: 90 90 30 pt t p +=°∴ =° − =°♠ 3 ˆ 2 x e = 1 2 z ˆ e = ˆ e ˆ y e ˆ e ˆ k i ˆ x e ˆ z e i ˆ y e 2
d. Given the incident and transmission angles and the medium properties, () 11 1 1 66 6 2 ˆ ˆ ˆ cos sin 21 3 ˆˆ ˆ cos60 sin 60 2 10 10 3 10 2 2 m ii i kk x z xz x z x z r a d ˆ π θθ λ ππ == + ⎛⎞ + ° = + = + ⎜⎟ ⎝⎠ k k 22 2 2 2 6 2 ˆ ˆ ˆ cos sin 23 1 3 cos30 sin30 2 10 3 10 3 3 10 2 2 m t t x z x z x z r a d ˆ ˆ + + ° =+ = + k k

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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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450fall09hw8_Sol - ECE 450 Fall 2009 Homework 8 Solutions...

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