LAMINATED TRANSMISSION LINES. I 929 



ill the numerical values of m and g^ for copper, we have 



Kjmhlc\i Umils f ^0 



For a plane Clogston 1 with stacks of equal thickness, the attenuation 

 constant is given by (174), and the fractional change in attenuation 

 with frequency is equal to the fractional change in resistance of either 

 stack, as calculated from (193). For a coaxial Clogston 1 with stacks 

 thin enough so that the plane approximation is valid we may also use 

 (193), but the fractional changes in resistance will be different for the 

 two stacks if these are of different thicknesses, and the fractional change 

 in the attenuation constant must be calculated from equation (17G). 

 If Rw and R20 are the dc resistances "per square" of the two stacks, and 

 ARi and ARo their increments as obtained from (193), then the fractional 

 increase in attenuation is given approximately by 



Aa ^ ARi/pi + AR2/P2 (200) 



oco Rio/ Pi + R20/P2 



For either plane or cylindrical geometry we find that if we scale up a 

 particular Clogston line by multiplying the thicknesses of the stacks and 

 the main dielectric by the same factor, then the low-frequency at- 

 tenuation constant will be divided by the square of the scale factor. 

 However, the permissible thickness of the individual conducting layers, 

 if we are to have the attenuation flat to a specified degree up to a fixed 

 frequency, is inversely proportional to the scale factor. Thus if we double 

 the overall dimensions of the line and double the amount of conducting 

 material in the stacks, we shall divide the low-frequency attenuation 

 constant by four, but we shall have to make the individual layers half as 

 thick in order to maintain the same relative increase in attenuation 

 constant at the same top frequency fm • In addition it is clear that if we 

 double the top frequency while maintaining the same requirement on 

 Aa/aa for a line of giv^en dimensions, we shall also have to cut the thick- 

 ness of the individual layers in half. 



As a numerical example, let us return to the cable whose specifications 

 were given })y (155) at the end of Section IV. For this cable we have: 



Pi = 55.49 mils dsi = 8.4(3 mils 



p.> =181.44 mils ds. = 4.04 mils (201) 



P2/P1 = 3.270 R20/R10 ^ S1/S2 = 2.094 



