216 BELL SYSTEM TECHNICAL JOURNAL 



where I' is circuit voltage. We can regard this in two ways. We can think 

 of —{V''L)M as the effective field at the location of the current /o , or we 

 can think of M'^h as the effective current referred to the circuit. 



If we have a broad beam of electrons and a constant current density /o 

 we compute (essentially as in Chapter III) a value of C^ by integrating 



a = (l/8/3-'Fo)/o(F/L)2 f AP da (4.78) 



where da is an element of area. We can think of the result in terms of an 

 effective field Ee 



El = (V/LY ^ (^-'^^ 



a 



where a is the total beam area, and a total current cr/o , or we can think of 

 the integral (4.77) in terms of an effective current /i, given by 



= Jo I M- da (4.80) 



and the voltage at the circuit. 



Of course, these same considerations apply to distributed circuits. Some- 

 times it is most convenient to think in terms of the total current and an 

 effective field (as we did in connection with helices in Chapter III) and 

 sometimes it is most convenient to think of the field at the circuit and an 

 effective current. Either concept refers to the same mathematics. 



4.6 Harmoxic Operatiox 



Of the field components making up E in (4.62) it is customary to regard 

 the m = component, for which (S — d/L, as the fundamental field com- 

 ponent, and the other components as harmonic components. These are some- 

 times called Hartree harmonics. If the electron speed is so adjusted that the 

 interaction is with the m — ox fundamental component we have funda- 

 mental operation; if the electron speed is adjusted so that we have interac- 

 tion with a harmonic component, we have harmonic operation. 



There are several reasons for using harmonic operation in connection 

 with filter-type circuits. For one thing the fundamental component may 

 appear to be traveling backwards. Thus, for circuits of the type shown in 

 Fig. 4.11, we see from Fig. 4.27 that d is always negative. Now, in terms of 

 the velocity v 



^ = a;/r = e^L (4.81) 



and if 6 is negative, v must be negative. However, consider the w = 1 

 component 



^ = 0,/^, = (27r + e)/L (4.82) 



