618 PROFESSOR CHRYSTAL ON THE DIFFERENTIAL TELEPHONE. 
For silence we must have y=z. We can express the mathematical con- 
ditions for this equality without solving the equations. We get, in fact, 
{ @L+M—K)D+2P+Q—gpy }y=Asinns, 
J2D? 4 
| (2L + N—K)D Pa R—-Hpat y= Asinnt . 
Hence we must have 
(A+ D +vD? + pD?)sinnt=0 ; ; (9) 
where 
\=(Q—R)ST, 
=(M—N)ST + (Q—R)(GT+HS), 
y=(Q—R)GH + (M—N)(GT+ HS) +8J2—TE, 
=(M—N)GH+GJ2—HL.. 
Hence the conditions for silence are 
coe . (10) 
p—7’p=0 
If there is to be silence for all frequencies, then we must have 
pret pede en eetie 
which require 
Q=Rk, 
M=N, 
SJ2=TI? , 
GJ?=HI?. 
Case of one Neighbouring Circuit.—If we make T=o, the conditions (10) 
become 
A—n*v=0, p=0 ; : 5 ; » i) 
where 
4=(Q—R)S, »=(M—N)S+(Q—R)G, v= (M—N)G_-L.. 
From these formule we’see that silence may be obtained for given fre- 
quency, but silence for all frequencies is impossible. 
Case of no Neighbouring Circuits.—The conditions reduce to 
M—N=0, Q—R=0, 
