prelim2Sol

# prelim2Sol - ECE 303 Electromagnetic Fields and Waves...

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ECE 303: Electromagnetic Fields and Waves Fall 2007 Solutions for Prelim 2 Reminder: Date: Tuesday 10/23 Problem 1: Oblique incidence on a dielectric with both TE and TM components. a. The transverse electric (TE) part is ~ E T E = 2 E 0 e - j ~ k i · ~ r ˆy and the transverse magnetic (TM) part is ~ E T M = - jE 0 e - j ~ k i · ~ r (ˆx - z). The magnitudes of the TE and TM components are the same and the TM component is lagging behind the TE component with a phase difference of 90 . This is a left-hand circularly polarized wave. b. The polarization direction of the TM wave is 1 2 ˆx - 3 2 ˆ z which must equal cos θ i ˆx - sin θ i ˆ z where sin θ i = 3 2 . One could use Snell’s law to find θ t , but note that the TM part sees an index of refraction of 3 while the TE part sees an index of q 2 3 . For the TM part, n i sin θ i = n t sin θ T M t = sin θ T M t = 1 3 3 2 = 1 2 = θ T M t = π 6 = 30 For the TE part, n i sin θ i = n t sin θ T E t = sin θ T E t = r 3 2 3 2 = r 9 8 > 1. So the TE part is totally reflected; only the TM part is transmitted. c. The transmitted electric field is ~ E t ( ~ r) = T T E [2 E 0 ˆy]e - j ~ k T E t · ~ r + E 0 [ - jT T M x ˆx + j 3 T T M z ˆ z]e - j ~ k T M t · ~ r where ~ k T M t = ω c 3[sin θ T M t ˆx + cos θ T M t ˆ z] and ~ k T E t = k T E tx ˆx + k T E tz ˆ z k T E tx = k T E ix = k iz = ω c sin θ i = ω c 3 2 = k T E tz = r ω 2 c 2 2 3 - ( k T E tx ) 2 = - j ω c 1 2 3 T T E = 2 k T E iz k T E iz + k T E tz and T T M x = 2 3 cot θ i + cos θ T M t and T T M z = 2 cos θ i 3 cos θ i + 3 cos θ T M t Problem 2: The total voltage along the line is V ( z ) = k + [e - jkz + Γ L e jkz ] where Γ L = | Γ L | e , and the magnitude of V ( z ) varies along the line as | V ( z ) | = | V + | p 1 + | Γ L | 2 + 2 | Γ L | cos(2 kz + φ ) (1) a. The variations in | V ( z ) | shown in the plot have a periodicity of half the wavelength, and noting ε = 3 ε 0 , λ = λ 0 3 = 8 mm = f = c λ 0 = 21 . 65 GHz b.

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