284 Mr. 0. Heaviside on the 



the speed is excessive. Now (77) may perhaps be put in a 

 finite form when B/ m is allowed to be different from B/, 

 though I do not see how to do it. But when B/ m = B/ and 

 L/ m =l/ it can of course be done, for we may then use the 

 finite solutions of (66) and (67). Thus, given Y = Y sin nt 

 at 2=0, and no impressed force elsewhere, find V and C 

 everywhere subject to (66) and (67) with e=0, and V=0 at 

 z=l. 

 Let 



F = (i$nf{{W+LVf-Vny, 1 (?8) 



Q=.... {( ) l +..-}% I ' 



tan^ 2 =sin2QZ-T-(6- 2Pi -cos2QZ), 1 (7d) 



tan(9 1 = (L^P-B'Q)-r-(RT + L' ? 2Q) j J 



then the finite V and C solutions are 

 V=V e- p *sin(^-Q£) 



y e Fg sin (nt + Qz + 6 2 ) — €- Pg sin (nt — Qz— 2 ) ,„„. 

 e pl (€ aK + €- 2K -2cos2QZ)* ? l ^ 



0=v »(^l)»[ rP ' !i,1, " i -^ 



_ 6 Pg sin(^+Qg— ^i + ^ 2 )+6- Pg sin(^— Qz— fli + fl 2 ) ~] /^ 

 e p *(6 2P * + e- 2P *-2cos2QQ* J"^ ' 



If we expand this last in cosines of mz we shall obtain (77), 

 with B/ m = B/. There are three waves ; the first is what 

 would represent the solution if the line were of infinite 

 length ; being of finite length there is a reflected wave (the 

 e Vz term), and another reflected at z = 0, the third and least 

 important. 



The amplitude of C anywhere is 



v (S»)* r 6 2P(*-*) + 6 -2P ( i- g ) + 2 cos 2Q(Z-s) -|* 

 ° (R' 2 + L Vf L e 2P ' + e~ 2P ' - 2 cos 2QZ J ' 



At the distant z = l end it is 



° 0=2Vo (R'* + L'W ( *" + 6 " 2P '~ 2 ° 0S 2Q0_J ' (82) 



I have already spoken of the apparent resistance of a line 

 as its impedance (from impede). The steady flow impedance 

 is the resistance. The short line impedance is (~R 2 + Un 2 )n 



