CURVED WA VE GUIDES 31 



Solving 7-5 for p one obtains 



p = ()A5^\^'\-' 9-1 



where p si the percentage increase in attenuation, and A the deviation angle 

 in degrees. 



Hence the average attenuation 



a^ = a(l + O-Ol/*) 9-2 



From 3-15 and 3-11 



. V Ri . 3 2-8 ^ - 



a = = 10 A, A a 9-3 



ai] 



Introducing the Ri value from 4-8 



a = 4.5 10-5piXi-6a-» 9-4 



where p is the high-frequency resistance of the wave guide relative to copper. 

 From 9-1, 2 and 4 



«A = 4.5'lO~Vxi-5a-3(l + qk-^a^) 9-5 



with 



q = 4.5 10-»A2 



The attenuation reaches a minimum when 



fix, a) = A^-^a~* + qX~--^a = minimum 

 C a^e /. X is given 



8f/8a = -3X'-^a-' + ^X-^-^ = 

 aopt = 1.32X?-"" = 5.2XA-"-s 

 From 9-5 



a^„p, = 4a = 1.29 10~V-'-'Ai-"^ 



Case 2. a w given 



bf/SX = 1.5X'^-5a-^ - 2.5qX-'-'a = 



X„^, = l.Uaq'-' = 0.294aA»-5 

 From 9-5 



-1.5. 0.75 



ctAopt ~ 1-6q: — 1.15 10 pa * A 

 Numerical Example 



Let A = 0.42° 

 a = 0.05 w 

 X = 0.01 m 



