PROPAGATION OF PERIODIC CURRENTS 513 



potential, at any point x of the wire. For generality, F{x) is assumed 

 to be discontinuous at any point x = n by the increment 



^F{u) = F(u +) - F(u -). 



The problem is to calculate the current I{x) produced at any 

 point X by the impressed field. 



rtl 



Positive 



directions 



//////A//////////////A//////////////////////A////// 



Fig. 1. 



Case 1 : General Case 



The current I{x) is the sum of two constituents: j{x) due to the 

 impressed axial electric force, and J{x) due to the impressed potential. 

 Formulas for these constituents will now be written down by aid of 

 the fundamental set of equivalent electromotive forces formulated in 

 the preceding subsection. Thus ^^ 



i(^) 



= {^ A{x,y)j{y)dy, (41) 



A{x, y) denoting the transfer admittance between points x and y. 

 Jix) itself consists of four constituents: /o(x) and J six), originating 

 in the terminal Impedances Zo and Z^ respectively, /os(x) originating 

 in the direct leakage admittance of the whole line 0-5, and Juix) 

 originating at the point x = u where F{x) is discontinuous. Thus 



/o(x) = - A{x, O)F(O), (42) 



Js{x) = A{x, s)F{s), (43) 



Josix) = I' Y'F(y)B(x,y)dy, (44) 



Ju{x) = - A{x,u)AF{u), (45) 



B(x, y) denoting a current transfer factor representing that fraction 



1^ With regard to the analytical evaluation of the integrals, attention should 

 perhaps be called to the fact that the integrand may be discontinuous or may change 

 its functional form at one or more points within the range of integration; whence 

 the integral must be broken up into a sum of integrals. 



