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EE540_S08_HW_01_Solution

# EE540_S08_HW_01_Solution - c 2 n m 2 a 2 p 2 b 2 q 2 d 2...

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Page 1 of 1 EE540 (S’ 08) Homework #1 Solutions Problem 7.1 Total number of modes for the given cavity between 0 and 10GHz is approximately N ( ! ) " ( cavity volume ) = 8 # " (10 " 10 9 ) 3 3(3 " 10 8 ) 3 " (6 " 10 \$ 5 ) % 18 modes. (1.2) Then, in order to count the exact number of allowed modes, we have to find all the modes for the given conditions; Only the m or p index of the TE m , p , q mode may be zero, but not both, and only the q index may be zero for the TM mode. Using the following equation, we find the appropriate frequencies listed in the table below ( a = 3 cm, b = 4 cm , and d = 5 cm ) ! m , p , q =
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Unformatted text preview: c 2 n [ m 2 a 2 + p 2 b 2 + q 2 d 2 ] 1/2 (1.3) When the index of reflection n=1, allowed modes by are listed as follows TM Modes TE Modes m p q m , p , q (GHz) m p q m , p , q (GHz) 1 1 6.2457 1 1 4.7991 1 1 1 6.9280 1 2 7.0706 1 1 2 8.6579 1 3 9.7433 1 2 9.0077 2 1 8.0722 1 2 1 9.4935 2 2 9.5981 1 0 1 5.8270 1 0 2 7.8049 1 1 1 6.9280 1 1 2 8.6579 1 2 1 9.4935 Therefore, exactly, 15 modes of TE and TM exist between 0 and 10 GHz in our given cavity. The graphs by exact counting and approximated formula are given below....
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