ELECTRIC CIRCUIT Til TORY 351 



circuinstances. Consequently from (252) 



6\i(/)-ao(/) + 2a2.(/) + 2a4.(/)+ - • • 



(267) 

 Svi{t)=2\as(t)+aUt)+aUt)+ . . .[. 



S22 is clearly e(iual to ^u by symmetry. 



Now suppose that an e.m.f. E = E{t) is impressed on the line al 

 x = (), 1 = 0, through a lerminal impedance Zi, and the distant end 

 {x = s) closed through an impedance Z2. We suppose these terminal 

 impedances and the actual impressed e.m.f. replaced by the actual 

 line voltages l\ and Fo, impressed directly on the line at x = () and at 

 x = s are 



W)=T^ ('Sn{t-T)Vx{T)dT 

 (It, Jo 



Is{t) = -'j-J\2(t-T)V,{r)dT 

 +^jf'^22(/-r)F2(r)rfr. 



(268) 



(269) 



But the current at x = s is also equal to the current in the terminal 

 impedance Z2 in response to the terminal voltage V2: denoting by 

 a2(0 the indicial admittance of Z2 it is 



Is{t)=J^j\{t-T)V2{T)dT. (270) 



Similarly the current entering the line at .\: = is the current flowing 

 in the terminal impedance Zi in response to the e.m.f. E— Vi. De- 

 noting by a\{t) the indicial admittance of Zi, it is 



Io{t) =j^ faiit - t) \ E{t) - Fi(r) (- dr. (271) 



Equating equation (268) and (271) and (269) and (270) we eliminate 

 Io{t) and /v(/) and get 



/'[^ii(/-r)+ai(/-r)] V,{r)dT- / '^i2(/- r) 1^2(r)^r 

 Jo Jo 



= j'ai{t-T)E{T)dT, (272) 



- rS,2{t-T)Vl{T)dT+ j'[S22{t-T)-a2{t-T)]V2{T)dT = 0. (273) 



Jo Jo 



