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Unformatted text preview: 34 Resonant modes and field fluctuations Since in a rectangular cavity the resonant frequencies f mnl = c 2 ( m a ) 2 + ( n b ) 2 + ( l d ) 2 2 f mnl c = ( m a ) 2 + ( n b ) 2 + ( l d ) 2 , we can consider 2 f mnl /c to be the length of a vector ( m a , n b , l d ) point ing away from the origin of a 3D Cartesian space where each lattice point , e.g., ( m a , n b , l d ) = ( 1 a , 2 b , 1 d ) , is associated with two resonant modes (TE and TM) of the cavity. In this space, volume per lattice point is 1 abd , and thus volume per resonant mode is 1 / 2 abd . Also, all the resonant modes with resonance frequencies f mnl less than a given frequency f can be associated with lattice points residing within one eight (an octant) of a sphere of radius 2 f/c centered about the origin of the same space only an octant is involved since the indices m , n , l employed are all nonnegative. Thus, the number of resonant modes with frequencies less than f , to be denoted as the cumulative distribution C ( f ) , is found to be C ( f ) = 1 8 ( sphere of radius 2 f/c ) 1 / 2 abd = 1 8 4 3 ( 2 f c ) 3 1 / 2 abd = 8 f 3 3 c 3 V 1 where V = abd is the physical volume of the cavity. Consequently, the number density N ( f ) of the available resonant modes in a cavity of volume V is obtained as N ( f ) = dC df = 8 f 2 c 3 V modes Hz which grows quadratically with frequency f . As illustrated later in this lecture, the distribution N ( f ) has deep theoretical implications. Example 1: Consider a rectangular cavity with dimensions a = b = d = 0 . 3 m. Determine N ( f ) for f = 50 GHz and the number of resonant modes to be found within a bandwidth of f = 1 GHz centered about f = 50 GHz. Solution: Using the density function derived above, we find that N (50 10 9 ) = 8 (50 10 9 ) 2 (3 10 8 ) 3 (3 10 1 ) 3 = 8 25 10 7 = 2 10 5 modes Hz ....
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 Fall '08
 Staff
 Electromagnet

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