TRANSIENTS IN GROUNDED WIRES 429 



III 



The equation above may be used to obtain a formula for voltage 

 due to suddenly applied current exp io^t; or the operational product, 

 of which it is an expression in terms of /, may be carried out directly 

 in terms of p. The current is expressed in terms of ^ by: 



P 

 exp ioot = — ■ 



The second term in ;//(«) is transformed by the operational equivalent 

 already developed: 



erf — p = 1 — exp ( — ayfp) . 



The last term in \J/{ii) is not known in closed form in p. 



The operational product of exp iuit and the second term is evaluated 

 by 



pl'l - exp (- q;V^)"| _ t> _ P exp ( - a^Jp) 

 p — 10} p — lb) p — Ico 



1 



exp loot — - 



n — / ^ P — " 



exp (;co/ — aV/co) erfc ( — p — yiosf 



+ exp {io:t + aV^) erfc ( — p + yuot j 



the last term of which is given by pair 819 (with /S = 0) in the tables 

 referred to. Erfc is the error function complement; 



erfc (s) = 1 - erf (2). 



The operational product of exp iwt and the last term in ^(11) may 

 be expressed in integral form by the formula : 



^^fu) = 



p — iO) 



p — tw 



= /(/) + io) exp io)t I exp (- iwt)f{t)dt. 

 Jo 



The complete expression for the voltage due to cisoidal current is 

 as follows: 



£12(0 = ttV [*(22 + a) - *(22 -a) - $(si + a) + $(01 - a)], (4) 



