8 26 solution for normal incident 1 23 1 24 1

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Unformatted text preview: 3ˆe e + ej 6z z √ √ = [ˆ(−j 10) sin 6z + z 10 3 cos 6z ]ej 6 3y V /m y ˆ √ j 6 3y = 5ˆe y (17) (18) (19) H1 = Hi + Hr (20) e−j 6z + ej 6z (21) −x ˆ e 12π √ j 6 3y = = −x ˆ √ 1 cos 6zej 6 3y A/m 6π (22) P.8-26 Solution For normal incident 1+Γ=τ (23) |Γ| ≤ 1. (24) 1 + |Γ | =3 1 − |Γ| (25) where when 1 |τ | = |Γ|, one solution is Γ = − 2 and τ = 1 . 2 Hence S= SdB = 20 log S = 9.54dB (26) σ ≫1 ωǫo (27) P.8-28 Given Ei = xEo e−jβo z ˆ Solution ηo = ηc = µo = 120π ǫo µ ≈ (1 + j ) ǫc = (28) πf µ σ j ωµ σ (29) (30) Hence |ηc | = ηo ω µ/σ = µ/ǫo ω ǫo ≪1 σ (31) ηc ηo ηc ηo (32) (a) 1− ηc − ηo Γ= =− ηc + ηo 1+ 2 ≈ −1 (b) Magnitude square |Γ|2 = 1− 1+ ηc 2 ηo ηc ηo (33) For small x (1 + x)−1...
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This document was uploaded on 01/17/2014.

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