122 BELL SYSTEM TECHNICAL JOURNAL 



the two lines equal so that 



£° X RnP = >L X 4^ = ^ = R^. 

 p jcoC 6 



Hence i?o = R^jRo- Substituting these values in equation (35), we 

 have 



I'm 



1 - [i?o/2i?](l - e-^n 



2Rll + [i?o/2i?](l - e-'P) 

 After some simple rearrangements, equation (36) takes the form 



1m 



2R 



4:R / 1 



- 1 



(36) 



(37) 



If m equals 1, the solution can be obtained exactly as discussed in the 

 first example in section (1), and it is 



^l 



2R 



[' 



ji^4LC - 2g-'i^i/^-^«+'''-i 



1 +icoVLC 



The solution for m sections of lattice network is discussed in the 

 Appendix, and it is there shown that the solution can be written in 

 the form 



I'm 



E^ 

 2R 



1 -jcoVLC y _ ^_naiy[Ed)+M 





iM'^ 



X 



(m-1)! 1+JcoVLC 



(m - 2) ! \ (1 + jo^^n^y ~ 1 + jcoViC / 



(i 



1 — jix>^LC\ 



-^^^rm) -^-^^' 



(38) 



Equation (38) represents the symbolic or complex algebra solution 

 for the current at the end of the mth section of a lattice network as 

 shown on Fig. 4. It is usually desirable to obtain the current due 



