prelim2Sol

prelim2Sol - ECE 303: Electromagnetic Fields and Waves Fall...

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Unformatted text preview: 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 TE = 2 E e- j ~ k i · ~ r ˆy and the transverse magnetic (TM) part is ~ E TM =- jE e- j ~ k i · ~ r (ˆx- √ 3ˆ 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 TM t = ⇒ sin θ TM t = 1 √ 3 √ 3 2 = 1 2 = ⇒ θ TM t = π 6 = 30 ◦ For the TE part, n i sin θ i = n t sin θ TE t = ⇒ sin θ TE 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 TE [2 E ˆy]e- j ~ k TE t · ~ r + E [- jT TM x ˆx + j √ 3 T TM z ˆ z]e- j ~ k TM t · ~ r where ~ k TM t = ω c √ 3[sin θ TM t ˆx + cos θ TM t ˆ z] and ~ k TE t = k TE tx ˆx + k TE tz ˆ z k TE tx = k TE ix = k iz = ω c sin θ i = ω c √ 3 2 = ⇒ k TE tz = r ω 2 c 2 2 3- ( k TE tx ) 2 =- j ω c 1 2 √ 3 T TE = 2 k TE iz k TE iz + k TE tz and T TM x = 2 √ 3cot θ i + cos θ TM t and T TM z = 2cos θ i 3cos θ i + √ 3cos θ TM 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 jφ , 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 , λ = λ √ 3 = 8 mm = ⇒ f = c λ = 21 . 65 GHz b. From the plot, V max = | V ( z =- 5mm) | = 6 V and V min = | V ( z =- 3mm) | = 2 V . So the standing wave ratio is SWR = V max V min = 6 2 = 3 = ⇒ | Γ L | = 1 2 c. The minimum value V min = | V ( z =- 3mm | occurs for the first time when the argument of the cosine function in Equation (1) is equal to- π , i.e., when 2 kz +...
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This note was uploaded on 09/27/2010 for the course ECE 4250 at Cornell.

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prelim2Sol - ECE 303: Electromagnetic Fields and Waves Fall...

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