Signalling with Condensers. 429 



subject to the first and second conditions. We thus obtain 



/(*)=V. ^+l_ u (2) 



for the final distribution. 



In expanding (2) in a series of sines we must remember that 



— =0 when x — l t and accordingly use the expansion 



which gives 



ft x_ v e* cl —> + €-*<*-* ) 



. ttV » 2i— 1 . (2« — lVa?. 

 =4,rV2 . (2M)V+W • Sln — « ' 



consequently the required solution is 



€« + €■ 





- 4?rV - e c * 2 » y-iiv+w ^hi ; < *" • (3) 



When the insulation is perfect and h = 0, this becomes 



_ r 4V_,« 1 . (2t-l)w# - (M ~ i y f 

 V=V -^^2i=l- SmA -2T-- € ^ ' ' W 



As the current equals — t"-tS we have, by differentiating (3) 

 and (4), 



VA 6 *(*-*)- 6 -^-*) 



2V _»_« (2i-l) 2 7r 2 (2i-l)w# , w-i)w 



+ 17 - 6 " 2 » (2,-l)V + tfy >CM 2/ ■' ^" : ?W 

 and when h=0, 



„ 2V V « (2i-l)7rar „(*-» 2 * 2 * 

 C =U 2 i cos 2/ * € " * • (6 ) 



We can employ (5) and (6) to determine the flow through the 

 receiving instrument, by giving x a value something less than /; 



CUD 



but it is preferable to use the series for -j obtained from (3) and 

 (4) by differentiation. 



