138 BELL SYSTEM TECHNICAL JOURNAL 



Examples 



The following examples are intended to give a more detailed picture 

 of certain rather simple special cases. They serve to illustrate the 

 previous discussion. In all the cases F is taken equal to ^/ so that 

 /o is equal to AG. This simplifies the discussion but does not detract 

 from the illustrative value. 



1. Let the network be pure resistance except for the distortionless 

 amplifier and a single bridged condenser, and let the amplifier be such 

 that there is no reversal. We have 



AJ{io^) = -^^, (41) 



where A and a are real positive constants. In (18) ^^ 



U = ^. r^"+'J»+i(ico)e*"'rf^"w (42) 



= Be-^'iB^'f^lnl). 

 s{t) = Be-^'il -]- Bt + BH^ll + • • •)• (43) 



The successive terms /o,/i, etc., represent the impressed wave and the 

 successive round trips. The whole series is the total current. 



It is suggested that the reader should sketch the first few terms 

 graphically for B = a, and sketch the admittance diagrams for B < a, 

 and B > a. 



The expression in parentheses equals e^' and 



s(t) = Be^^-"^'. (44) 



This expression will be seen to converge to as / increases or fail to do 

 so according to whether B < a or B ^ a. This will be found to check 

 the rule as applied to the admittance diagram. 



2. Let the network be as in 1 except that the amplifier is so arranged 

 that there is a reversal. Then 



AJiic.) = -^4- . (45) 



f^= {- l)-+'Be-"'{BV-lnl). (46) 



The solution is the same as in 1 except that every other term in the 

 series has its sign reversed : 



s(t) = - Be-"'il - Bt + BHyil + • • •) 



= - Be^-"-""^'. (47) 



1" Campbell, loc. cit. Pair 105. 



