TRANSMISSIOX IJXIi KQVATIOXS 153 



For symmetrical sections F21 = \\2, J '22 = ^n, Z21 — Z12, -2^22 = Zn 

 and equations (2.1) and (2.2) become 



v{n) = Zni(n) — Zrii{n + 1) + v°{n) 



v{n + 1) = Zi.i{n) - Zni(n + 1) + u°{n) 



i(u) = Ynv{n) + YMn + 1) + i°{n) 



-i{n + 1) = ri2^<«) + Vnvin + 1) - /(«) 



(2.5) 



(2.6) 



Eliminating i{ii) from (2.5) and v{n) from (2.6) and using, from (A4.4), 

 A = ZuZTi = — I T2 I^n leads to the difference equations 



v(n + 1) + v(n - 1) - 2.1K«) = B[i°(n) - f(n - 1)] (2.7) 

 i(n + 1) + /(;?. - 1) - 2A'i{ii) = C[v°{n) - h°(ii - 1)] (2.8) 



Since we also have B' — B, C = C, D' = A for symmetrical sections 

 equations (2.4) become 



v{n + 1) = AvCn) - Bi{n) + Bi°{n) 



(2.9) 

 i{n + 1) = -Cv{n) + .l'z"(") + Cz^°(«) 



We assume that the distribution of the sources in the branches of a sym- 

 metrical network need not be symmetrical with respect to the two ends, 

 even though the impedances of the branches are. 

 2.3 Slatetnent of Results for Infinite Symmetrical Section Line — Passive 

 When the sections are passive the equations to be solved are, from (2.9), 



v{n + 1) = Av{n) - Bi(n) 



(2.10) 

 i{n + 1) = -Cv(n) + A'i{n) 



If the line extends from n = to « = =© the solution is 



v(n) — e"" v{o) 



i(n) = e'"^'i{o) (2.11) 



v{n) = Zoi{n) 



where the matrix e^ is such that (a) the equation 



e^'^ + e"-^ = 2A (2.12) 



is satisfied, e^ being the inverse of e^^ , and (b) all the elements of the matrix 

 e~"^ approach zero as n ^ <x> . In dealing with sections we shall never 

 have occasion to consider T itself but only its exponential and associated 

 functions. The characteristic impedance matrix Zo is defined by the rela- 

 tion between the initial currents and voltages in an infinite line 



v(o) - Zoiio) (2.13) 



