Ass7sol - Assignment 7 Solutions November 7, 2009 1. a =...

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Unformatted text preview: Assignment 7 Solutions November 7, 2009 1. a = 6cm, b= 4cm (a) The fundemental mode is T E10 mode. The next mode is T E01 mode The cuto frequency of T E01 mode is f01 = (b) The next mode is T E11 and T M11 mode Their cuto frequencies are f11 = (c) β = mπ 2 nπ − a b α = 50.846 =⇒ l = 1.96cm ω2 µ − Ey 1 Ey 2 ZT E fc f 2 2 c 2 2 0 6e − 2 2 + 1 4e − 2 2 2 = 3.75GHz c 2 1 6e − 2 + 1 4e − 2 = 4.5GHz , imaginary part of β is attenuation coecient. 2. f = 10GHz,a = 2.25cm,b = 1cm P = 1 Ey Hx = 2 ZT E = η 1− = 377 1 − ( 6.6 )2 10 = 501.8Ω Note: In the question paper b was given as 100cm by mistake. In all the answer sheets the formula for fc = was used 2a to calculate cuto frequency without realizing that this is applicable only if a>b. 3. Cut o frequencies of : T E01 , T E10 , T E11 , T E02 , T E20 , T M11 , T M12 , T M22 (a) a = 2b fc = c 2 nπ 2 cπ √ 2 = m + 4 n2 b 4b In order of occurance: T E10 , T E20 , T E01 , T E11 , T M11 , T E02 , T M12 , T M22 mπ 2b 2 1 (3e5)2 = 89.677W/µm P= 2 501.8 P ≈ P ∗ 2.25e − 2 ∗ 1e − 2 = 20.177KW cπ + (b) a=b fc = nπ 2 cπ √ 2 = m + n2 b 2b In order of occurance:(T E01 , T E10 ), (T E, T M )11 , (T E02 , T E20 ), T M12 , T M22 c 2 mπ b 2 + 1 4. Considering θc1 and θc2 to be the angle wrt the normal of the interface (a) For interface between n1 and n2 critical angle θc1 = sin−1 n2 n1 For interface between n1 and n3 critical angle θc2 = sin−1 n3 n1 n3 > n2 =⇒ θc2 > θc1 so the minimum angle for guidance is θc2 If we consider angle wrt to the direction of propagation 90 − θc2 is maximum angle above which light will escape. (b) phase velocity vp = c nef f β k0 nef f cos(90 − θ2 ) = = k0 n1 k0 n1 nef f = n1 ∗ sin(θ2 ) = n3 2 ...
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This note was uploaded on 09/12/2010 for the course EE EE645 taught by Professor Ujin during the Spring '10 term at YTI Career Institute.

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Ass7sol - Assignment 7 Solutions November 7, 2009 1. a =...

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