Damped Electrical Oscillations along Parallel Wires. 301 



so that the change of amplitude is accompanied by a change 

 of phase. 



The values come out 



, 2 2 _ (p-g + vjBxY+ip-vax)* 

 J ~ ty " (-p + q + v&v^+ip + vax)* ^7 



x*v 2 (cc* + j3*) + 2xv{/3(q-p)+*p} + (q- P y+p* 

 g_ = 2xv{Pp-a{q-p)} 

 f xy*{a* + l&)-.(q--p)*--p* " 



(16) 



(17) 



In order that there should be complete absorption of the 

 incident waves it is necessary that the two squares in the 

 numerator of (15) should vanish separately. This requires 



va vB 



— = — - — = X, say. 



p q-p ' J 



If we substitute va=:p\ and v/3=(q — p)\ in equations (8) 

 and (9), and eliminate \ by division, we find the condition 

 reduces to 



(p-<r){(q- P y+p*}=0, .-. p=<r. 



Therefore complete absorption is only attainable in the 

 distortionless case. In general we can only reduce the 

 reflected amplitude to a minimum. 



We can write (16) and (17) in the forms 



/2 2 _ ax* — 2hx + b 



tanfl=|= W ^~ A2 (19) 



From (18) we see that to any value of reflected amplitude 

 correspond two values of the terminal resistance, say & l9 a? 2 . 

 We can show that the corresponding phase-differences 1? 2 

 are supplementary. 



For from (18) we have 



b 



^1^2=-; 



11 



.: tan *, + tan *i=2 V«&-A«[,£^ + ^TTjJ 



— _Q 1 + Ay)(a3? 1 ,r 2 -& ) _ n 

 =2V«6-A \ a tf-b)(a.*i-b) ~ U - 

 Phil. Mag. S. 5. Vol. 47. No. 286. March 1899. Y 



