TRANSIENT OSCILLATIONS IN ELECTRIC WAVE-FILTERS 31 

 where x = u c t. For the case w = 0, the solution can be recognized as 



To attack the problem by means of the integral equation II, we 

 write down the operational transfer impedance 



ih-rlvrhjp^^^^'-^y" (2 - 2) 



Writing u c t=x t and A n (t)= — F„(x), and substituting in II gives, 



k 



as the integral equation of the problem 



f" FMe-'-dx^ -^=(VF+T- I)" (2 . 3) 



The solution of this equation can be expressed in a number of differ- 

 ent forms, depending on the type of expansion of the right hand side 

 which we adopt. One form is as follows: 



Expansion of the bracketted expression on the right hand side by 

 the binomial theorem and rearrangement gives 



J Q K{x)e dx-^—^* 2 , p2n + 4| 



y (£+l)*_ "I pn 1 (2n)(2n-l)(2n-2) (ft 2 + l) -| 



£ 2 » 1" • • J [_ U pin-T 3| £2* T J- 



Recognizing that 



r x 1 



/ Jo(x)e-t*dx= . 



the solution, after rearrangement, becomes the terminating series 27 

 F n (x)=kA n (f) 



= (i | 4w r- 2 i 4tt 2 (4tt*-2 2 ) r _ 4 t 4tt 2 (4K 2 -2 2 )(4tt>-4 2 ) £> _ 6 ( _ \ JM 



n ( , (4w 2 -2 2 ) x 3 , (4« 2 -2 2 )(4n 2 -4 2 ) * B , \ 



where P _m indicates multiple integration, repeated w times. Thus 

 D-U (x) = fjoixOdx!-, D-Uo(x) = Afcc, r i Mx 2 )dx 2 \ etc. 



t/o «/o «/o 



27 This solution has been derived from the definite integral also. 



