FIELD SOLUTIONS 645 



We must identify this with (14.60). Thus, over the range considered, we 

 must have approximately 



(eW - i)/R = BoV^^iP + (tanh0)/0) (14.63) 



At = ^1 , we must have both sides zero, so that 



P = - (tanh 9i)/di and (14.64) 



(1 - (di/ey)/R = V^((tanh ed/ei - (tanh d)/d) (14.65) 



Taking the derivative with respect to 6 



2^' a rr f sech' , tanh e\ ,,.,,. 



These must be equal at = 0i , so that 



l/R = (1/2) (00 VeAi) (^-^ - sech^ d^^ (14.67) 



Thus, according to the methods of Chapter VI, our circuit equation should 

 be 



GT^)ii. = ('/«(T'--^'^-)^'-<^'/^>^^ '^^-^ 



Using (14.64), the correct equation (14.62) becomes 



tanh 01 tanh 6 





68) 



(14.69) 



71 U 



In a typical traveling-wave tube, we might have 



01 = 2.5 



In Fig. 14.8, the right-hand side of (14.69) is plotted as a solid line and 

 the right-hand side of (14.68) is plotted as a dashed line for dx = 2.5. 



14.5b Electronic Comparison 



Consider (14.25), which is the equation with electrons. For simplicity, 

 let do « 0, so that 



B ^ ^ ^ _ (6i/6)Uanh[(6i/6)i/^g] ^^^ ^^^ 



V c/m ^0 e 



For no electrons we would have 



B tanhe , ^ , 



= P = - -— — (14.71) 



0oVe/V 



