WAVES IN ELECTRON STREAMS AND CIRCUITS 645 



We have a relation between the 5*s 



02 = die 

 63 = die 

 From this we easily we see that a solution of the last two equations is 



Eio = E20 = E30 

 Accordingly, the first equation becomes 



Eioe-'^'\e'''^ -f- e^''^ + e^''^) = Ei 



J, Ete^^'^ (4.2) 



Let us now assume that the tube is very long. We easily see that in this 

 case 



I e'''^ I » 1 e'''^ I 



So very nearly 



\e'''^\»\e'''^ 



Eie'^''^ 

 Eio = E20 = E30 = — —7- (4.2) 



and the total field at the output end of the tube is 



E = E10 + £20 + £30 = SEce'^'^e-'"'^ (4.4)1 



This, however, is much smaller than the field Ee at the input end of the 

 tube. 



What is the physical picture? The electrons are injected into the circuit 

 as an unmodulated stream. In order to fit the boundary conditions at this 

 point, the three waves must have comparable magnitudes at the point of 

 injection. If this is the output, then any wave which ''grows" from input 

 toward output must be relatively very small at the input. 



If boundary conditions are fitted for other cases, as, for an electron speed 

 not equal to the circuit phase velocity O^o 9^ jSi), it may be found that the 

 output may be a little greater than the input under some circumstances; 

 this represents a small gain achieved through a spatial interference of the 

 three wave components. 



