42 BELL SYSTEM TECHNICAL JOURNAL 



From (3.42) and (3.43) we obtain 



Kt = V^e/C = Vu7e(e/C) 

 = 377 e/C 

 Now C is the charge Q divided by the voltage V. Hence 



Kt = 377 eV/Q (3.45) 



(3.44) 



In this case we have 



337e bi tan z 



J^t = 7i 



Zire 

 Kt = —60 In tSLUg 



(3.46) 



To obtain the impedance of the corresponding helically conducting sheet 

 we assume, following (3.30) 



Kt = (ym (7//?o) (30/7a) (3.47) 



and assuming a slow wave, let 7 = /3, so that 



K, = 30/|8oa (3.48) 



If we are to have n turns per wavelength, and the speed of light in the 

 direction of conduction, then we must have 



0oa = 1/w (3.49) 



whence 



Kt = 30n (3.50) 



For n = 2 (two turns per wavelength), K = 60. In Fig. 3.15, the charac- 

 teristic impedance Kt as obtained from (3.46) divided by 60 (from (3.50)) 

 is plotted vs. d/p. 



3.3b Four Turns per Wavelength 



In this case there are enough wires so that we can add a quadrature com- 

 ponent as in Fig. 3.10 and thus produce a traveling wave rather than a stand- 

 ing wave. Thus, we can make a more direct comparison between the de- 

 veloped sheet and the developed helix. 



For the developed helix we have 



V + j^|y = In tan (s + jx) + ^f^ . . (3.15) 



cos 2 (2 -\- JX) 



