440 



BELL SYSTEM TECHNICAL JOURNAL 



circular, its inherent double degeneracy will be lost and each of the original 

 modes (with minor exceptions) will split into two. 



Although the frequency and fields of the undisturbed mode are the 

 same, the Q is not necessarily so. For example, Fig. 14 shows a ""TE 01« 

 mode" in a half cylinder.* 



It is easy to calculate Q - for this case. The result is 



(1 + p'R'f" 



in which 





Ki = 1.290 A'2 = 0.653 



(17) 



Here A'l and K2 are constants which account for the resistance losses in 

 the flat side. For the full cavity, shown dotted in Fig. 14, eq. (17) holds 

 with A'l = A'2 = 0. If the circular cavity has a partition extending from 

 the center to the rim along the full length, (17) holds with the values 

 of A'l and A'2 halved. If a tin projects from the rim partway into the in- 

 terior, still other values of A'l and A'2 are required. It is a simple matter 

 to compute these for various immersions; Fig. 15 shows curves of A'l and 

 K2 . The following table gives an idea of the magnitudes involved: 



mode: r£ 0,1,12 R = 0.4 



The question now is asked, "Suppose a longitudinal fin were used, small 

 enough to cause only a tolerable reduction in the Q. Would such a fin 

 ameliorate the design difficulties due to extraneous modes?" 



Some of the effects seem predictable. All modes with ^ > will be split 

 to some extent, into two modes of different frequencies. Consider the 

 TE I2n mode, for example. There will be one mode, of the same frequency 

 as the original whose orientation must be such that its £-lines are perpendicu- 

 lar to the fin. The Q of this mode would be essentially unchanged. There 

 will be a second mode, oriented generally 90° from the first, whose £-lines 

 will be badly distorted (and the frequency thereby lowered) in the vicinity 



* Solutions for a cylinder of this cross-section are known and all the resonant fre- 

 quencies and Q values could be computed, if they had any application. 



