280 



BELL SYSTEM TECHNICAL JOURNAL 



while K, the iterative impedance, is the impedance measured looking into 

 an infinite sequence of such sections. The propagation constant V is 

 in general a complex number A + jB. The real part is the attenuation 

 constant and the complex part the phase constant. For a pass band, 

 A must be zero, which will occur when cosh T is between + 1 and — 1. 

 Hence to locate the pass band of the dissipationless filter, the value of 

 the first equation of (12) must lie between + 1 and — 1. 



In order to point out what types of filters result from a combination 

 of transmission lines, two simplifying cases are considered. The first 

 case is when h — h- Then the equation for the propagation constant 

 becomes 



cosh r 



1 +^ 



cos- 



Oj/i 



V 



— sm"" 



coZi 



V 



2w/i 

 = cos X 



V 



1 + 



The edges of the pass band occur when * 



tan 



; \2Zo 



2Z 



02 J 



4. ^oi 

 ~^2Zo,' 



(13) 



(14) 



and the centers of the bands occur when 



. co/i . w/i (2n + \)ir 



sm — = 1; ^=-^- ^ — '— 



V V 2 



and 



Jm — 



{2n + ^)v 

 4/1 



(15) 



where « = 0, 1, 2, etc. A plot of the propagation constant for several 

 ratios of ZoJZo^ is shown in Fig. 2. As is evident the filter is a multi- 

 band filter with bands centered around odd harmonic frequencies. 

 The ratio of Zo^ to Z02 determines the band width of the filter. This 

 ratio cannot be made very large because the characteristic impedance 

 of a coaxial line or a balanced line cannot be widely varied from the 

 mean value. For example when the ratio of b/a varies from 1.05 to 

 100 the characteristic impedance of a coaxial conductor changes 

 from about 3 ohms to 275 ohms. This represents about as extreme a 

 range as can be obtained. For a balanced conductor the impedance 

 may range from 90 to 1100 ohms by taking extreme values of the ratio 

 D/a. Taking 100 as the extreme range between Zoi and Z02 the band 

 width for this case cannot be made less than 20 per cent. 



We next consider the case given by taking Zoj = 2Zo2. For this 



