204 BELL SYSTEM TECHNICAL JOURNAL 



Eliminating ih, setting ia = "^and a = {R -{- Rc)jRRcC we have 



dq . E . , . Eo 



-J^aq = -^sm<.(f + tr) + -, 



whence 



q = EPF(t + /,) + :^ + ^ e-«S (7) 



aK 



where A is the constant of integration and 



P = \IR{a} -\- 00^), F{x) = a sin oox — o) cos wx. 



Since q obeys (7) from / = until t = h, we have for the initial and 

 final charges during the charging portion of the cycle: 



Q, = EPF{h) +^ + A 



aK 



Q, = EPF{h + /2) + ^ + Ae--'^'\ 

 aK 



Eliminating A 



Qi - <226«'^ = EP r F(/0 - e"^^Fih + h) + ^ [ 1 - ^"'^1 - (8) 



which is the additional equation required. 



It now merely remains to eliminate some of the unknowns. To this 

 end we first eliminate Qi and Q2 from (3) and (4) by means of (6) and, 

 solving for E, obtain 



£ = — Eo : . (9) 



D sin co(/i + ^2) — sin co/i 



A second expression for E involving the same variables is next obtained 

 by replacing Qi and Q2 in (8) by their values as given by (3) and (4). 

 Equating these two expressions for E leads to : 



G2 + G3 sin C0/2 — BG2 cos (joti 



(10) 



Referring to Fig. 26^1 it will be seen that t^ is if the input voltage 

 is just sufficient to make the grid positive, i.e., HE— — Eq; while on 



